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ale29559_IFC.qxd, , 07/11/2008, , 07:40 PM, , Page 2, , PRACTICAL APPLICATIONS, Each chapter devotes material to practical applications of the concepts covered in Fundamentals of Electric, Circuits to help the reader apply the concepts to real-life situations. Here is a sampling of the practical applications, found in the text:, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, , Rechargeable flashlight battery (Problem 1.11), Cost of operating toaster (Problem 1.25), Potentiometer (Section 2.8), Design a lighting system (Problem 2.61), Reading a voltmeter (Problem 2.66), Controlling speed of a motor (Problem 2.74), Electric pencil sharpener (Problem 2.78), Calculate voltage of transistor (Problem 3.86), Transducer modeling (Problem 4.87), Strain gauge (Problem 4.90), Wheatstone bridge (Problem 4.91), Design a six-bit DAC (Problem 5.83), Instrumentation amplifier (Problem 5.88), Design an analog computer circuit (Example 6.15), Design an op amp circuit (Problem 6.71), Design analog computer to solve differential equation (Problem 6.79), Electric power plant substation—capacitor bank (Problem 6.83), Electronic photo flash unit (Section 7.9), Automobile ignition circuit (Section 7.9), Welding machine (Problem 7.86), Airbag igniter (Problem 8.78), Electrical analog to bodily functions—study of convulsions (Problem 8.82), Electronic sensing device (Problem 9.87), Power transmission system (Problem 9.93), Design a Colpitts oscillator (Problem 10.94), Stereo amplifier circuit (Problem 13.85), Gyrator circuit (Problem 16.69), Calculate number of stations allowable in AM broadcast band (Problem 18.63), Voice signal—Nyquist rate (Problem 18.65)
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ale29559_IFC.qxd, , 07/11/2008, , 07:40 PM, , Page 3, , COMPUTER TOOLS promote flexibility and meet ABET requirements, • PSpice is introduced in Chapter 3 and appears in special sections throughout the text. Appendix D serves, as a tutorial on PSpice for Windows for readers not familiar with its use. The special sections contain examples and practice problems using PSpice. Additional homework problems at the end of each chapter also, provide an opportunity to use PSpice., • MATLAB® is introduced through a tutorial in Appendix E to show its usage in circuit analysis. A number, of examples and practice problems are presented throughout the book in a manner that will allow the student, to develop a facility with this powerful tool. A number of end-of-chapter problems will aid in understanding, how to effectively use MATLAB., • KCIDE for Circuits is a working software environment developed at Cleveland State University. It is, designed to help the student work through circuit problems in an organized manner following the process, on problem-solving discussed in Section 1.8. Appendix F contains a description of how to use the software., Additional examples can be found at the web site, http://kcide.fennresearch.org/. The actual software package can be downloaded for free from this site. One of the best benefits from using this package is that it, automatically generates a Word document and/or a PowerPoint presentation., , CAREERS AND HISTORY of electrical engineering pioneers, Since a course in circuit analysis may be a student’s first exposure to electrical engineering, each chapter opens, with discussions about how to enhance skills that contribute to successful problem-solving or career-oriented, talks on a sub-discipline of electrical engineering. The chapter openers are intended to help students grasp, the scope of electrical engineering and give thought to the various careers available to EE graduates. The opening boxes include information on careers in electronics, instrumentation, electromagnetics, control systems,, engineering education, and the importance of good communication skills. Historicals throughout the text, provide brief biological sketches of such engineering pioneers as Faraday, Ampere, Edison, Henry, Fourier,, Volta, and Bell.
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ale29559_IFC.qxd, , 07/11/2008, , 07:40 PM, , Page 4, , OUR COMMITMENT TO ACCURACY, You have a right to expect an accurate textbook, and McGraw-Hill Engineering invests, considerable time and effort to ensure that we deliver one. Listed below are the many, steps we take in this process., OUR ACCURACY VERIFICATION PROCESS, First Round, Step 1: Numerous college engineering instructors review the manuscript and report, errors to the editorial team. The authors review their comments and make the necessary, corrections in their manuscript., Second Round, Step 2: An expert in the field works through every example and exercise in the final, manuscript to verify the accuracy of the examples, exercises, and solutions. The authors, review any resulting corrections and incorporate them into the final manuscript and solutions manual., Step 3: The manuscript goes to a copyeditor, who reviews the pages for grammatical and, stylistic considerations. At the same time, the expert in the field begins a second accuracy, check. All corrections are submitted simultaneously to the authors, who review and integrate the editing, and then submit the manuscript pages for typesetting., Third Round, Step 4: The authors review their page proofs for a dual purpose: 1) to make certain that, any previous corrections were properly made, and 2) to look for any errors they might, have missed., Step 5: A proofreader is assigned to the project to examine the new page proofs, double, check the authors' work, and add a fresh, critical eye to the book. Revisions are incorporated into a new batch of pages which the authors check again., Fourth Round, Step 6: The author team submits the solutions manual to the expert in the field, who, checks text pages against the solutions manual as a final review., Step 7: The project manager, editorial team, and author team review the pages for a, final accuracy check., The resulting engineering textbook has gone through several layers of quality assurance, and is verified to be as accurate and error-free as possible. Our authors and publishing, staff are confident that through this process we deliver textbooks that are industry leaders, in their correctness and technical integrity.
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ale29559_fm.qxd, , 07/28/2008, , 11:54 AM, , Page i, , Fundamentals of, , Electric Circuits
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ale29559_fm.qxd, , 07/28/2008, , 11:54 AM, , fourth, , Page iii, , edition, , Fundamentals of, , Electric Circuits, Charles K. Alexander, Department of Electrical and, Computer Engineering, Cleveland State University, , Matthew N. O. Sadiku, Department of, Electrical Engineering, Prairie View A&M University
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ale29559_fm.qxd, , 07/28/2008, , 11:54 AM, , Page iv, , FUNDAMENTALS OF ELECTRIC CIRCUITS, FOURTH EDITION, Published by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc., 1221 Avenue of, the Americas, New York, NY 10020. Copyright © 2009 by The McGraw-Hill Companies, Inc., All rights reserved. Previous editions © 2007, 2004, and 2000. No part of this publication may be, reproduced or distributed in any form or by any means, or stored in a database or retrieval system,, without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to,, in any network or other electronic storage or transmission, or broadcast for distance learning., Some ancillaries, including electronic and print components, may not be available to customers, outside the United States., This book is printed on acid-free paper., 1 2 3 4 5 6 7 8 9 0 VNH/VNH 0 9 8, ISBN 978–0–07–352955–4, MHID 0–07–352955–9, Global Publisher: Raghothaman Srinivasan, Director of Development: Kristine Tibbetts, Developmental Editor: Lora Neyens, Senior Marketing Manager: Curt Reynolds, Project Manager: Joyce Watters, Senior Production Supervisor: Sherry L. Kane, Lead Media Project Manager: Stacy A. Patch, Associate Design Coordinator: Brenda A. Rolwes, Cover Designer: Studio Montage, St. Louis, Missouri, (USE) Cover Image: Astronauts Repairing Spacecraft: © StockTrek/Getty Images;, Printed Circuit Board: Photodisc Collection/Getty Images, Lead Photo Research Coordinator: Carrie K. Burger, Compositor: ICC Macmillan Inc., Typeface: 10/12 Times Roman, Printer: R. R. Donnelley, Jefferson City, MO, Library of Congress Cataloging-in-Publication Data, Alexander, Charles K., Fundamentals of electric circuits / Charles K. Alexander, Matthew N. O. Sadiku. — 4th ed., p. cm., Includes index., ISBN 978–0–07–352955–4 — ISBN 0–07–352955–9 (hard copy : alk. paper) 1. Electric circuits., I. Sadiku, Matthew N. O. II. Title., TK454.A452 2009, 621.319'24—dc22, , www.mhhe.com, , 2008023020
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ale29559_fm.qxd, , 07/28/2008, , 11:54 AM, , Page v, , Dedicated to our wives, Kikelomo and Hannah, whose understanding and, support have truly made this book possible., Matthew, and, Chuck
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Contents, Preface xiii, Acknowledgments xviii, Guided Tour xx, A Note to the Student xxv, About the Authors xxvii, , Chapter 3, 3.1, 3.2, 3.3, 3.4, 3.5, , PART 1, , DC Circuits 2, , Chapter 1, , Basic Concepts 3, , 3.6, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, , Introduction 4, Systems of Units 4, Charge and Current 6, Voltage 9, Power and Energy 10, Circuit Elements 15, †, Applications 17, , 3.7, 3.8, 3.9, 3.10, , Methods of Analysis 81, , Introduction 82, Nodal Analysis 82, Nodal Analysis with Voltage, Sources 88, Mesh Analysis 93, Mesh Analysis with Current, Sources 98, †, Nodal and Mesh Analyses, by Inspection 100, Nodal Versus Mesh Analysis 104, Circuit Analysis with PSpice 105, †, Applications: DC Transistor Circuits 107, Summary 112, Review Questions 113, Problems 114, Comprehensive Problem 126, , 1.7.1 TV Picture Tube, 1.7.2 Electricity Bills, , 1.8, 1.9, , †, , Problem Solving 20, Summary 23, , Chapter 4, , Review Questions 24, Problems 24, Comprehensive Problems 27, , 4.1, 4.2, 4.3, 4.4, 4.5, 4.6, 4.7, , Chapter 2, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, , Basic Laws 29, , Introduction 30, Ohm’s Law 30, †, Nodes, Branches, and Loops 35, Kirchhoff’s Laws 37, Series Resistors and Voltage, Division 43, Parallel Resistors and Current, Division 45, †, Wye-Delta Transformations 52, †, Applications 58, , 4.8, 4.9, 4.10, , Circuit Theorems 127, , Introduction 128, Linearity Property 128, Superposition 130, Source Transformation 135, Thevenin’s Theorem 139, Norton’s Theorem 145, †, Derivations of Thevenin’s and Norton’s, Theorems 149, Maximum Power Transfer 150, Verifying Circuit Theorems with, PSpice 152, †, Applications 155, 4.10.1 Source Modeling, 4.10.2 Resistance Measurement, , 4.11, , Summary, , 160, , Review Questions 161, Problems 162, Comprehensive Problems 173, , 2.8.1 Lighting Systems, 2.8.2 Design of DC Meters, , 2.9, , Summary 64, Review Questions 66, Problems 67, Comprehensive Problems 78, , Chapter 5, 5.1, 5.2, , Operational Amplifiers 175, , Introduction 176, Operational Amplifiers 176, vii
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Contents, , viii, , 5.3, 5.4, 5.5, 5.6, 5.7, 5.8, 5.9, 5.10, , Ideal Op Amp 179, Inverting Amplifier 181, Noninverting Amplifier 183, Summing Amplifier 185, Difference Amplifier 187, Cascaded Op Amp Circuits 191, Op Amp Circuit Analysis with PSpice 194, †, Applications 196, 5.10.1 Digital-to-Analog Converter, 5.10.2 Instrumentation Amplifiers, , 5.11, , Summary, , 199, , Review Questions 201, Problems 202, Comprehensive Problems 213, , Chapter 6, 6.1, 6.2, 6.3, 6.4, 6.5, 6.6, , Introduction 216, Capacitors 216, Series and Parallel Capacitors 222, Inductors 226, Series and Parallel Inductors 230, †, Applications 233, , Summary, , 240, , Review Questions 241, Problems 242, Comprehensive Problems 251, , Chapter 7, 7.1, 7.2, 7.3, 7.4, 7.5, 7.6, 7.7, 7.8, 7.9, , 7.10, , First-Order Circuits 253, , Introduction 254, The Source-Free RC Circuit 254, The Source-Free RL Circuit 259, Singularity Functions 265, Step Response of an RC Circuit 273, Step Response of an RL Circuit 280, †, First-Order Op Amp Circuits 284, Transient Analysis with PSpice 289, †, Applications 293, 7.9.1, 7.9.2, 7.9.3, 7.9.4, , 8.1, 8.2, 8.3, 8.4, 8.5, 8.6, 8.7, 8.8, 8.9, 8.10, 8.11, , Capacitors and, Inductors 215, , 6.6.1 Integrator, 6.6.2 Differentiator, 6.6.3 Analog Computer, , 6.7, , Chapter 8, , Delay Circuits, Photoflash Unit, Relay Circuits, Automobile Ignition Circuit, , Summary, , Introduction 314, Finding Initial and Final Values 314, The Source-Free Series, RLC Circuit 319, The Source-Free Parallel RLC, Circuit 326, Step Response of a Series RLC, Circuit 331, Step Response of a Parallel RLC, Circuit 336, General Second-Order Circuits 339, Second-Order Op Amp Circuits 344, PSpice Analysis of RLC Circuits 346, †, Duality 350, †, Applications 353, 8.11.1 Automobile Ignition System, 8.11.2 Smoothing Circuits, , 8.12, , Summary, , 356, , Review Questions 357, Problems 358, Comprehensive Problems 367, , PART 2, , AC Circuits 368, , Chapter 9, , Sinusoids and Phasors 369, , 9.1, 9.2, 9.3, 9.4, 9.5, 9.6, 9.7, 9.8, , Introduction 370, Sinusoids 371, Phasors 376, Phasor Relationships for, Circuit Elements 385, Impedance and Admittance 387, †, Kirchhoff’s Laws in the Frequency, Domain 389, Impedance Combinations 390, †, Applications 396, 9.8.1 Phase-Shifters, 9.8.2 AC Bridges, , 9.9, , Summary, , 402, , Review Questions 403, Problems 403, Comprehensive Problems 411, , Chapter 10, , 299, , Review Questions 300, Problems 301, Comprehensive Problems 311, , Second-Order Circuits 313, , 10.1, 10.2, 10.3, , Sinusoidal Steady-State, Analysis 413, , Introduction 414, Nodal Analysis 414, Mesh Analysis 417
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Contents, , 10.4, 10.5, 10.6, 10.7, 10.8, 10.9, , Superposition Theorem 421, Source Transformation 424, Thevenin and Norton Equivalent, Circuits 426, Op Amp AC Circuits 431, AC Analysis Using PSpice 433, †, Applications 437, 10.9.1 Capacitance Multiplier, 10.9.2 Oscillators, , 10.10 Summary, , 441, , Review Questions 441, Problems 443, , Chapter 11, 11.1, 11.2, 11.3, 11.4, 11.5, 11.6, 11.7, 11.8, 11.9, , AC Power Analysis 457, , Introduction 458, Instantaneous and Average, Power 458, Maximum Average Power, Transfer 464, Effective or RMS Value 467, Apparent Power and, Power Factor 470, Complex Power 473, †, Conservation of AC Power 477, Power Factor Correction 481, †, Applications 483, 11.9.1 Power Measurement, 11.9.2 Electricity Consumption Cost, , 11.10 Summary, , 488, , Review Questions 490, Problems 490, Comprehensive Problems 500, , Chapter 12, , 12.11 Summary, , Introduction 504, Balanced Three-Phase Voltages 505, Balanced Wye-Wye Connection 509, Balanced Wye-Delta Connection 512, Balanced Delta-Delta, Connection 514, 12.6 Balanced Delta-Wye Connection 516, 12.7 Power in a Balanced System 519, 12.8 †Unbalanced Three-Phase, Systems 525, 12.9 PSpice for Three-Phase Circuits 529, 12.10 †Applications 534, 12.10.1 Three-Phase Power Measurement, 12.10.2 Residential Wiring, , 543, , Review Questions 543, Problems 544, Comprehensive Problems 553, , Chapter 13, 13.1, 13.2, 13.3, 13.4, 13.5, 13.6, 13.7, 13.8, 13.9, , Magnetically Coupled, Circuits 555, , Introduction 556, Mutual Inductance 557, Energy in a Coupled Circuit 564, Linear Transformers 567, Ideal Transformers 573, Ideal Autotransformers 581, †, Three-Phase Transformers 584, PSpice Analysis of Magnetically, Coupled Circuits 586, †, Applications 591, 13.9.1 Transformer as an Isolation Device, 13.9.2 Transformer as a Matching Device, 13.9.3 Power Distribution, , 13.10 Summary, , 597, , Review Questions 598, Problems 599, Comprehensive Problems 611, , Chapter 14, 14.1, 14.2, 14.3, 14.4, 14.5, 14.6, 14.7, , 14.8, , Lowpass Filter, Highpass Filter, Bandpass Filter, Bandstop Filter, , Active Filters 642, 14.8.1, 14.8.2, 14.8.3, 14.8.4, , 14.9, , Frequency Response 613, , Introduction 614, Transfer Function 614, †, The Decibel Scale 617, Bode Plots 619, Series Resonance 629, Parallel Resonance 634, Passive Filters 637, 14.7.1, 14.7.2, 14.7.3, 14.7.4, , Three-Phase Circuits 503, , 12.1, 12.2, 12.3, 12.4, 12.5, , ix, , First-Order Lowpass Filter, First-Order Highpass Filter, Bandpass Filter, Bandreject (or Notch) Filter, , Scaling, , 648, , 14.9.1 Magnitude Scaling, 14.9.2 Frequency Scaling, 14.9.3 Magnitude and Frequency Scaling, , 14.10 Frequency Response Using, PSpice 652, 14.11 Computation Using MATLAB, , 655
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Contents, , x, , 14.12, , †, , Applications 657, , 17.3, , 14.12.1 Radio Receiver, 14.12.2 Touch-Tone Telephone, 14.12.3 Crossover Network, , 14.13 Summary, , 663, , Review Questions 664, Problems 665, Comprehensive Problems 673, , Symmetry Considerations 764, 17.3.1 Even Symmetry, 17.3.2 Odd Symmetry, 17.3.3 Half-Wave Symmetry, , 17.4, 17.5, 17.6, 17.7, , Circuit Applications 774, Average Power and RMS Values 778, Exponential Fourier Series 781, Fourier Analysis with PSpice 787, 17.7.1 Discrete Fourier Transform, 17.7.2 Fast Fourier Transform, , 17.8, , PART 3, , Advanced Circuit, Analysis 674, , Chapter 15, , Introduction to the Laplace, Transform 675, , 15.1, 15.2, 15.3, 15.4, , 15.5, 15.6, 15.7, , †, , Applications 793, , 17.8.1 Spectrum Analyzers, 17.8.2 Filters, , 17.9, , Summary, , 796, , Review Questions 798, Problems 798, Comprehensive Problems 807, , Introduction 676, Definition of the Laplace Transform 677, Properties of the Laplace Transform 679, The Inverse Laplace Transform 690, , Chapter 18, , 15.4.1 Simple Poles, 15.4.2 Repeated Poles, 15.4.3 Complex Poles, , 18.1, 18.2, 18.3, , The Convolution Integral 697, †, Application to Integrodifferential, Equations 705, Summary 708, , 18.4, 18.5, 18.6, , Review Questions 708, Problems 709, , 18.7, , Fourier Transform 809, , Introduction 810, Definition of the Fourier Transform 810, Properties of the Fourier, Transform 816, Circuit Applications 829, Parseval’s Theorem 832, Comparing the Fourier and Laplace, Transforms 835, †, Applications 836, 18.7.1 Amplitude Modulation, 18.7.2 Sampling, , Chapter 16, 16.1, 16.2, 16.3, 16.4, 16.5, 16.6, , Applications of the Laplace, Transform 715, , Introduction 716, Circuit Element Models 716, Circuit Analysis 722, Transfer Functions 726, State Variables 730, †, Applications 737, 16.6.1 Network Stability, 16.6.2 Network Synthesis, , 16.7, , Summary, , 745, , Review Questions 746, Problems 747, Comprehensive Problems 754, , 18.8, , 17.1, 17.2, , The Fourier Series 755, , Introduction 756, Trigonometric Fourier Series 756, , 839, , Review Questions 840, Problems 841, Comprehensive Problems 847, , Chapter 19, 19.1, 19.2, 19.3, 19.4, 19.5, 19.6, 19.7, 19.8, , Chapter 17, , Summary, , 19.9, , Two-Port Networks 849, , Introduction 850, Impedance Parameters 850, Admittance Parameters 855, Hybrid Parameters 858, Transmission Parameters 863, †, Relationships Between, Parameters 868, Interconnection of Networks 871, Computing Two-Port Parameters, Using PSpice 877, †, Applications 880, 19.9.1 Transistor Circuits, 19.9.2 Ladder Network Synthesis
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Contents, , 19.10 Summary, , 889, , Review Questions 890, Problems 890, Comprehensive Problems 901, , Appendix A, , xi, , Appendix D, , PSpice for Windows A-21, , Appendix E, , MATLAB A-46, , Appendix F, , KCIDE for Circuits A-65, , Appendix G, , Answers to Odd-Numbered, Problems A-75, , Simultaneous Equations and Matrix, Inversion A, , Selected Bibliography B-1, , Appendix B, , Complex Numbers A-9, , Index I-1, , Appendix C, , Mathematical Formulas A-16
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ale29559_fm.qxd, , 08/08/2008, , 10:16 AM, , Page xiii, , Preface, You may be wondering why we chose a photo of astronauts working, in space on the Space Station for the cover. We actually chose it for, several reasons. Obviously, it is very exciting; in fact, space represents, the most exciting frontier for the entire world! In addition, much of the, station itself consists of all kinds of circuits! One of the most significant circuits within the station is its power distribution system. It is a, complete and self contained, modern power generation and distribution, system. That is why NASA (especially NASA-Glenn) continues to be, at the forefront of both theoretical as well as applied power system, research and development. The technology that has gone into the development of space exploration continues to find itself impacting terrestrial technology in many important ways. For some of you, this will be, an important career path., , FEATURES, New to This Edition, A course in circuit analysis is perhaps the first exposure students have, to electrical engineering. This is also a place where we can enhance, some of the skills that they will later need as they learn how to design., In the fourth edition, we have included a very significant new, feature to help students enhance skills that are an important part of the, design process. We call this new feature, design a problem., We know it is not possible to fully develop a student’s design skills, in a fundamental course like circuits. To fully develop design skills a, student needs a design experience normally reserved for their senior, year. This does not mean that some of those skills cannot be developed, and exercised in a circuits course. The text already included openended questions that help students use creativity, which is an important part of learning how to design. We already have some questions, that are open desired to add much more into our text in this important, area and have developed an approach to do just that. When we develop, problems for the student to solve our goal is that in solving the problem the student learn more about the theory and the problem solving, process. Why not have the students design problems like we do? That, is exactly what we will do in each chapter. Within the normal problem, set, we have a set of problems where we ask the student to design a, problem. This will have two very important results. The first will be a, better understanding of the basic theory and the second will be the, enhancement of some of the student’s basic design skills., We are making effective use of the principle of learning by teaching. Essentially we all learn better when we teach a subject. Designing effective problems is a key part of the teaching process. Students, xiii
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ale29559_fm.qxd, , xiv, , 07/28/2008, , 11:54 AM, , Page xiv, , Preface, , should also be encouraged to develop problems, when appropriate,, which have nice numbers and do not necessarily overemphasize complicated mathematical manipulations., Additionally we have changed almost 40% of the Practice Problems with the idea to better reflect more real component values and to, help the student better understand the problem and have added 121, design a problem problems. We have also changed and added a total, of 357 end-of-chapter problems (this number contains the new design, a problem problems). This brings up a very important advantage to our, textbook, we have a total of 2404 Examples, Practice Problems,, Review Questions, and end-of-chapter problems!, , Retained from Previous Editions, The main objective of the fourth edition of this book remains the same, as the previous editions—to present circuit analysis in a manner that is, clearer, more interesting, and easier to understand than other circuit text,, and to assist the student in beginning to see the “fun” in engineering., This objective is achieved in the following ways:, • Chapter Openers and Summaries, Each chapter opens with a discussion about how to enhance skills, which contribute to successful problem solving as well as successful careers or a career-oriented talk on a sub-discipline of electrical engineering. This is followed by an introduction that links the, chapter with the previous chapters and states the chapter objectives., The chapter ends with a summary of key points and formulas., • Problem Solving Methodology, Chapter 1 introduces a six-step method for solving circuit problems, which is used consistently throughout the book and media supplements to promote best-practice problem-solving procedures., • Student Friendly Writing Style, All principles are presented in a lucid, logical, step-by-step manner., As much as possible, we avoid wordiness and giving too much, detail that could hide concepts and impede overall understanding of, the material., • Boxed Formulas and Key Terms, Important formulas are boxed as a means of helping students sort, out what is essential from what is not. Also, to ensure that students, clearly understand the key elements of the subject matter, key, terms are defined and highlighted., • Margin Notes, Marginal notes are used as a pedagogical aid. They serve multiple, uses such as hints, cross-references, more exposition, warnings,, reminders not to make some particular common mistakes, and, problem-solving insights., • Worked Examples, Thoroughly worked examples are liberally given at the end of every, section. The examples are regarded as a part of the text and are, clearly explained without asking the reader to fill in missing steps., Thoroughly worked examples give students a good understanding of the solution process and the confidence to solve problems
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ale29559_fm.qxd, , 07/28/2008, , 11:54 AM, , Page xv, , Preface, , •, , •, , •, , •, , •, , •, , •, , themselves. Some of the problems are solved in two or three different ways to facilitate a substantial comprehension of the subject, material as well as a comparison of different approaches., Practice Problems, To give students practice opportunity, each illustrative example is, immediately followed by a practice problem with the answer. The, student can follow the example step by step to aid in the solution, of the practice problem without flipping pages or looking at the, end of the book for answers. The practice problem is also intended, to test a student’s understanding of the preceding example. It will, reinforce their grasp of the material before the student can move, on to the next section. Complete solutions to the practice problems, are available to students on ARIS., Application Sections, The last section in each chapter is devoted to practical application, aspects of the concepts covered in the chapter. The material covered in the chapter is applied to at least one or two practical problems or devices. This helps students see how the concepts are, applied to real-life situations., Review Questions, Ten review questions in the form of multiple-choice objective items, are provided at the end of each chapter with answers. The review, questions are intended to cover the little “tricks” that the examples, and end-of-chapter problems may not cover. They serve as a selftest device and help students determine how well they have mastered the chapter., Computer Tools, In recognition of the requirements by ABET® on integrating computer tools, the use of PSpice, MATLAB, KCIDE for Circuits, and, developing design skills are encouraged in a student-friendly manner. PSpice is covered early on in the text so that students can, become familiar and use it throughout the text. Appendix D serves, as a tutorial on PSpice for Windows. MATLAB is also introduced, early in the book with a tutorial available in Appendix E. KCIDE, for Circuits is a brand new, state-of-the-art software system designed, to help the students maximize their chance of success in problem, solving. It is introduced in Appendix F. Finally, design a problem, problems have been introduced, for the first time. These are meant, to help the student develop skills that will be needed in the design, process., Historical Tidbits, Historical sketches throughout the text provide profiles of important, pioneers and events relevant to the study of electrical engineering., Early Op Amp Discussion, The operational amplifier (op amp) as a basic element is introduced, early in the text., Fourier and Laplace Transforms Coverage, To ease the transition between the circuit course and signals and, systems courses, Fourier and Laplace transforms are covered, lucidly and thoroughly. The chapters are developed in a manner, that the interested instructor can go from solutions of first-order, , xv
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ale29559_fm.qxd, , xvi, , 07/28/2008, , 11:54 AM, , Page xvi, , Preface, , •, , •, , •, , •, , •, , •, , circuits to Chapter 15. This then allows a very natural progression, from Laplace to Fourier to AC., Four Color Art Program, A completely redesigned interior design and four color art program, bring circuit drawings to life and enhance key pedagogical elements throughout the text., Extended Examples, Examples worked in detail according to the six-step problem solving method provide a roadmap for students to solve problems in a, consistent fashion. At least one example in each chapter is developed in this manner., EC 2000 Chapter Openers, Based on ABET’s new skill-based CRITERION 3, these chapter, openers are devoted to discussions as to how students can acquire, the skills that will lead to a significantly enhanced career as an, engineer. Because these skills are so very important to the student, while in college as well as in their career, we will use the heading, “Enhancing your Skills and your Career.”, Homework Problems, There are 358 new or changed end-of-chapter problems which will, provide students with plenty of practice as well as reinforce key, concepts., Homework Problem Icons, Icons are used to highlight problems that relate to engineering design, as well as problems that can be solved using PSpice or MATLAB., KCIDE for Circuits Appendix F, A new Appendix F provides a tutorial on the Knowledge Capturing Integrated Design Environment (KCIDE for Circuits) software,, available on ARIS., , Organization, This book was written for a two-semester or three-quarter course in, linear circuit analysis. The book may also be used for a one-semester, course by a proper selection of chapters and sections by the instructor., It is broadly divided into three parts., • Part 1, consisting of Chapters 1 to 8, is devoted to dc circuits. It, covers the fundamental laws and theorems, circuits techniques, and, passive and active elements., • Part 2, which contains Chapter 9 to 14, deals with ac circuits. It, introduces phasors, sinusoidal steady-state analysis, ac power, rms, values, three-phase systems, and frequency response., • Part 3, consisting of Chapters 15 to 19, is devoted to advanced, techniques for network analysis. It provides students with a solid, introduction to the Laplace transform, Fourier series, Fourier transform, and two-port network analysis., The material in three parts is more than sufficient for a two-semester, course, so the instructor must select which chapters or sections to cover., Sections marked with the dagger sign (†) may be skipped, explained, briefly, or assigned as homework. They can be omitted without loss of
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ale29559_fm.qxd, , 07/28/2008, , 11:54 AM, , Page xvii, , Preface, , continuity. Each chapter has plenty of problems grouped according to, the sections of the related material and diverse enough that the instructor can choose some as examples and assign some as homework. As, stated earlier, we are using three icons with this edition. We are using, to denote problems that either require PSpice in the solution process,, where the circuit complexity is such that PSpice would make the solution process easier, and where PSpice makes a good check to see if the, problem has been solved correctly. We are using, to denote problems, where MATLAB is required in the solution process, where MATLAB, makes sense because of the problem makeup and its complexity, and, where MATLAB makes a good check to see if the problem has been, solved correctly. Finally, we use, to identify problems that help the, student develop skills that are needed for engineering design. More difficult problems are marked with an asterisk (*). Comprehensive problems follow the end-of-chapter problems. They are mostly applications, problems that require skills learned from that particular chapter., , Prerequisites, As with most introductory circuit courses, the main prerequisites, for, a course using the text, are physics and calculus. Although familiarity, with complex numbers is helpful in the later part of the book, it is not, required. A very important asset of this text is that ALL the mathematical equations and fundamentals of physics needed by the student,, are included in the text., , Supplements, McGraw-Hill’s ARIS—Assessment, Review, and Instruction, System is a complete, online tutorial, electronic homework, and course, management system, designed for greater ease of use than any other, system available. Available on adoption, instructors can create and, share course materials and assignments with other instructors, edit questions and algorithms, import their own content, and create announcements and due dates for assignments. ARIS has automatic grading and, reporting of easy-to-assign algorithmically-generated homework,, quizzing, and testing. Once a student is registered in the course, all student activity within McGraw-Hill’s ARIS is automatically recorded and, available to the instructor through a fully integrated grade book that can, be downloaded to Excel. Also included on ARIS are a solutions manual, text image files, transition guides to instructors, and Network, Analysis Tutorials, software downloads, complete solutions to text, practice problems, FE Exam questions, flashcards, and web links to students. Visit www.mhhe.com/alexander., Knowledge Capturing Integrated Design Environment for Circuits, (KCIDE for Circuits) This new software, developed at Cleveland State, University and funded by NASA, is designed to help the student work, through a circuits problem in an organized manner using the six-step, problem-solving methodology in the text. KCIDE for Circuits allows, students to work a circuit problem in PSpice and MATLAB, track the, , xvii
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ale29559_fm.qxd, , xviii, , 07/28/2008, , 11:54 AM, , Page xviii, , Preface, , evolution of their solution, and save a record of their process for future, reference. In addition, the software automatically generates a Word, document and/or a PowerPoint presentation. Appendix F contains a, description of how to use the software. Additional examples can be, found at the web site, http://kcide.fennresearch.org/, which is linked, from ARIS. The software package can be downloaded for free., Problem Solving Made Almost Easy, a companion workbook to Fundamentals of Electric Circuits, is available on ARIS for students who, wish to practice their problem-solving techniques. The workbook contains a discussion of problem-solving strategies and 150 additional, problems with complete solutions provided., C.O.S.M.O.S This CD, available to instructors only, is a powerful solutions manual tool to help instructors streamline the creation of assignments, quizzes, and tests by using problems and solutions from the, textbook, as well as their own custom material. Instructors can edit, textbook end-of-chapter problems as well as track which problems have, been assigned., Although the textbook is meant to be self-explanatory and act as, a tutor for the student, the personal contact in teaching is not forgotten. It is hoped that the book and supplemental materials supply the, instructor with all the pedagogical tools necessary to effectively present the material., , Acknowledgements, We would like to express our appreciation for the loving support we, have received from our wives (Hannah and Kikelomo), daughters, (Christina, Tamara, Jennifer, Motunrayo, Ann, and Joyce), son (Baixi),, and our extended family members., At McGraw-Hill, we would like to thank the following editorial, and production staff: Raghu Srinivasan, publisher and senior sponsoring editor; Lora Kalb-Neyens, developmental editors; Joyce Watters,, project manager; Carrie Burger, photo researcher; and Brenda Rolwes,, designer. Also, we appreciate the hard work of Tom Hartley at the University of Akron for his very detailed evaluation of various elements, of the text and his many valued suggestions for continued improvement of this textbook., We wish to thank Yongjian Fu and his outstanding team of students, Bramarambha Elka and Saravaran Chinniah, for their efforts in, the development of KCIDE for Circuits. Their efforts to help us continue to improve this software are also appreciated., The fourth edition has benefited greatly from the many outstanding reviewers and symposium attendees who contributed to the success, of the first three editions! In addition, the following have made important contributions to the fourth edition (in alphabetical order):, Tom Brewer, Georgia Tech, Andy Chan, City University of Hong Kong, Alan Tan Wee Chiat, Multimedia University, Norman Cox, University of Missouri-Rolla, Walter L. Green, University of Tennessee
Page 23 : ale29559_fm.qxd, , 07/28/2008, , 11:54 AM, , Page xix, , Preface, , Dr. Gordon K. Lee, San Diego State University, Gary Perks, Cal Poly State University, San Luis Obispo, Dr. Raghu K. Settaluri, Oregon State University, Ramakant Srivastava, University of Florida, John Watkins, Wichita State University, Yik-Chung Wu, The University of Hong Kong, Xiao-Bang Xu, Clemson University, Finally, we appreciate the feedback received from instructors and students who used the previous editions. We want this to continue, so please, keep sending us emails or direct them to the publisher. We can be reached, at
[email protected] for Charles Alexander and
[email protected] for, Matthew Sadiku., C. K. Alexander and M.N.O. Sadiku, , xix
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ale29559_fm.qxd, , 07/28/2008, , 11:54 AM, , Page xx, , GUIDED TOUR, The main objective of this book is to present circuit analysis in a manner that is clearer, more interesting, and easier to understand than other, texts. For you, the student, here are some features to help you study, and be successful in this course., The four color art program brings circuit drawings to life and enhances key concepts throughout, the text., , 1.5, , Power and Energy, , 11, , To relate power and energy to voltage and current, we recall from, physics that:, Power is the time rate of expending or absorbing energy, measured in, watts (W)., , We write this relationship as, i, , p⫽, ¢, , dw, dt, , where p is power in watts (W), w is energy in joules (J), and t is time, in seconds (s). From Eqs. (1.1), (1.3), and (1.5), it follows that, p⫽, 20, , Chapter 1, , 1.8, , Basic Concepts, , p ⫽ vi, , Chapter 1, , (b), , When the voltage and current directions, conform to Fig. 1.8 (b), we have the active sign convention and p ⫽ ⫹vi., , 3A, , 3A, +, , −, , 4V, , 4V, , −, , +, (a), , (b), , Figure 1.9, Two cases of an element with an absorbing, power of 12 W: (a) p ⫽ 4 ⫻ 3 ⫽ 12 W,, (b) p ⫽ 4 ⫻ 3 ⫽ 12 W., , 3A, , 3A, , +, , −, , 4V, , 4V, , −, , +, (a), , (b), , Figure 1.10, Two cases of an element with a supplying, power of 12 W: (a) p ⫽ ⫺4 ⫻ 3 ⫽, ⫺12W, (b) p ⫽ ⫺4 ⫻ 3 ⫽ ⫺12 W., , Basic Concepts, , 2Ω, , Example 1.10, 2Ω, , Solution:, , 4Ω, 5V, , 5V, , +, −, , 8Ω, , i1, , v1, , + v −, 2Ω, , +, −, , Loop 1, , 3V, , i3, i2, , +, v8Ω, −, , 8Ω, , 4Ω, + v4Ω −, −, +, , Loop 2, , 3V, , 1.9, , Figure 1.19, , 2Ω, , 5V, , +, −, , Figure 1.20, Problem defintion., , Summary, , 23, , Figure 1.21, , Illustrative example., , Using nodal analysis., , Therefore,, we will solve for i8⍀ using nodal analysis., 4Ω, 4. Attempt a problem solution. We first write down all of the equationsi8Ωwe will need in order to find i8⍀., 8Ω, , −, +, , 3V, , i8⍀ ⫽ i2,, , i2 ⫽, , v1, ,, 8, , i8⍀ ⫽, , v1, 8, , v1 ⫺ 5, v1 ⫺ 0, v1 ⫹ 3, ⫹, ⫹, ⫽0, 2, 8, 4, , So we now have a very high degree of confidence in the accuracy, of our answer., 6. Has the problem been solved Satisfactorily? If so, present the solution; if not, then return to step 3 and continue through the process, again. This problem has been solved satisfactorily., , The current through the 8-⍀ resistor is 0.25 A flowing down through, the 8-⍀ resistor., , Now we can solve for v1., v1 ⫺ 5, v1 ⫺ 0, v1 ⫹ 3, 8c, ⫹, ⫹, d ⫽0, 2, 8, 4, leads to (4v1 ⫺ 20) ⫹ (v1) ⫹ (2v1 ⫹ 6) ⫽ 0, v1, 2, 7v1 ⫽ ⫹14,, v1 ⫽ ⫹2 V,, i8⍀ ⫽, ⫽ ⫽ 0.25 A, 8, 8, 5. Evaluate the solution and check for accuracy. We can now use, Kirchhoff’s voltage law (KVL) to check the results., v1 ⫺ 5, 2⫺5, 3, ⫽, ⫽ ⫺ ⫽ ⫺1.5 A, 2, 2, 2, i2 ⫽ i8⍀ ⫽ 0.25 A, v1 ⫹ 3, 2⫹3, 5, i3 ⫽, ⫽, ⫽ ⫽ 1.25 A, 4, 4, 4, i1 ⫹ i2 ⫹ i3 ⫽ ⫺1.5 ⫹ 0.25 ⫹ 1.25 ⫽ 0, (Checks.), i1 ⫽, , Applying KVL to loop 1,, ⫺5 ⫹ v2⍀ ⫹ v8⍀ ⫽ ⫺5 ⫹ (⫺i1 ⫻ 2) ⫹ (i2 ⫻ 8), ⫽ ⫺5 ⫹ (⫺(⫺1.5)2) ⫹ (0.25 ⫻ 8), ⫽ ⫺5 ⫹ 3 ⫹ 2 ⫽ 0, (Checks.), Applying KVL to loop 2,, , A six-step problem-solving methodology is introduced in Chapter 1 and, incorporated into worked examples, throughout the text to promote, sound, step-by-step problem-solving, practices., , (a), , analysis. To solve for i8⍀ using mesh analysis will require writing, two simultaneous equations to find the two loop currents indicated in, Fig. 1.21. Using nodal analysis requires solving for only one unknown., This is the easiest approach., , Solve for the current flowing through the 8-⍀ resistor in Fig. 1.19., , 1. Carefully Define the problem. This is only a simple example, but, we can already see that we do not know the polarity on the 3-V source., We have the following options. We can ask the professor what the, polarity should be. If we cannot ask, then we need to make a decision, on what to do next. If we have time to work the problem both ways,, we can solve for the current when the 3-V source is plus on top and, then plus on the bottom. If we do not have the time to work it both, ways, assume a polarity and then carefully document your decision., Let us assume that the professor tells us that the source is plus on the, bottom as shown in Fig. 1.20., 2. Present everything you know about the problem. Presenting all that, we know about the problem involves labeling the circuit clearly so that, we define what we seek., Given the circuit shown in Fig. 1.20, solve for i8⍀., We now check with the professor, if reasonable, to see if the problem is properly defined., 3. Establish a set of Alternative solutions and determine the one that, promises the greatest likelihood of success. There are essentially three, techniques that can be used to solve this problem. Later in the text you, will see that you can use circuit analysis (using Kirchhoff’s laws and, Ohm’s law), nodal analysis, and mesh analysis., To solve for i8⍀ using circuit analysis will eventually lead to a, solution, but it will likely take more work than either nodal or mesh, , Unless otherwise stated, we will follow the passive sign convention throughout this text. For example, the element in both circuits of, Fig. 1.9 has an absorbing power of ⫹12 W because a positive current, enters the positive terminal in both cases. In Fig. 1.10, however, the, element is supplying power of ⫹12 W because a positive current enters, the negative terminal. Of course, an absorbing power of ⫺12 W is, equivalent to a supplying power of ⫹12 W. In general,, ⫹Power absorbed ⫽ ⫺Power supplied, , reduce effort and increase accuracy. Again, Now, we want, to stress, let us, look atthat, thistime, process for a student taking an electrical, spent carefully defining the problemand, andcomputer, investigating, alternative, engineering, foundations course. (The basic process also, approaches to its solution will pay big applies, dividends, Evaluating, the, to later., almost, every engineering, course.) 22, Keep in mind that, alternatives and determining which promises, the the, greatest, although, steps likelihood, have been of, simplified to apply to academic types of, success may be difficult but will be well, worth the, problems,, the effort., processDocument, as stated always needs to be followed. We conthis process well since you will want sider, to come, back example., to it if the first, a simple, approach does not work., 4. Attempt a problem solution. Now is the time to actually begin, solving the problem. The process you follow must be well documented, , p = −vi, , (1.7), , Passive sign convention is satisfied when the current enters through, the positive terminal of an element and p ⫽ ⫹vi. If the current enters, through the negative terminal, p ⫽ ⫺vi., 21, , −, , Figure 1.8, , The power p in Eq. (1.7) is a time-varying quantity and is called the, instantaneous power. Thus, the power absorbed or supplied by an element is the product of the voltage across the element and the current, through it. If the power has a ⫹ sign, power is being delivered to or, absorbed by the element. If, on the other hand, the power has a ⫺ sign,, power is being supplied by the element. But how do we know when, the power has a negative or a positive sign?, Current direction and voltage polarity play a major role in determining the sign of power. It is therefore important that we pay attention to the relationship between current i and voltage v in Fig. 1.8(a)., The voltage polarity and current direction must conform with those, shown in Fig. 1.8(a) in order for the power to have a positive sign., This is known as the passive sign convention. By the passive sign convention, current enters through the positive polarity of the voltage. In, this case, p ⫽ ⫹vi or vi 7 0 implies that the element is absorbing, power. However, if p ⫽ ⫺vi or vi 6 0, as in Fig. 1.8(b), the element, is releasing or supplying power., , 1. Carefully Define the problem. This may be the most important part, of the process, because it becomes the foundation for all the rest of the, steps. In general, the presentation of engineering problems is somewhat, incomplete. You must do all you can to make sure you understand the, problem as thoroughly as the presenter of the problem understands it., Time spent at this point clearly identifying the problem will save you, 1.8 Problem Solving, considerable time and frustration later. As a student, you can clarify a, problem statement in a textbook by asking your professor. A problem, in that, orderyou, to consult, present several, a detailed, solution if successful, and to evaluate the, presented to you in industry may require, indiyou are not, viduals. At this step, it is important to process, developifquestions, thatsuccessful., need to This detailed evaluation may lead to, corrections, thatIf can, be addressed before continuing the solution, process., youthen, havelead, suchto a successful solution. It can also lead, alternatives, to try. Many, questions, you need to consult with to, thenew, appropriate, individuals, or times, it is wise to fully set up a solubefore With, putting, numbers, into equations. This will help in checking, resources to obtain the answers to thosetion, questions., those, answers,, yourthat, results., you can now refine the problem, and use, refinement as the probEvaluate the solution and check for accuracy. You now thoroughly, lem statement for the rest of the solution5.process., whatYou, youare, have, accomplished., Decide if you have an acceptable, 2. Present everything you know about evaluate, the problem., now, ready, solution,, one thatand, youitswant, to present to your team, boss, or professor., to write down everything you know about, the problem, possible, 6. Has, problem, beenlater., solved Satisfactorily? If so, present the solusolutions. This important step will save you, timethe, and, frustration, tion;and, if not,, then return, tothat, step 3 and continue through the process, 3. Establish a set of Alternative solutions, determine, the one, again., Now every, you need, to present, promises the greatest likelihood of success., Almost, problem, will your solution or try another alternative., point, presenting, have a number of possible paths that can, leadAttothis, a solution., It is highlyyour solution may bring closure to the, process., Often, however,, presentation of a solution leads to further, desirable to identify as many of those paths, as possible., At this point,, of the, problem, and the process continues. Folyou also need to determine what toolsrefinement, are available, to you,, suchdefinition,, as, lowing, this, process, will, eventually, lead to a satisfactory conclusion., PSpice and MATLAB and other software packages that can greatly, , v, , −, , Reference polarities for power using the, passive sign convention: (a) absorbing, power, (b) supplying power., , Problem Solving, , 1. Carefully Define the problem., 2. Present everything you know about the problem., 3. Establish a set of Alternative solutions and determine the one that, promises the greatest likelihood of success., 4. Attempt a problem solution., 5. Evaluate the solution and check for accuracy., 6. Has the problem been solved Satisfactorily? If so, present the, solution; if not, then return to step 3 and continue through the, process again., , +, , v, , p = +vi, , (1.6), , or, , Although the problems to be solved during one’s career will vary in, complexity and magnitude, the basic principles to be followed remain, the same. The process outlined here is the one developed by the, authors over many years of problem solving with students, for the, solution of engineering problems in industry, and for problem solving, in research., We will list the steps simply and then elaborate on them., , xx, , dw dq, dw, ⫽, ⴢ, ⫽ vi, dt, dq dt, , i, +, , (1.5), , ⫺v8⍀ ⫹ v4⍀ ⫺ 3 ⫽ ⫺(i2 ⫻ 8) ⫹ (i3 ⫻ 4) ⫺ 3, ⫽ ⫺(0.25 ⫻ 8) ⫹ (1.25 ⫻ 4) ⫺ 3, ⫽ ⫺2 ⫹ 5 ⫺ 3 ⫽ 0, (Checks.), , Try applying this process to some of the more difficult problems at the, end of the chapter., , 1.9, , Summary, , 1. An electric circuit consists of electrical elements connected, together., 2. The International System of Units (SI) is the international measurement language, which enables engineers to communicate their, results. From the six principal units, the units of other physical, quantities can be derived., 3. Current is the rate of charge flow., i⫽, , dq, dt, , 4. Voltage is the energy required to move 1 C of charge through an, element., v⫽, , dw, dq, , 5. Power is the energy supplied or absorbed per unit time. It is also, the product of voltage and current., p⫽, , dw, ⫽ vi, dt, , 6. According to the passive sign convention, power assumes a positive sign when the current enters the positive polarity of the voltage, across an element., 7. An ideal voltage source produces a specific potential difference, across its terminals regardless of what is connected to it. An ideal, current source produces a specific current through its terminals, regardless of what is connected to it., 8. Voltage and current sources can be dependent or independent. A, dependent source is one whose value depends on some other circuit variable., 9. Two areas of application of the concepts covered in this chapter, are the TV picture tube and electricity billing procedure., , Practice Problem 1.10
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ale29559_fm.qxd, , 07/28/2008, , 11:54 AM, , Page xxi, , Guided Tour, , Chapter 3, , 90, , Example 3.3, , 2V, , v1, , Solution:, The supernode contains the 2-V source, nodes 1 and 2, and the 10-⍀, resistor. Applying KCL to the supernode as shown in Fig. 3.10(a) gives, , v2, , +−, 2Ω, , 2A, , Each illustrative example is immediately followed by, a practice problem and answer to test understanding of, the preceding example., , Methods of Analysis, , For the circuit shown in Fig. 3.9, find the node voltages., , 10 Ω, , 2 ⫽ i1 ⫹ i2 ⫹ 7, 4Ω, , PSpice® for Windows is a student-friendly tool introduced to students early in the text and used throughout, with discussions and examples at the end of each, appropriate chapter., , Expressing i1 and i2 in terms of the node voltages, , 7A, , 2⫽, Figure 3.9, , v1 ⫺ 0, v2 ⫺ 0, ⫹, ⫹7, 2, 4, , 8 ⫽ 2v1 ⫹ v2 ⫹ 28, , 1, , or, , For Example 3.3., , v2 ⫽ ⫺20 ⫺ 2v1, , (3.3.1), , To get the relationship between v1 and v2, we apply KVL to the circuit, in Fig. 3.10(b). Going around the loop, we obtain, ⫺v1 ⫺ 2 ⫹ v2 ⫽ 0, , 1, , v2 ⫽ v1 ⫹ 2, , xxi, , (3.3.2), , From Eqs. (3.3.1) and (3.3.2), we write, v2 ⫽ v1 ⫹ 2 ⫽ ⫺20 ⫺ 2v1, or, 3v1 ⫽ ⫺22, , 1, , v1 ⫽ ⫺7.333 V, , and v2 ⫽ v1 ⫹ 2 ⫽ ⫺5.333 V. Note that the 10-⍀ resistor does not, make any difference because it is connected across the supernode., , 2 v2, , 4Ω, , 2V, , 1, , i2 7 A, , i1, 2Ω, , +, 7A, , +−, , 1 v1, 2A, 2A, , The last section in each chapter is devoted to applications of the concepts covered in the chapter to help, students apply the concepts to real-life situations., , 2, +, v2, , v1, −, , −, (b), , (a), , Figure 3.10, Applying: (a) KCL to the supernode, (b) KVL to the loop., , Practice Problem 3.3, , 3Ω, , +−, , 21 V +, −, , Find v and i in the circuit of Fig. 3.11., , 9V, , 4Ω, +, v, −, , 2Ω, , Answer: ⫺0.6 V, 4.2 A., i, 6Ω, , Figure 3.11, For Practice Prob. 3.3., Chapter 3, , 106, , Methods of Analysis, 120.0000, 1, , 3.9, , 81.2900, , R1, , 2, , 20, +, 120 V, −, , R3, , 89.0320, 3, , 10, IDC, , V1, , R2, , R4, , 30, , 40, , 3A, , I1, , 0, , Figure 3.32, For Example 3.10; the schematic of the circuit in Fig. 3.31., , are displayed on VIEWPOINTS and also saved in output file, exam310.out. The output file includes the following:, , E, , NODE VOLTAGE, NODE VOLTAGE NODE VOLTAGE, (1), 120.0000 (2), 81.2900 (3), 89.0320, R1, , 1, , 100 Ω, , R2, , +, 24 V, , 2, , 3, , −, , 2, , R3, , 60 Ω, , 50 Ω, , +, −, , 25 Ω, , 8, , R4, , 4, , V1, 1.333E + 00, , 30 Ω, , E1, , −+, , R6, , 4, , 1, , − +, , 2, , For the circuit in Fig. 3.33, use PSpice to find the node voltages., 2A, , 107, , R5, , indicating that V1 ⫽ 120 V, V2 ⫽ 81.29 V, V3 ⫽ 89.032 V., , Practice Problem 3.10, , Applications: DC Transistor Circuits, , Solution:, The schematic is shown in Fig. 3.35. (The schematic in Fig. 3.35, includes the output results, implying that it is the schematic displayed, on the screen after the simulation.) Notice that the voltage-controlled, voltage source E1 in Fig. 3.35 is connected so that its input is the, voltage across the 4-⍀ resistor; its gain is set equal to 3. In order to, display the required currents, we insert pseudocomponent IPROBES in, the appropriate branches. The schematic is saved as exam311.sch and, simulated by selecting Analysis/Simulate. The results are displayed on, IPROBES as shown in Fig. 3.35 and saved in output file exam311.out., From the output file or the IPROBES, we obtain i1 ⫽ i2 ⫽ 1.333 A and, i3 ⫽ 2.667 A., , 1.333E + 00, , 2.667E + 00, , 0, 200 V, , Figure 3.35, The schematic of the circuit in Fig. 3.34., , 0, , Use PSpice to determine currents i1, i2, and i3 in the circuit of Fig. 3.36., , Figure 3.33, , Practice Problem 3.11, , For Practice Prob. 3.10., , i1, , Answer: i1 ⫽ ⫺0.4286 A, i2 ⫽ 2.286 A, i3 ⫽ 2 A., , Answer: V1 ⫽ ⫺40 V, V2 ⫽ 57.14 V, V3 ⫽ 200 V., , 4Ω, 2A, , Example 3.11, , In the circuit of Fig. 3.34, determine the currents i1, i2, and i3., , 3.9, , p, , 2Ω, , Applications: DC Transistor Circuits, 10 V, , 1Ω, , 2Ω, , 3vo, +−, , 4Ω, , i2, , i1, 24 V +, −, , 2Ω, , Figure 3.34, For Example 3.11., , 8Ω, , i3, 4Ω, , +, vo, −, , i2, , Most of us deal with electronic products on a routine basis and have, some experience with personal computers. A basic component for, the integrated circuits found in these electronics and computers is the, active, three-terminal device known as the transistor. Understanding, the transistor is essential before an engineer can start an electronic circuit design., Figure 3.37 depicts various kinds of transistors commercially available. There are two basic types of transistors: bipolar junction transistors (BJTs) and field-effect transistors (FETs). Here, we consider only, the BJTs, which were the first of the two and are still used today. Our, objective is to present enough detail about the BJT to enable us to apply, the techniques developed in this chapter to analyze dc transistor circuits., , 1Ω, , i3, i1, , 2Ω, , +, −, , Figure 3.36, For Practice Prob. 3.11., , c h a p t e r, , 9, , Sinusoids and, Phasors, , Each chapter opens with a discussion about how to, enhance skills that contribute to successful problem, solving as well as successful careers or a careeroriented talk on a sub-discipline of electrical engineering to give students some real-world applications, of what they are learning., , He who knows not, and knows not that he knows not, is a fool—, shun him. He who knows not, and knows that he knows not, is a child—, teach him. He who knows, and knows not that he knows, is asleep—wake, him up. He who knows, and knows that he knows, is wise—follow him., —Persian Proverb, , Enhancing Your Skills and Your Career, ABET EC 2000 criteria (3.d), “an ability to function on, multi-disciplinary teams.”, The “ability to function on multidisciplinary teams” is inherently critical for the working engineer. Engineers rarely, if ever, work by themselves. Engineers will always be part of some team. One of the things, I like to remind students is that you do not have to like everyone on a, team; you just have to be a successful part of that team., Most frequently, these teams include individuals from of a variety, of engineering disciplines, as well as individuals from nonengineering, disciplines such as marketing and finance., Students can easily develop and enhance this skill by working in, study groups in every course they take. Clearly, working in study, groups in nonengineering courses as well as engineering courses outside your discipline will also give you experience with multidisciplinary teams., , Photo by Charles Alexander, , 369, , Icons next to the end-of-chapter homework problems, let students know which problems relate to engineering design and which problems can be solved using, PSpice or MATLAB. Appendices on these computer, programs provide tutorials for their use.
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ale29559_fm.qxd, , xxii, , 07/28/2008, , 11:54 AM, , Page xxii, , Guided Tour, , Supplements for Students and Instructors, McGraw-Hill’s ARIS—Assessment, Review,, and Instruction System is a complete, online, tutorial, electronic homework, and course management system, designed for greater ease of, use than any other system available. With ARIS, instructors can create, and share course materials and assignments with other instructors, edit, questions and algorithms, import their own content, and create, announcements and due dates for assignments. ARIS has automatic, grading and reporting of easy-to-assign algorithmically-generated, homework, quizzing, and testing. Once a student is registered in the, course, all student activity within McGraw-Hill’s ARIS is automatically, recorded and available to the instructor through a fully integrated grade, book that can be downloaded to Excel., www.mhhe.com/alexander, , Knowledge Capturing Integrated Design Environment, for Circuits (KCIDE for Circuits) software, linked, from ARIS, enhances student understanding of the, six-step problem-solving methodology in the book., KCIDE for Circuits allows students to work a circuit, problem in PSpice and MATLAB, track the evolution, of their solution, and save a record of their process for, future reference. Appendix F walks the user through, this program.
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ale29559_fm.qxd, , 07/28/2008, , 11:54 AM, , Page xxiii, , Guided Tour, , Other resources provided on ARIS., For Students:, — Network Analysis Tutorials—a series of interactive quizzes to help, students practice fundamental concepts in circuits., — FE Exam Interactive Review Quizzes—chapter based self-quizzes, provide hints for solutions and correct solution methods, and help, students prepare for the NCEES Fundamentals of Engineering, Examination., — Problem Solving Made Almost Easy—a companion workbook to the, text, featuring 150 additional problems with complete solutions., — Complete solutions to Practice Problems in the text, — Flashcards of key terms, — Web links, For Instructors:, — Image Sets—electronic files of text figures for easy integration into, your course presentations, exams, and assignments., — Transition Guides—compare coverage of the third edition to other, popular circuits books at the section level to aid transition to teaching from our text., , xxiii
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ale29559_fm.qxd, , 07/28/2008, , 11:54 AM, , Page xxv, , A Note to the Student, This may be your first course in electrical engineering. Although electrical engineering is an exciting and challenging discipline, the course, may intimidate you. This book was written to prevent that. A good textbook and a good professor are an advantage—but you are the one who, does the learning. If you keep the following ideas in mind, you will do, very well in this course., • This course is the foundation on which most other courses in the, electrical engineering curriculum rest. For this reason, put in as, much effort as you can. Study the course regularly., • Problem solving is an essential part of the learning process. Solve as, many problems as you can. Begin by solving the practice problem, following each example, and then proceed to the end-of-chapter, problems. The best way to learn is to solve a lot of problems. An, asterisk in front of a problem indicates a challenging problem., • Spice, a computer circuit analysis program, is used throughout the, textbook. PSpice, the personal computer version of Spice, is the, popular standard circuit analysis program at most universities., PSpice for Windows is described in Appendix D. Make an effort, to learn PSpice, because you can check any circuit problem with, PSpice and be sure you are handing in a correct problem solution., • MATLAB is another software that is very useful in circuit analysis, and other courses you will be taking. A brief tutorial on MATLAB, is given in Appendix E to get you started. The best way to learn, MATLAB is to start working with it once you know a few commands., • Each chapter ends with a section on how the material covered in, the chapter can be applied to real-life situations. The concepts in, this section may be new and advanced to you. No doubt, you will, learn more of the details in other courses. We are mainly interested, in gaining a general familiarity with these ideas., • Attempt the review questions at the end of each chapter. They, will help you discover some “tricks” not revealed in class or in the, textbook., • Clearly a lot of effort has gone into making the technical details in, this book easy to understand. It also contains all the mathematics, and physics necessary to understand the theory and will be very, useful in your other engineering courses. However, we have also, focused on creating a reference for you to use both in school as, well as when working in industry or seeking a graduate degree., • It is very tempting to sell your book after you have completed your, classroom experience; however, our advice to you is DO NOT SELL, YOUR ENGINEERING BOOKS! Books have always been expensive, however, the cost of this book is virtually the same as I paid, for my circuits text back in the early 60s in terms of real dollars., In fact, it is actually cheaper. In addition, engineering books of, the past are no where near as complete as what is available now., xxv
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ale29559_fm.qxd, , xxvi, , 07/28/2008, , 11:54 AM, , Page xxvi, , A Note to the Student, , When I was a student, I did not sell any of my engineering textbooks and was very glad I did not! I found that I needed most of, them throughout my career., A short review on finding determinants is covered in Appendix A,, complex numbers in Appendix B, and mathematical formulas in Appendix C. Answers to odd-numbered problems are given in Appendix G., Have fun!, C. K. A. and M. N. O. S.
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ale29559_fm.qxd, , 07/28/2008, , 11:54 AM, , Page xxvii, , About the Authors, Charles K. Alexander is professor of electrical and computer engineering of the Fenn College of Engineering at Cleveland State University, Cleveland, Ohio. He is also the Director of The Center for, Research in Electronics and Aerospace Technology (CREATE), and is, the Managing Director of the Wright Center for Sensor Systems, (WCSSE). From 2002 until 2006 he was Dean of the Fenn College of, Engineering. From 2004 until 2007, he was Director of Ohio ICE, a, research center in instrumentation, controls, electronics, and sensors (a, coalition of CSU, Case, the University of Akron, and a number of Ohio, industries). From 1998 until 2002, he was interim director (2000 and, 2001) of the Institute for Corrosion and Multiphase Technologies and, Stocker Visiting Professor of electrical engineering and computer science at Ohio University. From 1994–1996 he was dean of engineering, and computer science at California State University, Northridge., From 1989–1994 he was acting dean of the college of engineering at Temple University, and from 1986–1989 he was professor and, chairman of the department of electrical engineering at Temple. From, 1980–1986 he held the same positions at Tennessee Technological, University. He was an associate professor and a professor of electrical, engineering at Youngstown State University from 1972–1980, where, he was named Distinguished Professor in 1977 in recognition of, “outstanding teaching and research.” He was assistant professor of, electrical engineering at Ohio University in 1971–1972. He received, the Ph.D. (1971) and M.S.E.E. (1967) from Ohio University and the, B.S.E.E. (1965) from Ohio Northern University., Dr. Alexander has been a consultant to 23 companies and governmental organizations, including the Air Force and Navy and several law firms. He has received over $85 million in research and, development funds for projects ranging from solar energy to software, engineering. He has authored 40 publications, including a workbook, and a videotape lecture series, and is coauthor of Fundamentals of, Electric Circuits, Problem Solving Made Almost Easy, and the fifth, edition of the Standard Handbook of Electronic Engineering, with, McGraw-Hill. He has made more than 500 paper, professional, and, technical presentations., Dr. Alexander is a fellow of the IEEE and served as its president, and CEO in 1997. In 1993 and 1994 he was IEEE vice president, professional activities, and chair of the United States Activities Board, (USAB). In 1991–1992 he was region 2 director, serving on the, Regional Activities Board (RAB) and USAB. He has also been a member of the Educational Activities Board. He served as chair of the USAB, Member Activities Council and vice chair of the USAB Professional, Activities Council for Engineers, and he chaired the RAB Student, Activities Committee and the USAB Student Professional Awareness, Committee., , Charles K. Alexander, , xxvii
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ale29559_fm.qxd, , 07/28/2008, , 11:54 AM, , xxviii, , Page xxviii, , About the Authors, , In 1998 he received the Distinguished Engineering Education, Achievement Award from the Engineering Council, and in 1996 he, received the Distinguished Engineering Education Leadership Award, from the same group. When he became a fellow of the IEEE in 1994,, the citation read “for leadership in the field of engineering education, and the professional development of engineering students.” In 1984 he, received the IEEE Centennial Medal, and in 1983 he received the, IEEE/RAB Innovation Award, given to the IEEE member who best, contributes to RAB’s goals and objectives., , Matthew N. O. Sadiku, , Matthew N. O. Sadiku is presently a professor at Prairie View A&M, University. Prior to joining Prairie View, he taught at Florida Atlantic, University, Boca Raton, and Temple University, Philadelphia. He has, also worked for Lucent/Avaya and Boeing Satellite Systems., Dr. Sadiku is the author of over 170 professional papers and almost, 30 books including Elements of Electromagnetics (Oxford University, Press, 3rd ed., 2001), Numerical Techniques in Electromagnetics (2nd, ed., CRC Press, 2000), Simulation of Local Area Networks (with M., IIyas, CRC Press, 1994), Metropolitan Area Networks (CRC Press,, 1994), and Fundamentals of Electric Circuits (with C. K. Alexander,, McGraw-Hill, 3rd ed., 2007). His books are used worldwide, and some, of them have been translated into Korean, Chinese, Italian, and Spanish., He was the recipient of the 2000 McGraw-Hill/Jacob Millman Award for, outstanding contributions in the field of electrical engineering. He was, the IEEE region 2 Student Activities Committee chairman and is an associate editor for IEEE “Transactions on Education.” He received his Ph.D., at Tennessee Technological University, Cookeville.
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ale29559_fm.qxd, , 07/28/2008, , 11:54 AM, , Page 1, , Fundamentals of, , Electric Circuits
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ale29559_ch01.qxd, , 07/08/2008, , 10:37 AM, , Page 2, , P A R T, , O N E, , DC Circuits, OUTLINE, 1, , Basic Concepts, , 2, , Basic Laws, , 3, , Methods of Analysis, , 4, , Circuit Theorems, , 5, , Operational Amplifiers, , 6, , Capacitors and Inductors, , 7, , First-Order Circuits, , 8, , Second-Order Circuits
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ale29559_ch01.qxd, , 07/08/2008, , 10:37 AM, , Page 3, , c h a p t e r, , 1, , Basic Concepts, Some books are to be tasted, others to be swallowed, and some few to, be chewed and digested., —Francis Bacon, , Enhancing Your Skills and Your Career, ABET EC 2000 criteria (3.a), “an ability to apply knowledge, of mathematics, science, and engineering.”, As students, you are required to study mathematics, science, and engineering with the purpose of being able to apply that knowledge to the, solution of engineering problems. The skill here is the ability to apply, the fundamentals of these areas in the solution of a problem. So, how, do you develop and enhance this skill?, The best approach is to work as many problems as possible in all, of your courses. However, if you are really going to be successful with, this, you must spend time analyzing where and when and why you have, difficulty in easily arriving at successful solutions. You may be surprised to learn that most of your problem-solving problems are with, mathematics rather than your understanding of theory. You may also, learn that you start working the problem too soon. Taking time to think, about the problem and how you should solve it will always save you, time and frustration in the end., What I have found that works best for me is to apply our sixstep problem-solving technique. Then I carefully identify the areas, where I have difficulty solving the problem. Many times, my actual, deficiencies are in my understanding and ability to use correctly certain mathematical principles. I then return to my fundamental math, texts and carefully review the appropriate sections and in some cases, work some example problems in that text. This brings me to another, important thing you should always do: Keep nearby all your basic, mathematics, science, and engineering textbooks., This process of continually looking up material you thought you, had acquired in earlier courses may seem very tedious at first; however, as your skills develop and your knowledge increases, this process, will become easier and easier. On a personal note, it is this very process, that led me from being a much less than average student to someone, who could earn a Ph.D. and become a successful researcher., , Photo by Charles Alexander, , 3
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ale29559_ch01.qxd, , 07/08/2008, , 10:37 AM, , Page 4, , Chapter 1, , 4, , 1.1, , Basic Concepts, , Introduction, , Electric circuit theory and electromagnetic theory are the two fundamental theories upon which all branches of electrical engineering are, built. Many branches of electrical engineering, such as power, electric, machines, control, electronics, communications, and instrumentation,, are based on electric circuit theory. Therefore, the basic electric circuit, theory course is the most important course for an electrical engineering student, and always an excellent starting point for a beginning student in electrical engineering education. Circuit theory is also valuable, to students specializing in other branches of the physical sciences, because circuits are a good model for the study of energy systems in, general, and because of the applied mathematics, physics, and topology involved., In electrical engineering, we are often interested in communicating, or transferring energy from one point to another. To do this requires an, interconnection of electrical devices. Such interconnection is referred, to as an electric circuit, and each component of the circuit is known as, an element., An electric circuit is an interconnection of electrical elements., Current, , −, , +, , Battery, , Figure 1.1, A simple electric circuit., , Lamp, , A simple electric circuit is shown in Fig. 1.1. It consists of three, basic elements: a battery, a lamp, and connecting wires. Such a simple, circuit can exist by itself; it has several applications, such as a flashlight, a search light, and so forth., A complicated real circuit is displayed in Fig. 1.2, representing the, schematic diagram for a radio receiver. Although it seems complicated,, this circuit can be analyzed using the techniques we cover in this book., Our goal in this text is to learn various analytical techniques and, computer software applications for describing the behavior of a circuit, like this., Electric circuits are used in numerous electrical systems to accomplish different tasks. Our objective in this book is not the study of, various uses and applications of circuits. Rather our major concern is, the analysis of the circuits. By the analysis of a circuit, we mean a, study of the behavior of the circuit: How does it respond to a given, input? How do the interconnected elements and devices in the circuit, interact?, We commence our study by defining some basic concepts. These, concepts include charge, current, voltage, circuit elements, power, and, energy. Before defining these concepts, we must first establish a system of units that we will use throughout the text., , 1.2, , Systems of Units, , As electrical engineers, we deal with measurable quantities. Our measurement, however, must be communicated in a standard language that, virtually all professionals can understand, irrespective of the country, where the measurement is conducted. Such an international measurement
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ale29559_ch01.qxd, , 07/17/2008, , 11:15 AM, , Page 5, , 1.2, , L1, 0.445 H, Antenna, , C3, , R1 47, 8, 7, , U1, SBL-1, Mixer, , C1, 2200 pF, , 3, 4, , 2, 5, 6, , Oscillator R2, C, 10 k, B, R3, EQ1, 10 k, 2N2222A, , R6, 100 k, , R5, 100 k, , U2B, 1 2 TL072, C10 5, +, 7, 1.0 F, 16 V 6 −, , +, , +, , R7, 1M, C12, 0.0033, , R8, 15 k, C13, , C11, 100 F, 16 V, C15, U2A, 0.47, 1 2 TL072 16 V +, 3, 8, C14 +, 1, 0.0022 −, 4, 2, , R9, 15 k, , C6, , 0.1, , 5, , L2, 22.7 H, (see text), , C4, 910, , to, U1, Pin 8, , R11, 47, C8, 0.1, , +, , C9, 1.0 F, 16 V, , Y1, 7 MHz, , C5, 910, , R4, 220, , L3, 1 mH, , 5, , 0.1, 1, , C2, 2200 pF, , Systems of Units, , C7, 532, , R10, 10 k, GAIN, , 3, 2, , +, , +, , C16, 100 F, 16 V, , +, 12-V dc, Supply, −, , 6, , −, , 5, 4 R12, 10, , U3, C18, LM386N, Audio power amp 0.1, , Audio, +, Output, C17, 100 F, 16 V, , Figure 1.2, Electric circuit of a radio receiver., Reproduced with permission from QST, August 1995, p. 23., , language is the International System of Units (SI), adopted by the, General Conference on Weights and Measures in 1960. In this system,, there are six principal units from which the units of all other physical, quantities can be derived. Table 1.1 shows the six units, their symbols,, and the physical quantities they represent. The SI units are used, throughout this text., One great advantage of the SI unit is that it uses prefixes based on, the power of 10 to relate larger and smaller units to the basic unit., Table 1.2 shows the SI prefixes and their symbols. For example, the, following are expressions of the same distance in meters (m):, 600,000,000 mm, , 600,000 m, , 600 km, , TABLE 1.1, , Six basic SI units and one derived unit relevant to this text., Quantity, , Basic unit, , Length, Mass, Time, Electric current, Thermodynamic temperature, Luminous intensity, Charge, , meter, kilogram, second, ampere, kelvin, candela, coulomb, , Symbol, m, kg, s, A, K, cd, C, , TABLE 1.2, , The SI prefixes., Multiplier, 18, , 10, 1015, 1012, 109, 106, 103, 102, 10, 101, 102, 103, 106, 109, 1012, 1015, 1018, , Prefix, , Symbol, , exa, peta, tera, giga, mega, kilo, hecto, deka, deci, centi, milli, micro, nano, pico, femto, atto, , E, P, T, G, M, k, h, da, d, c, m, m, n, p, f, a
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ale29559_ch01.qxd, , 07/08/2008, , 10:38 AM, , Page 6, , Chapter 1, , 6, , 1.3, , Basic Concepts, , Charge and Current, , The concept of electric charge is the underlying principle for explaining all electrical phenomena. Also, the most basic quantity in an electric circuit is the electric charge. We all experience the effect of electric, charge when we try to remove our wool sweater and have it stick to, our body or walk across a carpet and receive a shock., Charge is an electrical property of the atomic particles of which matter consists, measured in coulombs (C)., , We know from elementary physics that all matter is made of fundamental building blocks known as atoms and that each atom consists of, electrons, protons, and neutrons. We also know that the charge e on an, electron is negative and equal in magnitude to 1.602 1019 C, while, a proton carries a positive charge of the same magnitude as the electron. The presence of equal numbers of protons and electrons leaves an, atom neutrally charged., The following points should be noted about electric charge:, 1. The coulomb is a large unit for charges. In 1 C of charge, there, are 1(1.602 1019) 6.24 1018 electrons. Thus realistic or, laboratory values of charges are on the order of pC, nC, or mC.1, 2. According to experimental observations, the only charges that, occur in nature are integral multiples of the electronic charge, e 1.602 1019 C., 3. The law of conservation of charge states that charge can neither, be created nor destroyed, only transferred. Thus the algebraic sum, of the electric charges in a system does not change., −, −, , I, , +, , −, −, , −, , Battery, , Figure 1.3, Electric current due to flow of electronic, charge in a conductor., , A convention is a standard way of, describing something so that others in, the profession can understand what, we mean. We will be using IEEE conventions throughout this book., , We now consider the flow of electric charges. A unique feature of, electric charge or electricity is the fact that it is mobile; that is, it can, be transferred from one place to another, where it can be converted to, another form of energy., When a conducting wire (consisting of several atoms) is connected to a battery (a source of electromotive force), the charges are, compelled to move; positive charges move in one direction while negative charges move in the opposite direction. This motion of charges, creates electric current. It is conventional to take the current flow as, the movement of positive charges. That is, opposite to the flow of negative charges, as Fig. 1.3 illustrates. This convention was introduced, by Benjamin Franklin (1706–1790), the American scientist and inventor. Although we now know that current in metallic conductors is due, to negatively charged electrons, we will follow the universally, accepted convention that current is the net flow of positive charges., Thus,, Electric current is the time rate of change of charge, measured in, amperes (A)., , 1, , However, a large power supply capacitor can store up to 0.5 C of charge.
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ale29559_ch01.qxd, , 07/16/2008, , 12:22 PM, , Page 7, , 1.3, , Charge and Current, , 7, , Historical, Andre-Marie Ampere (1775–1836), a French mathematician and, physicist, laid the foundation of electrodynamics. He defined the electric current and developed a way to measure it in the 1820s., Born in Lyons, France, Ampere at age 12 mastered Latin in a few, weeks, as he was intensely interested in mathematics and many of the, best mathematical works were in Latin. He was a brilliant scientist and, a prolific writer. He formulated the laws of electromagnetics. He invented the electromagnet and the ammeter. The unit of electric current,, the ampere, was named after him., , The Burndy Library Collection, at The Huntington Library,, San Marino, California., , Mathematically, the relationship between current i, charge q, and time t is, i, ¢, , dq, dt, , (1.1), , where current is measured in amperes (A), and, 1 ampere 1 coulomb/second, The charge transferred between time t0 and t is obtained by integrating both sides of Eq. (1.1). We obtain, I, , t, , Q, ¢, , i dt, , (1.2), , t0, , The way we define current as i in Eq. (1.1) suggests that current need, not be a constant-valued function. As many of the examples and problems in this chapter and subsequent chapters suggest, there can be several types of current; that is, charge can vary with time in several ways., If the current does not change with time, but remains constant, we, call it a direct current (dc)., , 0, , t, (a), i, , A direct current (dc) is a current that remains constant with time., , By convention the symbol I is used to represent such a constant current., A time-varying current is represented by the symbol i. A common, form of time-varying current is the sinusoidal current or alternating, current (ac)., , 0, , An alternating current (ac) is a current that varies sinusoidally with time., , Such current is used in your household, to run the air conditioner,, refrigerator, washing machine, and other electric appliances. Figure 1.4, , t, , (b), , Figure 1.4, Two common types of current: (a) direct, current (dc), (b) alternating current (ac).
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ale29559_ch01.qxd, , 07/08/2008, , 10:38 AM, , Page 8, , Chapter 1, , 8, , −5 A, , 5A, , (a), , (b), , Figure 1.5, Conventional current flow: (a) positive, current flow, (b) negative current flow., , Example 1.1, , Basic Concepts, , shows direct current and alternating current; these are the two most, common types of current. We will consider other types later in the, book., Once we define current as the movement of charge, we expect current to have an associated direction of flow. As mentioned earlier, the, direction of current flow is conventionally taken as the direction of positive charge movement. Based on this convention, a current of 5 A may, be represented positively or negatively as shown in Fig. 1.5. In other, words, a negative current of 5 A flowing in one direction as shown, in Fig. 1.5(b) is the same as a current of 5 A flowing in the opposite, direction., , How much charge is represented by 4,600 electrons?, Solution:, Each electron has 1.602 1019 C. Hence 4,600 electrons will have, 1.602 1019 C/electron 4,600 electrons 7.369 1016 C, , Practice Problem 1.1, , Calculate the amount of charge represented by four million protons., Answer: 6.408 1013 C., , Example 1.2, , The total charge entering a terminal is given by q 5t sin 4 p t mC., Calculate the current at t 0.5 s., Solution:, i, , dq, d, (5t sin 4 p t) mC/s (5 sin 4 p t 20 p t cos 4 p t) mA, dt, dt, , At t 0.5,, i 5 sin 2 p 10 p cos 2 p 0 10 p 31.42 mA, , Practice Problem 1.2, , If in Example 1.2, q (10 10e2t ) mC, find the current at t 0.5 s., Answer: 7.36 mA.
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ale29559_ch01.qxd, , 07/08/2008, , 10:38 AM, , Page 9, , 1.4, , Voltage, , 9, , Example 1.3, , Determine the total charge entering a terminal between t 1 s and, t 2 s if the current passing the terminal is i (3t 2 t) A., Solution:, Q, , , , 2, , i dt , , t1, , at 3 , , , , 2, , (3t 2 t) dt, , 1, , 2, , 2, , t, 1, b ` (8 2) a1 b 5.5 C, 2 1, 2, , Practice Problem 1.3, , The current flowing through an element is, i e, , 2 A,, 2t 2 A,, , 0 6 t 6 1, t 7 1, , Calculate the charge entering the element from t 0 to t 2 s., Answer: 6.667 C., , 1.4, , Voltage, , As explained briefly in the previous section, to move the electron in a, conductor in a particular direction requires some work or energy transfer. This work is performed by an external electromotive force (emf),, typically represented by the battery in Fig. 1.3. This emf is also known, as voltage or potential difference. The voltage vab between two points, a and b in an electric circuit is the energy (or work) needed to move, a unit charge from a to b; mathematically,, vab , ¢, , dw, dq, , (1.3), , where w is energy in joules (J) and q is charge in coulombs (C). The, voltage vab or simply v is measured in volts (V), named in honor of, the Italian physicist Alessandro Antonio Volta (1745–1827), who, invented the first voltaic battery. From Eq. (1.3), it is evident that, 1 volt 1 joule/coulomb 1 newton-meter/coulomb, Thus,, Voltage (or potential difference) is the energy required to move a unit, charge through an element, measured in volts (V)., , +, , a, , vab, , Figure 1.6 shows the voltage across an element (represented by a, rectangular block) connected to points a and b. The plus () and minus, () signs are used to define reference direction or voltage polarity. The, vab can be interpreted in two ways: (1) point a is at a potential of vab, , −, , Figure 1.6, Polarity of voltage vab., , b
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ale29559_ch01.qxd, , 07/16/2008, , 12:23 PM, , Page 10, , Chapter 1, , 10, , Basic Concepts, , Historical, Alessandro Antonio Volta (1745–1827), an Italian physicist,, invented the electric battery—which provided the first continuous flow, of electricity—and the capacitor., Born into a noble family in Como, Italy, Volta was performing, electrical experiments at age 18. His invention of the battery in 1796, revolutionized the use of electricity. The publication of his work in, 1800 marked the beginning of electric circuit theory. Volta received, many honors during his lifetime. The unit of voltage or potential difference, the volt, was named in his honor., , The Burndy Library Collection, at The Huntington Library,, San Marino, California., , volts higher than point b, or (2) the potential at point a with respect to, point b is vab. It follows logically that in general, +, , a, , (a), , +, b, , vab vba, , a, , −9 V, , 9V, −, , −, , b, (b), , Figure 1.7, Two equivalent representations of the, same voltage vab : (a) point a is 9 V above, point b, (b) point b is 9 V above point a., Keep in mind that electric current is, always through an element and that, electric voltage is always across the, element or between two points., , (1.4), , For example, in Fig. 1.7, we have two representations of the same voltage. In Fig. 1.7(a), point a is 9 V above point b; in Fig. 1.7(b), point b, is 9 V above point a. We may say that in Fig. 1.7(a), there is a 9-V, voltage drop from a to b or equivalently a 9-V voltage rise from b to, a. In other words, a voltage drop from a to b is equivalent to a voltage rise from b to a., Current and voltage are the two basic variables in electric circuits., The common term signal is used for an electric quantity such as a current or a voltage (or even electromagnetic wave) when it is used for, conveying information. Engineers prefer to call such variables signals, rather than mathematical functions of time because of their importance, in communications and other disciplines. Like electric current, a constant voltage is called a dc voltage and is represented by V, whereas a, sinusoidally time-varying voltage is called an ac voltage and is represented by v. A dc voltage is commonly produced by a battery; ac voltage is produced by an electric generator., , 1.5, , Power and Energy, , Although current and voltage are the two basic variables in an electric, circuit, they are not sufficient by themselves. For practical purposes,, we need to know how much power an electric device can handle. We, all know from experience that a 100-watt bulb gives more light than a, 60-watt bulb. We also know that when we pay our bills to the electric, utility companies, we are paying for the electric energy consumed over, a certain period of time. Thus, power and energy calculations are, important in circuit analysis.
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ale29559_ch01.qxd, , 07/08/2008, , 10:38 AM, , Page 11, , 1.5, , Power and Energy, , 11, , To relate power and energy to voltage and current, we recall from, physics that:, Power is the time rate of expending or absorbing energy, measured in, watts (W)., , We write this relationship as, i, , p, ¢, , dw, dt, , (1.5), , where p is power in watts (W), w is energy in joules (J), and t is time, in seconds (s). From Eqs. (1.1), (1.3), and (1.5), it follows that, p, , dw dq, dw, , , vi, dt, dq dt, , i, +, , +, , v, , v, , −, , −, , p = +vi, , p = −vi, , (a), , (b), , (1.6), , Figure 1.8, Reference polarities for power using the, passive sign convention: (a) absorbing, power, (b) supplying power., , or, p vi, , (1.7), , The power p in Eq. (1.7) is a time-varying quantity and is called the, instantaneous power. Thus, the power absorbed or supplied by an element is the product of the voltage across the element and the current, through it. If the power has a sign, power is being delivered to or, absorbed by the element. If, on the other hand, the power has a sign,, power is being supplied by the element. But how do we know when, the power has a negative or a positive sign?, Current direction and voltage polarity play a major role in determining the sign of power. It is therefore important that we pay attention to the relationship between current i and voltage v in Fig. 1.8(a)., The voltage polarity and current direction must conform with those, shown in Fig. 1.8(a) in order for the power to have a positive sign., This is known as the passive sign convention. By the passive sign convention, current enters through the positive polarity of the voltage. In, this case, p vi or vi 7 0 implies that the element is absorbing, power. However, if p vi or vi 6 0, as in Fig. 1.8(b), the element, is releasing or supplying power., Passive sign convention is satisfied when the current enters through, the positive terminal of an element and p vi. If the current enters, through the negative terminal, p vi., , Unless otherwise stated, we will follow the passive sign convention throughout this text. For example, the element in both circuits of, Fig. 1.9 has an absorbing power of 12 W because a positive current, enters the positive terminal in both cases. In Fig. 1.10, however, the, element is supplying power of 12 W because a positive current enters, the negative terminal. Of course, an absorbing power of 12 W is, equivalent to a supplying power of 12 W. In general,, Power absorbed Power supplied, , When the voltage and current directions, conform to Fig. 1.8 (b), we have the active sign convention and p vi., , 3A, , 3A, +, , −, , 4V, , 4V, , −, , +, (a), , (b), , Figure 1.9, Two cases of an element with an absorbing, power of 12 W: (a) p 4 3 12 W,, (b) p 4 3 12 W., , 3A, , 3A, , +, , −, , 4V, , 4V, , −, , +, (a), , (b), , Figure 1.10, Two cases of an element with a supplying, power of 12 W: (a) p 4 3 , 12W, (b) p 4 3 12 W.
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ale29559_ch01.qxd, , 07/08/2008, , 10:38 AM, , Page 12, , Chapter 1, , 12, , Basic Concepts, , In fact, the law of conservation of energy must be obeyed in any, electric circuit. For this reason, the algebraic sum of power in a circuit, at any instant of time, must be zero:, ap0, , (1.8), , This again confirms the fact that the total power supplied to the circuit, must balance the total power absorbed., From Eq. (1.6), the energy absorbed or supplied by an element, from time t0 to time t is, w, , t, , t, , t0, , t0, , p dt vi dt, , (1.9), , Energy is the capacity to do work, measured in joules (J)., , The electric power utility companies measure energy in watt-hours, (Wh), where, 1 Wh 3,600 J, , Example 1.4, , An energy source forces a constant current of 2 A for 10 s to flow, through a lightbulb. If 2.3 kJ is given off in the form of light and heat, energy, calculate the voltage drop across the bulb., Solution:, The total charge is, ¢q i ¢t 2 10 20 C, The voltage drop is, v, , Practice Problem 1.4, , ¢w, 2.3 103, , 115 V, ¢q, 20, , To move charge q from point a to point b requires 30 J. Find the, voltage drop vab if: (a) q 2 C, (b) q 6 C., Answer: (a) 15 V, (b) 5 V., , Example 1.5, , Find the power delivered to an element at t 3 ms if the current entering its positive terminal is, i 5 cos 60 p t A, and the voltage is: (a) v 3i, (b) v 3 didt.
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ale29559_ch01.qxd, , 07/08/2008, , 10:38 AM, , Page 13, , 1.5, , Power and Energy, , 13, , Solution:, (a) The voltage is v 3i 15 cos 60 p t; hence, the power is, p vi 75 cos2 60 p t W, , At t 3 ms,, , p 75 cos2 (60 p 3 103) 75 cos2 0.18 p 53.48 W, (b) We find the voltage and the power as, v3, , di, 3(60 p)5 sin 60 p t 900 p sin 60 p t V, dt, p vi 4500 p sin 60 p t cos 60 p t W, , At t 3 ms,, p 4500 p sin 0.18 p cos 0.18 p W, 14137.167 sin 32.4 cos 32.4 6.396 kW, , Find the power delivered to the element in Example 1.5 at t 5 ms, if the current remains the same but the voltage is: (a) v 2i V,, (b) v a10 5, , Practice Problem 1.5, , t, , i dtb V., 0, , Answer: (a) 17.27 W, (b) 29.7 W., , How much energy does a 100-W electric bulb consume in two hours?, , Example 1.6, , Solution:, w pt 100 (W) 2 (h) 60 (min/h) 60 (s/min), 720,000 J 720 kJ, This is the same as, w pt 100 W 2 h 200 Wh, , A stove element draws 15 A when connected to a 240-V line. How, long does it take to consume 60 kJ?, Answer: 16.667 s., , Practice Problem 1.6
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ale29559_ch01.qxd, , 07/08/2008, , 14, , 10:38 AM, , Page 14, , Chapter 1, , Basic Concepts, , Historical, 1884 Exhibition In the United States, nothing promoted the future, of electricity like the 1884 International Electrical Exhibition. Just, imagine a world without electricity, a world illuminated by candles and, gaslights, a world where the most common transportation was by walking and riding on horseback or by horse-drawn carriage. Into this world, an exhibition was created that highlighted Thomas Edison and reflected, his highly developed ability to promote his inventions and products., His exhibit featured spectacular lighting displays powered by an impressive 100-kW “Jumbo” generator., Edward Weston’s dynamos and lamps were featured in the United, States Electric Lighting Company’s display. Weston’s well known collection of scientific instruments was also shown., Other prominent exhibitors included Frank Sprague, Elihu Thompson,, and the Brush Electric Company of Cleveland. The American Institute, of Electrical Engineers (AIEE) held its first technical meeting on October 7–8 at the Franklin Institute during the exhibit. AIEE merged with, the Institute of Radio Engineers (IRE) in 1964 to form the Institute of, Electrical and Electronics Engineers (IEEE)., , Smithsonian Institution.
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ale29559_ch01.qxd, , 07/08/2008, , 10:38 AM, , Page 15, , 1.6, , 1.6, , Circuit Elements, , 15, , Circuit Elements, , As we discussed in Section 1.1, an element is the basic building block, of a circuit. An electric circuit is simply an interconnection of the elements. Circuit analysis is the process of determining voltages across, (or the currents through) the elements of the circuit., There are two types of elements found in electric circuits: passive elements and active elements. An active element is capable of, generating energy while a passive element is not. Examples of passive elements are resistors, capacitors, and inductors. Typical active, elements include generators, batteries, and operational amplifiers. Our, aim in this section is to gain familiarity with some important active, elements., The most important active elements are voltage or current, sources that generally deliver power to the circuit connected to, them. There are two kinds of sources: independent and dependent, sources., An ideal independent source is an active element that provides a, specified voltage or current that is completely independent of other, circuit elements., v, , In other words, an ideal independent voltage source delivers to the, circuit whatever current is necessary to maintain its terminal voltage. Physical sources such as batteries and generators may be, regarded as approximations to ideal voltage sources. Figure 1.11, shows the symbols for independent voltage sources. Notice that both, symbols in Fig. 1.11(a) and (b) can be used to represent a dc voltage source, but only the symbol in Fig. 1.11(a) can be used for a, time-varying voltage source. Similarly, an ideal independent current, source is an active element that provides a specified current completely independent of the voltage across the source. That is, the current source delivers to the circuit whatever voltage is necessary to, maintain the designated current. The symbol for an independent current source is displayed in Fig. 1.12, where the arrow indicates the, direction of current i., An ideal dependent (or controlled) source is an active element in, which the source quantity is controlled by another voltage or current., , Dependent sources are usually designated by diamond-shaped symbols,, as shown in Fig. 1.13. Since the control of the dependent source is, achieved by a voltage or current of some other element in the circuit,, and the source can be voltage or current, it follows that there are four, possible types of dependent sources, namely:, 1., 2., 3., 4., , A voltage-controlled voltage source (VCVS)., A current-controlled voltage source (CCVS)., A voltage-controlled current source (VCCS)., A current-controlled current source (CCCS)., , +, V, −, , +, −, , (b), , (a), , Figure 1.11, Symbols for independent voltage sources:, (a) used for constant or time-varying voltage, (b) used for constant voltage (dc)., , i, , Figure 1.12, Symbol for independent current source., , v, , +, −, , i, , (a), , (b), , Figure 1.13, Symbols for: (a) dependent voltage, source, (b) dependent current source.
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ale29559_ch01.qxd, , 07/08/2008, , 10:38 AM, , Page 16, , Chapter 1, , 16, B, , A, i, +, 5V, −, , +, −, , C, , 10i, , Figure 1.14, The source on the right-hand side is a, current-controlled voltage source., , Example 1.7, I=5A, , 20 V, , +, −, , p1, , Figure 1.15, For Example 1.7., , p3, , Dependent sources are useful in modeling elements such as transistors, operational amplifiers, and integrated circuits. An example of a, current-controlled voltage source is shown on the right-hand side of, Fig. 1.14, where the voltage 10i of the voltage source depends on, the current i through element C. Students might be surprised that, the value of the dependent voltage source is 10i V (and not 10i A), because it is a voltage source. The key idea to keep in mind is, that a voltage source comes with polarities ( ) in its symbol,, while a current source comes with an arrow, irrespective of what it, depends on., It should be noted that an ideal voltage source (dependent or independent) will produce any current required to ensure that the terminal, voltage is as stated, whereas an ideal current source will produce the, necessary voltage to ensure the stated current flow. Thus, an ideal, source could in theory supply an infinite amount of energy. It should, also be noted that not only do sources supply power to a circuit, they, can absorb power from a circuit too. For a voltage source, we know, the voltage but not the current supplied or drawn by it. By the same, token, we know the current supplied by a current source but not the, voltage across it., , Calculate the power supplied or absorbed by each element in Fig. 1.15., Solution:, We apply the sign convention for power shown in Figs. 1.8 and 1.9., For p1, the 5-A current is out of the positive terminal (or into the, negative terminal); hence,, , p2, −, +, 12 V, , Basic Concepts, , 6A, +, 8V, −, , p4, , 0.2 I, , p1 20(5) 100 W, , Supplied power, , For p2 and p3, the current flows into the positive terminal of the element in each case., p2 12(5) 60 W, p3 8(6) 48 W, , Absorbed power, Absorbed power, , For p4, we should note that the voltage is 8 V (positive at the top), the, same as the voltage for p3, since both the passive element and the, dependent source are connected to the same terminals. (Remember that, voltage is always measured across an element in a circuit.) Since the, current flows out of the positive terminal,, p4 8(0.2I ) 8(0.2 5) 8 W, , Supplied power, , We should observe that the 20-V independent voltage source and, 0.2I dependent current source are supplying power to the rest of, the network, while the two passive elements are absorbing power., Also,, p1 p2 p3 p4 100 60 48 8 0, In agreement with Eq. (1.8), the total power supplied equals the total, power absorbed.
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Page 17, , 1.7, , Applications, , Compute the power absorbed or supplied by each component of the, circuit in Fig. 1.16., , 17, , Practice Problem 1.7, 8A, , 1.7.1 TV Picture Tube, One important application of the motion of electrons is found in both, the transmission and reception of TV signals. At the transmission end,, a TV camera reduces a scene from an optical image to an electrical, signal. Scanning is accomplished with a thin beam of electrons in an, iconoscope camera tube., At the receiving end, the image is reconstructed by using a cathoderay tube (CRT) located in the TV receiver.3 The CRT is depicted in, Fig. 1.17. Unlike the iconoscope tube, which produces an electron, beam of constant intensity, the CRT beam varies in intensity according to the incoming signal. The electron gun, maintained at a high, potential, fires the electron beam. The beam passes through two sets, of plates for vertical and horizontal deflections so that the spot on the, screen where the beam strikes can move right and left and up and, down. When the electron beam strikes the fluorescent screen, it gives, off light at that spot. Thus, the beam can be made to “paint” a picture, on the TV screen., Horizontal, deflection, plates, , Bright spot on, fluorescent screen, Vertical, deflection, plates, , Electron, trajectory, , Figure 1.17, Cathode-ray tube., D. E. Tilley, Contemporary College Physics Menlo Park, CA: Benjamin/, Cummings, 1979, p. 319., , 2, , p1, , p3, , +, −, , Figure 1.16, , In this section, we will consider two practical applications of the, concepts developed in this chapter. The first one deals with the TV, picture tube and the other with how electric utilities determine your, electric bill., , Electron gun, , 3A, , p2, +, 5V, −, , Applications2, , I=5A, , +−, , Answer: p1 40 W, p2 16 W, p3 9 W, p4 15 W., , 1.7, , 2V, , The dagger sign preceding a section heading indicates the section that may be skipped,, explained briefly, or assigned as homework., 3, Modern TV tubes use a different technology., , For Practice Prob. 1.7., , 0.6I, , p4, , 3V, , −, , 10:38 AM, , +, −, , 07/08/2008, , +, , ale29559_ch01.qxd
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ale29559_ch01.qxd, , 07/08/2008, , 10:38 AM, , Page 18, , Chapter 1, , 18, , Basic Concepts, , Historical, Karl Ferdinand Braun and Vladimir K. Zworykin, , Zworykin with an iconoscope., © Bettmann/Corbis., , Example 1.8, , Karl Ferdinand Braun (1850–1918), of the University of Strasbourg,, invented the Braun cathode-ray tube in 1879. This then became the, basis for the picture tube used for so many years for televisions. It is, still the most economical device today, although the price of flat-screen, systems is rapidly becoming competitive. Before the Braun tube could, be used in television, it took the inventiveness of Vladimir K., Zworykin (1889–1982) to develop the iconoscope so that the modern, television would become a reality. The iconoscope developed into the, orthicon and the image orthicon, which allowed images to be captured, and converted into signals that could be sent to the television receiver., Thus, the television camera was born., , The electron beam in a TV picture tube carries 1015 electrons per second. As a design engineer, determine the voltage Vo needed to accelerate the electron beam to achieve 4 W., Solution:, The charge on an electron is, e 1.6 1019 C, If the number of electrons is n, then q ne and, , i, q, Vo, , Figure 1.18, A simplified diagram of the cathode-ray, tube; for Example 1.8., , i, , dq, dn, e, (1.6 1019)(1015) 1.6 104 A, dt, dt, , The negative sign indicates that the current flows in a direction, opposite to electron flow as shown in Fig. 1.18, which is a simplified, diagram of the CRT for the case when the vertical deflection plates, carry no charge. The beam power is, p Voi, , or, , Vo , , p, 4, 25,000 V, , i, 1.6 104, , Thus, the required voltage is 25 kV., , Practice Problem 1.8, , If an electron beam in a TV picture tube carries 1013 electrons/second, and is passing through plates maintained at a potential difference of, 30 kV, calculate the power in the beam., Answer: 48 mW.
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ale29559_ch01.qxd, , 07/17/2008, , 11:15 AM, , Page 19, , 1.7, , Applications, , 19, , TABLE 1.3, , Typical average monthly consumption of household, appliances., Appliance, , kWh consumed, , Water heater, Freezer, Lighting, Dishwasher, Electric iron, TV, Toaster, , 500, 100, 100, 35, 15, 10, 4, , Appliance, Washing machine, Stove, Dryer, Microwave oven, Personal computer, Radio, Clock, , kWh consumed, 120, 100, 80, 25, 12, 8, 2, , 1.7.2 Electricity Bills, The second application deals with how an electric utility company charges, their customers. The cost of electricity depends upon the amount of, energy consumed in kilowatt-hours (kWh). (Other factors that affect the, cost include demand and power factors; we will ignore these for now.), However, even if a consumer uses no energy at all, there is a minimum, service charge the customer must pay because it costs money to stay connected to the power line. As energy consumption increases, the cost per, kWh drops. It is interesting to note the average monthly consumption of, household appliances for a family of five, shown in Table 1.3., A homeowner consumes 700 kWh in January. Determine the electricity bill for the month using the following residential rate schedule:, , Example 1.9, , Base monthly charge of $12.00., First 100 kWh per month at 16 cents/kWh., Next 200 kWh per month at 10 cents/kWh., Over 300 kWh per month at 6 cents/kWh., Solution:, We calculate the electricity bill as follows., Base monthly charge $12.00, First 100 kWh @ $0.16/k Wh $16.00, Next 200 kWh @ $0.10/k Wh $20.00, Remaining 400 kWh @ $0.06/k Wh $24.00, Total charge $72.00, Average cost , , $72, 10.2 cents/kWh, 100 200 400, , Referring to the residential rate schedule in Example 1.9, calculate the, average cost per kWh if only 400 kWh are consumed in July when the, family is on vacation most of the time., Answer: 13.5 cents/kWh., , Practice Problem 1.9
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ale29559_ch01.qxd, , 20, , 07/08/2008, , 10:38 AM, , Page 20, , Chapter 1, , 1.8, , Basic Concepts, , Problem Solving, , Although the problems to be solved during one’s career will vary in, complexity and magnitude, the basic principles to be followed remain, the same. The process outlined here is the one developed by the, authors over many years of problem solving with students, for the, solution of engineering problems in industry, and for problem solving, in research., We will list the steps simply and then elaborate on them., 1. Carefully define the problem., 2. Present everything you know about the problem., 3. Establish a set of alternative solutions and determine the one that, promises the greatest likelihood of success., 4. Attempt a problem solution., 5. Evaluate the solution and check for accuracy., 6. Has the problem been solved satisfactorily? If so, present the solution; if not, then return to step 3 and continue through the process, again., 1. Carefully define the problem. This may be the most important part, of the process, because it becomes the foundation for all the rest of the, steps. In general, the presentation of engineering problems is somewhat, incomplete. You must do all you can to make sure you understand the, problem as thoroughly as the presenter of the problem understands it., Time spent at this point clearly identifying the problem will save you, considerable time and frustration later. As a student, you can clarify a, problem statement in a textbook by asking your professor. A problem, presented to you in industry may require that you consult several individuals. At this step, it is important to develop questions that need to, be addressed before continuing the solution process. If you have such, questions, you need to consult with the appropriate individuals or, resources to obtain the answers to those questions. With those answers,, you can now refine the problem, and use that refinement as the problem statement for the rest of the solution process., 2. Present everything you know about the problem. You are now ready, to write down everything you know about the problem and its possible, solutions. This important step will save you time and frustration later., 3. Establish a set of alternative solutions and determine the one that, promises the greatest likelihood of success. Almost every problem will, have a number of possible paths that can lead to a solution. It is highly, desirable to identify as many of those paths as possible. At this point,, you also need to determine what tools are available to you, such as, PSpice and MATLAB and other software packages that can greatly, reduce effort and increase accuracy. Again, we want to stress that time, spent carefully defining the problem and investigating alternative, approaches to its solution will pay big dividends later. Evaluating the, alternatives and determining which promises the greatest likelihood of, success may be difficult but will be well worth the effort. Document, this process well since you will want to come back to it if the first, approach does not work., 4. Attempt a problem solution. Now is the time to actually begin, solving the problem. The process you follow must be well documented
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ale29559_ch01.qxd, , 07/08/2008, , 10:38 AM, , Page 21, , 1.8, , Problem Solving, , 21, , in order to present a detailed solution if successful, and to evaluate the, process if you are not successful. This detailed evaluation may lead to, corrections that can then lead to a successful solution. It can also lead, to new alternatives to try. Many times, it is wise to fully set up a solution before putting numbers into equations. This will help in checking, your results., 5. Evaluate the solution and check for accuracy. You now thoroughly, evaluate what you have accomplished. Decide if you have an acceptable, solution, one that you want to present to your team, boss, or professor., 6. Has the problem been solved satisfactorily? If so, present the solution; if not, then return to step 3 and continue through the process, again. Now you need to present your solution or try another alternative. At this point, presenting your solution may bring closure to the, process. Often, however, presentation of a solution leads to further, refinement of the problem definition, and the process continues. Following this process will eventually lead to a satisfactory conclusion., Now let us look at this process for a student taking an electrical, and computer engineering foundations course. (The basic process also, applies to almost every engineering course.) Keep in mind that, although the steps have been simplified to apply to academic types of, problems, the process as stated always needs to be followed. We consider a simple example., , Example 1.10, , Solve for the current flowing through the 8- resistor in Fig. 1.19., 2Ω, , Solution:, 1. Carefully define the problem. This is only a simple example, but, we can already see that we do not know the polarity on the 3-V source., We have the following options. We can ask the professor what the, polarity should be. If we cannot ask, then we need to make a decision, on what to do next. If we have time to work the problem both ways,, we can solve for the current when the 3-V source is plus on top and, then plus on the bottom. If we do not have the time to work it both, ways, assume a polarity and then carefully document your decision., Let us assume that the professor tells us that the source is plus on the, bottom as shown in Fig. 1.20., 2. Present everything you know about the problem. Presenting all that, we know about the problem involves labeling the circuit clearly so that, we define what we seek., Given the circuit shown in Fig. 1.20, solve for i8., We now check with the professor, if reasonable, to see if the problem is properly defined., 3. Establish a set of alternative solutions and determine the one that, promises the greatest likelihood of success. There are essentially three, techniques that can be used to solve this problem. Later in the text you, will see that you can use circuit analysis (using Kirchhoff’s laws and, Ohm’s law), nodal analysis, and mesh analysis., To solve for i8 using circuit analysis will eventually lead to a, solution, but it will likely take more work than either nodal or mesh, , 5V, , +, −, , 4Ω, 8Ω, , 3V, , Figure 1.19, Illustrative example., , 2Ω, , 4Ω, i8Ω, , 5V, , +, −, , Figure 1.20, Problem defintion., , 8Ω, , −, +, , 3V
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ale29559_ch01.qxd, , 07/08/2008, , 10:38 AM, , Page 23, , 1.9, , Summary, , 23, , So we now have a very high degree of confidence in the accuracy, of our answer., 6. Has the problem been solved satisfactorily? If so, present the solution; if not, then return to step 3 and continue through the process, again. This problem has been solved satisfactorily., , The current through the 8- resistor is 0.25 A flowing down through, the 8- resistor., , Try applying this process to some of the more difficult problems at the, end of the chapter., , 1.9, , Summary, , 1. An electric circuit consists of electrical elements connected, together., 2. The International System of Units (SI) is the international measurement language, which enables engineers to communicate their, results. From the six principal units, the units of other physical, quantities can be derived., 3. Current is the rate of charge flow., i, , dq, dt, , 4. Voltage is the energy required to move 1 C of charge through an, element., v, , dw, dq, , 5. Power is the energy supplied or absorbed per unit time. It is also, the product of voltage and current., p, , dw, vi, dt, , 6. According to the passive sign convention, power assumes a positive sign when the current enters the positive polarity of the voltage, across an element., 7. An ideal voltage source produces a specific potential difference, across its terminals regardless of what is connected to it. An ideal, current source produces a specific current through its terminals, regardless of what is connected to it., 8. Voltage and current sources can be dependent or independent. A, dependent source is one whose value depends on some other circuit variable., 9. Two areas of application of the concepts covered in this chapter, are the TV picture tube and electricity billing procedure., , Practice Problem 1.10
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ale29559_ch01.qxd, , 07/08/2008, , 10:38 AM, , Page 24, , Chapter 1, , 24, , Basic Concepts, , Review Questions, 1.1, , One millivolt is one millionth of a volt., (a) True, , 1.2, , (b) False, , 1.5, , 1.6, , 1.7, , (d) 106, , 1.9, , The voltage 2,000,000 V can be expressed in powers, of 10 as:, (a) 2 mV, , 1.4, , (c) 103, , (b) 103, , (b) 2 kV, , The voltage across a 1.1-kW toaster that produces a, current of 10 A is:, (a) 11 kV, , The prefix micro stands for:, (a) 106, , 1.3, , 1.8, , (c) 2 MV, , (d) 2 GV, , (b) 1100 V, , (c) 110 V, , Which of these is not an electrical quantity?, (a) charge, , (b) time, , (d) current, , (e) power, , (c) voltage, , 1.10 The dependent source in Fig. 1.22 is:, , A charge of 2 C flowing past a given point each, second is a current of 2 A., , (a) voltage-controlled current source, , (a) True, , (c) current-controlled voltage source, , (b) voltage-controlled voltage source, , (b) False, , (d) current-controlled current source, , The unit of current is:, (a) coulomb, , (b) ampere, , (c) volt, , (d) joule, , io, vs, , Voltage is measured in:, (a) watts, , (b) amperes, , (c) volts, , (d) joules per second, , A 4-A current charging a dielectric material will, accumulate a charge of 24 C after 6 s., (a) True, , (d) 11 V, , (b) False, , +, −, , 6io, , Figure 1.22, For Review Question 1.10., , Answers: 1.1b, 1.2d, 1.3c, 1.4a, 1.5b, 1.6c, 1.7a, 1.8c,, 1.9b, 1.10d., , Problems, Section 1.3 Charge and Current, 1.1, , 1.2, , How many coulombs are represented by these, amounts of electrons?, (a) 6.482 1017, , (b) 1.24 1018, , (c) 2.46 1019, , (d) 1.628 10 20, , Determine the current flowing through an element if, the charge flow is given by, (a) q(t) (3t 8) mC, (b) q(t) (8t2 4t 2) C, , 1.4, , A current of 3.2 A flows through a conductor., Calculate how much charge passes through any, cross-section of the conductor in 20 s., , 1.5, , Determine the total charge transferred over the time, interval of 0 t 10 s when i(t) 12 t A., , 1.6, , The charge entering a certain element is shown in, Fig. 1.23. Find the current at:, (a) t 1 ms, , (b) t 6 ms, , (c) t 10 ms, , q(t) (mC), , (c) q(t) (3et 5e2t ) nC, , 80, , (d) q(t) 10 sin 120 p t pC, (e) q(t) 20e4t cos 50t m C, 1.3, , Find the charge q(t) flowing through a device if the, current is:, (a) i(t) 3 A, q(0) 1 C, (b) i(t) (2t 5) mA, q(0) 0, (c) i(t) 20 cos(10t p6) mA, q(0) 2 mC, (d) i(t) 10e30t sin 40t A, q(0) 0, , 0, , Figure 1.23, For Prob. 1.6., , 2, , 4, , 6, , 8, , 10, , 12, , t (ms)
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ale29559_ch01.qxd, , 07/08/2008, , 10:38 AM, , Page 25, , Problems, , 1.7, , The charge flowing in a wire is plotted in Fig. 1.24., Sketch the corresponding current., , 25, , 1.13 The charge entering the positive terminal of an, element is, q 10 sin 4 p t mC, , q (C), , while the voltage across the element (plus to minus) is, , 50, , v 2 cos 4 p t V, 0, , 2, , 4, , 8, , 6, , t (s), , −50, , (b) Calculate the energy delivered to the element, between 0 and 0.6 s., , Figure 1.24, , 1.14 The voltage v across a device and the current i, through it are, , For Prob. 1.7., 1.8, , (a) Find the power delivered to the element at, t 0.3 s., , The current flowing past a point in a device is shown in, Fig. 1.25. Calculate the total charge through the point., , i(t) 10 (1 e 0.5t ) A, , v(t) 5 cos 2t V,, Calculate:, , i (mA), , (a) the total charge in the device at t 1 s, (b) the power consumed by the device at t 1 s., , 10, , 0, , 2, , 1, , t (ms), , Figure 1.25, , (a) Find the charge delivered to the device between, t 0 and t 2 s., , For Prob. 1.8., 1.9, , 1.15 The current entering the positive terminal of a device, is i(t) 3e2t A and the voltage across the device is, v(t) 5didt V., , The current through an element is shown in Fig. 1.26., Determine the total charge that passed through the, element at:, (a) t 1 s, , (b) t 3 s, , (c) t 5 s, , (b) Calculate the power absorbed., (c) Determine the energy absorbed in 3 s., , Section 1.6 Circuit Elements, 1.16 Find the power absorbed by each element in, Fig. 1.27., , i (A), 10, 5, , 4A, 0, , 1, , 2, , 3, , 4, , −3 A, , 2A, , 5 t (s), , Figure 1.26, , +, , +, , −, , 10 V, −, , 12 V, −, , 5V, +, , For Prob. 1.9., , Sections 1.4 and 1.5 Voltage, Power, and Energy, 1.10 A lightning bolt with 8 kA strikes an object for 15 ms., How much charge is deposited on the object?, 1.11 A rechargeable flashlight battery is capable of, delivering 85 mA for about 12 h. How much charge, can it release at that rate? If its terminal voltage is, 1.2 V, how much energy can the battery deliver?, , Figure 1.27, For Prob. 1.16., 1.17 Figure 1.28 shows a circuit with five elements. If, p1 205 W, p2 60 W, p4 45 W, p5 30 W,, calculate the power p3 received or delivered by, element 3., , 1.12 If the current flowing through an element is given by, 3tA,, 0, 18A,, 6, i(t) µ, 12A, 10, 0,, , t 6 6s, t 6 10 s, t 6 15 s, t, 15 s, , Plot the charge stored in the element over, 0 6 t 6 20 s., , 2, 1, , Figure 1.28, For Prob. 1.17., , 4, 3, , 5
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ale29559_ch01.qxd, , 07/08/2008, , 10:38 AM, , Page 26, , Chapter 1, , 26, , 1.18 Calculate the power absorbed or supplied by each, element in Fig. 1.29., 4A, , 4A, , 9V +, −, , 1.22 A lightning bolt strikes an airplane with 30 kA for, 2 ms. How many coulombs of charge are deposited, on the plane?, , +, 3V, −, , 2, , Section 1.7 Applications, 1.21 A 60-W incandescent bulb operates at 120 V. How, many electrons and coulombs flow through the bulb, in one day?, , + 6V −, 1, , Basic Concepts, , 1.23 A 1.8-kW electric heater takes 15 min to boil a, quantity of water. If this is done once a day and, power costs 10 cents/kWh, what is the cost of its, operation for 30 days?, , (a), Io = 3 A + 10 V −, , 1.24 A utility company charges 8.5 cents/kWh. If a, consumer operates a 40-W light bulb continuously, for one day, how much is the consumer charged?, , 1, 24 V +, −, , 3Io, , +, −, , 1.25 A 1.2-kW toaster takes roughly 4 minutes to heat, four slices of bread. Find the cost of operating the, toaster once per day for 1 month (30 days). Assume, energy costs 9 cents/kWh., , 2, − 5V +, , 3A, , 1.26 A 12-V car battery supported a current of 150 mA to, a bulb. Calculate:, , (b), , Figure 1.29, , (a) the power absorbed by the bulb,, , For Prob. 1.18., , (b) the energy absorbed by the bulb over an interval, of 20 minutes., , 1.19 Find I in the network of Fig. 1.30., I, , 1A, , +, 9V, −, , 4A, , +, 3V, −, , +, 9V, −, , +, −, , 6V, , For Prob. 1.19., 1.20 Find Vo in the circuit of Fig. 1.31., , 12 V, +, −, , 1A, 3A, , +, −, , 6A, , Figure 1.31, For Prob. 1.20., , (b) how much energy is expended?, (c) how much does the charging cost? Assume, electricity costs 9 cents/kWh., , (a) the current through the lamp., , Io = 2 A, , 30 V, , (a) how much charge is transported as a result of the, charging?, , 1.28 A 30-W incandescent lamp is connected to a 120-V, source and is left burning continuously in an, otherwise dark staircase. Determine:, , Figure 1.30, , 6A, , 1.27 A constant current of 3 A for 4 hours is required, to charge an automotive battery. If the terminal, voltage is 10 t2 V, where t is in hours,, , +, Vo, −, , (b) the cost of operating the light for one non-leap, year if electricity costs 12 cents per kWh., , +, −, 28 V, , 1.29 An electric stove with four burners and an oven is, used in preparing a meal as follows., , +, −, 28 V, –, +, , 5Io, 3A, , Burner 1: 20 minutes, , Burner 2: 40 minutes, , Burner 3: 15 minutes, , Burner 4: 45 minutes, , Oven: 30 minutes, If each burner is rated at 1.2 kW and the oven at, 1.8 kW, and electricity costs 12 cents per kWh,, calculate the cost of electricity used in preparing, the meal.
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ale29559_ch01.qxd, , 07/08/2008, , 10:38 AM, , Page 27, , Comprehensive Problems, , 1.30 Reliant Energy (the electric company in Houston,, Texas) charges customers as follows:, Monthly charge $6, First 250 kWh @ $0.02/kWh, All additional kWh @ $0.07/kWh, , 27, , 1.31 In a household, a 120-W personal computer (PC) is, run for 4 h/day, while a 60-W bulb runs for 8 h/day., If the utility company charges $0.12/kWh, calculate, how much the household pays per year on the PC, and the bulb., , If a customer uses 1,218 kWh in one month, how, much will Reliant Energy charge?, , Comprehensive Problems, 1.32 A telephone wire has a current of 20 mA flowing, through it. How long does it take for a charge of, 15 C to pass through the wire?, 1.33 A lightning bolt carried a current of 2 kA and lasted, for 3 ms. How many coulombs of charge were, contained in the lightning bolt?, , p (MW), 8, 5, 4, 3, 8.00, , 1.34 Figure 1.32 shows the power consumption of a, certain household in 1 day. Calculate:, , 8.05, , 8.10, , 8.15, , 8.20, , 8.25, , 8.30 t, , Figure 1.33, For Prob. 1.35., , (a) the total energy consumed in kWh,, 1.36 A battery may be rated in ampere-hours (Ah). A, lead-acid battery is rated at 160 Ah., , (b) the average power per hour., 1200 W, , (a) What is the maximum current it can supply for, 40 h?, , p, 800 W, , (b) How many days will it last if it is discharged at, 1 mA?, 200 W, t (h), , 12, , 2, , 4, , 6, , 8 10 12 2, noon, , 4, , 6, , 8 10 12, , Figure 1.32, For Prob. 1.34., 1.35 The graph in Fig. 1.33 represents the power drawn, by an industrial plant between 8:00 and 8:30 A.M., Calculate the total energy in MWh consumed by the, plant., , 1.37 A unit of power often used for electric motors is the, horsepower (hp), which equals 746 W. A small, electric car is equipped with a 40-hp electric motor., How much energy does the motor deliver in one, hour, assuming the motor is operating at maximum, power for the whole time?, 1.38 How much energy does a 10-hp motor deliver in, 30 minutes? Assume that 1 horsepower 746 W., 1.39 A 600-W TV receiver is turned on for 4 h with, nobody watching it. If electricity costs 10 cents/kWh,, how much money is wasted?
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ale29559_ch02.qxd, , 07/21/2008, , 10:49 AM, , Page 29, , c h a p t e r, , 2, , Basic Laws, There are too many people praying for mountains of difficulty to be, removed, when what they really need is the courage to climb them!, —Unknown, , Enhancing Your Skills and Your Career, ABET EC 2000 criteria (3.b), “an ability to design and conduct experiments, as well as to analyze and interpret data., Engineers must be able to design and conduct experiments, as well as, analyze and interpret data. Most students have spent many hours performing experiments in high school and in college. During this time,, you have been asked to analyze the data and to interpret the data., Therefore, you should already be skilled in these two activities. My, recommendation is that, in the process of performing experiments in, the future, you spend more time in analyzing and interpreting the data, in the context of the experiment. What does this mean?, If you are looking at a plot of voltage versus resistance or current, versus resistance or power versus resistance, what do you actually see?, Does the curve make sense? Does it agree with what the theory tells, you? Does it differ from expectation, and, if so, why? Clearly, practice, with analyzing and interpreting data will enhance this skill., Since most, if not all, the experiments you are required to do as a, student involve little or no practice in designing the experiment, how, can you develop and enhance this skill?, Actually, developing this skill under this constraint is not as difficult as it seems. What you need to do is to take the experiment and, analyze it. Just break it down into its simplest parts, reconstruct it trying to understand why each element is there, and finally, determine, what the author of the experiment is trying to teach you. Even though, it may not always seem so, every experiment you do was designed by, someone who was sincerely motivated to teach you something., , Photo by Charles Alexander, , 29
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ale29559_ch02.qxd, , 07/17/2008, , 12:29 PM, , Page 30, , Chapter 2, , 30, , Basic Laws, , 2.1, , Introduction, , Chapter 1 introduced basic concepts such as current, voltage, and, power in an electric circuit. To actually determine the values of these, variables in a given circuit requires that we understand some fundamental laws that govern electric circuits. These laws, known as Ohm’s, law and Kirchhoff’s laws, form the foundation upon which electric circuit analysis is built., In this chapter, in addition to these laws, we shall discuss some, techniques commonly applied in circuit design and analysis. These techniques include combining resistors in series or parallel, voltage division,, current division, and delta-to-wye and wye-to-delta transformations. The, application of these laws and techniques will be restricted to resistive, circuits in this chapter. We will finally apply the laws and techniques to, real-life problems of electrical lighting and the design of dc meters., , 2.2, , l, , i, , Material with, resistivity , , +, v, −, , Cross-sectional, area A, (a), , Figure 2.1, (a) Resistor, (b) Circuit symbol for, resistance., , Ohm’s Law, , Materials in general have a characteristic behavior of resisting the flow, of electric charge. This physical property, or ability to resist current, is, known as resistance and is represented by the symbol R. The resistance of any material with a uniform cross-sectional area A depends on, A and its length /, as shown in Fig. 2.1(a). We can represent resistance, (as measured in the laboratory), in mathematical form,, , R, , Rr, , (b), , /, A, , (2.1), , where r is known as the resistivity of the material in ohm-meters. Good, conductors, such as copper and aluminum, have low resistivities, while, insulators, such as mica and paper, have high resistivities. Table 2.1, presents the values of r for some common materials and shows which, materials are used for conductors, insulators, and semiconductors., The circuit element used to model the current-resisting behavior of a, material is the resistor. For the purpose of constructing circuits, resistors, are usually made from metallic alloys and carbon compounds. The circuit, TABLE 2.1, , Resistivities of common materials., Material, Silver, Copper, Aluminum, Gold, Carbon, Germanium, Silicon, Paper, Mica, Glass, Teflon, , Resistivity (m), 8, , 1.64 10, 1.72 108, 2.8 108, 2.45 108, 4 105, 47 102, 6.4 102, 1010, 5 1011, 1012, 3 1012, , Usage, Conductor, Conductor, Conductor, Conductor, Semiconductor, Semiconductor, Semiconductor, Insulator, Insulator, Insulator, Insulator
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ale29559_ch02.qxd, , 07/21/2008, , 10:49 AM, , Page 31, , 2.2, , Ohm’s Law, , symbol for the resistor is shown in Fig. 2.1(b), where R stands for the, resistance of the resistor. The resistor is the simplest passive element., Georg Simon Ohm (1787–1854), a German physicist, is credited, with finding the relationship between current and voltage for a resistor. This relationship is known as Ohm’s law., Ohm’s law states that the voltage v across a resistor is directly proportional to the current i flowing through the resistor., , That is,, v r i, , (2.2), , Ohm defined the constant of proportionality for a resistor to be the, resistance, R. (The resistance is a material property which can change, if the internal or external conditions of the element are altered, e.g., if, there are changes in the temperature.) Thus, Eq. (2.2) becomes, v iR, , (2.3), , which is the mathematical form of Ohm’s law. R in Eq. (2.3) is measured in the unit of ohms, designated . Thus,, The resistance R of an element denotes its ability to resist the flow of, electric current; it is measured in ohms ()., , We may deduce from Eq. (2.3) that, R, , v, i, , (2.4), , so that, 1 1 V/A, To apply Ohm’s law as stated in Eq. (2.3), we must pay careful, attention to the current direction and voltage polarity. The direction of, current i and the polarity of voltage v must conform with the passive, , Historical, Georg Simon Ohm (1787–1854), a German physicist, in 1826, experimentally determined the most basic law relating voltage and current for a resistor. Ohm’s work was initially denied by critics., Born of humble beginnings in Erlangen, Bavaria, Ohm threw himself into electrical research. His efforts resulted in his famous law., He was awarded the Copley Medal in 1841 by the Royal Society of, London. In 1849, he was given the Professor of Physics chair by the, University of Munich. To honor him, the unit of resistance was named, the ohm., , 31
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ale29559_ch02.qxd, , 07/21/2008, , 10:49 AM, , Page 32, , Chapter 2, , 32, , +, , i, , v=0 R=0, −, , Basic Laws, , sign convention, as shown in Fig. 2.1(b). This implies that current flows, from a higher potential to a lower potential in order for v i R. If current flows from a lower potential to a higher potential, v i R., Since the value of R can range from zero to infinity, it is important that we consider the two extreme possible values of R. An element, with R 0 is called a short circuit, as shown in Fig. 2.2(a). For a short, circuit,, v iR 0, , (a), , +, v, , i=0, R=∞, , −, , showing that the voltage is zero but the current could be anything. In, practice, a short circuit is usually a connecting wire assumed to be a, perfect conductor. Thus,, A short circuit is a circuit element with resistance approaching zero., , Similarly, an element with R is known as an open circuit, as, shown in Fig. 2.2(b). For an open circuit,, i lim, , (b), , RS, , Figure 2.2, , (a) Short circuit (R 0), (b) Open circuit, (R )., , (2.5), , v, 0, R, , (2.6), , indicating that the current is zero though the voltage could be anything., Thus,, An open circuit is a circuit element with resistance approaching infinity., , A resistor is either fixed or variable. Most resistors are of the fixed, type, meaning their resistance remains constant. The two common types, of fixed resistors (wirewound and composition) are shown in Fig. 2.3., The composition resistors are used when large resistance is needed., The circuit symbol in Fig. 2.1(b) is for a fixed resistor. Variable resistors have adjustable resistance. The symbol for a variable resistor is, shown in Fig. 2.4(a). A common variable resistor is known as a potentiometer or pot for short, with the symbol shown in Fig. 2.4(b). The, pot is a three-terminal element with a sliding contact or wiper. By sliding the wiper, the resistances between the wiper terminal and the fixed, terminals vary. Like fixed resistors, variable resistors can be of either, wirewound or composition type, as shown in Fig. 2.5. Although resistors, like those in Figs. 2.3 and 2.5 are used in circuit designs, today most, , (a), , (b), , Figure 2.3, Fixed resistors: (a) wirewound type,, (b) carbon film type., Courtesy of Tech America., , (a), , (a), , (b), , (b), , Figure 2.4, , Figure 2.5, , Circuit symbol for: (a) a variable resistor, in general, (b) a potentiometer., , Variable resistors: (a) composition type, (b) slider pot., Courtesy of Tech America.
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 33, , 2.2, , Ohm’s Law, , circuit components including resistors are either surface mounted or, integrated, as typically shown in Fig. 2.6., It should be pointed out that not all resistors obey Ohm’s law. A, resistor that obeys Ohm’s law is known as a linear resistor. It has a, constant resistance and thus its current-voltage characteristic is as illustrated in Fig. 2.7(a): its i-v graph is a straight line passing through the, origin. A nonlinear resistor does not obey Ohm’s law. Its resistance, varies with current and its i-v characteristic is typically shown in, Fig. 2.7(b). Examples of devices with nonlinear resistance are the lightbulb and the diode. Although all practical resistors may exhibit nonlinear behavior under certain conditions, we will assume in this book that, all elements actually designated as resistors are linear., A useful quantity in circuit analysis is the reciprocal of resistance, R, known as conductance and denoted by G:, , 33, , Figure 2.6, Resistors in a thick-film circuit., , 1, i, G , v, R, , (2.7), , The conductance is a measure of how well an element will conduct electric current. The unit of conductance is the mho (ohm spelled, backward) or reciprocal ohm, with symbol , the inverted omega., Although engineers often use the mho, in this book we prefer to use, the siemens (S), the SI unit of conductance:, , , , , 1S1, , 1 A/ V, , G. Daryanani, Principles of Active Network, Synthesis and Design (New York: John Wiley,, 1976), p. 461c., , v, , (2.8), , Slope = R, , Thus,, i, , Conductance is the ability of an element to conduct electric current;, it is measured in mhos ( ) or siemens (S)., , (a), , , , v, , The same resistance can be expressed in ohms or siemens. For, example, 10 is the same as 0.1 S. From Eq. (2.7), we may write, i Gv, , (2.9), Slope = R, , The power dissipated by a resistor can be expressed in terms of R., Using Eqs. (1.7) and (2.3),, , i, , 2, , p vi i 2R , , v, R, , (b), , (2.10), , The power dissipated by a resistor may also be expressed in terms of, G as, p vi v2G , , i2, G, , (2.11), , We should note two things from Eqs. (2.10) and (2.11):, 1. The power dissipated in a resistor is a nonlinear function of either, current or voltage., 2. Since R and G are positive quantities, the power dissipated in a, resistor is always positive. Thus, a resistor always absorbs power, from the circuit. This confirms the idea that a resistor is a passive, element, incapable of generating energy., , Figure 2.7, The i-v characteristic of: (a) a linear, resistor, (b) a nonlinear resistor.
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 34, , Chapter 2, , 34, , Example 2.1, , Basic Laws, , An electric iron draws 2 A at 120 V. Find its resistance., Solution:, From Ohm’s law,, R, , Practice Problem 2.1, , v, 120, , 60 , i, 2, , The essential component of a toaster is an electrical element (a resistor) that converts electrical energy to heat energy. How much current, is drawn by a toaster with resistance 10 at 110 V?, Answer: 11 A., , Example 2.2, , In the circuit shown in Fig. 2.8, calculate the current i, the conductance, G, and the power p., i, , 30 V +, −, , 5 kΩ, , Solution:, The voltage across the resistor is the same as the source voltage (30 V), because the resistor and the voltage source are connected to the same, pair of terminals. Hence, the current is, , +, v, −, , Figure 2.8, For Example 2.2., , i, , v, 30, , 6 mA, R, 5 103, , G, , 1, 1, , 0.2 mS, R, 5 103, , The conductance is, , We can calculate the power in various ways using either Eqs. (1.7),, (2.10), or (2.11)., p vi 30(6 103) 180 mW, or, p i 2R (6 103)25 103 180 mW, or, p v2G (30)20.2 103 180 mW, , Practice Problem 2.2, , For the circuit shown in Fig. 2.9, calculate the voltage v, the conductance G, and the power p., , i, 2 mA, , Figure 2.9, For Practice Prob. 2.2, , 10 kΩ, , +, v, −, , Answer: 20 V, 100 mS, 40 mW.
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 35, , 2.3, , Nodes, Branches, and Loops, , 35, , Example 2.3, , A voltage source of 20 sin p t V is connected across a 5-k resistor., Find the current through the resistor and the power dissipated., Solution:, i, , v, 20 sin p t, , 4 sin p t mA, R, 5 103, , Hence,, p vi 80 sin2 p t mW, , Practice Problem 2.3, , A resistor absorbs an instantaneous power of 20 cos2 t mW when connected to a voltage source v 10 cos t V. Find i and R., Answer: 2 cos t mA, 5 k., , 2.3, , Nodes, Branches, and Loops, , Since the elements of an electric circuit can be interconnected in several ways, we need to understand some basic concepts of network, topology. To differentiate between a circuit and a network, we may, regard a network as an interconnection of elements or devices, whereas, a circuit is a network providing one or more closed paths. The convention, when addressing network topology, is to use the word network, rather than circuit. We do this even though the words network and circuit mean the same thing when used in this context. In network topology, we study the properties relating to the placement of elements in, the network and the geometric configuration of the network. Such elements include branches, nodes, and loops., A branch represents a single element such as a voltage source or a, resistor., , 5Ω, , a, , 10 V +, −, , b, , 2Ω, , 3Ω, , 2A, , c, , In other words, a branch represents any two-terminal element. The circuit in Fig. 2.10 has five branches, namely, the 10-V voltage source,, the 2-A current source, and the three resistors., , Figure 2.10, Nodes, branches, and loops., , A node is the point of connection between two or more branches., b, , A node is usually indicated by a dot in a circuit. If a short circuit (a, connecting wire) connects two nodes, the two nodes constitute a single node. The circuit in Fig. 2.10 has three nodes a, b, and c. Notice, that the three points that form node b are connected by perfectly conducting wires and therefore constitute a single point. The same is true, of the four points forming node c. We demonstrate that the circuit in, Fig. 2.10 has only three nodes by redrawing the circuit in Fig. 2.11., The two circuits in Figs. 2.10 and 2.11 are identical. However, for the, sake of clarity, nodes b and c are spread out with perfect conductors, as in Fig. 2.10., , 5Ω, 2Ω, a, , 3Ω, , +, −, 10 V, , c, , Figure 2.11, The three-node circuit of Fig. 2.10 is, redrawn., , 2A
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 36, , Chapter 2, , 36, , Basic Laws, , A loop is any closed path in a circuit., , A loop is a closed path formed by starting at a node, passing through a, set of nodes, and returning to the starting node without passing through, any node more than once. A loop is said to be independent if it contains, at least one branch which is not a part of any other independent loop., Independent loops or paths result in independent sets of equations., It is possible to form an independent set of loops where one of the, loops does not contain such a branch. In Fig. 2.11, abca with the 2, resistor is independent. A second loop with the 3 resistor and the current source is independent. The third loop could be the one with the 2, resistor in parallel with the 3 resistor. This does form an independent, set of loops., A network with b branches, n nodes, and l independent loops will, satisfy the fundamental theorem of network topology:, bln1, , (2.12), , As the next two definitions show, circuit topology is of great value, to the study of voltages and currents in an electric circuit., Two or more elements are in series if they exclusively share a single, node and consequently carry the same current., Two or more elements are in parallel if they are connected to the same, two nodes and consequently have the same voltage across them., , Elements are in series when they are chain-connected or connected, sequentially, end to end. For example, two elements are in series if, they share one common node and no other element is connected to, that common node. Elements in parallel are connected to the same pair, of terminals. Elements may be connected in a way that they are neither in series nor in parallel. In the circuit shown in Fig. 2.10, the voltage source and the 5- resistor are in series because the same current, will flow through them. The 2- resistor, the 3- resistor, and the current source are in parallel because they are connected to the same two, nodes b and c and consequently have the same voltage across them., The 5- and 2- resistors are neither in series nor in parallel with, each other., , Example 2.4, , Determine the number of branches and nodes in the circuit shown in, Fig. 2.12. Identify which elements are in series and which are in, parallel., Solution:, Since there are four elements in the circuit, the circuit has four, branches: 10 V, 5 , 6 , and 2 A. The circuit has three nodes as, identified in Fig. 2.13. The 5- resistor is in series with the 10-V, voltage source because the same current would flow in both. The 6-, resistor is in parallel with the 2-A current source because both are, connected to the same nodes 2 and 3.
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 37, , 2.4, 5Ω, , 10 V, , 1, , +, −, , 6Ω, , 10 V, , 2A, , 5Ω, , 37, , 2, , +, −, , Figure 2.12, , Kirchhoff’s Laws, , 6Ω, , 2A, , 3, , For Example 2.4., , Figure 2.13, The three nodes in the circuit of, Fig. 2.12., , Practice Problem 2.4, , How many branches and nodes does the circuit in Fig. 2.14 have? Identify the elements that are in series and in parallel., Answer: Five branches and three nodes are identified in Fig. 2.15. The, 1- and 2- resistors are in parallel. The 4- resistor and 10-V source, are also in parallel., 5Ω, , 1Ω, , 2Ω, , 3Ω, , 1, , + 10 V, −, , 4Ω, , 1Ω, , + 10 V, −, , 2Ω, , Figure 2.14, , 2, , 3, , For Practice Prob. 2.4., , Figure 2.15, Answer for Practice Prob. 2.4., , 2.4, , Kirchhoff’s Laws, , Ohm’s law by itself is not sufficient to analyze circuits. However, when, it is coupled with Kirchhoff’s two laws, we have a sufficient, powerful, set of tools for analyzing a large variety of electric circuits. Kirchhoff’s, laws were first introduced in 1847 by the German physicist Gustav, Robert Kirchhoff (1824–1887). These laws are formally known as, Kirchhoff’s current law (KCL) and Kirchhoff’s voltage law (KVL)., Kirchhoff’s first law is based on the law of conservation of charge,, which requires that the algebraic sum of charges within a system cannot, change., Kirchhoff’s current law (KCL) states that the algebraic sum of currents, entering a node (or a closed boundary) is zero., , Mathematically, KCL implies that, N, , a in 0, , (2.13), , n1, , where N is the number of branches connected to the node and in is, the nth current entering (or leaving) the node. By this law, currents, , 4Ω
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ale29559_ch02.qxd, , 07/21/2008, , 10:49 AM, , Page 38, , Chapter 2, , 38, , Basic Laws, , Historical, Gustav Robert Kirchhoff (1824–1887), a German physicist, stated, two basic laws in 1847 concerning the relationship between the currents and voltages in an electrical network. Kirchhoff’s laws, along, with Ohm’s law, form the basis of circuit theory., Born the son of a lawyer in Konigsberg, East Prussia, Kirchhoff, entered the University of Konigsberg at age 18 and later became a lecturer in Berlin. His collaborative work in spectroscopy with German, chemist Robert Bunsen led to the discovery of cesium in 1860 and, rubidium in 1861. Kirchhoff was also credited with the Kirchhoff law, of radiation. Thus Kirchhoff is famous among engineers, chemists, and, physicists., , entering a node may be regarded as positive, while currents leaving the, node may be taken as negative or vice versa., To prove KCL, assume a set of currents ik (t), k 1, 2, p , flow, into a node. The algebraic sum of currents at the node is, iT (t) i1(t) i2(t) i3(t) p, i1, , Integrating both sides of Eq. (2.14) gives, , i5, , qT (t) q1(t) q2(t) q3(t) p, i4, , i2, , (2.14), , i3, , Figure 2.16, Currents at a node illustrating KCL., , where qk (t) ik (t) d t and qT (t) iT (t) d t. But the law of conservation of electric charge requires that the algebraic sum of electric, charges at the node must not change; that is, the node stores no net, charge. Thus qT (t) 0 S iT (t) 0, confirming the validity of KCL., Consider the node in Fig. 2.16. Applying KCL gives, i1 (i2) i3 i4 (i5) 0, , Closed boundary, , (2.15), , (2.16), , since currents i1, i3, and i4 are entering the node, while currents i2 and, i5 are leaving it. By rearranging the terms, we get, i1 i3 i4 i2 i5, , (2.17), , Equation (2.17) is an alternative form of KCL:, The sum of the currents entering a node is equal to the sum of the currents leaving the node., , Figure 2.17, Applying KCL to a closed boundary., Two sources (or circuits in general) are, said to be equivalent if they have the, same i-v relationship at a pair of, terminals., , Note that KCL also applies to a closed boundary. This may be, regarded as a generalized case, because a node may be regarded as a, closed surface shrunk to a point. In two dimensions, a closed boundary is the same as a closed path. As typically illustrated in the circuit, of Fig. 2.17, the total current entering the closed surface is equal to the, total current leaving the surface., A simple application of KCL is combining current sources in parallel. The combined current is the algebraic sum of the current supplied, by the individual sources. For example, the current sources shown in
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 39, , 2.4, , Kirchhoff’s Laws, , Fig. 2.18(a) can be combined as in Fig. 2.18(b). The combined or, equivalent current source can be found by applying KCL to node a., , 39, IT, a, , IT I2 I1 I3, IT I1 I2 I3, , I2, , I1, , or, (2.18), , I3, , b, (a), , A circuit cannot contain two different currents, I1 and I2, in series,, unless I1 I2; otherwise KCL will be violated., Kirchhoff’s second law is based on the principle of conservation, of energy:, , IT, a, IT = I1 – I2 + I3, , Kirchhoff’s voltage law (KVL) states that the algebraic sum of all voltages around a closed path (or loop) is zero., , b, (b), , Figure 2.18, Expressed mathematically, KVL states that, , Current sources in parallel: (a) original, circuit, (b) equivalent circuit., , M, , a vm 0, , (2.19), , m1, , where M is the number of voltages in the loop (or the number of, branches in the loop) and vm is the mth voltage., To illustrate KVL, consider the circuit in Fig. 2.19. The sign on, each voltage is the polarity of the terminal encountered first as we, travel around the loop. We can start with any branch and go around, the loop either clockwise or counterclockwise. Suppose we start with, the voltage source and go clockwise around the loop as shown; then, voltages would be v1, v2, v3, v4, and v5, in that order. For, example, as we reach branch 3, the positive terminal is met first; hence,, we have v3. For branch 4, we reach the negative terminal first; hence,, v4. Thus, KVL yields, (2.20), , −, +, , v1 +, −, , Rearranging terms gives, (2.21), , which may be interpreted as, , −, , v2 v3 v5 v1 v4, , + v3 −, , + v2 −, , v5, , v4, , +, , v1 v2 v3 v4 v5 0, , KVL can be applied in two ways: by, taking either a clockwise or a counterclockwise trip around the loop. Either, way, the algebraic sum of voltages, around the loop is zero., , Figure 2.19, A single-loop circuit illustrating KVL., , Sum of voltage drops Sum of voltage rises, , (2.22), , This is an alternative form of KVL. Notice that if we had traveled, counterclockwise, the result would have been v1, v5, v4, v3,, and v2, which is the same as before except that the signs are reversed., Hence, Eqs. (2.20) and (2.21) remain the same., When voltage sources are connected in series, KVL can be applied, to obtain the total voltage. The combined voltage is the algebraic sum, of the voltages of the individual sources. For example, for the voltage, sources shown in Fig. 2.20(a), the combined or equivalent voltage, source in Fig. 2.20(b) is obtained by applying KVL., Vab V1 V2 V3 0
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 41, , 2.4, , Kirchhoff’s Laws, , 41, , Practice Problem 2.5, , Find v1 and v2 in the circuit of Fig. 2.22., , 4Ω, , Answer: 12 V, 6 V., , + v1 −, +, −, , +, −, , 10 V, , 8V, , + v2 −, 2Ω, , Figure 2.22, For Practice Prob. 2.5., , Example 2.6, , Determine vo and i in the circuit shown in Fig. 2.23(a)., i, , 12 V, , 4Ω, , 4Ω, , 2vo, +−, , +, −, , 4V, , −, +, , 2vo, +−, i, , 12 V +, −, , −, 4V +, , 6Ω, , 6Ω, , + vo −, , + vo −, , (a), , (b), , Figure 2.23, For Example 2.6., , Solution:, We apply KVL around the loop as shown in Fig. 2.23(b). The result is, 12 4i 2 vo 4 6i 0, , (2.6.1), , Applying Ohm’s law to the 6- resistor gives, vo 6i, , (2.6.2), , Substituting Eq. (2.6.2) into Eq. (2.6.1) yields, 16 10i 12i 0, , 1, , i 8 A, , and vo 48 V., , Find vx and vo in the circuit of Fig. 2.24., , Practice Problem 2.6, , Answer: 10 V, 5 V., , 10 Ω, + vx −, 35 V +, −, , 5Ω, + vo −, , Figure 2.24, For Practice Prob. 2.6., , +, −, , 2vx
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 43, , 2.5, , Series Resistors and Voltage Division, , 43, , Since the voltage and current of each resistor are related by Ohm’s, law as shown, we are really looking for three things: (v1, v2, v3) or, (i1, i2, i3). At node a, KCL gives, i1 i2 i3 0, , (2.8.2), , Applying KVL to loop 1 as in Fig. 2.27(b),, 30 v1 v2 0, We express this in terms of i1 and i2 as in Eq. (2.8.1) to obtain, 30 8i1 3i2 0, or, i1 , , (30 3i2), 8, , (2.8.3), , Applying KVL to loop 2,, v2 v3 0, , 1, , v3 v2, , (2.8.4), , as expected since the two resistors are in parallel. We express v1 and, v2 in terms of i1 and i2 as in Eq. (2.8.1). Equation (2.8.4) becomes, 6i3 3i2, , 1, , i3 , , i2, 2, , (2.8.5), , Substituting Eqs. (2.8.3) and (2.8.5) into (2.8.2) gives, 30 3i2, i2, i2 0, 8, 2, or i2 2 A. From the value of i2, we now use Eqs. (2.8.1) to (2.8.5), to obtain, i1 3 A,, , i3 1 A,, , v1 24 V,, , v2 6 V,, , v3 6 V, , Practice Problem 2.8, , Find the currents and voltages in the circuit shown in Fig. 2.28., Answer: v1 3 V, v2 2 V, v3 5 V, i1 1.5 A, i2 0.25 A,, i3 1.25 A., , 2Ω, + v1 −, 5V, , 2.5, , Series Resistors and Voltage Division, , The need to combine resistors in series or in parallel occurs so frequently that it warrants special attention. The process of combining the, resistors is facilitated by combining two of them at a time. With this, in mind, consider the single-loop circuit of Fig. 2.29. The two resistors, , i1, , +, −, , i3, , 4Ω, , i2 + v3 −, +, v2, −, , Figure 2.28, For Practice Prob. 2.8., , 8Ω, , −, +, , 3V
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 44, , Chapter 2, , 44, i, , v, , R1, , R2, , + v1 −, , + v2 −, , a, , Basic Laws, , are in series, since the same current i flows in both of them. Applying, Ohm’s law to each of the resistors, we obtain, v1 iR1,, , +, −, , v2 iR2, , (2.24), , If we apply KVL to the loop (moving in the clockwise direction), we, have, , b, , v v1 v2 0, , Figure 2.29, A single-loop circuit with two resistors in, series., , (2.25), , Combining Eqs. (2.24) and (2.25), we get, v v1 v2 i(R1 R2), , (2.26), , or, i, , v, R1 R2, , (2.27), , Notice that Eq. (2.26) can be written as, v iReq, i, , a, , Req, + v −, , v, , +, −, , b, , Figure 2.30, Equivalent circuit of the Fig. 2.29 circuit., , (2.28), , implying that the two resistors can be replaced by an equivalent resistor Req; that is,, Req R1 R2, , (2.29), , Thus, Fig. 2.29 can be replaced by the equivalent circuit in Fig. 2.30., The two circuits in Figs. 2.29 and 2.30 are equivalent because they, exhibit the same voltage-current relationships at the terminals a-b. An, equivalent circuit such as the one in Fig. 2.30 is useful in simplifying, the analysis of a circuit. In general,, The equivalent resistance of any number of resistors connected in, series is the sum of the individual resistances., , Resistors in series behave as a single, resistor whose resistance is equal to, the sum of the resistances of the, individual resistors., , For N resistors in series then,, N, , Req R1 R2 p RN a Rn, , (2.30), , n1, , To determine the voltage across each resistor in Fig. 2.29, we substitute Eq. (2.26) into Eq. (2.24) and obtain, v1 , , R1, v,, R1 R2, , v2 , , R2, v, R1 R2, , (2.31), , Notice that the source voltage v is divided among the resistors in direct, proportion to their resistances; the larger the resistance, the larger the, voltage drop. This is called the principle of voltage division, and the, circuit in Fig. 2.29 is called a voltage divider. In general, if a voltage, divider has N resistors (R1, R2, . . . , RN) in series with the source voltage v, the nth resistor (Rn ) will have a voltage drop of, vn , , Rn, v, R1 R2 p RN, , (2.32)
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 45, , 2.6, , 2.6, , Parallel Resistors and Current Division, , 45, , Parallel Resistors, and Current Division, , Consider the circuit in Fig. 2.31, where two resistors are connected, in parallel and therefore have the same voltage across them. From, Ohm’s law,, v i1R1 i2R2, , i, , Node a, i2, , i1, v, , +, −, , R1, , R2, , or, i1 , , v, ,, R1, , i2 , , v, R2, , (2.33), , Two resistors in parallel., , Applying KCL at node a gives the total current i as, i i1 i2, , (2.34), , Substituting Eq. (2.33) into Eq. (2.34), we get, i, , v, v, 1, 1, v, , va b, R1, R2, R1, R2, Req, , (2.35), , where Req is the equivalent resistance of the resistors in parallel:, 1, 1, 1, , , Req, R1, R2, , (2.36), , or, R1 R2, 1, , Req, R1R2, or, Req , , R1R2, R1 R2, , (2.37), , Thus,, The equivalent resistance of two parallel resistors is equal to the product of their resistances divided by their sum., , It must be emphasized that this applies only to two resistors in parallel. From Eq. (2.37), if R1 R2, then Req R12., We can extend the result in Eq. (2.36) to the general case of a circuit with N resistors in parallel. The equivalent resistance is, 1, 1, 1, 1, , , p, Req, R1, R2, RN, , (2.38), , Note that Req is always smaller than the resistance of the smallest resistor in the parallel combination. If R1 R2 p RN R, then, Req , , R, N, , Node b, , Figure 2.31, , (2.39)
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 46, , Chapter 2, , 46, , Conductances in parallel behave as a, single conductance whose value is, equal to the sum of the individual, conductances., , Basic Laws, , For example, if four 100- resistors are connected in parallel, their, equivalent resistance is 25 ., It is often more convenient to use conductance rather than resistance when dealing with resistors in parallel. From Eq. (2.38), the equivalent conductance for N resistors in parallel is, Geq G1 G2 G3 p GN, , i, , v, , where Geq 1Req, G1 1R1, G2 1R2, G3 1R3, p , GN 1RN., Equation (2.40) states:, , a, , +, −, , v, , (2.40), , Req or Geq, , The equivalent conductance of resistors connected in parallel is the, sum of their individual conductances., , This means that we may replace the circuit in Fig. 2.31 with that in, Fig. 2.32. Notice the similarity between Eqs. (2.30) and (2.40). The, equivalent conductance of parallel resistors is obtained the same way, as the equivalent resistance of series resistors. In the same manner,, the equivalent conductance of resistors in series is obtained just, the same way as the resistance of resistors in parallel. Thus the, equivalent conductance Geq of N resistors in series (such as shown in, Fig. 2.29) is, , b, , Figure 2.32, Equivalent circuit to Fig. 2.31., , 1, 1, 1, 1, 1, , , , p, Geq, G1, G2, G3, GN, , (2.41), , Given the total current i entering node a in Fig. 2.31, how do we, obtain current i1 and i2? We know that the equivalent resistor has the, same voltage, or, v iReq , , i, i1 = 0, R1, , i2 = i, , iR1 R2, R1 R2, , (2.42), , Combining Eqs. (2.33) and (2.42) results in, , R2 = 0, , i1 , , R2 i, ,, R1 R2, , i2 , , R1 i, R1 R2, , (2.43), , (a), i, i1 = i, R1, , i2 = 0, R2 = ∞, , (b), , Figure 2.33, (a) A shorted circuit, (b) an open circuit., , which shows that the total current i is shared by the resistors in, inverse proportion to their resistances. This is known as the principle of current division, and the circuit in Fig. 2.31 is known as a, current divider. Notice that the larger current flows through the, smaller resistance., As an extreme case, suppose one of the resistors in Fig. 2.31 is, zero, say R2 0; that is, R2 is a short circuit, as shown in, Fig. 2.33(a). From Eq. (2.43), R2 0 implies that i1 0, i2 i. This, means that the entire current i bypasses R1 and flows through the, short circuit R2 0, the path of least resistance. Thus when a circuit
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 47, , 2.6, , Parallel Resistors and Current Division, , 47, , is short circuited, as shown in Fig. 2.33(a), two things should be kept, in mind:, 1. The equivalent resistance Req 0. [See what happens when, R2 0 in Eq. (2.37).], 2. The entire current flows through the short circuit., As another extreme case, suppose R2 , that is, R2 is an open, circuit, as shown in Fig. 2.33(b). The current still flows through the, path of least resistance, R1. By taking the limit of Eq. (2.37) as R2 S ,, we obtain Req R1 in this case., If we divide both the numerator and denominator by R1R2, Eq. (2.43), becomes, i1 , , G1, i, G1 G2, , (2.44a), , i2 , , G2, i, G1 G2, , (2.44b), , Thus, in general, if a current divider has N conductors (G1, G2, p , GN), in parallel with the source current i, the nth conductor (Gn) will have, current, in , , Gn, i, G1 G2 p GN, , (2.45), , In general, it is often convenient and possible to combine resistors in series and parallel and reduce a resistive network to a single, equivalent resistance Req. Such an equivalent resistance is the resistance between the designated terminals of the network and must, exhibit the same i-v characteristics as the original network at the, terminals., , Example 2.9, , Find Req for the circuit shown in Fig. 2.34., Solution:, To get Req, we combine resistors in series and in parallel. The 6- and, 3- resistors are in parallel, so their equivalent resistance is, 63, 2, 6 3 , 63, (The symbol is used to indicate a parallel combination.) Also, the 1-, and 5- resistors are in series; hence their equivalent resistance is, 156, Thus the circuit in Fig. 2.34 is reduced to that in Fig. 2.35(a). In, Fig. 2.35(a), we notice that the two 2- resistors are in series, so the, equivalent resistance is, 224, , 4Ω, , 1Ω, 2Ω, , Req, , 5Ω, 8Ω, , Figure 2.34, For Example 2.9., , 6Ω, , 3Ω
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 48, , Chapter 2, , 48, , Basic Laws, , This 4- resistor is now in parallel with the 6- resistor in Fig. 2.35(a);, their equivalent resistance is, , 4Ω, 2Ω, , Req, , 46, , 6Ω, 2Ω, , 8Ω, , The circuit in Fig. 2.35(a) is now replaced with that in Fig. 2.35(b). In, Fig. 2.35(b), the three resistors are in series. Hence, the equivalent, resistance for the circuit is, , (a), 4Ω, Req, , 46, 2.4 , 46, , Req 4 2.4 8 14.4 , 2.4 Ω, , 8Ω, (b), , Figure 2.35, Equivalent circuits for Example 2.9., , Practice Problem 2.9, , By combining the resistors in Fig. 2.36, find Req., Answer: 6 ., , 2Ω, Req, , 3Ω, , 6Ω, 1Ω, , 4Ω, , 4Ω, , 5Ω, , 3Ω, , Figure 2.36, For Practice Prob. 2.9., , Example 2.10, , Calculate the equivalent resistance Rab in the circuit in Fig. 2.37., 10 Ω, a, Rab, , c, , 1Ω, , 1Ω, , d, , 6Ω, 4Ω, , 3Ω, , 5Ω, , 12 Ω, b, b, , b, , Figure 2.37, For Example 2.10., , Solution:, The 3- and 6- resistors are in parallel because they are connected, to the same two nodes c and b. Their combined resistance is, 36, , 36, 2, 36, , (2.10.1)
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ale29559_ch02.qxd, , 07/09/2008, , 01:44 PM, , Page 49, , 2.6, , Parallel Resistors and Current Division, , Similarly, the 12- and 4- resistors are in parallel since they are, connected to the same two nodes d and b. Hence, 12 4 , , 12 4, 3, 12 4, , 49, 10 Ω, a, , c 1Ω d, , 2Ω, , (2.10.2), , Also the 1- and 5- resistors are in series; hence, their equivalent, resistance is, 156, (2.10.3), , b, , With these three combinations, we can replace the circuit in Fig. 2.37 with, that in Fig. 2.38(a). In Fig. 2.38(a), 3- in parallel with 6- gives 2-,, as calculated in Eq. (2.10.1). This 2- equivalent resistance is now in series, with the 1- resistance to give a combined resistance of 1 2 3 ., Thus, we replace the circuit in Fig. 2.38(a) with that in Fig. 2.38(b). In, Fig. 2.38(b), we combine the 2- and 3- resistors in parallel to get, , a, , 23, 23, 1.2 , 23, , b, , 3Ω, , 6Ω, , b, , b, , (a), 10 Ω, , c, 3Ω, , 2Ω, b, b, , b, , (b), , Figure 2.38, Equivalent circuits for Example 2.10., , This 1.2- resistor is in series with the 10- resistor, so that, Rab 10 1.2 11.2 , , Find Rab for the circuit in Fig. 2.39., , Practice Problem 2.10, 20 Ω, , Answer: 11 ., , 5Ω, , 8Ω, a, Rab, , 18 Ω, , 20 Ω, 9Ω, , 1Ω, , 2Ω, b, , Figure 2.39, For Practice Prob. 2.10., , Find the equivalent conductance Geq for the circuit in Fig. 2.40(a)., Solution:, The 8-S and 12-S resistors are in parallel, so their conductance is, 8 S 12 S 20 S, This 20-S resistor is now in series with 5 S as shown in Fig. 2.40(b), so that the combined conductance is, 20 5, 4S, 20 5, This is in parallel with the 6-S resistor. Hence,, Geq 6 4 10 S, We should note that the circuit in Fig. 2.40(a) is the same as that, in Fig. 2.40(c). While the resistors in Fig. 2.40(a) are expressed in, , Example 2.11
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 50, , Chapter 2, , 50, , siemens, those in Fig. 2.40(c) are expressed in ohms. To show that the, circuits are the same, we find Req for the circuit in Fig. 2.40(c)., , 5S, Geq, , Basic Laws, , 12 S, , 8S, , 6S, , Req , , 1, 1, 1 1, 1, 1, 1, 1 1, ga g b ga b g, 6, 5, 8 12, 6, 5, 20, 6 4, , (a), , , , 5S, Geq, , 6S, , Req, , 1, 6, , 14, 14, , , , 1, , 10, Geq , , 20 S, , 1, 10 S, Req, , This is the same as we obtained previously., , (b), 1, 5, , 1, 6, 1, 6, , Ω, , 1, 8, , Ω, , Ω, , 1, 12, , Ω, , (c), , Figure 2.40, For Example 2.11: (a) original circuit,, (b) its equivalent circuit, (c) same circuit as, in (a) but resistors are expressed in ohms., , Practice Problem 2.11, , Calculate Geq in the circuit of Fig. 2.41., Answer: 4 S., , 8S, , 4S, , Geq, 2S, , 12 S, , 6S, , Figure 2.41, For Practice Prob. 2.11., , Example 2.12, , Find io and vo in the circuit shown in Fig. 2.42(a). Calculate the power, dissipated in the 3- resistor., Solution:, The 6- and 3- resistors are in parallel, so their combined resistance is, 63, , 63, 2, 63, , Thus our circuit reduces to that shown in Fig. 2.42(b). Notice that vo is, not affected by the combination of the resistors because the resistors are, in parallel and therefore have the same voltage vo. From Fig. 2.42(b),, we can obtain vo in two ways. One way is to apply Ohm’s law to get, i, , 12, 2A, 42
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 51, , 2.6, , Parallel Resistors and Current Division, , and hence, vo 2i 2 2 4 V. Another way is to apply voltage, division, since the 12 V in Fig. 2.42(b) is divided between the 4- and, 2- resistors. Hence,, 2, vo , (12 V) 4 V, 24, , 51, i, , 4Ω, , 12 V +, −, , 1, , io , , 3Ω, , b, (a), i, , 4, A, 3, , 4Ω, , 4, po vo io 4 a b 5.333 W, 3, , Find v1 and v2 in the circuit shown in Fig. 2.43. Also calculate i1 and, i2 and the power dissipated in the 12- and 40- resistors., , 2Ω, , b, , 6, 2, 4, i (2 A) A, 63, 3, 3, , The power dissipated in the 3- resistor is, , a, +, vo, −, , 12 V +, −, , Another approach is to apply current division to the circuit in Fig. 2.42(a), now that we know i, by writing, io , , +, vo, −, , 6Ω, , Similarly, io can be obtained in two ways. One approach is to apply, Ohm’s law to the 3- resistor in Fig. 2.42(a) now that we know vo; thus,, vo 3io 4, , io, , a, , (b), , Figure 2.42, For Example 2.12: (a) original circuit,, (b) its equivalent circuit., , Practice Problem 2.12, i1, , Answer: v1 5 V, i1 416.7 mA, p1 2.083 W, v2 10 V, i2 , 250 mA, p2 2.5 W., , 12 Ω, + v1 −, 6Ω, i2, , 15 V, , +, −, , 10 Ω, , +, v2, −, , 40 Ω, , Figure 2.43, For Practice Prob. 2.12., , For the circuit shown in Fig. 2.44(a), determine: (a) the voltage vo,, (b) the power supplied by the current source, (c) the power absorbed, by each resistor., Solution:, (a) The 6-k and 12-k resistors are in series so that their combined, value is 6 12 18 k. Thus the circuit in Fig. 2.44(a) reduces to, that shown in Fig. 2.44(b). We now apply the current division technique, to find i1 and i2., 18,000, (30 mA) 20 mA, 9,000 18,000, 9,000, (30 mA) 10 mA, i2 , 9,000 18,000, i1 , , Example 2.13
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 52, , Chapter 2, , 52, , Notice that the voltage across the 9-k and 18-k resistors is the same,, and vo 9,000i1 18,000i2 180 V, as expected., (b) Power supplied by the source is, , 6 kΩ, , 30 mA, , +, vo, −, , 12 kΩ, , 9 kΩ, , po voio 180(30) mW 5.4 W, (c) Power absorbed by the 12-k resistor is, , (a), , p iv i2 (i2 R) i 22 R (10 103)2 (12,000) 1.2 W, , i2, , io, , 30 mA, , Basic Laws, , +, vo, −, , Power absorbed by the 6-k resistor is, , i1, 9 kΩ, , p i 22 R (10 103)2 (6,000) 0.6 W, , 18 kΩ, , Power absorbed by the 9-k resistor is, , (b), , p, , Figure 2.44, For Example 2.13: (a) original circuit,, (b) its equivalent circuit., , v2o, (180)2, , 3.6 W, R, 9,000, , or, p voi1 180(20) mW 3.6 W, Notice that the power supplied (5.4 W) equals the power absorbed, (1.2 0.6 3.6 5.4 W). This is one way of checking results., , Practice Problem 2.13, , For the circuit shown in Fig. 2.45, find: (a) v1 and v2, (b) the power, dissipated in the 3-k and 20-k resistors, and (c) the power supplied, by the current source., 1 kΩ, , 3 kΩ, , +, v1, −, , 10 mA, , 5 kΩ, , +, v2, −, , 20 kΩ, , Figure 2.45, For Practice Prob. 2.13., , Answer: (a) 15 V, 20 V, (b) 75 mW, 20 mW, (c) 200 mW., R1, , R2, , R3, , 2.7, , R4, , vs +, −, R5, , Figure 2.46, The bridge network., , R6, , Wye-Delta Transformations, , Situations often arise in circuit analysis when the resistors are neither in, parallel nor in series. For example, consider the bridge circuit in Fig. 2.46., How do we combine resistors R1 through R6 when the resistors are neither, in series nor in parallel? Many circuits of the type shown in Fig. 2.46, can be simplified by using three-terminal equivalent networks. These are
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 53, , 2.7, , Wye-Delta Transformations, , the wye (Y) or tee (T) network shown in Fig. 2.47 and the delta ( ¢ ) or, pi ( ß ) network shown in Fig. 2.48. These networks occur by themselves, or as part of a larger network. They are used in three-phase networks,, electrical filters, and matching networks. Our main interest here is in how, to identify them when they occur as part of a network and how to apply, wye-delta transformation in the analysis of that network., , 53, Rc, 3, , 1, Rb, , Ra, , 2, , 4, (a), Rc, , 3, , 1, R1, , R1, , R2, , R2, 3, , 1, , Rb, , R3, , R3, 2, , 2, , 4, , 3, , 1, , 4, , 2, , (b), , (a), , Figure 2.48, , Two forms of the same network: (a) Y, (b) T., , Two forms of the same network: (a) ¢ ,, (b) ß ., , Delta to Wye Conversion, Suppose it is more convenient to work with a wye network in a place, where the circuit contains a delta configuration. We superimpose a wye, network on the existing delta network and find the equivalent resistances in the wye network. To obtain the equivalent resistances in the, wye network, we compare the two networks and make sure that the, resistance between each pair of nodes in the ¢ (or ß ) network is the, same as the resistance between the same pair of nodes in the Y (or T), network. For terminals 1 and 2 in Figs. 2.47 and 2.48, for example,, (2.46), , Setting R12(Y) R12 (¢) gives, , R12 R1 R3 , , Rb (Ra Rc), Ra Rb Rc, , (2.47a), , R13 R1 R2 , , Rc (Ra Rb), Ra Rb Rc, , (2.47b), , R34 R2 R3 , , Ra (Rb Rc), Ra Rb Rc, , (2.47c), , Similarly,, , Subtracting Eq. (2.47c) from Eq. (2.47a), we get, R1 R2 , , Rc (Rb Ra), Ra Rb Rc, , (2.48), , Adding Eqs. (2.47b) and (2.48) gives, R1 , , Rb Rc, Ra Rb Rc, , 4, (b), , Figure 2.47, , R12 (Y) R1 R3, R12 (¢) Rb 7 (Ra Rc), , Ra, , (2.49)
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 54, , Chapter 2, , 54, , Basic Laws, , and subtracting Eq. (2.48) from Eq. (2.47b) yields, , R2 , , Rc Ra, Ra Rb Rc, , (2.50), , Subtracting Eq. (2.49) from Eq. (2.47a), we obtain, , R3 , , Rc, a, , b, , Ra Rb, Ra Rb Rc, , (2.51), , We do not need to memorize Eqs. (2.49) to (2.51). To transform a ¢ network to Y, we create an extra node n as shown in Fig. 2.49 and follow, this conversion rule:, , R2, , R1, , Each resistor in the Y network is the product of the resistors in the two, adjacent ¢ branches, divided by the sum of the three ¢ resistors., , n, Rb, , Ra, , One can follow this rule and obtain Eqs. (2.49) to (2.51) from Fig. 2.49., , R3, , Wye to Delta Conversion, c, , Figure 2.49, Superposition of Y and ¢ networks as an, aid in transforming one to the other., , To obtain the conversion formulas for transforming a wye network to, an equivalent delta network, we note from Eqs. (2.49) to (2.51) that, R1 R2 R2 R3 R3 R1 , , Ra Rb Rc (Ra Rb Rc), (Ra Rb Rc)2, , Ra Rb Rc, , Ra Rb Rc, , (2.52), , Dividing Eq. (2.52) by each of Eqs. (2.49) to (2.51) leads to the following equations:, Ra , , R1 R2 R2 R3 R3 R1, R1, , (2.53), , Rb , , R1 R2 R2 R3 R3 R1, R2, , (2.54), , Rc , , R1 R2 R2 R3 R3 R1, R3, , (2.55), , From Eqs. (2.53) to (2.55) and Fig. 2.49, the conversion rule for Y to, ¢ is as follows:, Each resistor in the ¢ network is the sum of all possible products of Y, resistors taken two at a time, divided by the opposite Y resistor.
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 55, , 2.7, , Wye-Delta Transformations, , 55, , The Y and ¢ networks are said to be balanced when, R1 R2 R3 RY,, , Ra Rb Rc R¢, , (2.56), , Under these conditions, conversion formulas become, , RY , , R¢, 3, , R¢ 3RY, , or, , (2.57), , One may wonder why RY is less than R¢. Well, we notice that the Yconnection is like a “series” connection while the ¢ -connection is like, a “parallel” connection., Note that in making the transformation, we do not take anything, out of the circuit or put in anything new. We are merely substituting, different but mathematically equivalent three-terminal network patterns, to create a circuit in which resistors are either in series or in parallel,, allowing us to calculate Req if necessary., , Example 2.14, , Convert the ¢ network in Fig. 2.50(a) to an equivalent Y network., , Rc, , a, , b, , a, , b, , 25 Ω, 5Ω, , 7.5 Ω, R2, , R1, Rb, , 10 Ω, , 15 Ω, , Ra, R3, , c, , (a), , 3Ω, , c, , (b), , Figure 2.50, For Example 2.14: (a) original ¢ network, (b) Y equivalent network., , Solution:, Using Eqs. (2.49) to (2.51), we obtain, Rb Rc, 10 25, 250, , , 5, Ra Rb Rc, 15 10 25, 50, Rc Ra, 25 15, R2 , , 7.5 , Ra Rb Rc, 50, Ra Rb, 15 10, R3 , , 3, Ra Rb Rc, 50, R1 , , The equivalent Y network is shown in Fig. 2.50(b).
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 56, , Chapter 2, , 56, , Practice Problem 2.14, R1, , R2, , a, , b, 10 Ω, , Basic Laws, , Transform the wye network in Fig. 2.51 to a delta network., Answer: Ra 140 , Rb 70 , Rc 35 ., , 20 Ω, 40 Ω, , R3, , c, , Figure 2.51, For Practice Prob. 2.14., , Example 2.15, i, , Obtain the equivalent resistance Rab for the circuit in Fig. 2.52 and use, it to find current i., , 10 Ω, , 12.5 Ω, 120 V +, −, , Solution:, , a, , a, , c, , 5Ω, , n, 20 Ω, , 15 Ω, , b, , Figure 2.52, For Example 2.15., , b, , 30 Ω, , 1. Define. The problem is clearly defined. Please note, this part, normally will deservedly take much more time., 2. Present. Clearly, when we remove the voltage source, we end, up with a purely resistive circuit. Since it is composed of deltas, and wyes, we have a more complex process of combining the, elements together. We can use wye-delta transformations as one, approach to find a solution. It is useful to locate the wyes (there, are two of them, one at n and the other at c) and the deltas, (there are three: can, abn, cnb)., 3. Alternative. There are different approaches that can be used to, solve this problem. Since the focus of Sec. 2.7 is the wye-delta, transformation, this should be the technique to use. Another, approach would be to solve for the equivalent resistance by, injecting one amp into the circuit and finding the voltage, between a and b; we will learn about this approach in Chap. 4., The approach we can apply here as a check would be to use, a wye-delta transformation as the first solution to the problem., Later we can check the solution by starting with a delta-wye, transformation., 4. Attempt. In this circuit, there are two Y networks and three ¢, networks. Transforming just one of these will simplify the circuit., If we convert the Y network comprising the 5-, 10-, and, 20- resistors, we may select, R1 10 ,, , R2 20 ,, , R3 5 , , Thus from Eqs. (2.53) to (2.55) we have, R1 R2 R2 R3 R3 R1, 10 20 20 5 5 10, , R1, 10, 350, , 35 , 10, , Ra , , Rb , , R1 R2 R2 R3 R3 R1, 350, 17.5 , , R2, 20, , Rc , , R1 R2 R2 R3 R3 R1, 350, , 70 , R3, 5
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 57, , 2.7, , Wye-Delta Transformations, , 57, a, 4.545 Ω, , a, d, 12.5 Ω, , 17.5 Ω, , 2.273 Ω, , a, 70 Ω, , 30 Ω, , c, , 7.292 Ω, 21 Ω, , 35 Ω, , 15 Ω, , (b), , Figure 2.53, Equivalent circuits to Fig. 2.52, with the voltage source removed., , With the Y converted to ¢, the equivalent circuit (with the, voltage source removed for now) is shown in Fig. 2.53(a)., Combining the three pairs of resistors in parallel, we obtain, 70 30 , , 70 30, 21 , 70 30, , 12.5 17.5, 7.292 , 12.5 17.5, 15 35, 15 35 , 10.5 , 15 35, , 12.5 17.5 , , so that the equivalent circuit is shown in Fig. 2.53(b). Hence, we, find, Rab (7.292 10.5) 21 , , 17.792 21, 9.632 , 17.792 21, , Then, i, , 20 Ω, , b, , b, (a), , n, , 15 Ω, , 10.5 Ω, , b, , 1.8182 Ω, , vs, 120, , 12.458 A, Rab, 9.632, , We observe that we have successfully solved the problem., Now we must evaluate the solution., 5. Evaluate. Now we must determine if the answer is correct and, then evaluate the final solution., It is relatively easy to check the answer; we do this by, solving the problem starting with a delta-wye transformation. Let, us transform the delta, can, into a wye., Let Rc 10 , Ra 5 , and Rn 12.5 . This will lead, to (let d represent the middle of the wye):, Rad , , Rc Rn, 10 12.5, , 4.545 , Ra Rc Rn, 5 10 12.5, , Rcd , , Ra Rn, 5 12.5, , 2.273 , 27.5, 27.5, , Rnd , , Ra Rc, 5 10, , 1.8182 , 27.5, 27.5, , (c), , 30 Ω
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 58, , Chapter 2, , 58, , Basic Laws, , This now leads to the circuit shown in Figure 2.53(c). Looking, at the resistance between d and b, we have two series, combination in parallel, giving us, Rdb , , (2.273 15)(1.8182 20), 376.9, , 9.642 , 2.273 15 1.8182 20, 39.09, , This is in series with the 4.545- resistor, both of which are in, parallel with the 30- resistor. This then gives us the equivalent, resistance of the circuit., Rab , , (9.642 4.545)30, 425.6, , 9.631 , 9.642 4.545 30, 44.19, , This now leads to, i, , vs, 120, , 12.46 A, Rab, 9.631, , We note that using two variations on the wye-delta transformation, leads to the same results. This represents a very good check., 6. Satisfactory? Since we have found the desired answer by, determining the equivalent resistance of the circuit first and the, answer checks, then we clearly have a satisfactory solution. This, represents what can be presented to the individual assigning the, problem., , Practice Problem 2.15, i, , a, , 100 V, , Answer: 40 , 2.5 A., , 13 Ω, 24 Ω, , +, −, 30 Ω, , For the bridge network in Fig. 2.54, find Rab and i., , 20 Ω, , 10 Ω, , 2.8, 50 Ω, , b, , Figure 2.54, For Practice Prob. 2.15., , Applications, , Resistors are often used to model devices that convert electrical energy, into heat or other forms of energy. Such devices include conducting, wire, lightbulbs, electric heaters, stoves, ovens, and loudspeakers. In, this section, we will consider two real-life problems that apply the concepts developed in this chapter: electrical lighting systems and design, of dc meters., , 2.8.1. Lighting Systems, So far, we have assumed that connecting wires are perfect conductors (i.e.,, conductors of zero resistance). In real, physical systems, however, the resistance of the connecting wire may be, appreciably large, and the modeling, of the system must include that, resistance., , Lighting systems, such as in a house or on a Christmas tree, often consist of N lamps connected either in parallel or in series, as shown in, Fig. 2.55. Each lamp is modeled as a resistor. Assuming that all the lamps, are identical and Vo is the power-line voltage, the voltage across each, lamp is Vo for the parallel connection and VoN for the series connection. The series connection is easy to manufacture but is seldom used, in practice, for at least two reasons. First, it is less reliable; when a lamp, fails, all the lamps go out. Second, it is harder to maintain; when a lamp, is bad, one must test all the lamps one by one to detect the faulty one.
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ale29559_ch02.qxd, , 07/14/2008, , 02:40 PM, , Page 59, , 2.8, , Applications, , 59, , Historical, Thomas Alva Edison (1847–1931) was perhaps the greatest, American inventor. He patented 1093 inventions, including such, history-making inventions as the incandescent electric bulb, the phonograph, and the first commercial motion pictures., Born in Milan, Ohio, the youngest of seven children, Edison received, only three months of formal education because he hated school. He was, home-schooled by his mother and quickly began to read on his own. In, 1868, Edison read one of Faraday’s books and found his calling. He, moved to Menlo Park, New Jersey, in 1876, where he managed a wellstaffed research laboratory. Most of his inventions came out of this, laboratory. His laboratory served as a model for modern research organizations. Because of his diverse interests and the overwhelming number, of his inventions and patents, Edison began to establish manufacturing, companies for making the devices he invented. He designed the first electric power station to supply electric light. Formal electrical engineering, education began in the mid-1880s with Edison as a role model and leader., , Library of Congress, , 1, 2, , +, Vo, −, +, Vo, −, Power, plug, , 1, , 2, , 3, , N, , 3, , N, Lamp, , (a), , (b), , Figure 2.55, (a) Parallel connection of lightbulbs, (b) series connection of lightbulbs., , Three lightbulbs are connected to a 9-V battery as shown in Fig. 2.56(a)., Calculate: (a) the total current supplied by the battery, (b) the current, through each bulb, (c) the resistance of each bulb., I, , 9V, , 15 W, 20 W, , 9V, 10 W, , (a), , I1, I2, , +, V2, −, , R2, , +, V3, −, , R3, , +, V1, −, , R1, , (b), , Figure 2.56, (a) Lighting system with three bulbs, (b) resistive circuit equivalent model., , Example 2.16
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 60, , Chapter 2, , 60, , Basic Laws, , Solution:, (a) The total power supplied by the battery is equal to the total power, absorbed by the bulbs; that is,, p 15 10 20 45 W, Since p V I, then the total current supplied by the battery is, I, , p, 45, , 5A, V, 9, , (b) The bulbs can be modeled as resistors as shown in Fig. 2.56(b)., Since R1 (20-W bulb) is in parallel with the battery as well as the series, combination of R2 and R3,, V1 V2 V3 9 V, The current through R1 is, I1 , , p1, 20, , 2.222 A, V1, 9, , By KCL, the current through the series combination of R2 and R3 is, I2 I I1 5 2.222 2.778 A, (c) Since p I 2R,, R1 , R2 , R3 , , Practice Problem 2.16, , p1, I 12, p2, I 22, p3, I 32, , , , 20, 4.05 , 2.222 2, , , , 15, 1.945 , 2.777 2, , , , 10, 1.297 , 2.777 2, , Refer to Fig. 2.55 and assume there are 10 lightbulbs that can be connected in parallel and 10 lightbulbs that can be connected in series,, each with a power rating of 40 W. If the voltage at the plug is 110 V, for the parallel and series connections, calculate the current through, each bulb for both cases., Answer: 0.364 A (parallel), 3.64 A (series)., , 2.8.2 Design of DC Meters, a, Max, Vin, , b, , +, −, , +, Vout, Min −, c, , By their nature, resistors are used to control the flow of current. We, take advantage of this property in several applications, such as in a, potentiometer (Fig. 2.57). The word potentiometer, derived from the, words potential and meter, implies that potential can be metered out., The potentiometer (or pot for short) is a three-terminal device that operates on the principle of voltage division. It is essentially an adjustable, voltage divider. As a voltage regulator, it is used as a volume or level, control on radios, TVs, and other devices. In Fig. 2.57,, , Figure 2.57, The potentiometer controlling potential, levels., , Vout Vbc , , Rbc, Vin, Rac, , (2.58)
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 61, , 2.8, , Applications, , where Rac Rab Rbc. Thus, Vout decreases or increases as the sliding, contact of the pot moves toward c or a, respectively., Another application where resistors are used to control current flow, is in the analog dc meters—the ammeter, voltmeter, and ohmmeter,, which measure current, voltage, and resistance, respectively. Each of, these meters employs the d’Arsonval meter movement, shown in, Fig. 2.58. The movement consists essentially of a movable iron-core coil, mounted on a pivot between the poles of a permanent magnet. When, current flows through the coil, it creates a torque which causes the pointer, to deflect. The amount of current through the coil determines the deflection of the pointer, which is registered on a scale attached to the meter, movement. For example, if the meter movement is rated 1 mA, 50 , it, would take 1 mA to cause a full-scale deflection of the meter movement., By introducing additional circuitry to the d’Arsonval meter movement,, an ammeter, voltmeter, or ohmmeter can be constructed., Consider Fig. 2.59, where an analog voltmeter and ammeter are, connected to an element. The voltmeter measures the voltage across a, load and is therefore connected in parallel with the element. As shown, , 61, , An instrument capable of measuring, voltage, current, and resistance is, called a multimeter or a volt-ohm, meter (VOM)., , A load is a component that is receiving, energy (an energy sink), as opposed, to a generator supplying energy (an, energy source). More about loading, will be discussed in Section 4.9.1., Ammeter, , scale, , I, , A, spring, , +, Voltmeter V V, −, , pointer, , Element, , S, , Figure 2.59, , spring, , Connection of a voltmeter and an ammeter to an element., , permanent magnet, , N, , rotating coil, stationary iron core, , Figure 2.58, A d’Arsonval meter movement., , in Fig. 2.60(a), the voltmeter consists of a d’Arsonval movement in, series with a resistor whose resistance Rm is deliberately made very, large (theoretically, infinite), to minimize the current drawn from the, circuit. To extend the range of voltage that the meter can measure,, series multiplier resistors are often connected with the voltmeters, as, shown in Fig. 2.60(b). The multiple-range voltmeter in Fig. 2.60(b) can, measure voltage from 0 to 1 V, 0 to 10 V, or 0 to 100 V, depending on, whether the switch is connected to R1, R2, or R3, respectively., Let us calculate the multiplier resistor Rn for the single-range voltmeter in Fig. 2.60(a), or Rn R1, R2, or R3 for the multiple-range, voltmeter in Fig. 2.60(b). We need to determine the value of Rn to be, connected in series with the internal resistance Rm of the voltmeter. In, any design, we consider the worst-case condition. In this case, the, worst case occurs when the full-scale current Ifs Im flows through, the meter. This should also correspond to the maximum voltage reading or the full-scale voltage Vfs. Since the multiplier resistance Rn is in, series with the internal resistance Rm,, Vfs I fs (Rn Rm), , (2.59)
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 62, , Chapter 2, , 62, , Basic Laws, , Meter, , Multiplier, Rn, +, Probes, , Im, , Rm, , V, −, , (a), R1, 1V, R2, , 10 V, , +, Probes V, −, , Meter, Switch, , 100 V, , R3, , Im, , Rm, , (b), , Figure 2.60, Voltmeters: (a) single-range type, (b) multiple-range type., , From this, we obtain, , Rn, , In, , Rn , , Meter, Im, Rm, I, Probes, (a), R1, 10 mA, R2, , 100 mA, , Switch, 1A, , R3, Meter, , (2.60), , Similarly, the ammeter measures the current through the load and, is connected in series with it. As shown in Fig. 2.61(a), the ammeter, consists of a d’Arsonval movement in parallel with a resistor whose, resistance Rm is deliberately made very small (theoretically, zero) to, minimize the voltage drop across it. To allow multiple ranges, shunt, resistors are often connected in parallel with Rm as shown in, Fig. 2.61(b). The shunt resistors allow the meter to measure in the, range 0–10 mA, 0–100 mA, or 0–1 A, depending on whether the switch, is connected to R1, R2, or R3, respectively., Now our objective is to obtain the multiplier shunt Rn for the singlerange ammeter in Fig. 2.61(a), or Rn R1, R2, or R3 for the multiplerange ammeter in Fig. 2.61(b). We notice that Rm and Rn are in parallel, and that at full-scale reading I Ifs Im In, where In is the current, through the shunt resistor Rn. Applying the current division principle, yields, , Im, I, , Vfs, Rm, Ifs, , Rm, , Im , , Rn, Ifs, Rn Rm, , Rn , , Im, Rm, Ifs Im, , or, Probes, (b), , Figure 2.61, Ammeters: (a) single-range type,, (b) multiple-range type., , (2.61), , The resistance Rx of a linear resistor can be measured in two ways., An indirect way is to measure the current I that flows through it by
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ale29559_ch02.qxd, , 07/14/2008, , 02:40 PM, , Page 63, , 2.8, , Applications, , 63, , connecting an ammeter in series with it and the voltage V across it by, connecting a voltmeter in parallel with it, as shown in Fig. 2.62(a)., Then, Rx , , A, I, , V, I, , (2.62), , Rx, , The direct method of measuring resistance is to use an ohmmeter. An, ohmmeter consists basically of a d’Arsonval movement, a variable, resistor or potentiometer, and a battery, as shown in Fig. 2.62(b)., Applying KVL to the circuit in Fig. 2.62(b) gives, , +, V, −, , V, , (a), Ohmmeter, , E (R Rm Rx) Im, Im, , or, Rx , , R, , Rm, , E, (R Rm), Im, , (2.63), , Rx, , E, , The resistor R is selected such that the meter gives a full-scale deflection; that is, Im Ifs when Rx 0. This implies that, E (R Rm) Ifs, , (2.64), , Substituting Eq. (2.64) into Eq. (2.63) leads to, Rx a, , Ifs, 1b (R Rm), Im, , (2.65), , (b), , Figure 2.62, Two ways of measuring resistance:, (a) using an ammeter and a voltmeter,, (b) using an ohmmeter., , As mentioned, the types of meters we have discussed are known, as analog meters and are based on the d’Arsonval meter movement., Another type of meter, called a digital meter, is based on active circuit, elements such as op amps. For example, a digital multimeter displays, measurements of dc or ac voltage, current, and resistance as discrete, numbers, instead of using a pointer deflection on a continuous scale as, in an analog multimeter. Digital meters are what you would most likely, use in a modern lab. However, the design of digital meters is beyond, the scope of this book., , Historical, Samuel F. B. Morse (1791–1872), an American painter, invented, the telegraph, the first practical, commercialized application of, electricity., Morse was born in Charlestown, Massachusetts and studied at Yale, and the Royal Academy of Arts in London to become an artist. In the, 1830s, he became intrigued with developing a telegraph. He had a, working model by 1836 and applied for a patent in 1838. The U.S., Senate appropriated funds for Morse to construct a telegraph line, between Baltimore and Washington, D.C. On May 24, 1844, he sent, the famous first message: “What hath God wrought!” Morse also developed a code of dots and dashes for letters and numbers, for sending, messages on the telegraph. The development of the telegraph led to the, invention of the telephone., , Library of Congress
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 64, , Chapter 2, , 64, , Example 2.17, , Basic Laws, , Following the voltmeter setup of Fig. 2.60, design a voltmeter for the, following multiple ranges:, (a) 0–1 V, (b) 0–5 V, (c) 0–50 V, (d) 0–100 V, Assume that the internal resistance Rm 2 k and the full-scale current Ifs 100 mA., Solution:, We apply Eq. (2.60) and assume that R1, R2, R3, and R4 correspond, with ranges 0–1 V, 0–5 V, 0–50 V, and 0–100 V, respectively., (a) For range 0–1 V,, R1 , , 1, 2000 10,000 2000 8 k, 100 106, , (b) For range 0–5 V,, R2 , , 5, 2000 50,000 2000 48 k, 100 106, , (c) For range 0–50 V,, R3 , , 50, 2000 500,000 2000 498 k, 100 106, , (d) For range 0–100 V,, R4 , , 100 V, 2000 1,000,000 2000 998 k, 100 106, , Note that the ratio of the total resistance (Rn Rm) to the full-scale, voltage Vfs is constant and equal to 1Ifs for the four ranges. This ratio, (given in ohms per volt, or /V) is known as the sensitivity of the, voltmeter. The larger the sensitivity, the better the voltmeter., , Practice Problem 2.17, , Following the ammeter setup of Fig. 2.61, design an ammeter for the, following multiple ranges:, (a) 0–1 A, (b) 0–100 mA, (c) 0–10 mA, Take the full-scale meter current as Im 1 mA and the internal resistance of the ammeter as Rm 50 ., Answer: Shunt resistors: 0.05 , 0.505 , 5.556 ., , 2.9, , Summary, , 1. A resistor is a passive element in which the voltage v across it is, directly proportional to the current i through it. That is, a resistor, is a device that obeys Ohm’s law,, v iR, where R is the resistance of the resistor.
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 65, , 2.9, , Summary, , 2. A short circuit is a resistor (a perfectly conducting wire) with zero, resistance (R 0). An open circuit is a resistor with infinite resistance (R )., 3. The conductance G of a resistor is the reciprocal of its resistance:, G, , 1, R, , 4. A branch is a single two-terminal element in an electric circuit. A, node is the point of connection between two or more branches. A, loop is a closed path in a circuit. The number of branches b, the, number of nodes n, and the number of independent loops l in a, network are related as, bln1, 5. Kirchhoff’s current law (KCL) states that the currents at any node, algebraically sum to zero. In other words, the sum of the currents, entering a node equals the sum of currents leaving the node., 6. Kirchhoff’s voltage law (KVL) states that the voltages around a, closed path algebraically sum to zero. In other words, the sum of, voltage rises equals the sum of voltage drops., 7. Two elements are in series when they are connected sequentially,, end to end. When elements are in series, the same current flows, through them (i1 i2). They are in parallel if they are connected, to the same two nodes. Elements in parallel always have the same, voltage across them (v1 v2)., 8. When two resistors R1 (1G1) and R2 (1G2) are in series, their, equivalent resistance Req and equivalent conductance Geq are, Req R1 R2,, , Geq , , G1G2, G1 G2, , 9. When two resistors R1 (1G1) and R2 (1G2) are in parallel,, their equivalent resistance Req and equivalent conductance Geq are, Req , , R1R2, ,, R1 R2, , Geq G1 G2, , 10. The voltage division principle for two resistors in series is, v1 , , R1, v,, R1 R2, , v2 , , R2, v, R1 R2, , 11. The current division principle for two resistors in parallel is, i1 , , R2, i,, R1 R2, , i2 , , R1, i, R1 R2, , 12. The formulas for a delta-to-wye transformation are, R1 , , Rb Rc, ,, Ra Rb Rc, R3 , , R2 , , Rc Ra, Ra Rb Rc, , Ra Rb, Ra Rb Rc, , 65
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 66, , Chapter 2, , 66, , Basic Laws, , 13. The formulas for a wye-to-delta transformation are, Ra , , R1 R2 R2 R3 R3 R1, ,, R1, Rc , , Rb , , R1 R2 R2 R3 R3 R1, R2, , R1 R2 R2 R3 R3 R1, R3, , 14. The basic laws covered in this chapter can be applied to the problems of electrical lighting and design of dc meters., , Review Questions, 2.1, , 2.2, , 2.3, , 2.4, , 2.5, , The reciprocal of resistance is:, , The current Io of Fig. 2.64 is:, (a) 4 A, , (a) voltage, , (b) current, , (c) conductance, , (d) coulombs, , (b) 2 A, , (c) 4 A, , (d) 16 A, , An electric heater draws 10 A from a 120-V line. The, resistance of the heater is:, (a) 1200 , , (b) 120 , , (c) 12 , , (d) 1.2 , , 10 A, , The voltage drop across a 1.5-kW toaster that draws, 12 A of current is:, (a) 18 kV, , (b) 125 V, , (c) 120 V, , (d) 10.42 V, , 4A, , 2A, , The maximum current that a 2W, 80 k resistor can, safely conduct is:, (a) 160 kA, , (b) 40 kA, , (c) 5 mA, , (d) 25 mA, , (b) 17, , (c) 5, , Io, , Figure 2.64, For Review Question 2.7., , A network has 12 branches and 8 independent, loops. How many nodes are there in the, network?, (a) 19, , 2.6, , 2.7, , (d) 4, , 2.8, , In the circuit in Fig. 2.65, V is:, (a) 30 V, , (b) 14 V, , (c) 10 V, , (d) 6 V, , The current I in the circuit of Fig. 2.63 is:, (a) 0.8 A, , (b) 0.2 A, , (c) 0.2 A, , (d) 0.8 A, 10 V, +, −, 4Ω, , 3V +, −, , I, 12 V +, −, +, −, , +, −, , 5V, , 6Ω, , +, V, , Figure 2.63, , Figure 2.65, , For Review Question 2.6., , For Review Question 2.8., , −, , 8V
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 67, , Problems, , 2.9, , Which of the circuits in Fig. 2.66 will give you, Vab 7 V?, 5V, , 2.10 In the circuit of Fig. 2.67, a decrease in R3 leads to a, decrease of:, (a) current through R3, , 5V, , +−, , −+, , a, , 67, , (b) voltage across R3, , a, , (c) voltage across R1, 3V +, −, , (d) power dissipated in R2, , 3V +, −, +−, , (e) none of the above, +−, , b, , 1V, , 1V, , (a), , (b), , 5V, , 5V, , +−, , −+, , a, , b, , R1, , Vs, , +, −, , R2, , R3, , a, , Figure 2.67, 3V +, −, , For Review Question 2.10., , 3V +, −, −+, , −+, , b, , 1V, , b, , Answers: 2.1c, 2.2c, 2.3b, 2.4c, 2.5c, 2.6b, 2.7a, 2.8d,, 2.9d, 2.10b, d., , 1V, , (c), , (d), , Figure 2.66, For Review Question 2.9., , Problems, Section 2.2 Ohm’s Law, 2.1, , Design a problem, complete with a solution, to help, students to better understand Ohm’s Law. Use at, least two resistors and one voltage source. Hint, you, could use both resistors at once or one at a time, it is, up to you. Be creative., , 2.2, , Find the hot resistance of a lightbulb rated 60 W, 120 V., , 2.3, , A bar of silicon is 4 cm long with a circular cross section. If the resistance of the bar is 240 at room temperature, what is the cross-sectional radius of the bar?, , 2.4, , (a) Calculate current i in Fig. 2.68 when the switch is, in position 1., (b) Find the current when the switch is in position 2., 1, , 100 Ω, , Figure 2.69, For Prob. 2.5., , 2.6, , In the network graph shown in Fig. 2.70, determine, the number of branches and nodes., , 2, , i, +, −, , 15 V, , 150 Ω, , Figure 2.68, For Prob. 2.4., , Section 2.3 Nodes, Branches, and Loops, 2.5, , For the network graph in Fig. 2.69, find the number, of nodes, branches, and loops., , Figure 2.70, For Prob. 2.6.
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 69, , Problems, , 2.15 Calculate v and ix in the circuit of Fig. 2.79., , +, , 12 Ω, , −, , ix, , +−, , +, 2V, −, , +, −, , 2.19 From the circuit in Fig. 2.83, find I, the power, dissipated by the resistor, and the power absorbed by, each source., , 10 V, , + v −, 12 V, , 8V, , 69, , +, −, , 3ix, , I, , 20 V +, −, , 3Ω, , Figure 2.79, , +−, , For Prob. 2.15., , −4 V, , Figure 2.83, For Prob. 2.19., 2.16 Determine Vo in the circuit of Fig. 2.80., 2.20 Determine io in the circuit of Fig. 2.84., 2Ω, , 6Ω, +, 9V +, −, , io, +, −, , Vo, , 4Ω, , 3V, 36 V, , −, , Figure 2.80, , +, −, , +, −, , 5io, , Figure 2.84, , For Prob. 2.16., , For Prob. 2.20., , 2.21 Find Vx in the circuit of Fig. 2.85., , + v1 −, , 2 Vx, , 1Ω, , +, v2 −, +, , 24 V +, −, , +, v3, −, , +, −, , 10 V, , 15 V +, −, , 5Ω, , −+, 12 V, , Figure 2.85, , For Prob. 2.17., , For Prob. 2.21., , 2.18 Find I and Vab in the circuit of Fig. 2.82., , 10 V, +−, , 3Ω, , a, +, , +, −, , +, Vx, −, , 2Ω, , Figure 2.81, , 30 V, , −, , 2.17 Obtain v1 through v3 in the circuit of Fig. 2.81., , Vab, −, , 2.22 Find Vo in the circuit of Fig. 2.86 and the power, dissipated by the controlled source., , 5Ω, , 4Ω, I, + 8V, −, , + V −, o, 6Ω, , b, , Figure 2.82, , Figure 2.86, , For Prob. 2.18., , For Prob. 2.22., , 10 A, , 2Vo
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 70, , Chapter 2, , 70, , Basic Laws, , 2.23 In the circuit shown in Fig. 2.87, determine vx and, the power absorbed by the 12- resistor., 1Ω, , 1.2 Ω, , +v –, x, , 4Ω, +, , 4Ω, 8Ω, , 2Ω, , 6A, , 2.27 Calculate Vo in the circuit of Fig. 2.91., , Vo, , −, , 16 V +, −, , 12 Ω, , 6Ω, , 6Ω, , 3Ω, , Figure 2.91, For Prob. 2.27., , Figure 2.87, For Prob. 2.23., 2.24 For the circuit in Fig. 2.88, find VoVs in terms of, a, R1, R2, R3, and R4. If R1 R2 R3 R4, what, value of a will produce |Vo Vs | 10?, Io, , Vs, , 2.28 Design a problem, using Fig. 2.92, to help other, students better understand series and parallel, circuits., , R1, , +, −, , R1, , R2, , ␣Io, , R3, , R4, , +, Vo, −, , + v1 −, +, v2, −, , Vs +, −, , R2, , +, v3, −, , R3, , Figure 2.88, For Prob. 2.24., , Figure 2.92, For Prob. 2.28., , 2.25 For the network in Fig. 2.89, find the current,, voltage, and power associated with the 20-k, resistor., , 10 kΩ, , 5 mA, , +, Vo, −, , 0.01Vo, , 5 kΩ, , 2.29 All resistors in Fig. 2.93 are 1 each. Find Req., , 20 kΩ, Req, , Figure 2.89, For Prob. 2.25., , Figure 2.93, For Prob. 2.29., , Sections 2.5 and 2.6 Series and Parallel Resistors, 2.26 For the circuit in Fig. 2.90, io 2 A. Calculate ix, and the total power dissipated by the circuit., , 2.30 Find Req for the circuit of Fig. 2.94., , ix, , 6Ω, , 6Ω, , io, 2Ω, , 4Ω, , 8Ω, , Req, , 16 Ω, , Figure 2.90, , Figure 2.94, , For Prob. 2.26., , For Prob. 2.30., , 2Ω, , 2Ω
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 74, , Chapter 2, , 74, , Basic Laws, , 2.50 Design a problem to help other students better, understand wye-delta transformations using, Fig. 2.114., , R, , *2.53 Obtain the equivalent resistance Rab in each of the, circuits of Fig. 2.117. In (b), all resistors have a, value of 30 ., , 40 Ω, , 30 Ω, , R, 20 Ω, , R, 9 mA, , 10 Ω, , a, R, , R, , 60 Ω, , 80 Ω, , 50 Ω, , b, , Figure 2.114, , (a), , For Prob. 2.50., 2.51 Obtain the equivalent resistance at the terminals a-b, for each of the circuits in Fig. 2.115., , a, 30 Ω, , a, 20 Ω, , 10 Ω, , b, , 10 Ω, , 30 Ω, 10 Ω, , (b), 20 Ω, , Figure 2.117, For Prob. 2.53., , b, (a), , 2.54 Consider the circuit in Fig. 2.118. Find the, equivalent resistance at terminals: (a) a-b, (b) c-d., , 30 Ω, 25 Ω, , 10 Ω, , 20 Ω, , a, 5Ω, , 15 Ω, , 150 Ω, , 50 Ω, , a, , (b), , Figure 2.115, For Prob. 2.51., , b, , d, 150 Ω, , Figure 2.118, For Prob. 2.54., , *2.52 For the circuit shown in Fig. 2.116, find the, equivalent resistance. All resistors are 1 ., 2.55 Calculate Io in the circuit of Fig. 2.119., , Io, , 20 Ω, 24 V, , +, −, , 20 Ω, , For Prob. 2.52., , Figure 2.119, For Prob. 2.55., , 60 Ω, , 40 Ω, 10 Ω, , Req, , Figure 2.116, , * An asterisk indicates a challenging problem., , c, , 100 Ω, , 100 Ω, , b, , 60 Ω, , 50 Ω
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 75, , Problems, , 2.56 Determine V in the circuit of Fig. 2.120., , 100 V, , 15 Ω, +, V, −, , +, −, , 50 W, , 10 Ω, , 35 Ω, , 20 Ω, , 12 Ω, , Figure 2.123, For Prob. 2.59., 2.60 If the three bulbs of Prob. 2.59 are connected in, parallel to the 100-V battery, calculate the current, through each bulb., , For Prob. 2.56., , *2.57 Find Req and I in the circuit of Fig. 2.121., , I, , 4Ω, , 2Ω, , 1Ω, , 6Ω, 12 Ω, 8Ω, , +, −, , 2.61 As a design engineer, you are asked to design a, lighting system consisting of a 70-W power supply, and two lightbulbs as shown in Fig. 2.124. You must, select the two bulbs from the following three, available bulbs., R1 80 , cost $0.60 (standard size), R2 90 , cost $0.90 (standard size), R3 100 , cost $0.75 (nonstandard size), The system should be designed for minimum cost, such that lies within the range I 1.2 A 5 percent., , 2Ω, 4Ω, 3Ω, , 10 Ω, 5Ω, , I, +, 70 W, Power, Supply, , Rx, , Ry, , −, , Req, , Figure 2.124, , Figure 2.121, , For Prob. 2.61., , For Prob. 2.57., , Section 2.8 Applications, 2.58 The lightbulb in Fig. 2.122 is rated 120 V, 0.75 A., Calculate Vs to make the lightbulb operate at the, rated conditions., , 40 Ω, , Vs, , 40 W, , 100 V +, −, , Figure 2.120, , 20 V, , 30 W, , I, , 30 Ω, 16 Ω, , 75, , +, −, , Bulb, , 80 Ω, , 2.62 A three-wire system supplies two loads A and B as, shown in Fig. 2.125. Load A consists of a motor, drawing a current of 8 A, while load B is a PC, drawing 2 A. Assuming 10 h/day of use for 365 days, and 6 cents/kWh, calculate the annual energy cost of, the system., , +, 110 V –, , A, , 110 V +, –, , B, , Figure 2.122, For Prob. 2.58., , Figure 2.125, For Prob. 2.62., , 2.59 Three lightbulbs are connected in series to a 100-V, battery as shown in Fig. 2.123. Find the current I, through the bulbs., , 2.63 If an ammeter with an internal resistance of 100 , and a current capacity of 2 mA is to measure 5 A,, determine the value of the resistance needed.
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 76, , Chapter 2, , 76, , Basic Laws, , Calculate the power dissipated in the shunt, resistor., 2.64 The potentiometer (adjustable resistor) Rx in Fig. 2.126, is to be designed to adjust current ix from 1 A to, 10 A. Calculate the values of R and Rx to achieve this., ix, , 2.68 (a) Find the current I in the circuit of Fig. 2.128(a)., (b) An ammeter with an internal resistance of 1 is, inserted in the network to measure I¿ as shown in, Fig. 2.128(b). What is I¿?, (c) Calculate the percent error introduced by the, meter as, `, , R, , I I¿, ` 100%, I, , Rx, , 110 V +, −, , ix, I, , Figure 2.126, , 16 Ω, , For Prob. 2.64., 4V +, −, , 2.65 A d’Arsonval meter with an internal resistance of, 1 k requires 10 mA to produce full-scale deflection., Calculate the value of a series resistance needed to, measure 50 V of full scale., , 40 Ω, , 60 Ω, , (a), , 2.66 A 20-k/V voltmeter reads 10 V full scale., I' 16 Ω, , (a) What series resistance is required to make the, meter read 50 V full scale?, (b) What power will the series resistor dissipate, when the meter reads full scale?, , Ammeter, , 4V +, −, , 40 Ω, , 60 Ω, , 2.67 (a) Obtain the voltage Vo in the circuit of Fig. 2.127(a)., (b) Determine the voltage Vo¿ measured when a, voltmeter with 6-k internal resistance is, connected as shown in Fig. 2.127(b)., , (b), , Figure 2.128, For Prob. 2.68., , (c) The finite resistance of the meter introduces an, error into the measurement. Calculate the percent, error as, `, , Vo Vo¿, ` 100%, Vo, , (d) Find the percent error if the internal resistance, were 36 k., , (a) R2 1 k, , 1 kΩ, , 2 mA, , 5 kΩ, , 2.69 A voltmeter is used to measure Vo in the circuit in, Fig. 2.129. The voltmeter model consists of an ideal, voltmeter in parallel with a 100-k resistor. Let, Vs 40 V, Rs 10 k, and R1 20 k. Calculate, Vo with and without the voltmeter when, , 4 kΩ, , (b) R2 10 k, , (c) R2 100 k, , +, Vo, −, , (a), , Rs, , 1 kΩ, , 2 mA, , 5 kΩ, , 4 kΩ, , +, Vo, −, , R1, Voltmeter, , Vs, , +, −, R2, , (b), , Figure 2.127, , Figure 2.129, , For Prob. 2.67., , For Prob. 2.69., , +, Vo, −, , 100 kΩ, , V
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 77, , Problems, , 77, , 2.70 (a) Consider the Wheatstone bridge shown in, Fig. 2.130. Calculate va, vb, and vab., 20 Ω, , (b) Rework part (a) if the ground is placed at, a instead of o., , Ammeter, model, , A, 8 kΩ, 25 V +, –, , a, , R, Rx, , b, , 12 kΩ, , o, , I, , 15 kΩ, , 10 kΩ, , Figure 2.133, For Prob. 2.73., , Figure 2.130, For Prob. 2.70., 2.71 Figure 2.131 represents a model of a solar, photovoltaic panel. Given that Vs 30 V,, R1 20 , and iL 1 A, find RL., , 2.74 The circuit in Fig. 2.134 is to control the speed of a, motor such that the motor draws currents 5 A, 3 A,, and 1 A when the switch is at high, medium, and low, positions, respectively. The motor can be modeled as, a load resistance of 20 m. Determine the series, dropping resistances R1, R2, and R3., , R1, iL, Vs +, −, , Low, R1, 10-A, 0.01-Ω fuse, , RL, , Medium, , Figure 2.131, , High, , R2, , For Prob. 2.71., 6V, , 2.72 Find Vo in the two-way power divider circuit in, Fig. 2.132., , R3, Motor, , 1Ω, , Figure 2.134, For Prob. 2.74., , 1Ω, , 2Ω, , Vo, 10 V +, −, , 1Ω, , 1Ω, , 2.75 Find Rab in the four-way power divider circuit in, Fig. 2.135. Assume each element is 1 ., , 1Ω, 1, , 1, , Figure 2.132, , 1, , 1, , For Prob. 2.72., , 1, , 1, , 1, , 1, , 1, a, , 2.73 An ammeter model consists of an ideal ammeter, in series with a 20- resistor. It is connected, with a current source and an unknown resistor, Rx as shown in Fig. 2.133. The ammeter reading, is noted. When a potentiometer R is added and, adjusted until the ammeter reading drops to one, half its previous reading, then R 65 . What, is the value of Rx?, , 1, , 1, , 1, 1, , b, , Figure 2.135, For Prob. 2.75., , 1
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 78, , Chapter 2, , 78, , Basic Laws, , Comprehensive Problems, 2.76 Repeat Prob. 2.75 for the eight-way divider shown in, Fig. 2.136., 1, , 1, , 2.79 An electric pencil sharpener rated 240 mW, 6 V is, connected to a 9-V battery as shown in Fig. 2.138., Calculate the value of the series-dropping resistor Rx, needed to power the sharpener., , 1, , 1, , 1, , 1, , 1, , 1, , 1, , Rx, , Switch, , 9V, 1, , 1, , 1, 1, , 1, , 1, , 1, , Figure 2.138, , 1, a, , 1, , 1, , For Prob. 2.79., , 1, , 1, , 1, , 1, , 1, , 1, , 1, , 1, , 2.80 A loudspeaker is connected to an amplifier as shown, in Fig. 2.139. If a 10- loudspeaker draws the, maximum power of 12 W from the amplifier,, determine the maximum power a 4- loudspeaker, will draw., , 1, 1, , 1, , b, , Figure 2.136, Amplifier, , For Prob. 2.76., 2.77 Suppose your circuit laboratory has the following, standard commercially available resistors in large, quantities:, 1.8 , , 20 , , 300 , , 24 k, , (b) 311.8 , , (c) 40 k, , (d) 52.32 k, , For Prob. 2.80., , 56 k, , Using series and parallel combinations and a, minimum number of available resistors, how would, you obtain the following resistances for an electronic, circuit design?, (a) 5 , , Loudspeaker, , Figure 2.139, , 2.81 In a certain application, the circuit in Fig. 2.140, must be designed to meet these two criteria:, (a) VoVs 0.05, , (b) Req 40 k, , If the load resistor 5 k is fixed, find R1 and R2 to, meet the criteria., , 2.78 In the circuit in Fig. 2.137, the wiper divides the, potentiometer resistance between aR and (1 a)R,, 0 a 1. Find vovs., R1, R, +, vs, , +, −, , Vs, R, , vo, , +, −, , R2, , ␣R, −, , Req, , Figure 2.137, , Figure 2.140, , For Prob. 2.78., , For Prob. 2.81., , +, Vo, −, , 5 kΩ
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ale29559_ch02.qxd, , 07/09/2008, , 11:19 AM, , Page 79, , Comprehensive Problems, , 2.82 The pin diagram of a resistance array is shown in, Fig. 2.141. Find the equivalent resistance between, the following:, , 79, , 2.83 Two delicate devices are rated as shown in Fig., 2.142. Find the values of the resistors R1 and R2, needed to power the devices using a 24-V battery., , (a) 1 and 2, (b) 1 and 3, , 60-mA, 2-Ω fuse, , (c) 1 and 4, 4, , 3, 20 Ω, , R1, 20 Ω, , Figure 2.142, 80 Ω, 1, , Figure 2.141, For Prob. 2.82., , For Prob. 2.83., , 2, , Device 1, 9 V, 45 mW, , 40 Ω, , 10 Ω, , Device 2, , 24 V, R2, , 10 Ω, , 24 V, 480 mW
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ale29559_ch03.qxd, , 07/14/2008, , 02:42 PM, , Page 81, , c h a p t e r, , 3, , Methods of, Analysis, No great work is ever done in a hurry. To develop a great scientific, discovery, to print a great picture, to write an immortal poem, to, become a minister, or a famous general—to do anything great requires, time, patience, and perseverance. These things are done by degrees,, “little by little.”, —W. J. Wilmont Buxton, , Enhancing Your Career, Career in Electronics, One area of application for electric circuit analysis is electronics. The, term electronics was originally used to distinguish circuits of very low, current levels. This distinction no longer holds, as power semiconductor devices operate at high levels of current. Today, electronics is, regarded as the science of the motion of charges in a gas, vacuum, or, semiconductor. Modern electronics involves transistors and transistor, circuits. The earlier electronic circuits were assembled from components. Many electronic circuits are now produced as integrated circuits,, fabricated in a semiconductor substrate or chip., Electronic circuits find applications in many areas, such as automation, broadcasting, computers, and instrumentation. The range of devices, that use electronic circuits is enormous and is limited only by our imagination. Radio, television, computers, and stereo systems are but a few., An electrical engineer usually performs diverse functions and is likely, to use, design, or construct systems that incorporate some form of electronic circuits. Therefore, an understanding of the operation and analysis, of electronics is essential to the electrical engineer. Electronics has, become a specialty distinct from other disciplines within electrical engineering. Because the field of electronics is ever advancing, an electronics, engineer must update his/her knowledge from time to time. The best way, to do this is by being a member of a professional organization such as, the Institute of Electrical and Electronics Engineers (IEEE). With a membership of over 300,000, the IEEE is the largest professional organization, in the world. Members benefit immensely from the numerous magazines,, journals, transactions, and conference/symposium proceedings published, yearly by IEEE. You should consider becoming an IEEE member., , Troubleshooting an electronic circuit, board., © BrandX Pictures/Punchstock, , 81
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ale29559_ch03.qxd, , 07/08/2008, , 10:46 AM, , Page 82, , Chapter 3, , 82, , 3.1, , Methods of Analysis, , Introduction, , Having understood the fundamental laws of circuit theory (Ohm’s law, and Kirchhoff’s laws), we are now prepared to apply these laws to, develop two powerful techniques for circuit analysis: nodal analysis,, which is based on a systematic application of Kirchhoff’s current law, (KCL), and mesh analysis, which is based on a systematic application, of Kirchhoff’s voltage law (KVL). The two techniques are so important that this chapter should be regarded as the most important in the, book. Students are therefore encouraged to pay careful attention., With the two techniques to be developed in this chapter, we can analyze any linear circuit by obtaining a set of simultaneous equations that, are then solved to obtain the required values of current or voltage. One, method of solving simultaneous equations involves Cramer’s rule, which, allows us to calculate circuit variables as a quotient of determinants. The, examples in the chapter will illustrate this method; Appendix A also, briefly summarizes the essentials the reader needs to know for applying, Cramer’s rule. Another method of solving simultaneous equations is to, use MATLAB, a computer software discussed in Appendix E., Also in this chapter, we introduce the use of PSpice for Windows,, a circuit simulation computer software program that we will use, throughout the text. Finally, we apply the techniques learned in this, chapter to analyze transistor circuits., , 3.2, Nodal analysis is also known as the, node-voltage method., , Nodal Analysis, , Nodal analysis provides a general procedure for analyzing circuits, using node voltages as the circuit variables. Choosing node voltages, instead of element voltages as circuit variables is convenient and, reduces the number of equations one must solve simultaneously., To simplify matters, we shall assume in this section that circuits, do not contain voltage sources. Circuits that contain voltage sources, will be analyzed in the next section., In nodal analysis, we are interested in finding the node voltages., Given a circuit with n nodes without voltage sources, the nodal analysis of the circuit involves taking the following three steps., , Steps to Determine Node Voltages:, 1. Select a node as the reference node. Assign voltages v1,, v2, p , vn1 to the remaining n 1 nodes. The voltages are, referenced with respect to the reference node., 2. Apply KCL to each of the n 1 nonreference nodes. Use, Ohm’s law to express the branch currents in terms of node, voltages., 3. Solve the resulting simultaneous equations to obtain the, unknown node voltages., We shall now explain and apply these three steps., The first step in nodal analysis is selecting a node as the reference, or datum node. The reference node is commonly called the ground
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ale29559_ch03.qxd, , 07/08/2008, , 10:46 AM, , Page 83, , 3.2, , Nodal Analysis, , since it is assumed to have zero potential. A reference node is indicated, by any of the three symbols in Fig. 3.1. The type of ground in Fig. 3.1(c), is called a chassis ground and is used in devices where the case, enclosure, or chassis acts as a reference point for all circuits. When the, potential of the earth is used as reference, we use the earth ground in, Fig. 3.1(a) or (b). We shall always use the symbol in Fig. 3.1(b)., Once we have selected a reference node, we assign voltage designations to nonreference nodes. Consider, for example, the circuit in, Fig. 3.2(a). Node 0 is the reference node (v 0), while nodes 1 and, 2 are assigned voltages v1 and v2, respectively. Keep in mind that the, node voltages are defined with respect to the reference node. As illustrated in Fig. 3.2(a), each node voltage is the voltage rise from the reference node to the corresponding nonreference node or simply the, voltage of that node with respect to the reference node., As the second step, we apply KCL to each nonreference node in, the circuit. To avoid putting too much information on the same circuit,, the circuit in Fig. 3.2(a) is redrawn in Fig. 3.2(b), where we now add, i1, i2, and i3 as the currents through resistors R1, R2, and R3, respectively. At node 1, applying KCL gives, I1 I2 i1 i2, , 83, , The number of nonreference nodes is, equal to the number of independent, equations that we will derive., , (a), , (c), , (b), , Figure 3.1, Common symbols for indicating a, reference node, (a) common ground,, (b) ground, (c) chassis ground., , (3.1), , At node 2,, , I2, , I2 i2 i3, , (3.2), , We now apply Ohm’s law to express the unknown currents i1, i2, and, i3 in terms of node voltages. The key idea to bear in mind is that, since, resistance is a passive element, by the passive sign convention, current, must always flow from a higher potential to a lower potential., , I1, , +, v1, −, , 2, +, v2, −, , R1, , R3, , 0, , Current flows from a higher potential to a lower potential in a resistor., , We can express this principle as, i, , R2, , 1, , (a), , vhigher vlower, R, , I2, , (3.3), v1, , Note that this principle is in agreement with the way we defined resistance in Chapter 2 (see Fig. 2.1). With this in mind, we obtain from, Fig. 3.2(b),, v1 0, or, i1 G1v1, R1, v1 v2, i2 , or, i2 G2 (v1 v2), R2, v2 0, i3 , or, i3 G3v2, R3, , I1, , i2, , R2, , i2, , v2, , i1, , i3, , R1, , R3, , i1 , , (b), , (3.4), , Typical circuit for nodal analysis., , Substituting Eq. (3.4) in Eqs. (3.1) and (3.2) results, respectively, in, v1, v1 v2, , R1, R2, v1 v2, v2, , I2 , R2, R3, , I1 I2 , , Figure 3.2, , (3.5), (3.6)
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ale29559_ch03.qxd, , 07/08/2008, , 10:46 AM, , Page 84, , Chapter 3, , 84, , Methods of Analysis, , In terms of the conductances, Eqs. (3.5) and (3.6) become, I1 I2 G1v1 G2(v1 v2), I2 G2(v1 v2) G3v2, , (3.7), (3.8), , The third step in nodal analysis is to solve for the node voltages., If we apply KCL to n 1 nonreference nodes, we obtain n 1 simultaneous equations such as Eqs. (3.5) and (3.6) or (3.7) and (3.8). For, the circuit of Fig. 3.2, we solve Eqs. (3.5) and (3.6) or (3.7) and (3.8), to obtain the node voltages v1 and v2 using any standard method, such, as the substitution method, the elimination method, Cramer’s rule, or, matrix inversion. To use either of the last two methods, one must cast, the simultaneous equations in matrix form. For example, Eqs. (3.7) and, (3.8) can be cast in matrix form as, , Appendix A discusses how to use, Cramer’s rule., , c, , G1 G 2, G 2, , G 2, v1, I1 I2, d c d c, d, G 2 G 3 v2, I2, , (3.9), , which can be solved to get v1 and v2. Equation 3.9 will be generalized, in Section 3.6. The simultaneous equations may also be solved using, calculators or with software packages such as MATLAB, Mathcad,, Maple, and Quattro Pro., , Example 3.1, , Calculate the node voltages in the circuit shown in Fig. 3.3(a)., , 5A, , 4Ω, , 2, , 1, 2Ω, , 6Ω, , 10 A, , Solution:, Consider Fig. 3.3(b), where the circuit in Fig. 3.3(a) has been prepared, for nodal analysis. Notice how the currents are selected for the, application of KCL. Except for the branches with current sources, the, labeling of the currents is arbitrary but consistent. (By consistent, we, mean that if, for example, we assume that i2 enters the 4- resistor, from the left-hand side, i2 must leave the resistor from the right-hand, side.) The reference node is selected, and the node voltages v1 and v2, are now to be determined., At node 1, applying KCL and Ohm’s law gives, i1 i2 i3, , (a), , i1 = 5, v1, i3, 2Ω, , 5, , v1 v2, v1 0, , 4, 2, , Multiplying each term in the last equation by 4, we obtain, , 5A, , i2, , 1, , 20 v1 v2 2v1, , i1 = 5, 4Ω, , v2, , or, , i4 = 10, , 3v1 v2 20, , i2 i, 5, 6Ω, , (3.1.1), , At node 2, we do the same thing and get, 10 A, , i2 i4 i1 i5, , 1, , v2 0, v1 v2, 10 5 , 4, 6, , Multiplying each term by 12 results in, 3v1 3v2 120 60 2v2, , (b), , Figure 3.3, For Example 3.1: (a) original circuit,, (b) circuit for analysis., , or, 3v1 5v2 60, , (3.1.2)
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ale29559_ch03.qxd, , 07/08/2008, , 10:46 AM, , Page 85, , 3.2, , Nodal Analysis, , 85, , Now we have two simultaneous Eqs. (3.1.1) and (3.1.2). We can solve, the equations using any method and obtain the values of v1 and v2., , ■ METHOD 1 Using the elimination technique, we add Eqs. (3.1.1), and (3.1.2)., 4v2 80, , 1, , v2 20 V, , Substituting v2 20 in Eq. (3.1.1) gives, 3v1 20 20, , 1, , v1 , , 40, 13.333 V, 3, , ■ METHOD 2 To use Cramer’s rule, we need to put Eqs. (3.1.1), and (3.1.2) in matrix form as, c, , 3 1 v1, 20, d c d c d, 3, 5 v2, 60, , (3.1.3), , The determinant of the matrix is, ¢ `, , 3 1, ` 15 3 12, 3, 5, , We now obtain v1 and v2 as, 20 1, `, ¢1, 60, 5, 100 60, v1 , , , 13.333 V, ¢, ¢, 12, 3 20, `, `, ¢2, 3 60, 180 60, , , 20 V, v2 , ¢, ¢, 12, `, , giving us the same result as did the elimination method., If we need the currents, we can easily calculate them from the, values of the nodal voltages., i1 5 A,, , v1 v2, v1, i3 , 1.6668 A,, 6.666 A, 4, 2, v2, i4 10 A,, i5 , 3.333 A, 6, , i2 , , The fact that i2 is negative shows that the current flows in the direction, opposite to the one assumed., , Practice Problem 3.1, , Obtain the node voltages in the circuit of Fig. 3.4., Answer: v1 2 V, v2 14 V., , 6Ω, , 1, , 1A, , 2Ω, , Figure 3.4, For Practice Prob. 3.1., , 2, , 7Ω, , 4A
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ale29559_ch03.qxd, , 07/08/2008, , 10:46 AM, , Page 87, , 3.2, , Nodal Analysis, , Substituting Eq. (3.2.5) into Eq. (3.2.4) yields, 2v2 v2 2.4, , v2 2.4,, , 1, , v1 2v2 4.8 V, , From Eq. (3.2.3), we get, v3 3v2 2v1 3v2 4v2 v2 2.4 V, Thus,, v1 4.8 V,, , v2 2.4 V,, , v3 2.4 V, , ■ METHOD 2 To use Cramer’s rule, we put Eqs. (3.2.1) to (3.2.3), in matrix form., 3 2 1, v1, 12, £ 4, 7 1 § £ v2 § £ 0 §, 2 3, 1, v3, 0, , (3.2.6), , From this, we obtain, v1 , , ¢1, ,, ¢, , v2 , , ¢2, ,, ¢, , v3 , , ¢3, ¢, , where ¢, ¢ 1, ¢ 2, and ¢ 3 are the determinants to be calculated as, follows. As explained in Appendix A, to calculate the determinant of, a 3 by 3 matrix, we repeat the first two rows and cross multiply., 3 2 1, 3 2 1, 4, 7 1, ¢ 3 4, 7 1 3 , 5 2 3, 15, 2 3, 1, , 3 2 1 , 4, 7 1 , , , 21 12 4 14 9 8 10, Similarly, we obtain, , ¢1 , , , , , ¢2 , , , , , ¢3 , , , , , 12 2 1, 0, 7 1, 5 0 3, 15, 12 2 1, 0, 7 1, 3 12 1, 4 0 1, 5 2 0, 15, 3 12 1, 4 0 1, 3 2 12, 4, 7 0, 5 2 3 0 5, 3 2 12, 4, 7 0, , 84 0 0 0 36 0 48, , , , , 0 0 24 0 0 48 24, , , , , 0 144 0 168 0 0 24, , , , , 87
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ale29559_ch03.qxd, , 07/08/2008, , 10:46 AM, , Page 88, , Chapter 3, , 88, , Methods of Analysis, , Thus, we find, v1 , , ¢1, ¢2, 48, 24, , 4.8 V,, v2 , , 2.4 V, ¢, 10, ¢, 10, ¢3, 24, v3 , , 2.4 V, ¢, 10, , as we obtained with Method 1., , ■ METHOD 3 We now use MATLAB to solve the matrix. Equation (3.2.6) can be written as, AV B, , 1, , V A1B, , where A is the 3 by 3 square matrix, B is the column vector, and V is, a column vector comprised of v1, v2, and v3 that we want to determine., We use MATLAB to determine V as follows:, A [3 2 1;, B [12 0 0];, V inv(A) * B, 4.8000, V 2.4000, 2.4000, , 4, , 7, , 1; 2, , 3, , 1];, , Thus, v1 4.8 V, v2 2.4 V, and v3 2.4 V, as obtained previously., , Practice Problem 3.2, , Find the voltages at the three nonreference nodes in the circuit of, Fig. 3.6., , 2Ω, 3Ω, 1, , Answer: v1 80 V, v2 64 V, v3 156 V., , 4ix, 2, , 3, ix, , 10 A, , 4Ω, , 6Ω, , 3.3, Figure 3.6, , Nodal Analysis with Voltage Sources, , We now consider how voltage sources affect nodal analysis. We use the, circuit in Fig. 3.7 for illustration. Consider the following two possibilities., , For Practice Prob. 3.2., , ■ CASE 1 If a voltage source is connected between the reference, node and a nonreference node, we simply set the voltage at the nonreference node equal to the voltage of the voltage source. In Fig. 3.7,, for example,, v1 10 V, , (3.10), , Thus, our analysis is somewhat simplified by this knowledge of the voltage at this node., , ■ CASE 2 If the voltage source (dependent or independent) is connected between two nonreference nodes, the two nonreference nodes
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ale29559_ch03.qxd, , 07/08/2008, , 10:46 AM, , Page 89, , 3.3, , Nodal Analysis with Voltage Sources, , 89, , 4Ω, Supernode, i4, 2Ω, , v1, , i1, , 5V, , v2, , +−, , i2, 10 V +, −, , v3, i3, , 8Ω, , 6Ω, , Figure 3.7, A circuit with a supernode., , form a generalized node or supernode; we apply both KCL and KVL, to determine the node voltages., , A supernode may be regarded as a, closed surface enclosing the voltage, source and its two nodes., , A supernode is formed by enclosing a (dependent or independent), voltage source connected between two nonreference nodes and any, elements connected in parallel with it., , In Fig. 3.7, nodes 2 and 3 form a supernode. (We could have more, than two nodes forming a single supernode. For example, see the circuit in Fig. 3.14.) We analyze a circuit with supernodes using the, same three steps mentioned in the previous section except that the, supernodes are treated differently. Why? Because an essential component of nodal analysis is applying KCL, which requires knowing, the current through each element. There is no way of knowing the, current through a voltage source in advance. However, KCL must, be satisfied at a supernode like any other node. Hence, at the supernode in Fig. 3.7,, i1 i4 i2 i3, , (3.11a), , v1 v3, v3 0, v1 v2, v2 0, , , , 2, 4, 8, 6, , (3.11b), , or, , To apply Kirchhoff’s voltage law to the supernode in Fig. 3.7, we, redraw the circuit as shown in Fig. 3.8. Going around the loop in the, clockwise direction gives, v2 5 v3 0, , 1, , v2 v3 5, , 5V, +, , (3.12), , From Eqs. (3.10), (3.11b), and (3.12), we obtain the node voltages., Note the following properties of a supernode:, 1. The voltage source inside the supernode provides a constraint, equation needed to solve for the node voltages., 2. A supernode has no voltage of its own., 3. A supernode requires the application of both KCL and KVL., , +−, , +, , v2, , v3, , −, , −, , Figure 3.8, Applying KVL to a supernode.
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ale29559_ch03.qxd, , 92, , 07/08/2008, , 10:46 AM, , Page 92, , Chapter 3, , Methods of Analysis, , We now apply KVL to the branches involving the voltage sources as, shown in Fig. 3.13(b). For loop 1,, v1 20 v2 0, , 1, , v1 v2 20, , (3.4.3), , For loop 2,, v3 3vx v4 0, But vx v1 v4 so that, 3v1 v3 2v4 0, , (3.4.4), , For loop 3,, vx 3vx 6i3 20 0, But 6i3 v3 v2 and vx v1 v4. Hence,, 2v1 v2 v3 2v4 20, , (3.4.5), , We need four node voltages, v1, v2, v3, and v4, and it requires only, four out of the five Eqs. (3.4.1) to (3.4.5) to find them. Although the fifth, equation is redundant, it can be used to check results. We can solve, Eqs. (3.4.1) to (3.4.4) directly using MATLAB. We can eliminate one, node voltage so that we solve three simultaneous equations instead of, four. From Eq. (3.4.3), v2 v1 20. Substituting this into Eqs. (3.4.1), and (3.4.2), respectively, gives, 6v1 v3 2v4 80, , (3.4.6), , 6v1 5v3 16v4 40, , (3.4.7), , and, , Equations (3.4.4), (3.4.6), and (3.4.7) can be cast in matrix form as, v1, 0, 3 1 2, £ 6 1 2 § £ v3 § £ 80 §, v4, 40, 6 5 16, Using Cramer’s rule gives, 3 1 2, 0, ¢ † 6 1 2 † 18,, ¢ 1 † 80, 6 5 16, 40, 3 0 2, ¢ 3 † 6 80 2 † 3120,, ¢4 , 6 40 16, , 1 2, 1 2 † 480,, 5 16, 3 1 0, † 6 1 80 † 840, 6 5 40, , Thus, we arrive at the node voltages as, v1 , , ¢3, ¢1, 480, 3120, , 26.67 V,, v3 , , 173.33 V,, ¢, 18, ¢, 18, ¢4, 840, v4 , , 46.67 V, ¢, 18, , and v2 v1 20 6.667 V. We have not used Eq. (3.4.5); it can be, used to cross check results.
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07/08/2008, , 10:46 AM, , Page 93, , 3.4, , Mesh Analysis, , 93, , Practice Problem 3.4, , Find v1, v2, and v3 in the circuit of Fig. 3.14 using nodal analysis., , 6Ω, , Answer: v1 3.043 V, v2 6.956 V, v3 0.6522 V., 10 V, v1, , +−, , 5i, , v2, , +−, , ale29559_ch03.qxd, , i, 2Ω, , 3.4, , v3, , 4Ω, , 3Ω, , Mesh Analysis, , Mesh analysis provides another general procedure for analyzing circuits, using mesh currents as the circuit variables. Using mesh currents, instead of element currents as circuit variables is convenient and, reduces the number of equations that must be solved simultaneously., Recall that a loop is a closed path with no node passed more than once., A mesh is a loop that does not contain any other loop within it., Nodal analysis applies KCL to find unknown voltages in a given, circuit, while mesh analysis applies KVL to find unknown currents., Mesh analysis is not quite as general as nodal analysis because it is, only applicable to a circuit that is planar. A planar circuit is one that, can be drawn in a plane with no branches crossing one another; otherwise it is nonplanar. A circuit may have crossing branches and still, be planar if it can be redrawn such that it has no crossing branches., For example, the circuit in Fig. 3.15(a) has two crossing branches, but, it can be redrawn as in Fig. 3.15(b). Hence, the circuit in Fig. 3.15(a), is planar. However, the circuit in Fig. 3.16 is nonplanar, because there, is no way to redraw it and avoid the branches crossing. Nonplanar circuits can be handled using nodal analysis, but they will not be considered in this text., , Figure 3.14, For Practice Prob. 3.4., , Mesh analysis is also known as loop, analysis or the mesh-current method., , 1A, , 2Ω, , 5Ω, , 1Ω, , 6Ω, , 3Ω, , 4Ω, 7Ω, , 8Ω, 1Ω, (a), 5Ω, 4Ω, 6Ω, , 7Ω, , 1A, , 2Ω, , 3Ω, , 2Ω, , 13 Ω, 5A, , 12 Ω, , 11 Ω, , 9Ω, , 1Ω, 8Ω, , 3Ω, 4Ω, , 5Ω, 8Ω, , 6Ω, 7Ω, , 10 Ω, , Figure 3.16, , (b), , A nonplanar circuit., , Figure 3.15, To understand mesh analysis, we should first explain more about, what we mean by a mesh., A mesh is a loop which does not contain any other loops within it., , (a) A planar circuit with crossing branches,, (b) the same circuit redrawn with no crossing branches.
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ale29559_ch03.qxd, , 07/08/2008, , 10:46 AM, , Page 94, , Chapter 3, , 94, , Methods of Analysis, , a, , I1, , R1, , b, , I2, , R2, , c, , I3, V1 +, −, , i2, , i1, , R3, , e, , f, , + V, 2, −, , d, , Figure 3.17, A circuit with two meshes., , Although path abcdefa is a loop and, not a mesh, KVL still holds. This is the, reason for loosely using the terms, loop analysis and mesh analysis to, mean the same thing., , In Fig. 3.17, for example, paths abefa and bcdeb are meshes, but path, abcdefa is not a mesh. The current through a mesh is known as mesh, current. In mesh analysis, we are interested in applying KVL to find, the mesh currents in a given circuit., In this section, we will apply mesh analysis to planar circuits that, do not contain current sources. In the next section, we will consider, circuits with current sources. In the mesh analysis of a circuit with n, meshes, we take the following three steps., , Steps to Determine Mesh Currents:, 1. Assign mesh currents i1, i2, p , in to the n meshes., 2. Apply KVL to each of the n meshes. Use Ohm’s law to, express the voltages in terms of the mesh currents., 3. Solve the resulting n simultaneous equations to get the mesh, currents., The direction of the mesh current is, arbitrary—(clockwise or counterclockwise)—and does not affect the validity, of the solution., , To illustrate the steps, consider the circuit in Fig. 3.17. The first, step requires that mesh currents i1 and i2 are assigned to meshes 1 and, 2. Although a mesh current may be assigned to each mesh in an arbitrary direction, it is conventional to assume that each mesh current, flows clockwise., As the second step, we apply KVL to each mesh. Applying KVL, to mesh 1, we obtain, V1 R1i1 R3 (i1 i2) 0, or, (R1 R3) i1 R3i2 V1, , (3.13), , For mesh 2, applying KVL gives, R2 i2 V2 R3(i2 i1) 0, or, R3 i1 (R2 R3)i2 V2, The shortcut way will not apply if one, mesh current is assumed clockwise, and the other assumed counterclockwise, although this is permissible., , (3.14), , Note in Eq. (3.13) that the coefficient of i1 is the sum of the resistances, in the first mesh, while the coefficient of i2 is the negative of the resistance common to meshes 1 and 2. Now observe that the same is true, in Eq. (3.14). This can serve as a shortcut way of writing the mesh, equations. We will exploit this idea in Section 3.6.
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ale29559_ch03.qxd, , 07/08/2008, , 10:46 AM, , Page 95, , 3.4, , Mesh Analysis, , 95, , The third step is to solve for the mesh currents. Putting Eqs. (3.13), and (3.14) in matrix form yields, c, , R1 R3, R3, i1, V1, d c d c, d, R3, R2 R3 i2, V2, , (3.15), , which can be solved to obtain the mesh currents i1 and i2. We are at, liberty to use any technique for solving the simultaneous equations., According to Eq. (2.12), if a circuit has n nodes, b branches, and l independent loops or meshes, then l b n 1. Hence, l independent, simultaneous equations are required to solve the circuit using mesh, analysis., Notice that the branch currents are different from the mesh currents unless the mesh is isolated. To distinguish between the two types, of currents, we use i for a mesh current and I for a branch current. The, current elements I1, I2, and I3 are algebraic sums of the mesh currents., It is evident from Fig. 3.17 that, I1 i1,, , I2 i2,, , I3 i1 i2, , (3.16), , Example 3.5, , For the circuit in Fig. 3.18, find the branch currents I1, I2, and I3 using, mesh analysis., I1, , 5Ω, , Solution:, We first obtain the mesh currents using KVL. For mesh 1,, , 10 Ω, 15 V +, −, , or, , i1, , (3.5.1), , 6i2 4i2 10(i2 i1) 10 0, , Figure 3.18, For Example 3.5., , or, (3.5.2), , ■ METHOD 1 Using the substitution method, we substitute, Eq. (3.5.2) into Eq. (3.5.1), and write, 6i2 3 2i2 1, , 1, , i2 1 A, , From Eq. (3.5.2), i1 2i2 1 2 1 1 A. Thus,, I1 i1 1 A,, , I2 i2 1 A,, , I3 i1 i2 0, , ■ METHOD 2 To use Cramer’s rule, we cast Eqs. (3.5.1) and, (3.5.2) in matrix form as, c, , 3 2 i1, 1, d c d c d, 1, 2 i2, 1, , i2, + 10 V, −, , For mesh 2,, , i1 2i2 1, , 6Ω, , I3, , 15 5i1 10(i1 i2) 10 0, 3i1 2i2 1, , I2, , 4Ω
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ale29559_ch03.qxd, , 07/08/2008, , 10:46 AM, , Page 97, , 3.4, , Mesh Analysis, , But at node A, Io i1 i2, so that, 4(i1 i2) 12(i3 i1) 4(i3 i2) 0, or, i1 i2 2i3 0, , (3.6.3), , In matrix form, Eqs. (3.6.1) to (3.6.3) become, 11 5 6, i1, 12, £ 5 19 2 § £ i2 § £ 0 §, 1 1, 2, i3, 0, We obtain the determinants as, , ¢, , ¢1, , ¢2, , ¢3, , 11 5 6, 5 19 2, , 5 1 1, 25, , 11 5 6 , , 5 19 2 , , , 418 30 10 114 22 50 192, 12 5 6, 0 19 2, , 5 0 1, 25, 456 24 432, , 12 5 6, , , 0 19 2, , , , 11 12 6, 5 0 2, , 5 1 0, 25, 24 120 144, , 11 12 6, , , 5 0 2, , , , 11 5 12, 5 19 0, , 5 1 1 0 5, 60 228 288, , 11 5 12 , 5 19 0 , , , , We calculate the mesh currents using Cramer’s rule as, i1 , , ¢1, 432, , 2.25 A,, ¢, 192, i3 , , Thus, Io i1 i2 1.5 A., , i2 , , ¢2, 144, , 0.75 A,, ¢, 192, , ¢3, 288, , 1.5 A, ¢, 192, , 97
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ale29559_ch03.qxd, , 07/08/2008, , 10:46 AM, , Page 98, , Chapter 3, , 98, , Practice Problem 3.6, , Using mesh analysis, find Io in the circuit of Fig. 3.21., , 6Ω, , Io, 20 V +, −, , Answer: 5 A., , i3, , 4Ω, , 8Ω, –, +, , 2Ω, , i1, , i2, , 3.5, , 10io, , For Practice Prob. 3.6., , 4Ω, , ■ CASE 1 When a current source exists only in one mesh: Consider, , 3Ω, , 6Ω, , i1, , Mesh Analysis with Current Sources, , Applying mesh analysis to circuits containing current sources (dependent, or independent) may appear complicated. But it is actually much easier, than what we encountered in the previous section, because the presence, of the current sources reduces the number of equations. Consider the, following two possible cases., , Figure 3.21, , 10 V +, −, , Methods of Analysis, , the circuit in Fig. 3.22, for example. We set i2 5 A and write a, mesh equation for the other mesh in the usual way; that is,, , 5A, , i2, , 10 4i1 6(i1 i2) 0, , i1 2 A, , 1, , (3.17), , ■ CASE 2 When a current source exists between two meshes: Con-, , Figure 3.22, A circuit with a current source., , sider the circuit in Fig. 3.23(a), for example. We create a supermesh, by excluding the current source and any elements connected in series, with it, as shown in Fig. 3.23(b). Thus,, A supermesh results when two meshes have a (dependent or independent) current source in common., , 6Ω, , 10 Ω, 6Ω, , 10 Ω, , 2Ω, 20 V, , +, −, , i1, , i2, , 4Ω, , 6A, , i1, , 0, (a), , i2, , Exclude these, elements, , 20 V +, −, , i1, , i2, , 4Ω, , (b), , Figure 3.23, (a) Two meshes having a current source in common, (b) a supermesh, created by excluding the current, source., , As shown in Fig. 3.23(b), we create a supermesh as the periphery of, the two meshes and treat it differently. (If a circuit has two or more, supermeshes that intersect, they should be combined to form a larger, supermesh.) Why treat the supermesh differently? Because mesh analysis applies KVL—which requires that we know the voltage across each, branch—and we do not know the voltage across a current source in, advance. However, a supermesh must satisfy KVL like any other mesh., Therefore, applying KVL to the supermesh in Fig. 3.23(b) gives, 20 6i1 10i2 4i2 0
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ale29559_ch03.qxd, , 07/08/2008, , 10:47 AM, , Page 101, , 3.6, , Nodal and Mesh Analyses by Inspection, , or simply, Gv i, , (3.23), , where, Gkk Sum of the conductances connected to node k, Gk j Gjk Negative of the sum of the conductances directly, connecting nodes k and j, k j, vk Unknown voltage at node k, ik Sum of all independent current sources directly connected, to node k, with currents entering the node treated as positive, G is called the conductance matrix; v is the output vector; and i is the, input vector. Equation (3.22) can be solved to obtain the unknown node, voltages. Keep in mind that this is valid for circuits with only independent current sources and linear resistors., Similarly, we can obtain mesh-current equations by inspection, when a linear resistive circuit has only independent voltage sources., Consider the circuit in Fig. 3.17, shown again in Fig. 3.26(b) for convenience. The circuit has two nonreference nodes and the node equations were derived in Section 3.4 as, c, , R1 R3, R3, i1, v1, d c d c, d, R3, R2 R3 i2, v2, , (3.24), , We notice that each of the diagonal terms is the sum of the resistances, in the related mesh, while each of the off-diagonal terms is the negative of the resistance common to meshes 1 and 2. Each term on the, right-hand side of Eq. (3.24) is the algebraic sum taken clockwise of, all independent voltage sources in the related mesh., In general, if the circuit has N meshes, the mesh-current equations, can be expressed in terms of the resistances as, ≥, , R11 R12, R21 R22, o, , o, , RN1 RN2, , p, p, , R1N, R2N, , o, p, , o, , ¥ ≥, , RNN, , i1, i2, o, iN, , ¥ ≥, , v1, v2, o, vN, , ¥, , (3.25), , or simply, Ri v, , (3.26), , where, Rkk Sum of the resistances in mesh k, Rkj Rjk Negative of the sum of the resistances in common, with meshes k and j, k j, ik Unknown mesh current for mesh k in the clockwise direction, vk Sum taken clockwise of all independent voltage sources in, mesh k, with voltage rise treated as positive, R is called the resistance matrix; i is the output vector; and v is, the input vector. We can solve Eq. (3.25) to obtain the unknown mesh, currents., , 101
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ale29559_ch03.qxd, , 07/08/2008, , 102, , Example 3.8, , 10:47 AM, , Page 102, , Chapter 3, , Methods of Analysis, , Write the node-voltage matrix equations for the circuit in Fig. 3.27 by, inspection., , 2A, , 1Ω, , 10 Ω, , 3A, , 8Ω, , 5 Ω v2, , v1, , 8Ω, , v3, , 4Ω, , 1A, , v4, , 2Ω, , 4A, , Figure 3.27, For Example 3.8., , Solution:, The circuit in Fig. 3.27 has four nonreference nodes, so we need four, node equations. This implies that the size of the conductance matrix, G, is 4 by 4. The diagonal terms of G, in siemens, are, 1, 1, , 0.3,, 5, 10, 1, 1, 1, 0.5,, 8, 8, 4, , G11 , G33, , 1, 1, 1, 1.325, 5, 8, 1, 1, 1, 1, 1.625, 8, 2, 1, , G22 , G44, , The off-diagonal terms are, 1, G12 0.2,, 5, G21 0.2,, G31 0,, , G13 G14 0, , 1, G23 0.125,, 8, G32 0.125,, , G41 0,, , G42 1,, , G34, , 1, G24 1, 1, 1, 0.125, 8, , G43 0.125, , The input current vector i has the following terms, in amperes:, i1 3,, , i2 1 2 3,, , i3 0,, , i4 2 4 6, , Thus the node-voltage equations are, 0.3 0.2, 0, 0, v1, 3, 0.2, 1.325 0.125 1, 3, v2, ≥, ¥ ≥ ¥ ≥, ¥, 0, 0.125, 0.5, 0.125, v3, 0, 0, 1, 0.125, 1.625, v4, 6, which can be solved using MATLAB to obtain the node voltages v1, v2,, v3, and v4.
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ale29559_ch03.qxd, , 07/08/2008, , 10:47 AM, , 104, , Page 104, , Chapter 3, , Methods of Analysis, , Thus, the mesh-current equations are:, 9, 2, E2, 0, 0, , 2 2, 0, 0, i1, 4, 10 4 1 1, i2, 6, 4, 9, 0, 0U Ei3U E6U, 1, 0, 8 3, i4, 0, 1, 0 3, 4, i5, 6, , From this, we can use MATLAB to obtain mesh currents i1, i2, i3, i4,, and i5., , Practice Problem 3.9, , By inspection, obtain the mesh-current equations for the circuit in, Fig. 3.30., 50 Ω, , 40 Ω, i2, 24 V +, −, , i1, , 10 Ω, , i3, 20 Ω, , i4, 80 Ω, , + 12 V, −, , 30 Ω, , i5, −, + 10 V, , 60 Ω, , Figure 3.30, For Practice Prob. 3.9., , Answer:, 170 40, 0 80, 0, i1, 24, 40, 80 30 10, 0, i2, 0, E 0 30, 50, 0 20U Ei3U E12U, 80 10, 0, 90, 0, i4, 10, 0, 0 20, 0, 80, i5, 10, , 3.7, , Nodal Versus Mesh Analysis, , Both nodal and mesh analyses provide a systematic way of analyzing, a complex network. Someone may ask: Given a network to be analyzed, how do we know which method is better or more efficient? The, choice of the better method is dictated by two factors.
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ale29559_ch03.qxd, , 07/08/2008, , 10:47 AM, , Page 105, , 3.8, , Circuit Analysis with PSpice, , 105, , The first factor is the nature of the particular network. Networks, that contain many series-connected elements, voltage sources, or supermeshes are more suitable for mesh analysis, whereas networks with, parallel-connected elements, current sources, or supernodes are more, suitable for nodal analysis. Also, a circuit with fewer nodes than, meshes is better analyzed using nodal analysis, while a circuit with, fewer meshes than nodes is better analyzed using mesh analysis. The, key is to select the method that results in the smaller number of, equations., The second factor is the information required. If node voltages are, required, it may be expedient to apply nodal analysis. If branch or mesh, currents are required, it may be better to use mesh analysis., It is helpful to be familiar with both methods of analysis, for at, least two reasons. First, one method can be used to check the results, from the other method, if possible. Second, since each method has its, limitations, only one method may be suitable for a particular problem., For example, mesh analysis is the only method to use in analyzing transistor circuits, as we shall see in Section 3.9. But mesh analysis cannot easily be used to solve an op amp circuit, as we shall see in Chapter 5,, because there is no direct way to obtain the voltage across the op amp, itself. For nonplanar networks, nodal analysis is the only option,, because mesh analysis only applies to planar networks. Also, nodal, analysis is more amenable to solution by computer, as it is easy to program. This allows one to analyze complicated circuits that defy hand, calculation. A computer software package based on nodal analysis is, introduced next., , 3.8, , Circuit Analysis with PSpice, , PSpice is a computer software circuit analysis program that we will, gradually learn to use throughout the course of this text. This section, illustrates how to use PSpice for Windows to analyze the dc circuits we, have studied so far., The reader is expected to review Sections D.1 through D.3 of, Appendix D before proceeding in this section. It should be noted that, PSpice is only helpful in determining branch voltages and currents, when the numerical values of all the circuit components are known., , Appendix D provides a tutorial on, using PSpice for Windows., , Example 3.10, , Use PSpice to find the node voltages in the circuit of Fig. 3.31., Solution:, The first step is to draw the given circuit using Schematics. If one, follows the instructions given in Appendix sections D.2 and D.3, the, schematic in Fig. 3.32 is produced. Since this is a dc analysis, we use, voltage source VDC and current source IDC. The pseudocomponent, VIEWPOINTS are added to display the required node voltages. Once, the circuit is drawn and saved as exam310.sch, we run PSpice by, selecting Analysis/Simulate. The circuit is simulated and the results, , 1, 120 V +, −, , 20 Ω, , 2, , 30 Ω, , 10 Ω, 40 Ω, , 0, , Figure 3.31, For Example 3.10., , 3, 3A
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ale29559_ch03.qxd, , 07/21/2008, , 11:21 AM, , 106, , Page 106, , Chapter 3, , Methods of Analysis, 120.0000, 1, , R1 81.2900, , 2, , 20, +, 120 V, −, , R3 89.0320, , 3, , 10, IDC, , V1, , R2, , R4, , 30, , 40, , 3A, , I1, , 0, , Figure 3.32, For Example 3.10; the schematic of the circuit in Fig. 3.31., , are displayed on VIEWPOINTS and also saved in output file, exam310.out. The output file includes the following:, NODE VOLTAGE, NODE VOLTAGE NODE VOLTAGE, (1), 120.0000 (2), 81.2900 (3), 89.0320, indicating that V1 120 V, V2 81.29 V, V3 89.032 V., , Practice Problem 3.10, , For the circuit in Fig. 3.33, use PSpice to find the node voltages., 2A, , 1, , 60 Ω, , 30 Ω, , 2, , 50 Ω, , 100 Ω, , 3, , +, −, , 25 Ω, , 200 V, , 0, , Figure 3.33, For Practice Prob. 3.10., , Answer: V1 40 V, V2 57.14 V, V3 200 V., , In the circuit of Fig. 3.34, determine the currents i1, i2, and i3., 1Ω, , 4Ω, , 2Ω, , 3vo, +−, , Example 3.11, , i2, , i1, 24 V +, −, , 2Ω, , Figure 3.34, For Example 3.11., , 8Ω, , 4Ω, , i3, +, vo, −
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ale29559_ch03.qxd, , 07/08/2008, , 10:47 AM, , Page 107, , 3.9, , Applications: DC Transistor Circuits, , 107, , Solution:, The schematic is shown in Fig. 3.35. (The schematic in Fig. 3.35, includes the output results, implying that it is the schematic displayed, on the screen after the simulation.) Notice that the voltage-controlled, voltage source E1 in Fig. 3.35 is connected so that its input is the, voltage across the 4- resistor; its gain is set equal to 3. In order to, display the required currents, we insert pseudocomponent IPROBES in, the appropriate branches. The schematic is saved as exam311.sch and, simulated by selecting Analysis/Simulate. The results are displayed on, IPROBES as shown in Fig. 3.35 and saved in output file exam311.out., From the output file or the IPROBES, we obtain i1 i2 1.333 A and, i3 2.667 A., E, − +, , 2, , E1, , −+, , R5, R1, , 1, R6, , 4, R2, , +, 24 V, , −, , 2, , R3, , 8, , R4, , 4, , V1, 1.333E + 00, , 1.333E + 00, , 2.667E + 00, , 0, , Figure 3.35, The schematic of the circuit in Fig. 3.34., , Use PSpice to determine currents i1, i2, and i3 in the circuit of Fig. 3.36., , Practice Problem 3.11, i1, , Answer: i1 0.4286 A, i2 2.286 A, i3 2 A., , 4Ω, 2A, i2, , 2Ω, , 3.9, , Applications: DC Transistor Circuits, 10 V, , Most of us deal with electronic products on a routine basis and have, some experience with personal computers. A basic component for, the integrated circuits found in these electronics and computers is the, active, three-terminal device known as the transistor. Understanding, the transistor is essential before an engineer can start an electronic circuit design., Figure 3.37 depicts various kinds of transistors commercially available. There are two basic types of transistors: bipolar junction transistors (BJTs) and field-effect transistors (FETs). Here, we consider only, the BJTs, which were the first of the two and are still used today. Our, objective is to present enough detail about the BJT to enable us to apply, the techniques developed in this chapter to analyze dc transistor circuits., , 1Ω, +, −, , Figure 3.36, For Practice Prob. 3.11., , i3, i1, , 2Ω
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ale29559_ch03.qxd, , 07/08/2008, , 10:47 AM, , Page 108, , Chapter 3, , 108, , Methods of Analysis, , Historical, William Schockley (1910–1989), John Bardeen (1908–1991), and, Walter Brattain (1902–1987) co-invented the transistor., Nothing has had a greater impact on the transition from the “Industrial Age” to the “Age of the Engineer” than the transistor. I am sure, that Dr. Shockley, Dr. Bardeen, and Dr. Brattain had no idea they would, have this incredible effect on our history. While working at Bell Laboratories, they successfully demonstrated the point-contact transistor,, invented by Bardeen and Brattain in 1947, and the junction transistor,, which Shockley conceived in 1948 and successfully produced in 1951., It is interesting to note that the idea of the field-effect transistor,, the most commonly used one today, was first conceived in 1925–1928, by J. E. Lilienfeld, a German immigrant to the United States. This is, evident from his patents of what appears to be a field-effect transistor., Unfortunately, the technology to realize this device had to wait until, 1954 when Shockley’s field-effect transistor became a reality. Just think, what today would be like if we had this transistor 30 years earlier!, For their contributions to the creation of the transistor, Dr. Shockley,, Dr. Bardeen, and Dr. Brattain received, in 1956, the Nobel Prize in, physics. It should be noted that Dr. Bardeen is the only individual to, win two Nobel prizes in physics; the second came later for work in, superconductivity at the University of Illinois., , Courtesy of Lucent, Technologies/Bell Labs, , Collector, , C, , n, p, , Base, , B, , n, , E, , Emitter, (a), Collector, , C, , Figure 3.37, Various types of transistors., (Courtesy of Tech America.), , p, Base, , B, , n, p, , E, , Emitter, , There are two types of BJTs: npn and pnp, with their circuit symbols as shown in Fig. 3.38. Each type has three terminals, designated, as emitter (E), base (B), and collector (C). For the npn transistor, the, currents and voltages of the transistor are specified as in Fig. 3.39., Applying KCL to Fig. 3.39(a) gives, , (b), , Figure 3.38, Two types of BJTs and their circuit, symbols: (a) npn, (b) pnp., , IE IB IC, , (3.27)
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ale29559_ch03.qxd, , 07/08/2008, , 10:47 AM, , Page 109, , 3.9, , Applications: DC Transistor Circuits, , 109, , where IE, IC, and IB are emitter, collector, and base currents, respectively. Similarly, applying KVL to Fig. 3.39(b) gives, VCE VEB VBC 0, , C, IC, IB, , (3.28), B, , where VCE, VEB, and VBC are collector-emitter, emitter-base, and basecollector voltages. The BJT can operate in one of three modes: active,, cutoff, and saturation. When transistors operate in the active mode, typically VBE 0.7 V,, IC a IE, , IE, E, , (3.29), , (a), , where a is called the common-base current gain. In Eq. (3.29),, a denotes the fraction of electrons injected by the emitter that are collected by the collector. Also,, , C, , IC bIB, , (3.30), , +, , +, VCB, B, , −, , VCE, , +, , where b is known as the common-emitter current gain. The a and b, are characteristic properties of a given transistor and assume constant, values for that transistor. Typically, a takes values in the range of 0.98 to, 0.999, while b takes values in the range of 50 to 1000. From Eqs. (3.27), to (3.30), it is evident that, IE (1 b)IB, , (3.31), , and, b, , a, 1a, , IB, , IC, C, , B, +, IB, B, , +, , VCE, , VBE, −, , +, VBE, −, , +, bIB, VCE, , −, E, , (a), , −, , −, E, (b), , Figure 3.39, The terminal variables of an npn transistor:, (a) currents, (b) voltages., , (3.32), , These equations show that, in the active mode, the BJT can be modeled, as a dependent current-controlled current source. Thus, in circuit analysis, the dc equivalent model in Fig. 3.40(b) may be used to replace the, npn transistor in Fig. 3.40(a). Since b in Eq. (3.32) is large, a small base, current controls large currents in the output circuit. Consequently, the, bipolar transistor can serve as an amplifier, producing both current gain, and voltage gain. Such amplifiers can be used to furnish a considerable, amount of power to transducers such as loudspeakers or control motors., C, , VBE, , −, E, (b), , Figure 3.40, (a) An npn transistor, (b) its dc equivalent model., , It should be observed in the following examples that one cannot, directly analyze transistor circuits using nodal analysis because of the, potential difference between the terminals of the transistor. Only when the, transistor is replaced by its equivalent model can we apply nodal analysis., , In fact, transistor circuits provide motivation to study dependent sources.
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ale29559_ch03.qxd, , 07/08/2008, , 10:47 AM, , Page 111, , 3.9, , Applications: DC Transistor Circuits, , 111, , Example 3.13, , For the BJT circuit in Fig. 3.43, b 150 and VBE 0.7 V. Find vo., Solution:, , 1 kΩ, , 1. Define. The circuit is clearly defined and the problem is clearly, stated. There appear to be no additional questions that need to, be asked., 2. Present. We are to determine the output voltage of the circuit, shown in Fig. 3.43. The circuit contains an ideal transistor with, b 150 and VBE 0.7 V., 3. Alternative. We can use mesh analysis to solve for vo. We can, replace the transistor with its equivalent circuit and use nodal, analysis. We can try both approaches and use them to check, each other. As a third check, we can use the equivalent circuit, and solve it using PSpice., 4. Attempt., , ■ METHOD 1 Working with Fig. 3.44(a), we start with the first loop., 2 100kI1 200k(I1 I2) 0, , 3I1 2I2 2 105, (3.13.1), , or, , 1 kΩ, +, 100 kΩ, vo, +, 2V, −, , −, , 200 kΩ, , I1, , I2, , (a), 100 kΩ, , V1, , 1 kΩ, , IB, 150IB, , +, 2V, −, , +, 16 V, −, , I3, , +, , +, 0.7 V, , 200 kΩ, , −, , −, , +, 16 V, −, , vo, , (b), R1, , 700.00mV, , 14.58 V, , 100k, , 1k, , +, 2V, , −, , R3, , +, R2, , 200k, , 0.7 V, , F1, , −, F, , (c), , Figure 3.44, Solution of the problem in Example 3.13: (a) Method 1, (b) Method 2,, (c) Method 3., , +, 16 V, −, , +, 100 kΩ, +, 2V, −, , vo, 200 kΩ, , Figure 3.43, For Example 3.13., , −, , +, 16 V, −
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ale29559_ch03.qxd, , 07/08/2008, , 10:47 AM, , Page 112, , Chapter 3, , 112, , Methods of Analysis, , Now for loop 2., 200k(I2 I1) VBE 0, , or, , 2I1 2I2 0.7 105, (3.13.2), , Since we have two equations and two unknowns, we can solve for I1, and I2. Adding Eq. (3.13.1) to (3.13.2) we get;, I1 1.3 105A, , I2 (0.7 2.6)1052 9.5 mA, , and, , Since I3 150I2 1.425 mA, we can now solve for vo using loop 3:, vo 1kI3 16 0, , or, , vo 1.425 16 14.575 V, , ■ METHOD 2 Replacing the transistor with its equivalent circuit, produces the circuit shown in Fig. 3.44(b). We can now use nodal, analysis to solve for vo., At node number 1: V1 0.7 V, (0.7 2)100k 0.7200k IB 0, , or, , IB 9.5 mA, , At node number 2 we have:, 150IB (vo 16)1k 0, or, vo 16 150 103 9.5 106 14.575 V, 5. Evaluate. The answers check, but to further check we can use, PSpice (Method 3), which gives us the solution shown in, Fig. 3.44(c)., 6. Satisfactory? Clearly, we have obtained the desired answer with, a very high confidence level. We can now present our work as a, solution to the problem., , Practice Problem 3.13, , The transistor circuit in Fig. 3.45 has b 80 and VBE 0.7 V. Find vo, and Io., , 20 kΩ, , Answer: 3 V, 150 m A., Io +, 120 kΩ, +, 1V, −, , +, , 20 kΩ, , vo, , VBE, −, , Figure 3.45, For Practice Prob. 3.13., , −, , +, 10 V, −, , 3.10, , Summary, , 1. Nodal analysis is the application of Kirchhoff’s current law at the, nonreference nodes. (It is applicable to both planar and nonplanar, circuits.) We express the result in terms of the node voltages. Solving the simultaneous equations yields the node voltages., 2. A supernode consists of two nonreference nodes connected by a, (dependent or independent) voltage source., 3. Mesh analysis is the application of Kirchhoff’s voltage law around, meshes in a planar circuit. We express the result in terms of mesh, currents. Solving the simultaneous equations yields the mesh, currents.
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ale29559_ch03.qxd, , 07/08/2008, , 10:47 AM, , Page 114, , Chapter 3, , 114, , 3.7, , Methods of Analysis, , In the circuit of Fig. 3.49, current i1 is:, (a) 4 A, , (b) 3 A, , (c) 2 A, , 3.9, (d) 1 A, , The PSpice part name for a current-controlled, voltage source is:, (a) EX, , 20 V, , +, −, , v, −, , (b) It plots the branch current., , i2, , 2A, , (c) It displays the current through the branch in, which it is connected., , 3Ω, , 4Ω, , (d) It can be used to display voltage by connecting it, in parallel., , Figure 3.49, For Review Questions 3.7 and 3.8., , 3.8, , (e) It is used only for dc analysis., , The voltage v across the current source in the circuit, of Fig. 3.49 is:, (a) 20 V, , (b) 15 V, , (d) GX, , (a) It must be connected in series., , +, , i1, , (c) HX, , 3.10 Which of the following statements are not true of the, pseudocomponent IPROBE:, , 1Ω, , 2Ω, , (b) FX, , (c) 10 V, , (d) 5 V, , (f) It does not correspond to a particular circuit, element., , Answers: 3.1a, 3.2c, 3.3a, 3.4c, 3.5c, 3.6a, 3.7d, 3.8b,, 3.9c, 3.10b,d., , Problems, Sections 3.2 and 3.3 Nodal Analysis, 3.1, , 3.3, , Using Fig. 3.50, design a problem to help other, students better understand nodal analysis., , Find the currents I1 through I4 and the voltage vo in, the circuit of Fig. 3.52., , vo, R1, , R2, Ix, , 12 V +, −, , I2, , I1, 10 A, + 9V, −, , R3, , Figure 3.50, , 10 Ω, , 20 Ω, , I3, , 30 Ω, , I4, 60 Ω, , 2A, , Figure 3.52, , For Prob. 3.1., , For Prob. 3.3., 3.2, , For the circuit in Fig. 3.51, obtain v1 and v2., 3.4, 2Ω, 12 A, , v1, , 10 Ω, , 5Ω, , Given the circuit in Fig. 3.53, calculate the currents, I1 through I4., , v2, , 4Ω, , 2A, I1, , I2, , I3, , I4, , 6A, 4A, , 5Ω, , Figure 3.51, , Figure 3.53, , For Prob. 3.2., , For Prob. 3.4., , 10 Ω, , 10 Ω, , 5Ω, , 5A
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ale29559_ch03.qxd, , 07/08/2008, , 10:47 AM, , Page 115, , Problems, , 3.5, , Obtain vo in the circuit of Fig. 3.54., , 30 V, , +, −, , 4 kΩ, 5 kΩ, , 2 kΩ, , Determine Ib in the circuit of Fig. 3.58 using nodal, analysis., , Ib, , +, −, , 20 V, , 3.9, , 115, , +, vo, −, , 60Ib, , 250 Ω, , +−, , 12 V +, −, , 50 Ω, , 150 Ω, , Figure 3.54, Figure 3.58, , For Prob. 3.5., , For Prob. 3.9., 3.6, , Use nodal analysis to obtain vo in the circuit of, Fig. 3.55., , 3.10 Find Io in the circuit of Fig. 3.59., 1Ω, , 4Ω, I1, 12 V, , 10 V, , vo, , +−, , I2, +, −, , 6Ω, , 2 Io, , 4A, , I3, 2Ω, , Io, 8Ω, , Figure 3.55, , Figure 3.59, , For Prob. 3.6., , For Prob. 3.10., , 3.7, , Apply nodal analysis to solve for Vx in the circuit of, Fig. 3.56., , 4Ω, , 2Ω, , 3.11 Find Vo and the power dissipated in all the resistors, in the circuit of Fig. 3.60., , 1Ω, , 4Ω, , Vo, , +, 10 Ω, , 2A, , 20 Ω, , Vx, , 0.2Vx, , 36 V +, −, , −, +, , 2Ω, , 12 V, , −, , Figure 3.56, , Figure 3.60, , For Prob. 3.7., , 3.8, , For Prob. 3.11., , Using nodal analysis, find vo in the circuit of Fig. 3.57., , 3Ω, , +, vo, −, , 3.12 Using nodal analysis, determine Vo in the circuit in, Fig. 3.61., 10 Ω, , 5Ω, +, −, , Ix, , 3V, , 2Ω, 1Ω, , 1Ω, , +, −, , 4vo, , 30 V, , Figure 3.57, , Figure 3.61, , For Prob. 3.8., , For Prob. 3.12., , +, −, , 2Ω, , 5Ω, 4 Ix, , +, Vo, −
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ale29559_ch03.qxd, , 07/08/2008, , 10:47 AM, , Page 126, , Chapter 3, , 126, , Methods of Analysis, , *3.88 Determine the gain vo vs of the transistor amplifier, circuit in Fig. 3.124., 200 Ω, , vs, , +, −, , 100 Ω, , 3.91 For the transistor circuit of Fig. 3.127, find IB, VCE,, and vo. Take b 200, VBE 0.7 V., , Io, , 2 kΩ, , vo, 1000, , 5 kΩ, , +, −, , +, vo, −, , 40Io, , +, 10 kΩ, , 6 kΩ, , IB, VCE, , +, 9V, −, , −, , Figure 3.124, For Prob. 3.88., , 3V, , 2 kΩ, 400 Ω, , 3.89 For the transistor circuit shown in Fig. 3.125, find IB, and VCE. Let b 100, and VBE 0.7 V., , +, vo, −, , Figure 3.127, , 0.7 V 100 kΩ, − +, , + 15 V −, , 3V, , For Prob. 3.91., 1 kΩ, , +, −, , 3.92 Using Fig. 3.128, design a problem to help other, students better understand transistors. Make sure you, use reasonable numbers!, , Figure 3.125, For Prob. 3.89., , R2, R1, , 3.90 Calculate vs for the transistor in Fig. 3.126 given that, vo 4 V, b 150, VBE 0.7 V., , VC, , 1 kΩ, , IB, R3, , 10 kΩ, , vs, 500 Ω, , +, V1, −, , +, vo, −, , +, 18 V, −, , Figure 3.126, For Prob. 3.90., , Comprehensive Problem, *3.93 Rework Example 3.11 with hand calculation., , Figure 3.128, For Prob. 3.92.
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ale29559_ch04.qxd, , 07/14/2008, , 02:51 PM, , Page 127, , c h a p t e r, , Circuit Theorems, , 4, , Your success as an engineer will be directly proportional to your ability, to communicate!, —Charles K. Alexander, , Enhancing Your Skills and Your Career, Enhancing Your Communication Skills, Taking a course in circuit analysis is one step in preparing yourself for, a career in electrical engineering. Enhancing your communication skills, while in school should also be part of that preparation, as a large part, of your time will be spent communicating., People in industry have complained again and again that graduating engineers are ill-prepared in written and oral communication. An, engineer who communicates effectively becomes a valuable asset., You can probably speak or write easily and quickly. But how effectively do you communicate? The art of effective communication is of, the utmost importance to your success as an engineer., For engineers in industry, communication is key to promotability., Consider the result of a survey of U.S. corporations that asked what, factors influence managerial promotion. The survey includes a listing, of 22 personal qualities and their importance in advancement. You may, be surprised to note that “technical skill based on experience” placed, fourth from the bottom. Attributes such as self-confidence, ambition,, flexibility, maturity, ability to make sound decisions, getting things, done with and through people, and capacity for hard work all ranked, higher. At the top of the list was “ability to communicate.” The higher, your professional career progresses, the more you will need to communicate. Therefore, you should regard effective communication as an, important tool in your engineering tool chest., Learning to communicate effectively is a lifelong task you should, always work toward. The best time to begin is while still in school., Continually look for opportunities to develop and strengthen your reading, writing, listening, and speaking skills. You can do this through, classroom presentations, team projects, active participation in student, organizations, and enrollment in communication courses. The risks are, less now than later in the workplace., , Ability to communicate effectively is regarded by many as the most important, step to an executive promotion., © IT Stock/Punchstock, , 127
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ale29559_ch04.qxd, , 128, , 07/08/2008, , 10:55 AM, , Page 128, , Chapter 4, , 4.1, , Circuit Theorems, , Introduction, , A major advantage of analyzing circuits using Kirchhoff’s laws as we, did in Chapter 3 is that we can analyze a circuit without tampering, with its original configuration. A major disadvantage of this approach, is that, for a large, complex circuit, tedious computation is involved., The growth in areas of application of electric circuits has led to an, evolution from simple to complex circuits. To handle the complexity,, engineers over the years have developed some theorems to simplify circuit analysis. Such theorems include Thevenin’s and Norton’s theorems., Since these theorems are applicable to linear circuits, we first discuss the, concept of circuit linearity. In addition to circuit theorems, we discuss the, concepts of superposition, source transformation, and maximum power, transfer in this chapter. The concepts we develop are applied in the last, section to source modeling and resistance measurement., , 4.2, , Linearity Property, , Linearity is the property of an element describing a linear relationship, between cause and effect. Although the property applies to many circuit elements, we shall limit its applicability to resistors in this chapter. The property is a combination of both the homogeneity (scaling), property and the additivity property., The homogeneity property requires that if the input (also called the, excitation) is multiplied by a constant, then the output (also called the, response) is multiplied by the same constant. For a resistor, for example, Ohm’s law relates the input i to the output v,, v iR, , (4.1), , If the current is increased by a constant k, then the voltage increases, correspondingly by k; that is,, kiR kv, , (4.2), , The additivity property requires that the response to a sum of, inputs is the sum of the responses to each input applied separately., Using the voltage-current relationship of a resistor, if, v1 i1R, , (4.3a), , v2 i2R, , (4.3b), , v (i1 i2)R i1R i2R v1 v2, , (4.4), , and, then applying (i1 i2) gives, , We say that a resistor is a linear element because the voltage-current, relationship satisfies both the homogeneity and the additivity properties., In general, a circuit is linear if it is both additive and homogeneous. A linear circuit consists of only linear elements, linear dependent sources, and independent sources.
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ale29559_ch04.qxd, , 07/08/2008, , 10:55 AM, , Page 129, , 4.2, , Linearity Property, , A linear circuit is one whose output is linearly related (or directly proportional) to its input., , Throughout this book we consider only linear circuits. Note that since, p i2R v2R (making it a quadratic function rather than a linear one),, the relationship between power and voltage (or current) is nonlinear., Therefore, the theorems covered in this chapter are not applicable to power., To illustrate the linearity principle, consider the linear circuit, shown in Fig. 4.1. The linear circuit has no independent sources inside, it. It is excited by a voltage source vs, which serves as the input. The, circuit is terminated by a load R. We may take the current i through R, as the output. Suppose vs 10 V gives i 2 A. According to the linearity principle, vs 1 V will give i 0.2 A. By the same token,, i 1 mA must be due to vs 5 mV., , 129, , For example, when current i1 flows, through resistor R, the power is p1 Ri 21,, and when current i2 flows through R, the, power is p2 Ri 22. If current i1 i2 flows, through R, the power absorbed is p3 , R (i1 i2)2 Ri 12 Ri 22 2Ri 1i 2 p1 , p2. Thus, the power relation is nonlinear., i, +, −, , vs, , Figure 4.1, A linear circuit with input vs and output i., , Example 4.1, , For the circuit in Fig. 4.2, find Io when vs 12 V and vs 24 V., 2Ω, , Solution:, Applying KVL to the two loops, we obtain, (4.1.1), , 4i1 16i2 3vx vs 0, , (4.1.2), , 6Ω, , i1, , 4Ω, , i2, vs, , 10i1 16i2 vs 0, , Io, , 4Ω, , But vx 2i1. Equation (4.1.2) becomes, , +, −, , –, +, , (4.1.3), Figure 4.2, , Adding Eqs. (4.1.1) and (4.1.3) yields, 1, , 8Ω, , + vx −, , 12i1 4i2 vs 0, , 2i1 12i2 0, , R, , Linear circuit, , For Example 4.1., , i1 6i2, , Substituting this in Eq. (4.1.1), we get, 76i2 vs 0, , 1, , i2 , , vs, 76, , When vs 12 V,, Io i2 , , 12, A, 76, , Io i2 , , 24, A, 76, , When vs 24 V,, , showing that when the source value is doubled, Io doubles., For the circuit in Fig. 4.3, find vo when is 15 and is 30 A., , Practice Problem 4.1, 12 Ω, , Answer: 20 V, 40 V., is, , 4Ω, , Figure 4.3, For Practice Prob. 4.1., , 8Ω, , +, vo, −, , 3vx
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ale29559_ch04.qxd, , 07/08/2008, , 10:55 AM, , Page 130, , Chapter 4, , 130, , Example 4.2, , Circuit Theorems, , Assume Io 1 A and use linearity to find the actual value of Io in the, circuit of Fig. 4.4., I4, , 6Ω, , 2 V I, 2, 2, , 2Ω, , 1 V, 1, , I3, 7Ω, , I s = 15 A, , 3Ω, Io, , I1, 5Ω, , 4Ω, , Figure 4.4, For Example 4.2., , Solution:, If Io 1 A, then V1 (3 5)Io 8 V and I1 V14 2 A. Applying, KCL at node 1 gives, I2 I1 Io 3 A, V2 V1 2I2 8 6 14 V,, , I3 , , V2, 2A, 7, , Applying KCL at node 2 gives, I4 I3 I2 5 A, Therefore, Is 5 A. This shows that assuming Io 1 gives Is 5 A,, the actual source current of 15 A will give Io 3 A as the actual value., , Practice Problem 4.2, 12 Ω, , 30 V, , +, −, , 5Ω, , 8Ω, , +, Vo, −, , Assume that Vo 1 V and use linearity to calculate the actual value, of Vo in the circuit of Fig. 4.5., Answer: 12 V., , Figure 4.5, For Practice Prob. 4.2., , 4.3, , Superposition is not limited to circuit, analysis but is applicable in many, fields where cause and effect bear a, linear relationship to one another., , Superposition, , If a circuit has two or more independent sources, one way to determine, the value of a specific variable (voltage or current) is to use nodal or, mesh analysis as in Chapter 3. Another way is to determine the contribution of each independent source to the variable and then add them, up. The latter approach is known as the superposition., The idea of superposition rests on the linearity property., The superposition principle states that the voltage across (or current, through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone.
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ale29559_ch04.qxd, , 07/08/2008, , 10:55 AM, , Page 131, , 4.3, , Superposition, , 131, , The principle of superposition helps us to analyze a linear circuit with, more than one independent source by calculating the contribution of, each independent source separately. However, to apply the superposition principle, we must keep two things in mind:, 1. We consider one independent source at a time while all other independent sources are turned off. This implies that we replace every, voltage source by 0 V (or a short circuit), and every current source, by 0 A (or an open circuit). This way we obtain a simpler and more, manageable circuit., 2. Dependent sources are left intact because they are controlled by, circuit variables., , Other terms such as killed, made inactive, deadened, or set equal to zero, are often used to convey the same, idea., , With these in mind, we apply the superposition principle in three, steps:, , Steps to Apply Superposition Principle:, 1. Turn off all independent sources except one source. Find the, output (voltage or current) due to that active source using, the techniques covered in Chapters 2 and 3., 2. Repeat step 1 for each of the other independent sources., 3. Find the total contribution by adding algebraically all the, contributions due to the independent sources., Analyzing a circuit using superposition has one major disadvantage: it may very likely involve more work. If the circuit has three independent sources, we may have to analyze three simpler circuits each, providing the contribution due to the respective individual source., However, superposition does help reduce a complex circuit to simpler, circuits through replacement of voltage sources by short circuits and, of current sources by open circuits., Keep in mind that superposition is based on linearity. For this, reason, it is not applicable to the effect on power due to each source,, because the power absorbed by a resistor depends on the square of, the voltage or current. If the power value is needed, the current, through (or voltage across) the element must be calculated first using, superposition., , Example 4.3, , Use the superposition theorem to find v in the circuit of Fig. 4.6., , 8Ω, , Solution:, Since there are two sources, let, v v1 v2, , 6V, , where v1 and v2 are the contributions due to the 6-V voltage source, and the 3-A current source, respectively. To obtain v1, we set the current, source to zero, as shown in Fig. 4.7(a). Applying KVL to the loop in, Fig. 4.7(a) gives, 12i1 6 0, , 1, , i1 0.5 A, , +, −, , Figure 4.6, For Example 4.3., , 4Ω, , +, v, −, , 3A
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ale29559_ch04.qxd, , 07/08/2008, , 10:55 AM, , Page 132, , Chapter 4, , 132, 8Ω, , 6V, , +, −, , Circuit Theorems, , Thus,, 4Ω, , i1, , v1 4i1 2 V, , +, v1, −, , We may also use voltage division to get v1 by writing, v1 , , (a), 8Ω, , To get v2, we set the voltage source to zero, as in Fig. 4.7(b). Using, current division,, , i2, i3, +, v2, −, , 4Ω, , 4, (6) 2 V, 48, , i3 , , 3A, , 8, (3) 2 A, 48, , Hence,, v2 4i3 8 V, , (b), , Figure 4.7, , And we find, , For Example 4.3: (a) calculating v1,, (b) calculating v2., , v v1 v2 2 8 10 V, , Practice Problem 4.3, 5Ω, , 3Ω, +, vo, −, , Using the superposition theorem, find vo in the circuit of Fig. 4.8., , 2Ω, , 4A, , Answer: 6 V., +, −, , 10 V, , Figure 4.8, For Practice Prob. 4.3., , Example 4.4, , Find io in the circuit of Fig. 4.9 using superposition., 2Ω, , 3Ω, , Solution:, The circuit in Fig. 4.9 involves a dependent source, which must be left, intact. We let, io i¿o i–o, , 5io, , 1Ω, , +−, , 4A, io, 5Ω, , 4Ω, +−, , where i¿o and i–o are due to the 4-A current source and 20-V voltage, source respectively. To obtain i¿o, we turn off the 20-V source so that, we have the circuit in Fig. 4.10(a). We apply mesh analysis in order to, obtain i¿o. For loop 1,, , 20 V, , Figure 4.9, For Example 4.4., , (4.4.1), , i1 4 A, , (4.4.2), , 3i1 6i2 1i3 5i¿o 0, , (4.4.3), , For loop 2,
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ale29559_ch04.qxd, , 07/08/2008, , 10:55 AM, , Page 133, , 4.3, , Superposition, , 133, 2Ω, , 2Ω, , 3Ω, , i1, , 5io′, , 1Ω, , i o′, , i1, , i3, , 5i o′′, , 1Ω, , +−, , 4A, 5Ω, , i4, , 3Ω, , i2, , +−, , i o′′, i5, , 5Ω, , 4Ω, , i3, , +−, 20 V, , 0, (a), , (b), , Figure 4.10, For Example 4.4: Applying superposition to (a) obtain i¿o, (b) obtain i–o., , For loop 3,, 5i1 1i2 10i3 5i¿o 0, , (4.4.4), , i3 i1 i¿o 4 i¿o, , (4.4.5), , But at node 0,, , Substituting Eqs. (4.4.2) and (4.4.5) into Eqs. (4.4.3) and (4.4.4) gives, two simultaneous equations, 3i2 2i¿o 8, , (4.4.6), , i2 5i¿o 20, , (4.4.7), , which can be solved to get, i¿o , , 52, A, 17, , (4.4.8), , To obtain i–o, we turn off the 4-A current source so that the circuit, becomes that shown in Fig. 4.10(b). For loop 4, KVL gives, 6i4 i5 5i–o 0, , (4.4.9), , i4 10i5 20 5i–o 0, , (4.4.10), , and for loop 5,, But i5 i–o. Substituting this in Eqs. (4.4.9) and (4.4.10) gives, 6i4 4i–o 0, , (4.4.11), , i4 5i–o 20, , (4.4.12), , which we solve to get, i–o , , 60, A, 17, , (4.4.13), , Now substituting Eqs. (4.4.8) and (4.4.13) into Eq. (4.4.1) gives, io , , 8, 0.4706 A, 17, , 4Ω
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ale29559_ch04.qxd, , 07/08/2008, , 10:55 AM, , Page 135, , 4.4, , Source Transformation, , 135, , 8Ω, 4Ω, , 3Ω, , 4Ω, , i1, , i1, +, −, , 12 V, , 3Ω, , 12 V, , +, −, , 3Ω, , (a), 24 V, , 8Ω, , +−, 4Ω, , ia, , ib, , 8Ω, , 4Ω, , 4Ω, , 4Ω, , v1, , v2, , i2, , i3, , 3Ω, , 3Ω, , (b), , 3A, , (c), , Figure 4.13, For Example 4.5., , Find I in the circuit of Fig. 4.14 using the superposition principle., , 6Ω, , 16 V, , +, −, , 2Ω, , I, , 8Ω, , 4A, , + 12 V, −, , Figure 4.14, For Practice Prob. 4.5., , Answer: 0.75 A., , 4.4, , Source Transformation, , We have noticed that series-parallel combination and wye-delta transformation help simplify circuits. Source transformation is another tool, for simplifying circuits. Basic to these tools is the concept of equivalence. We recall that an equivalent circuit is one whose v-i characteristics are identical with the original circuit., In Section 3.6, we saw that node-voltage (or mesh-current) equations can be obtained by mere inspection of a circuit when the sources, are all independent current (or all independent voltage) sources. It is, therefore expedient in circuit analysis to be able to substitute a voltage, source in series with a resistor for a current source in parallel with a, , Practice Problem 4.5
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ale29559_ch04.qxd, , 136, , 07/08/2008, , 10:55 AM, , Page 136, , Chapter 4, , Circuit Theorems, , resistor, or vice versa, as shown in Fig. 4.15. Either substitution is, known as a source transformation., R, a, , a, vs, , +, −, , is, , R, , b, , b, , Figure 4.15, Transformation of independent sources., A source transformation is the process of replacing a voltage source, vs in series with a resistor R by a current source is in parallel with a resistor R, or vice versa., , The two circuits in Fig. 4.15 are equivalent—provided they have the, same voltage-current relation at terminals a-b. It is easy to show that, they are indeed equivalent. If the sources are turned off, the equivalent, resistance at terminals a-b in both circuits is R. Also, when terminals, a-b are short-circuited, the short-circuit current flowing from a to b is, isc vsR in the circuit on the left-hand side and isc is for the circuit, on the right-hand side. Thus, vsR is in order for the two circuits to, be equivalent. Hence, source transformation requires that, vs is R, , is , , or, , vs, R, , (4.5), , Source transformation also applies to dependent sources, provided, we carefully handle the dependent variable. As shown in Fig. 4.16, a, dependent voltage source in series with a resistor can be transformed, to a dependent current source in parallel with the resistor or vice versa, where we make sure that Eq. (4.5) is satisfied., R, a, vs, , +, −, , a, is, , b, , R, b, , Figure 4.16, Transformation of dependent sources., , Like the wye-delta transformation we studied in Chapter 2, a, source transformation does not affect the remaining part of the circuit., When applicable, source transformation is a powerful tool that allows, circuit manipulations to ease circuit analysis. However, we should keep, the following points in mind when dealing with source transformation., 1. Note from Fig. 4.15 (or Fig. 4.16) that the arrow of the current source, is directed toward the positive terminal of the voltage source., 2. Note from Eq. (4.5) that source transformation is not possible when, R 0, which is the case with an ideal voltage source. However, for, a practical, nonideal voltage source, R 0. Similarly, an ideal current source with R cannot be replaced by a finite voltage source., More will be said on ideal and nonideal sources in Section 4.10.1.
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ale29559_ch04.qxd, , 07/08/2008, , 10:55 AM, , Page 137, , 4.4, , Source Transformation, , 137, , Use source transformation to find vo in the circuit of Fig. 4.17., , Example 4.6, , Solution:, We first transform the current and voltage sources to obtain the circuit, in Fig. 4.18(a). Combining the 4- and 2- resistors in series and, transforming the 12-V voltage source gives us Fig. 4.18(b). We now, combine the 3- and 6- resistors in parallel to get 2-. We also, combine the 2-A and 4-A current sources to get a 2-A source. Thus,, by repeatedly applying source transformations, we obtain the circuit in, Fig. 4.18(c)., , 2Ω, , 3A, , 8Ω, , +, vo, −, , + 12 V, −, , Figure 4.17, For Example 4.6., , 2Ω, , 4Ω, −, +, , 12 V, , 4Ω, , 3Ω, , +, vo, −, , 8Ω, , 3Ω, , 4A, , (a), , 6Ω, , 2A, , i, , +, vo, −, , 8Ω, , 3Ω, , 4A, , 8Ω, , (b), , +, vo, −, , 2Ω, , 2A, , (c), , Figure 4.18, For Example 4.6., , We use current division in Fig. 4.18(c) to get, i, , 2, (2) 0.4 A, 28, , and, vo 8i 8(0.4) 3.2 V, Alternatively, since the 8- and 2- resistors in Fig. 4.18(c) are, in parallel, they have the same voltage vo across them. Hence,, vo (8 2)(2 A) , , 82, (2) 3.2 V, 10, , Find io in the circuit of Fig. 4.19 using source transformation., 5V, , 1Ω, , −+, 6Ω, , 5A, , 3Ω, , Figure 4.19, For Practice Prob. 4.6., , Answer: 1.78 A., , io, 7Ω, , 3A, , 4Ω, , Practice Problem 4.6
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ale29559_ch04.qxd, , 07/08/2008, , 10:55 AM, , Page 138, , Chapter 4, , 138, , Example 4.7, , Find vx in Fig. 4.20 using source transformation., 4Ω, 0.25vx, , 2Ω, , 6V, , +, −, , +, vx, −, , 2Ω, , Circuit Theorems, , + 18 V, −, , Figure 4.20, For Example 4.7., , Solution:, The circuit in Fig. 4.20 involves a voltage-controlled dependent current, source. We transform this dependent current source as well as the 6-V, independent voltage source as shown in Fig. 4.21(a). The 18-V voltage, source is not transformed because it is not connected in series with any, resistor. The two 2- resistors in parallel combine to give a 1-, resistor, which is in parallel with the 3-A current source. The current, source is transformed to a voltage source as shown in Fig. 4.21(b)., Notice that the terminals for vx are intact. Applying KVL around the, loop in Fig. 4.21(b) gives, 3 5i vx 18 0, , vx, , 4Ω, , 3A, , 2Ω, , 2Ω, , 1Ω, , +−, , (4.7.1), , vx, , 4Ω, , +−, , +, , +, vx, −, , + 18 V, −, , 3V +, −, , vx, , i, , + 18 V, −, , −, (b), , (a), , Figure 4.21, For Example 4.7: Applying source transformation to the circuit in Fig. 4.20., , Applying KVL to the loop containing only the 3-V voltage source, the, 1- resistor, and vx yields, 3 1i vx 0, , vx 3 i, , 1, , (4.7.2), , Substituting this into Eq. (4.7.1), we obtain, 15 5i 3 i 0, , 1, , i 4.5 A, , Alternatively, we may apply KVL to the loop containing vx, the 4-, resistor, the voltage-controlled dependent voltage source, and the 18-V, voltage source in Fig. 4.21(b). We obtain, vx 4i vx 18 0, , 1, , i 4.5 A, , Thus, vx 3 i 7.5 V., , Practice Problem 4.7, , Use source transformation to find ix in the circuit shown in Fig. 4.22., , 5Ω, , Answer: 7.056 mA., , ix, 24 mA, , 10 Ω, , Figure 4.22, For Practice Prob. 4.7., , –, +, , 2ix
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ale29559_ch04.qxd, , 07/08/2008, , 10:55 AM, , Page 139, , 4.5, , 4.5, , Thevenin’s Theorem, , 139, I, , Thevenin’s Theorem, , It often occurs in practice that a particular element in a circuit is variable (usually called the load) while other elements are fixed. As a typical example, a household outlet terminal may be connected to different, appliances constituting a variable load. Each time the variable element, is changed, the entire circuit has to be analyzed all over again. To avoid, this problem, Thevenin’s theorem provides a technique by which the, fixed part of the circuit is replaced by an equivalent circuit., According to Thevenin’s theorem, the linear circuit in Fig. 4.23(a), can be replaced by that in Fig. 4.23(b). (The load in Fig. 4.23 may be, a single resistor or another circuit.) The circuit to the left of the terminals a-b in Fig. 4.23(b) is known as the Thevenin equivalent circuit;, it was developed in 1883 by M. Leon Thevenin (1857–1926), a French, telegraph engineer., Thevenin’s theorem states that a linear two-terminal circuit can be, replaced by an equivalent circuit consisting of a voltage source VTh in, series with a resistor RTh, where VTh is the open-circuit voltage at the, terminals and RTh is the input or equivalent resistance at the terminals, when the independent sources are turned off., , The proof of the theorem will be given later, in Section 4.7. Our, major concern right now is how to find the Thevenin equivalent voltage VTh and resistance RTh. To do so, suppose the two circuits in, Fig. 4.23 are equivalent. Two circuits are said to be equivalent if they, have the same voltage-current relation at their terminals. Let us find, out what will make the two circuits in Fig. 4.23 equivalent. If the terminals a-b are made open-circuited (by removing the load), no current, flows, so that the open-circuit voltage across the terminals a-b in, Fig. 4.23(a) must be equal to the voltage source VTh in Fig. 4.23(b),, since the two circuits are equivalent. Thus VTh is the open-circuit voltage across the terminals as shown in Fig. 4.24(a); that is,, VTh voc, Linear, two-terminal, circuit, , a, +, voc, −, b, , (4.6), , Linear circuit with, all independent, sources set equal, to zero, , V Th = voc, , RTh = R in, , (a), , (b), , a, R in, b, , Figure 4.24, Finding VTh and RTh., , Again, with the load disconnected and terminals a-b opencircuited, we turn off all independent sources. The input resistance, (or equivalent resistance) of the dead circuit at the terminals a-b in, Fig. 4.23(a) must be equal to RTh in Fig. 4.23(b) because the two circuits, are equivalent. Thus, RTh is the input resistance at the terminals when the, independent sources are turned off, as shown in Fig. 4.24(b); that is,, RTh Rin, , (4.7), , a, +, V, −, , Linear, two-terminal, circuit, , Load, , b, (a), R Th, , VTh, , I, , a, +, V, −, , +, −, , Load, , b, (b), , Figure 4.23, Replacing a linear two-terminal circuit, by its Thevenin equivalent: (a) original, circuit, (b) the Thevenin equivalent, circuit.
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 140, , Chapter 4, , 140, , Circuit with, all independent, sources set equal, to zero, RTh, , vo, =, io, , To apply this idea in finding the Thevenin resistance RTh, we need, to consider two cases., , io, , a, , +, −, , vo, , ■ CASE 1 If the network has no dependent sources, we turn off all, independent sources. RTh is the input resistance of the network looking between terminals a and b, as shown in Fig. 4.24(b)., , b, , ■ CASE 2 If the network has dependent sources, we turn off all, , (a), , independent sources. As with superposition, dependent sources are not, to be turned off because they are controlled by circuit variables. We, apply a voltage source vo at terminals a and b and determine the resulting current io. Then RTh voio, as shown in Fig. 4.25(a). Alternatively, we may insert a current source io at terminals a-b as shown in, Fig. 4.25(b) and find the terminal voltage vo. Again RTh voio. Either, of the two approaches will give the same result. In either approach we, may assume any value of vo and io. For example, we may use vo 1 V, or io 1 A, or even use unspecified values of vo or io., , a, Circuit with, all independent, sources set equal, to zero, RTh =, , vo, io, , Circuit Theorems, , +, vo, −, , io, , b, , (b), , Figure 4.25, Finding RTh when circuit has dependent, sources., Later we will see that an alternative way, of finding RTh is RTh vocisc., a, IL, Linear, circuit, , RL, , b, (a), R Th, , a, IL, , VTh, , +, −, , RL, , b, , VL RLIL , , (b), , Figure 4.26, A circuit with a load: (a) original circuit,, (b) Thevenin equivalent., , 32 V +, −, , 12 Ω, , 1Ω, , RL, b, , Figure 4.27, For Example 4.8., , (4.8b), , Find the Thevenin equivalent circuit of the circuit shown in Fig. 4.27, to, the left of the terminals a-b. Then find the current through RL 6, 16,, and 36 ., , a, , 2A, , RL, VTh, RTh RL, , Note from Fig. 4.26(b) that the Thevenin equivalent is a simple voltage divider, yielding VL by mere inspection., , Example 4.8, 4Ω, , It often occurs that RTh takes a negative value. In this case, the, negative resistance (v iR) implies that the circuit is supplying, power. This is possible in a circuit with dependent sources; Example 4.10, will illustrate this., Thevenin’s theorem is very important in circuit analysis. It helps, simplify a circuit. A large circuit may be replaced by a single independent voltage source and a single resistor. This replacement technique, is a powerful tool in circuit design., As mentioned earlier, a linear circuit with a variable load can be, replaced by the Thevenin equivalent, exclusive of the load. The equivalent network behaves the same way externally as the original circuit., Consider a linear circuit terminated by a load RL, as shown in Fig. 4.26(a)., The current IL through the load and the voltage VL across the load are, easily determined once the Thevenin equivalent of the circuit at the, load’s terminals is obtained, as shown in Fig. 4.26(b). From Fig. 4.26(b),, we obtain, VTh, IL , (4.8a), RTh RL, , Solution:, We find RTh by turning off the 32-V voltage source (replacing it, with a short circuit) and the 2-A current source (replacing it with an
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 141, , 4.5, , Thevenin’s Theorem, , 141, , open circuit). The circuit becomes what is shown in Fig. 4.28(a)., Thus,, RTh 4 12 1 , , 4Ω, , 4 12, 14, 16, , 1Ω, , 4Ω, , 1Ω, , VTh, , a, , a, +, , R Th, , 12 Ω, , 32 V, , +, −, , i1, , 12 Ω, , i2, , 2A, , VTh, −, , b, (a), , b, , (b), , Figure 4.28, For Example 4.8: (a) finding RTh, (b) finding VTh., , To find VTh, consider the circuit in Fig. 4.28(b). Applying mesh, analysis to the two loops, we obtain, 32 4i1 12(i1 i2) 0,, , i2 2 A, , Solving for i1, we get i1 0.5 A. Thus,, VTh 12(i1 i2) 12(0.5 2.0) 30 V, Alternatively, it is even easier to use nodal analysis. We ignore the, 1- resistor since no current flows through it. At the top node, KCL, gives, 32 VTh, VTh, 2, 4, 12, or, 96 3VTh 24 VTh, , 1, , VTh 30 V, , as obtained before. We could also use source transformation to find VTh., The Thevenin equivalent circuit is shown in Fig. 4.29. The current, through RL is, IL , , 4Ω, , IL, 30 V, , +, −, , VTh, 30, , RTh RL, 4 RL, , When RL 6,, , a, , RL, , b, , Figure 4.29, 30, IL , 3A, 10, , When RL 16,, IL , , 30, 1.5 A, 20, , IL , , 30, 0.75 A, 40, , When RL 36,, , The Thevenin equivalent circuit for, Example 4.8.
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 143, , 4.5, , Thevenin’s Theorem, , 143, , Solving these equations gives, 1, i3 A, 6, But io i3 16 A. Hence,, RTh , , 1V, 6, io, , To get VTh, we find voc in the circuit of Fig. 4.32(b). Applying, mesh analysis, we get, i1 5, , (4.9.4), , 2vx 2(i3 i2) 0, 1, vx i3 i2, 4(i2 i1) 2(i2 i3) 6i2 0, , (4.9.5), , 12i2 4i1 2i3 0, , (4.9.6), , or, , 6Ω, a, , But 4(i1 i2) vx. Solving these equations leads to i2 103., Hence,, VTh voc 6i2 20 V, The Thevenin equivalent is as shown in Fig. 4.33., , 20 V, , +, −, b, , Figure 4.33, The Thevenin equivalent of the circuit in, Fig. 4.31., , Practice Problem 4.9, , Find the Thevenin equivalent circuit of the circuit in Fig. 4.34 to the, left of the terminals., , 5Ω, , Answer: VTh 5.33 V, RTh 0.44 ., , Ix, , 3Ω, a, , 6V, , +, −, , 1.5Ix, , 4Ω, b, , Figure 4.34, For Practice Prob. 4.9., , Determine the Thevenin equivalent of the circuit in Fig. 4.35(a) at, terminals a-b., Solution:, 1. Define. The problem is clearly defined; we are to determine the, Thevenin equivalent of the circuit shown in Fig. 4.35(a)., 2. Present. The circuit contains a 2- resistor in parallel with a, 4- resistor. These are, in turn, in parallel with a dependent, current source. It is important to note that there are no, independent sources., 3. Alternative. The first thing to consider is that, since we have no, independent sources in this circuit, we must excite the circuit, externally. In addition, when you have no independent, sources you will not have a value for VTh; you will only have, to find RTh., , Example 4.10
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 144, , Chapter 4, , 144, a, ix, 4Ω, , 2ix, , 2Ω, b, (a), vo, , a, ix, , 4Ω, , 2ix, , 2Ω, , io, , (b), , ix (0 vo)2 vo2, , ix, 8ix, , −, +, , 2Ω, , i1, , i2, , + 10 V, −, , (c), a, , b, (d), , Figure 4.35, For Example 4.10., , 9Ω, , + 10 V, −, , i, , (4.10.2), , Substituting Eq. (4.10.2) into Eq. (4.10.1) yields, 2(vo2) (vo 0)4 (vo 0)2 (1) 0, or, vo 4 V, (1 14 12)vo 1, , b, , −4 Ω, , (4.10.1), , Since we have two unknowns and only one equation, we will, need a constraint equation., , 9Ω, , a, , The simplest approach is to excite the circuit with either a, 1-V voltage source or a 1-A current source. Since we will end, up with an equivalent resistance (either positive or negative), I, prefer to use the current source and nodal analysis which will, yield a voltage at the output terminals equal to the resistance, (with 1 A flowing in, vo is equal to 1 times the equivalent, resistance)., As an alternative, the circuit could also be excited by a 1-V, voltage source and mesh analysis could be used to find the, equivalent resistance., 4. Attempt. We start by writing the nodal equation at a in Fig. 4.35(b), assuming io 1 A., 2ix (vo 0)4 (vo 0)2 (1) 0, , b, , 4Ω, , Circuit Theorems, , Since vo 1 RTh, then RTh vo1 4 ., The negative value of the resistance tells us that, according, to the passive sign convention, the circuit in Fig. 4.35(a) is, supplying power. Of course, the resistors in Fig. 4.35(a) cannot, supply power (they absorb power); it is the dependent source, that supplies the power. This is an example of how a, dependent source and resistors could be used to simulate, negative resistance., 5. Evaluate. First of all, we note that the answer has a negative, value. We know this is not possible in a passive circuit, but in, this circuit we do have an active device (the dependent current, source). Thus, the equivalent circuit is essentially an active, circuit that can supply power., Now we must evaluate the solution. The best way to do this, is to perform a check, using a different approach, and see if we, obtain the same solution. Let us try connecting a 9- resistor in, series with a 10-V voltage source across the output terminals of, the original circuit and then the Thevenin equivalent. To make, the circuit easier to solve, we can take and change the parallel, current source and 4- resistor to a series voltage source and, 4- resistor by using source transformation. This, with the new, load, gives us the circuit shown in Fig. 4.35(c)., We can now write two mesh equations., 8ix 4i1 2(i1 i2) 0, 2(i2 i1) 9i2 10 0, Note, we only have two equations but have 3 unknowns, so we, need a constraint equation. We can use, ix i2 i1
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 145, , 4.6, , Norton’s Theorem, , 145, , This leads to a new equation for loop 1. Simplifying leads to, (4 2 8)i1 (2 8)i2 0, or, 2i1 6i2 0, or, i1 3i2, 2i1 11i2 10, Substituting the first equation into the second gives, 6i2 11i2 10, , or, , i2 105 2 A, , Using the Thevenin equivalent is quite easy since we have only, one loop, as shown in Fig. 4.35(d)., 4i 9i 10 0, , or, , i 105 2 A, , 6. Satisfactory? Clearly we have found the value of the equivalent, circuit as required by the problem statement. Checking does, validate that solution (we compared the answer we obtained by, using the equivalent circuit with one obtained by using the load, with the original circuit). We can present all this as a solution to, the problem., , Practice Problem 4.10, , Obtain the Thevenin equivalent of the circuit in Fig. 4.36., Answer: VTh 0 V, RTh 7.5 ., , 10 Ω, , 4vx, +−, , a, , +, 5Ω, , vx, −, , 4.6, , b, , Norton’s Theorem, , Figure 4.36, , In 1926, about 43 years after Thevenin published his theorem, E. L., Norton, an American engineer at Bell Telephone Laboratories, proposed a similar theorem., , For Practice Prob. 4.10., , Norton’s theorem states that a linear two-terminal circuit can be, replaced by an equivalent circuit consisting of a current source IN in, parallel with a resistor RN, where IN is the short-circuit current through, the terminals and RN is the input or equivalent resistance at the terminals when the independent sources are turned off., , Thus, the circuit in Fig. 4.37(a) can be replaced by the one in Fig. 4.37(b)., The proof of Norton’s theorem will be given in the next section., For now, we are mainly concerned with how to get RN and IN. We find, RN in the same way we find RTh. In fact, from what we know about, source transformation, the Thevenin and Norton resistances are equal;, that is,, RN RTh, , 15 Ω, , Linear, two-terminal, circuit, , a, b, , (a), a, IN, , RN, b, , (4.9), (b), , To find the Norton current IN, we determine the short-circuit current, flowing from terminal a to b in both circuits in Fig. 4.37. It is evident, , Figure 4.37, (a) Original circuit, (b) Norton equivalent, circuit.
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 146, , Chapter 4, , 146, , that the short-circuit current in Fig. 4.37(b) is IN. This must be the same, short-circuit current from terminal a to b in Fig. 4.37(a), since the two, circuits are equivalent. Thus,, , a, Linear, two-terminal, circuit, , Circuit Theorems, , isc = IN, , IN isc, , b, , (4.10), , shown in Fig. 4.38. Dependent and independent sources are treated the, same way as in Thevenin’s theorem., Observe the close relationship between Norton’s and Thevenin’s, theorems: RN RTh as in Eq. (4.9), and, , Figure 4.38, Finding Norton current IN., , IN , , VTh, RTh, , (4.11), , This is essentially source transformation. For this reason, source transformation is often called Thevenin-Norton transformation., Since VTh, IN, and RTh are related according to Eq. (4.11), to determine the Thevenin or Norton equivalent circuit requires that we find:, , The Thevenin and Norton equivalent, circuits are related by a source, transformation., , • The open-circuit voltage voc across terminals a and b., • The short-circuit current isc at terminals a and b., • The equivalent or input resistance Rin at terminals a and b when, all independent sources are turned off., We can calculate any two of the three using the method that takes the, least effort and use them to get the third using Ohm’s law. Example 4.11, will illustrate this. Also, since, VTh voc, , (4.12a), , IN isc, , (4.12b), , RTh , , voc, RN, isc, , (4.12c), , the open-circuit and short-circuit tests are sufficient to find any Thevenin, or Norton equivalent, of a circuit which contains at least one independent source., , Example 4.11, , Find the Norton equivalent circuit of the circuit in Fig. 4.39 at, terminals a-b., , 8Ω, a, 4Ω, 5Ω, , 2A, + 12 V, −, , b, 8Ω, , Solution:, We find RN in the same way we find RTh in the Thevenin equivalent, circuit. Set the independent sources equal to zero. This leads to the, circuit in Fig. 4.40(a), from which we find RN. Thus,, RN 5 (8 4 8) 5 20 , , Figure 4.39, For Example 4.11., , 20 5, 4, 25, , To find IN, we short-circuit terminals a and b, as shown in Fig. 4.40(b)., We ignore the 5- resistor because it has been short-circuited., Applying mesh analysis, we obtain, i1 2 A,, , 20i2 4i1 12 0, , From these equations, we obtain, i2 1 A isc IN
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 148, , Chapter 4, , 148, , Example 4.12, , Circuit Theorems, , Using Norton’s theorem, find RN and IN of the circuit in Fig. 4.43 at, terminals a-b., , 2 ix, , Solution:, To find RN, we set the independent voltage source equal to zero and, connect a voltage source of vo 1 V (or any unspecified voltage vo), to the terminals. We obtain the circuit in Fig. 4.44(a). We ignore the, 4- resistor because it is short-circuited. Also due to the short circuit,, the 5- resistor, the voltage source, and the dependent current source, 1v, 0.2 A, and, are all in parallel. Hence, ix 0. At node a, io 5, , 5Ω, ix, , a, + 10 V, −, , 4Ω, , b, , Figure 4.43, RN , , For Example 4.12., , vo, 1, , 5, io, 0.2, , To find IN, we short-circuit terminals a and b and find the current, isc, as indicated in Fig. 4.44(b). Note from this figure that the 4-, resistor, the 10-V voltage source, the 5- resistor, and the dependent, current source are all in parallel. Hence,, ix , , 10, 2.5 A, 4, , At node a, KCL gives, isc , , 10, 2ix 2 2(2.5) 7 A, 5, , Thus,, IN 7 A, 2ix, , 2ix, , 5Ω, , 5Ω, , a, , ix, , io, +, −, , 4Ω, , vo = 1 V, , a, , ix, 4Ω, , isc = IN, , + 10 V, −, , b, (a), , b, (b), , Figure 4.44, For Example 4.12: (a) finding RN, (b) finding IN., , Practice Problem 4.12, , Find the Norton equivalent circuit of the circuit in Fig. 4.45 at, terminals a-b., , 2vx, + −, 6Ω, , 10 A, , a, 2Ω, , +, vx, −, b, , Figure 4.45, For Practice Prob. 4.12., , Answer: RN 1 , IN 10 A.
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , 4.7, , Page 149, , Derivations of Thevenin’s and Norton’s Theorems, , 149, a, , 4.7, , Derivations of Thevenin’s and, Norton’s Theorems, , In this section, we will prove Thevenin’s and Norton’s theorems using, the superposition principle., Consider the linear circuit in Fig. 4.46(a). It is assumed that the, circuit contains resistors, and dependent and independent sources. We, have access to the circuit via terminals a and b, through which current, from an external source is applied. Our objective is to ensure that the, voltage-current relation at terminals a and b is identical to that of the, Thevenin equivalent in Fig. 4.46(b). For the sake of simplicity, suppose the linear circuit in Fig. 4.46(a) contains two independent voltage, sources vs1 and vs2 and two independent current sources is1 and is2. We, may obtain any circuit variable, such as the terminal voltage v, by, applying superposition. That is, we consider the contribution due to, each independent source including the external source i. By superposition, the terminal voltage v is, v A 0 i A1vs1 A 2 vs2 A 3 i s1 A 4 i s2, , +, v, −, , i, , Linear, circuit, , b, (a), R Th, , a, +, i, , + V, Th, −, , v, −, b, (b), , Figure 4.46, Derivation of Thevenin equivalent: (a) a, current-driven circuit, (b) its Thevenin, equivalent., , (4.13), , where A0, A1, A2, A3, and A4 are constants. Each term on the right-hand, side of Eq. (4.13) is the contribution of the related independent source;, that is, A0i is the contribution to v due to the external current source i,, A1vs1 is the contribution due to the voltage source vs1, and so on. We, may collect terms for the internal independent sources together as B0,, so that Eq. (4.13) becomes, v A 0 i B0, , (4.14), , where B0 A1vs1 A 2 vs2 A3 i s1 A 4 i s2. We now want to evaluate the values of constants A0 and B0. When the terminals a and b are, open-circuited, i 0 and v B0. Thus, B0 is the open-circuit voltage, voc, which is the same as VTh, so, B0 VTh, , (4.15), , When all the internal sources are turned off, B0 0. The circuit can, then be replaced by an equivalent resistance Req, which is the same as, RTh, and Eq. (4.14) becomes, v A 0 i RThi, , 1, , A0 RTh, , (4.16), , i, v, , Linear, circuit, , +, −, , Substituting the values of A0 and B0 in Eq. (4.14) gives, v RTh i VTh, , b, , (4.17), , which expresses the voltage-current relation at terminals a and b of the, circuit in Fig. 4.46(b). Thus, the two circuits in Fig. 4.46(a) and 4.46(b), are equivalent., When the same linear circuit is driven by a voltage source v as, shown in Fig. 4.47(a), the current flowing into the circuit can be, obtained by superposition as, i C0 v D0, , a, , (a), i, , v, , +, −, , RN, , IN, , b, , (4.18), , where C0 v is the contribution to i due to the external voltage source v, and D0 contains the contributions to i due to all internal independent, sources. When the terminals a-b are short-circuited, v 0 so that, , a, , (b), , Figure 4.47, Derivation of Norton equivalent: (a) a, voltage-driven circuit, (b) its Norton, equivalent.
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 150, , Chapter 4, , 150, , Circuit Theorems, , i D0 isc, where isc is the short-circuit current flowing out of terminal a, which is the same as the Norton current IN, i.e.,, D0 IN, , (4.19), , When all the internal independent sources are turned off, D0 0 and, the circuit can be replaced by an equivalent resistance Req (or an equivalent conductance Geq 1Req), which is the same as RTh or RN. Thus, Eq. (4.19) becomes, i, , v, IN, RTh, , (4.20), , This expresses the voltage-current relation at terminals a-b of the circuit in Fig. 4.47(b), confirming that the two circuits in Fig. 4.47(a) and, 4.47(b) are equivalent., , 4.8, , RTh, , In many practical situations, a circuit is designed to provide power to, a load. There are applications in areas such as communications where, it is desirable to maximize the power delivered to a load. We now, address the problem of delivering the maximum power to a load when, given a system with known internal losses. It should be noted that this, will result in significant internal losses greater than or equal to the, power delivered to the load., The Thevenin equivalent is useful in finding the maximum power, a linear circuit can deliver to a load. We assume that we can adjust the, load resistance RL. If the entire circuit is replaced by its Thevenin, equivalent except for the load, as shown in Fig. 4.48, the power delivered to the load is, , a, i, , VTh +, −, , Maximum Power Transfer, , RL, , b, , Figure 4.48, The circuit used for maximum power, transfer., , p i 2RL a, p, , 2, VTh, b RL, RTh RL, , (4.21), , For a given circuit, VTh and RTh are fixed. By varying the load resistance RL, the power delivered to the load varies as sketched in Fig. 4.49., We notice from Fig. 4.49 that the power is small for small or large values of RL but maximum for some value of RL between 0 and . We, now want to show that this maximum power occurs when RL is equal, to RTh. This is known as the maximum power theorem., , pmax, , 0, , RTh, , RL, , Figure 4.49, Power delivered to the load as a function, of RL., , Maximum power is transferred to the load when the load resistance, equals the Thevenin resistance as seen from the load (RL RTh)., , To prove the maximum power transfer theorem, we differentiate p, in Eq. (4.21) with respect to RL and set the result equal to zero. We, obtain, dp, (RTh RL )2 2RL(RTh RL ), V 2Th c, d, dRL, (RTh RL )4, (RTh RL 2RL ), V 2Th c, d 0, (RTh RL )3
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 152, , Chapter 4, , 152, , Circuit Theorems, , To get VTh, we consider the circuit in Fig. 4.51(b). Applying mesh, analysis gives, 12 18i1 12i2 0,, , i2 2 A, , Solving for i1, we get i1 23. Applying KVL around the outer loop, to get VTh across terminals a-b, we obtain, 12 6i1 3i2 2(0) VTh 0, , 1, , VTh 22 V, , For maximum power transfer,, RL RTh 9 , and the maximum power is, pmax , , Practice Problem 4.13, 2Ω, , +, −, , Determine the value of RL that will draw the maximum power from, the rest of the circuit in Fig. 4.52. Calculate the maximum power., , 4Ω, , Answer: 4.22 , 2.901 W., , + vx −, 9V, , V 2Th, 222, , 13.44 W, 4RL, 49, , 1Ω, RL, +, −, , Figure 4.52, For Practice Prob. 4.13., , 3vx, , 4.9, , Verifying Circuit Theorems with PSpice, , In this section, we learn how to use PSpice to verify the theorems covered in this chapter. Specifically, we will consider using DC Sweep analysis to find the Thevenin or Norton equivalent at any pair of nodes in a, circuit and the maximum power transfer to a load. The reader is advised, to read Section D.3 of Appendix D in preparation for this section., To find the Thevenin equivalent of a circuit at a pair of open terminals using PSpice, we use the schematic editor to draw the circuit, and insert an independent probing current source, say, Ip, at the terminals. The probing current source must have a part name ISRC. We then, perform a DC Sweep on Ip, as discussed in Section D.3. Typically, we, may let the current through Ip vary from 0 to 1 A in 0.1-A increments., After saving and simulating the circuit, we use Probe to display a plot, of the voltage across Ip versus the current through Ip. The zero intercept of the plot gives us the Thevenin equivalent voltage, while the, slope of the plot is equal to the Thevenin resistance., To find the Norton equivalent involves similar steps except that we, insert a probing independent voltage source (with a part name VSRC),, say, Vp, at the terminals. We perform a DC Sweep on Vp and let Vp, vary from 0 to 1 V in 0.1-V increments. A plot of the current through, Vp versus the voltage across Vp is obtained using the Probe menu after, simulation. The zero intercept is equal to the Norton current, while the, slope of the plot is equal to the Norton conductance., To find the maximum power transfer to a load using PSpice, involves performing a DC parametric Sweep on the component value, of RL in Fig. 4.48 and plotting the power delivered to the load as a, function of RL. According to Fig. 4.49, the maximum power occurs
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 153, , 4.9, , Verifying Circuit Theorems with PSpice, , 153, , when RL RTh. This is best illustrated with an example, and Example 4.15 provides one., We use VSRC and ISRC as part names for the independent voltage and current sources, respectively., , Example 4.14, , Consider the circuit in Fig. 4.31 (see Example 4.9). Use PSpice to find, the Thevenin and Norton equivalent circuits., Solution:, (a) To find the Thevenin resistance RTh and Thevenin voltage VTh at, the terminals a-b in the circuit in Fig. 4.31, we first use Schematics to, draw the circuit as shown in Fig. 4.53(a). Notice that a probing current, source I2 is inserted at the terminals. Under Analysis/Setput, we select, DC Sweep. In the DC Sweep dialog box, we select Linear for the, Sweep Type and Current Source for the Sweep Var. Type. We enter I2, under the Name box, 0 as Start Value, 1 as End Value, and 0.1 as, Increment. After simulation, we add trace V(I2:–) from the PSpice A/D, window and obtain the plot shown in Fig. 4.53(b). From the plot, we, obtain, VTh Zero intercept 20 V,, , RTh Slope , , 26 20, 6, 1, , These agree with what we got analytically in Example 4.9., 26 V, , I1, , R4, , 4, , E1, + +, −, −, GAIN=2, , R2, , R4, , 2, , 2, , R3, , 6, , 24 V, , I2, , 0, , 22 V, , 20 V, 0 A, 0.2 A, = V(I2:_), , (a), , Figure 4.53, For Example 4.14: (a) schematic and (b) plot for finding RTh and VTh., , (b) To find the Norton equivalent, we modify the schematic in Fig. 4.53(a), by replaying the probing current source with a probing voltage source, V1. The result is the schematic in Fig. 4.54(a). Again, in the DC Sweep, dialog box, we select Linear for the Sweep Type and Voltage Source, for the Sweep Var. Type. We enter V1 under Name box, 0 as Start Value,, 1 as End Value, and 0.1 as Increment. Under the PSpice A/D Window,, we add trace I (V1) and obtain the plot in Fig. 4.54(b). From the plot,, we obtain, IN Zero intercept 3.335 A, 3.335 3.165, 0.17 S, GN Slope , 1, , 0.4 A, (b), , 0.6 A, , 0.8 A, , 1.0 A
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 154, , Chapter 4, , 154, , Circuit Theorems, 3.4 A, , I1, , R4, , 4, , R2, , R1, , 2, , 2, , E1, + +, −, −, GAIN=2, , R3, , 6, , 3.3 A, V1 +, −, , 3.2 A, , 3.1 A, 0 V, 0, , 0.2 V, I(V1), , 0.4 V, 0.6 V, V_V1, , 0.8 V, , 1.0 V, , (b), , (a), , Figure 4.54, For Example 4.14: (a) schematic and (b) plot for finding GN and IN., , Practice Problem 4.14, , Rework Practice Prob. 4.9 using PSpice., Answer: VTh 5.33 V, RTh 0.44 ., , Example 4.15, , Refer to the circuit in Fig. 4.55. Use PSpice to find the maximum, power transfer to RL., , 1 kΩ, , 1V, , +, −, , Solution:, We need to perform a DC Sweep on RL to determine when the power, across it is maximum. We first draw the circuit using Schematics as, shown in Fig. 4.56. Once the circuit is drawn, we take the following, three steps to further prepare the circuit for a DC Sweep., The first step involves defining the value of RL as a parameter,, since we want to vary it. To do this:, , RL, , Figure 4.55, For Example 4.15., , 1. DCLICKL the value 1k of R2 (representing RL) to open up the, Set Attribute Value dialog box., 2. Replace 1k with {RL} and click OK to accept the change., , PARAMETERS:, RL, 2k, R1, , Note that the curly brackets are necessary., The second step is to define parameter. To achieve this:, , 1k, V1, DC=1 V, , +, −, , R2, , {RL}, , 0, , Figure 4.56, Schematic for the circuit in Fig. 4.55., , Select Draw/Get New Part/Libraries p /special.slb., Type PARAM in the PartName box and click OK., DRAG the box to any position near the circuit., CLICKL to end placement mode., DCLICKL to open up the PartName: PARAM dialog box., CLICKL on NAME1 and enter RL (with no curly brackets), in the Value box, and CLICKL Save Attr to accept change., 7. CLICKL on VALUE1 and enter 2k in the Value box, and, CLICKL Save Attr to accept change., 8. Click OK., , 1., 2., 3., 4., 5., 6.
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 155, , 4.10, , Applications, , The value 2k in item 7 is necessary for a bias point calculation; it, cannot be left blank., The third step is to set up the DC Sweep to sweep the parameter., To do this:, 1. Select Analysis/Setput to bring up the DC Sweep dialog box., 2. For the Sweep Type, select Linear (or Octave for a wide range, of RL)., 3. For the Sweep Var. Type, select Global Parameter., 4. Under the Name box, enter RL., 5. In the Start Value box, enter 100., 6. In the End Value box, enter 5k., 7. In the Increment box, enter 100., 8. Click OK and Close to accept the parameters., After taking these steps and saving the circuit, we are ready to, simulate. Select Analysis/Simulate. If there are no errors, we select, Add Trace in the PSpice A/D window and type V(R2:2)*I(R2) in, the Trace Command box. [The negative sign is needed since I(R2) is, negative.] This gives the plot of the power delivered to RL as RL varies, from 100 to 5 k. We can also obtain the power absorbed by RL by, typing V(R2:2)*V(R2:2)/RL in the Trace Command box. Either way,, we obtain the plot in Fig. 4.57. It is evident from the plot that the, maximum power is 250 mW. Notice that the maximum occurs when, RL 1 k, as expected analytically., , Find the maximum power transferred to RL if the 1-k resistor in, Fig. 4.55 is replaced by a 2-k resistor., , 155, 250 uW, , 200 uW, , 150 uW, , 100 uW, , 50 uW, 0, , 2.0 K, 4.0 K, –V(R2:2)*I(R2), RL, , 6.0 K, , Figure 4.57, For Example 4.15: the plot of power, across RL., , Practice Problem 4.15, , Answer: 125 mW., Rs, , 4.10, , Applications, , vs, , +, −, , In this section we will discuss two important practical applications of, the concepts covered in this chapter: source modeling and resistance, measurement., , (a), , 4.10.1 Source Modeling, Source modeling provides an example of the usefulness of the, Thevenin or the Norton equivalent. An active source such as a battery, is often characterized by its Thevenin or Norton equivalent circuit. An, ideal voltage source provides a constant voltage irrespective of the current drawn by the load, while an ideal current source supplies a constant current regardless of the load voltage. As Fig. 4.58 shows,, practical voltage and current sources are not ideal, due to their internal resistances or source resistances Rs and Rp. They become ideal as, Rs S 0 and Rp S . To show that this is the case, consider the effect, , Rp, , is, , (b), , Figure 4.58, (a) Practical voltage source, (b) practical, current source.
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 156, , Chapter 4, , 156, , Circuit Theorems, , of the load on voltage sources, as shown in Fig. 4.59(a). By the voltage division principle, the load voltage is, vL , , RL, vs, Rs RL, , (4.25), , As RL increases, the load voltage approaches a source voltage vs, as, illustrated in Fig. 4.59(b). From Eq. (4.25), we should note that:, 1. The load voltage will be constant if the internal resistance Rs of, the source is zero or, at least, Rs V RL. In other words, the, smaller Rs is compared with RL, the closer the voltage source is to, being ideal., vL, , Rs, , vs, , +, −, , Ideal source, , vs, , +, vL, , Practical source, , RL, , −, 0, , (a), , (b), , RL, , Figure 4.59, (a) Practical voltage source connected to a load RL, (b) load voltage decreases as RL decreases., , 2. When the load is disconnected (i.e., the source is open-circuited so, that RL S ), voc vs. Thus, vs may be regarded as the unloaded, source voltage. The connection of the load causes the terminal voltage to drop in magnitude; this is known as the loading effect., The same argument can be made for a practical current source when, connected to a load as shown in Fig. 4.60(a). By the current division, principle,, , IL, , Rp, , is, , RL, , IL, Ideal source, , Practical source, 0, , Rp, Rp RL, , is, , (4.26), , Figure 4.60(b) shows the variation in the load current as the load resistance increases. Again, we notice a drop in current due to the load (loading effect), and load current is constant (ideal current source) when the, internal resistance is very large (i.e., Rp S or, at least, Rp W RL)., Sometimes, we need to know the unloaded source voltage vs and, the internal resistance Rs of a voltage source. To find vs and Rs, we follow the procedure illustrated in Fig. 4.61. First, we measure the opencircuit voltage voc as in Fig. 4.61(a) and set, , (a), , is, , iL , , RL, (b), , Figure 4.60, (a) Practical current source connected to a, load RL, (b) load current decreases as RL, increases., , vs voc, , (4.27), , Then, we connect a variable load RL across the terminals as in, Fig. 4.61(b). We adjust the resistance RL until we measure a load voltage of exactly one-half of the open-circuit voltage, vL voc 2,, because now RL RTh Rs. At that point, we disconnect RL and, measure it. We set, Rs RL, , (4.28), , For example, a car battery may have vs 12 V and Rs 0.05 .
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 157, , 4.10, , +, Signal, source, , +, vL, , Signal, source, , voc, −, , Applications, , 157, , RL, , −, (b), , (a), , Figure 4.61, (a) Measuring voc, (b) measuring vL., , Example 4.16, , The terminal voltage of a voltage source is 12 V when connected to a, 2-W load. When the load is disconnected, the terminal voltage rises to, 12.4 V. (a) Calculate the source voltage vs and internal resistance Rs., (b) Determine the voltage when an 8- load is connected to the source., Solution:, (a) We replace the source by its Thevenin equivalent. The terminal, voltage when the load is disconnected is the open-circuit voltage,, vs voc 12.4 V, When the load is connected, as shown in Fig. 4.62(a), vL 12 V and, pL 2 W. Hence,, pL , , v2L, RL, , 1, , RL , , v2L, pL, , Rs, , 2, , , , 12, 72 , 2, , +, vs, , +, −, , vL, , RL, , −, , The load current is, iL , , vL, 12, 1, , A, RL, 72, 6, , (a), , The voltage across Rs is the difference between the source voltage vs, and the load voltage vL, or, 12.4 12 0.4 R s iL,, , Rs , , 0.4, 2.4 , IL, , 2.4 Ω, +, 12.4 V +, −, , v, , 8, (12.4) 9.538 V, 8 2.4, , The measured open-circuit voltage across a certain amplifier is 9 V., The voltage drops to 8 V when a 20- loudspeaker is connected to the, amplifier. Calculate the voltage when a 10- loudspeaker is used, instead., , v, , 8Ω, , −, , (b) Now that we have the Thevenin equivalent of the source, we, connect the 8- load across the Thevenin equivalent as shown in, Fig. 4.62(b). Using voltage division, we obtain, , Answer: 7.2 V., , iL, , (b), , Figure 4.62, For Example 4.16., , Practice Problem 4.16
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 158, , Chapter 4, , 158, , Circuit Theorems, , 4.10.2 Resistance Measurement, , Historical note: The bridge was, invented by Charles Wheatstone, (1802–1875), a British professor who, also invented the telegraph, as Samuel, Morse did independently in the, United States., , R1, v, , +, −, R2, , R3, , Galvanometer, +, v1, −, , Although the ohmmeter method provides the simplest way to measure, resistance, more accurate measurement may be obtained using the, Wheatstone bridge. While ohmmeters are designed to measure resistance in low, mid, or high range, a Wheatstone bridge is used to measure resistance in the mid range, say, between 1 and 1 M. Very low, values of resistances are measured with a milliohmmeter, while very, high values are measured with a Megger tester., The Wheatstone bridge (or resistance bridge) circuit is used in a, number of applications. Here we will use it to measure an unknown, resistance. The unknown resistance Rx is connected to the bridge as, shown in Fig. 4.63. The variable resistance is adjusted until no current, flows through the galvanometer, which is essentially a d’Arsonval, movement operating as a sensitive current-indicating device like an, ammeter in the microamp range. Under this condition v1 v2, and the, bridge is said to be balanced. Since no current flows through the galvanometer, R1 and R2 behave as though they were in series; so do R3, and Rx. The fact that no current flows through the galvanometer also, implies that v1 v2. Applying the voltage division principle,, v1 , , +, v2, −, , Rx, , Rx, R2, v v2 , v, R1 R2, R3 Rx, , (4.29), , Hence, no current flows through the galvanometer when, , Figure 4.63, , Rx, R2, , R1 R2, R3 Rx, , The Wheatstone bridge; Rx is the, resistance to be measured., , 1, , R2 R3 R1Rx, , or, Rx , , R3, R2, R1, , (4.30), , If R1 R3, and R2 is adjusted until no current flows through the galvanometer, then Rx R2., How do we find the current through the galvanometer when the, Wheatstone bridge is unbalanced? We find the Thevenin equivalent, (VTh and RTh) with respect to the galvanometer terminals. If Rm is the, resistance of the galvanometer, the current through it under the unbalanced condition is, VTh, I, (4.31), RTh Rm, Example 4.18 will illustrate this., , Example 4.17, , In Fig. 4.63, R1 500 and R3 200 . The bridge is balanced when, R2 is adjusted to be 125 . Determine the unknown resistance Rx., Solution:, Using Eq. (4.30) gives, Rx , , R3, 200, R2 , 125 50 , R1, 500
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 159, , 4.10, , Applications, , A Wheatstone bridge has R1 R3 1 k. R2 is adjusted until no current flows through the galvanometer. At that point, R2 3.2 k. What, is the value of the unknown resistance?, , 159, , Practice Problem 4.17, , Answer: 3.2 k., , The circuit in Fig. 4.64 represents an unbalanced bridge. If the galvanometer has a resistance of 40 , find the current through the, galvanometer., , 400 Ω, , 3 kΩ, 220 V, , 40 Ω, , a, , +, −, , b, G, 600 Ω, , 1 kΩ, , Figure 4.64, Unbalanced bridge of Example 4.18., , Solution:, We first need to replace the circuit by its Thevenin equivalent at, terminals a and b. The Thevenin resistance is found using the circuit, in Fig. 4.65(a). Notice that the 3-k and 1-k resistors are in parallel;, so are the 400- and 600- resistors. The two parallel combinations, form a series combination with respect to terminals a and b. Hence,, RTh 3000 1000 400 600, 400 600, 3000 1000, , 750 240 990 , , 3000 1000, 400 600, To find the Thevenin voltage, we consider the circuit in Fig. 4.65(b)., Using the voltage division principle gives, v1 , , 1000, (220) 55 V,, 1000 3000, , v2 , , 600, (220) 132 V, 600 400, , Applying KVL around loop ab gives, v1 VTh v2 0, , or, , VTh v1 v2 55 132 77 V, , Having determined the Thevenin equivalent, we find the current, through the galvanometer using Fig. 4.65(c)., IG , , VTh, 77, 74.76 mA, , RTh Rm, 990 40, , The negative sign indicates that the current flows in the direction, opposite to the one assumed, that is, from terminal b to terminal a., , Example 4.18
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 160, , Chapter 4, , 160, , Circuit Theorems, , 400 Ω, , 3 kΩ, , a, , RTh, , +, , 220 V +, −, , b, 600 Ω, , 1 kΩ, , 400 Ω, , 3 kΩ, , 1 kΩ, , +, v1, −, , (a), , a, , −, VTh, , b, , +, v2, −, , 600 Ω, , (b), RTh, , a, IG, 40 Ω, , VTh, , +, −, G, , b, (c), , Figure 4.65, For Example 4.18: (a) Finding RTh, (b) finding VTh, (c) determining the current through the galvanometer., , Practice Problem 4.18, 20 Ω, , 30 Ω, , Obtain the current through the galvanometer, having a resistance of, 14 , in the Wheatstone bridge shown in Fig. 4.66., Answer: 64 mA., , G, , 14 Ω, 40 Ω, , 60 Ω, , 16 V, , Figure 4.66, For Practice Prob. 4.18., , 4.11, , Summary, , 1. A linear network consists of linear elements, linear dependent, sources, and linear independent sources., 2. Network theorems are used to reduce a complex circuit to a simpler one, thereby making circuit analysis much simpler., 3. The superposition principle states that for a circuit having multiple independent sources, the voltage across (or current through) an, element is equal to the algebraic sum of all the individual voltages, (or currents) due to each independent source acting one at a time., 4. Source transformation is a procedure for transforming a voltage, source in series with a resistor to a current source in parallel with, a resistor, or vice versa., 5. Thevenin’s and Norton’s theorems allow us to isolate a portion of, a network while the remaining portion of the network is replaced, by an equivalent network. The Thevenin equivalent consists of a, voltage source VTh in series with a resistor RTh, while the Norton, equivalent consists of a current source IN in parallel with a resistor RN. The two theorems are related by source transformation., RN RTh,, , IN , , VTh, RTh
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 161, , Review Questions, , 161, , 6. For a given Thevenin equivalent circuit, maximum power transfer, occurs when RL RTh; that is, when the load resistance is equal, to the Thevenin resistance., 7. The maximum power transfer theorem states that the maximum, power is delivered by a source to the load RL when RL is equal to, RTh, the Thevenin resistance at the terminals of the load., 8. PSpice can be used to verify the circuit theorems covered in this, chapter., 9. Source modeling and resistance measurement using the Wheatstone bridge provide applications for Thevenin’s theorem., , Review Questions, 4.1, , 4.2, , The current through a branch in a linear network is, 2 A when the input source voltage is 10 V. If the, voltage is reduced to 1 V and the polarity is reversed,, the current through the branch is:, (a) 2 A, , (b) 0.2 A, , (d) 2 A, , (e) 20 A, , Which pair of circuits in Fig. 4.68 are equivalent?, (a) a and b, , (b) b and d, , (c) a and c, , (d) c and d, , 20 V, , The superposition principle applies to power, calculation., , +, −, , Refer to Fig. 4.67. The Thevenin resistance at, terminals a and b is:, (b) 20 , , (c) 5 , , (d) 4 , , 5Ω, 4A, , (a), , (b) False, , (a) 25 , , (b) False, , 5Ω, , (b) False, , (a) True, 4.4, , 4.8, , (c) 0.2 A, , The Norton resistance RN is exactly equal to the, Thevenin resistance RTh., (a) True, , For superposition, it is not required that only one, independent source be considered at a time; any, number of independent sources may be considered, simultaneously., (a) True, , 4.3, , 4.7, , (b), , 5Ω, , 4A, , 20 V, , +, −, , (c), , 5Ω, , 5Ω, , (d), , Figure 4.68, For Review Question 4.8., , 50 V, , +, −, , a, b, , 20 Ω, , Figure 4.67, For Review Questions 4.4 to 4.6., 4.5, , 4.6, , The Thevenin voltage across terminals a and b of the, circuit in Fig. 4.67 is:, (a) 50 V, , (b) 40 V, , (c) 20 V, , (d) 10 V, , The Norton current at terminals a and b of the circuit, in Fig. 4.67 is:, (a) 10 A, , (b) 2.5 A, , (c) 2 A, , (d) 0 A, , 4.9, , A load is connected to a network. At the terminals to, which the load is connected, RTh 10 and, VTh 40 V. The maximum possible power supplied, to the load is:, (a) 160 W, , (b) 80 W, , (c) 40 W, , (d) 1 W, , 4.10 The source is supplying the maximum power to the, load when the load resistance equals the source, resistance., (a) True, , (b) False, , Answers: 4.1b, 4.2a, 4.3b, 4.4d, 4.5b, 4.6a, 4.7a, 4.8c,, 4.9c, 4.10a.
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 162, , Chapter 4, , 162, , Circuit Theorems, , Problems, Section 4.2 Linearity Property, 4.1, , 4.5, , Calculate the current io in the current of Fig. 4.69., What does this current become when the input, voltage is raised to 10 V?, 1Ω, , For the circuit in Fig. 4.73, assume vo 1 V, and, use linearity to find the actual value of vo., 2Ω, +, −, , 15 V, , 5Ω, , 3Ω, , vo, , 2Ω, 6Ω, , 6Ω, , 4Ω, , io, 1V, , +, −, , 8Ω, , 3Ω, , Figure 4.73, For Prob. 4.5., 4.6, , Figure 4.69, For Prob. 4.1., 4.2, , Using Fig. 4.70, design a problem to help other, students better understand linearity., R2, , I, , R1, , For the linear circuit shown in Fig. 4.74, use linearity, to complete the following table., , Experiment, , Vs, , Vo, , 1, 2, 3, 4, , 12 V, , 4V, 16 V, , R4, , R3, , 1V, 2 V, , +, vo, −, , R5, , +, −, , Vs, , Figure 4.70, For Prob. 4.2., 4.3, , (a) In the circuit of Fig. 4.71, calculate vo and io, when vs 1 V., (b) Find vo and io when vs 10 V., (c) What are vo and io when each of the 1-, resistors is replaced by a 10- resistor and, vs 10 V?, , +, Vo, –, , Linear, circuit, , Figure 4.74, For Prob. 4.6., 4.7, , Use linearity and the assumption that Vo 1 V to, find the actual value of Vo in Fig. 4.75., 1Ω, , 4Ω, , 1Ω, 1Ω, , 1Ω, , vs, , +, 4V −, , +, −, , 1Ω, , +, vo, −, , 3Ω, , 2Ω, , +, Vo, –, , io, 1Ω, , Figure 4.75, For Prob. 4.7., , Section 4.3 Superposition, Figure 4.71, 4.8, , For Prob. 4.3., 4.4, , Using superposition, find Vo in the circuit of Fig. 4.76., Check with PSpice., , Use linearity to determine io in the circuit of Fig. 4.72., 3Ω, , 2Ω, 5Ω, , io, 6Ω, , 4Ω, , 4Ω, , 9A, , Figure 4.72, , Figure 4.76, , For Prob. 4.4., , For Prob. 4.8., , Vo, , 1Ω, 3Ω, , + 9V, −, , + 3V, −
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 163, , Problems, , 4.9, , Use superposition to find vo in the circuit of Fig. 4.77., , 163, , 4.13 Use superposition to find vo in the circuit of Fig. 4.81., 4A, , 6A, , 2Ω, , 4Ω, , 8Ω, , −+, , 2Ω, +, vo, −, , + 18 V, −, , 1Ω, , Figure 4.77, , Figure 4.81, , For Prob. 4.9., , For Prob. 4.13., , 4.10 Using Fig. 4.78, design a problem to help other, students better understand superposition. Note, the, letter k is a gain you can specify to make the, problem easier to solve but must not be zero., , 12 V, , 10 Ω, , 2A, , +, vo, −, , 5Ω, , 4.14 Apply the superposition principle to find vo in the, circuit of Fig. 4.82., 6Ω, 4A, , kVab, , R, , V, , +−, , +, −, , a, , 4Ω, , +, I, , Vab, −, , b, , +, vo, −, , 2A, , 3Ω, , Figure 4.82, , For Prob. 4.10., , For Prob. 4.14., , 4.11 Use the superposition principle to find io and vo in, the circuit of Fig. 4.79., io 10 Ω, , 40 Ω, , 4.15 For the circuit in Fig. 4.83, use superposition to find i., Calculate the power delivered to the 3- resistor., , 20 Ω, , + vo −, 6A, , +, −, , 40 V, , Figure 4.78, , 2Ω, , 20 V, 4io, , 1Ω, , +, −, , 2A, , − 30 V, +, , i, 2Ω, , 4Ω, − 16 V, +, , 3Ω, , Figure 4.79, For Prob. 4.11., , Figure 4.83, For Probs. 4.15 and 4.56., , 4.12 Determine vo in the circuit of Fig. 4.80 using the, superposition principle., , 4.16 Given the circuit in Fig. 4.84, use superposition to, get io., 2A, , 4A, , 6Ω, , 5Ω, , 4Ω, , io, , 4Ω, , 3Ω, , 2Ω, , + v −, o, 24 V, , +, −, , 3Ω, , 12 Ω, , + 38 V, −, , 12 V, , +, −, , Figure 4.80, , Figure 4.84, , For Probs. 4.12 and 4.35., , For Prob. 4.16., , 10 Ω, , 5Ω, , 4A
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 164, , Chapter 4, , 164, , Circuit Theorems, , 4.17 Use superposition to obtain vx in the circuit of, Fig. 4.85. Check your result using PSpice., 30 Ω, , 10 Ω, , 4.21 Using Fig. 4.89, design a problem to help other, students better understand source transformation., , 20 Ω, , io, , R1, , + vx −, 90 V, , +, −, , 60 Ω, , 30 Ω, , 6A, , +, −, , V +, −, , 40 V, , R2, , +, vo, −, , I, , Figure 4.89, , Figure 4.85, , For Prob. 4.21., , For Prob. 4.17., , 4.18 Use superposition to find Vo in the circuit of Fig. 4.86., , 4.22 For the circuit in Fig. 4.90, use source, transformation to find i., , 1Ω, , 5Ω, i, , 0.5Vo, , 2Ω, , 5Ω, , 2A, 10 V +, −, , 4Ω, , 2A, , 10 Ω, , +, Vo, −, , 4Ω, , +, −, , 20 V, , Figure 4.90, For Prob. 4.22., , Figure 4.86, For Prob. 4.18., , 4.19 Use superposition to solve for vx in the circuit of, Fig. 4.87., , 4.23 Referring to Fig. 4.91, use source transformation to, determine the current and power in the 8- resistor., 8Ω, , ix, 2Ω, , 6A, , 8Ω, , 4A, , +, vx, −, , 10 Ω, , 9A, , 3Ω, , 6Ω, , +, − 45 V, , − +, , Figure 4.91, , 4ix, , For Prob. 4.23., , Figure 4.87, For Prob. 4.19., , 4.24 Use source transformation to find the voltage Vx in, the circuit of Fig. 4.92., , Section 4.4 Source Transformation, 4.20 Use source transformations to reduce the circuit in, Fig. 4.88 to a single voltage source in series with a, single resistor., , 3A, , 8Ω, , 3A, , 10 Ω, , 20 Ω, 12 V +, −, , 10 Ω, , + Vx −, , 40 Ω, , 40 V, + 16 V, −, , +, −, , Figure 4.88, , Figure 4.92, , For Prob. 4.20., , For Prob. 4.24., , 10 Ω, , 2Vx
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 165, , Problems, , 4.25 Obtain vo in the circuit of Fig. 4.93 using source, transformation. Check your result using PSpice., , 165, , 4.29 Use source transformation to find vo in the circuit of, Fig. 4.97., , 2A, , 4 kΩ, , 4Ω, , 3A, , 5Ω, , + vo −, 2Ω, , 3vo, , 2 kΩ, , 9Ω, , − +, 1 kΩ, , 3 mA, , 6A, , +−, , Figure 4.97, , 30 V, , For Prob. 4.29., , +, vo, −, , Figure 4.93, For Prob. 4.25., , 4.30 Use source transformation on the circuit shown in, Fig 4.98 to find ix., , 4.26 Use source transformation to find io in the circuit of, Fig. 4.94., ix, , 24 Ω, , 60 Ω, , 5Ω, 12 V, 3A, , io, , 4Ω, , +, −, , 2Ω, , 6A, , 20 V, , Figure 4.94, , 30 Ω, , +, −, , 10 Ω, , 0.7ix, , Figure 4.98, For Prob. 4.30., 4.31 Determine vx in the circuit of Fig. 4.99 using source, transformation., , For Prob. 4.26., 4.27 Apply source transformation to find vx in the circuit, of Fig. 4.95., , 3Ω, , 6Ω, , + vx −, 10 Ω, , a, , 12 Ω, , b, , 20 Ω, , 12 V, , +, −, , +, −, , 8Ω, , + vx −, +, −, , 50 V, , 40 Ω, , 8A, , +, −, , 40 V, , 2vx, , Figure 4.99, For Prob. 4.31., , Figure 4.95, , 4.32 Use source transformation to find ix in the circuit of, Fig. 4.100., , For Probs. 4.27 and 4.40., 4.28 Use source transformation to find Io in Fig. 4.96., , 10 Ω, 1Ω, , Io, , 4Ω, ix, , + Vo −, 8V, , +, −, , 3Ω, , 1, V, 3 o, , 60 V, , +, −, , Figure 4.96, , Figure 4.100, , For Prob. 4.28., , For Prob. 4.32., , 15 Ω, , 0.5ix, , 50 Ω, , 40 Ω
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 166, , Chapter 4, , 166, , Circuit Theorems, , Sections 4.5 and 4.6 Thevenin’s and Norton’s, Theorems, , 4.37 Find the Norton equivalent with respect to terminals, a-b in the circuit shown in Fig. 4.104., , 4.33 Determine RTh and VTh at terminals 1-2 of each of, the circuits in Fig. 4.101., , 3A, 20 Ω, , a, , 10 Ω, 1, 20 V, , +, −, , 180 V, , 40 Ω, , +, −, , 12 Ω, , 40 Ω, b, , 2, , Figure 4.104, , (a), , For Prob. 4.37., 60 Ω, , 4.38 Apply Thevenin’s theorem to find Vo in the circuit of, Fig. 4.105., , 1, +, −, , 30 Ω, , 2A, , 30 V, , 1Ω, , 4Ω, , 2, , 5Ω, , (b), 16 Ω, , 3A, , Figure 4.101, For Probs. 4.33 and 4.46., , +, −, , 4.34 Using Fig. 4.102, design a problem that will help, other students better understand Thevenin equivalent, circuits., , +, Vo, –, , 10 Ω, , 12 V, , Figure 4.105, For Prob. 4.38., 4.39 Obtain the Thevenin equivalent at terminals a-b of, the circuit in Fig. 4.106., , I, , 1A, R1, , R3, a, , V +, −, , 10 Ω, , 16 Ω, a, , R2, , 10 Ω, b, , 5Ω, , 8V +, −, , Figure 4.102, For Probs. 4.34 and 4.49., , b, , Figure 4.106, 4.35 Use Thevenin’s theorem to find vo in Prob. 4.12., 4.36 Solve for the current i in the circuit of Fig. 4.103, using Thevenin’s theorem. (Hint: Find the Thevenin, equivalent seen by the 12- resistor.), , For Prob. 4.39., 4.40 Find the Thevenin equivalent at terminals a-b of the, circuit in Fig. 4.107., + V −, o, , i, 10 Ω, 50 V, , 12 Ω, +, −, , +, −, , 10 kΩ, 40 Ω, , 70 V, , +, −, , 20 kΩ, a, b, , 30 V, , Figure 4.103, , Figure 4.107, , For Prob. 4.36., , For Prob. 4.40., , +, −, , 4Vo
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 167, , Problems, , 4.41 Find the Thevenin and Norton equivalents at, terminals a-b of the circuit shown in Fig. 4.108., 14 V, , 14 Ω, , −+, 6Ω, , 1A, , 167, , 4.45 Find the Norton equivalent of the circuit in, Fig. 4.112., 6Ω, , a, , a, 6Ω, , 6A, , 5Ω, , 3A, , 4Ω, b, , b, , Figure 4.112, For Prob. 4.45., , Figure 4.108, For Prob. 4.41., *4.42 For the circuit in Fig. 4.109, find the Thevenin, equivalent between terminals a and b., , 4.46 Using Fig. 4.113, design a problem to help other, students better understand Norton equivalent circuits., , 20 Ω, , R2, −, +, , 20 Ω, , 10 Ω, , a, , 40 V, , a, , b, , I, , R1, , R3, , 10 Ω, , b, 10 Ω, , 10 Ω, , 10 A, , Figure 4.113, For Prob. 4.46., , 60 V +, −, , 4.47 Obtain the Thevenin and Norton equivalent circuits, of the circuit in Fig. 4.114 with respect to terminals a, and b., , Figure 4.109, For Prob. 4.42., 4.43 Find the Thevenin equivalent looking into terminals, a-b of the circuit in Fig. 4.110 and solve for ix., 10 Ω, , 20 V, , +, −, , 6Ω, , a, , 10 Ω, , 12 Ω, a, , b, , ix, , 50 V, 5Ω, , 2A, , +, −, , +, Vx, –, , 60 Ω, , 2Vx, , b, , Figure 4.110, , Figure 4.114, , For Prob. 4.43., , For Prob. 4.47., , 4.44 For the circuit in Fig. 4.111, obtain the Thevenin, equivalent as seen from terminals:, (a) a-b, , (b) b-c, 3Ω, , 4.48 Determine the Norton equivalent at terminals a-b for, the circuit in Fig. 4.115., , 1Ω, , 10io, a, , 24 V, , +, −, , 4Ω, , 5Ω, , 4A, , 2A, , a, , 8Ω, , b, c, , Figure 4.111, , 4Ω, , io, b, , 2Ω, , + −, , Figure 4.115, For Prob. 4.48., , For Prob. 4.44., * An asterisk indicates a challenging problem., , 4.49 Find the Norton equivalent looking into terminals, a-b of the circuit in Fig. 4.102.
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 168, , Chapter 4, , 168, , 4.50 Obtain the Norton equivalent of the circuit in, Fig. 4.116 to the left of terminals a-b. Use the, result to find current i., 12 V, , 6Ω, , Circuit Theorems, , 4.54 Find the Thevenin equivalent between terminals a-b, of the circuit in Fig. 4.120., , 1 kΩ, , a, , +−, , a, i, 5Ω, , 4Ω, , 2A, , 4A, , 3V, , Io, , +, −, , +, −, , 2Vx, , 40Io +, Vx, –, , 50 Ω, b, , Figure 4.120, , b, , For Prob. 4.54., , Figure 4.116, For Prob. 4.50., 4.51 Given the circuit in Fig. 4.117, obtain the Norton, equivalent as viewed from terminals:, (a) a-b, , *4.55 Obtain the Norton equivalent at terminals a-b of the, circuit in Fig. 4.121., , (b) c-d, a, , b, , 6Ω, , I, , 8 kΩ, , 4Ω, , a, c, 2V, , +, −, , 120 V, , 3Ω, , 6A, , 2Ω, , +, −, , 0.001Vab, , +, −, , 80I, , 50 kΩ, , +, Vab, −, b, , d, , Figure 4.121, For Prob. 4.55., , Figure 4.117, For Prob. 4.51., 4.52 For the transistor model in Fig. 4.118, obtain the, Thevenin equivalent at terminals a-b., , 4.56 Use Norton’s theorem to find Vo in the circuit of, Fig. 4.122., , 3 kΩ, a, , Io, 12 V, , +, −, , 20Io, , 12 kΩ, , 2 kΩ, , 10 kΩ, , 2 kΩ, +, b, , +, 360 V −, , 30 mA 1 kΩ, , 24 kΩ, , Figure 4.118, For Prob. 4.52., , Figure 4.122, , 4.53 Find the Norton equivalent at terminals a-b of the, circuit in Fig. 4.119., , For Prob. 4.56., , 4.57 Obtain the Thevenin and Norton equivalent circuits, at terminals a-b for the circuit in Fig. 4.123., , 0.25vo, , 2Ω, , 6Ω, , +, −, , 3Ω, , 2Ω, , 3Ω, , a, 18 V, , Vo, −, , a, , +, vo, −, , 50 V, , +, −, , 6Ω, , +, vx, −, , 0.5vx, , 10 Ω, b, , b, , Figure 4.119, , Figure 4.123, , For Prob. 4.53., , For Probs. 4.57 and 4.79.
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 169, , Problems, , 4.58 The network in Fig. 4.124 models a bipolar transistor, common-emitter amplifier connected to a load. Find, the Thevenin resistance seen by the load., ib, , vs, , *4.62 Find the Thevenin equivalent of the circuit in, Fig. 4.128., , bib, , R1, , +, −, , 169, , 0.1io, a, , R2, , +, vo, −, , 10 Ω, , RL, , io, , Figure 4.124, , 40 Ω, , For Prob. 4.58., , 20 Ω, +−, , 4.59 Determine the Thevenin and Norton equivalents at, terminals a-b of the circuit in Fig. 4.125., , b, , 2vo, , Figure 4.128, For Prob. 4.62., 20 Ω, , 10 Ω, a, , 8A, , 4.63 Find the Norton equivalent for the circuit in, Fig. 4.129., , b, , 50 Ω, , 40 Ω, , 10 Ω, , Figure 4.125, For Probs. 4.59 and 4.80., +, vo, −, , *4.60 For the circuit in Fig. 4.126, find the Thevenin and, Norton equivalent circuits at terminals a-b., , 20 Ω, , 0.5vo, , Figure 4.129, , 2A, , For Prob. 4.63., , 18 V, +−, , a, , 4Ω, , 4.64 Obtain the Thevenin equivalent seen at terminals a-b, of the circuit in Fig. 4.130., , 6Ω, b, , 3A, , 4Ω, , 5Ω, , 1Ω, a, ix, , +−, 10 V, , 10ix, , Figure 4.126, , +, −, , 2Ω, , For Probs. 4.60 and 4.81., , b, , *4.61 Obtain the Thevenin and Norton equivalent circuits, at terminals a-b of the circuit in Fig. 4.127., 2Ω, a, 12 V, , +, −, , 6Ω, , 2Ω, , 6Ω, , 6Ω, −, + 12 V, , Figure 4.130, For Prob. 4.64., 4.65 For the circuit shown in Fig. 4.131, determine the, relationship between Vo and Io., , + 12 V, −, , 4Ω, , 2Ω, , Io, +, , 2Ω, , 64 V, , +, −, , b, , Figure 4.127, , Figure 4.131, , For Prob. 4.61., , For Prob. 4.65., , 12 Ω, , Vo, −
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 170, , Chapter 4, , 170, , Circuit Theorems, , Section 4.8 Maximum Power Transfer, , 4.70 Determine the maximum power delivered to the, variable resistor R shown in the circuit of Fig. 4.136., , 4.66 Find the maximum power that can be delivered to, the resistor R in the circuit of Fig. 4.132., , 3Ω, , 20 V, , 3 Vx, , 10 V, , 2Ω, , −+, 5Ω, , R, , +, −, , 5Ω, , 5Ω, , 6A, 4V, , +, −, , 15 Ω, , R, , 6Ω, , Figure 4.132, For Prob. 4.66., , +, , 4.67 The variable resistor R in Fig. 4.133 is adjusted until, it absorbs the maximum power from the circuit., (a) Calculate the value of R for maximum power., (b) Determine the maximum power absorbed by R., 80 Ω, , Vx, , −, , Figure 4.136, For Prob. 4.70., , 4.71 For the circuit in Fig. 4.137, what resistor connected, across terminals a-b will absorb maximum power, from the circuit? What is that power?, , 20 Ω, 40 V, +−, , R, 3 kΩ, , 10 Ω, , 10 kΩ, a, , 90 Ω, , +, , Figure 4.133, , +, −, , 8V, , For Prob. 4.67., , vo, −, , 1 kΩ, , –, +, , 40 kΩ, , 120vo, , b, , Figure 4.137, , *4.68 Compute the value of R that results in maximum, power transfer to the 10- resistor in Fig. 4.134., Find the maximum power., , For Prob. 4.71., , R, , +, −, , 12 V, , 10 Ω, +, −, , 4.72 (a) For the circuit in Fig. 4.138, obtain the Thevenin, equivalent at terminals a-b., (b) Calculate the current in RL 8 ., , 20 Ω, , (c) Find RL for maximum power deliverable to RL., , 8V, , (d) Determine that maximum power., , Figure 4.134, For Prob. 4.68., 2A, , 4.69 Find the maximum power transferred to resistor R in, the circuit of Fig. 4.135., 10 kΩ, , 100 V +, −, , +, vo, −, , 4Ω, , 22 kΩ, 4A, 40 kΩ 0.003v, o, , 30 kΩ, , R, , 6Ω, , a, , 2Ω, , RL, +−, 20 V, , Figure 4.135, , Figure 4.138, , For Prob. 4.69., , For Prob. 4.72., , b
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 171, , Problems, , 4.73 Determine the maximum power that can be delivered, to the variable resistor R in the circuit of Fig. 4.139., , 171, , 4.80 Use PSpice to find the Thevenin equivalent circuit at, terminals a-b of the circuit in Fig. 4.125., 4.81 For the circuit in Fig. 4.126, use PSpice to find the, Thevenin equivalent at terminals a-b., , 10 Ω, , 60 V, , 25 Ω, , Section 4.10 Applications, , R, , +, −, 20 Ω, , 4.82 A battery has a short-circuit current of 20 A and an, open-circuit voltage of 12 V. If the battery is, connected to an electric bulb of resistance 2 ,, calculate the power dissipated by the bulb., , 5Ω, , Figure 4.139, , 4.83 The following results were obtained from, measurements taken between the two terminals of a, resistive network., , For Prob. 4.73., 4.74 For the bridge circuit shown in Fig. 4.140, find the, load RL for maximum power transfer and the, maximum power absorbed by the load., , Terminal Voltage, Terminal Current, , 12 V, 0A, , 0V, 1.5 A, , Find the Thevenin equivalent of the network., R1, vs, , +, −, , 4.84 When connected to a 4- resistor, a battery has a, terminal voltage of 10.8 V but produces 12 V on an, open circuit. Determine the Thevenin equivalent, circuit for the battery., , R3, , RL, , R2, , R4, , Figure 4.140, For Prob. 4.74., *4.75 Looking into terminals of the circuit shown in, Fig. 4.141, from the right (the RL side), determine, the Thevenin equivalent circuit. What value of RL, produces maximum power to RL?, , 4.85 The Thevenin equivalent at terminals a-b of the, linear network shown in Fig. 4.142 is to be, determined by measurement. When a 10-k resistor, is connected to terminals a-b, the voltage Vab is, measured as 6 V. When a 30-k resistor is connected, to the terminals, Vab is measured as 12 V. Determine:, (a) the Thevenin equivalent at terminals a-b, (b) Vab, when a 20-k resistor is connected to terminals a-b., a, Linear, , 10 Ω, , 20I, – +, , network, , I, , b, , a, , Figure 4.142, +, 10 V −, , RL, b, , Figure 4.141, For Prob. 4.75., , For Prob. 4.85., 4.86 A black box with a circuit in it is connected to a, variable resistor. An ideal ammeter (with zero, resistance) and an ideal voltmeter (with infinite, resistance) are used to measure current and voltage, as shown in Fig. 4.143. The results are shown in the, table on the next page., , Section 4.9 Verifying Circuit Theorems, with PSpice, , i, A, , 4.76 Solve Prob. 4.34 using PSpice., , Black, box, , 4.77 Use PSpice to solve Prob. 4.44., 4.78 Use PSpice to solve Prob. 4.52., 4.79 Obtain the Thevenin equivalent of the circuit in, Fig. 4.123 using PSpice., , Figure 4.143, For Prob. 4.86., , V, , R
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 172, , Chapter 4, , 172, , Circuit Theorems, , (a) Find i when R 4 ., (b) Determine the maximum power from the box., , R(), , V(V), , i(A), , 2, 8, 14, , 3, 8, 10.5, , 1.5, 1.0, 0.75, , 4.90 The Wheatstone bridge circuit shown in Fig. 4.146 is, used to measure the resistance of a strain gauge. The, adjustable resistor has a linear taper with a maximum, value of 100 . If the resistance of the strain gauge, is found to be 42.6 , what fraction of the full slider, travel is the slider when the bridge is balanced?, Rs, , 4.87 A transducer is modeled with a current source Is and, a parallel resistance Rs. The current at the terminals, of the source is measured to be 9.975 mA when an, ammeter with an internal resistance of 20 is used., (a) If adding a 2-k resistor across the source, terminals causes the ammeter reading to fall to, 9.876 mA, calculate Is and Rs., (b) What will the ammeter reading be if the, resistance between the source terminals is, changed to 4 k?, , 2 kΩ, , 4 mA, , +, −, , For Prob. 4.90., 4.91 (a) In the Wheatstone bridge circuit of Fig. 4.147,, select the values of R1 and R3 such that the bridge, can measure Rx in the range of 0–10 ., , R1, , 5 kΩ, , b, , V, , 20 kΩ, , G, , 100 Ω, , Rx, , Io, 30 kΩ, , vs, , 4 kΩ, , Figure 4.146, , 4.88 Consider the circuit in Fig. 4.144. An ammeter with, internal resistance Ri is inserted between A and B to, measure Io. Determine the reading of the ammeter if:, (a) Ri 500 , (b) Ri 0 . (Hint: Find the, Thevenin equivalent circuit at terminals a-b.), , a, , 2 kΩ, , +, −, , 60 V, , R3, G, , +, −, , 50 Ω, , Rx, , 10 kΩ, , Figure 4.144, , Figure 4.147, , For Prob. 4.88., , For Prob. 4.91., (b) Repeat for the range of 0–100 ., , 4.89 Consider the circuit in Fig. 4.145. (a) Replace the, resistor RL by a zero resistance ammeter and, determine the ammeter reading. (b) To verify the, reciprocity theorem, interchange the ammeter and, the 12-V source and determine the ammeter reading, again., , *4.92 Consider the bridge circuit of Fig. 4.148. Is the, bridge balanced? If the 10-k resistor is replaced by, an 18-k resistor, what resistor connected between, terminals a-b absorbs the maximum power? What is, this power?, 2 kΩ, 6 kΩ, , 3 kΩ, 10 kΩ, , 20 kΩ, RL, , 12 V, , 220 V, , +, −, 12 kΩ, , +, −, , a, 5 kΩ, , 15 kΩ, , Figure 4.145, , Figure 4.148, , For Prob. 4.89., , For Prob. 4.92., , b, 10 kΩ
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ale29559_ch04.qxd, , 07/08/2008, , 10:56 AM, , Page 173, , Comprehensive Problems, , 173, , Comprehensive Problems, 4.93 The circuit in Fig. 4.149 models a common-emitter, transistor amplifier. Find ix using source, transformation., ix, , (a) Find the value of R such that Vo 1.8 V., (b) Calculate the value of R that will draw the, maximum current. What is the maximum current?, , Rs, , +, −, , vs, , *4.96 A resistance array is connected to a load resistor R, and a 9-V battery as shown in Fig. 4.151., , R, , bix, , Ro, , + V −, o, , 3, , Figure 4.149, , 5Ω, , For Prob. 4.93., , 30 Ω, , 4.94 An attenuator is an interface circuit that reduces the, voltage level without changing the output resistance., (a) By specifying Rs and Rp of the interface circuit in, Fig. 4.150, design an attenuator that will meet the, following requirements:, Vo, 0.125,, Vg, , Vg, , 4Ω, , 4, 5Ω, , 20 Ω, 1, + 9V −, , Figure 4.151, For Prob. 4.96., 4.97 A common-emitter amplifier circuit is shown in, Fig. 4.152. Obtain the Thevenin equivalent to the, left of points B and E., , Rs, , +, −, , 2, 4Ω, , Req RTh Rg 100 , , (b) Using the interface designed in part (a), calculate, the current through a load of RL 50 when, Vg 12 V., , Rg, , 5Ω, , Rp, , +, Vo, −, , RL, , RL, 6 kΩ, , Attenuator, , +, , B, , Load, , −, , Req, , Figure 4.150, , 12 V, , 4 kΩ, , For Prob. 4.94., , Rc, E, , *4.95 A dc voltmeter with a sensitivity of 20 k/V is used, to find the Thevenin equivalent of a linear network., Readings on two scales are as follows:, (a) 0–10 V scale: 4 V, , (b) 0–50 V scale: 5 V, , Obtain the Thevenin voltage and the Thevenin, resistance of the network., , Figure 4.152, For Prob. 4.97., *4.98 For Practice Prob. 4.18, determine the current, through the 40- resistor and the power dissipated, by the resistor.
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 175, , c h a p t e r, , 5, , Operational, Amplifiers, He who will not reason is a bigot; he who cannot is a fool; and he, who dares not is a slave, —Lord Byron, , Enhancing Your Career, Career in Electronic Instrumentation, Engineering involves applying physical principles to design devices for, the benefit of humanity. But physical principles cannot be understood, without measurement. In fact, physicists often say that physics is the, science that measures reality. Just as measurements are a tool for understanding the physical world, instruments are tools for measurement., The operational amplifier introduced in this chapter is a building block, of modern electronic instrumentation. Therefore, mastery of operational, amplifier fundamentals is paramount to any practical application of, electronic circuits., Electronic instruments are used in all fields of science and engineering. They have proliferated in science and technology to the extent, that it would be ridiculous to have a scientific or technical education, without exposure to electronic instruments. For example, physicists,, physiologists, chemists, and biologists must learn to use electronic, instruments. For electrical engineering students in particular, the skill, in operating digital and analog electronic instruments is crucial. Such, instruments include ammeters, voltmeters, ohmmeters, oscilloscopes,, spectrum analyzers, and signal generators., Beyond developing the skill for operating the instruments, some, electrical engineers specialize in designing and constructing electronic, instruments. These engineers derive pleasure in building their own, instruments. Most of them invent and patent their inventions. Specialists in electronic instruments find employment in medical schools, hospitals, research laboratories, aircraft industries, and thousands of other, industries where electronic instruments are routinely used., , Electronic Instrumentation used in, medical research., © Royalty-Free/Corbis, , 175
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , 176, , Page 176, , Chapter 5, , 5.1, The term operational amplifier was introduced in 1947 by John Ragazzini, and his colleagues, in their work on, analog computers for the National, Defense Research Council after World, War II. The first op amps used vacuum, tubes rather than transistors., An op amp may also be regarded as a, voltage amplifier with very high gain., , Operational Amplifiers, , Introduction, , Having learned the basic laws and theorems for circuit analysis, we are, now ready to study an active circuit element of paramount importance:, the operational amplifier, or op amp for short. The op amp is a versatile circuit building block., The op amp is an electronic unit that behaves like a voltage-controlled, voltage source., , It can also be used in making a voltage- or current-controlled current, source. An op amp can sum signals, amplify a signal, integrate it, or, differentiate it. The ability of the op amp to perform these mathematical operations is the reason it is called an operational amplifier. It is, also the reason for the widespread use of op amps in analog design., Op amps are popular in practical circuit designs because they are versatile, inexpensive, easy to use, and fun to work with., We begin by discussing the ideal op amp and later consider the, nonideal op amp. Using nodal analysis as a tool, we consider ideal op, amp circuits such as the inverter, voltage follower, summer, and difference amplifier. We will also analyze op amp circuits with PSpice., Finally, we learn how an op amp is used in digital-to-analog converters and instrumentation amplifiers., , 5.2, , Operational Amplifiers, , An operational amplifier is designed so that it performs some mathematical operations when external components, such as resistors and, capacitors, are connected to its terminals. Thus,, An op amp is an active circuit element designed to perform mathematical operations of addition, subtraction, multiplication, division, differentiation, and integration., , Figure 5.1, A typical operational amplifier., Courtesy of Tech America., , The pin diagram in Fig. 5.2(a), corresponds to the 741 generalpurpose op amp made by Fairchild, Semiconductor., , The op amp is an electronic device consisting of a complex, arrangement of resistors, transistors, capacitors, and diodes. A full discussion of what is inside the op amp is beyond the scope of this book., It will suffice to treat the op amp as a circuit building block and simply study what takes place at its terminals., Op amps are commercially available in integrated circuit packages, in several forms. Figure 5.1 shows a typical op amp package. A typical, one is the eight-pin dual in-line package (or DIP), shown in Fig. 5.2(a)., Pin or terminal 8 is unused, and terminals 1 and 5 are of little concern, to us. The five important terminals are:, 1., 2., 3., 4., 5., , The, The, The, The, The, , inverting input, pin 2., noninverting input, pin 3., output, pin 6., positive power supply V , pin 7., negative power supply V , pin 4., , The circuit symbol for the op amp is the triangle in Fig. 5.2(b); as, shown, the op amp has two inputs and one output. The inputs are
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 177, , 5.2, , Operational Amplifiers, , 177, , V+, 7, Balance, , 1, , 8, , Inverting input 2, , −, , Noninverting input 3, , +, , No connection, +, , Inverting input, , 2, , 7, , V, , Noninverting input, , 3, , 6, , Output, , V−, , 4, , 5, , Balance, , 6 Output, , 415, V−, Offset Null, (b), , (a), , Figure 5.2, A typical op amp: (a) pin configuration, (b) circuit symbol., , marked with minus () and plus () to specify inverting and noninverting inputs, respectively. An input applied to the noninverting terminal will appear with the same polarity at the output, while an input, applied to the inverting terminal will appear inverted at the output., As an active element, the op amp must be powered by a voltage, supply as typically shown in Fig. 5.3. Although the power supplies are, often ignored in op amp circuit diagrams for the sake of simplicity, the, power supply currents must not be overlooked. By KCL,, io i1 i2 i i, , (5.3), , A is called the open-loop voltage gain because it is the gain of the op, amp without any external feedback from output to input. Table 5.1, TABLE 5.1, , Typical ranges for op amp parameters., Parameter, , Typical range, , Open-loop gain, A, Input resistance, Ri, Output resistance, Ro, Supply voltage, VCC, , 105 to 108, 105 to 1013 , 10 to 100 , 5 to 24 V, , 2, , io, 6, , 3, , +, VCC, −, , 4, , i2, , i−, , Figure 5.3, Powering the op amp., , v1, −, vd, +, , (5.2), , where v1 is the voltage between the inverting terminal and ground and, v2 is the voltage between the noninverting terminal and ground. The, op amp senses the difference between the two inputs, multiplies it by, the gain A, and causes the resulting voltage to appear at the output., Thus, the output vo is given by, vo Avd A(v2 v1), , 7, , (5.1), , The equivalent circuit model of an op amp is shown in Fig. 5.4., The output section consists of a voltage-controlled source in series with, the output resistance Ro. It is evident from Fig. 5.4 that the input resistance Ri is the Thevenin equivalent resistance seen at the input terminals, while the output resistance Ro is the Thevenin equivalent resistance, seen at the output. The differential input voltage vd is given by, vd v2 v1, , +, VCC, −, , i+, , i1, , Ideal values, , , 0, , Ro, , Ri, +, −, , Avd, , v2, , Figure 5.4, The equivalent circuit of the nonideal, op amp., , Sometimes, voltage gain is expressed, in decibels (dB), as discussed in, Chapter 14., A dB 20 log10 A, , vo
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 178, , Chapter 5, , 178, , vo, , Positive saturation, , VCC, , vd, , 0, Negative saturation, , Operational Amplifiers, , shows typical values of voltage gain A, input resistance Ri, output, resistance Ro, and supply voltage VCC., The concept of feedback is crucial to our understanding of op amp, circuits. A negative feedback is achieved when the output is fed back, to the inverting terminal of the op amp. As Example 5.1 shows, when, there is a feedback path from output to input, the ratio of the output, voltage to the input voltage is called the closed-loop gain. As a result, of the negative feedback, it can be shown that the closed-loop gain is, almost insensitive to the open-loop gain A of the op amp. For this reason, op amps are used in circuits with feedback paths., A practical limitation of the op amp is that the magnitude of its, output voltage cannot exceed |VCC |. In other words, the output voltage, is dependent on and is limited by the power supply voltage. Figure 5.5, illustrates that the op amp can operate in three modes, depending on, the differential input voltage vd :, 1. Positive saturation, vo VCC., 2. Linear region, VCC vo Avd VCC., 3. Negative saturation, vo VCC., , −VCC, , If we attempt to increase vd beyond the linear range, the op amp, becomes saturated and yields vo VCC or vo VCC. Throughout, this book, we will assume that our op amps operate in the linear mode., This means that the output voltage is restricted by, , Figure 5.5, Op amp output voltage vo as a function of, the differential input voltage vd., , VCC vo VCC, , (5.4), , Throughout this book, we assume that, an op amp operates in the linear range., Keep in mind the voltage constraint on, the op amp in this mode., , Although we shall always operate the op amp in the linear region, the, possibility of saturation must be borne in mind when one designs with, op amps, to avoid designing op amp circuits that will not work in the, laboratory., , Example 5.1, , A 741 op amp has an open-loop voltage gain of 2 105, input resistance of 2 M, and output resistance of 50 . The op amp is used in, the circuit of Fig. 5.6(a). Find the closed-loop gain vovs. Determine, current i when vs 2 V., , 20 kΩ, 20 kΩ, 10 kΩ, , i, 10 kΩ, , i, , 1, , −, 741, +, , vs +, −, , 1, O, , +, vo, −, , vs, , +, −, , Ro = 50 Ω v, o, , v1, −, vd, +, , (a), , Figure 5.6, For Example 5.1: (a) original circuit, (b) the equivalent circuit., , i, Ri = 2 MΩ, , (b), , +, −, , Avd, , O
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 179, , 5.3, , Ideal Op Amp, , 179, , Solution:, Using the op amp model in Fig. 5.4, we obtain the equivalent circuit, of Fig. 5.6(a) as shown in Fig. 5.6(b). We now solve the circuit in, Fig. 5.6(b) by using nodal analysis. At node 1, KCL gives, vs v1, 10 10, , 3, , , , v1, 2000 10, , 3, , , , v1 vo, 20 103, , Multiplying through by 2000 103, we obtain, 200vs 301v1 100vo, or, 2vs 3v1 vo, , 1, , v1 , , , , vo Avd, 50, , 2vs vo, 3, , (5.1.1), , At node O,, v1 vo, 20 10, , 3, , But vd v1 and A 200,000. Then, v1 vo 400(vo 200,000v1), , (5.1.2), , Substituting v1 from Eq. (5.1.1) into Eq. (5.1.2) gives, 0 26,667,067vo 53,333,333vs, , 1, , vo, 1.9999699, vs, , This is closed-loop gain, because the 20-k feedback resistor closes, the loop between the output and input terminals. When vs 2 V, vo , 3.9999398 V. From Eq. (5.1.1), we obtain v1 20.066667 V. Thus,, i , , v1 vo, 20 103, , 0.19999 mA, , It is evident that working with a nonideal op amp is tedious, as we are, dealing with very large numbers., , Practice Problem 5.1, , If the same 741 op amp in Example 5.1 is used in the circuit of Fig. 5.7,, calculate the closed-loop gain vovs. Find io when vs 1 V., , +, 741, −, , Answer: 9.00041, 0.657 mA., vs, , +, −, , 40 kΩ, 5 kΩ, , 5.3, , Ideal Op Amp, , To facilitate the understanding of op amp circuits, we will assume ideal, op amps. An op amp is ideal if it has the following characteristics:, 1. Infinite open-loop gain, A ., 2. Infinite input resistance, Ri ., 3. Zero output resistance, Ro 0., , io, , Figure 5.7, For Practice Prob. 5.1., , 20 kΩ, , +, vo, −
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 180, , Chapter 5, , 180, , Operational Amplifiers, , An ideal op amp is an amplifier with infinite open-loop gain, infinite, input resistance, and zero output resistance., , Although assuming an ideal op amp provides only an approximate analysis, most modern amplifiers have such large gains and, input impedances that the approximate analysis is a good one. Unless, stated otherwise, we will assume from now on that every op amp is, ideal., For circuit analysis, the ideal op amp is illustrated in Fig. 5.8,, which is derived from the nonideal model in Fig. 5.4. Two important, characteristics of the ideal op amp are:, , i1 = 0, i2 = 0, v1, , −, , −, vd, +, , +, , +, , +, , +, v 2 = v1, −, , −, , 1. The currents into both input terminals are zero:, , vo, , i1 0,, , −, , i2 0, , (5.5), , This is due to infinite input resistance. An infinite resistance, between the input terminals implies that an open circuit exists there, and current cannot enter the op amp. But the output current is not, necessarily zero according to Eq. (5.1)., 2. The voltage across the input terminals is equal to zero; i.e.,, , Figure 5.8, Ideal op amp model., , vd v2 v1 0, , (5.6), , v1 v2, , (5.7), , or, , The two characteristics can be exploited by noting that for voltage calculations the input port behaves as a, short circuit, while for current calculations the input port behaves as an, open circuit., , Example 5.2, +, , v1, i1 = 0, vs, , Rework Practice Prob. 5.1 using the ideal op amp model., , i2 = 0, , v2, , +, −, , −, , 40 kΩ, 5 kΩ, , Figure 5.9, For Example 5.2., , Thus, an ideal op amp has zero current into its two input terminals and the voltage between the two input terminals is equal, to zero. Equations (5.5) and (5.7) are extremely important, and should be regarded as the key handles to analyzing op amp, circuits., , i0, , O, +, vo, −, , Solution:, We may replace the op amp in Fig. 5.7 by its equivalent model in, Fig. 5.9 as we did in Example 5.1. But we do not really need to do, this. We just need to keep Eqs. (5.5) and (5.7) in mind as we analyze, the circuit in Fig. 5.7. Thus, the Fig. 5.7 circuit is presented as in, Fig. 5.9. Notice that, v2 vs, , 20 kΩ, , (5.2.1), , Since i1 0, the 40-k and 5-k resistors are in series; the same, current flows through them. v1 is the voltage across the 5-k resistor., Hence, using the voltage division principle,, v1 , , vo, 5, vo , 5 40, 9, , (5.2.2)
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 181, , 5.4, , Inverting Amplifier, , 181, , According to Eq. (5.7),, v2 v1, , (5.2.3), , Substituting Eqs. (5.2.1) and (5.2.2) into Eq. (5.2.3) yields the closedloop gain,, vs , , vo, 9, , 1, , vo, 9, vs, , (5.2.4), , which is very close to the value of 9.00041 obtained with the nonideal, model in Practice Prob. 5.1. This shows that negligibly small error, results from assuming ideal op amp characteristics., At node O,, io , , vo, vo, , mA, 40 5, 20, , (5.2.5), , From Eq. (5.2.4), when vs 1 V, vo 9 V. Substituting for vo 9 V, in Eq. (5.2.5) produces, io 0.2 0.45 0.65 mA, This, again, is close to the value of 0.657 mA obtained in Practice, Prob. 5.1 with the nonideal model., , Practice Problem 5.2, , Repeat Example 5.1 using the ideal op amp model., Answer: 2, 0.2 mA., , 5.4, , i2, , Inverting Amplifier, , In this and the following sections, we consider some useful op amp, circuits that often serve as modules for designing more complex circuits. The first of such op amp circuits is the inverting amplifier shown, in Fig. 5.10. In this circuit, the noninverting input is grounded, vi is, connected to the inverting input through R1, and the feedback resistor, Rf is connected between the inverting input and output. Our goal is to, obtain the relationship between the input voltage vi and the output voltage vo. Applying KCL at node 1,, i1 i2, , 1, , vi v1, v1 vo, , R1, Rf, , R1, , v1, , 0A, − −, 0V, v2 +, +, , 1, vi, , +, −, , +, vo, −, , Figure 5.10, The inverting amplifier., , (5.8), , But v1 v2 0 for an ideal op amp, since the noninverting terminal, is grounded. Hence,, vi, vo, , R1, Rf, , i1, , Rf, , A key feature of the inverting amplifier, is that both the input signal and the, feedback are applied at the inverting, terminal of the op amp.
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 182, , Chapter 5, , 182, , Operational Amplifiers, , or, , vo , Note there are two types of gains: the, one here is the closed-loop voltage, gain Av , while the op amp itself has an, open-loop voltage gain A., , Rf, R1, , vi, , (5.9), , The voltage gain is Av vovi RfR1. The designation of the circuit in Fig. 5.10 as an inverter arises from the negative sign. Thus,, An inverting amplifier reverses the polarity of the input signal while, amplifying it., , +, vi, , R1, , −, , –, +, , Rf, v, R1 i, , +, , Notice that the gain is the feedback resistance divided by the, input resistance which means that the gain depends only on the, external elements connected to the op amp. In view of Eq. (5.9), an, equivalent circuit for the inverting amplifier is shown in Fig. 5.11., The inverting amplifier is used, for example, in a current-to-voltage, converter., , vo, −, , Figure 5.11, An equivalent circuit for the inverter in, Fig. 5.10., , Example 5.3, , Refer to the op amp in Fig. 5.12. If vi 0.5 V, calculate: (a) the output, voltage vo, and (b) the current in the 10-k resistor., , 25 kΩ, 10 kΩ, , vi, , Solution:, (a) Using Eq. (5.9),, , −, +, , +, vo, −, , +, −, , Rf, vo, 25, 2.5, vi, R1, 10, vo 2.5vi 2.5(0.5) 1.25 V, , Figure 5.12, , (b) The current through the 10-k resistor is, , For Example 5.3., , i, , Practice Problem 5.3, , Find the output of the op amp circuit shown in Fig. 5.13. Calculate the, current through the feedback resistor., , 120 kΩ, 3 kΩ, 30 mV, , +, −, , Figure 5.13, For Practice Prob. 5.3., , −, +, , vi 0, 0.5 0, , 50 mA, R1, 10 103, , Answer: 1.2 V, 10 A., +, vo, −
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 183, , 5.5, , Noninverting Amplifier, , 183, , Example 5.4, , Determine vo in the op amp circuit shown in Fig. 5.14., , 40 kΩ, , Solution:, Applying KCL at node a,, , 20 kΩ, , va vo, 6 va, , 40 k, 20 k, va vo 12 2va, 1, vo 3va 12, , a, , −, +, , b, 6V, , But va vb 2 V for an ideal op amp, because of the zero voltage, drop across the input terminals of the op amp. Hence,, vo 6 12 6 V, , +, −, , +, −, , 2V, , +, vo, −, , Figure 5.14, For Example 5.4., , Notice that if vb 0 va, then vo 12, as expected from Eq. (5.9)., , Two kinds of current-to-voltage converters (also known as transresistance amplifiers) are shown in Fig. 5.15., , Practice Problem 5.4, , (a) Show that for the converter in Fig. 5.15(a),, vo, R, is, (b) Show that for the converter in Fig. 5.15(b),, R3, R3, vo, R1a1 , b, is, R1, R2, Answer: Proof., , R1, , R, −, +, is, , +, vo, −, , R2, R3, , −, +, , +, vo, −, , is, , (a), , (b), , Figure 5.15, i2, , For Practice Prob. 5.4., R1, , i1, , v1, v2, , vi, , 5.5, , Noninverting Amplifier, , Another important application of the op amp is the noninverting amplifier shown in Fig. 5.16. In this case, the input voltage vi is applied, directly at the noninverting input terminal, and resistor R1 is connected, , +, −, , Rf, , −, +, , +, vo, −, , Figure 5.16, The noninverting amplifier.
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 184, , Chapter 5, , 184, , Operational Amplifiers, , between the ground and the inverting terminal. We are interested in the, output voltage and the voltage gain. Application of KCL at the inverting terminal gives, i1 i2 1, , v1 vo, 0 v1, , R1, Rf, , (5.10), , But v1 v2 vi. Equation (5.10) becomes, vi, vi vo, , R1, Rf, or, , vo a1 , , Rf, R1, , b vi, , (5.11), , The voltage gain is Av vovi 1 Rf R1, which does not have a, negative sign. Thus, the output has the same polarity as the input., A noninverting amplifier is an op amp circuit designed to provide a, positive voltage gain., , −, +, vi, , +, , +, −, , vo = vi, −, , Figure 5.17, The voltage follower., , First, stage, , +, vi, −, , −, +, , vo vi, +, vo, −, , Second, stage, , Figure 5.18, A voltage follower used to isolate two, cascaded stages of a circuit., , Example 5.5, , Again we notice that the gain depends only on the external resistors., Notice that if feedback resistor Rf 0 (short circuit) or R1 , (open circuit) or both, the gain becomes 1. Under these conditions, (Rf 0 and R1 ), the circuit in Fig. 5.16 becomes that shown, in Fig. 5.17, which is called a voltage follower (or unity gain, amplifier) because the output follows the input. Thus, for a voltage, follower, (5.12), , Such a circuit has a very high input impedance and is therefore useful as an intermediate-stage (or buffer) amplifier to isolate one circuit, from another, as portrayed in Fig. 5.18. The voltage follower minimizes interaction between the two stages and eliminates interstage, loading., , For the op amp circuit in Fig. 5.19, calculate the output voltage vo., Solution:, We may solve this in two ways: using superposition and using nodal, analysis., , ■ METHOD 1 Using superposition, we let, vo vo1 vo2
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 185, , 5.6, , Summing Amplifier, , 185, , where vo1 is due to the 6-V voltage source, and vo2 is due to the 4-V, input. To get vo1, we set the 4-V source equal to zero. Under this, condition, the circuit becomes an inverter. Hence Eq. (5.9) gives, vo1, , 10, (6) 15 V, 4, , 10 kΩ, 4 kΩ, , a, , −, +, , b, +, −, , 6V, , +, , +, −, , 4V, , vo, −, , To get vo2, we set the 6-V source equal to zero. The circuit becomes, a noninverting amplifier so that Eq. (5.11) applies., vo2 a1 , , 10, b 4 14 V, 4, , Figure 5.19, For Example 5.5., , Thus,, vo vo1 vo2 15 14 1 V, , ■ METHOD 2 Applying KCL at node a,, 6 va, va vo, , 4, 10, But va vb 4, and so, 4 vo, 64, , 4, 10, , 1, , 5 4 vo, , or vo 1 V, as before., , Practice Problem 5.5, , Calculate vo in the circuit of Fig. 5.20., , 4 kΩ, , Answer: 7 V., , 3V, , +, −, , 8 kΩ, , +, −, , +, , 5 kΩ, , vo, , 2 kΩ, −, , 5.6, , Summing Amplifier, , Besides amplification, the op amp can perform addition and subtraction. The addition is performed by the summing amplifier covered in, this section; the subtraction is performed by the difference amplifier, covered in the next section., , A summing amplifier is an op amp circuit that combines several inputs, and produces an output that is the weighted sum of the inputs., , Figure 5.20, For Practice Prob. 5.5., , v1, v2, v3, , R1, R2, , i1, i2, , Rf, i, , 0, −, , a, R3, , i, , i3, , +, 0, , +, vo, −, , The summing amplifier, shown in Fig. 5.21, is a variation of the, inverting amplifier. It takes advantage of the fact that the inverting configuration can handle many inputs at the same time. We keep in mind, , Figure 5.21, The summing amplifier.
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 187, , 5.7, , Difference Amplifier, , Practice Problem 5.6, , Find vo and io in the op amp circuit shown in Fig. 5.23., 20 kΩ, , 8 kΩ, , 10 kΩ, , −, +, , 6 kΩ, −, +, , 2V, , −, +, , −, +, , 1.5 V, , 187, , io, , +, vo, −, , 4 kΩ, , 1.2 V, , Figure 5.23, For Practice Prob. 5.6., , Answer: 3.8 V, 1.425 mA., , 5.7, , Difference Amplifier, , Difference (or differential) amplifiers are used in various applications, where there is need to amplify the difference between two input signals. They are first cousins of the instrumentation amplifier, the most, useful and popular amplifier, which we will discuss in Section 5.10., A difference amplifier is a device that amplifies the difference between, two inputs but rejects any signals common to the two inputs., , Consider the op amp circuit shown in Fig. 5.24. Keep in mind that, zero currents enter the op amp terminals. Applying KCL to node a,, v1 va, va vo, , R1, R2, or, vo a, , R2, R2, 1b va , v1, R1, R1, , (5.16), , R2, R1, R3, v1, , +, −, , 0, , va, , + v, − 2, , Figure 5.24, Difference amplifier., , 0, , vb, , −, +, , R4, , +, vo, −, , The difference amplifier is also known, as the subtractor, for reasons to be, shown later.
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 191, , 5.8, , 5.8, , Cascaded Op Amp Circuits, , 191, , Cascaded Op Amp Circuits, , As we know, op amp circuits are modules or building blocks for, designing complex circuits. It is often necessary in practical applications to connect op amp circuits in cascade (i.e., head to tail) to achieve, a large overall gain. In general, two circuits are cascaded when they, are connected in tandem, one behind another in a single file., A cascade connection is a head-to-tail arrangement of two or more op, amp circuits such that the output of one is the input of the next., , When op amp circuits are cascaded, each circuit in the string is, called a stage; the original input signal is increased by the gain of the, individual stage. Op amp circuits have the advantage that they can be, cascaded without changing their input-output relationships. This is due to, the fact that each (ideal) op amp circuit has infinite input resistance and, zero output resistance. Figure 5.28 displays a block diagram representation of three op amp circuits in cascade. Since the output of one stage, is the input to the next stage, the overall gain of the cascade connection, is the product of the gains of the individual op amp circuits, or, A A1 A2 A3, , (5.22), , Although the cascade connection does not affect the op amp inputoutput relationships, care must be exercised in the design of an actual, op amp circuit to ensure that the load due to the next stage in the cascade does not saturate the op amp., , +, v1, −, , Stage 1, A1, , +, v 2 = A1v 1, −, , Stage 2, A2, , +, v 3 = A2v 2, −, , Stage 3, A3, , +, vo = A3v 3, −, , Figure 5.28, A three-stage cascaded connection., , Example 5.9, , Find vo and io in the circuit in Fig. 5.29., , +, −, , Solution:, This circuit consists of two noninverting amplifiers cascaded. At the, output of the first op amp,, 12, va a1 b (20) 100 mV, 3, , 12 kΩ, 20 mV +, −, 3 kΩ, , 10, b va (1 2.5)100 350 mV, 4, , The required current io is the current through the 10-k resistor., io , , vo vb, mA, 10, , +, −, , +, io, , b, 10 kΩ, , vo, 4 kΩ, −, , At the output of the second op amp,, vo a1 , , a, , Figure 5.29, For Example 5.9.
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 192, , Chapter 5, , 192, , Operational Amplifiers, , But vb va 100 mV. Hence,, io , , Practice Problem 5.9, +, −, , 8V, , +, −, , (350 100) 103, 25 mA, 10 103, , Determine vo and io in the op amp circuit in Fig. 5.30., , +, −, , +, , 8 kΩ, , vo, , io, , −, , Answer: 24 V, 2 mA., , 4 kΩ, , Figure 5.30, For Practice Prob. 5.9., , Example 5.10, , If v1 1 V and v2 2 V, find vo in the op amp circuit of Fig. 5.31., A, 6 kΩ, 2 kΩ, , −, +, , v1, , 5 kΩ, a, , 10 kΩ, , B, , −, +, , 8 kΩ, 4 kΩ, v2, , −, +, , C, , vo, , 15 kΩ, b, , Figure 5.31, For Example 5.10., , Solution:, 1. Define. The problem is clearly defined., 2. Present. With an input of v1 of 1 V and of v2 of 2 V, determine, the output voltage of the circuit shown in Figure 5.31. The op, amp circuit is actually composed of three circuits. The first, circuit acts as an amplifier of gain 3(6 k2 k) for v1 and, the second functions as an amplifier of gain 2(8 k4 k), for v2. The last circuit serves as a summer of two different gains, for the output of the other two circuits., 3. Alternative. There are different ways of working with this circuit., Since it involves ideal op amps, then a purely mathematical
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 193, , 5.8, , Cascaded Op Amp Circuits, , 193, , approach will work quite easily. A second approach would be to, use PSpice as a confirmation of the math., 4. Attempt. Let the output of the first op amp circuit be designated, as v11 and the output of the second op amp circuit be designated, as v22. Then we get, v11 3v1 3 1 3 V,, v22 2v2 2 2 4 V, In the third circuit we have, vo (10 k5 k) v11 3(10 k15 k) v22 4, 2(3) (23)(4), 6 2.667 8.667 V, 5. Evaluate. In order to properly evaluate our solution, we need to, identify a reasonable check. Here we can easily use PSpice to, provide that check., Now we can simulate this in PSpice. We see the results are, shown in Fig. 5.32., , R4, R6, + v1, 1V, , −3.000, , 6 kΩ, OPAMP, −, , 2 kΩ, +, , −, , R2, 5 kΩ, , U1, , R1, , 8.667 V, , 10 kΩ, OPAMP, −, R5, R7, + v2, 2V, , −4.000, +, , 8 kΩ, OPAMP, −, , 4 kΩ, , −, , +, , R3, 15 kΩ, , U2, , Figure 5.32, For Example 5.10., , We note that we obtain the same results using two entirely, different techniques (the first is to treat the op amp circuits as, just gains and a summer and the second is to use circuit analysis, with PSpice). This is a very good method of assuring that we, have the correct answer., 6. Satisfactory? We are satisfied we have obtained the asked for, results. We can now present our work as a solution to the, problem., , U3
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 194, , Chapter 5, , 194, , Practice Problem 5.10, , Operational Amplifiers, , If v1 4 V and v2 3 V, find vo in the op amp circuit of, Fig. 5.33., , 60 kΩ, 20 kΩ, , −, +, , −, +, , vo, , v1 +, −, 50 kΩ, 10 kΩ, v2, , 30 kΩ, , −, +, , +, −, , Figure 5.33, For Practice Prob. 5.10., , Answer: 18 V., , Op Amp Circuit Analysis with PSpice, , 5.9, , PSpice for Windows does not have a model for an ideal op amp, although, one may create one as a subcircuit using the Create Subcircuit line in, the Tools menu. Rather than creating an ideal op amp, we will use one, of the four nonideal, commercially available op amps supplied in the, PSpice library eval.slb. The op amp models have the part names LF411,, LM111, LM324, and uA741, as shown in Fig. 5.34. Each of them can, be obtained from Draw/Get New Part/libraries . . . /eval.lib or by simply selecting Draw/Get New Part and typing the part name in the, PartName dialog box, as usual. Note that each of them requires dc supplies, without which the op amp will not work. The dc supplies should, be connected as shown in Fig. 5.3., , U2, , U4, 3, , +, , 7, 5, V+ B2, B1, , 2, , +, , U3, 85, V+, , 6, , 3, 6, BB ⁄S, , 4, , V− G, 1, −, 4, LM111, , (a) JFET–input op, amp subcircuit, , (b) Op amp, subcircuit, , 2, , −, , V−, , 1, , LF411, , +, , 7, , 3, , Figure 5.34, Nonideal op amp model available in PSpice., , 2, , −, , 4 U1A, V+, 1, V−, , 11, LM324, (c) Five–, connection, op amp subcircuit, , 3, , 2, , +, , −, , 7, 5, V+ 052, V−, , 6, , 051, , 1, , 4, , uA741, (d) Five–connection, op amp subcircuit
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 195, , 5.9, , Op Amp Circuit Analysis with PSpice, , 195, , Example 5.11, , Use PSpice to solve the op amp circuit for Example 5.1., Solution:, Using Schematics, we draw the circuit in Fig. 5.6(a) as shown in, Fig. 5.35. Notice that the positive terminal of the voltage source vs is, connected to the inverting terminal (pin 2) via the 10-k resistor, while, the noninverting terminal (pin 3) is grounded as required in Fig. 5.6(a)., Also, notice how the op amp is powered; the positive power supply, terminal V (pin 7) is connected to a 15-V dc voltage source, while, the negative power supply terminal V (pin 4) is connected to 15 V., Pins 1 and 5 are left floating because they are used for offset null, adjustment, which does not concern us in this chapter. Besides adding, the dc power supplies to the original circuit in Fig. 5.6(a), we have also, added pseudocomponents VIEWPOINT and IPROBE to respectively, measure the output voltage vo at pin 6 and the required current i, through the 20-k resistor., , 0, −, VS +, , V2, +, , U1, 2V, , 3, R1, , 10 K, , 2, , +, −, , 7, 5, V+ 052, V−, 4, , 6, , –3.9983, , −, , 15 V, , +, , 051, , 1, , −, , uA741, , 15 V, , 0, , V3, , 1.999E–04, R2, 20 K, , Figure 5.35, Schematic for Example 5.11., , After saving the schematic, we simulate the circuit by selecting, Analysis/Simulate and have the results displayed on VIEWPOINT and, IPROBE. From the results, the closed-loop gain is, vo, 3.9983, 1.99915, , vs, 2, and i 0.1999 mA, in agreement with the results obtained analytically, in Example 5.1., , Rework Practice Prob. 5.1 using PSpice., Answer: 9.0027, 0.6502 mA., , Practice Problem 5.11
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 196, , Chapter 5, , 196, , 5.10, , Digital, input, (0000 –1111), , Analog, output, , Four-bit, DAC, , V3, , V2, , V4, Rf, , R1, , R2, , MSB, , R3, , Applications, , The op amp is a fundamental building block in modern electronic, instrumentation. It is used extensively in many devices, along with, resistors and other passive elements. Its numerous practical applications, include instrumentation amplifiers, digital-to-analog converters, analog, computers, level shifters, filters, calibration circuits, inverters, summers, integrators, differentiators, subtractors, logarithmic amplifiers,, comparators, gyrators, oscillators, rectifiers, regulators, voltage-tocurrent converters, current-to-voltage converters, and clippers. Some of, these we have already considered. We will consider two more applications here: the digital-to-analog converter and the instrumentation, amplifier., , 5.10.1 Digital-to-Analog Converter, , (a), V1, , Operational Amplifiers, , R4, LSB, , −, +, , (b), , Figure 5.36, Four-bit DAC: (a) block diagram, (b), binary weighted ladder type., , In practice, the voltage levels may be, typically 0 and ; 5 V., , Example 5.12, , Vo, , The digital-to-analog converter (DAC) transforms digital signals into, analog form. A typical example of a four-bit DAC is illustrated in, Fig. 5.36(a). The four-bit DAC can be realized in many ways. A simple realization is the binary weighted ladder, shown in Fig. 5.36(b)., The bits are weights according to the magnitude of their place value,, by descending value of RfRn so that each lesser bit has half the, weight of the next higher. This is obviously an inverting summing, amplifier. The output is related to the inputs as shown in Eq. (5.15)., Thus,, Vo , , Rf, R1, , V1 , , Rf, R2, , V2 , , Rf, R3, , V3 , , Rf, R4, , V4, , (5.23), , Input V1 is called the most significant bit (MSB), while input V4 is the, least significant bit (LSB). Each of the four binary inputs V1, . . . , V4, can assume only two voltage levels: 0 or 1 V. By using the proper input, and feedback resistor values, the DAC provides a single output that is, proportional to the inputs., , In the op amp circuit of Fig. 5.36(b), let Rf 10 k, R1 10 k,, R2 20 k, R3 40 k, and R4 80 k. Obtain the analog output, for binary inputs [0000], [0001], [0010], . . . , [1111]., Solution:, Substituting the given values of the input and feedback resistors in, Eq. (5.23) gives, Vo , , Rf, R1, , V1 , , Rf, R2, , V2 , , Rf, R3, , V3 , , Rf, R4, , V4, , V1 0.5V2 0.25V3 0.125V4, Using this equation, a digital input [V1V2V3V4] [0000] produces an analog output of Vo 0 V; [V1V2V3V4] [0001] gives Vo 0.125 V.
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 197, , 5.10, , Applications, , 197, , Similarly,, 3 V1 V2 V3 V4 4 300104, 3V1 V2 V3 V4 4 300114, 3V1 V2 V3 V4 4 301004, , 1, 1, 1, , Vo 0.25 V, Vo 0.25 0.125 0.375 V, Vo 0.5 V, , o, 3V1 V2 V3 V4 4 311114, , 1, , Vo 1 0.5 0.25 0.125, 1.875 V, , Table 5.2 summarizes the result of the digital-to-analog conversion., Note that we have assumed that each bit has a value of 0.125 V. Thus,, in this system, we cannot represent a voltage between 1.000 and 1.125,, for example. This lack of resolution is a major limitation of digital-toanalog conversions. For greater accuracy, a word representation with a, greater number of bits is required. Even then a digital representation, of an analog voltage is never exact. In spite of this inexact representation, digital representation has been used to accomplish remarkable, things such as audio CDs and digital photography., , TABLE 5.2, , Input and output values of the four-bit DAC., Binary input, [V1V2V3V4], , Decimal value, , Output, Vo, , 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, , 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, , 0, 0.125, 0.25, 0.375, 0.5, 0.625, 0.75, 0.875, 1.0, 1.125, 1.25, 1.375, 1.5, 1.625, 1.75, 1.875, , Practice Problem 5.12, , A three-bit DAC is shown in Fig. 5.37., (a), (b), (c), (d), , Determine |Vo| for [V1V2V3] [010]., Find |Vo| if [V1V2V3] [110]., If |Vo| 1.25 V is desired, what should be [V1V2V3]?, To get |Vo| 1.75 V, what should be [V1V2V3]?, , Answer: 0.5 V, 1.5 V, [101], [111]., , v1, v2, v3, , 10 kΩ, 20 kΩ, , 10 kΩ, , −, +, , 40 kΩ, , Figure 5.37, Three-bit DAC; for Practice Prob. 5.12., , vo
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 198, , Chapter 5, , 198, , Operational Amplifiers, , 5.10.2 Instrumentation Amplifiers, One of the most useful and versatile op amp circuits for precision, measurement and process control is the instrumentation amplifier (IA),, so called because of its widespread use in measurement systems. Typical applications of IAs include isolation amplifiers, thermocouple, amplifiers, and data acquisition systems., The instrumentation amplifier is an extension of the difference, amplifier in that it amplifies the difference between its input signals., As shown in Fig. 5.26 (see Example 5.8), an instrumentation amplifier, typically consists of three op amps and seven resistors. For convenience, the amplifier is shown again in Fig. 5.38(a), where the resistors are, made equal except for the external gain-setting resistor RG, connected, between the gain set terminals. Figure 5.38(b) shows its schematic, symbol. Example 5.8 showed that, vo Av(v2 v1), , Inverting input v, 1, Gain set, , R, , +, −1, , R, , R, RG, , Noninverting input, , −, +3, , R, , Gain set, v2, , (5.24), , vo, , Output, , R, , −, +2, , −, , R, , +, (a), , (b), , Figure 5.38, (a) The instrumentation amplifier with an external resistance to adjust the gain, (b) schematic diagram., , where the voltage gain is, Av 1 , , 2R, RG, , (5.25), , As shown in Fig. 5.39, the instrumentation amplifier amplifies small, differential signal voltages superimposed on larger common-mode, , −, RG, +, Small differential signals riding on larger, common-mode signals, , Instrumentation amplifier, , Amplified differential signal,, No common-mode signal, , Figure 5.39, The IA rejects common voltages but amplifies small signal voltages., T. L. Floyd, Electronic Devices, 2nd ed., Englewood Cliffs, NJ: Prentice Hall, 1996, p. 795.
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 199, , 5.11, , Summary, , 199, , voltages. Since the common-mode voltages are equal, they cancel each, other., The IA has three major characteristics:, 1. The voltage gain is adjusted by one external resistor RG., 2. The input impedance of both inputs is very high and does not vary, as the gain is adjusted., 3. The output vo depends on the difference between the inputs v1, and v2, not on the voltage common to them (common-mode, voltage)., Due to the widespread use of IAs, manufacturers have developed, these amplifiers on single-package units. A typical example is the, LH0036, developed by National Semiconductor. The gain can be varied from 1 to 1,000 by an external resistor whose value may vary from, 100 to 10 k., , In Fig. 5.38, let R 10 k, v1 2.011 V, and v2 2.017 V. If RG, is adjusted to 500 , determine: (a) the voltage gain, (b) the output, voltage vo., , Example 5.13, , Solution:, (a) The voltage gain is, Av 1 , , 2R, 2 10,000, 1, 41, RG, 500, , (b) The output voltage is, vo Av(v2 v1) 41(2.017 2.011) 41(6) mV 246 mV, , Determine the value of the external gain-setting resistor RG required, for the IA in Fig. 5.38 to produce a gain of 142 when R 25 k., Answer: 354.6 ., , 5.11, , Summary, , 1. The op amp is a high-gain amplifier that has high input resistance, and low output resistance., 2. Table 5.3 summarizes the op amp circuits considered in this chapter. The expression for the gain of each amplifier circuit holds, whether the inputs are dc, ac, or time-varying in general., , Practice Problem 5.13
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ale29559_ch05.qxd, , 200, , 07/16/2008, , 05:19 PM, , Page 200, , Chapter 5, , Operational Amplifiers, , TABLE 5.3, , Summary of basic op amp circuits., Op amp circuit, , Name/output-input relationship, Inverting amplifier, , R2, vi, , R1, , vo , , −, +, , vo, , Noninverting amplifier, R2, vo a1 b vi, R1, , R2, R1, vi, , vi, , v1, v2, v3, v1, , −, +, , vo, , −, +, , vo a, −, +, , R3, , R1, , vo, , Rf, R1, , v1 , , Rf, R2, , v2 , , Rf, R3, , v3 b, , Difference amplifier, , R2, −, +, , v2, , Summer, , Rf, , R2, , R1, , Voltage follower, vo vi, , vo, , R1, , R2, vi, R1, , vo , vo, , R2, (v2 v1), R1, , R2, , 3. An ideal op amp has an infinite input resistance, a zero output, resistance, and an infinite gain., 4. For an ideal op amp, the current into each of its two input terminals is zero, and the voltage across its input terminals is negligibly small., 5. In an inverting amplifier, the output voltage is a negative multiple, of the input., 6. In a noninverting amplifier, the output is a positive multiple of the, input., 7. In a voltage follower, the output follows the input., 8. In a summing amplifier, the output is the weighted sum of the, inputs., 9. In a difference amplifier, the output is proportional to the difference of the two inputs., 10. Op amp circuits may be cascaded without changing their inputoutput relationships., 11. PSpice can be used to analyze an op amp circuit., 12. Typical applications of the op amp considered in this chapter include, the digital-to-analog converter and the instrumentation amplifier.
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 201, , Review Questions, , 201, , Review Questions, 5.1, , (a) high and low., , If vs 8 mV in the circuit of Fig. 5.41, the output, voltage is:, , (b) positive and negative., , (a) 44 mV, , (b) 8 mV, , (c) inverting and noninverting., , (c) 4 mV, , (d) 7 mV, , The two input terminals of an op amp are labeled as:, , 5.6, , (d) differential and nondifferential., 5.2, , For an ideal op amp, which of the following statements, are not true?, , 5.7, , (a) The differential voltage across the input terminals, , Refer to Fig. 5.41. If vs 8 mV voltage va is:, (a) 8 mV, , (b) 0 mV, , (c) 103 mV, , (d) 8 mV, , is zero., (b) The current into the input terminals is zero., , 5.8, , (c) The current from the output terminal is zero., (d) The input resistance is zero., (e) The output resistance is zero., 5.3, , The power absorbed by the 4-k resistor in, Fig. 5.42 is:, (a) 9 mW, , (b) 4 mW, , (c) 2 mW, , (d) 1 mW, , For the circuit in Fig. 5.40, voltage vo is:, (a) 6 V, , (b) 5 V, , (c) 1.2 V, , (d) 0.2 V, 10 kΩ, , +, −, ix, , 6V, , 4 kΩ, , +, −, , +, 2 kΩ, , vo, −, , 2 kΩ, , −, +, , 1V +, −, , +, vo, −, , 3 kΩ, , Figure 5.42, For Review Questions 5.8., , Figure 5.40, For Review Questions 5.3 and 5.4., 5.9, 5.4, , 5.5, , For the circuit in Fig. 5.40, current ix is:, , Which of these amplifiers is used in a digital-to-analog, converter?, , (a) 0.6 mA, , (b) 0.5 mA, , (a) noninverter, , (c) 0.2 mA, , (d) 112 mA, , (b) voltage follower, , If vs 0 in the circuit of Fig. 5.41, current io is:, (a) 10 mA, , (b) 2.5 mA, , (c) 1012 mA, , (d) 1014 mA, , (c) summer, (d) difference amplifier, 5.10 Difference amplifiers are used in:, (a) instrumentation amplifiers, (b) voltage followers, , 8 kΩ, , (c) voltage regulators, 4 kΩ, a, , (d) buffers, , −, +, , 10 mV, , +, −, , vs, , +, −, , Figure 5.41, For Review Questions 5.5, 5.6, and 5.7., , 2 kΩ, , io, +, vo, −, , (e) summing amplifiers, (f ) subtracting amplifiers, , Answers: 5.1c, 5.2c,d, 5.3b, 5.4b, 5.5a, 5.6c, 5.7d, 5.8b,, 5.9c, 5.10a,f.
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 202, , Chapter 5, , 202, , Operational Amplifiers, , Problems, Section 5.2 Operational Amplifiers, 5.1, , +, 741, −, , The equivalent model of a certain op amp is shown, in Fig. 5.43. Determine:, , +−, , (a) the input resistance, , 1 mV, , (b) the output resistance, , Figure 5.45, , (c) the voltage gain in dB., , For Prob. 5.6., , 60 Ω, −, vd, , 1.5 MΩ, , vo, , +, −, , 5.7, , 8 × 104v d, , The op amp in Fig. 5.46 has Ri 100 k,, Ro 100 , A 100,000. Find the differential, voltage vd and the output voltage vo., , +, , Figure 5.43, , − +, vd, +, −, , For Prob. 5.1., , 5.2, , 5.3, , 5.4, , 5.5, , The open-loop gain of an op amp is 100,000. Calculate, the output voltage when there are inputs of 10 V, on the inverting terminal and 20 V on the, noninverting terminal., Determine the output voltage when 20 V is, applied to the inverting terminal of an op amp and, 30 V to its noninverting terminal. Assume that, the op amp has an open-loop gain of 200,000., The output voltage of an op amp is 4 V when the, noninverting input is 1 mV. If the open-loop gain, of the op amp is 2 106, what is the inverting, input?, For the op amp circuit of Fig. 5.44, the op amp has, an open-loop gain of 100,000, an input resistance of, 10 k, and an output resistance of 100 . Find the, voltage gain vovi using the nonideal model of the, op amp., , 10 kΩ, , 1 mV, , 100 kΩ, +, vo, −, , +, −, , Figure 5.46, For Prob. 5.7., , Section 5.3 Ideal Op Amp, 5.8, , Obtain vo for the op amp circuit in Fig. 5.47., , 10 kΩ, 5 kΩ, 3V, , −, +, vi, , +, −, , −, +, +, , 1 mA, , vo, , −, +, −, +, , +, vo, −, , 7V, , +, −, , 2 kΩ, , −, , Figure 5.44, For Prob. 5.5., , (a), , (b), , Figure 5.47, For Prob. 5.8., , 5.6, , Using the same parameters for the 741 op amp in, Example 5.1, find vo in the op amp circuit of, Fig. 5.45., , 5.9, , Determine vo for each of the op amp circuits in, Fig. 5.48., , +, vo, −
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 204, , Chapter 5, , 204, , Operational Amplifiers, , 5.16 Using Fig. 5.55, design a problem to help students, better understand inverting op amps., , 5.19 Determine io in the circuit of Fig. 5.58., 2 kΩ, , 4 kΩ, , 10 kΩ, , R3, R1, , ix, , +, −, , 1V, , iy, , io, , −, +, , 4 kΩ, , 5 kΩ, , –, +, , +, V −, , Figure 5.58, , R4, , For Prob. 5.19., , R2, , 5.20 In the circuit of Fig. 5.59, calculate vo of vs 0., 8 kΩ, , Figure 5.55, For Prob. 5.16., , 4 kΩ, , 5.17 Calculate the gain vovi when the switch in Fig. 5.56, is in:, (a) position 1, , (b) position 2, , 9V, , 2 kΩ, , 4 kΩ, , +, −, , −, +, vs, , (c) position 3, , +, vo, , +, −, , −, , 12 kΩ, , Figure 5.59, , 1, , For Prob. 5.20., , 80 kΩ, 2, 2 MΩ, 10 kΩ, , vi, , 5.21 Calculate vo in the op amp circuit of Fig. 5.60., , 3, , 10 kΩ, , −, +, , +, −, , 4 kΩ, 10 kΩ, , +, vo, −, , −, +, , 3V, , +, −, , 1V, , +, vo, , +, −, , −, , Figure 5.56, For Prob. 5.17., , Figure 5.60, For Prob. 5.21., , *5.18 For the circuit shown in Figure 5.57, solve for the, Thevenin equivalent circuit looking into terminals A, and B., , 5.23 For the op amp circuit in Fig. 5.61, find the voltage, gain vovs., , 10 kΩ, 10 kΩ, , 5V +, −, , 5.22 Design an inverting amplifier with a gain of 15., , Rf, , −, , a, , +, , R1, , 10 Ω, b, , –, +, vs +, −, , Figure 5.57, , +, vo, −, , For Prob. 5.18., , Figure 5.61, * An asterisk indicates a challenging problem., , R2, , For Prob. 5.23.
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 205, , Problems, , 5.24 In the circuit shown in Fig. 5.62, find k in the voltage, transfer function vo kvs., , 205, , 5.28 Find io in the op amp circuit of Fig. 5.66., 50 kΩ, , Rf, R1, , −, +, , R2, , vs, , −, +, , −+, , 10 kΩ, , + 10 V, −, , +, , 20 kΩ, , vo, , R4, , R3, , io, , −, , Figure 5.66, For Prob. 5.28., , Figure 5.62, For Prob. 5.24., , 5.29 Determine the voltage gain vovi of the op amp, circuit in Fig. 5.67., , Section 5.5 Noninverting Amplifier, R1, , 5.25 Calculate vo in the op amp circuit of Fig. 5.63., , +, −, , 12 kΩ, , −, +, 2V +, −, , vi +, −, 20 kΩ, , R2, , +, vo, , R2, R1, , +, vo, −, , –, , Figure 5.67, For Prob. 5.29., , Figure 5.63, For Prob. 5.25., , 5.30 In the circuit shown in Fig. 5.68, find ix and the, power absorbed by the 20-k resistor., , 5.26 Using Fig. 5.64, design a problem to help other, students better understand noninverting op amps., +, −, , io, 2.4 V, , V +, −, , ix, , +, −, , 30 kΩ, , 20 kΩ, , R3, , R2, , R1, , 60 kΩ, , −, +, , Figure 5.68, For Prob. 5.30., , Figure 5.64, For Prob. 5.26., , 5.31 For the circuit in Fig. 5.69, find ix., , 5.27 Find vo in the op amp circuit of Fig. 5.65., 16 Ω, , 5V, , +, −, , v1, , −, +, 24 Ω, , 12 kΩ, 6 kΩ, , v2 8 Ω, , ix, 12 Ω, , +, vo, −, , 4 mA, , 3 kΩ, , 6 kΩ, , +, −, , +, vo, −, , Figure 5.65, , Figure 5.69, , For Prob. 5.27., , For Prob. 5.31.
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 206, , Chapter 5, , 206, , Operational Amplifiers, +, −, , 5.32 Calculate ix and vo in the circuit of Fig. 5.70. Find, the power dissipated by the 30-k resistor., vs, ix, , +, −, , 4 mV, , +, −, , +, −, , a, , R2, R1, , 48 kΩ, , b, , 50 kΩ, , 60 kΩ, , 30 kΩ, , 10 kΩ, , +, vo, −, , Figure 5.73, For Prob. 5.36., , Section 5.6 Summing Amplifier, 5.37 Determine the output of the summing amplifier in, Fig. 5.74., , Figure 5.70, 1V, , For Prob. 5.32., , −+, , 5.33 Refer to the op amp circuit in Fig. 5.71. Calculate ix, and the power dissipated by the 3-k resistor., , 2V, −+, 3V, , 1 kΩ, , +, −, 4 kΩ, , 3 mA, , 2 kΩ, , 10 kΩ, 30 kΩ, 20 kΩ, , −, +, , 30 kΩ, , +−, ix, , +, vo, −, , Figure 5.74, For Prob. 5.37., , 3 kΩ, , 5.38 Using Fig. 5.75, design a problem to help other, students better understand summing amplifiers., V1, −+, , Figure 5.71, , R1, , For Prob. 5.33., V2, +−, , 5.34 Given the op amp circuit shown in Fig. 5.72, express, vo in terms of v1 and v2., , V3, −+, V4, , R1, v1, v2, , v in, , +−, , +, −, R4, , R2, , +, vo, , R3, , –, , R3, , +, vo, −, , R5, R4, , For Prob. 5.38., 5.39 For the op amp circuit in Fig. 5.76, determine the, value of v2 in order to make vo 16.5 V., 10 kΩ, , 50 kΩ, , 20 kΩ, , −, +, , +2 V, v2, , 5.35 Design a noninverting amplifier with a gain of 10., , 50 kΩ, –1 V, , 5.36 For the circuit shown in Fig. 5.73, find the Thevenin, equivalent at terminals a-b. (Hint: To find RTh, apply, a current source io and calculate vo.), , −, , Figure 5.75, , Figure 5.72, For Prob. 5.34., , +, R2, , Figure 5.76, For Prob. 5.39., , vo
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 207, , Problems, , 5.40 Find vo in terms of v1, v2, and v3 in the circuit of, Fig. 5.77., , 207, , 5.46 Using only two op amps, design a circuit to solve, vout , , v3, v1 v2, , 3, 2, , +, vo, , −, R, , R, , R, , R1, , Section 5.7 Difference Amplifier, 5.47 The circuit in Fig. 5.79 is for a difference amplifier., Find vo given that v1 1 V and v2 2 V., , v1, , +, −, , v2, , +, −, , + v, − 3, , R2, , 30 kΩ, , Figure 5.77, For Prob. 5.40., , 2 kΩ, , 5.41 An averaging amplifier is a summer that provides, an output equal to the average of the inputs. By, using proper input and feedback resistor values,, one can get, vout , , 1, 4 (v1, , −, 2 kΩ, , v1 +, −, , v2 +, −, , vo, , 20 kΩ, , −, , v2 v3 v4), , Using a feedback resistor of 10 k design an, averaging amplifier with four inputs., 5.42 A three-input summing amplifier has input resistors, with R1 R2 R3 30 k. To produce an, averaging amplifier, what value of feedback resistor, is needed?, , Figure 5.79, For Prob. 5.47., , 5.48 The circuit in Fig. 5.80 is a differential amplifier, driven by a brige. Find vo., , 5.43 A four-input summing amplifier has R1 R2 , R3 R4 12 k. What value of feedback resistor, is needed to make it an averaging amplifier?, , 20 kΩ, , 5.44 Show that the output voltage vo of the circuit in, Fig. 5.78 is, vo , , +, , +, , (R3 R4), (R2v1 R1v2), R3(R1 R2), , 10 kΩ, , 80 kΩ, , 30 kΩ, −, +, , + 5 mV, 40 kΩ, , vo, , 60 kΩ, , R4, 20 kΩ, R3, , v1, v2, , 80 kΩ, −, vo, , R1, +, , Figure 5.80, For Prob. 5.48., , R2, , Figure 5.78, , 5.49 Design a difference amplifier to have a gain of 2 and, a common-mode input resistance of 10 k at each, input., , For Prob. 5.44., , 5.45 Design an op amp circuit to perform the following, operation:, vo 3v1 2v2, All resistances must be 100 k., , 5.50 Design a circuit to amplify the difference between, two inputs by 2., (a) Use only one op amp., (b) Use two op amps.
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ale29559_ch05.qxd, , 07/16/2008, , 05:19 PM, , Page 208, , Chapter 5, , 208, , Operational Amplifiers, , 5.51 Using two op amps, design a subtractor., , R2, 2, R1, , *5.52 Design an op amp circuit such that, vo 2v1 4v2 5v3 v4, Let all the resistors be in the range of 5 to 100 k., , R2, 2, , RG, , −, , +, vi, −, , +, +, R1, , *5.53 The ordinary difference amplifier for fixed-gain, operation is shown in Fig. 5.81(a). It is simple and, reliable unless gain is made variable. One way of, providing gain adjustment without losing simplicity, and accuracy is to use the circuit in Fig. 5.81(b)., Another way is to use the circuit in Fig. 5.81(c)., Show that:, , vo, R2, 2, , R2, 2, , −, , (c), , Figure 5.81, For Prob. 5.53., , (a) for the circuit in Fig. 5.81(a),, vo, R2, , vi, R1, , Section 5.8 Cascaded Op Amp Circuits, 5.54 Determine the voltage transfer ratio vovs in the op, amp circuit of Fig. 5.82, where R 10 k., , (b) for the circuit in Fig. 5.81(b),, vo, R2, , vi, R1, , 1, 1, , R1, 2RG, , R, R, , (c) for the circuit in Fig. 5.81(c),, R, , vo, R2, R2, , a1 , b, vi, R1, 2RG, , −, +, , +, , +, −, , vs, −, , −, , −, vi, +, , R, , −, , Figure 5.82, +, , For Prob. 5.54., , +, R1, , vo, , R2, , −, , (a), R1, 2, , R2, , R1, 2, , 5.55 In a certain electronic device, a three-stage amplifier, is desired, whose overall voltage gain is 42 dB. The, individual voltage gains of the first two stages are to, be equal, while the gain of the third is to be onefourth of each of the first two. Calculate the voltage, gain of each., 5.56 Using Fig. 5.83, design a problem to help other, students better understand cascaded op amps., , −, , −, vi, +, , RG, , R2, , +, , R4, , +, R1, 2, , vo, , R, , R2, R1, , +, , R1, 2, , R2, , vo, −, , R1, +, vi, −, , Figure 5.83, (b), , For Prob. 5.56., , −, +, , R3, , −, +
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ale29559_ch05.qxd, , 07/16/2008, , 05:20 PM, , Page 212, , Chapter 5, , 212, , Operational Amplifiers, , 5.80 Use PSpice to solve Prob. 5.70., 5.81 Use PSpice to verify the results in Example 5.9., Assume nonideal op amps LM324., , Section 5.10 Applications, 5.82 A five-bit DAC covers a voltage range of 0 to 7.75 V., Calculate how much voltage each bit is worth., 5.83 Design a six-bit digital-to-analog converter., , 5.86 Design a voltage controlled ideal current source, (within the operating limits of the op amp) where the, output current is equal to 200 vs(t) mA.., , 5.87 Figure 5.105 displays a two-op-amp instrumentation, amplifier. Derive an expression for vo in terms of v1, and v2. How can this amplifier be used as a, subtractor?, , (a) If |Vo| 1.1875 V is desired, what should, [V1V2V3V4V5V6] be?, (b) Calculate |Vo| if [V1V2V3V4V5V6] [011011]., (c) What is the maximum value |Vo| can assume?, *5.84 A four-bit R-2R ladder DAC is presented in Fig. 5.103., , v1, , −, , (a) Show that the output voltage is given by, Vo Rf a, , R4, , +, R2, , V3, V1, V2, V4, , , , b, 2R, 4R, 8R, 16R, , v2, , R1, , (b) If Rf 12 k and R 10 k, find |Vo| for, [V1V2V3V4] [1011] and [V1V2V3V4] [0101]., , R3, , −, , vo, , +, , Figure 5.105, For Prob. 5.87., , Rf, 2R, , −, +, , V1, , Vo, , R, 2R, V2, , *5.88 Figure 5.106 shows an instrumentation amplifier, driven by a bridge. Obtain the gain vovi of the, amplifier., , R, 2R, V3, R, 2R, V4, R, , 20 kΩ, , Figure 5.103, For Prob. 5.84., , 30 kΩ, , vi, , 5.85 In the op amp circuit of Fig. 5.104, find the value of, R so that the power absorbed by the 10-k resistor is, 10 mW. Take vs 2 V., , 25 kΩ, , 40 kΩ, , 80 kΩ, 2 kΩ, 10 kΩ, −, +, , 10 kΩ, , 25 kΩ, , R, 40 kΩ, , 500 kΩ, , Figure 5.104, , Figure 5.106, , For Prob. 5.85., , For Prob. 5.88., , 500 kΩ, , −, +, , 10 kΩ, , +, −, +, − vs, , +, −, , vo
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ale29559_ch05.qxd, , 07/16/2008, , 05:20 PM, , Page 213, , Comprehensive Problems, , 213, , Comprehensive Problems, 5.89 Design a circuit that provides a relationship between, output voltage vo and input voltage vs such that, vo 12vs 10. Two op amps, a 6-V battery, and, several resistors are available., , 5.92 Refer to the bridge amplifier shown in Fig. 5.109., Determine the voltage gain vovi., 60 kΩ, , 5.90 The op amp circuit in Fig. 5.107 is a current, amplifier. Find the current gain iois of the amplifier., , 30 kΩ, , −, +, 50 kΩ, , 20 kΩ, 20 kΩ, −, +, , vi, , RL, , +, vo, −, , −, +, , +, −, , 4 kΩ, io, is, , 5 kΩ, , Figure 5.109, For Prob. 5.92., , 2 kΩ, , *5.93 A voltage-to-current converter is shown in Fig. 5.110,, which means that iL Avi if R1R2 R3R4. Find the, constant term A., , Figure 5.107, For Prob. 5.90., , R3, R1, , 5.91 A noninverting current amplifier is portrayed in, Fig. 5.108. Calculate the gain iois. Take R1 8 k, and R2 1 k., , +, , R4, vi, , −, +, R1, , R2, −, R2, , Figure 5.110, For Prob. 5.93., , Figure 5.108, For Prob. 5.91., , iL, R2, , io, is, , −, +, , RL
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ale29559_ch06.qxd, , 07/08/2008, , 10:59 AM, , Page 215, , c h a p t e r, , 6, , Capacitors and, Inductors, But in science the credit goes to the man who convinces the world, not, to the man whom the idea first occurs., —Francis Darwin, , Enhancing Your Skills and Your Career, ABET EC 2000 criteria (3.c), “an ability to design a system,, component, or process to meet desired needs.”, The “ability to design a system, component, or process to meet, desired needs” is why engineers are hired. That is why this is the, most important technical skill that an engineer has. Interestingly, your, success as an engineer is directly proportional to your ability to communicate but your being able to design is why you will be hired in, the first place., Design takes place when you have what is termed an open-ended, problem that eventually is defined by the solution. Within the context, of this course or textbook, we can only explore some of the elements, of design. Pursuing all of the steps of our problem-solving technique, teaches you several of the most important elements of the design, process., Probably the most important part of design is clearly defining what, the system, component, process, or, in our case, problem is. Rarely is, an engineer given a perfectly clear assignment. Therefore, as a student,, you can develop and enhance this skill by asking yourself, your colleagues, or your professors questions designed to clarify the problem, statement., Exploring alternative solutions is another important part of the, design process. Again, as a student, you can practice this part of the, design process on almost every problem you work., Evaluating your solutions is critical to any engineering assignment., Again, this is a skill that you as a student can practice on every problem you work., , Photo by Charles Alexander, , 215
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ale29559_ch06.qxd, , 07/08/2008, , 10:59 AM, , Page 216, , Chapter 6, , 216, , 6.1, , In contrast to a resistor, which spends, or dissipates energy irreversibly, an, inductor or capacitor stores or releases, energy (i.e., has a memory)., , Dielectric with permittivity , Metal plates,, each with area A, , Capacitors and Inductors, , Introduction, , So far we have limited our study to resistive circuits. In this chapter,, we shall introduce two new and important passive linear circuit elements: the capacitor and the inductor. Unlike resistors, which dissipate, energy, capacitors and inductors do not dissipate but store energy,, which can be retrieved at a later time. For this reason, capacitors and, inductors are called storage elements., The application of resistive circuits is quite limited. With the introduction of capacitors and inductors in this chapter, we will be able to, analyze more important and practical circuits. Be assured that the circuit analysis techniques covered in Chapters 3 and 4 are equally applicable to circuits with capacitors and inductors., We begin by introducing capacitors and describing how to combine them in series or in parallel. Later, we do the same for inductors., As typical applications, we explore how capacitors are combined with, op amps to form integrators, differentiators, and analog computers., , 6.2, , Capacitors, , A capacitor is a passive element designed to store energy in its electric field. Besides resistors, capacitors are the most common electrical, components. Capacitors are used extensively in electronics, communications, computers, and power systems. For example, they are used in, the tuning circuits of radio receivers and as dynamic memory elements, in computer systems., A capacitor is typically constructed as depicted in Fig. 6.1., , d, , Figure 6.1, A typical capacitor., , A capacitor consists of two conducting plates separated by an insulator (or dielectric)., −, , +, +, +q, , −, , +, +, , −, , +, +, +, v, , −q, , −, , In many practical applications, the plates may be aluminum foil while, the dielectric may be air, ceramic, paper, or mica., When a voltage source v is connected to the capacitor, as in, Fig. 6.2, the source deposits a positive charge q on one plate and a negative charge q on the other. The capacitor is said to store the electric, charge. The amount of charge stored, represented by q, is directly proportional to the applied voltage v so that, , Figure 6.2, A capacitor with applied voltage v., , Alternatively, capacitance is the amount, of charge stored per plate for a unit, voltage difference in a capacitor., , q Cv, , (6.1), , where C, the constant of proportionality, is known as the capacitance, of the capacitor. The unit of capacitance is the farad (F), in honor of, the English physicist Michael Faraday (1791–1867). From Eq. (6.1),, we may derive the following definition., Capacitance is the ratio of the charge on one plate of a capacitor to, the voltage difference between the two plates, measured in farads (F)., , Note from Eq. (6.1) that 1 farad 1 coulomb/volt.
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ale29559_ch06.qxd, , 07/16/2008, , 12:41 PM, , Page 217, , 6.2, , Capacitors, , 217, , Historical, Michael Faraday (1791–1867), an English chemist and physicist,, was probably the greatest experimentalist who ever lived., Born near London, Faraday realized his boyhood dream by working with the great chemist Sir Humphry Davy at the Royal Institution, where he worked for 54 years. He made several contributions, in all areas of physical science and coined such words as electrolysis, anode, and cathode. His discovery of electromagnetic induction, in 1831 was a major breakthrough in engineering because it provided, a way of generating electricity. The electric motor and generator operate on this principle. The unit of capacitance, the farad, was named, in his honor., The Burndy Library Collection, at The Huntington Library,, San Marino, California., , Although the capacitance C of a capacitor is the ratio of the charge, q per plate to the applied voltage v, it does not depend on q or v. It, depends on the physical dimensions of the capacitor. For example, for, the parallel-plate capacitor shown in Fig. 6.1, the capacitance is given by, C, , A, d, , (6.2), , Capacitor voltage rating and capacitance are typically inversely rated due, to the relationships in Eqs. (6.1) and, (6.2). Arcing occurs if d is small and V, is high., , where A is the surface area of each plate, d is the distance between, the plates, and is the permittivity of the dielectric material between, the plates. Although Eq. (6.2) applies to only parallel-plate capacitors,, we may infer from it that, in general, three factors determine the value, of the capacitance:, 1. The surface area of the plates—the larger the area, the greater the, capacitance., 2. The spacing between the plates—the smaller the spacing, the greater, the capacitance., 3. The permittivity of the material—the higher the permittivity, the, greater the capacitance., Capacitors are commercially available in different values and types., Typically, capacitors have values in the picofarad (pF) to microfarad (mF), range. They are described by the dielectric material they are made of and, by whether they are of fixed or variable type. Figure 6.3 shows the circuit symbols for fixed and variable capacitors. Note that according to the, passive sign convention, if v 7 0 and i 7 0 or if v 6 0 and i 6 0, the, capacitor is being charged, and if v i 6 0, the capacitor is discharging., Figure 6.4 shows common types of fixed-value capacitors. Polyester capacitors are light in weight, stable, and their change with temperature is predictable. Instead of polyester, other dielectric materials, such as mica and polystyrene may be used. Film capacitors are rolled, and housed in metal or plastic films. Electrolytic capacitors produce, very high capacitance. Figure 6.5 shows the most common types of, variable capacitors. The capacitance of a trimmer (or padder) capacitor, , i, , C, + v −, (a), , i, , C, + v −, (b), , Figure 6.3, Circuit symbols for capacitors: (a) fixed, capacitor, (b) variable capacitor.
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ale29559_ch06.qxd, , 07/08/2008, , 10:59 AM, , Page 218, , Chapter 6, , 218, , Capacitors and Inductors, , (b), , (a), , (c), , Figure 6.4, Fixed capacitors: (a) polyester capacitor, (b) ceramic capacitor, (c) electrolytic capacitor., Courtesy of Tech America., , is often placed in parallel with another capacitor so that the equivalent, capacitance can be varied slightly. The capacitance of the variable air, capacitor (meshed plates) is varied by turning the shaft. Variable capacitors are used in radio receivers allowing one to tune to various stations. In addition, capacitors are used to block dc, pass ac, shift phase,, store energy, start motors, and suppress noise., To obtain the current-voltage relationship of the capacitor, we take, the derivative of both sides of Eq. (6.1). Since, , (a), , i, , dq, dt, , (6.3), , differentiating both sides of Eq. (6.1) gives, , iC, , (b), , dv, dt, , (6.4), , Figure 6.5, Variable capacitors: (a) trimmer capacitor,, (b) filmtrim capacitor., Courtesy of Johanson., , According to Eq. (6.4), for a capacitor, to carry current, its voltage must vary, with time. Hence, for constant voltage,, i 0., , This is the current-voltage relationship for a capacitor, assuming the, passive sign convention. The relationship is illustrated in Fig. 6.6 for, a capacitor whose capacitance is independent of voltage. Capacitors, that satisfy Eq. (6.4) are said to be linear. For a nonlinear capacitor,, the plot of the current-voltage relationship is not a straight line., Although some capacitors are nonlinear, most are linear. We will, assume linear capacitors in this book., The voltage-current relation of the capacitor can be obtained by, integrating both sides of Eq. (6.4). We get, v, , i, , 1, C, , , , t, , i dt, , (6.5), , , , or, Slope = C, , 0, , dv ⁄dt, , Figure 6.6, Current-voltage relationship of a capacitor., , v, , 1, C, , t, , i dt v(t ), 0, , (6.6), , t0, , where v(t0) q(t0)C is the voltage across the capacitor at time t0., Equation (6.6) shows that capacitor voltage depends on the past history
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ale29559_ch06.qxd, , 07/08/2008, , 10:59 AM, , Page 219, , 6.2, , Capacitors, , 219, , of the capacitor current. Hence, the capacitor has memory—a property, that is often exploited., The instantaneous power delivered to the capacitor is, p vi C v, , dv, dt, , (6.7), , The energy stored in the capacitor is therefore, w, , , , t, , , , p dt C, , , , t, , , , v, , dv, dt C, dt, , v(t), 1, v dv Cv2 `, 2, v(), v(), , , , v(t), , (6.8), , We note that v() 0, because the capacitor was uncharged at, t . Thus,, 1, w Cv2, 2, , (6.9), , Using Eq. (6.1), we may rewrite Eq. (6.9) as, w, , q2, 2C, , (6.10), , Equation (6.9) or (6.10) represents the energy stored in the electric field, that exists between the plates of the capacitor. This energy can be, retrieved, since an ideal capacitor cannot dissipate energy. In fact, the, word capacitor is derived from this element’s capacity to store energy, in an electric field., We should note the following important properties of a capacitor:, 1. Note from Eq. (6.4) that when the voltage across a capacitor is not, changing with time (i.e., dc voltage), the current through the capacitor is zero. Thus,, , v, , v, , t, , t, , (a), , (b), , Figure 6.7, A capacitor is an open circuit to dc., , However, if a battery (dc voltage) is connected across a capacitor,, the capacitor charges., 2. The voltage on the capacitor must be continuous., The voltage on a capacitor cannot change abruptly., , The capacitor resists an abrupt change in the voltage across it., According to Eq. (6.4), a discontinuous change in voltage requires, an infinite current, which is physically impossible. For example,, the voltage across a capacitor may take the form shown in, Fig. 6.7(a), whereas it is not physically possible for the capacitor, voltage to take the form shown in Fig. 6.7(b) because of the abrupt, changes. Conversely, the current through a capacitor can change, instantaneously., 3. The ideal capacitor does not dissipate energy. It takes power from, the circuit when storing energy in its field and returns previously, stored energy when delivering power to the circuit., 4. A real, nonideal capacitor has a parallel-model leakage resistance,, as shown in Fig. 6.8. The leakage resistance may be as high as, , Voltage across a capacitor: (a) allowed,, (b) not allowable; an abrupt change is not, possible., , An alternative way of looking at this is, using Eq. (6.9), which indicates that, energy is proportional to voltage, squared. Since injecting or extracting, energy can only be done over some, finite time, voltage cannot change, instantaneously across a capacitor., , Leakage resistance, , Capacitance, , Figure 6.8, Circuit model of a nonideal capacitor.
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ale29559_ch06.qxd, , 07/08/2008, , 10:59 AM, , 220, , Page 220, , Chapter 6, , Capacitors and Inductors, , 100 M and can be neglected for most practical applications. For, this reason, we will assume ideal capacitors in this book., , Example 6.1, , (a) Calculate the charge stored on a 3-pF capacitor with 20 V across it., (b) Find the energy stored in the capacitor., Solution:, (a) Since q Cv,, q 3 1012 20 60 pC, (b) The energy stored is, 1, 1, w Cv2 3 1012 400 600 pJ, 2, 2, , Practice Problem 6.1, , What is the voltage across a 3-mF capacitor if the charge on one plate, is 0.12 mC? How much energy is stored?, Answer: 40 V, 2.4 mJ., , Example 6.2, , The voltage across a 5-mF capacitor is, v(t) 10 cos 6000t V, Calculate the current through it., Solution:, By definition, the current is, dv, d, 5 106 (10 cos 6000t), dt, dt, 5 106 6000 10 sin 6000t 0.3 sin 6000t A, , i(t) C, , Practice Problem 6.2, , If a 10-mF capacitor is connected to a voltage source with, v(t) 50 sin 2000t V, determine the current through the capacitor., Answer: cos 2000t A., , Example 6.3, , Determine the voltage across a 2-mF capacitor if the current through it is, i(t) 6e3000t mA, Assume that the initial capacitor voltage is zero.
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ale29559_ch06.qxd, , 07/08/2008, , 10:59 AM, , Page 221, , 6.2, , Capacitors, , 221, , Solution:, Since v , , 1, C, , t, , i dt v(0) and v(0) 0,, 0, , v, , 1, 2 106, , t, , 6e, , 3000t, , dt 103, , 0, , 3 10 3000t t, e, ` (1 e3000t ) V, 3000, 0, 3, , , , Practice Problem 6.3, , The current through a 100-mF capacitor is i(t) 50 sin 120 p t mA., Calculate the voltage across it at t 1 ms and t 5 ms. Take v(0) 0., Answer: 93.14 mV, 1.736 V., , Determine the current through a 200-mF capacitor whose voltage is, shown in Fig. 6.9., Solution:, The voltage waveform can be described mathematically as, 50t V, 100 50t V, v(t) d, 200 50t V, 0, , 0 6 t 6 1, 1 6 t 6 3, 3 6 t 6 4, otherwise, , Since i C dvdt and C 200 mF, we take the derivative of v to obtain, 50, 50, i(t) 200 106 d, 50, 0, 10 mA, 10 mA, d, 10 mA, 0, , 0 6 t 6 1, 1 6 t 6 3, 3 6 t 6 4, otherwise, , 0 6 t 6 1, 1 6 t 6 3, 3 6 t 6 4, otherwise, , Thus the current waveform is as shown in Fig. 6.10., , An initially uncharged 1-mF capacitor has the current shown in, Fig. 6.11 across it. Calculate the voltage across it at t 2 ms and, t 5 ms., , Example 6.4, v (t), 50, , 0, , 1, , 2, , 3, , 4, , t, , 3, , 4, , t, , −50, , Figure 6.9, For Example 6.4., i (mA), 10, , 0, 1, , 2, , −10, , Figure 6.10, For Example 6.4., , Practice Problem 6.4, i (mA), 100, , Answer: 100 mV, 400 mV., 0, , 2, , Figure 6.11, For Practice Prob. 6.4., , 4, , 6, , t (ms)
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ale29559_ch06.qxd, , 07/08/2008, , 10:59 AM, , Page 222, , Chapter 6, , 222, , Example 6.5, , Capacitors and Inductors, , Obtain the energy stored in each capacitor in Fig. 6.12(a) under dc, conditions., + v1 −, , 2 mF, , i, , 2 kΩ, 2 kΩ, 5 kΩ, 6 mA, , 5 kΩ, 6 mA, , 3 kΩ, , 3 kΩ, , +, v2, −, , 4 kΩ, 4 mF, , 4 kΩ, , (b), , (a), , Figure 6.12, For Example 6.5., , Solution:, Under dc conditions, we replace each capacitor with an open circuit,, as shown in Fig. 6.12(b). The current through the series combination, of the 2-k and 4-k resistors is obtained by current division as, i, , 3, (6 mA) 2 mA, 324, , Hence, the voltages v1 and v2 across the capacitors are, v1 2000i 4 V, , v2 4000i 8 V, , and the energies stored in them are, 1, 1, w1 C1v21 (2 103)(4)2 16 mJ, 2, 2, 1, 1, w2 C2v22 (4 103)(8)2 128 mJ, 2, 2, , Practice Problem 6.5, 3 kΩ, , Under dc conditions, find the energy stored in the capacitors in Fig. 6.13., Answer: 810 mJ, 135 mJ., , 1 kΩ, , 10 V, , +, −, , 30 F, 20 F, , 6 kΩ, , 6.3, Figure 6.13, For Practice Prob. 6.5., , Series and Parallel Capacitors, , We know from resistive circuits that the series-parallel combination is a, powerful tool for reducing circuits. This technique can be extended to, series-parallel connections of capacitors, which are sometimes encountered. We desire to replace these capacitors by a single equivalent, capacitor Ceq., In order to obtain the equivalent capacitor Ceq of N capacitors in, parallel, consider the circuit in Fig. 6.14(a). The equivalent circuit is
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ale29559_ch06.qxd, , 07/08/2008, , 224, , 10:59 AM, , Page 224, , Chapter 6, , Capacitors and Inductors, , The initial voltage v(t0) across Ceq is required by KVL to be the sum, of the capacitor voltages at t0. Or according to Eq. (6.15),, v(t0) v1(t0) v2(t0) p vN (t0), Thus, according to Eq. (6.16),, The equivalent capacitance of series-connected capacitors is the, reciprocal of the sum of the reciprocals of the individual capacitances., , Note that capacitors in series combine in the same manner as resistors, in parallel. For N 2 (i.e., two capacitors in series), Eq. (6.16), becomes, 1, 1, 1, , , Ceq, C1, C2, or, , Ceq , , Example 6.6, , C1C2, C1 C2, , (6.17), , Find the equivalent capacitance seen between terminals a and b of the, circuit in Fig. 6.16., 5 F, , 60 F, a, , 20 F, , 6 F, , 20 F, , Ceq, b, , Figure 6.16, For Example 6.6., , Solution:, The 20-mF and 5-mF capacitors are in series; their equivalent capacitance is, 20 5, 4 mF, 20 5, This 4-mF capacitor is in parallel with the 6-mF and 20-mF capacitors;, their combined capacitance is, 4 6 20 30 mF, This 30-mF capacitor is in series with the 60-mF capacitor. Hence, the, equivalent capacitance for the entire circuit is, Ceq , , 30 60, 20 mF, 30 60
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ale29559_ch06.qxd, , 07/08/2008, , 10:59 AM, , Page 225, , 6.3, , Series and Parallel Capacitors, , 225, , Practice Problem 6.6, , Find the equivalent capacitance seen at the terminals of the circuit in, Fig. 6.17., , 50 F, 60 F, , Answer: 40 mF., Ceq, , 70 F, , 20 F, , 120 F, , Figure 6.17, For Practice Prob. 6.6., , Example 6.7, , For the circuit in Fig. 6.18, find the voltage across each capacitor., Solution:, We first find the equivalent capacitance Ceq, shown in Fig. 6.19. The two, parallel capacitors in Fig. 6.18 can be combined to get 40 20 60 mF., This 60-mF capacitor is in series with the 20-mF and 30-mF capacitors., Thus,, 1, Ceq 1, 1, 1 mF 10 mF, 60 30 20, , 20 mF, , 30 mF, , + v1 −, , + v2 −, , 30 V +, −, , 40 mF, , +, v3, −, , 20 mF, , Figure 6.18, For Example 6.7., , The total charge is, q Ceq v 10 103 30 0.3 C, This is the charge on the 20-mF and 30-mF capacitors, because they are, in series with the 30-V source. (A crude way to see this is to imagine, that charge acts like current, since i dqdt.) Therefore,, v1 , , q, 0.3, , 15 V, C1, 20 103, , v2 , , q, 0.3, , 10 V, C2, 30 103, , 30 V +, −, , Ceq, , Figure 6.19, Equivalent circuit for Fig. 6.18., , Having determined v1 and v2, we now use KVL to determine v3 by, v3 30 v1 v2 5 V, Alternatively, since the 40-mF and 20-mF capacitors are in parallel,, they have the same voltage v3 and their combined capacitance is 40 , 20 60 mF. This combined capacitance is in series with the 20-mF and, 30-mF capacitors and consequently has the same charge on it. Hence,, v3 , , q, 0.3, , 5V, 60 mF, 60 103, , Find the voltage across each of the capacitors in Fig. 6.20., Answer: v1 30 V, v2 30 V, v3 10 V, v4 20 V., , Practice Problem 6.7, 40 F, + v1 −, +, v2, 60 V +, −, −, , 60 F, + v3 −, 20 F, , Figure 6.20, For Practice Prob. 6.7., , +, v4, −, , 30 F
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ale29559_ch06.qxd, , 07/08/2008, , 10:59 AM, , Page 226, , Chapter 6, , 226, , Length, ᐍ, Cross-sectional area, A, , Core material, Number of turns, N, , Figure 6.21, Typical form of an inductor., , 6.4, , Capacitors and Inductors, , Inductors, , An inductor is a passive element designed to store energy in its magnetic field. Inductors find numerous applications in electronic and, power systems. They are used in power supplies, transformers, radios,, TVs, radars, and electric motors., Any conductor of electric current has inductive properties and may, be regarded as an inductor. But in order to enhance the inductive effect,, a practical inductor is usually formed into a cylindrical coil with many, turns of conducting wire, as shown in Fig. 6.21., An inductor consists of a coil of conducting wire., , If current is allowed to pass through an inductor, it is found that the, voltage across the inductor is directly proportional to the time rate of, change of the current. Using the passive sign convention,, , In view of Eq. (6.18), for an inductor, to have voltage across its terminals, its, current must vary with time. Hence,, v 0 for constant current through, the inductor., , vL, , di, dt, , (6.18), , where L is the constant of proportionality called the inductance of the, inductor. The unit of inductance is the henry (H), named in honor of, the American inventor Joseph Henry (1797–1878). It is clear from, Eq. (6.18) that 1 henry equals 1 volt-second per ampere., Inductance is the property whereby an inductor exhibits opposition, to the change of current flowing through it, measured in henrys (H)., , (a), , The inductance of an inductor depends on its physical dimension, and construction. Formulas for calculating the inductance of inductors, of different shapes are derived from electromagnetic theory and can be, found in standard electrical engineering handbooks. For example, for, the inductor, (solenoid) shown in Fig. 6.21,, L, , (b), , (c), , Figure 6.22, Various types of inductors: (a) solenoidal, wound inductor, (b) toroidal inductor,, (c) chip inductor., Courtesy of Tech America., , N 2mA, /, , (6.19), , where N is the number of turns, / is the length, A is the cross-sectional, area, and m is the permeability of the core. We can see from Eq. (6.19), that inductance can be increased by increasing the number of turns of, coil, using material with higher permeability as the core, increasing the, cross-sectional area, or reducing the length of the coil., Like capacitors, commercially available inductors come in different values and types. Typical practical inductors have inductance values, ranging from a few microhenrys (mH), as in communication systems,, to tens of henrys (H) as in power systems. Inductors may be fixed or, variable. The core may be made of iron, steel, plastic, or air. The terms, coil and choke are also used for inductors. Common inductors are, shown in Fig. 6.22. The circuit symbols for inductors are shown in, Fig. 6.23, following the passive sign convention., Equation (6.18) is the voltage-current relationship for an inductor., Figure 6.24 shows this relationship graphically for an inductor whose
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ale29559_ch06.qxd, , 07/08/2008, , 10:59 AM, , Page 227, , 6.4, , Inductors, , 227, , Historical, Joseph Henry (1797–1878), an American physicist, discovered inductance and constructed an electric motor., Born in Albany, New York, Henry graduated from Albany Academy and taught philosophy at Princeton University from 1832 to 1846., He was the first secretary of the Smithsonian Institution. He conducted, several experiments on electromagnetism and developed powerful electromagnets that could lift objects weighing thousands of pounds. Interestingly, Joseph Henry discovered electromagnetic induction before, Faraday but failed to publish his findings. The unit of inductance, the, henry, was named after him., , inductance is independent of current. Such an inductor is known as a, linear inductor. For a nonlinear inductor, the plot of Eq. (6.18) will, not be a straight line because its inductance varies with current. We, will assume linear inductors in this textbook unless stated otherwise., The current-voltage relationship is obtained from Eq. (6.18) as, 1, v dt, L, , di , , i, , i, +, v, −, , L, , (a), , Integrating gives, , i, +, v, −, , L, , +, v, −, , (b), , L, , (c), , Figure 6.23, 1, i, L, , , , t, , v (t) dt, , (6.20), , Circuit symbols for inductors: (a) air-core,, (b) iron-core, (c) variable iron-core., , , , or, v, , i, , 1, L, , t, , v(t) dt i(t ), 0, , (6.21), , t0, , Slope = L, , where i(t0) is the total current for 6 t 6 t0 and i() 0. The, idea of making i() 0 is practical and reasonable, because there, must be a time in the past when there was no current in the inductor., The inductor is designed to store energy in its magnetic field. The, energy stored can be obtained from Eq. (6.18). The power delivered to, the inductor is, p vi aL, , di, bi, dt, , (6.22), , The energy stored is, w, , , , t, , p dt , , , , , , , , t, , , , t, , aL, , di, bi dt, dt, , 1, 1, L, i di Li2(t) Li2(), 2, 2, , , (6.23), , 0, , di ⁄dt, , Figure 6.24, Voltage-current relationship of an inductor.
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ale29559_ch06.qxd, , 07/08/2008, , 10:59 AM, , Page 228, , Chapter 6, , 228, , Capacitors and Inductors, , Since i () 0,, 1, w Li2, 2, , (6.24), , We should note the following important properties of an inductor., 1. Note from Eq. (6.18) that the voltage across an inductor is zero, when the current is constant. Thus,, An inductor acts like a short circuit to dc., , 2. An important property of the inductor is its opposition to the, change in current flowing through it., i, , i, , The current through an inductor cannot change instantaneously., , t, , t, (a), , (b), , Figure 6.25, Current through an inductor: (a) allowed,, (b) not allowable; an abrupt change is not, possible., , Since an inductor is often made of a, highly conducting wire, it has a very, small resistance., , L, , Rw, , Cw, , Figure 6.26, Circuit model for a practical inductor., , Example 6.8, , According to Eq. (6.18), a discontinuous change in the current, through an inductor requires an infinite voltage, which is not physically possible. Thus, an inductor opposes an abrupt change in the, current through it. For example, the current through an inductor, may take the form shown in Fig. 6.25(a), whereas the inductor current cannot take the form shown in Fig. 6.25(b) in real-life situations due to the discontinuities. However, the voltage across an, inductor can change abruptly., 3. Like the ideal capacitor, the ideal inductor does not dissipate, energy. The energy stored in it can be retrieved at a later time. The, inductor takes power from the circuit when storing energy and, delivers power to the circuit when returning previously stored, energy., 4. A practical, nonideal inductor has a significant resistive component,, as shown in Fig. 6.26. This is due to the fact that the inductor is, made of a conducting material such as copper, which has some, resistance. This resistance is called the winding resistance Rw, and, it appears in series with the inductance of the inductor. The presence of Rw makes it both an energy storage device and an energy, dissipation device. Since Rw is usually very small, it is ignored in, most cases. The nonideal inductor also has a winding capacitance, Cw due to the capacitive coupling between the conducting coils. Cw, is very small and can be ignored in most cases, except at high frequencies. We will assume ideal inductors in this book., , The current through a 0.1-H inductor is i(t) 10te5t A. Find the voltage across the inductor and the energy stored in it., Solution:, Since v L didt and L 0.1 H,, v 0.1, , d, (10te5t ) e5t t(5)e5t e5t(1 5t) V, dt
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ale29559_ch06.qxd, , 07/08/2008, , 10:59 AM, , Page 229, , 6.4, , Inductors, , 229, , The energy stored is, w, , 1 2 1, Li (0.1)100t 2e10t 5t 2e10t J, 2, 2, , If the current through a 1-mH inductor is i(t) 20 cos 100t mA, find, the terminal voltage and the energy stored., , Practice Problem 6.8, , Answer: 2 sin 100t mV, 0.2 cos2 100t mJ., , Find the current through a 5-H inductor if the voltage across it is, 30t 2,, v(t) b, 0,, , Example 6.9, , t 7 0, t 6 0, , Also, find the energy stored at t 5 s. Assume i(v) 7 0., Solution:, Since i , , 1, L, , t, , v(t) dt i (t ) and L 5 H,, 0, , t0, , 1, 5, , i, , t, , 30t, , 2, , dt 0 6 , , 0, , t3, 2t 3 A, 3, , The power p vi 60t 5, and the energy stored is then, w, , , , p dt , , , , 5, , 0, , 60t 5 dt 60, , t6 5, 2 156.25 kJ, 6 0, , Alternatively, we can obtain the energy stored using Eq. (6.24), by, writing, 1, 1, 1, w 0 50 Li2(5) Li(0) (5)(2 53)2 0 156.25 kJ, 2, 2, 2, as obtained before., , The terminal voltage of a 2-H inductor is v 10(1 t) V. Find the, current flowing through it at t 4 s and the energy stored in it at t 4 s., Assume i(0) 2 A., Answer: 18 A, 320 J., , Practice Problem 6.9
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ale29559_ch06.qxd, , 07/08/2008, , 10:59 AM, , Page 230, , Chapter 6, , 230, , Example 6.10, 1Ω, , i, , Consider the circuit in Fig. 6.27(a). Under dc conditions, find: (a) i, vC,, and iL, (b) the energy stored in the capacitor and inductor., 5Ω, iL, , 12 V, , 4Ω, , +, −, , Capacitors and Inductors, , (a) Under dc conditions, we replace the capacitor with an open circuit, and the inductor with a short circuit, as in Fig. 6.27(b). It is evident, from Fig. 6.27(b) that, , 2H, , +, vC, −, , Solution:, , 1F, , i iL , , (a), 1Ω, , i, , 5Ω, , The voltage vC is the same as the voltage across the 5- resistor. Hence,, iL, , 12 V, , 4Ω, , +, −, , vC 5i 10 V, (b) The energy in the capacitor is, , +, vC, −, , 1, 1, wC Cv2C (1)(102) 50 J, 2, 2, (b), , and that in the inductor is, , Figure 6.27, For Example 6.10., , 1, 1, wL Li2L (2)(22) 4 J, 2, 2, , Practice Problem 6.10, iL, , Answer: 6 V, 3 A, 72 J, 27 J., , +, vC, −, , 2Ω, , 4F, , Figure 6.28, , 6.5, , For Practice Prob. 6.10., , i, +, , L1, , Determine vC, iL, and the energy stored in the capacitor and inductor, in the circuit of Fig. 6.28 under dc conditions., , 6H, 6Ω, , 4A, , 12, 2A, 15, , L3, , L2, , +v − +v − +v −, 1, 2, 3, , LN, ..., , +v −, N, , v, −, (a), , Series and Parallel Inductors, , Now that the inductor has been added to our list of passive elements, it is, necessary to extend the powerful tool of series-parallel combination. We, need to know how to find the equivalent inductance of a series-connected, or parallel-connected set of inductors found in practical circuits., Consider a series connection of N inductors, as shown in Fig. 6.29(a),, with the equivalent circuit shown in Fig. 6.29(b). The inductors have, the same current through them. Applying KVL to the loop,, v v1 v2 v3 p vN, (6.25), Substituting vk Lk didt results in, , i, , di, di, di, di, L2 L3 p LN, dt, dt, dt, dt, di, (L 1 L 2 L 3 p L N), dt, , v L1, , +, L eq, , v, −, (b), , N, di, di, a a L k b Leq, dt, dt, k1, , Figure 6.29, (a) A series connection of N inductors,, (b) equivalent circuit for the series, inductors., , (6.26), , where, Leq L 1 L 2 L 3 p L N, , (6.27)
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ale29559_ch06.qxd, , 07/17/2008, , 11:59 AM, , Page 232, , Chapter 6, , 232, , Capacitors and Inductors, , TABLE 6.1, , Important characteristics of the basic elements.†, Relation, , Resistor (R), , v-i:, i-v:, , 1, C, , Inductor (L), , t, , i dt v(t ), , v, , i vR, , dv, iC, dt, , i, , 1, w Cv2, 2, C1C2, Ceq , C1 C2, , 1, w Li2, 2, , Ceq C1 C2, Open circuit, , L1L2, L1 L2, Short circuit, , v, , i, , v2, R, , p i2R , , Series:, , Req R1 R2, , Parallel:, , Req , , At dc:, , Same, , R1R2, R1 R2, , Circuit variable, that cannot, change abruptly: Not applicable, , 0, , vL, , di, dt, , v iR, , p or w:, , †, , Capacitor (C), , t0, , 1, L, , t, , v dt i(t ), 0, , t0, , Leq L1 L2, Leq , , Passive sign convention is assumed., , It is appropriate at this point to summarize the most important, characteristics of the three basic circuit elements we have studied. The, summary is given in Table 6.1., The wye-delta transformation discussed in Section 2.7 for resistors, can be extended to capacitors and inductors., , Example 6.11, , Find the equivalent inductance of the circuit shown in Fig. 6.31., 20 H, , 4H, L eq, 7H, , 8H, , 12 H, , Solution:, The 10-H, 12-H, and 20-H inductors are in series; thus, combining, them gives a 42-H inductance. This 42-H inductor is in parallel with, the 7-H inductor so that they are combined, to give, 7 42, 6H, 7 42, , 10 H, , Figure 6.31, For Example 6.11., , This 6-H inductor is in series with the 4-H and 8-H inductors. Hence,, Leq 4 6 8 18 H, , Practice Problem 6.11, , Calculate the equivalent inductance for the inductive ladder network in, Fig. 6.32., 20 mH, , 100 mH, , 40 mH, , L eq, 50 mH, , Figure 6.32, For Practice Prob. 6.11., , Answer: 25 mH., , 40 mH, , 30 mH, , 20 mH
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ale29559_ch06.qxd, , 07/08/2008, , 11:00 AM, , Page 234, , Chapter 6, , 234, , Capacitors and Inductors, , come in discrete form and tend to be more bulky and expensive. For, this reason, inductors are not as versatile as capacitors and resistors,, and they are more limited in applications. However, there are several, applications in which inductors have no practical substitute. They are, routinely used in relays, delays, sensing devices, pick-up heads, telephone circuits, radio and TV receivers, power supplies, electric motors,, microphones, and loudspeakers, to mention a few., Capacitors and inductors possess the following three special properties that make them very useful in electric circuits:, 1. The capacity to store energy makes them useful as temporary voltage or current sources. Thus, they can be used for generating a large, amount of current or voltage for a short period of time., 2. Capacitors oppose any abrupt change in voltage, while inductors, oppose any abrupt change in current. This property makes inductors useful for spark or arc suppression and for converting pulsating dc voltage into relatively smooth dc voltage., 3. Capacitors and inductors are frequency sensitive. This property, makes them useful for frequency discrimination., The first two properties are put to use in dc circuits, while the third, one is taken advantage of in ac circuits. We will see how useful these, properties are in later chapters. For now, consider three applications, involving capacitors and op amps: integrator, differentiator, and analog, computer., , 6.6.1 Integrator, , Rf, , i2, i1, , R1, , v1, , +, vi, , Important op amp circuits that use energy-storage elements include, integrators and differentiators. These op amp circuits often involve, resistors and capacitors; inductors (coils) tend to be more bulky and, expensive., The op amp integrator is used in numerous applications, especially, in analog computers, to be discussed in Section 6.6.3., , 0A, −, 1 −, 0V, v2 + +, , +, vo, , −, , An integrator is an op amp circuit whose output is proportional to the, integral of the input signal., , −, , If the feedback resistor R f in the familiar inverting amplifier of, Fig. 6.35(a) is replaced by a capacitor, we obtain an ideal integrator,, as shown in Fig. 6.35(b). It is interesting that we can obtain a mathematical representation of integration this way. At node a in Fig. 6.35(b),, , (a), C, , iC, iR, , iR iC, , R, , +, vi, , a, , −, , +, , But, +, vo, −, , −, , (b), , Figure 6.35, Replacing the feedback resistor in the, inverting amplifier in (a) produces an, integrator in (b)., , (6.32), , iR , , vi, ,, R, , iC C, , dvo, dt, , Substituting these in Eq. (6.32), we obtain, dvo, vi, C, R, dt, dvo , , 1, vi dt, RC, , (6.33a), (6.33b)
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ale29559_ch06.qxd, , 07/08/2008, , 11:00 AM, , Page 235, , 6.6, , Applications, , 235, , Integrating both sides gives, vo (t) vo (0) , , 1, RC, , t, , v (t) dt, i, , (6.34), , 0, , To ensure that vo (0) 0, it is always necessary to discharge the integrator’s capacitor prior to the application of a signal. Assuming vo (0) 0,, vo , , t, , v (t) dt, , 1, RC, , i, , (6.35), , 0, , which shows that the circuit in Fig. 6.35(b) provides an output voltage, proportional to the integral of the input. In practice, the op amp integrator requires a feedback resistor to reduce dc gain and prevent saturation. Care must be taken that the op amp operates within the linear, range so that it does not saturate., , Example 6.13, , If v1 10 cos 2t mV and v2 0.5t mV, find vo in the op amp circuit, in Fig. 6.36. Assume that the voltage across the capacitor is initially zero., 3 MΩ, , Solution:, This is a summing integrator, and, 1, vo , R1C, , , , 1, v1 dt , R2C, , , , 1, , 6, 3 10 2 106, , , 2 F, , v1, , v2 dt, , −, +, , v2, 100 kΩ, , t, , 10 cos 2t dt, , Figure 6.36, For Example 6.13., , 0, , 1, 3, 100 10 2 106, , t, , 0.5t dt, 0, , 2, , , , 1 10, 1 0.5t, sin 2t , 0.833 sin 2t 1.25t 2 mV, 6 2, 0.2 2, , The integrator in Fig. 6.35(b) has R 100 k, C 20 mF. Determine, the output voltage when a dc voltage of 10 mV is applied at t 0., Assume that the op amp is initially nulled., Answer: 5t mV., , 6.6.2 Differentiator, A differentiator is an op amp circuit whose output is proportional to, the rate of change of the input signal., , In Fig. 6.35(a), if the input resistor is replaced by a capacitor, the, resulting circuit is a differentiator, shown in Fig. 6.37. Applying KCL, at node a,, iR iC, , (6.36), , Practice Problem 6.13, , vo
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ale29559_ch06.qxd, , 07/08/2008, , 11:00 AM, , Page 236, , Chapter 6, , 236, , Capacitors and Inductors, , But, vo, iR ,, R, , iC C, , dvi, dt, , Substituting these in Eq. (6.36) yields, R, , iR, C, , iC, , −, +, , a, , +, vi, −, , vo RC, +, vo, −, , Figure 6.37, An op amp differentiator., , Example 6.14, , dvi, dt, , (6.37), , showing that the output is the derivative of the input. Differentiator circuits are electronically unstable because any electrical noise within the, circuit is exaggerated by the differentiator. For this reason, the differentiator circuit in Fig. 6.37 is not as useful and popular as the integrator. It is seldom used in practice., , Sketch the output voltage for the circuit in Fig. 6.38(a), given the input, voltage in Fig. 6.38(b). Take vo 0 at t 0., 5 kΩ, , Solution:, This is a differentiator with, , 0.2 F, −, +, vi, , +, vo, −, , +, −, , RC 5 103 0.2 106 103 s, For 0 6 t 6 4 ms, we can express the input voltage in Fig. 6.38(b) as, vi e, , (a), , 2000t, 8 2000t, , 0 6 t 6 2 ms, 2 6 t 6 4 ms, , This is repeated for 4 6 t 6 8 ms. Using Eq. (6.37), the output is, obtained as, , vo(V), 4, , vo RC, 0, , 2, , 4, , 6, , 8, , t (ms), , (b), , Figure 6.38, For Example 6.14., , dvi, 2 V, e, dt, 2V, , 0 6 t 6 2 ms, 2 6 t 6 4 ms, , Thus, the output is as sketched in Fig. 6.39., vo (V), 2, , 0, 2, , 4, , 6, , 8, , t (ms), , −2, , Figure 6.39, Output of the circuit in Fig. 6.38(a)., , Practice Problem 6.14, , The differentiator in Fig. 6.37 has R 100 k and C 0.1 mF. Given, that vi 3t V, determine the output vo., Answer: 30 mV.
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ale29559_ch06.qxd, , 07/08/2008, , 11:00 AM, , Page 237, , 6.6, , Applications, , 237, , 6.6.3 Analog Computer, Op amps were initially developed for electronic analog computers., Analog computers can be programmed to solve mathematical models of, mechanical or electrical systems. These models are usually expressed in, terms of differential equations., To solve simple differential equations using the analog computer, requires cascading three types of op amp circuits: integrator circuits,, summing amplifiers, and inverting/noninverting amplifiers for negative/, positive scaling. The best way to illustrate how an analog computer solves, a differential equation is with an example., Suppose we desire the solution x(t) of the equation, a, , d 2x, dx, b cx f (t),, 2, dt, dt, , t 7 0, , (6.38), , where a, b, and c are constants, and f (t) is an arbitrary forcing function. The solution is obtained by first solving the highest-order derivative term. Solving for d 2xdt 2 yields, f (t), b dx, d 2x, c, , , x, 2, a, a dt, a, dt, , (6.39), , To obtain dxdt, the d 2xdt 2 term is integrated and inverted. Finally,, to obtain x, the dxdt term is integrated and inverted. The forcing function is injected at the proper point. Thus, the analog computer for solving Eq. (6.38) is implemented by connecting the necessary summers,, inverters, and integrators. A plotter or oscilloscope may be used to view, the output x, or dxdt, or d 2xdt 2, depending on where it is connected, in the system., Although the above example is on a second-order differential equation, any differential equation can be simulated by an analog computer, comprising integrators, inverters, and inverting summers. But care must, be exercised in selecting the values of the resistors and capacitors, to, ensure that the op amps do not saturate during the solution time interval., The analog computers with vacuum tubes were built in the 1950s and, 1960s. Recently their use has declined. They have been superseded by, modern digital computers. However, we still study analog computers for, two reasons. First, the availability of integrated op amps has made it possible to build analog computers easily and cheaply. Second, understanding analog computers helps with the appreciation of the digital computers., , Design an analog computer circuit to solve the differential equation:, 2, , d vo, dt, , 2, , 2, , dvo, vo 10 sin 4t,, dt, , t 7 0, , subject to vo(0) 4, v¿o(0) 1, where the prime refers to the time, derivative., Solution:, 1. Define. We have a clearly defined problem and expected solution., I might remind the student that many times the problem is not so, well defined and this portion of the problem-solving process could, , Example 6.15
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ale29559_ch06.qxd, , 238, , 07/08/2008, , 11:00 AM, , Page 238, , Chapter 6, , Capacitors and Inductors, , require much more effort. If this is so, then you should always, keep in mind that time spent here will result in much less effort, later and most likely save you a lot of frustration in the process., 2. Present. Clearly, using the devices developed in Section 6.6.3, will allow us to create the desired analog computer circuit. We, will need the integrator circuits (possibly combined with a, summing capability) and one or more inverter circuits., 3. Alternative. The approach for solving this problem is straightforward. We will need to pick the correct values of resistances, and capacitors to allow us to realize the equation we are representing. The final output of the circuit will give the desired, result., 4. Attempt. There are an infinite number of possibilities for, picking the resistors and capacitors, many of which will result, in correct solutions. Extreme values of resistors and capacitors, will result in incorrect outputs. For example, low values of, resistors will overload the electronics. Picking values of, resistors that are too large will cause the op amps to stop, functioning as ideal devices. The limits can be determined from, the characteristics of the real op amp., We first solve for the second derivative as, d2vo, dt, , 2, , 10 sin 4t 2, , dvo, vo, dt, , (6.15.1), , Solving this requires some mathematical operations, including, summing, scaling, and integration. Integrating both sides of, Eq. (6.15.1) gives, dvo, , dt, , t, , a10 sin 4t 2 dt, , dvo, , 0, , vo b dt v¿o (0) (6.15.2), , where v¿o (0) 1. We implement Eq. (6.15.2) using the summing, integrator shown in Fig. 6.40(a). The values of the resistors and, capacitors have been chosen so that RC 1 for the term, , , 1, RC, , t, , v, , o, , dt, , 0, , Other terms in the summing integrator of Eq. (6.15.2) are, implemented accordingly. The initial condition dvo(0)dt 1 is, implemented by connecting a 1-V battery with a switch across the, capacitor as shown in Fig. 6.40(a)., The next step is to obtain vo by integrating dvodt and, inverting the result,, t, , vo , , a dt b dt v(0), dvo, , (6.15.3), , 0, , This is implemented with the circuit in Fig. 6.40(b) with the, battery giving the initial condition of 4 V. We now combine the, two circuits in Fig. 6.40(a) and (b) to obtain the complete circuit, shown in Fig. 6.40(c). When the input signal 10 sin 4t is applied,, we open the switches at t 0 to obtain the output waveform vo,, which may be viewed on an oscilloscope.
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ale29559_ch06.qxd, , 07/08/2008, , 11:00 AM, , Page 239, , 6.6, −, , 1V, , 1 MΩ, , 239, , +, t=0, , −, , 4V, , +, t=0, , 1 F, , −10 sin (4t), , 1 F, , 1 MΩ, , −, +, , vo, dvo, dt, , Applications, , 0.5 MΩ, , 1 MΩ, , dvo, dt, , 1 MΩ, , dvo, dt, , 10 sin (4t), , +, −, , −, , 1V, , +, , −, , t=0, , 4V, , 1 V 1 F, 1 MΩ, , vo, , −, +, , 1 F, 1 MΩ, , −, +, , (c), , Figure 6.40, For Example 6.15., , 5. Evaluate. The answer looks correct, but is it? If an actual, solution for vo is desired, then a good check would be to first, find the solution by realizing the circuit in PSpice. This result, could then be compared with a solution using the differential, solution capability of MATLAB., Since all we need to do is check the circuit and confirm that, it represents the equation, we have an easier technique to use., We just go through the circuit and see if it generates the desired, equation., However, we still have choices to make. We could go through, the circuit from left to right but that would involve differentiating, the result to obtain the original equation. An easier approach, would be to go from right to left. This is the approach we will, use to check the answer., Starting with the output, vo, we see that the right-hand op, amp is nothing more than an inverter with a gain of one. This, means that the output of the middle circuit is vo. The following, represents the action of the middle circuit., t, , t, dvo, dt vo(0)b avo 2 vo (0)b, dt, 0, 0, (vo(t) vo(0) vo(0)), , where vo(0) 4 V is the initial voltage across the capacitor., We check the circuit on the left the same way., dvo, a, dt, , , , t, , 0, , , , d 2vo, dt, , 2, , vo, , t=0, , dvo, dt, , , , −, +, , +, , 0.5 MΩ, , vo a, , −vo, (b), , (a), 1 MΩ, , 1 MΩ, , −, +, , dt v¿o(0)b a, , dvo, v¿o(0) v¿o(0)b, dt, , 1 MΩ, 1 MΩ, , −, +, , vo
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ale29559_ch06.qxd, , 07/08/2008, , 11:00 AM, , 240, , Page 240, , Chapter 6, , Capacitors and Inductors, , Now all we need to verify is that the input to the first op amp is, d 2vodt 2., Looking at the input we see that it is equal to, 10 sin(4t) vo , , dvo, 1106 dvo, 10 sin(4t) vo 2, 0.5 M dt, dt, , which does produce d2vodt2 from the original equation., 6. Satisfactory? The solution we have obtained is satisfactory. We, can now present this work as a solution to the problem., , Practice Problem 6.15, , Design an analog computer circuit to solve the differential equation:, d 2vo, dt, , 2, , 3, , dvo, 2vo 4 cos 10t,, dt, , t 7 0, , subject to vo(0) 2, v¿o(0) 0., Answer: See Fig. 6.41, where RC 1 s., , 2V, , t=0, , C, R, , C, R, , −, +, , d 2v, , R, R, 2, , −, +, , o, dt 2, , vo, , R, R, , −, +, , d 2vo, dt 2, , R, 3, , −, +, R, R, , cos (10t), , +, −, , R, 4, , −, +, , Figure 6.41, For Practice Prob. 6.15., , 6.7, , Summary, , 1. The current through a capacitor is directly proportional to the time, rate of change of the voltage across it., iC, , dv, dt, , The current through a capacitor is zero unless the voltage is changing. Thus, a capacitor acts like an open circuit to a dc source.
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ale29559_ch06.qxd, , 07/08/2008, , 11:00 AM, , Page 241, , Review Questions, , 241, , 2. The voltage across a capacitor is directly proportional to the time, integral of the current through it., v, , 1, C, , , , t, , i dt , , , , t, , i dt v(t ), , 1, C, , 0, , t0, , The voltage across a capacitor cannot change instantly., 3. Capacitors in series and in parallel are combined in the same way, as conductances., 4. The voltage across an inductor is directly proportional to the time, rate of change of the current through it., vL, , di, dt, , The voltage across the inductor is zero unless the current is changing. Thus, an inductor acts like a short circuit to a dc source., 5. The current through an inductor is directly proportional to the time, integral of the voltage across it., i, , 1, L, , , , t, , v dt , , , , 1, L, , t, , v dt i(t ), 0, , t0, , The current through an inductor cannot change instantly., 6. Inductors in series and in parallel are combined in the same way, resistors in series and in parallel are combined., 7. At any given time t, the energy stored in a capacitor is 12 Cv2, while, the energy stored in an inductor is 12 Li2., 8. Three application circuits, the integrator, the differentiator, and the, analog computer, can be realized using resistors, capacitors, and, op amps., , Review Questions, 6.1, , 6.2, , 6.3, , 6.4, , What charge is on a 5-F capacitor when it is, connected across a 120-V source?, (a) 600 C, , (b) 300 C, , (c) 24 C, , (d) 12 C, , v (t), 10, , 0, , 1, , Capacitance is measured in:, (a) coulombs, , (b) joules, , (c) henrys, , (d) farads, , When the total charge in a capacitor is doubled, the, energy stored:, (a) remains the same, , (b) is halved, , (c) is doubled, , (d) is quadrupled, , Can the voltage waveform in Fig. 6.42 be associated, with a real capacitor?, (a) Yes, , (b) No, , 2, , t, , −10, , Figure 6.42, For Review Question 6.4., , 6.5, , The total capacitance of two 40-mF series-connected, capacitors in parallel with a 4-mF capacitor is:, (a) 3.8 mF, , (b) 5 mF, , (d) 44 mF, , (e) 84 mF, , (c) 24 mF
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ale29559_ch06.qxd, , 07/08/2008, , 11:00 AM, , Page 242, , Chapter 6, , 242, , 6.6, , Capacitors and Inductors, , In Fig. 6.43, if i cos 4t and v sin 4t, the, element is:, (a) a resistor, , (b) a capacitor, , 6.9, , (c) an inductor, , +, −, , (a) True, , (b) False, , 6.10 For the circuit in Fig. 6.44, the voltage divider, formula is:, , i, v, , Inductors in parallel can be combined just like, resistors in parallel., , Element, , Figure 6.43, , (a) v1 , , L1 L2, vs, L1, , (b) v1 , , L1 L2, vs, L2, , (c) v1 , , L2, vs, L1 L2, , (d) v1 , , L1, vs, L1 L2, , For Review Question 6.6., L1, , 6.7, , 6.8, , A 5-H inductor changes its current by 3 A in 0.2 s. The, voltage produced at the terminals of the inductor is:, (a) 75 V, , (b) 8.888 V, , (c) 3 V, , (d) 1.2 V, , vs, , If the current through a 10-mH inductor increases, from zero to 2 A, how much energy is stored in the, inductor?, (a) 40 mJ, , (b) 20 mJ, , (c) 10 mJ, , (d) 5 mJ, , + v −, 1, +, −, , +, v2, −, , L2, , Figure 6.44, For Review Question 6.10., , Answers: 6.1a, 6.2d, 6.3d, 6.4b, 6.5c, 6.6b, 6.7a, 6.8b,, 6.9a, 6.10d., , Problems, Section 6.2 Capacitors, , 6.6, 3t, , 6.1, , If the voltage across a 5-F capacitor is 2te, the current and the power., , V, find, , 6.2, , A 20-mF capacitor has energy w(t) 10 cos2 377t J., Determine the current through the capacitor., , 6.3, , Design a problem to help other students better, understand how capacitors work., , 6.4, , A current of 6 sin 4t A flows through a 2-F capacitor., Find the voltage v(t) across the capacitor given that, v(0) 1 V., , 6.5, , The voltage across a 4-mF capacitor is shown in, Fig. 6.45. Find the current waveform., , v (t) V, 10, , 0, 2, , 0, , 4, , 6, , 8, , t (ms), , 6, , 8, , 10, , 12 t (ms), , Figure 6.46, For Prob. 6.6., 6.7, , At t 0, the voltage across a 50-mF capacitor is 10 V., Calculate the voltage across the capacitor for t 7 0, when current 4t mA flows through it., , 6.8, , A 4-mF capacitor has the terminal voltage, vb, , 2, , 4, , −10, , v(t) V, 10, , The voltage waveform in Fig. 6.46 is applied across, a 30-mF capacitor. Draw the current waveform, through it., , 50 V,, Ae100t Be600t V,, , t0, t0, , If the capacitor has an initial current of 2 A, find:, −10, , (a) the constants A and B,, , Figure 6.45, , (b) the energy stored in the capacitor at t 0,, , For Prob. 6.5., , (c) the capacitor current for t 7 0.
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ale29559_ch06.qxd, , 07/08/2008, , 11:00 AM, , Page 243, , Problems, , 6.9, , The current through a 0.5-F capacitor is 6(1 et) A., Determine the voltage and power at t 2 s. Assume, v(0) 0., , 6.10 The voltage across a 2-mF capacitor is shown in, Fig. 6.47. Determine the current through the capacitor., v (t) (V), , 243, , 6.15 Two capacitors (20 mF and 30 mF) are connected, to a 100-V source. Find the energy stored in each, capacitor if they are connected in:, (a) parallel, , (b) series, , 6.16 The equivalent capacitance at terminals a-b in the, circuit of Fig. 6.50 is 30 mF. Calculate the value of C., a, , 16, , C, 0, , 1, , 2, , 3, , 14 F, , t (s), , 4, , Figure 6.47, , 80 F, , For Prob. 6.10., b, , 6.11 A 4-mF capacitor has the current waveform shown in, Fig. 6.48. Assuming that v(0) 10 V, sketch the, voltage waveform v(t)., i(t) (mA), , Figure 6.50, For Prob. 6.16., 6.17 Determine the equivalent capacitance for each of the, circuits of Fig. 6.51., 12 F, , 4F, , 15, 10, , 6F, , 3F, , 5, 0, , 2, , −5, , 6, , 4, , t (s), , 8, , 4F, (a), , −10, , 6F, , Figure 6.48, For Prob. 6.11., , 5F, , 6.12 A voltage of 6e2000t V appears across a parallel, combination of a 100-mF capacitor and a 12-, resistor. Calculate the power absorbed by the parallel, combination., , 4F, , (b), 3F, , 6F, , 2F, , 6.13 Find the voltage across the capacitors in the circuit, of Fig. 6.49 under dc conditions., 4F, 50 Ω, , 10 Ω, , 30 Ω, , C1, , +, v1, −, , 20 Ω, +, −, , 60 V, , 2F, , 3F, , (c), , Figure 6.51, +, v2, −, , For Prob. 6.17., C2, , 6.18 Find Ceq in the circuit of Fig. 6.52 if all capacitors, are 4 mF., , Figure 6.49, For Prob. 6.13., , Section 6.3 Series and Parallel Capacitors, 6.14 Series-connected 20-pF and 60-pF capacitors are, placed in parallel with series-connected 30-pF and, 70-pF capacitors. Determine the equivalent, capacitance., , Ceq, , Figure 6.52, For Prob. 6.18.
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ale29559_ch06.qxd, , 07/08/2008, , 11:00 AM, , Page 244, , Chapter 6, , 244, , Capacitors and Inductors, 40 F, , 6.19 Find the equivalent capacitance between terminals, a and b in the circuit of Fig. 6.53. All capacitances, are in mF., , 10 F, , 10 F, , 35 F, , 80, , 5 F, 20 F, , 12, , 40, , 15 F, , a, , 15 F, , 20, , 50, 12, , 10, , 30, , a, , b, , Figure 6.56, , b, , For Prob. 6.22., 60, , Figure 6.53, , 6.23 Using Fig. 6.57, design a problem that will help, other students better understand how capacitors work, together when connected in series and in parallel., , For Prob. 6.19., 6.20 Find the equivalent capacitance at terminals a-b of, the circuit in Fig. 6.54., , C1, , a, , 1 F, , V, , +, −, , C3, , C2, , C4, , 1 F, , Figure 6.57, For Prob. 6.23., 2 F, , 2 F, , 6.24 Repeat Prob. 6.23 for the circuit of Fig. 6.58., , 2 F, , 60 F, , 3 F, , 3 F, , 3 F, , 90 V +, −, , 3 F, , 20 F, , 30 F, , 80 F, , 14 F, , Figure 6.58, For Prob. 6.24., , b, , Figure 6.54, , 6.25 (a) Show that the voltage-division rule for two, capacitors in series as in Fig. 6.59(a) is, , For Prob. 6.20., 6.21 Determine the equivalent capacitance at terminals, a-b of the circuit in Fig. 6.55., , v1 , , C2, vs,, C1 C2, , v2 , , C1, vs, C1 C2, , assuming that the initial conditions are zero., 5 F, , 6 F, , 4 F, C1, , a, 2 F, , 3 F, , 12 F, , b, , Figure 6.55, , vs, , +, −, , + v1 −, +, v2, −, , C2, , is, , For Prob. 6.21., (a), , 6.22 Obtain the equivalent capacitance of the circuit in, Fig. 6.56., , Figure 6.59, For Prob. 6.25., , (b), , i1, , i2, , C1, , C2
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ale29559_ch06.qxd, , 07/08/2008, , 11:00 AM, , Page 245, , Problems, , (b) For two capacitors in parallel as in Fig. 6.59(b),, show that the current-division rule is, i1 , , C1, is,, C1 C2, , i2 , , 245, , 6.30 Assuming that the capacitors are initially uncharged,, find vo(t) in the circuit of Fig. 6.62., , C2, is, C1 C2, , assuming that the initial conditions are zero., 6.26 Three capacitors, C1 5 mF, C2 10 mF, and, C3 20 mF, are connected in parallel across a, 150-V source. Determine:, , is (mA), , 6 F, , 60, is, 0, , (a) the total capacitance,, , 2 t (s), , 1, , (b) the charge on each capacitor,, , Figure 6.62, , (c) the total energy stored in the parallel, combination., , For Prob. 6.30., , 6.27 Given that four 4-mF capacitors can be connected in, series and in parallel, find the minimum and, maximum values that can be obtained by such, series/parallel combinations., , +, vo (t), −, , 3 F, , 6.31 If v(0) 0, find v(t), i1(t), and i2(t) in the circuit of, Fig. 6.63., , *6.28 Obtain the equivalent capacitance of the network, shown in Fig. 6.60., is (mA), 20, 40 F, , 50 F, , 30 F, , 0, 10 F, , 20 F, , 1, , 2, , 3, , 5, , 4, , t, , −20, , Figure 6.60, i1, , For Prob. 6.28., 6 F, , is, , 6.29 Determine Ceq for each circuit in Fig. 6.61., , i2, +, v, −, , 4 F, , Figure 6.63, , C, , For Prob. 6.31., C eq, , C, , C, C, , C, , 6.32 In the circuit of Fig. 6.64, let is 30e2t mA and, v1(0) 50 V, v2(0) 20 V. Determine: (a) v1(t), and v2(t), (b) the energy in each capacitor at, t 0.5 s., , (a), , C, , C, , C, , C, , C eq, 12 F, , (b), , Figure 6.61, , +, is, , For Prob. 6.29., , Figure 6.64, * An asterisk indicates a challenging problem., , For Prob. 6.32., , v1, , –, 20 F, , v2, , +, –, , 40 F
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ale29559_ch06.qxd, , 07/08/2008, , 11:00 AM, , Page 246, , Chapter 6, , 246, , Capacitors and Inductors, , 6.33 Obtain the Thevenin equivalent at the terminals, a-b,, of the circuit shown in Fig. 6.65. Please note that, Thevenin equivalent circuits do not generally exist, for circuits involving capacitors and resistors. This is, a special case where the Thevenin equivalent circuit, does exist., , 6.41 The voltage across a 2-H inductor is 20 (1 e2t) V., If the initial current through the inductor is 0.3 A,, find the current and the energy stored in the inductor, at t 1 s., 6.42 If the voltage waveform in Fig. 6.67 is applied, across the terminals of a 10-H inductor, calculate the, current through the inductor. Assume i(0) 1 A., v (t) (V), , 5F, +, 15 V −, , 30, , a, 3F, , 2F, 0, , b, , 1, , Figure 6.65, , Figure 6.67, , For Prob. 6.33., , For Prob. 6.42., , Section 6.4 Inductors, 6.34 The current through a 10-mH inductor is 6et2 A., Find the voltage and the power at t 3 s., 6.35 An inductor has a linear change in current from, 50 mA to 100 mA in 2 ms and induces a voltage of, 160 mV. Calculate the value of the inductor., 6.36 Design a problem to help other students better, understand how inductors work., 6.37 The current through a 12-mH inductor is 4 sin 100t A., Find the voltage, across the inductor for 0 6 t 6, p, p200 s, and the energy stored at t 200, s., , 3, , 2, , t, , 5, , 4, , 6.43 The current in an 80-mH inductor increases from 0, to 60 mA. How much energy is stored in the, inductor?, *6.44 A 100-mH inductor is connected in parallel with a, 2-k resistor. The current through the inductor is, i(t) 50e400t mA. (a) Find the voltage vL across, the inductor. (b) Find the voltage vR across the, resistor. (c) Does vR(t) vL(t) 0? (d) Calculate, the energy in the inductor at t 0., 6.45 If the voltage waveform in Fig. 6.68 is applied to a, 50-mH inductor, find the inductor current i(t)., Assume i(0) 0., v (t) (V), , 6.38 The current through a 40-mH inductor is, , 10, , 0,, i(t) b 2t, te A,, , t 6 0, t 7 0, 0, , Find the voltage v(t)., , 1, , 2, , t, , 6.39 The voltage across a 200-mH inductor is given by, –10, , v(t) 3t2 2t 4 V, , for t 7 0., , Determine the current i(t) through the inductor., Assume that i(0) 1 A., 6.40 The current through a 10-mH inductor is shown in, Fig. 6.66. Determine the voltage across the inductor, at t 1, 3, and 5 ms., , Figure 6.68, For Prob. 6.45., 6.46 Find vC, iL, and the energy stored in the capacitor, and inductor in the circuit of Fig. 6.69 under dc, conditions., 2Ω, , i(t) (A), 20, 6A, 0, , 4Ω, , +, vC, −, , 2F, , 0.5 H, 5Ω, , 2, , 4, , 6, , t (ms), , Figure 6.66, , Figure 6.69, , For Prob. 6.40., , For Prob. 6.46., , iL
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ale29559_ch06.qxd, , 07/08/2008, , 11:00 AM, , Page 247, , Problems, , 6.47 For the circuit in Fig. 6.70, calculate the value of R that, will make the energy stored in the capacitor the same, as that stored in the inductor under dc conditions., R, , 247, , 6.52 Using Fig. 6.74, design a problem to help other, students better understand how inductors behave, when connected in series and when connected in, parallel., , 160 F, 2Ω, , 5A, , L4, 4 mH, , L2, , Figure 6.70, , Leq, , For Prob. 6.47., 6.48 Under steady-state dc conditions, find i and v in the, circuit of Fig. 6.71., i, , L1, , L5, , L6, , Figure 6.74, For Prob. 6.52., , 2 mH, , 30 kΩ, , 10 mA, , L3, , +, v, −, , 6 F, , 20 kΩ, , 6.53 Find Leq at the terminals of the circuit in Fig. 6.75., , Figure 6.71, For Prob. 6.48., , Section 6.5 Series and Parallel Inductors, , 6 mH, , 6.49 Find the equivalent inductance of the circuit in, Fig. 6.72. Assume all inductors are 10 mH., , 8 mH, , a, 5 mH, , 12 mH, , 8 mH, 6 mH, 4 mH, b, 8 mH, , 10 mH, , Figure 6.75, For Prob. 6.53., , Figure 6.72, For Prob. 6.49., 6.50 An energy-storage network consists of seriesconnected 16-mH and 14-mH inductors in parallel, with series-connected 24-mH and 36-mH inductors., Calculate the equivalent inductance., , 6.54 Find the equivalent inductance looking into the, terminals of the circuit in Fig. 6.76., , 6.51 Determine Leq at terminals a-b of the circuit in, Fig. 6.73., , 9H, 10 H, , 10 mH, , 12 H, , 60 mH, 25 mH, , 4H, , 20 mH, , a, , 6H, , b, 30 mH, a, , Figure 6.73, , Figure 6.76, , For Prob. 6.51., , For Prob. 6.54., , b, , 3H
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ale29559_ch06.qxd, , 07/08/2008, , 11:00 AM, , Page 248, , Chapter 6, , 248, , 6.55 Find Leq in each of the circuits in Fig. 6.77., , Capacitors and Inductors, , 6.58 The current waveform in Fig. 6.80 flows through a, 3-H inductor. Sketch the voltage across the inductor, over the interval 0 6 t 6 6 s., , L, i(t), , L, Leq, L, , 2, , L, , L, 0, , 1, , 2, , 3, , 4, , 5, , 6, , t, , Figure 6.80, , (a), , For Prob. 6.58., L, L, , L, , L, , 6.59 (a) For two inductors in series as in Fig. 6.81(a),, show that the voltage division principle is, , L, Leq, , v1 , , (b), , Figure 6.77, , L1, vs,, L1 L2, , v2 , , L2, vs, L1 L2, , assuming that the initial conditions are zero., , For Prob. 6.55., , (b) For two inductors in parallel as in Fig. 6.81(b),, show that the current-division principle is, 6.56 Find Leq in the circuit of Fig. 6.78., , i1 , , L2, is,, L1 L 2, , i2 , , L1, is, L1 L 2, , assuming that the initial conditions are zero., L, , L, , L, , L1, , L, , L, , L, , + v −, 1, L, , L, , vs, , +, v2, −, , +, −, , is, , L2, , i1, , i2, , L1, , L2, , L eq, (a), , Figure 6.78, , (b), , Figure 6.81, , For Prob. 6.56., , For Prob. 6.59., , *6.57 Determine Leq that may be used to represent the, inductive network of Fig. 6.79 at the terminals., , i, , 2, 4H, , 6.60 In the circuit of Fig. 6.82, io(0) 2 A. Determine, io(t) and vo(t) for t 7 0., , di, dt, io (t), , +−, , a, L eq, 3H, , 5H, , 4e–2t V, , b, , Figure 6.79, , Figure 6.82, , For Prob. 6.57., , For Prob. 6.60., , 3H, , 5H, , +, vo, −
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ale29559_ch06.qxd, , 07/08/2008, , 11:00 AM, , Page 249, , Problems, , 6.61 Consider the circuit in Fig. 6.83. Find: (a) Leq, i1(t),, and i2(t) if is 3et mA, (b) vo(t), (c) energy stored, in the 20-mH inductor at t 1 s., , i1, , 249, , 6.64 The switch in Fig. 6.86 has been in position A for a, long time. At t 0, the switch moves from position, A to B. The switch is a make-before-break type so, that there is no interruption in the inductor current., Find: (a) i(t) for t 6 0, (b) v just after the switch has, been moved to position B, (c) v(t) long after the, switch is in position B., , i2, 4 mH, , +, vo, –, , is, , 20 mH, , 4Ω, , 6 mH, , B, , t=0, , A, , i, L eq, , 40 V, , Figure 6.83, , +, –, , 0.5 H, , For Prob. 6.61., , +, v, –, , 5Ω, , 20 A, , Figure 6.86, For Prob. 6.64., 6.62 Consider the circuit in Fig. 6.84. Given that, v(t) 12e3t mV for t 7 0 and i1(0) 10 mA,, find: (a) i2(0), (b) i1(t) and i2(t)., , 6.65 The inductors in Fig. 6.87 are initially charged and are, connected to the black box at t 0. If i1(0) 4 A,, i2(0) 2 A, and v(t) 50e200t mV, t 0, find:, (a) the energy initially stored in each inductor,, , 25 mH, +, , i1(t), , i2(t), , v(t), , 20 mH, , 60 mH, , (b) the total energy delivered to the black box from, t 0 to t ,, (c) i1(t) and i2(t), t 0,, (d) i(t), t 0., , –, , Figure 6.84, For Prob. 6.62., , i(t), +, Black box v, , 6.63 In the circuit of Fig. 6.85, sketch vo., , i1, , i2, , 5H, , 20 H, , t=0, , −, , Figure 6.87, For Prob. 6.65., +, vo, –, , i1(t), , i2(t), , 2H, , 6.66 The current i(t) through a 40-mH inductor is equal,, in magnitude, to the voltage across it for all values of, time. If i(0) 5 A, find i(t)., , i2(t) (A), 4, , i1(t) (A), 3, , Section 6.6 Applications, 0, , Figure 6.85, For Prob. 6.63., , 3, , 6 t (s), , 0, , 2, , 4, , 6 t (s), , 6.67 An op amp integrator has R 100 k and C , 0.01 mF. If the input voltage is vi 10 sin 50t mV,, obtain the output voltage.
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ale29559_ch06.qxd, , 07/08/2008, , 11:00 AM, , Page 250, , Chapter 6, , 250, , Capacitors and Inductors, , 6.68 A 10-V dc voltage is applied to an integrator with, R 50 k, C 100 mF at t 0. How long will it, take for the op amp to saturate if the saturation, voltages are 12 V and 12 V? Assume that the, initial capacitor voltage was zero., , 6.73 Show that the circuit in Fig. 6.90 is a noninverting, integrator., , R, , 6.69 An op amp integrator with R 4 M and, C 1 mF has the input waveform shown in, Fig. 6.88. Plot the output waveform., , R, , −, +, R, , vi, , vi (mV), , +, , R, , vo, , +, −, , C, , −, , 20, , Figure 6.90, , 10, 0, , For Prob. 6.73., 1 2, , 3, , 4 5, , 6, , t (ms), , 6.74 The triangular waveform in Fig. 6.91(a) is applied to, the input of the op amp differentiator in Fig. 6.91(b)., Plot the output., , –10, –20, , Figure 6.88, vi (t), , For Prob. 6.69., , 2, , 6.70 Using a single op amp, a capacitor, and resistors of, 100 k or less, design a circuit to implement, , 0, , 1, , 2, , 3, , 4, , t (s), , t, , vo 50, , v (t) dt, i, , −2, , 0, , Assume vo 0 at t 0., , (a), , 6.71 Show how you would use a single op amp to generate, 100 kΩ, , t, , vo , , (v, , 1, , 4v2 10v3) dt, , 0.01 F, , 0, , If the integrating capacitor is C 2 mF, obtain the, other component values., , −, +, vi, , 6.72 At t 1.5 ms, calculate vo due to the cascaded, integrators in Fig. 6.89. Assume that the integrators, are reset to 0 V at t 0., , +, −, , +, vo, −, , (b), , Figure 6.91, For Prob. 6.74., 2 F, 10 kΩ, , 1V, , +, −, , Figure 6.89, For Prob. 6.72., , −, +, , 0.5 F, 20 kΩ, , −, +, , +, vo, −, , 6.75 An op amp differentiator has R 250 k and C , 10 mF. The input voltage is a ramp r(t) 12t mV., Find the output voltage., 6.76 A voltage waveform has the following characteristics:, a positive slope of 20 V/s for 5 ms followed by a, negative slope of 10 V/s for 10 ms. If the waveform, is applied to a differentiator with R 50 k,, C 10 mF, sketch the output voltage waveform.
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ale29559_ch06.qxd, , 07/08/2008, , 11:00 AM, , Page 251, , Comprehensive Problems, , *6.77 The output vo of the op amp circuit in Fig. 6.92(a) is, shown in Fig. 6.92(b). Let Ri Rf 1 M and, C 1 mF. Determine the input voltage waveform, and sketch it., , 251, , 6.79 Design an analog computer circuit to solve the, following ordinary differential equation., dy(t), 4y(t) f (t), dt, where y(0) 1 V., 6.80 Figure 6.93 presents an analog computer designed, to solve a differential equation. Assuming f (t) is, known, set up the equation for f (t)., , Rf, C, Ri, vi, , −, +, , +, vo, −, , +, −, , 1 F, 1 MΩ, , 1 F, 1 MΩ, , −, +, , 1 MΩ, , 500 kΩ, , −, +, , (a), , −, +, v o (t), , 100 kΩ, , vo, 100 kΩ, , 4, , −, +, , 200 kΩ, −f (t), , 0, , 1, , 2, , 3, , 4, , t (s), , Figure 6.93, For Prob. 6.80., , −4, (b), , Figure 6.92, 6.81 Design an analog computer to simulate the following, equation:, , For Prob. 6.77., , d 2v, 5v 2f (t), dt 2, 6.82 Design an op amp circuit such that, 6.78 Design an analog computer to simulate, d 2vo, dt, , 2, , 2, , dvo, vo 10 sin 2t, dt, , where v0(0) 2 and v¿0(0) 0., , vo 10vs 2, , v dt, s, , where vs and vo are the input voltage and output, voltage, respectively., , Comprehensive Problems, 6.83 Your laboratory has available a large number of, 10-mF capacitors rated at 300 V. To design a, capacitor bank of 40 mF rated at 600 V, how many, 10-mF capacitors are needed and how would you, connect them?, , 6.84 An 8-mH inductor is used in a fusion power, experiment. If the current through the inductor is, i(t) 5 sin2 p t mA, t 7 0, find the power being, delivered to the inductor and the energy stored in it, at t 0.5 s.
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ale29559_ch06.qxd, , 07/08/2008, , 11:00 AM, , Page 252, , Chapter 6, , 252, , Capacitors and Inductors, , 6.85 A square-wave generator produces the voltage, waveform shown in Fig. 6.94(a). What kind of a, circuit component is needed to convert the voltage, waveform to the triangular current waveform shown, in Fig. 6.94(b)? Calculate the value of the, component, assuming that it is initially uncharged., , i (A), 4, , 0, , 1, , 3, , 2, , 4, , t (ms), , (b), v (V), , Figure 6.94, , 5, , For Prob. 6.85., , 0, 1, , 2, , 3, , −5, (a), , 4, , t (ms), , 6.86 An electric motor can be modeled as a series, combination of a 12- resistor and 200-mH inductor., If a current i(t) 2te10t A flows through the series, combination, find the voltage across the combination.
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 253, , c h a p t e r, , First-Order Circuits, , 7, , We live in deeds, not years; in thoughts, not breaths; in feelings, not in, figures on a dial. We should count time in heart-throbs. He most lives, who thinks most, feels the noblest, acts the best., —F. J. Bailey, , Enhancing Your Career, Careers in Computer Engineering, Electrical engineering education has gone through drastic changes in, recent decades. Most departments have come to be known as Department, of Electrical and Computer Engineering, emphasizing the rapid changes, due to computers. Computers occupy a prominent place in modern society and education. They have become commonplace and are helping to, change the face of research, development, production, business, and entertainment. The scientist, engineer, doctor, attorney, teacher, airline pilot,, businessperson—almost anyone benefits from a computer’s abilities to, store large amounts of information and to process that information in very, short periods of time. The internet, a computer communication network,, is essential in business, education, and library science. Computer usage, continues to grow by leaps and bounds., An education in computer engineering should provide breadth in software, hardware design, and basic modeling techniques. It should include, courses in data structures, digital systems, computer architecture, microprocessors, interfacing, software engineering, and operating systems., Electrical engineers who specialize in computer engineering find, jobs in computer industries and in numerous fields where computers, are being used. Companies that produce software are growing rapidly, in number and size and providing employment for those who are skilled, in programming. An excellent way to advance one’s knowledge of, computers is to join the IEEE Computer Society, which sponsors, diverse magazines, journals, and conferences., , Computer design of very large scale, integrated (VLSI) circuits., Courtesy Brian Fast, Cleveland State, University, , 253
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 254, , Chapter 7, , 254, , 7.1, , First-Order Circuits, , Introduction, , Now that we have considered the three passive elements (resistors,, capacitors, and inductors) and one active element (the op amp) individually, we are prepared to consider circuits that contain various combinations of two or three of the passive elements. In this chapter, we, shall examine two types of simple circuits: a circuit comprising a resistor and capacitor and a circuit comprising a resistor and an inductor., These are called RC and RL circuits, respectively. As simple as these, circuits are, they find continual applications in electronics, communications, and control systems, as we shall see., We carry out the analysis of RC and RL circuits by applying, Kirchhoff’s laws, as we did for resistive circuits. The only difference, is that applying Kirchhoff’s laws to purely resistive circuits results in, algebraic equations, while applying the laws to RC and RL circuits produces differential equations, which are more difficult to solve than, algebraic equations. The differential equations resulting from analyzing RC and RL circuits are of the first order. Hence, the circuits are, collectively known as first-order circuits., A first-order circuit is characterized by a first-order differential, equation., , In addition to there being two types of first-order circuits (RC and, RL), there are two ways to excite the circuits. The first way is by initial conditions of the storage elements in the circuits. In these so-called, source-free circuits, we assume that energy is initially stored in the, capacitive or inductive element. The energy causes current to flow in, the circuit and is gradually dissipated in the resistors. Although sourcefree circuits are by definition free of independent sources, they may, have dependent sources. The second way of exciting first-order circuits, is by independent sources. In this chapter, the independent sources we, will consider are dc sources. (In later chapters, we shall consider sinusoidal and exponential sources.) The two types of first-order circuits, and the two ways of exciting them add up to the four possible situations we will study in this chapter., Finally, we consider four typical applications of RC and RL circuits: delay and relay circuits, a photoflash unit, and an automobile, ignition circuit., iC, C, , +, v, , iR, R, , −, , Figure 7.1, A source-free RC circuit., , A circuit response is the manner in, which the circuit reacts to an, excitation., , 7.2, , The Source-Free RC Circuit, , A source-free RC circuit occurs when its dc source is suddenly disconnected. The energy already stored in the capacitor is released to the, resistors., Consider a series combination of a resistor and an initially charged, capacitor, as shown in Fig. 7.1. (The resistor and capacitor may be the, equivalent resistance and equivalent capacitance of combinations of, resistors and capacitors.) Our objective is to determine the circuit, response, which, for pedagogic reasons, we assume to be the voltage
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 255, , 7.2, , The Source-Free RC Circuit, , 255, , v(t) across the capacitor. Since the capacitor is initially charged, we, can assume that at time t 0, the initial voltage is, v(0) V0, , (7.1), , with the corresponding value of the energy stored as, w(0) , , 1, CV 20, 2, , (7.2), , Applying KCL at the top node of the circuit in Fig. 7.1 yields, iC iR 0, , (7.3), , By definition, iC C dvdt and iR vR. Thus,, dv, v, 0, dt, R, , (7.4a), , dv, v, , 0, dt, RC, , (7.4b), , C, or, , This is a first-order differential equation, since only the first derivative, of v is involved. To solve it, we rearrange the terms as, dv, 1, , dt, v, RC, , (7.5), , Integrating both sides, we get, ln v , , t, ln A, RC, , where ln A is the integration constant. Thus,, ln, , v, t, , A, RC, , (7.6), , Taking powers of e produces, v(t) AetRC, But from the initial conditions, v(0) A V0. Hence,, v(t) V0 etRC, , (7.7), , This shows that the voltage response of the RC circuit is an exponential decay of the initial voltage. Since the response is due to the initial, energy stored and the physical characteristics of the circuit and not due, to some external voltage or current source, it is called the natural, response of the circuit., The natural response of a circuit refers to the behavior (in terms of, voltages and currents) of the circuit itself, with no external sources of, excitation., , The natural response is illustrated graphically in Fig. 7.2. Note that at, t 0, we have the correct initial condition as in Eq. (7.1). As t, increases, the voltage decreases toward zero. The rapidity with which, , The natural response depends on the, nature of the circuit alone, with no external sources. In fact, the circuit has a, response only because of the energy, initially stored in the capacitor.
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ale29559_ch07.qxd, , 07/17/2008, , 11:23 AM, , Page 256, , Chapter 7, , 256, v, , the voltage decreases is expressed in terms of the time constant,, denoted by t, the lowercase Greek letter tau., , V0, V0e−t ⁄ , , 0.368V0, , The time constant of a circuit is the time required for the response to, decay to a factor of 1e or 36.8 percent of its initial value.1, , , , 0, , First-Order Circuits, , t, , Figure 7.2, The voltage response of the RC circuit., , This implies that at t t, Eq. (7.7) becomes, V0etRC V0e1 0.368V0, or, t RC, , (7.8), , In terms of the time constant, Eq. (7.7) can be written as, v(t) V0ett, , TABLE 7.1, , Values of v (t)V0 et., t, , v(t)V0, , t, 2t, 3t, 4t, 5t, , 0.36788, 0.13534, 0.04979, 0.01832, 0.00674, , v, V0, , With a calculator it is easy to show that the value of v(t)V0 is as, shown in Table 7.1. It is evident from Table 7.1 that the voltage v(t), is less than 1 percent of V0 after 5t (five time constants). Thus, it is, customary to assume that the capacitor is fully discharged (or charged), after five time constants. In other words, it takes 5t for the circuit to, reach its final state or steady state when no changes take place with, time. Notice that for every time interval of t, the voltage is reduced, by 36.8 percent of its previous value, v(t t) v(t)e 0.368v(t),, regardless of the value of t., Observe from Eq. (7.8) that the smaller the time constant, the more, rapidly the voltage decreases, that is, the faster the response. This is, illustrated in Fig. 7.4. A circuit with a small time constant gives a fast, response in that it reaches the steady state (or final state) quickly due, to quick dissipation of energy stored, whereas a circuit with a large, time constant gives a slow response because it takes longer to reach, steady state. At any rate, whether the time constant is small or large,, the circuit reaches steady state in five time constants., With the voltage v(t) in Eq. (7.9), we can find the current iR(t),, iR(t) , , 1.0, , (7.9), , V0, v(t), ett, R, R, , (7.10), , 0.75, 1, The time constant may be viewed from another perspective. Evaluating the derivative, of v(t) in Eq. (7.7) at t 0, we obtain, , Tangent at t = 0, , 0.50, 0.37, , d v, 1, 1, a b2, e tt 2, , dt V0 t0, t, t, t0, , 0.25, , 0, , , , 2, , 3, , 4, , 5 t (s), , Figure 7.3, Graphical determination of the time, constant t from the response curve., , Thus, the time constant is the initial rate of decay, or the time taken for vV0 to decay, from unity to zero, assuming a constant rate of decay. This initial slope interpretation of, the time constant is often used in the laboratory to find t graphically from the response, curve displayed on an oscilloscope. To find t from the response curve, draw the tangent, to the curve at t 0, as shown in Fig. 7.3. The tangent intercepts with the time axis at, t t.
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 257, , 7.2, , The Source-Free RC Circuit, , 257, , v = e−t ⁄, V0, 1, =2, , =1, = 0.5, 0, , 1, , 2, , 3, , 4, , 5, , t, , Figure 7.4, , Plot of vV0 e tt for various values of the time constant., , The power dissipated in the resistor is, p(t) viR , , V 20 2tt, e, R, , (7.11), , The energy absorbed by the resistor up to time t is, wR(t) , , t, , t, , 0, , 0, , p dt , , V 20 2tt, e, dt, R, , tV 20 2tt t, 1, , e, 2 CV 20 (1 e2tt),, 2R, 2, 0, , (7.12), t RC, , Notice that as t S , wR() S 12CV 20, which is the same as wC (0),, the energy initially stored in the capacitor. The energy that was initially, stored in the capacitor is eventually dissipated in the resistor., In summary:, , The Key to Working with a Source-free RC Circuit, Is Finding:, , The time constant is the same regardless of what the output is defined, to be., , 1. The initial voltage v(0) V0 across the capacitor., 2. The time constant t., , With these two items, we obtain the response as the capacitor voltage, vC (t) v(t) v(0)ett. Once the capacitor voltage is first obtained,, other variables (capacitor current iC, resistor voltage vR, and resistor current iR) can be determined. In finding the time constant t RC, R is, often the Thevenin equivalent resistance at the terminals of the capacitor;, that is, we take out the capacitor C and find R RTh at its terminals., , In Fig. 7.5, let vC (0) 15 V. Find vC, vx, and ix for t 7 0., Solution:, We first need to make the circuit in Fig. 7.5 conform with the standard, RC circuit in Fig. 7.1. We find the equivalent resistance or the Thevenin, , When a circuit contains a single, capacitor and several resistors and, dependent sources, the Thevenin, equivalent can be found at the, terminals of the capacitor to form a, simple RC circuit. Also, one can use, Thevenin’s theorem when several, capacitors can be combined to form, a single equivalent capacitor., , Example 7.1
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 258, , Chapter 7, , 258, 8Ω, ix, 5Ω, , +, vC, −, , 0.1 F, , 12 Ω, , +, vx, −, , First-Order Circuits, , resistance at the capacitor terminals. Our objective is always to first, obtain capacitor voltage vC. From this, we can determine vx and ix., The 8- and 12- resistors in series can be combined to give a, 20- resistor. This 20- resistor in parallel with the 5- resistor can, be combined so that the equivalent resistance is, Req , , Figure 7.5, For Example 7.1., , 20 5, 4, 20 5, , Hence, the equivalent circuit is as shown in Fig. 7.6, which is analogous, to Fig. 7.1. The time constant is, t ReqC 4(0.1) 0.4 s, , +, v, , Req, , Thus,, , 0.1 F, , v v(0)ett 15et0.4 V,, vC v 15e2.5t V, From Fig. 7.5, we can use voltage division to get vx; so, , −, , Figure 7.6, , vx , , Equivalent circuit for the circuit in, Fig. 7.5., , 12, v 0.6(15e2.5t ) 9e2.5t V, 12 8, , Finally,, ix , , Practice Problem 7.1, io, , 12 Ω, , 6Ω, , Refer to the circuit in Fig. 7.7. Let vC (0) 45 V. Determine vC , vx ,, and io for t 0., , 8Ω, , +, vx, −, , 1, 3, , F, , vx, 0.75e2.5t A, 12, , +, vC, −, , Answer: 45e0.25t V, 15e0.25t V, 3.75e0.25t A., , Figure 7.7, For Practice Prob. 7.1., , Example 7.2, 3Ω, , 20 V, , +, −, , Figure 7.8, For Example 7.2., , t=0, , 9Ω, , The switch in the circuit in Fig. 7.8 has been closed for a long time,, and it is opened at t 0. Find v(t) for t 0. Calculate the initial, energy stored in the capacitor., , 1Ω, +, v, −, , 20 mF, , Solution:, For t 6 0, the switch is closed; the capacitor is an open circuit to dc,, as represented in Fig. 7.9(a). Using voltage division, vC (t) , , 9, (20) 15 V,, 93, , t 6 0, , Since the voltage across a capacitor cannot change instantaneously, the, voltage across the capacitor at t 0 is the same at t 0, or, vC (0) V0 15 V
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 259, , 7.3, , The Source-Free RL Circuit, , For t 7 0, the switch is opened, and we have the RC circuit, shown in Fig. 7.9(b). [Notice that the RC circuit in Fig. 7.9(b) is, source free; the independent source in Fig. 7.8 is needed to provide, V0 or the initial energy in the capacitor.] The 1- and 9- resistors, in series give, , 259, 3Ω, , 1Ω, +, , 20 V, , +, −, , 9Ω, , vC (0), −, (a), , Req 1 9 10 , , 1Ω, , The time constant is, t ReqC 10 20 103 0.2 s, , +, Vo = 15 V, −, , 9Ω, , Thus, the voltage across the capacitor for t 0 is, v(t) vC (0)ett 15et0.2 V, , 20 mF, , (b), , Figure 7.9, , or, , For Example 7.2: (a) t 6 0, (b) t 7 0., , v(t) 15e5t V, The initial energy stored in the capacitor is, 1 2, 1, CvC (0) 20 103 152 2.25 J, 2, 2, , Practice Problem 7.2, , If the switch in Fig. 7.10 opens at t 0, find v(t) for t 0 and wC (0)., Answer: 8e2t V, 5.33 J., , 6Ω, , 24 V, , 7.3, , 1, 6, , +, −, , +, v, −, , F, , For Practice Prob. 7.2., , Consider the series connection of a resistor and an inductor, as shown, in Fig. 7.11. Our goal is to determine the circuit response, which we, will assume to be the current i(t) through the inductor. We select the, inductor current as the response in order to take advantage of the idea, that the inductor current cannot change instantaneously. At t 0, we, assume that the inductor has an initial current I0, or, L, , vL, , +, , (7.13), , i, , with the corresponding energy stored in the inductor as, 1, w(0) L I 20, 2, , (7.14), , But vL L didt and vR iR. Thus,, L, , di, Ri 0, dt, , Figure 7.11, A source-free RL circuit., , Applying KVL around the loop in Fig. 7.11,, vL vR 0, , 12 Ω, , Figure 7.10, , The Source-Free RL Circuit, , i(0) I0, , t=0, , −, , wC (0) , , (7.15), , R, , +, vR, −, , 4Ω
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 260, , Chapter 7, , 260, , First-Order Circuits, , or, di, R, i0, dt, L, , (7.16), , Rearranging terms and integrating gives, , , , i(t), , I0, , ln i 2, , i(t), I0, , , , di, , i, , t, , Rt, 2, L 0, , , , t, , 0, , R, dt, L, , ln i(t) ln I0 , , 1, , Rt, 0, L, , or, ln, , i(t), Rt, , I0, L, , (7.17), , Taking the powers of e, we have, i(t) I0eRtL, , (7.18), , This shows that the natural response of the RL circuit is an exponential decay of the initial current. The current response is shown in, Fig. 7.12. It is evident from Eq. (7.18) that the time constant for the, RL circuit is, , i(t), I0, , Tangent at t = 0, 0.368I0, , t, , I 0 e −t ⁄ , , L, R, , (7.19), , with t again having the unit of seconds. Thus, Eq. (7.18) may be, written as, 0, , , , t, , Figure 7.12, , i(t) I0ett, , The current response of the RL circuit., , (7.20), , With the current in Eq. (7.20), we can find the voltage across the, resistor as, The smaller the time constant t of a, circuit, the faster the rate of decay of, the response. The larger the time constant, the slower the rate of decay of, the response. At any rate, the response, decays to less than 1 percent of its, initial value (i.e., reaches steady state), after 5t., , vR (t) i R I0 Rett, , (7.21), , The power dissipated in the resistor is, p vR i I 20 Re2tt, , (7.22), , The energy absorbed by the resistor is, wR(t) , , , , t, , 0, , p dt , , , , t, , 0, , t, 1, I 20 Re2tt dt t I 20 Re2tt 2 ,, 2, 0, , t, , L, R, , or, 1, wR (t) L I 20 (1 e2tt ), 2, , Figure 7.12 shows an initial slope interpretation may be given to t., , (7.23), , Note that as t S , wR() S 12 L I 20, which is the same as wL(0),, the initial energy stored in the inductor as in Eq. (7.14). Again, the, energy initially stored in the inductor is eventually dissipated in, the resistor.
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 261, , The Source-Free RL Circuit, , 7.3, , 261, , In summary:, , The Key to Working with a Source-free RL Circuit, Is to Find:, 1. The initial current i(0) I0 through the inductor., 2. The time constant t of the circuit., When a circuit has a single inductor, and several resistors and dependent, sources, the Thevenin equivalent can, be found at the terminals of the inductor to form a simple RL circuit. Also,, one can use Thevenin’s theorem when, several inductors can be combined to, form a single equivalent inductor., , With the two items, we obtain the response as the inductor current, iL(t) i(t) i(0)ett. Once we determine the inductor current iL,, other variables (inductor voltage vL, resistor voltage vR, and resistor, current iR) can be obtained. Note that in general, R in Eq. (7.19) is the, Thevenin resistance at the terminals of the inductor., , Example 7.3, , Assuming that i(0) 10 A, calculate i(t) and ix (t) in the circuit of, Fig. 7.13., , 4Ω, , Solution:, There are two ways we can solve this problem. One way is to obtain, the equivalent resistance at the inductor terminals and then use, Eq. (7.20). The other way is to start from scratch by using Kirchhoff’s, voltage law. Whichever approach is taken, it is always better to first, obtain the inductor current., , ix, , i, , +, −, , 2Ω, , 0.5 H, , Figure 7.13, For Example 7.3., , ■ METHOD 1 The equivalent resistance is the same as the, Thevenin resistance at the inductor terminals. Because of the dependent source, we insert a voltage source with vo 1 V at the inductor, terminals a-b, as in Fig. 7.14(a). (We could also insert a 1-A current, source at the terminals.) Applying KVL to the two loops results in, 2(i1 i2) 1 0, , i1 i2 , , 1, , 6i2 2i1 3i1 0, , 1, , 5, i2 i1, 6, , 1, 2, , (7.3.1), (7.3.2), , Substituting Eq. (7.3.2) into Eq. (7.3.1) gives, i1 3 A,, io, , io i1 3 A, 4Ω, , a, , 4Ω, vo = 1 V +, −, , 2Ω, , i1, , i2, , +, −, , 3i1, 0.5 H, , i1, , 2Ω, , i2, , b, (a), , Figure 7.14, Solving the circuit in Fig. 7.13., , (b), , +, −, , 3i, , 3i
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ale29559_ch07.qxd, , 262, , 07/08/2008, , 11:49 AM, , Page 262, , Chapter 7, , First-Order Circuits, , Hence,, Req RTh , , vo, 1, , io, 3, , The time constant is, t, , L, , Req, , 1, 2, 1, 3, , , , 3, s, 2, , Thus, the current through the inductor is, i(t) i(0)ett 10e(23)t A,, , t 7 0, , ■ METHOD 2 We may directly apply KVL to the circuit as in, Fig. 7.14(b). For loop 1,, 1 di1, 2(i1 i2) 0, 2 dt, or, di1, 4i1 4i2 0, dt, , (7.3.3), , For loop 2,, 6i2 2i1 3i1 0, , 1, , 5, i2 i1, 6, , Substituting Eq. (7.3.4) into Eq. (7.3.3) gives, di1, 2, i1 0, dt, 3, Rearranging terms,, di1, 2, dt, i1, 3, Since i1 i, we may replace i1 with i and integrate:, ln i 2, , 2 t, t2, 3 0, i(0), , i(t), , or, ln, , i(t), 2, t, i(0), 3, , Taking the powers of e, we finally obtain, i(t) i(0)e(23)t 10e(23)t A,, , t 7 0, , which is the same as by Method 1., The voltage across the inductor is, vL, , di, 2, 10, 0.5(10) a b e(23)t e(23)t V, dt, 3, 3, , (7.3.4)
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 263, , 7.3, , The Source-Free RL Circuit, , 263, , Since the inductor and the 2- resistor are in parallel,, ix (t) , , v, 1.6667e(23)t A,, 2, , t 7 0, , Practice Problem 7.3, , Find i and vx in the circuit of Fig. 7.15. Let i(0) 5 A., , 4Ω, , Answer: 5e4t V, 20e4t V., , + vx −, , i, , 1Ω, 4Ω, , 2H, +, −, , 2vx, , Figure 7.15, For Practice Prob. 7.3., , Example 7.4, , The switch in the circuit of Fig. 7.16 has been closed for a long time., At t 0, the switch is opened. Calculate i(t) for t 7 0., Solution:, When t 6 0, the switch is closed, and the inductor acts as a short, circuit to dc. The 16- resistor is short-circuited; the resulting circuit, is shown in Fig. 7.17(a). To get i1 in Fig. 7.17(a), we combine the 4-, and 12- resistors in parallel to get, 4 12, 3, 4 12, , 2Ω, , t=0, , 4Ω, i(t), , +, −, , 12 Ω, , 40 V, , 16 Ω, , 2H, , Figure 7.16, For Example 7.4., , Hence,, i1 , , 40, 8A, 23, , i1, , 12, i1 6 A,, 12 4, , 4Ω, i(t), , We obtain i(t) from i1 in Fig. 7.17(a) using current division, by, writing, i(t) , , 2Ω, , 40 V, , +, −, , 12 Ω, , t 6 0, , (a), 4Ω, , Since the current through an inductor cannot change instantaneously,, i(0) i(0) 6 A, When t 7 0, the switch is open and the voltage source is, disconnected. We now have the source-free RL circuit in Fig. 7.17(b)., Combining the resistors, we have, Req (12 4) 16 8 , , i(t), 12 Ω, , 16 Ω, , (b), , Figure 7.17, Solving the circuit of Fig. 7.16: (a) for, t 6 0, (b) for t 7 0., , The time constant is, t, , L, 2, 1, s, Req, 8, 4, , Thus,, i(t) i(0)ett 6e4t A, , 2H
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 264, , Chapter 7, , 264, , Practice Problem 7.4, , For the circuit in Fig. 7.18, find i(t) for t 7 0., Answer: 2e2t A, t 7 0., , t=0, , 8Ω, , 12 Ω, 24 Ω, , 6A, , First-Order Circuits, , 5Ω, , i(t), 2H, , Figure 7.18, For Practice Prob. 7.4., , Example 7.5, 2Ω, , 10 V, , In the circuit shown in Fig. 7.19, find io, vo, and i for all time, assuming that the switch was open for a long time., , 3Ω, , +, −, , + v −, o, , io, , i, , t=0, , 6Ω, , 2H, , Figure 7.19, For Example 7.5., , Solution:, It is better to first find the inductor current i and then obtain other, quantities from it., For t 6 0, the switch is open. Since the inductor acts like a short, circuit to dc, the 6- resistor is short-circuited, so that we have the, circuit shown in Fig. 7.20(a). Hence, io 0, and, 10, 2 A,, 23, vo (t) 3i(t) 6 V,, , i(t) , 2Ω, , 3Ω, + v −, o, , 10 V, , +, −, , io, , t 6 0, t 6 0, , i, , Thus, i(0) 2., For t 7 0, the switch is closed, so that the voltage source is shortcircuited. We now have a source-free RL circuit as shown in, Fig. 7.20(b). At the inductor terminals,, , 6Ω, , (a), , R Th 3 6 2 , , 3Ω, + v −, o, , i, , io, , so that the time constant is, , +, , 6Ω, , vL, −, , 2H, , (b), , t, Hence,, , Figure 7.20, The circuit in Fig. 7.19 for: (a) t 6 0,, (b) t 7 0., , L, 1s, RTh, , i(t) i(0)ett 2et A,, , t 7 0, , Since the inductor is in parallel with the 6- and 3- resistors,, vo(t) vL L, , di, 2(2et) 4et V,, dt, , and, io(t) , , vL, 2, et A,, 6, 3, , t 7 0, , t 7 0
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 265, , 7.4, , Singularity Functions, , 265, , Thus, for all time,, 0 A,, io(t) c 2 t, e A,, 3, , t 6 0, t 7 0, , i(t) b, , vo(t) b, , ,, , 2 A,, 2et A,, , 6 V,, 4et V,, , t 6 0, t 7 0, , 2, , i(t), , t 6 0, t0, , t, , We notice that the inductor current is continuous at t 0, while the, current through the 6- resistor drops from 0 to 23 at t 0, and, the voltage across the 3- resistor drops from 6 to 4 at t 0. We also, notice that the time constant is the same regardless of what the output, is defined to be. Figure 7.21 plots i and io., , Determine i, io, and vo for all t in the circuit shown in Fig. 7.22., Assume that the switch was closed for a long time. It should be noted, that opening a switch in series with an ideal current source creates an, infinite voltage at the current source terminals. Clearly this is impossible. For the purposes of problem solving, we can place a shunt resistor in parallel with the source (which now makes it a voltage source, in series with a resistor). In more practical circuits, devices that act like, current sources are, for the most part, electronic circuits. These circuits, will allow the source to act like an ideal current source over its operating range but voltage-limit it when the load resistor becomes too large, (as in an open circuit)., , −2, 3, , io(t), , Figure 7.21, A plot of i and io., , Practice Problem 7.5, 3Ω, t=0, , i, io, , 18 A, , 4Ω, , Figure 7.22, For Practice Prob. 7.5., , Answer:, ib, , 12 A,, 12e2t A,, , t 6 0, ,, t0, vo b, , 7.4, , io b, , 24 V,, 8e2t V,, , 6 A,, 4e2t A,, , t 6 0, ,, t 7 0, , t 6 0, t 7 0, , Singularity Functions, , Before going on with the second half of this chapter, we need to digress, and consider some mathematical concepts that will aid our understanding of transient analysis. A basic understanding of singularity, functions will help us make sense of the response of first-order circuits, to a sudden application of an independent dc voltage or current source., Singularity functions (also called switching functions) are very useful in circuit analysis. They serve as good approximations to the, switching signals that arise in circuits with switching operations. They, are helpful in the neat, compact description of some circuit phenomena, especially the step response of RC or RL circuits to be discussed, in the next sections. By definition,, Singularity functions are functions that either are discontinuous or have, discontinuous derivatives., , 1H, , 2Ω, , +, vo, −
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 266, , Chapter 7, , 266, , First-Order Circuits, , The three most widely used singularity functions in circuit analysis are the unit step, the unit impulse, and the unit ramp functions., , u(t), , The unit step function u (t ) is 0 for negative values of t and 1 for positive values of t., , 1, , In mathematical terms,, 0, , t, , Figure 7.23, The unit step function., , u (t) b, , 0,, 1,, , t 6 0, t 7 0, , (7.24), , The unit step function is undefined at t 0, where it changes abruptly, from 0 to 1. It is dimensionless, like other mathematical functions such, as sine and cosine. Figure 7.23 depicts the unit step function. If the, abrupt change occurs at t t0 (where t0 7 0) instead of t 0, the unit, step function becomes, , u(t − t0), , 1, , 0, , t0, , u (t t0) b, , t, , (a), , 1, , u (t t0) b, 0, , t 6 t0, t 7 t0, , (7.25), , which is the same as saying that u (t) is delayed by t0 seconds, as shown, in Fig. 7.24(a). To get Eq. (7.25) from Eq. (7.24), we simply replace, every t by t t0. If the change is at t t0, the unit step function, becomes, , u(t + t0), , −t0, , 0,, 1,, , t, (b), , Figure 7.24, (a) The unit step function delayed by t0,, (b) the unit step advanced by t0., , 0,, 1,, , t 6 t0, t 7 t0, , (7.26), , meaning that u (t) is advanced by t0 seconds, as shown in Fig. 7.24(b)., We use the step function to represent an abrupt change in voltage, or current, like the changes that occur in the circuits of control systems, and digital computers. For example, the voltage, v(t) b, , 0,, V0,, , t 6 t0, t 7 t0, , (7.27), , may be expressed in terms of the unit step function as, v(t) V0 u (t t0), Alternatively, we may derive, Eqs. (7.25) and (7.26) from Eq. (7.24), by writing u [f (t )] 1, f (t ) 7 0,, where f (t ) may be t t 0 or t t 0., , (7.28), , If we let t0 0, then v(t) is simply the step voltage V0 u (t). A voltage, source of V0 u (t) is shown in Fig. 7.25(a); its equivalent circuit is shown, in Fig. 7.25(b). It is evident in Fig. 7.25(b) that terminals a-b are shortcircuited (v 0) for t 6 0 and that v V0 appears at the terminals, , t=0, a, , a, V0 u(t), , =, , +, −, , V0 +, −, b, , b, (a), , (b), , Figure 7.25, (a) Voltage source of V0u(t), (b) its equivalent circuit.
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 267, , 7.4, , Singularity Functions, , 267, , for t 7 0. Similarly, a current source of I0 u (t) is shown in Fig. 7.26(a),, while its equivalent circuit is in Fig. 7.26(b). Notice that for t 6 0,, there is an open circuit (i 0), and that i I0 flows for t 7 0., , t=0, , i, a, , a, , =, , I0 u(t), , I0, b, , b, (b), , (a), , Figure 7.26, (a) Current source of I0u (t), (b) its equivalent circuit., , The derivative of the unit step function u (t) is the unit impulse, function d(t), which we write as, 0,, d, d(t) u (t) c Undefined,, dt, 0,, , t 6 0, t0, t 7 0, , (t), , (7.29), , The unit impulse function—also known as the delta function—is, shown in Fig. 7.27., , (∞), , 0, , t, , Figure 7.27, The unit impulse function., , The unit impulse function d(t ) is zero everywhere except at t 0,, where it is undefined., , Impulsive currents and voltages occur in electric circuits as a result of, switching operations or impulsive sources. Although the unit impulse, function is not physically realizable (just like ideal sources, ideal, resistors, etc.), it is a very useful mathematical tool., The unit impulse may be regarded as an applied or resulting shock., It may be visualized as a very short duration pulse of unit area. This, may be expressed mathematically as, , , , 0, , d(t) dt 1, , (7.30), , 0, , where t 0 denotes the time just before t 0 and t 0 is the time, just after t 0. For this reason, it is customary to write 1 (denoting, unit area) beside the arrow that is used to symbolize the unit impulse, function, as in Fig. 7.27. The unit area is known as the strength of the, impulse function. When an impulse function has a strength other than, unity, the area of the impulse is equal to its strength. For example, an, impulse function 10d (t) has an area of 10. Figure 7.28 shows the, impulse functions 5d (t 2), 10d(t), and 4d (t 3)., To illustrate how the impulse function affects other functions, let, us evaluate the integral, , , , f (t)d (t t0) dt, , −2, , −1, , 0, , 1, , (7.31), , 2, , 3, −4(t − 3), , Figure 7.28, , b, , a, , 10(t), 5(t + 2), , Three impulse functions., , t
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 268, , Chapter 7, , 268, , First-Order Circuits, , where a 6 t0 6 b. Since d(t t0) 0 except at t t0, the integrand, is zero except at t0. Thus,, , , , b, , , , f (t)d(t t0) dt , , a, , b, , f (t0) d(t t0) dt, , a, , f (t0), , , , b, , d(t t0) dt f (t0), , a, , or, , , , b, , f (t)d(t t0) dt f (t0), , (7.32), , a, , This shows that when a function is integrated with the impulse function, we obtain the value of the function at the point where the impulse, occurs. This is a highly useful property of the impulse function known, as the sampling or sifting property. The special case of Eq. (7.31) is, for t0 0. Then Eq. (7.32) becomes, , r(t), , , , 1, , 0, , f (t) d(t) dt f (0), , (7.33), , 0, , Integrating the unit step function u (t) results in the unit ramp function r (t); we write, 0, , t, , 1, , r (t) , , Figure 7.29, , , , t, , u (t) dt tu (t), , (7.34), , , , The unit ramp function., , or, , r (t) b, , r (t − t0), 1, , t0, t0, , 0,, t,, , (7.35), , The unit ramp function is zero for negative values of t and has a unit, slope for positive values of t., 0 t0, , t0 + 1 t, , Figure 7.29 shows the unit ramp function. In general, a ramp is a function that changes at a constant rate., The unit ramp function may be delayed or advanced as shown in, Fig. 7.30. For the delayed unit ramp function,, , (a), r(t + t0), , r (t t0) b, 1, , t t0, t t0, , (7.36), , t t0, t t0, , (7.37), , 0,, t t0,, , and for the advanced unit ramp function,, r (t t0) b, −t0, , −t0 + 1 0, , t, , (b), , We should keep in mind that the three singularity functions, (impulse, step, and ramp) are related by differentiation as, , Figure 7.30, The unit ramp function: (a) delayed by t0,, (b) advanced by t0., , 0,, t t0,, , d(t) , , du (t), ,, dt, , u (t) , , dr (t), dt, , (7.38)
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 269, , 7.4, , Singularity Functions, , 269, , or by integration as, u (t) , , , , t, , d(t) dt,, , r (t) , , , , , , t, , u (t) dt, , (7.39), , , , Although there are many more singularity functions, we are only interested in these three (the impulse function, the unit step function, and, the ramp function) at this point., , Example 7.6, , Express the voltage pulse in Fig. 7.31 in terms of the unit step. Calculate its derivative and sketch it., Solution:, The type of pulse in Fig. 7.31 is called the gate function. It may be, regarded as a step function that switches on at one value of t and, switches off at another value of t. The gate function shown in Fig. 7.31, switches on at t 2 s and switches off at t 5 s. It consists of the, sum of two unit step functions as shown in Fig. 7.32(a). From the, figure, it is evident that, , Gate functions are used along, with switches to pass or block, another signal., v (t), 10, , v(t) 10u (t 2) 10u (t 5) 10[u (t 2) u (t 5)], Taking the derivative of this gives, , 0, , dv, 10[d(t 2) d(t 5)], dt, , 1, , 2, , 3, , 4, , Figure 7.31, For Example 7.6., , which is shown in Fig. 7.32(b). We can obtain Fig. 7.32(b) directly, from Fig. 7.31 by simply observing that there is a sudden increase by, 10 V at t 2 s leading to 10d(t 2). At t 5 s, there is a sudden, decrease by 10 V leading to 10 V d(t 5)., 10u(t − 2), , −10u(t − 5), , 10, , 10, , +, 0, , 1, , 2, , 0, , t, , 1, , 2, , 3, , 4, , 5, , −10, (a), dv, dt, 10, , 0, , 1, , 2, , 3, , 4, , 5, , t, , −10, (b), , Figure 7.32, (a) Decomposition of the pulse in Fig. 7.31, (b) derivative of the pulse in Fig. 7.31., , t, , 5, , t
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 270, , Chapter 7, , 270, , Practice Problem 7.6, , First-Order Circuits, , Express the current pulse in Fig. 7.33 in terms of the unit step. Find, its integral and sketch it., Answer: 10[u (t) 2u (t 2) u (t 4)], 10[r (t) 2r (t 2) , r (t 4)]. See Fig. 7.34., i(t), , ∫ i dt, 10, 20, 0, , 2, , t, , 4, , −10, , Example 7.7, , 0, , 2, , 4, , Figure 7.33, , Figure 7.34, , For Practice Prob. 7.6., , Integral of i(t) in Fig. 7.33., , t, , Express the sawtooth function shown in Fig. 7.35 in terms of singularity functions., , v(t), , Solution:, There are three ways of solving this problem. The first method is by, mere observation of the given function, while the other methods, involve some graphical manipulations of the function., , 10, , ■ METHOD 1 By looking at the sketch of v(t) in Fig. 7.35, it is, 0, , not hard to notice that the given function v(t) is a combination of singularity functions. So we let, , t, , 2, , Figure 7.35, For Example 7.7., , v(t) v1(t) v2(t) p, , (7.7.1), , The function v1(t) is the ramp function of slope 5, shown in Fig. 7.36(a);, that is,, v1(t) 5r (t), , v1(t), , v1 + v2, , 10, , 10, , 0, , 2, , t, , +, , v2(t), 0, 2, , t, , =, , (7.7.2), , 0, , 2, , −10, (a), , Figure 7.36, Partial decomposition of v(t) in Fig. 7.35., , (b), , (c), , t
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 271, , 7.4, , Singularity Functions, , 271, , Since v1(t) goes to infinity, we need another function at t 2 s in order to, get v(t). We let this function be v2, which is a ramp function of slope 5,, as shown in Fig. 7.36(b); that is,, v2(t) 5r (t 2), , (7.7.3), , Adding v1 and v2 gives us the signal in Fig. 7.36(c). Obviously, this is, not the same as v(t) in Fig. 7.35. But the difference is simply a constant, 10 units for t 7 2 s. By adding a third signal v3, where, v3 10u (t 2), , (7.7.4), , we get v(t), as shown in Fig. 7.37. Substituting Eqs. (7.7.2) through, (7.7.4) into Eq. (7.7.1) gives, v(t) 5r (t) 5r (t 2) 10u (t 2), , v1 + v2, , v(t), , +, , 10, , 0, , 2, , =, , v3(t), 0, , t, , 2, , t, , 10, , 2, , 0, , −10, (a), , (b), , Figure 7.37, Complete decomposition of v(t) in Fig. 7.35., , ■ METHOD 2 A close observation of Fig. 7.35 reveals that v(t) is, a multiplication of two functions: a ramp function and a gate function., Thus,, v(t) 5t[u (t) u (t 2)], 5tu (t) 5tu (t 2), 5r (t) 5(t 2 2)u (t 2), 5r (t) 5(t 2)u (t 2) 10u (t 2), 5r (t) 5r (t 2) 10u (t 2), the same as before., , ■ METHOD 3 This method is similar to Method 2. We observe, from Fig. 7.35 that v(t) is a multiplication of a ramp function and a, unit step function, as shown in Fig. 7.38. Thus,, v(t) 5r (t)u (t 2), If we replace u (t) by 1 u (t), then we can replace u (t 2) by, 1 u (t 2). Hence,, v(t) 5r (t)[1 u (t 2)], which can be simplified as in Method 2 to get the same result., , (c), , t
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 272, , Chapter 7, , 272, , First-Order Circuits, , 5r(t), , 10, , u(−t + 2), , ×, 0, , 2, , t, , 1, 0, , 2, , t, , Figure 7.38, Decomposition of v(t) in Fig. 7.35., , Practice Problem 7.7, , Refer to Fig. 7.39. Express i(t) in terms of singularity functions., Answer: 2u (t) 2r (t) 4r (t 2) 2r (t 3)., , i(t) (A), 2, , 0, , 1, , 2, , 3, , t (s), , −2, , Figure 7.39, For Practice Prob. 7.7., , Example 7.8, , Given the signal, 3,, g(t) c 2,, 2t 4,, , t 6 0, 0 6 t 6 1, t 7 1, , express g(t) in terms of step and ramp functions., Solution:, The signal g(t) may be regarded as the sum of three functions specified, within the three intervals t 6 0, 0 6 t 6 1, and t 7 1., For t 6 0, g(t) may be regarded as 3 multiplied by u (t), where, u (t) 1 for t 6 0 and 0 for t 7 0. Within the time interval, 0 6 t 6 1, the function may be considered as 2 multiplied by a, gated function [u (t) u (t 1)]. For t 7 1, the function may be, regarded as 2t 4 multiplied by the unit step function u (t 1). Thus,, g(t) 3u (t), 3u (t), 3u (t), 3u (t), , , , , , , 2[u (t) u (t 1)] (2t 4)u (t 1), 2u (t) (2t 4 2)u (t 1), 2u (t) 2(t 1)u (t 1), 2u (t) 2r (t 1), , One may avoid the trouble of using u (t) by replacing it with, 1 u (t). Then, g(t) 3[1 u (t)] 2u (t) 2r (t 1) 3 5u (t) 2r (t 1), Alternatively, we may plot g(t) and apply Method 1 from Example 7.7.
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 273, , Step Response of an RC Circuit, , 7.5, , 273, , Practice Problem 7.8, , If, 0,, 8,, h (t) d, 2t 6,, 0,, , t 6 0, 0 6 t 6 2, 2 6 t 6 6, t 7 6, , express h (t) in terms of the singularity functions., Answer: 8u (t) 2u (t 2) 2r (t 2) 18u(t 6) 2r(t 6)., , Evaluate the following integrals involving the impulse function:, , , , Example 7.9, , 10, , (t2 4t 2) d (t 2) dt, , 0, , , , , , [d (t 1)et cos t d(t 1)et sin t]dt, , , , Solution:, For the first integral, we apply the sifting property in Eq. (7.32)., , , , 10, , 0, , (t2 4t 2)d(t 2) dt (t2 4t 2) 0 t2 4 8 2 10, , Similarly, for the second integral,, , , , , , [d(t 1)et cos t d(t 1)et sin t] dt, , , , et cos t 0 t1 et sin t 0 t1, , e1 cos 1 e1 sin (1) 0.1988 2.2873 2.0885, , Practice Problem 7.9, , Evaluate the following integrals:, , , , , , (t3 5t2 10)d(t 3) dt,, , , , , , 10, , d(t p) cos 3t dt, , 0, , Answer: 28, 1., , 7.5, , Step Response of an RC Circuit, , When the dc source of an RC circuit is suddenly applied, the voltage, or current source can be modeled as a step function, and the response, is known as a step response., The step response of a circuit is its behavior when the excitation is the, step function, which may be a voltage or a current source.
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 274, , Chapter 7, , 274, , (a), , The step response is the response of the circuit due to a sudden application of a dc voltage or current source., Consider the RC circuit in Fig. 7.40(a) which can be replaced by, the circuit in Fig. 7.40(b), where Vs is a constant dc voltage source., Again, we select the capacitor voltage as the circuit response to be, determined. We assume an initial voltage V0 on the capacitor, although, this is not necessary for the step response. Since the voltage of a capacitor cannot change instantaneously,, , R, , v(0) v(0 ) V0, , t=0, , R, , Vs, , Vs u(t), , First-Order Circuits, , +, −, , C, , +, −, , C, , +, v, −, , +, v, −, , (7.40), , where v(0) is the voltage across the capacitor just before switching and, v(0 ) is its voltage immediately after switching. Applying KCL, we have, C, , v Vsu (t), dv, , 0, dt, R, , or, Vs, dv, v, , , u (t), dt, RC, RC, , (b), , Figure 7.40, An RC circuit with voltage step input., , (7.41), , where v is the voltage across the capacitor. For t 7 0, Eq. (7.41) becomes, Vs, dv, v, , , dt, RC, RC, , (7.42), , Rearranging terms gives, v Vs, dv, , dt, RC, or, dv, dt, , v Vs, RC, , (7.43), , Integrating both sides and introducing the initial conditions,, v(t), , , , ln(v Vs)2, V0, , t t, 2, RC 0, , ln(v(t) Vs) ln(V0 Vs) , or, ln, , t, 0, RC, , v Vs, t, , V0 Vs, RC, , (7.44), , Taking the exponential of both sides, v Vs, ett,, t RC, V0 Vs, v Vs (V0 Vs)ett, or, , v(t) Vs (V0 Vs)ett,, , t 7 0, , (7.45), , V0,, v(t) b, Vs (V0 Vs)et/t,, , t 6 0, t 7 0, , (7.46), , Thus,
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 275, , 7.5, , Step Response of an RC Circuit, , 275, v(t), , This is known as the complete response (or total response) of the RC, circuit to a sudden application of a dc voltage source, assuming the, capacitor is initially charged. The reason for the term “complete” will, become evident a little later. Assuming that Vs 7 V0, a plot of v(t) is, shown in Fig. 7.41., If we assume that the capacitor is uncharged initially, we set, V0 0 in Eq. (7.46) so that, 0,, v(t) b, Vs (1 ett ),, , t 6 0, t 7 0, , Vs, , V0, , (7.47), 0, , t, , Figure 7.41, , which can be written alternatively as, v(t) Vs(1 ett)u(t), , (7.48), , Response of an RC circuit with initially, charged capacitor., , This is the complete step response of the RC circuit when the capacitor is initially uncharged. The current through the capacitor is obtained, from Eq. (7.47) using i(t) C dvdt. We get, i(t) C, , dv, C, Vsett,, t, dt, , t RC,, , t 7 0, , or, i (t) , , Vs tt, e, u (t), R, , (7.49), , Figure 7.42 shows the plots of capacitor voltage v(t) and capacitor current i(t)., Rather than going through the derivations above, there is a systematic approach—or rather, a short-cut method—for finding the step, response of an RC or RL circuit. Let us reexamine Eq. (7.45), which is, more general than Eq. (7.48). It is evident that v(t) has two components., Classically there are two ways of decomposing this into two components. The first is to break it into a “natural response and a forced, response’’ and the second is to break it into a “transient response and, a steady-state response.’’ Starting with the natural response and forced, response, we write the total or complete response as, , Vs, , 0, , t, (a), , i(t), Vs, R, , Complete response natural response forced response, stored energy, , v(t), , independent source, , or, v vn vf, , (7.50), , where, vn Voett, , 0, , t, (b), , and, vf Vs(1 ett), We are familiar with the natural response vn of the circuit, as discussed, in Section 7.2. vf is known as the forced response because it is produced by the circuit when an external “force’’ (a voltage source in this, case) is applied. It represents what the circuit is forced to do by the, input excitation. The natural response eventually dies out along with, the transient component of the forced response, leaving only the steadystate component of the forced response., , Figure 7.42, Step response of an RC circuit with, initially uncharged capacitor: (a) voltage, response, (b) current response.
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 276, , Chapter 7, , 276, , First-Order Circuits, , Another way of looking at the complete response is to break into, two components—one temporary and the other permanent, i.e.,, Complete response transient response steady-state response, temporary part, , permanent part, , or, v vt vss, , (7.51), , vt (Vo Vs)ett, , (7.52a), , vss Vs, , (7.52b), , where, , and, , The transient response vt is temporary; it is the portion of the complete response that decays to zero as time approaches infinity. Thus,, The transient response is the circuit’s temporary response that will die, out with time., , The steady-state response vss is the portion of the complete response, that remains after the transient reponse has died out. Thus,, The steady-state response is the behavior of the circuit a long time, after an external excitation is applied., , This is the same as saying that the complete response is the sum of the transient response and the steady-state, response., , The first decomposition of the complete response is in terms of the, source of the responses, while the second decomposition is in terms of, the permanency of the responses. Under certain conditions, the natural, response and transient response are the same. The same can be said, about the forced response and steady-state response., Whichever way we look at it, the complete response in Eq. (7.45), may be written as, v(t) v() [v(0) v()]ett, , (7.53), , where v(0) is the initial voltage at t 0 and v() is the final or steadystate value. Thus, to find the step response of an RC circuit requires, three things:, , Once we know x (0), x ( ), and t,, almost all the circuit problems in this, chapter can be solved using the, formula, , x(t) x() 3 x(0) x()4 ett, , 1. The initial capacitor voltage v(0)., 2. The final capacitor voltage v()., 3. The time constant t., , We obtain item 1 from the given circuit for t 6 0 and items 2 and 3, from the circuit for t 7 0. Once these items are determined, we obtain
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 277, , 7.5, , Step Response of an RC Circuit, , 277, , the response using Eq. (7.53). This technique equally applies to RL circuits, as we shall see in the next section., Note that if the switch changes position at time t t0 instead of, at t 0, there is a time delay in the response so that Eq. (7.53), becomes, v(t) v() [v(t0) v()]e(tt0)t, , (7.54), , where v(t0) is the initial value at t t0 . Keep in mind that Eq. (7.53), or (7.54) applies only to step responses, that is, when the input excitation is constant., , The switch in Fig. 7.43 has been in position A for a long time. At t 0,, the switch moves to B. Determine v(t) for t 7 0 and calculate its value, at t 1 s and 4 s., 3 kΩ, , A, , B, , 4 kΩ, , t=0, 24 V +, −, , 5 kΩ, , +, v, −, , 0.5 mF, , + 30 V, −, , Figure 7.43, For Example 7.10., , Solution:, For t 6 0, the switch is at position A. The capacitor acts like an open, circuit to dc, but v is the same as the voltage across the 5-k resistor., Hence, the voltage across the capacitor just before t 0 is obtained, by voltage division as, v(0) , , 5, (24) 15 V, 53, , Using the fact that the capacitor voltage cannot change instantaneously,, v(0) v(0) v(0 ) 15 V, For t 7 0, the switch is in position B. The Thevenin resistance, connected to the capacitor is RTh 4 k, and the time constant is, t RThC 4 103 0.5 103 2 s, Since the capacitor acts like an open circuit to dc at steady state,, v() 30 V. Thus,, v(t) v() [v(0) v()]ett, 30 (15 30)et2 (30 15e0.5t ) V, At t 1,, v(1) 30 15e0.5 20.9 V, At t 4,, v(4) 30 15e2 27.97 V, , Example 7.10
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Chapter 7, , 278, , Practice Problem 7.10, t=0, , +, v, −, , +, −, , 1, 3, , 6Ω, , F, , First-Order Circuits, , Find v(t) for t 7 0 in the circuit of Fig. 7.44. Assume the switch has, been open for a long time and is closed at t 0. Calculate v(t) at, t 0.5., Answer: (6.25 3.75e2t) V for all t 7 0, 7.63 V., , +, −, , 2Ω, , 10 V, , Page 278, , 5V, , Figure 7.44, For Practice Prob. 7.10., , Example 7.11, , In Fig. 7.45, the switch has been closed for a long time and is opened, at t 0. Find i and v for all time., , 30u(t) V, , +, −, , t=0, , i, , 10 Ω, , +, v, −, , 20 Ω, , 1, 4, , F, , + 10 V, −, , Figure 7.45, For Example 7.11., , Solution:, The resistor current i can be discontinuous at t 0, while the capacitor, voltage v cannot. Hence, it is always better to find v and then obtain i, from v., By definition of the unit step function,, 0,, 30u(t) b, 30,, , For t 6 0, the switch is closed and 30u(t) 0, so that the 30u(t), voltage source is replaced by a short circuit and should be regarded as, contributing nothing to v. Since the switch has been closed for a long, time, the capacitor voltage has reached steady state and the capacitor, acts like an open circuit. Hence, the circuit becomes that shown in, Fig. 7.46(a) for t 6 0. From this circuit we obtain, , i, , 10 Ω, , +, v, −, , 20 Ω, , + 10 V, −, , (a), 10 Ω, , 30 V, , +, −, , t 6 0, t 7 0, , i, , v, 1 A, 10, , Since the capacitor voltage cannot change instantaneously,, v(0) v(0) 10 V, , i, , 20 Ω, , v 10 V,, , +, v, −, , 1, 4, , F, , (b), , For t 7 0, the switch is opened and the 10-V voltage source is, disconnected from the circuit. The 30u(t) voltage source is now operative,, so the circuit becomes that shown in Fig. 7.46(b). After a long time, the, circuit reaches steady state and the capacitor acts like an open circuit, again. We obtain v() by using voltage division, writing, , Figure 7.46, Solution of Example 7.11: (a) for t 6 0,, (b) for t 7 0., , v() , , 20, (30) 20 V, 20 10
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 279, , 7.5, , Step Response of an RC Circuit, , 279, , The Thevenin resistance at the capacitor terminals is, 10 20, 20, , , 30, 3, , RTh 10 20 , and the time constant is, t RTh C , , 20 1, 5, s, 3 4, 3, , Thus,, v(t) v() [v(0) v()]ett, 20 (10 20)e(35)t (20 10e0.6t) V, To obtain i, we notice from Fig. 7.46(b) that i is the sum of the currents, through the 20- resistor and the capacitor; that is,, v, dv, C, 20, dt, 1 0.5e0.6t 0.25(0.6)(10)e0.6t (1 e0.6t) A, , i, , Notice from Fig. 7.46(b) that v 10i 30 is satisfied, as expected., Hence,, 10 V,, vb, (20 10e0.6t ) V,, ib, , 1 A,, (1 e0.6t) A,, , t 6 0, t 0, t 6 0, t 7 0, , Notice that the capacitor voltage is continuous while the resistor current, is not., , The switch in Fig. 7.47 is closed at t 0. Find i(t) and v(t) for all time., Note that u(t) 1 for t 6 0 and 0 for t 7 0. Also, u(t) 1 u(t)., , 5Ω, , 20u(−t) V +, −, , t=0, , i, +, v, −, , 0.2 F, , 10 Ω, , Figure 7.47, For Practice Prob. 7.11., , Answer: i (t) b, vb, , 0,, 2(1 e1.5t ) A,, , 20 V,, 10(1 e1.5t ) V,, , t 6 0, t 7 0, , t 6 0, ,, t 7 0, , 3A, , Practice Problem 7.11
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 280, , Chapter 7, , 280, R, i, , t=0, Vs, , Vs u(t), , +, v (t), −, , 7.6, , First-Order Circuits, , Step Response of an RL Circuit, , (a), , Consider the RL circuit in Fig. 7.48(a), which may be replaced by the, circuit in Fig. 7.48(b). Again, our goal is to find the inductor current i, as the circuit response. Rather than apply Kirchhoff’s laws, we will use, the simple technique in Eqs. (7.50) through (7.53). Let the response be, the sum of the transient response and the steady-state response,, , R, , i it iss, , +, −, , L, , +, −, , i, +, v (t), −, , L, , (b), , Figure 7.48, An RL circuit with a step input voltage., , (7.55), , We know that the transient response is always a decaying exponential,, that is,, it Aett,, , t, , L, R, , (7.56), , where A is a constant to be determined., The steady-state response is the value of the current a long time after, the switch in Fig. 7.48(a) is closed. We know that the transient response, essentially dies out after five time constants. At that time, the inductor, becomes a short circuit, and the voltage across it is zero. The entire, source voltage Vs appears across R. Thus, the steady-state response is, iss , , Vs, R, , (7.57), , Substituting Eqs. (7.56) and (7.57) into Eq. (7.55) gives, i Aett , , Vs, R, , (7.58), , We now determine the constant A from the initial value of i. Let I0 be, the initial current through the inductor, which may come from a source, other than Vs. Since the current through the inductor cannot change, instantaneously,, i(0 ) i(0) I0, , (7.59), , Thus, at t 0, Eq. (7.58) becomes, I0 A , , Vs, R, , A I0 , , Vs, R, , From this, we obtain A as, i(t), I0, , Substituting for A in Eq. (7.58), we get, i(t) , , Vs, R, , Vs, Vs, aI0 bett, R, R, , (7.60), , This is the complete response of the RL circuit. It is illustrated in, Fig. 7.49. The response in Eq. (7.60) may be written as, 0, , t, , Figure 7.49, Total response of the RL circuit with, initial inductor current I0., , i(t) i() [i(0) i()]ett, , (7.61)
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 281, , 7.6, , Step Response of an RL Circuit, , where i(0) and i() are the initial and final values of i, respectively., Thus, to find the step response of an RL circuit requires three things:, , 1. The initial inductor current i(0) at t 0., 2. The final inductor current i()., 3. The time constant t., , We obtain item 1 from the given circuit for t 6 0 and items 2 and 3, from the circuit for t 7 0. Once these items are determined, we obtain, the response using Eq. (7.61). Keep in mind that this technique applies, only for step responses., Again, if the switching takes place at time t t0 instead of t 0,, Eq. (7.61) becomes, i(t) i() [i(t0) i()]e(tt0)t, , (7.62), , If I0 0, then, 0,, i(t) c Vs, (1 ett ),, R, , t 6 0, (7.63a), , t 7 0, , or, i(t) , , Vs, (1 ett)u(t), R, , (7.63b), , This is the step response of the RL circuit with no initial inductor current. The voltage across the inductor is obtained from Eq. (7.63) using, v L didt. We get, v(t) L, , di, L, Vs ett,, dt, tR, , t, , L, ,, R, , t 7 0, , or, v(t) Vsettu(t), , (7.64), , Figure 7.50 shows the step responses in Eqs. (7.63) and (7.64)., i(t), , v(t), , Vs, R, , Vs, , 0, , t, (a), , 0, , t, (b), , Figure 7.50, Step responses of an RL circuit with no initial inductor, current: (a) current response, (b) voltage response., , 281
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 282, , Chapter 7, , 282, , Example 7.12, , First-Order Circuits, , Find i(t) in the circuit of Fig. 7.51 for t 7 0. Assume that the switch, has been closed for a long time., , t=0, , 2Ω, , Solution:, When t 6 0, the 3- resistor is short-circuited, and the inductor acts, like a short circuit. The current through the inductor at t 0 (i.e., just, before t 0) is, , 3Ω, i, , 10 V, , +, −, , 1, 3, , H, , i(0) , Figure 7.51, For Example 7.12., , 10, 5A, 2, , Since the inductor current cannot change instantaneously,, i(0) i(0 ) i(0) 5 A, When t 7 0, the switch is open. The 2- and 3- resistors are in series,, so that, i() , , 10, 2A, 23, , The Thevenin resistance across the inductor terminals is, RTh 2 3 5 , For the time constant,, 1, , t, , L, 1, 3, , s, RTh, 5, 15, , Thus,, i(t) i() [i(0) i()]ett, 2 (5 2)e15t 2 3e15t A,, , t 7 0, , Check: In Fig. 7.51, for t 7 0, KVL must be satisfied; that is,, di, dt, di, 1, 5i L [10 15e15t] c (3)(15)e15t d 10, dt, 3, 10 5i L, , This confirms the result., , Practice Problem 7.12, i, , 5Ω, , 1.5 H, , t=0, , Figure 7.52, For Practice Prob. 7.12., , The switch in Fig. 7.52 has been closed for a long time. It opens at, t 0. Find i(t) for t 7 0., Answer: (6 3e10t) A for all t 7 0., , 10 Ω, , 9A
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 283, , Step Response of an RL Circuit, , 7.6, , At t 0, switch 1 in Fig. 7.53 is closed, and switch 2 is closed 4 s later., Find i(t) for t 7 0. Calculate i for t 2 s and t 5 s., 4Ω, , S1 t = 0, , P, , 6Ω, , S2, , i, t=4, , 40 V, , +, −, , 2Ω, , 5H, , 10 V +, −, , Figure 7.53, For Example 7.13., , Solution:, We need to consider the three time intervals t 0, 0 t 4, and, t 4 separately. For t 6 0, switches S1 and S2 are open so that i 0., Since the inductor current cannot change instantly,, i(0) i(0) i(0 ) 0, For 0 t 4, S1 is closed so that the 4- and 6- resistors are, in series. (Remember, at this time, S2 is still open.) Hence, assuming, for now that S1 is closed forever,, i() , , 40, 4 A,, RTh 4 6 10 , 46, L, 5, 1, t, , s, RTh, 10, 2, , Thus,, i(t) i() [i(0) i()]ett, 4 (0 4)e2t 4(1 e2t) A,, , 0t4, , For t 4, S2 is closed; the 10-V voltage source is connected, and, the circuit changes. This sudden change does not affect the inductor, current because the current cannot change abruptly. Thus, the initial, current is, i(4) i(4) 4(1 e8) 4 A, To find i(), let v be the voltage at node P in Fig. 7.53. Using KCL,, 40 v, 10 v, v, 180, , , 1, v, V, 4, 2, 6, 11, v, 30, i() , 2.727 A, 6, 11, The Thevenin resistance at the inductor terminals is, RTh 4 2 6 , and, t, , 42, 22, 6, , 6, 3, , 5, 15, L, 22 , s, RTh, 22, 3, , 283, , Example 7.13
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 284, , Chapter 7, , 284, , First-Order Circuits, , Hence,, i(t) i() [i(4) i()]e(t4)t,, , t4, , We need (t 4) in the exponential because of the time delay. Thus,, i(t) 2.727 (4 2.727)e(t4)t,, 2.727 1.273e1.4667(t4),, Putting all this together,, , t, , 15, 22, , t4, , 0,, i(t) c 4(1 e2t),, 2.727 1.273e1.4667(t4),, , t 0, 0t4, t 4, , At t 2,, At t 5,, , i(2) 4(1 e4) 3.93 A, i(5) 2.727 1.273e1.4667 3.02 A, , Practice Problem 7.13, , Switch S1 in Fig. 7.54 is closed at t 0, and switch S2 is closed at, t 2 s. Calculate i(t) for all t. Find i(1) and i(3)., , t=2, , Answer:, S1, , 10 Ω, , t=0, 6A, , 15 Ω, , S2, 20 Ω, , i(t), 5H, , 0,, i(t) c 2(1 e9t),, 3.6 1.6e5(t2),, i(1) 1.9997 A, i(3) 3.589 A., , t 6 0, 0 6 t 6 2, t 7 2, , Figure 7.54, For Practice Prob. 7.13., , 7.7, , First-Order Op Amp Circuits, , An op amp circuit containing a storage element will exhibit first-order, behavior. Differentiators and integrators treated in Section 6.6 are, examples of first-order op amp circuits. Again, for practical reasons,, inductors are hardly ever used in op amp circuits; therefore, the op amp, circuits we consider here are of the RC type., As usual, we analyze op amp circuits using nodal analysis. Sometimes, the Thevenin equivalent circuit is used to reduce the op amp circuit to one that we can easily handle. The following three examples, illustrate the concepts. The first one deals with a source-free op amp, circuit, while the other two involve step responses. The three examples, have been carefully selected to cover all possible RC types of op amp, circuits, depending on the location of the capacitor with respect to the, op amp; that is, the capacitor can be located in the input, the output,, or the feedback loop.
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 285, , 7.7, , First-Order Op Amp Circuits, , 285, , Example 7.14, , For the op amp circuit in Fig. 7.55(a), find vo for t 7 0, given that, v(0) 3 V. Let Rf 80 k, R1 20 k, and C 5 mF., Rf, , 1, , C, + v −, , 2, 3, , 80 kΩ, , 80 kΩ, , 1, , −, +, , + 3V −, , +, , R1, , vo, , C, , 2, 3, , 20 kΩ, , 1A, −, +, vo, , −, , (0+), , (a), , (b), , Solution:, This problem can be solved in two ways:, , ■ METHOD 1 Consider the circuit in Fig. 7.55(a). Let us derive the, appropriate differential equation using nodal analysis. If v1 is the voltage at node 1, at that node, KCL gives, 0 v1, dv, C, R1, dt, , (7.14.1), , Since nodes 2 and 3 must be at the same potential, the potential at node, 2 is zero. Thus, v1 0 v or v1 v and Eq. (7.14.1) becomes, dv, v, , 0, dt, CR1, , (7.14.2), , This is similar to Eq. (7.4b) so that the solution is obtained the same, way as in Section 7.2, i.e.,, t R1C, , (7.14.3), , where V0 is the initial voltage across the capacitor. But v(0) 3 V0, and t 20 103 5 106 0.1. Hence,, v(t) 3e10t, , (7.14.4), , Applying KCL at node 2 gives, 0 vo, dv, , dt, Rf, , or, vo Rf C, , dv, dt, , (7.14.5), , Now we can find v0 as, vo 80 103 5 106(30e10t) 12e10t V,, , t 7 0, , +, vo, −, , (c), , For Example 7.14., , C, , 20 kΩ, , −, , Figure 7.55, , v(t) V0ett,, , −, +, , +v −, , +
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 286, , Chapter 7, , 286, , First-Order Circuits, , ■ METHOD 2 Let us apply the short-cut method from Eq. (7.53)., , We need to find vo(0 ), vo(), and t. Since v(0 ) v(0) 3 V, we, apply KCL at node 2 in the circuit of Fig. 7.55(b) to obtain, 0 vo(0 ), 3, , 0, 20,000, 80,000, , or vo(0 ) 12 V. Since the circuit is source free, v() 0 V. To find, t, we need the equivalent resistance Req across the capacitor terminals., If we remove the capacitor and replace it by a 1-A current source, we, have the circuit shown in Fig. 7.55(c). Applying KVL to the input loop, yields, 20,000(1) v 0, , 1, , v 20 kV, , Then, Req , , v, 20 k, 1, , and t ReqC 0.1. Thus,, vo(t) vo() [vo(0) vo()]ett, 0 (12 0)e10t 12e10t V,, , t 7 0, , as before., , Practice Problem 7.14, , For the op amp circuit in Fig. 7.56, find vo for t 7 0 if v(0) 4 V., Assume that Rf 50 k, R1 10 k, and C 10 mF., , C, , Answer: 4e2t V, t 7 0 ., , + v −, Rf, −, +, R1, , +, vo, −, , Figure 7.56, For Practice Prob. 7.14., , Example 7.15, , Determine v(t) and vo(t) in the circuit of Fig. 7.57., Solution:, This problem can be solved in two ways, just like the previous example., However, we will apply only the second method. Since what we are, looking for is the step response, we can apply Eq. (7.53) and write, v(t) v() [v(0) v()]ett,, , t 7 0, , (7.15.1)
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 287, , 7.7, , First-Order Op Amp Circuits, , 287, + v −, , where we need only find the time constant t, the initial value v(0), and, the final value v(). Notice that this applies strictly to the capacitor, voltage due a step input. Since no current enters the input terminals of, the op amp, the elements on the feedback loop of the op amp constitute, an RC circuit, with, t RC 50 103 106 0.05, , 50 kΩ, , t=0, , 10 kΩ, , v1, , +, −, , (7.15.2), , +, , For t 6 0, the switch is open and there is no voltage across the, capacitor. Hence, v(0) 0. For t 7 0, we obtain the voltage at node, 1 by voltage division as, v1 , , 1 F, , 20, 32V, 20 10, , (7.15.3), , 3V, , +, −, , 20 kΩ, , 20 kΩ, , vo, −, , Figure 7.57, For Example 7.15., , Since there is no storage element in the input loop, v1 remains constant, for all t. At steady state, the capacitor acts like an open circuit so that, the op amp circuit is a noninverting amplifier. Thus,, vo() a1 , , 50, b v1 3.5 2 7 V, 20, , (7.15.4), , But, v1 vo v, , (7.15.5), , so that, v() 2 7 5 V, Substituting t, v(0), and v() into Eq. (7.15.1) gives, v(t) 5 [0 (5)]e20t 5(e20t 1) V,, , t 7 0, , (7.15.6), , From Eqs. (7.15.3), (7.15.5), and (7.15.6), we obtain, vo(t) v1(t) v(t) 7 5e20t V,, , t 7 0, , (7.15.7), , Find v(t) and vo(t) in the op amp circuit of Fig. 7.58., , Practice Problem 7.15, 100 kΩ, , Answer: (Note, the voltage across the capacitor and the output voltage, must be both equal to zero, for t 6 0, since the input was zero for all, t 6 0.) 40(1 e10t) u(t) mV, 40 (e10t 1) u(t) mV., , 1 F, , 10 kΩ, , 4 mV, , t=0, , +, −, , + v −, −, +, , +, vo, −, , Figure 7.58, For Practice Prob. 7.15.
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 288, , Chapter 7, , 288, , Example 7.16, , Find the step response vo(t) for t 7 0 in the op amp circuit of Fig. 7.59., Let vi 2u (t) V, R1 20 k, Rf 50 k, R2 R3 10 k, C , 2 mF., , Rf, , R1, , vi, , R2, , −, +, , +, −, , First-Order Circuits, , R3, , C, , +, vo, −, , Figure 7.59, , Solution:, Notice that the capacitor in Example 7.14 is located in the input loop,, while the capacitor in Example 7.15 is located in the feedback loop. In, this example, the capacitor is located in the output of the op amp. Again,, we can solve this problem directly using nodal analysis. However, using, the Thevenin equivalent circuit may simplify the problem., We temporarily remove the capacitor and find the Thevenin, equivalent at its terminals. To obtain VTh, consider the circuit in, Fig. 7.60(a). Since the circuit is an inverting amplifier,, , For Example 7.16., , Vab , , Rf, R1, , vi, , By voltage division,, VTh , , Rf, R3, R3, Vab , vi, R2 R3, R2 R3 R1, , Rf, R1, , vi, , −, +, , a, +, , +, −, , Vab, −, , R2, , R2, +, R3, , VTh, , R3, , Ro, , RTh, , −, , b, (a), , (b), , Figure 7.60, Obtaining VTh and RTh across the capacitor in Fig. 7.59., , To obtain RTh, consider the circuit in Fig. 7.60(b), where Ro is the, output resistance of the op amp. Since we are assuming an ideal op amp,, Ro 0, and, R2R3, RTh R2 R3 , R2 R3, Substituting the given numerical values,, VTh , 5 kΩ, , −2.5u(t) +, −, , 2 F, , Figure 7.61, Thevenin equivalent circuit of the circuit, in Fig. 7.59., , Rf, R3, 10 50, vi , 2u(t) 2.5u(t), R2 R3 R1, 20 20, R2R3, RTh , 5 k, R2 R3, , The Thevenin equivalent circuit is shown in Fig. 7.61, which is similar, to Fig. 7.40. Hence, the solution is similar to that in Eq. (7.48); that is,, vo(t) 2.5(1 ett)u(t), where t RThC 5 103 2 106 0.01. Thus, the step response, for t 7 0 is, vo(t) 2.5(e100t 1)u(t) V
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 289, , Transient Analysis with PSpice, , 7.8, , Obtain the step response vo(t) for the circuit in Fig. 7.62. Let, vi 3u(t) V, R1 20 k, Rf 40 k, R2 R3 10 k, C 2 mF., , 289, , Practice Problem 7.16, Rf, , 50t, , Answer: 9(1 e, , )u(t) V., R1, , 7.8, , vi, , Transient Analysis with PSpice, , As we discussed in Section 7.5, the transient response is the temporary, response of the circuit that soon disappears. PSpice can be used to, obtain the transient response of a circuit with storage elements. Section D.4 in Appendix D provides a review of transient analysis using, PSpice for Windows. It is recommended that you read Section D.4, before continuing with this section., If necessary, dc PSpice analysis is first carried out to determine the, initial conditions. Then the initial conditions are used in the transient, PSpice analysis to obtain the transient responses. It is recommended, but not necessary that during this dc analysis, all capacitors should be, open-circuited while all inductors should be short-circuited., , R2, , +, −, , +, vo, −, , C, , Figure 7.62, For Practice Prob. 7.16., , PSpice uses “transient” to mean “function of time.” Therefore, the transient, response in PSpice may not actually, die out as expected., , Example 7.17, , Use PSpice to find the response i(t) for t 7 0 in the circuit of Fig. 7.63., Solution:, Solving this problem by hand, RTh 6, t 36 0.5 s, so that, , R3, , −, +, , 4Ω, , gives, , i(0) 0, i() 2 A,, , i(t) i() 3i(0) i()4ett 2(1 e2t),, , t 7 0, , To use PSpice, we first draw the schematic as shown in Fig. 7.64., We recall from Appendix D that the part name for a closed switch is, Sw_tclose. We do not need to specify the initial condition of the, inductor because PSpice will determine that from the circuit. By, selecting Analysis/Setup/Transient, we set Print Step to 25 ms and, Final Step to 5t 2.5 s. After saving the circuit, we simulate by, selecting Analysis/Simulate. In the PSpice A/D window, we select, Trace/Add and display –I(L1) as the current through the inductor., Figure 7.65 shows the plot of i(t), which agrees with that obtained by, hand calculation., , i(t), t=0, 2Ω, , 6A, , 3H, , Figure 7.63, For Example 7.17., , 2.0 A, , 1.5 A, , 1.0 A, tClose = 0, 1, 2, U1, 6A, , IDC, R1, , 2, , R2, 0.5 A, 4, L1, , 3H, , 0, , Figure 7.64, The schematic of the circuit in Fig. 7.63., , 0 A, 0 s, , 1.0 s, 2.0 s, -I(L1), Time, , 3.0 s, , Figure 7.65, For Example 7.17; the response of the, circuit in Fig. 7.63.
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 290, , Chapter 7, , 290, , First-Order Circuits, , Note that the negative sign on I(L1) is needed because the current, enters through the upper terminal of the inductor, which happens to be, the negative terminal after one counterclockwise rotation. A way to avoid, the negative sign is to ensure that current enters pin 1 of the inductor., To obtain this desired direction of positive current flow, the initially, horizontal inductor symbol should be rotated counterclockwise 270, and placed in the desired location., , Practice Problem 7.17, 3Ω, , 12 V +, −, , Answer: v(t) 8(1 et) V, t 7 0. The response is similar in shape, to that in Fig. 7.65., , t=0, , 6Ω, , For the circuit in Fig. 7.66, use Pspice to find v(t) for t 7 0., , 0.5 F, , +, v (t), −, , Figure 7.66, For Practice Prob. 7.17., , Example 7.18, , In the circuit of Fig. 7.67(a), determine the response v(t)., 12 Ω, , t=0, , t=0, , + v(t) −, 0.1 F, , 30 V +, −, , 6Ω, , 6Ω, , 3Ω, , 4A, , (a), + v(t) −, , 12 Ω, , 0.1 F, 30 V +, −, , 6Ω, , 6Ω, , (b), + v(t) −, , 10 Ω, , 0.1 F, 10 V +, −, , (c), , Figure 7.67, For Example 7.18. Original circuit (a), circuit for t 7 0 (b), and, reduced circuit for t 7 0 (c).
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 291, , 7.8, , Transient Analysis with PSpice, , Solution:, 1. Define. The problem is clearly stated and the circuit is clearly, labeled., 2. Present. Given the circuit shown in Fig. 7.67(a), determine the, response v(t)., 3. Alternative. We can solve this circuit using circuit analysis, techniques, nodal analysis, mesh analysis, or PSpice. Let us, solve the problem using circuit analysis techniques (this time, Thevenin equivalent circuits) and then check the answer using, two methods of PSpice., 4. Attempt. For time 6 0, the switch on the left is open and the, switch on the right is closed. Assume that the switch on the right, has been closed long enough for the circuit to reach steady state;, then the capacitor acts like an open circuit and the current from, the 4-A source flows through the parallel combination of the 6-, and 3- resistors (6 3 189 2), producing a voltage equal, to 2 4 8 V v(0)., At t 0, the switch on the left closes and the switch on, the right opens, producing the circuit shown in Fig. 7.67(b)., The easiest way to complete the solution is to find the, Thevenin equivalent circuit as seen by the capacitor. The opencircuit voltage (with the capacitor removed) is equal to the, voltage drop across the 6- resistor on the left, or 10 V (the, voltage drops uniformly across the 12- resistor, 20 V, and, across the 6- resistor, 10 V). This is VTh. The resistance, looking in where the capacitor was is equal to 12 6 6 , 7218 6 10 , which is Req. This produces the Thevenin, equivalent circuit shown in Fig. 7.67(c). Matching up the, boundary conditions (v(0) 8 V and v() 10 V) and t , RC 1, we get, v(t) 10 18et V, 5. Evaluate. There are two ways of solving the problem using, PSpice., , ■ METHOD 1 One way is to first do the dc PSpice analysis to, determine the initial capacitor voltage. The schematic of the revelant, circuit is in Fig. 7.68(a). Two pseudocomponent VIEWPOINTs are, inserted to measure the voltages at nodes 1 and 2. When the circuit, is simulated, we obtain the displayed values in Fig. 7.68(a) as, V1 0 V and V2 8 V. Thus, the initial capacitor voltage is v(0) , V1 V2 8 V. The PSpice transient analysis uses this value along, with the schematic in Fig. 7.68(b). Once the circuit in Fig. 7.68(b), is drawn, we insert the capacitor initial voltage as IC 8. We, select Analysis/Setup/Transient and set Print Step to 0.1 s and, Final Step to 4t 4 s. After saving the circuit, we select Analysis/, Simulate to simulate the circuit. In the PSpice A/D window, we, select Trace/Add and display V(R2:2) V(R3:2) or V(C1:1) , V(C1:2) as the capacitor voltage v(t). The plot of v(t) is shown in, Fig. 7.69. This agrees with the result obtained by hand calculation,, v(t) 10 18 et V., , 291
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 292, , Chapter 7, , 292, 0.0000, 1, , C1, , 8.0000, , First-Order Circuits, 10 V, , 2, , 0.1, 5 V, R2, , 6, , R3, , 6, , R4, , 3, , 4A, , I1, , 0 V, 0, −5 V, , (a), , 30 V, , +, −, , V1, , R1, , C1, , 12, , 0.1, , −10 V, 6, , R2, , R3, , 6, , 0s, , 1.0 s, 2.0 s, 3.0 s, V(R2:2) − V(R3:2), Time, , 4.0 s, , Figure 7.69, Response v(t) for the circuit in Fig. 7.67., 0, (b), , Figure 7.68, (a) Schematic for dc analysis to get v(0), (b) schematic, for transient analysis used in getting the response v(t)., , ■ METHOD 2 We can simulate the circuit in Fig. 7.67 directly,, since PSpice can handle the open and closed switches and determine, the initial conditions automatically. Using this approach, the schematic, is drawn as shown in Fig. 7.70. After drawing the circuit, we select, Analysis/Setup/Transient and set Print Step to 0.1 s and Final Step, to 4t 4 s. We save the circuit, then select Analysis/Simulate to simulate the circuit. In the PSpice A/D window, we select Trace/Add and, display V(R2:2) V(R3:2) as the capacitor voltage v(t). The plot of, v(t) is the same as that shown in Fig. 7.69., , tClose = 0, 1, 2, U1, 12, , tOpen = 0, 1, 2, U2, , C1, , R1, , 0.1, , V1, 30 V, , +, −, , R2, , 6, , R3, , 6, , R4, , 3, , I1, , 4 A, , 0, , Figure 7.70, For Example 7.18., , 6. Satisfactory? Clearly, we have found the value of the output, response v(t), as required by the problem statement. Checking, does validate that solution. We can present all this as a complete, solution to the problem.
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 293, , 7.9, , Applications, , The switch in Fig. 7.71 was open for a long time but closed at t 0., If i(0) 10 A, find i(t) for t 7 0 by hand and also by PSpice., , 293, , Practice Problem 7.18, 5Ω, , Answer: i(t) 6 4e5t A. The plot of i(t) obtained by PSpice analysis is shown in Fig. 7.72., 12 A, , 30 Ω, , 10 A, , Figure 7.71, 9 A, , For Practice Prob. 7.18., , 8 A, , 7 A, , 6 A, 0 s, , 0.5 s, I(L1), Time, , 1.0 s, , Figure 7.72, For Practice Prob. 7.18., , 7.9, , Applications, , The various devices in which RC and RL circuits find applications, include filtering in dc power supplies, smoothing circuits in digital communications, differentiators, integrators, delay circuits, and relay circuits., Some of these applications take advantage of the short or long time constants of the RC or RL circuits. We will consider four simple applications here. The first two are RC circuits, the last two are RL circuits., , 7.9.1 Delay Circuits, An RC circuit can be used to provide various time delays. Figure 7.73, shows such a circuit. It basically consists of an RC circuit with the capacitor connected in parallel with a neon lamp. The voltage source can provide enough voltage to fire the lamp. When the switch is closed, the, capacitor voltage increases gradually toward 110 V at a rate determined, by the circuit’s time constant, (R1 R2)C. The lamp will act as an open, R1, +, 110 V, −, , Figure 7.73, An RC delay circuit., , S, , R2, , C, , 0.1 F, , 70 V, Neon, lamp, , i(t), , t=0, 6Ω, , 2H
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ale29559_ch07.qxd, , 07/08/2008, , 294, , 11:49 AM, , Page 294, , Chapter 7, , First-Order Circuits, , circuit and not emit light until the voltage across it exceeds a particular, level, say 70 V. When the voltage level is reached, the lamp fires (goes, on), and the capacitor discharges through it. Due to the low resistance, of the lamp when on, the capacitor voltage drops fast and the lamp turns, off. The lamp acts again as an open circuit and the capacitor recharges., By adjusting R2, we can introduce either short or long time delays into, the circuit and make the lamp fire, recharge, and fire repeatedly every, time constant t (R1 R2)C, because it takes a time period t to get, the capacitor voltage high enough to fire or low enough to turn off., The warning blinkers commonly found on road construction sites, are one example of the usefulness of such an RC delay circuit., , Example 7.19, , Consider the circuit in Fig. 7.73, and assume that R1 1.5 M,, 0 6 R2 6 2.5 M. (a) Calculate the extreme limits of the time constant of the circuit. (b) How long does it take for the lamp to glow for, the first time after the switch is closed? Let R2 assume its largest value., Solution:, (a) The smallest value for R2 is 0 , and the corresponding time constant, for the circuit is, t (R1 R2)C (1.5 106 0) 0.1 106 0.15 s, The largest value for R2 is 2.5 M, and the corresponding time constant, for the circuit is, t (R1 R2)C (1.5 2.5) 106 0.1 106 0.4 s, Thus, by proper circuit design, the time constant can be adjusted to, introduce a proper time delay in the circuit., (b) Assuming that the capacitor is initially uncharged, vC (0) 0, while, vC () 110. But, vC (t) vC () [vC (0) vC ()]ett 110[1 ett], where t 0.4 s, as calculated in part (a). The lamp glows when, vC 70 V. If vC (t) 70 V at t t0, then, 70 110[1 et0t], , 1, , 7, 1 et0t, 11, , or, et0t , , 4, 11, , 1, , et0t , , 11, 4, , Taking the natural logarithm of both sides gives, t0 t ln, , 11, 0.4 ln 2.75 0.4046 s, 4, , A more general formula for finding t0 is, t0 t ln, , v(), v(t0) v(), , The lamp will fire repeatedly every t0 seconds if and only if v (t0) 6 v ().
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 295, , 7.9, , Applications, , The RC circuit in Fig. 7.74 is designed to operate an alarm which, activates when the current through it exceeds 120 mA. If 0 R , 6 k, find the range of the time delay that the variable resistor can, create., , 295, , Practice Problem 7.19, R, , S, , 10 kΩ, , Answer: Between 47.23 ms and 124 ms., , +, 9V, −, , 7.9.2 Photoflash Unit, , Figure 7.74, , 80 F, , 4 kΩ, , Alarm, , For Practice Prob. 7.19., , An electronic flash unit provides a common example of an RC circuit., This application exploits the ability of the capacitor to oppose any, abrupt change in voltage. Figure 7.75 shows a simplified circuit. It consists essentially of a high-voltage dc supply, a current-limiting large, resistor R1, and a capacitor C in parallel with the flashlamp of low, resistance R2. When the switch is in position 1, the capacitor charges, slowly due to the large time constant (t1 R1C ). As shown in, Fig. 7.76(a), the capacitor voltage rises gradually from zero to Vs, while, its current decreases gradually from I1 VsR1 to zero. The charging, time is approximately five times the time constant,, tcharge 5R1C, , (7.65), , With the switch in position 2, the capacitor voltage is discharged., The low resistance R2 of the photolamp permits a high discharge, current with peak I2 VsR2 in a short duration, as depicted in, Fig. 7.76(b). Discharging takes place in approximately five times the, time constant,, tdischarge 5R2C, , (7.66), , i, v, , I1, , Vs, 0, , t, , 0, (a), , −I2, (b), , Figures 7.76, (a) Capacitor voltage showing slow charge and fast discharge, (b) capacitor, current showing low charging current I1 VsR1 and high discharge current, I2 VsR2., , Thus, the simple RC circuit of Fig. 7.75 provides a short-duration, highcurrent pulse. Such a circuit also finds applications in electric spot, welding and the radar transmitter tube., , R1, , 1, i, , High, voltage, dc supply, , 2, +, −, , vs, , R2, , C, , Figure 7.75, Circuit for a flash unit providing slow, charge in position 1 and fast discharge in, position 2., , +, v, −
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , 296, , Example 7.20, , Page 296, , Chapter 7, , First-Order Circuits, , An electronic flashgun has a current-limiting 6-k resistor and 2000-mF, electrolytic capacitor charged to 240 V. If the lamp resistance is 12 ,, find: (a) the peak charging current, (b) the time required for the capacitor to fully charge, (c) the peak discharging current, (d) the total energy, stored in the capacitor, and (e) the average power dissipated by the, lamp., Solution:, (a) The peak charging current is, I1 , , Vs, 240, , 40 mA, R1, 6 103, , (b) From Eq. (7.65),, tcharge 5R1C 5 6 103 2000 106 60 s 1 minute, (c) The peak discharging current is, I2 , , Vs, 240, , 20 A, R2, 12, , (d) The energy stored is, 1, 1, W CV 2s 2000 106 2402 57.6 J, 2, 2, (e) The energy stored in the capacitor is dissipated across the lamp, during the discharging period. From Eq. (7.66),, tdischarge 5R2C 5 12 2000 106 0.12 s, Thus, the average power dissipated is, p, , Practice Problem 7.20, , 57.6, W, , 480 watts, tdischarge, 0.12, , The flash unit of a camera has a 2-mF capacitor charged to 80 V., (a) How much charge is on the capacitor?, (b) What is the energy stored in the capacitor?, (c) If the flash fires in 0.8 ms, what is the average current through, the flashtube?, (d) How much power is delivered to the flashtube?, (e) After a picture has been taken, the capacitor needs to be, recharged by a power unit that supplies a maximum of 5 mA., How much time does it take to charge the capacitor?, Answer: (a) 0.16 C, (b) 6.4 J, (c) 200 A, (d) 8 kW, (e) 32 s., , 7.9.3 Relay Circuits, A magnetically controlled switch is called a relay. A relay is essentially an electromagnetic device used to open or close a switch that, controls another circuit. Figure 7.77(a) shows a typical relay circuit.
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 297, , 7.9, , Applications, , 297, , The coil circuit is an RL circuit like that in Fig. 7.77(b), where R and, L are the resistance and inductance of the coil. When switch S1 in, Fig. 7.77(a) is closed, the coil circuit is energized. The coil current, gradually increases and produces a magnetic field. Eventually the magnetic field is sufficiently strong to pull the movable contact in the other, circuit and close switch S2. At this point, the relay is said to be pulled, in. The time interval td between the closure of switches S1 and S2 is, called the relay delay time., Relays were used in the earliest digital circuits and are still used, for switching high-power circuits., , S2, , S1, , Magnetic field, S1, , Vs, , R, , Coil, , Vs, L, , (a), , (b), , Figure 7.77, A relay circuit., , The coil of a certain relay is operated by a 12-V battery. If the coil has, a resistance of 150 and an inductance of 30 mH and the current, needed to pull in is 50 mA, calculate the relay delay time., Solution:, The current through the coil is given by, i(t) i() [i(0) i()]ett, where, i(0) 0,, t, , i() , , 12, 80 mA, 150, , L, 30 103, , 0.2 ms, R, 150, , Thus,, i(t) 80[1 ett] mA, If i(td) 50 mA, then, 50 80[1 etdt], , 1, , 5, 1 etdt, 8, , or, etdt , , 3, 8, , 1, , etdt , , 8, 3, , Example 7.21
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 298, , Chapter 7, , 298, , First-Order Circuits, , By taking the natural logarithm of both sides, we get, 8, 8, td t ln 0.2 ln ms 0.1962 ms, 3, 3, Alternatively, we may find td using, td t ln, , Practice Problem 7.21, , i(0) i(), i(td) i(), , A relay has a resistance of 200 and an inductance of 500 mH. The, relay contacts close when the current through the coil reaches 350 mA., What time elapses between the application of 110 V to the coil and, contact closure?, Answer: 2.529 ms., , 7.9.4 Automobile Ignition Circuit, , R, i, Vs, , +, v, −, , L, , Spark, plug, Air gap, , Figure 7.78, Circuit for an automobile ignition system., , The ability of inductors to oppose rapid change in current makes them, useful for arc or spark generation. An automobile ignition system takes, advantage of this feature., The gasoline engine of an automobile requires that the fuel-air, mixture in each cylinder be ignited at proper times. This is achieved, by means of a spark plug (Fig. 7.78), which essentially consists of a, pair of electrodes separated by an air gap. By creating a large voltage, (thousands of volts) between the electrodes, a spark is formed across, the air gap, thereby igniting the fuel. But how can such a large voltage be obtained from the car battery, which supplies only 12 V? This, is achieved by means of an inductor (the spark coil) L. Since the voltage across the inductor is v L didt, we can make didt large by creating a large change in current in a very short time. When the ignition, switch in Fig. 7.78 is closed, the current through the inductor increases, gradually and reaches the final value of i VsR, where Vs 12 V., Again, the time taken for the inductor to charge is five times the time, constant of the circuit (t LR),, tcharge 5, , L, R, , (7.67), , Since at steady state, i is constant, didt 0 and the inductor voltage, v 0. When the switch suddenly opens, a large voltage is developed, across the inductor (due to the rapidly collapsing field) causing a spark, or arc in the air gap. The spark continues until the energy stored in the, inductor is dissipated in the spark discharge. In laboratories, when one, is working with inductive circuits, this same effect causes a very nasty, shock, and one must exercise caution.
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 299, , 7.10, , Summary, , A solenoid with resistance 4 and inductance 6 mH is used in an automobile ignition circuit similar to that in Fig. 7.78. If the battery supplies 12 V, determine: the final current through the solenoid when the, switch is closed, the energy stored in the coil, and the voltage across, the air gap, assuming that the switch takes 1 ms to open., , 299, , Example 7.22, , Solution:, The final current through the coil is, I, , Vs, 12, , 3A, R, 4, , The energy stored in the coil is, 1, 1, W L I 2 6 103 32 27 mJ, 2, 2, The voltage across the gap is, VL, , ¢I, 3, 6 103 , 18 kV, ¢t, 1 106, , The spark coil of an automobile ignition system has a 20-mH inductance, and a 5- resistance. With a supply voltage of 12 V, calculate: the, time needed for the coil to fully charge, the energy stored in the coil,, and the voltage developed at the spark gap if the switch opens in 2 ms., Answer: 20 ms, 57.6 mJ, and 24 kV., , 7.10, , Summary, , 1. The analysis in this chapter is applicable to any circuit that can be, reduced to an equivalent circuit comprising a resistor and a single, energy-storage element (inductor or capacitor). Such a circuit is, first-order because its behavior is described by a first-order differential equation. When analyzing RC and RL circuits, one must, always keep in mind that the capacitor is an open circuit to steadystate dc conditions while the inductor is a short circuit to steadystate dc conditions., 2. The natural response is obtained when no independent source is, present. It has the general form, x(t) x(0)ett, where x represents current through (or voltage across) a resistor, a, capacitor, or an inductor, and x(0) is the initial value of x. Because, most practical resistors, capacitors, and inductors always have losses,, the natural response is a transient response, i.e. it dies out with time., 3. The time constant t is the time required for a response to decay, to le of its initial value. For RC circuits, t RC and for RL circuits, t LR., , Practice Problem 7.22
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 300, , Chapter 7, , 300, , First-Order Circuits, , 4. The singularity functions include the unit step, the unit ramp function, and the unit impulse functions. The unit step function u (t) is, u (t) b, , 0,, 1,, , t 6 0, t 7 0, , The unit impulse function is, 0,, d (t) cUndefined,, 0,, , t 6 0, t 0, t 7 0, , The unit ramp function is, r (t) b, , 0,, t,, , t0, t0, , 5. The steady-state response is the behavior of the circuit after an, independent source has been applied for a long time. The transient, response is the component of the complete response that dies out, with time., 6. The total or complete response consists of the steady-state, response and the transient response., 7. The step response is the response of the circuit to a sudden application of a dc current or voltage. Finding the step response of a, first-order circuit requires the initial value x(0 ), the final value, x(), and the time constant t. With these three items, we obtain, the step response as, x(t) x() [x(0 ) x()]ett, A more general form of this equation is, x(t) x() [x(t0 ) x()]e(tt0)t, Or we may write it as, Instantaneous value Final [Initial Final]e(tt0)t, 8. PSpice is very useful for obtaining the transient response of a circuit., 9. Four practical applications of RC and RL circuits are: a delay circuit,, a photoflash unit, a relay circuit, and an automobile ignition circuit., , Review Questions, 7.1, , 7.2, , 7.3, , An RC circuit has R 2 and C 4 F. The time, constant is:, , capacitor voltage to reach 63.2 percent of its steadystate value is:, , (a) 0.5 s, , (b) 2 s, , (a) 2 s, , (b) 4 s, , (d) 8 s, , (e) 15 s, , (d) 16 s, , (e) none of the above, , (c) 4 s, , The time constant for an RL circuit with R 2 , and L 4 H is:, (a) 0.5 s, , (b) 2 s, , (d) 8 s, , (e) 15 s, , (c) 4 s, , A capacitor in an RC circuit with R 2 and, C 4 F is being charged. The time required for the, , 7.4, , (c) 8 s, , An RL circuit has R 2 and L 4 H. The time, needed for the inductor current to reach 40 percent, of its steady-state value is:, (a) 0.5 s, , (b) 1 s, , (c) 2 s, , (d) 4 s, , (e) none of the above
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 301, , Problems, , 7.5, , 301, , In the circuit of Fig. 7.79, the capacitor voltage just, before t 0 is:, (a) 10 V, , (b) 7 V, , (d) 4 V, , (e) 0 V, , i(t), , (c) 6 V, , 5H, , 2Ω, , 10 A, , 3Ω, , t=0, , Figure 7.80, 3Ω, , 10 V, , For Review Questions 7.7 and 7.8., 2Ω, , +, v(t), −, , +, −, , 7.8, , 7F, t=0, , 7.9, , Figure 7.79, For Review Questions 7.5 and 7.6., , 7.6, , 7.7, , (b) 7 V, , (d) 4 V, , (e) 0 V, , (c) 6 V, , (b) 6 A, , (d) 2 A, , (e) 0 A, , (c) 4 A, , (b) 6 A, , (d) 2 A, , (e) 0 A, , (c) 4 A, , If vs changes from 2 V to 4 V at t 0, we may, express vs as:, (a) d(t) V, , (b) 2u(t) V, , (c) 2u(t) 4u(t) V, , (d) 2 2u(t) V, , 7.10 The pulse in Fig. 7.116(a) can be expressed in terms, of singularity functions as:, , For the circuit in Fig. 7.80, the inductor current just, before t 0 is:, (a) 8 A, , (a) 10 A, , (e) 4u(t) 2 V, , In the circuit in Fig. 7.79, v() is:, (a) 10 V, , In the circuit of Fig. 7.80, i() is:, , (a) 2u(t) 2u(t 1) V, , (b) 2u(t) 2u(t 1) V, , (c) 2u(t) 4u(t 1) V, , (d) 2u(t) 4u(t 1) V, , Answers: 7.1d, 7.2b, 7.3c, 7.4b, 7.5d, 7.6a, 7.7c, 7.8e,, 7.9c,d, 7.10b., , Problems, Section 7.2 The Source-Free RC Circuit, 7.1, , 7.2, , Find the time constant for the RC circuit in Fig. 7.82., , In the circuit shown in Fig. 7.81, , 120 Ω, , v(t) 56e200t V, t 7 0, i(t) 8e200t mA, t 7 0, , 50 V, , (a) Find the values of R and C., , +, −, , 12 Ω, , 80 Ω, , 0.5 mF, , (b) Calculate the time constant t., (c) Determine the time required for the voltage to, decay half its initial value at t 0., , Figure 7.82, For Prob. 7.2., 7.3, , Determine the time constant for the circuit in Fig. 7.83., , i, 10 kΩ, , R, , +, v, −, , C, , 100 pF, , Figure 7.81, , Figure 7.83, , For Prob. 7.1., , For Prob. 7.3., , 20 kΩ, , 40 kΩ, , 30 kΩ
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ale29559_ch07.qxd, , 07/08/2008, , 11:49 AM, , Page 302, , Chapter 7, , 302, , 7.4, , First-Order Circuits, , The switch in Fig. 7.84 has been in position A for a, long time. Assume the switch moves instantaneously, from A to B at t 0. Find v for t 7 0., , 7.8, , For the circuit in Fig. 7.88, if, v 10e4t V, , i 0.2 e4t A,, , and, , t 7 0, , (a) Find R and C., (b) Determine the time constant., , 5 kΩ A, , 40 V, , 10 F, , B, , +, −, , 2 kΩ, , (c) Calculate the initial energy in the capacitor., , +, v, −, , (d) Obtain the time it takes to dissipate 50 percent of, the initial energy., i, , Figure 7.84, For Prob. 7.4., 7.5, , +, v, −, , C, , R, , Using Fig. 7.85, design a problem to help other, students better understand source-free RC circuits., , Figure 7.88, For Prob. 7.8., , t=0, , R1, , 7.9, , i, , R2, v +, −, , The switch in Fig. 7.89 opens at t 0. Find vo for, t 7 0., , R3, , Figure 7.85, , +, vo, −, , 6V +, −, , For Prob. 7.5., , 7.6, , t=0, , 2 kΩ, , C, , The switch in Fig. 7.86 has been closed for a long, time, and it opens at t 0. Find v(t) for t 0., , 4 kΩ, , 50 F, , Figure 7.89, For Prob. 7.9., 7.10 For the circuit in Fig. 7.90, find vo(t) for t 7 0., Determine the time necessary for the capacitor, voltage to decay to one-third of its value at t 0., , t=0, 10 kΩ, , t=0, 9 kΩ, 24 V, , +, −, , +, v (t), –, , 2 kΩ, , 40 F, 60 V +, −, , 3 kΩ, , +, vo, −, , 20 F, , Figure 7.86, For Prob. 7.6., , Figure 7.90, For Prob. 7.10., , 7.7, , Assuming that the switch in Fig. 7.87 has been in, position A for a long time and is moved to position B, at t 0, find vo(t) for t 0., , Section 7.3 The Source-Free RL Circuit, 7.11 For the circuit in Fig. 7.91, find io for t 7 0., , 20 kΩ, , 3Ω, , t=0, 12 V, , +, −, , A, 40 kΩ, , B, , 2 mF, 20 kΩ, , +, vo(t), −, , t=0, 4H, io, , 24 V +, −, , Figure 7.87, , Figure 7.91, , For Prob. 7.7., , For Prob. 7.11., , 4Ω, , 8Ω
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ale29559_ch07.qxd, , 07/08/2008, , 11:50 AM, , Page 303, , Problems, , 7.12 Using Fig. 7.92, design a problem to help other, students better understand source-free RL circuits., t=0, , 303, , 7.16 Determine the time constant for each of the circuits, in Fig. 7.96., , R1, , L1, i(t), R1, , v +, −, , R3, , R3, , L, , R2, , L2, , R2, R1, , L, , R2, , (a), , Figure 7.92, , (b), , Figure 7.96, , For Prob. 7.12., , For Prob. 7.16., , 7.13 In the circuit of Fig. 7.93,, 7.17 Consider the circuit of Fig. 7.97. Find vo(t) if, i(0) 2 A and v(t) 0., , v(t) 20e10 t V, t 7 0, 3, , i(t) 4e10 t mA, t 7 0, 3, , (a) Find R, L, and t., , 1Ω, , (b) Calculate the energy dissipated in the resistance, for 0 6 t 6 0.5 ms., i, , R, , +, , i(t), , vo(t), , H, , −, , +, −, , v(t), +, v, −, , 3Ω, , 1, 4, , L, , Figure 7.97, For Prob. 7.17., , Figure 7.93, For Prob. 7.13., 7.14 Calculate the time constant of the circuit in Fig. 7.94., 20 kΩ, , 7.18 For the circuit in Fig. 7.98, determine vo(t) when, i(0) 1 A and v(t) 0., 2Ω, , 10 kΩ, , 0.4 H, 40 kΩ, , 5 mH, , 30 kΩ, , +, , i(t), v(t), , 3Ω, , +, −, , vo(t), −, , Figure 7.94, For Prob. 7.14., 7.15 Find the time constant for each of the circuits in, Fig. 7.95., 10 Ω, , Figure 7.98, For Prob. 7.18., 7.19 In the circuit of Fig. 7.99, find i(t) for t 7 0 if, i(0) 2 A., , 40 Ω, 8Ω, 12 Ω, , 5H, , i, , 160 Ω, , 40 Ω, , 20 mH, 10 Ω, , (a), , 6H, , (b), , Figure 7.95, , Figure 7.99, , For Prob. 7.15., , For Prob. 7.19., , 0.5i, , 40 Ω
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ale29559_ch07.qxd, , 07/08/2008, , 11:50 AM, , Page 304, , Chapter 7, , 304, , First-Order Circuits, , Section 7.4 Singularity Functions, , 7.20 For the circuit in Fig. 7.100,, 50t, , v 150e, , V, , 7.24 Express the following signals in terms of singularity, functions., , and, i 30e50t A,, , (a) v(t) e, , t 7 0, , (a) Find L and R., (c) Calculate the initial energy in the inductor., (d) What fraction of the initial energy is dissipated, in 10 ms?, , +, v, −, , For Prob. 7.20., 7.21 In the circuit of Fig. 7.101, find the value of R for, which the steady-state energy stored in the inductor, will be 0.25 J., , 1 6 t 6 2, 2 6 t 6 3, 3 6 t 6 4, Otherwise, t 6 0, 0 6 t 6 1, t 7 1, , 2,, (d) y(t) c 5,, 0,, , Figure 7.100, , 40 Ω, , t 6 1, 1 6 t 6 3, 3 6 t 6 5, t 7 5, , t 1,, 1,, (c) x(t) d, 4 t,, 0,, , i, , L, , t 6 0, t 7 0, , 0,, 10,, (b) i(t) d, 10,, 0,, , (b) Determine the time constant., , R, , 0,, 5,, , 7.25 Design a problem to help other students better, understand singularity functions., 7.26 Express the signals in Fig. 7.104 in terms of, singularity functions., , R, , 30 V +, −, , 80 Ω, , 2H, v1(t), 1, v 2(t), , Figure 7.101, For Prob. 7.21., , 1, −1, , 7.22 Find i(t) and v(t) for t 7 0 in the circuit of Fig. 7.102, if i(0) 20 A., , 0, −1, , 2H, , 0, , 20 Ω, , 1Ω, , +, v(t), −, , 4, , 2, , t, , (b), , (a), , i(t), 5Ω, , 2, t, , v 3(t), 4, , 2, , v 4(t), , Figure 7.102, For Prob. 7.22., 0, , 2, , 4, (c), , 7.23 Consider the circuit in Fig. 7.103. Given that, vo(0) 2 V, find vo and vx for t 7 0., 3Ω, +, vx, −, , 1Ω, , 6, , t, , 0, 1, , 2, , t, , −1, −2, , 1, 3, , H, , 2Ω, , +, vo, −, , (d), , Figure 7.104, For Prob. 7.26., , Figure 7.103, For Prob. 7.23., , 7.27 Express v(t) in Fig. 7.105 in terms of step functions.
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ale29559_ch07.qxd, , 07/08/2008, , 11:50 AM, , Page 305, , Problems, v(t), , 7.35 Find the solution to the following differential, equations:, , 30, 20, , (a), , dv, 2v 0,, dt, , 10, , (b) 2, , −1, , 305, , 0, , 1, , 2, , 3, , t, , −10, −20, , v(0) 1 V, , di, 3i 0,, dt, , i(0) 2, , 7.36 Solve for v in the following differential equations,, subject to the stated initial condition., (a) dvdt v u(t),, , Figure 7.105, , v(0) 0, , (b) 2 dvdt v 3u(t),, , For Prob. 7.27., , v(0) 6, , 7.37 A circuit is described by, 7.28 Sketch the waveform represented by, i(t) r (t) r (t 1) u(t 2) r (t 2), r (t 3) u(t 4), 7.29 Sketch the following functions:, (a) x(t) 5etu(t 1), (b) y(t) 20e(t1)u(t), (c) z(t) 5 cos 4td(t 1), 7.30 Evaluate the following integrals involving the, impulse functions:, (a), , , , , , dv, v 10, dt, , (a) What is the time constant of the circuit?, (b) What is v(), the final value of v?, (c) If v(0) 2, find v(t) for t 0., 7.38 A circuit is described by, di, 3i 2u(t), dt, Find i(t) for t 7 0 given that i(0) 0., , 4t2d(t 1) dt, , Section 7.5 Step Response of an RC Circuit, , , , , (b), , 4, , , , 4t cos 2p td(t 0.5) dt, 2, , , , 7.39 Calculate the capacitor voltage for t 6 0 and t 7 0, for each of the circuits in Fig. 7.106., , 7.31 Evaluate the following integrals:, (a), , , , , , e4t d(t 2) dt, 2, , , , , (b), , , , 4Ω, , [5d(t) etd(t) cos 2p td(t)] dt, , , , 20 V, , 7.32 Evaluate the following integrals:, (a), , , , t, , , , 4, , +, −, , , , t=0, (a), , r (t 1) dt, , 2F, , 0, 5, , (c), , 2F, , u(l) dl, , 1, , (b), , 1Ω, , +, v, −, , (t 6)2d(t 2) dt, , + v −, , 1, , 7.33 The voltage across a 10-mH inductor is, 20d(t 2) mV. Find the inductor current, assuming, that the inductor is initially uncharged., , 12 V, , +, −, , 4Ω, , 2A, , 3Ω, , 7.34 Evaluate the following derivatives:, d, (a) [u(t 1)u(t 1)], dt, d, (b) [r (t 6)u(t 2)], dt, d, (c) [sin 4tu(t 3)], dt, , t=0, , (b), , Figure 7.106, For Prob. 7.39., , 7.40 Find the capacitor voltage for t 6 0 and t 7 0 for, each of the circuits in Fig. 7.107.
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ale29559_ch07.qxd, , 07/08/2008, , 11:50 AM, , Page 306, , Chapter 7, , 306, 3Ω, , First-Order Circuits, , 2Ω, t=0, , +, −, , 12 V, , 4V, , +, −, , +, v, −, , 3F, , 7.44 The switch in Fig. 7.111 has been in position a for a, long time. At t 0, it moves to position b. Calculate, i(t) for all t 7 0., a, , t=0, b, , (a), 30 V +, −, , t=0, , 4Ω, 2Ω, , 6A, , 6Ω, i, , 12 V +, −, , 3Ω, , 2F, , Figure 7.111, +, v, −, , For Prob. 7.44., 5F, , 7.45 Find vo in the circuit of Fig. 7.112 when vs 6u(t)., Assume that vo(0) 1 V., , (b), , Figure 7.107, For Prob. 7.40., , 20 kΩ, , 7.41 Using Fig. 7.108, design a problem to help other, students better understand the step response of an RC, circuit., R1, , v +, −, , vs, , +, −, , 10 kΩ, , t=0, , R2, , C, , Figure 7.112, , +, vo, −, , For Prob. 7.45., 7.46 For the circuit in Fig. 7.113, is(t) 5u(t). Find v(t)., , Figure 7.108, For Prob. 7.41., 7.42 (a) If the switch in Fig. 7.109 has been open for a, long time and is closed at t 0, find vo(t)., (b) Suppose that the switch has been closed for a, long time and is opened at t 0. Find vo(t)., 2Ω, 18 V +, −, , 4Ω, , 3F, , +, vo, −, , is, , 3F, , v, , +, –, , 0.25 F, , For Prob. 7.46., 7.47 Determine v(t) for t 7 0 in the circuit of Fig. 7.114, if v(0) 0., + v −, , 30 Ω, 0.1 F, , i, 80 V +, −, , 6Ω, , Figure 7.113, , For Prob. 7.42., 7.43 Consider the circuit in Fig. 7.110. Find i(t) for t 6 0, and t 7 0., t=0, , 2Ω, , t=0, , Figure 7.109, , 40 Ω, , +, vo, −, , 3 F, , 40 Ω, , 0.5i, , 50 Ω, , 6u(t − 1) A, , Figure 7.110, , Figure 7.114, , For Prob. 7.43., , For Prob. 7.47., , 2Ω, , 8Ω, , 6u(t) A
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ale29559_ch07.qxd, , 07/08/2008, , 11:50 AM, , Page 307, , Problems, , 307, , 7.48 Find v(t) and i(t) in the circuit of Fig. 7.115., , R1, i, , 20 Ω, , t=0, v +, −, , i, 10 Ω, , 6(1 − u(t)) A, , 0.1 F, , +, v, −, , L, R2, , Figure 7.118, For Prob. 7.52., , Figure 7.115, For Prob. 7.48., 7.49 If the waveform in Fig. 7.116(a) is applied to the, circuit of Fig. 7.116(b), find v(t). Assume v(0) 0., , 7.53 Determine the inductor current i(t) for both t 6 0, and t 7 0 for each of the circuits in Fig. 7.119., 2Ω, , 3Ω, is (A), , i, , 2, , 25 V, , 0, , 1, (a), , +, −, , 4H, , t=0, , (a), , t (s), , t=0, 6Ω, , is, , 4Ω, , i, , 0.5 F, , 4Ω, , 6A, , +, v, −, , 3H, , (b), , (b), , Figure 7.119, , Figure 7.116, , For Prob. 7.53., , For Prob. 7.49 and Review Question 7.10., *7.50 In the circuit of Fig. 7.117, find ix for t 7 0. Let, R1 R2 1 k, R3 2 k, and C 0.25 mF., t=0, , 7.54 Obtain the inductor current for both t 6 0 and t 7 0, in each of the circuits in Fig. 7.120., , R2, , i, , ix, 30 mA, , 2Ω, , R1, , C, , R3, , 4Ω, , 3A, , 12 Ω, , 4Ω, , t=0, , 3.5 H, , (a), , Figure 7.117, For Prob. 7.50., , i, , Section 7.6 Step Response of an RL Circuit, 7.51 Rather than applying the short-cut technique used in, Section 7.6, use KVL to obtain Eq. (7.60)., 7.52 Using Fig. 7.118, design a problem to help other, students better understand the step response of an RL, circuit., , 20 V, , 2H, , t=0, 2Ω, , 6Ω, (b), , Figure 7.120, * An asterisk indicates a challenging problem., , +, −, , +, −, , 48 V, , For Prob. 7.54., , 3Ω
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ale29559_ch07.qxd, , 07/08/2008, , 11:50 AM, , Page 308, , Chapter 7, , 308, , First-Order Circuits, , 7.55 Find v(t) for t 6 0 and t 7 0 in the circuit of, Fig. 7.121., io, , 7.60 Find v(t) for t 7 0 in the circuit of Fig. 7.125 if the, initial current in the inductor is zero., , 0.5 H, , 3Ω, , 24 V, , 8Ω, +, −, , 20 V, , 5Ω, , 4u(t) A, , t=0, , +, −, , 4io, , +, v, −, , 2Ω, , +, −, , 20 Ω, , 8H, , +, v, −, , Figure 7.125, For Prob. 7.60., 7.61 In the circuit of Fig. 7.126, is changes from 5 A to 10 A, at t 0; that is, is (5 5u(t)) A. Find v and i., , Figure 7.121, For Prob. 7.55., , i, , 7.56 For the network shown in Fig. 7.122, find v(t) for, t 7 0., 5Ω, t=0, , 20 Ω, , 12 Ω, , 2A, , For Prob. 7.61., , +, v, −, , +, −, , 20 V, , 7.62 For the circuit in Fig. 7.127, calculate i(t) if i(0) 0., 3Ω, , 6Ω, i, , 10u(t − 1) V +, −, , Figure 7.122, For Prob. 7.56., *7.57 Find i1(t) and i2(t) for t 7 0 in the circuit of, Fig. 7.123., i1, 6Ω, , 10 A, , t=0, , 50 mH, , Figure 7.126, , 6Ω, , 0.5 H, , 4Ω, , is, , +, v, −, , 2.5 H, , 10u(t) V, , Figure 7.127, For Prob. 7.62., i2, , 7.63 Obtain v(t) and i(t) in the circuit of Fig. 7.128., , 20 Ω, , 5Ω, , +, −, , 2H, , i, , 5Ω, , 4H, 20u(−t) V, , Figure 7.123, , +, −, , 20 Ω, , 0.5 H, , +, v, −, , For Prob. 7.57., 7.58 Rework Prob. 7.17 if i(0) 10 A and, v(t) 20u(t) V., 7.59 Determine the step response vo(t) to vs 9u (t) V in, the circuit of Fig. 7.124., , Figure 7.128, For Prob. 7.63., 7.64 Find vo(t) for t 7 0 in the circuit of Fig. 7.129., 6Ω, , 6Ω, 10 V, 4Ω, vs, , +, −, , 3Ω, 1.5 H, , +, vo, −, , +, −, , 3Ω, , +, vo, −, 4H, 2Ω, , t=0, , Figure 7.124, , Figure 7.129, , For Prob. 7.59., , For Prob. 7.64.
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ale29559_ch07.qxd, , 07/08/2008, , 11:50 AM, , Page 309, , Problems, , 309, , 7.65 If the input pulse in Fig. 7.130(a) is applied to the, circuit in Fig. 7.130(b), determine the response i(t)., , vs (V), , t=0, +, −, , 5Ω, , 6V, , +, −, , 10 kΩ, 10 kΩ, , i, , 20, +, −, , vs, , 20 Ω, , +, vo, −, , 25 F, , 2H, , Figure 7.133, 0, , 1, , t (s), , For Prob. 7.68., 7.69 For the op amp circuit in Fig. 7.134, find vo(t) for, t 7 0., , (b), , (a), , Figure 7.130, For Prob. 7.65., , 25 mF, , 10 kΩ, , t=0, , 20 kΩ, , 100 kΩ, , Section 7.7 First-order Op Amp Circuits, 7.66 Using Fig. 7.131, design a problem to help other, students better understand first-order op amp, circuits., , 4V, , −, +, , +, −, , +, vo, −, , R2, , Figure 7.134, C, R1, , vs, , For Prob. 7.69., 7.70 Determine vo for t 7 0 when vs 20 mV in the op, amp circuit of Fig. 7.135., , −, +, , t=0, , +, , +, −, , +, −, , vo, , vo, , −, +, −, , vs, , Figure 7.131, , 5 F, 20 kΩ, , For Prob. 7.66., , 7.67 If v(0) 10 V, find vo(t) for t 7 0 in the op amp, circuit of Fig. 7.132. Let R 10 k and C 1 mF., , Figure 7.135, For Prob. 7.70., 7.71 For the op amp circuit in Fig. 7.136, suppose v0 0, and vs 3 V. Find v(t) for t 7 0., , R, , 10 kΩ, −, +, , R, , R, , +, v, −, , vo, 10 kΩ, , −, +, , C, , vs, , Figure 7.132, For Prob. 7.67., , Figure 7.136, 7.68 Obtain vo for t 7 0 in the circuit of Fig. 7.133., , For Prob. 7.71., , +, −, , 20 kΩ, , 10 F, , +, v, −
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ale29559_ch07.qxd, , 07/08/2008, , 11:50 AM, , Page 310, , Chapter 7, , 310, , First-Order Circuits, , 7.72 Find io in the op amp circuit in Fig. 7.137. Assume, that v(0) 2 V, R 10 k, and C 10 mF., , 7.76 Repeat Prob. 7.49 using PSpice., , C, , −, +, , Section 7.8 Transient Analysis with PSpice, , 7.77 The switch in Fig. 7.141 opens at t 0. Use PSpice, to determine v(t) for t 7 0., , io, , + v −, , 3u(t) +, −, , R, t=0, , + v −, , 5Ω, , 100 mF, , Figure 7.137, , 6Ω, , 4Ω, , 5A, , For Prob. 7.72., , +, −, , 20 Ω, , 30 V, , 7.73 For the circuit shown in Fig. 7.138, solve for io(t)., , Figure 7.141, For Prob. 7.77., , 100 F, 10 kΩ, , io(t), , −, +, , 5u(t) V +, −, , 10 mF, , +, vo(t), , 7.78 The switch in Fig. 7.142 moves from position a to b, at t 0. Use PSpice to find i(t) for t 7 0., , −, , 6Ω, , a, , 4Ω, t=0, , i(t), , Figure 7.138, For Prob. 7.73., , 108 V, , 7.74 Determine vo(t) for t 7 0 in the circuit of Fig. 7.139., Let is 10u (t) mA and assume that the capacitor is, initially uncharged., , −, +, , 6Ω, , 2H, , Figure 7.142, , 7.79 In the circuit of Fig. 7.143, the switch has been in, position a for a long time but moves instantaneously, to position b at t 0. Determine io(t)., , +, vo, , 50 kΩ, , is, , 3Ω, , For Prob. 7.78., , 10 kΩ, , 2 F, , b, , +, −, , −, a, , Figure 7.139, , t=0, , 3Ω, , b, , io, , For Prob. 7.74., 7.75 In the circuit of Fig. 7.140, find vo and io, given that, vs 4u(t) V and v(0) 1 V., +, −, , −, +, , 4V, , Figure 7.143, io, , +, −, , 12 V, , 4Ω, 0.1 H, , vo, , 10 kΩ, vs, , 5Ω, +, −, , For Prob. 7.79., , 2 F, 20 kΩ, , + v −, , 7.80 In the circuit of Fig. 7.144, assume that the switch, has been in position a for a long time, find:, (a) i1(0), i2 (0), and vo(0), , Figure 7.140, , (b) iL(t), , For Prob. 7.75., , (c) i1(), i2(), and vo().
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ale29559_ch07.qxd, , 07/08/2008, , 11:50 AM, , Page 311, , Comprehensive Problems, 10 Ω, , a, i1, , 30 V, , 4 MΩ, , t=0, i2, , b, , 5Ω, , +, −, , 311, , 3Ω, , 6Ω, , iL, +, vo, –, , 4H, , +, 120 V, −, , 6 F, , Neon lamp, , Figure 7.145, For Prob. 7.85., , Figure 7.144, For Prob. 7.80., , 7.81 Repeat Prob. 7.65 using PSpice., , 7.86 Figure 7.146 shows a circuit for setting the length of, time voltage is applied to the electrodes of a welding, machine. The time is taken as how long it takes the, capacitor to charge from 0 to 8 V. What is the time, range covered by the variable resistor?, , Section 7.9 Applications, 7.82 In designing a signal-switching circuit, it was found, that a 100-mF capacitor was needed for a time, constant of 3 ms. What value resistor is necessary for, the circuit?, 7.83 An RC circuit consists of a series connection of a, 120-V source, a switch, a 34-M resistor, and a, 15-mF capacitor. The circuit is used in estimating the, speed of a horse running a 4-km racetrack. The, switch closes when the horse begins and opens when, the horse crosses the finish line. Assuming that the, capacitor charges to 85.6 V, calculate the speed of, the horse., 7.84 The resistance of a 160-mH coil is 8 . Find the, time required for the current to build up to 60, percent of its final value when voltage is applied to, the coil., 7.85 A simple relaxation oscillator circuit is shown in, Fig. 7.145. The neon lamp fires when its voltage, reaches 75 V and turns off when its voltage drops to, 30 V. Its resistance is 120 when on and infinitely, high when off., (a) For how long is the lamp on each time the, capacitor discharges?, (b) What is the time interval between light flashes?, , 100 kΩ to 1 MΩ, , 2 F, , 12 V, , Welding, control, unit, Electrode, , Figure 7.146, For Prob. 7.86., 7.87 A 120-V dc generator energizes a motor whose coil, has an inductance of 50 H and a resistance of 100 ., A field discharge resistor of 400 is connected in, parallel with the motor to avoid damage to the, motor, as shown in Fig. 7.147. The system is at, steady state. Find the current through the discharge, resistor 100 ms after the breaker is tripped., Circuit breaker, , 120 V, , +, −, , Motor, , 400 Ω, , Figure 7.147, For Prob. 7.87., , Comprehensive Problems, 7.88 The circuit in Fig. 7.148(a) can be designed as, an approximate differentiator or an integrator,, depending on whether the output is taken across, the resistor or the capacitor, and also on the time, constant t RC of the circuit and the width T of, the input pulse in Fig. 7.148(b). The circuit is a, differentiator if t V T, say t 6 0.1T, or an, integrator if t W T, say t 7 10T., , (a) What is the minimum pulse width that will allow, a differentiator output to appear across the, capacitor?, (b) If the output is to be an integrated form of the, input, what is the maximum value the pulse, width can assume?
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ale29559_ch07.qxd, , 07/08/2008, , 11:50 AM, , Page 312, , Chapter 7, , 312, , First-Order Circuits, , vi, 300 kΩ, , vi, , +, −, , Vm, 200 pF, 0, , (a), , T, , t, , 7.91 The circuit in Fig. 7.150 is used by a biology student, to study “frog kick.” She noticed that the frog kicked, a little when the switch was closed but kicked, violently for 5 s when the switch was opened. Model, the frog as a resistor and calculate its resistance., Assume that it takes 10 mA for the frog to kick, violently., , (b), , Figure 7.148, , 50 Ω, , Switch, Frog, , For Prob. 7.88., , 7.89 An RL circuit may be used as a differentiator if the, output is taken across the inductor and t V T (say, t 6 0.1T ), where T is the width of the input pulse., If R is fixed at 200 k, determine the maximum, value of L required to differentiate a pulse with, T 10 ms., 7.90 An attenuator probe employed with oscilloscopes, was designed to reduce the magnitude of the input, voltage vi by a factor of 10. As shown in Fig. 7.149,, the oscilloscope has internal resistance Rs and, capacitance Cs, while the probe has an internal, resistance Rp. If Rp is fixed at 6 M, find Rs and Cs, for the circuit to have a time constant of 15 ms., , +, 12 V, −, , 2H, , Figure 7.150, For Prob. 7.91., 7.92 To move a spot of a cathode-ray tube across the, screen requires a linear increase in the voltage across, the deflection plates, as shown in Fig. 7.151. Given, that the capacitance of the plates is 4 nF, sketch the, current flowing through the plates., v (V), , Probe, +, vi, −, , Scope, , 10, +, , Rp, Rs, , Cs, , vo, −, , Rise time = 2 ms, , Drop time = 5 s, , (not to scale), , Figure 7.149, , Figure 7.151, , For Prob. 7.90., , For Prob. 7.92., , t
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ale29559_ch08.qxd, , 07/08/2008, , 11:15 AM, , Page 313, , c h a p t e r, , Second-Order, Circuits, , 8, , Everyone who can earn a masters degree in engineering must earn a, masters degree in engineering in order to maximize the success of their, career! If you want to do research, state-of-the-art engineering, teach in, a university, or start your own business, you really need to earn a doctoral, degree!, —Charles K. Alexander, , Enhancing Your Career, To increase your engineering career opportunities after graduation,, develop a strong fundamental understanding in a broad set of engineering areas. When possible, this might best be accomplished by working, toward a graduate degree immediately upon receiving your undergraduate degree., Each degree in engineering represents certain skills the student, acquires. At the Bachelor degree level, you learn the language of engineering and the fundamentals of engineering and design. At the Master’s level, you acquire the ability to do advanced engineering projects, and to communicate your work effectively both orally and in writing., The Ph.D. represents a thorough understanding of the fundamentals of, electrical engineering and a mastery of the skills necessary both for, working at the frontiers of an engineering area and for communicating, one’s effort to others., If you have no idea what career you should pursue after graduation, a graduate degree program will enhance your ability to explore, career options. Since your undergraduate degree will only provide you, with the fundamentals of engineering, a Master’s degree in engineering supplemented by business courses benefits more engineering students than does getting a Master’s of Business Administration (MBA)., The best time to get your MBA is after you have been a practicing, engineer for some years and decide your career path would be, enhanced by strengthening your business skills., Engineers should constantly educate themselves, formally and, informally, taking advantage of all means of education. Perhaps there, is no better way to enhance your career than to join a professional society such as IEEE and be an active member., , Enhancing your career involves understanding your goals, adapting to changes,, anticipating opportunities, and planning, your own niche., © 2005 Institute of Electrical and, Electronics Engineers (IEEE)., , 313
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ale29559_ch08.qxd, , 07/08/2008, , 11:15 AM, , Page 314, , Chapter 8, , 314, , 8.1, , R, , vs, , L, , +, −, , C, , (a), , is, , R, , C, , L, , vs, , +, −, , R2, , L1, , L2, , (c), R, , is, , C1, , Introduction, , In the previous chapter we considered circuits with a single storage element (a capacitor or an inductor). Such circuits are first-order because, the differential equations describing them are first-order. In this chapter we will consider circuits containing two storage elements. These are, known as second-order circuits because their responses are described, by differential equations that contain second derivatives., Typical examples of second-order circuits are RLC circuits, in, which the three kinds of passive elements are present. Examples of, such circuits are shown in Fig. 8.1(a) and (b). Other examples are RL, and RC circuits, as shown in Fig. 8.1(c) and (d). It is apparent from, Fig. 8.1 that a second-order circuit may have two storage elements of, different type or the same type (provided elements of the same type, cannot be represented by an equivalent single element). An op amp circuit with two storage elements may also be a second-order circuit. As, with first-order circuits, a second-order circuit may contain several, resistors and dependent and independent sources., A second-order circuit is characterized by a second-order differential equation. It consists of resistors and the equivalent of two energy, storage elements., , (b), R1, , Second-Order Circuits, , C2, , (d), , Figure 8.1, Typical examples of second-order circuits:, (a) series RLC circuit, (b) parallel RLC, circuit, (c) RL circuit, (d) RC circuit., , Our analysis of second-order circuits will be similar to that used for, first-order. We will first consider circuits that are excited by the initial conditions of the storage elements. Although these circuits may, contain dependent sources, they are free of independent sources., These source-free circuits will give natural responses as expected., Later we will consider circuits that are excited by independent, sources. These circuits will give both the transient response and the, steady-state response. We consider only dc independent sources in, this chapter. The case of sinusoidal and exponential sources is deferred, to later chapters., We begin by learning how to obtain the initial conditions for the, circuit variables and their derivatives, as this is crucial to analyzing, second-order circuits. Then we consider series and parallel RLC circuits such as shown in Fig. 8.1 for the two cases of excitation: by, initial conditions of the energy storage elements and by step inputs., Later we examine other types of second-order circuits, including op, amp circuits. We will consider PSpice analysis of second-order circuits. Finally, we will consider the automobile ignition system and, smoothing circuits as typical applications of the circuits treated in this, chapter. Other applications such as resonant circuits and filters will, be covered in Chapter 14., , 8.2, , Finding Initial and Final Values, , Perhaps the major problem students face in handling second-order circuits is finding the initial and final conditions on circuit variables., Students are usually comfortable getting the initial and final values, of v and i but often have difficulty finding the initial values of their
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ale29559_ch08.qxd, , 07/08/2008, , 11:15 AM, , Page 315, , 8.2, , Finding Initial and Final Values, , 315, , derivatives: dvdt and didt. For this reason, this section is explicitly, devoted to the subtleties of getting v(0), i(0), dv(0)dt, di(0)dt, i(),, and v(). Unless otherwise stated in this chapter, v denotes capacitor, voltage, while i is the inductor current., There are two key points to keep in mind in determining the initial conditions., First—as always in circuit analysis—we must carefully handle the, polarity of voltage v(t) across the capacitor and the direction of the current i(t) through the inductor. Keep in mind that v and i are defined, strictly according to the passive sign convention (see Figs. 6.3 and 6.23)., One should carefully observe how these are defined and apply them, accordingly., Second, keep in mind that the capacitor voltage is always continuous so that, v(0) v(0), , (8.1a), , and the inductor current is always continuous so that, i(0) i(0), , (8.1b), , where t 0 denotes the time just before a switching event and, t 0 is the time just after the switching event, assuming that the, switching event takes place at t 0., Thus, in finding initial conditions, we first focus on those variables that cannot change abruptly, capacitor voltage and inductor current, by applying Eq. (8.1). The following examples illustrate these, ideas., , Example 8.1, , The switch in Fig. 8.2 has been closed for a long time. It is open at, t 0. Find: (a) i(0), v(0), (b) di(0)dt, dv(0)dt, (c) i(), v()., , 4Ω, , i, , 0.25 H, , Solution:, (a) If the switch is closed a long time before t 0, it means that the, circuit has reached dc steady state at t 0. At dc steady state, the, inductor acts like a short circuit, while the capacitor acts like an open, circuit, so we have the circuit in Fig. 8.3(a) at t 0. Thus,, 12, i(0) , 2 A,, 42, , 4Ω, , 12 V, , +, −, , 2Ω, , (a), , Figure 8.3, , 4Ω, +, v, −, , i, , 2Ω, , +, −, , t=0, , Figure 8.2, , 4Ω, , 0.25 H, , i, , + vL −, 12 V, , +, −, , +, v, −, , 0.1 F, , For Example 8.1., , v(0) 2i(0) 4 V, , i, , 12 V, , 0.1 F, , +, +, v, −, , 12 V, , +, −, , v, −, , (b), , Equivalent circuit of that in Fig. 8.2 for: (a) t 0 , (b) t 0 , (c) t S ., , (c)
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ale29559_ch08.qxd, , 07/08/2008, , 11:15 AM, , 316, , Page 316, , Chapter 8, , Second-Order Circuits, , As the inductor current and the capacitor voltage cannot change, abruptly,, i(0) i(0) 2 A,, , v(0) v(0) 4 V, , (b) At t 0, the switch is open; the equivalent circuit is as shown in, Fig. 8.3(b). The same current flows through both the inductor and, capacitor. Hence,, iC (0) i(0) 2 A, Since C dvdt iC, dvdt iCC, and, iC (0), dv(0), 2, , , 20 V/s, dt, C, 0.1, Similarly, since L didt vL , didt vLL. We now obtain vL by, applying KVL to the loop in Fig. 8.3(b). The result is, 12 4i(0) vL(0) v(0) 0, or, vL(0) 12 8 4 0, Thus,, vL(0), di(0), 0, , , 0 A/s, dt, L, 0.25, (c) For t 7 0, the circuit undergoes transience. But as t S , the, circuit reaches steady state again. The inductor acts like a short circuit, and the capacitor like an open circuit, so that the circuit in Fig. 8.3(b), becomes that shown in Fig. 8.3(c), from which we have, v() 12 V, , i() 0 A,, , Practice Problem 8.1, , The switch in Fig. 8.4 was open for a long time but closed at t 0., Determine: (a) i(0), v(0), (b) di(0)dt, dv(0)dt, (c) i(), v()., t=0, 10 Ω, , 0.4 H, , i, , +, 2Ω, , v, , −, , 1, 20, , F, , +, −, , 12 V, , Figure 8.4, For Practice Prob. 8.1., , Answer: (a) 1 A, 2 V, (b) 25 A/s, 0 V/s, (c) 6 A, 12 V.
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ale29559_ch08.qxd, , 07/08/2008, , 11:15 AM, , Page 317, , 8.2, , Finding Initial and Final Values, , 317, , In the circuit of Fig. 8.5, calculate: (a) iL (0), vC (0), vR(0),, (b) diL(0)dt, dvC (0)dt, dvR(0)dt, (c) iL(), vC (), vR()., , Example 8.2, , 4Ω, , 3u(t) A, , 2Ω, , 1, 2, , +, vR, −, , +, vC, −, , F, +, −, , iL, 0.6 H, , 20 V, , Figure 8.5, For Example 8.2., , Solution:, (a) For t 6 0, 3u(t) 0. At t 0, since the circuit has reached, steady state, the inductor can be replaced by a short circuit, while the, capacitor is replaced by an open circuit as shown in Fig. 8.6(a). From, this figure we obtain, iL(0) 0,, , vR(0) 0,, , vC (0) 20 V, , (8.2.1), , Although the derivatives of these quantities at t 0 are not required,, it is evident that they are all zero, since the circuit has reached steady, state and nothing changes., 4Ω, , a, , +, , vR, , +, vC, −, , 2Ω, +, −, , + vo −, , iL, , 3A, , 2Ω, , 20 V, , (a), , +, vR, −, , 1, 2, , The circuit in Fig. 8.5 for: (a) t 0, (b) t 0., , For t 7 0, 3u(t) 3, so that the circuit is now equivalent to that, in Fig. 8.6(b). Since the inductor current and capacitor voltage cannot, change abruptly,, vC (0) vC (0) 20 V, , (8.2.2), , Although the voltage across the 4- resistor is not required, we will, use it to apply KVL and KCL; let it be called vo. Applying KCL at, node a in Fig. 8.6(b) gives, vo(0), vR(0), , 2, 4, , (8.2.3), , Applying KVL to the middle mesh in Fig. 8.6(b) yields, vR(0) vo(0) vC (0) 20 0, , (8.2.4), , F, +, −, , (b), , Figure 8.6, , 3, , iC, +, vC, −, , 4Ω, , −, , iL (0) iL (0) 0,, , b, , 20 V, , iL, +, vL, −, , 0.6 H
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ale29559_ch08.qxd, , 07/08/2008, , 11:15 AM, , Page 319, , The Source-Free Series RLC Circuit, , 8.3, , 319, , We can find diR(0)dt although it is not required. Since vR 5iR,, diR(0), 1 dvR(0), 12, 2, , , , A/s, dt, 5, dt, 53, 15, (c) As t S , the circuit reaches steady state. We have the equivalent, circuit in Fig. 8.6(a) except that the 3-A current source is now, operative. By current division principle,, iL() , , 2, 3A1A, 24, , 4, vR() , 3 A 2 4 V,, 24, , (8.2.12), , vC () 20 V, , For the circuit in Fig. 8.7, find: (a) iL(0), vC (0), vR(0),, (b) diL(0)dt, dvC (0)dt, dvR(0)dt, (c) iL(), vC (), vR()., + vR −, , iR, , 5Ω, , iC, F, , −, , iL, +, vL, −, , +, 1, 5, , 4u(t) A, , Practice Problem 8.2, , vC, , 2H, , 6A, , Figure 8.7, For Practice Prob. 8.2., , Answer: (a) 6 A, 0, 0, (b) 0, 20 V/s, 0, (c) 2 A, 20 V, 20 V., , 8.3, , The Source-Free Series RLC Circuit, , An understanding of the natural response of the series RLC circuit is, a necessary background for future studies in filter design and communications networks., Consider the series RLC circuit shown in Fig. 8.8. The circuit is, being excited by the energy initially stored in the capacitor and inductor. The energy is represented by the initial capacitor voltage V0 and, initial inductor current I0. Thus, at t 0,, v(0) , , 1, C, , , , i dt V0, , i, , (8.2b), , Applying KVL around the loop in Fig. 8.8,, , , , t, , , , i dt 0, , +, V0, −, , (8.2a), , , , 1, di, , dt, C, , L, I0, , 0, , i(0) I0, , Ri L, , R, , (8.3), , Figure 8.8, A source-free series RLC circuit., , C
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ale29559_ch08.qxd, , 07/08/2008, , 11:15 AM, , 320, , Page 320, , Chapter 8, , Second-Order Circuits, , To eliminate the integral, we differentiate with respect to t and, rearrange terms. We get, d 2i, R di, i, , , 0, 2, L dt, LC, dt, , (8.4), , This is a second-order differential equation and is the reason for calling the RLC circuits in this chapter second-order circuits. Our goal is, to solve Eq. (8.4). To solve such a second-order differential equation, requires that we have two initial conditions, such as the initial value, of i and its first derivative or initial values of some i and v. The initial value of i is given in Eq. (8.2b). We get the initial value of the, derivative of i from Eqs. (8.2a) and (8.3); that is,, Ri(0) L, , di(0), V0 0, dt, , or, di(0), 1, (RI0 V0), dt, L, , (8.5), , With the two initial conditions in Eqs. (8.2b) and (8.5), we can now, solve Eq. (8.4). Our experience in the preceding chapter on first-order, circuits suggests that the solution is of exponential form. So we let, i Aest, , (8.6), , where A and s are constants to be determined. Substituting Eq. (8.6), into Eq. (8.4) and carrying out the necessary differentiations, we obtain, As2est , , AR st, A st, se , e 0, L, LC, , or, Aest as2 , , R, 1, s, b0, L, LC, , (8.7), , Since i Aest is the assumed solution we are trying to find, only the, expression in parentheses can be zero:, s2 , , See Appendix C.1 for the formula to, find the roots of a quadratic equation., , R, 1, s, 0, L, LC, , (8.8), , This quadratic equation is known as the characteristic equation of the, differential Eq. (8.4), since the roots of the equation dictate the character of i. The two roots of Eq. (8.8) are, s1 , , R 2, 1, R, a b , 2L B 2L, LC, , (8.9a), , s2 , , R, R 2, 1, a b , 2L B 2L, LC, , (8.9b), , A more compact way of expressing the roots is, s1 a 2a2 20,, , s2 a 2a2 20, , (8.10)
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ale29559_ch08.qxd, , 07/08/2008, , 11:15 AM, , Page 321, , The Source-Free Series RLC Circuit, , 8.3, , 321, , where, a, , R, ,, 2L, , 0 , , 1, 2LC, , (8.11), , The roots s1 and s2 are called natural frequencies, measured in, nepers per second (Np/s), because they are associated with the natural, response of the circuit; 0 is known as the resonant frequency or, strictly as the undamped natural frequency, expressed in radians per, second (rad/s); and a is the neper frequency or the damping factor,, expressed in nepers per second. In terms of a and 0, Eq. (8.8) can be, written as, s2 2a s 20 0, , (8.8a), , The variables s and 0 are important quantities we will be discussing, throughout the rest of the text., The two values of s in Eq. (8.10) indicate that there are two possible solutions for i, each of which is of the form of the assumed solution in Eq. (8.6); that is,, i1 A1es1t,, , i2 A2es2t, , The neper (Np) is a dimensionless unit, named after John Napier (1550–1617),, a Scottish mathematician., , The ratio a0 is known as the damping ratio z., , (8.12), , Since Eq. (8.4) is a linear equation, any linear combination of the, two distinct solutions i1 and i2 is also a solution of Eq. (8.4). A complete or total solution of Eq. (8.4) would therefore require a linear, combination of i1 and i2. Thus, the natural response of the series RLC, circuit is, i(t) A1es1t A2es2t, , (8.13), , where the constants A1 and A2 are determined from the initial values, i(0) and di(0)dt in Eqs. (8.2b) and (8.5)., From Eq. (8.10), we can infer that there are three types of solutions:, 1. If a 7 0, we have the overdamped case., 2. If a 0, we have the critically damped case., 3. If a 6 0, we have the underdamped case., We will consider each of these cases separately., , Overdamped Case (A 0), , From Eqs. (8.9) and (8.10), a 7 0 implies C 7 4LR2. When this, happens, both roots s1 and s2 are negative and real. The response is, i(t) A1es1t A2es2t, , (8.14), , which decays and approaches zero as t increases. Figure 8.9(a) illustrates a typical overdamped response., , Critically Damped Case (A 0), When a 0, C 4LR2 and, , s1 s2 a , , R, 2L, , (8.15), , The response is overdamped when, the roots of the circuit’s characteristic, equation are unequal and real, critically, damped when the roots are equal and, real, and underdamped when the, roots are complex.
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ale29559_ch08.qxd, , 07/08/2008, , 11:15 AM, , Page 322, , 322, , Chapter 8, , Second-Order Circuits, , i(t), , For this case, Eq. (8.13) yields, i(t) A1eat A2eat A3eat, , 0, , t, , where A3 A1 A2 . This cannot be the solution, because the two initial conditions cannot be satisfied with the single constant A3. What, then could be wrong? Our assumption of an exponential solution is, incorrect for the special case of critical damping. Let us go back to, Eq. (8.4). When a 0 R2L, Eq. (8.4) becomes, d 2i, di, 2a a2i 0, 2, dt, dt, , (a), , or, , i(t), , d di, di, a aib a a aib 0, dt dt, dt, , (8.16), , If we let, f, 0, , 1, , , di, ai, dt, , (8.17), , t, , then Eq. (8.16) becomes, df, af 0, dt, , (b), , which is a first-order differential equation with solution f A1eat,, where A1 is a constant. Equation (8.17) then becomes, , i(t), e –t, , 0, , di, ai A1eat, dt, , t, 2, d, , or, eat, , (c), , Figure 8.9, (a) Overdamped response, (b) critically, damped response, (c) underdamped, response., , di, eatai A1, dt, , (8.18), , d at, (e i) A1, dt, , (8.19), , This can be written as, , Integrating both sides yields, eati A1t A2, or, i (A1t A2)eat, , (8.20), , where A2 is another constant. Hence, the natural response of the critically damped circuit is a sum of two terms: a negative exponential and, a negative exponential multiplied by a linear term, or, i(t) (A2 A1t)eat, , (8.21), , A typical critically damped response is shown in Fig. 8.9(b). In fact,, Fig. 8.9(b) is a sketch of i(t) teat, which reaches a maximum value of, e1a at t 1a, one time constant, and then decays all the way to zero.
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ale29559_ch08.qxd, , 07/08/2008, , 11:15 AM, , Page 323, , 8.3, , The Source-Free Series RLC Circuit, , 323, , Underdamped Case (A 0), , For a 6 0, C 6 4LR2. The roots may be written as, s1 a 2(20 a2) a jd, s2 a , , 2(20, , a ) a jd, 2, , (8.22a), (8.22b), , where j 21 and d 220 a2, which is called the damping, frequency. Both 0 and d are natural frequencies because they help, determine the natural response; while 0 is often called the undamped, natural frequency, d is called the damped natural frequency. The natural, response is, i(t) A1e(ajd)t A2e(ajd)t, (8.23), ea t(A1e jd t A2ejd t ), Using Euler’s identities,, e ju cos u j sin u,, , eju cos u j sin u, , (8.24), , we get, i(t) ea t[A1(cos d t j sin d t) A2(cos d t j sin d t)], (8.25), ea t[(A1 A2) cos d t j(A1 A2) sin d t], Replacing constants (A1 A2) and j(A1 A2) with constants B1 and B2,, we write, i(t) ea t(B1 cos d t B2 sin d t), , (8.26), , With the presence of sine and cosine functions, it is clear that the natural response for this case is exponentially damped and oscillatory in, nature. The response has a time constant of 1a and a period of, T 2pd. Figure 8.9(c) depicts a typical underdamped response., [Figure 8.9 assumes for each case that i(0) 0.], Once the inductor current i(t) is found for the RLC series circuit, as shown above, other circuit quantities such as individual element, voltages can easily be found. For example, the resistor voltage is, vR Ri, and the inductor voltage is vL L didt. The inductor current i(t) is selected as the key variable to be determined first in order, to take advantage of Eq. (8.1b)., We conclude this section by noting the following interesting, peculiar properties of an RLC network:, 1. The behavior of such a network is captured by the idea of damping,, which is the gradual loss of the initial stored energy, as evidenced by, the continuous decrease in the amplitude of the response. The damping effect is due to the presence of resistance R. The damping factor, a determines the rate at which the response is damped. If R 0,, then a 0, and we have an LC circuit with 11LC as the, undamped natural frequency. Since a 6 0 in this case, the response, is not only undamped but also oscillatory. The circuit is said to be, loss-less, because the dissipating or damping element (R) is absent., By adjusting the value of R, the response may be made undamped,, overdamped, critically damped, or underdamped., 2. Oscillatory response is possible due to the presence of the two, types of storage elements. Having both L and C allows the flow of, , R 0 produces a perfectly sinusoidal, response. This response cannot be, practically accomplished with L and C, because of the inherent losses in them., See Figs 6.8 and 6.26. An electronic, device called an oscillator can produce a perfectly sinusoidal response., Examples 8.5 and 8.7 demonstrate the, effect of varying R., The response of a second-order circuit, with two storage elements of the same, type, as in Fig. 8.1(c) and (d), cannot, be oscillatory.
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ale29559_ch08.qxd, , 07/08/2008, , 11:15 AM, , 324, , What this means in most practical circuits is that we seek an overdamped, circuit that is as close as possible to a, critically damped circuit., , Example 8.3, , Page 324, , Chapter 8, , Second-Order Circuits, , energy back and forth between the two. The damped oscillation, exhibited by the underdamped response is known as ringing. It, stems from the ability of the storage elements L and C to transfer, energy back and forth between them., 3. Observe from Fig. 8.9 that the waveforms of the responses differ., In general, it is difficult to tell from the waveforms the difference, between the overdamped and critically damped responses. The critically damped case is the borderline between the underdamped and, overdamped cases and it decays the fastest. With the same initial, conditions, the overdamped case has the longest settling time,, because it takes the longest time to dissipate the initial stored, energy. If we desire the response that approaches the final value, most rapidly without oscillation or ringing, the critically damped, circuit is the right choice., , In Fig. 8.8, R 40 , L 4 H, and C 14 F. Calculate the characteristic roots of the circuit. Is the natural response overdamped, underdamped, or critically damped?, Solution:, We first calculate, a, , R, 40, , 5,, 2L, 2(4), , 0 , , 1, 2LC, , , , 1, 24 14, , 1, , The roots are, s1,2 a 2a2 20 5 225 1, or, s1 0.101,, , s2 9.899, , Since a 7 0, we conclude that the response is overdamped. This is, also evident from the fact that the roots are real and negative., , Practice Problem 8.3, , If R 10 , L 5 H, and C 2 mF in Fig. 8.8, find a, 0, s1, and s2., What type of natural response will the circuit have?, Answer: 1, 10, 1 j 9.95, underdamped., , Example 8.4, , Find i(t) in the circuit of Fig. 8.10. Assume that the circuit has reached, steady state at t 0., Solution:, For t 6 0, the switch is closed. The capacitor acts like an open circuit, while the inductor acts like a shunted circuit. The equivalent circuit is, shown in Fig. 8.11(a). Thus, at t 0,, i(0) , , 10, 1 A,, 46, , v(0) 6i(0) 6 V
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ale29559_ch08.qxd, , 07/08/2008, , 11:15 AM, , Page 325, , The Source-Free Series RLC Circuit, , 8.3, , 4Ω, , t=0, , i, , 4Ω, , +, v, −, , 0.02 F, 10 V, , i, , i, , +, −, , 325, , 6Ω, +, −, , 10 V, 3Ω, , +, v, −, , 0.5 H, , Figure 8.10, , 6Ω, , 0.02 F, , (a), , For Example 8.4., , Figure 8.11, The circuit in Fig. 8.10: (a) for t 6 0, (b) for t 7 0., , where i(0) is the initial current through the inductor and v(0) is the, initial voltage across the capacitor., For t 7 0, the switch is opened and the voltage source is disconnected. The equivalent circuit is shown in Fig. 8.11(b), which is a sourcefree series RLC circuit. Notice that the 3- and 6- resistors, which are, in series in Fig. 8.10 when the switch is opened, have been combined to, give R 9 in Fig. 8.11(b). The roots are calculated as follows:, a, , R, 9, 1 9,, 2L, 2(2), , 0 , , 1, 2LC, , , , 1, 212, , 501, , 10, , s1,2 a 2a2 20 9 281 100, or, s1,2 9 j 4.359, Hence, the response is underdamped (a 6 ); that is,, i(t) e9t(A1 cos 4.359t A2 sin 4.359 t), , (8.4.1), , We now obtain A1 and A2 using the initial conditions. At t 0,, i(0) 1 A1, , (8.4.2), , From Eq. (8.5),, di, 1, 2, [Ri(0) v(0)] 2[9(1) 6] 6 A/s, dt t0, L, , (8.4.3), , Note that v(0) V0 6 V is used, because the polarity of v in, Fig. 8.11(b) is opposite that in Fig. 8.8. Taking the derivative of i(t) in, Eq. (8.4.1),, di, 9e9t(A1 cos 4.359t A2 sin 4.359t), dt, e9t(4.359)(A1 sin 4.359t A2 cos 4.359t), Imposing the condition in Eq. (8.4.3) at t 0 gives, 6 9(A1 0) 4.359(0 A2), But A1 1 from Eq. (8.4.2). Then, 6 9 4.359A2, , 1, , A2 0.6882, , Substituting the values of A1 and A2 in Eq. (8.4.1) yields the, complete solution as, i(t) e9t( cos 4.359t 0.6882 sin 4.359t) A, , 9Ω, , +, v, −, , 0.5 H, , (b)
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ale29559_ch08.qxd, , 07/08/2008, , 11:15 AM, , Page 326, , Chapter 8, , 326, , The circuit in Fig. 8.12 has reached steady state at t 0. If the makebefore-break switch moves to position b at t 0, calculate i(t) for, t 7 0., , Practice Problem 8.4, 10 Ω, , a, , b, , 1, 9, , F, , Answer: e2.5t(5 cos 1.6583t 7.5378 sin 1.6583t) A., , t=0, , 50 V, , Second-Order Circuits, , i(t), , +, −, , 5Ω, 1H, , 8.4, , Figure 8.12, For Practice Prob. 8.4., , Parallel RLC circuits find many practical applications, notably in communications networks and filter designs., Consider the parallel RLC circuit shown in Fig. 8.13. Assume initial inductor current I0 and initial capacitor voltage V0,, , v, +, R, , v, −, , +, L, , I0 v, , The Source-Free Parallel RLC Circuit, , C, , +, V0, −, , i(0) I0 , , A source-free parallel RLC circuit., , , , 0, , v(t) dt, , (8.27a), , , , v(0) V0, , −, , Figure 8.13, , 1, L, , (8.27b), , Since the three elements are in parallel, they have the same voltage v, across them. According to passive sign convention, the current is entering each element; that is, the current through each element is leaving, the top node. Thus, applying KCL at the top node gives, v, 1, , R, L, , , , t, , , , v dt C, , dv, 0, dt, , (8.28), , Taking the derivative with respect to t and dividing by C results in, d 2v, 1 dv, 1, , , v0, 2, RC dt, LC, dt, , (8.29), , We obtain the characteristic equation by replacing the first derivative, by s and the second derivative by s2. By following the same reasoning, used in establishing Eqs. (8.4) through (8.8), the characteristic equation is obtained as, s2 , , 1, 1, s, 0, RC, LC, , (8.30), , The roots of the characteristic equation are, s1,2 , , 1, 1 2, 1, a, b , 2RC, B 2RC, LC, , or, s1,2 a 2a2 20, , (8.31), , where, a, , 1, ,, 2RC, , 0 , , 1, 2LC, , (8.32)
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ale29559_ch08.qxd, , 07/08/2008, , 11:15 AM, , Page 327, , 8.4, , The Source-Free Parallel RLC Circuit, , The names of these terms remain the same as in the preceding section,, as they play the same role in the solution. Again, there are three possible solutions, depending on whether a 7 0, a 0, or a 6 0., Let us consider these cases separately., , Overdamped Case (A 0), , From Eq. (8.32), a 7 0 when L 7 4R2C. The roots of the characteristic equation are real and negative. The response is, v(t) A1es1t A2es2t, , (8.33), , Critically Damped Case (A 0), , For a 0, L 4R2C. The roots are real and equal so that the, response is, v(t) (A1 A2t)ea t, , (8.34), , Underdamped Case (A 0), , When a 6 0, L 6 4R2C. In this case the roots are complex and may, be expressed as, s1,2 a jd, , (8.35), , d 220 a2, , (8.36), , v(t) ea t(A1 cos dt A2 sin dt), , (8.37), , where, , The response is, , The constants A1 and A2 in each case can be determined from the, initial conditions. We need v(0) and dv(0)dt. The first term is known, from Eq. (8.27b). We find the second term by combining Eqs. (8.27), and (8.28), as, V0, dv(0), I0 C, 0, R, dt, or, (V0 RI0), dv(0), , dt, RC, , (8.38), , The voltage waveforms are similar to those shown in Fig. 8.9 and will, depend on whether the circuit is overdamped, underdamped, or critically damped., Having found the capacitor voltage v(t) for the parallel RLC circuit as shown above, we can readily obtain other circuit quantities such, as individual element currents. For example, the resistor current is, iR vR and the capacitor voltage is vC C dvdt. We have selected, the capacitor voltage v(t) as the key variable to be determined first in, order to take advantage of Eq. (8.1a). Notice that we first found the, inductor current i(t) for the RLC series circuit, whereas we first found, the capacitor voltage v(t) for the parallel RLC circuit., , 327
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ale29559_ch08.qxd, , 07/08/2008, , 328, , Example 8.5, , 11:15 AM, , Page 328, , Chapter 8, , Second-Order Circuits, , In the parallel circuit of Fig. 8.13, find v(t) for t 7 0, assuming, v(0) 5 V, i(0) 0, L 1 H, and C 10 mF. Consider these cases:, R 1.923 , R 5 , and R 6.25 ., Solution:, , ■ CASE 1 If R 1.923 ,, 1, 1, , 26, 2RC, 2 1.923 10 103, 1, 1, 0 , , 10, 2LC, 21 10 103, , a, , Since a 7 0 in this case, the response is overdamped. The roots of, the characteristic equation are, s1,2 a 2a2 20 2, 50, and the corresponding response is, v(t) A1e2t A2e50t, , (8.5.1), , We now apply the initial conditions to get A1 and A2., v(0) 5 A1 A2, , (8.5.2), , dv(0), v(0) Ri(0), 50, , , 260, dt, RC, 1.923 10 103, But differentiating Eq. (8.5.1),, dv, 2A1e2t 50A2e50t, dt, At t 0,, 260 2A1 50A2, , (8.5.3), , From Eqs. (8.5.2) and (8.5.3), we obtain A1 0.2083 and A2 5.208., Substituting A1 and A2 in Eq. (8.5.1) yields, v(t) 0.2083e2t 5.208e50t, , (8.5.4), , ■ CASE 2 When R 5 ,, a, , 1, 1, , 10, 2RC, 2 5 10 103, , while 0 10 remains the same. Since a 0 10, the response is, critically damped. Hence, s1 s2 10, and, v(t) (A1 A2t)e10t, , (8.5.5), , To get A1 and A2, we apply the initial conditions, v(0) 5 A1, dv(0), v(0) Ri(0), 50, 100, , , dt, RC, 5 10 103, But differentiating Eq. (8.5.5),, dv, (10A1 10A2t A2)e10t, dt, , (8.5.6)
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ale29559_ch08.qxd, , 07/08/2008, , 11:15 AM, , Page 329, , 8.4, , The Source-Free Parallel RLC Circuit, , At t 0,, 100 10A1 A2, , (8.5.7), , From Eqs. (8.5.6) and (8.5.7), A1 5 and A2 50. Thus,, v(t) (5 50t)e10t V, , (8.5.8), , ■ CASE 3 When R 6.25 ,, a, , 1, 1, , 8, 2RC, 2 6.25 10 103, , while 0 10 remains the same. As a 6 0 in this case, the response, is underdamped. The roots of the characteristic equation are, s1,2 a 2a2 20 8 j6, Hence,, v(t) (A1 cos 6t A2 sin 6t)e8t, , (8.5.9), , We now obtain A1 and A2, as, v(0) 5 A1, , (8.5.10), , dv(0), v(0) Ri(0), 50, , , 80, dt, RC, 6.25 10 103, But differentiating Eq. (8.5.9),, dv, (8A1 cos 6t 8A2 sin 6t 6A1 sin 6t 6A2 cos 6t)e8t, dt, At t 0,, 80 8A1 6A2, , (8.5.11), , From Eqs. (8.5.10) and (8.5.11), A1 5 and A2 6.667. Thus,, v(t) (5 cos 6t 6.667 sin 6t)e8t, , (8.5.12), , Notice that by increasing the value of R, the degree of damping, decreases and the responses differ. Figure 8.14 plots the three cases., v (t) V, 5, 4, , 3, , 2, , 1, , Overdamped, Critically damped, , 0, Underdamped, , –1, 0, , 0.5, , 1, , Figure 8.14, For Example 8.5: responses for three degrees of damping., , 1.5 t (s), , 329
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ale29559_ch08.qxd, , 07/08/2008, , 11:15 AM, , 330, , Practice Problem 8.5, , Page 330, , Chapter 8, , Second-Order Circuits, , In Fig. 8.13, let R 2 , L 0.4 H, C 25 mF, v(0) 0, i(0) 10 mA., Find v(t) for t 7 0., Answer: 400te10t u(t) mV., , Example 8.6, , Find v(t) for t 7 0 in the RLC circuit of Fig. 8.15., 30 Ω, , 40 V, , +, −, , 0.4 H, , i, , 50 Ω, , t=0, , 20 F, , +, v, −, , Figure 8.15, For Example 8.6., , Solution:, When t 6 0, the switch is open; the inductor acts like a short circuit, while the capacitor behaves like an open circuit. The initial voltage across, the capacitor is the same as the voltage across the 50- resistor; that is,, v(0) , , 50, 5, (40) 40 25 V, 30 50, 8, , (8.6.1), , The initial current through the inductor is, i(0) , , 40, 0.5 A, 30 50, , The direction of i is as indicated in Fig. 8.15 to conform with the, direction of I0 in Fig. 8.13, which is in agreement with the convention, that current flows into the positive terminal of an inductor (see Fig. 6.23)., We need to express this in terms of dvdt, since we are looking for v., dv(0), v(0) Ri(0), 25 50 0.5, 0, , , dt, RC, 50 20 106, , (8.6.2), , When t 7 0, the switch is closed. The voltage source along with the, 30- resistor is separated from the rest of the circuit. The parallel RLC, circuit acts independently of the voltage source, as illustrated in Fig. 8.16., Next, we determine that the roots of the characteristic equation are, 1, 1, , 500, 2RC, 2 50 20 106, 1, 1, 0 , , 354, 2LC, 20.4 20 106, a, , s1,2 a 2a2 20, 500 2250,000 124,997.6 500 354, or, s1 854,, , s2 146
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 331, , Step Response of a Series RLC Circuit, , 8.5, 30 Ω, , 40 V, , 331, , 0.4 H, , +, −, , 20 F, , 50 Ω, , Figure 8.16, The circuit in Fig. 8.15 when t 7 0. The parallel, RLC circuit on the right-hand side acts independently, of the circuit on the left-hand side of the junction., , Since a 7 0, we have the overdamped response, v(t) A1e854t A2e146t, , (8.6.3), , At t 0, we impose the condition in Eq. (8.6.1),, v(0) 25 A1 A2, , 1, , A2 25 A1, , (8.6.4), , Taking the derivative of v(t) in Eq. (8.6.3),, dv, 854A1e854t 146A2e146t, dt, Imposing the condition in Eq. (8.6.2),, dv(0), 0 854A1 146A2, dt, or, 0 854A1 146A2, , (8.6.5), , Solving Eqs. (8.6.4) and (8.6.5) gives, A1 5.156,, , A2 30.16, , Thus, the complete solution in Eq. (8.6.3) becomes, v(t) 5.156e854t 30.16e146t V, , Refer to the circuit in Fig. 8.17. Find v(t) for t 7 0., , Practice Problem 8.6, , Answer: 100(e10t e2.5t) V., , t=0, , 20 Ω, , 3A, , 8.5, , 4 mF, , 10 H, , +, v, −, , Step Response of a Series RLC Circuit, , As we learned in the preceding chapter, the step response is obtained, by the sudden application of a dc source. Consider the series RLC circuit shown in Fig. 8.18. Applying KVL around the loop for t 7 0,, L, , di, Ri v Vs, dt, , Figure 8.17, For Practice Prob. 8.6., t=0, , dv, dt, , L, , i, , (8.39), Vs, , +, −, , But, iC, , R, , C, , +, v, −, , Figure 8.18, Step voltage applied to a series RLC circuit.
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ale29559_ch08.qxd, , 332, , 07/08/2008, , 11:16 AM, , Page 332, , Chapter 8, , Second-Order Circuits, , Substituting for i in Eq. (8.39) and rearranging terms,, Vs, d 2v, R dv, v, , , , 2, L dt, LC, LC, dt, , (8.40), , which has the same form as Eq. (8.4). More specifically, the coefficients are the same (and that is important in determining the frequency, parameters) but the variable is different. (Likewise, see Eq. (8.47).), Hence, the characteristic equation for the series RLC circuit is not, affected by the presence of the dc source., The solution to Eq. (8.40) has two components: the transient, response vt(t) and the steady-state response vss(t); that is,, v(t) vt (t) vss (t), , (8.41), , The transient response vt (t) is the component of the total response that, dies out with time. The form of the transient response is the same as the, form of the solution obtained in Section 8.3 for the source-free circuit,, given by Eqs. (8.14), (8.21), and (8.26). Therefore, the transient repsonse, vt (t) for the overdamped, underdamped, and critically damped cases are:, vt (t) A1es1t A2es2t, at, , vt (t) (A1 A2t)e, , (Overdamped), , (8.42a), , (Critically damped), , (8.42b), , vt (t) (A1 cos d t A2 sin d t)eat, , (Underdamped), , (8.42c), , The steady-state response is the final value of v(t). In the circuit in, Fig. 8.18, the final value of the capacitor voltage is the same as the, source voltage Vs. Hence,, vss(t) v() Vs, , (8.43), , Thus, the complete solutions for the overdamped, underdamped, and, critically damped cases are:, v(t) Vs A1es1t A2es2t, v(t) Vs (A1 A2t)ea t, , (Overdamped), , (8.44a), , (Critically damped), , (8.44b), , at, , v(t) Vs (A1 cos d t A2 sin d t)e, , (Underdamped), , (8.44c), , The values of the constants A1 and A2 are obtained from the initial conditions: v(0) and dv(0)dt. Keep in mind that v and i are, respectively,, the voltage across the capacitor and the current through the inductor., Therefore, Eq. (8.44) only applies for finding v. But once the capacitor voltage vC v is known, we can determine i C dvdt, which is, the same current through the capacitor, inductor, and resistor. Hence,, the voltage across the resistor is vR iR, while the inductor voltage is, vL L didt., Alternatively, the complete response for any variable x(t) can be, found directly, because it has the general form, x(t) xss(t) xt(t), , (8.45), , where the xss x() is the final value and xt(t) is the transient response., The final value is found as in Section 8.2. The transient response has, the same form as in Eq. (8.42), and the associated constants are determined from Eq. (8.44) based on the values of x(0) and dx(0)dt.
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 333, , 8.5, , Step Response of a Series RLC Circuit, , 333, , Example 8.7, , For the circuit in Fig. 8.19, find v(t) and i(t) for t 7 0 . Consider these, cases: R 5 , R 4 , and R 1 ., , R, , Solution:, , i, , ■ CASE 1 When R 5 . For t 6 0, the switch is closed for a, long time. The capacitor behaves like an open circuit while the, inductor acts like a short circuit. The initial current through the, inductor is, i(0) , , 24, 4A, 51, , and the initial voltage across the capacitor is the same as the voltage, across the 1- resistor; that is,, v(0) 1i(0) 4 V, For t 7 0, the switch is opened, so that we have the 1- resistor, disconnected. What remains is the series RLC circuit with the voltage, source. The characteristic roots are determined as follows:, 5, R, , 2.5,, 2L, 21, , a, , 0 , , 1, 2LC, , , , 1, 21 0.25, , 2, , s1,2 a 2a2 20 1, 4, Since a 7 0, we have the overdamped natural response. The total, response is therefore, v(t) vss (A1et A2e4t), where vss is the steady-state response. It is the final value of the, capacitor voltage. In Fig. 8.19, vf 24 V. Thus,, v(t) 24 (A1et A2e4t), , (8.7.1), , We now need to find A1 and A2 using the initial conditions., v(0) 4 24 A1 A2, or, 20 A1 A2, , (8.7.2), , The current through the inductor cannot change abruptly and is the, same current through the capacitor at t 0 because the inductor and, capacitor are now in series. Hence,, i(0) C, , 1H, , dv(0), 4, dt, , 1, , dv(0), 4, 4, , 16, dt, C, 0.25, , Before we use this condition, we need to take the derivative of v in, Eq. (8.7.1)., dv, A1et 4A2e4t, dt, , (8.7.3), , dv(0), 16 A1 4A2, dt, , (8.7.4), , At t 0,, , 24 V, , +, −, , Figure 8.19, For Example 8.7., , 0.25 F, , t=0, +, v, −, , 1Ω
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ale29559_ch08.qxd, , 334, , 07/08/2008, , 11:16 AM, , Page 334, , Chapter 8, , Second-Order Circuits, , From Eqs. (8.7.2) and (8.7.4), A1 643 and A2 43. Substituting, A1 and A2 in Eq. (8.7.1), we get, 4, v(t) 24 (16et e4t) V, 3, , (8.7.5), , Since the inductor and capacitor are in series for t 7 0, the inductor, current is the same as the capacitor current. Hence,, i(t) C, , dv, dt, , Multiplying Eq. (8.7.3) by C 0.25 and substituting the values of A1, and A2 gives, 4, i(t) (4et e4t) A, 3, , (8.7.6), , Note that i(0) 4 A, as expected., , ■ CASE 2 When R 4 . Again, the initial current through the, inductor is, i(0) , , 24, 4.8 A, 41, , and the initial capacitor voltage is, v(0) 1i(0) 4.8 V, For the characteristic roots,, a, , R, 4, , 2, 2L, 21, , while 0 2 remains the same. In this case, s1 s2 a 2, and, we have the critically damped natural response. The total response is, therefore, v(t) vss (A1 A2t)e2t, and, as before vss 24 V,, v(t) 24 (A1 A2t)e2t, , (8.7.7), , To find A1 and A2, we use the initial conditions. We write, v(0) 4.8 24 A1, , 1, , A1 19.2, , (8.7.8), , Since i(0) C dv(0)dt 4.8 or, dv(0), 4.8, , 19.2, dt, C, From Eq. (8.7.7),, dv, (2A1 2tA2 A2)e2t, dt, , (8.7.9), , dv(0), 19.2 2A1 A2, dt, , (8.7.10), , At t 0,
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 335, , Step Response of a Series RLC Circuit, , 8.5, , From Eqs. (8.7.8) and (8.7.10), A1 19.2 and A2 19.2. Thus,, Eq. (8.7.7) becomes, v(t) 24 19.2(1 t)e2t V, , (8.7.11), , The inductor current is the same as the capacitor current; that is,, i(t) C, , dv, dt, , Multiplying Eq. (8.7.9) by C 0.25 and substituting the values of A1, and A2 gives, i(t) (4.8 9.6t)e2t A, (8.7.12), Note that i(0) 4.8 A, as expected., , ■ CASE 3 When R 1 . The initial inductor current is, i(0) , , 24, 12 A, 11, , and the initial voltage across the capacitor is the same as the voltage, across the 1- resistor,, v(0) 1i(0) 12 V, R, 1, a, , 0.5, 2L, 21, Since a 0.5 6 0 2, we have the underdamped response, s1,2 a 2a2 20 0.5 j1.936, The total response is therefore, v(t) 24 (A1 cos 1.936t A2 sin 1.936t)e0.5t, , (8.7.13), , We now determine A1 and A2. We write, v(0) 12 24 A1, , 1, , A1 12, , (8.7.14), , Since i(0) C dv(0)dt 12,, dv(0), 12, , 48, dt, C, , (8.7.15), , But, dv, e0.5t(1.936A1 sin 1.936t 1.936 A2 cos 1.936t), dt, (8.7.16), 0.5e0.5t(A1 cos 1.936t A2 sin 1.936t), At t 0,, dv(0), 48 (0 1.936 A2) 0.5(A1 0), dt, Substituting A1 12 gives A2 21.694, and Eq. (8.7.13) becomes, v(t) 24 (21.694 sin 1.936t 12 cos 1.936t)e0.5t V (8.7.17), The inductor current is, i(t) C, , dv, dt, , 335
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 336, , Chapter 8, , 336, , Second-Order Circuits, , Multiplying Eq. (8.7.16) by C 0.25 and substituting the values of A1, and A2 gives, i(t) (3.1 sin 1.936t 12 cos 1.936t)e0.5t A, , (8.7.18), , Note that i(0) 12 A, as expected., Figure 8.20 plots the responses for the three cases. From this, figure, we observe that the critically damped response approaches the, step input of 24 V the fastest., v (t) V, 40, Underdamped, , 35, 30, , Critically damped, , 35, 20, 15, Overdamped, , 10, 5, 0, 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , t (s), , Figure 8.20, For Example 8.7: response for three degrees of damping., , Practice Problem 8.7, , Having been in position a for a long time, the switch in Fig. 8.21 is, moved to position b at t 0. Find v(t) and vR(t) for t 7 0., 1Ω, , 12 V, , +, −, , a, , 2Ω, , 1, 40, , F, , b, , 2.5 H, , t=0, +, v, −, , 10 Ω, − vR +, 10 V, , +, −, , Figure 8.21, For Practice Prob. 8.7., , Answer: 10 (1.1547 sin 3.464t 2 cos 3.464t)e2t V,, 2.31e2t sin 3.464t V., , i, Is, , t=0, , R, , L, , C, , Figure 8.22, Parallel RLC circuit with an applied, current., , +, v, −, , 8.6, , Step Response of a Parallel RLC Circuit, , Consider the parallel RLC circuit shown in Fig. 8.22. We want to find, i due to a sudden application of a dc current. Applying KCL at the top, node for t 7 0,, dv, v, iC, Is, R, dt, , (8.46)
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 337, , Step Response of a Parallel RLC Circuit, , 8.6, , 337, , But, vL, , di, dt, , Substituting for v in Eq. (8.46) and dividing by LC, we get, Is, d 2i, 1 di, i, , , , RC dt, LC, LC, dt 2, , (8.47), , which has the same characteristic equation as Eq. (8.29)., The complete solution to Eq. (8.47) consists of the transient, response it(t) and the steady-state response iss; that is,, i(t) it (t) iss (t), , (8.48), , The transient response is the same as what we had in Section 8.4. The, steady-state response is the final value of i. In the circuit in Fig. 8.22,, the final value of the current through the inductor is the same as the, source current Is. Thus,, i(t) Is A1es1t A2es2t, a t, , i(t) Is (A1 A2t)e, , (Overdamped), (Critically damped), a t, , i(t) Is (A1 cos d t A2 sin d t)e, , (8.49), , (Underdamped), , The constants A1 and A2 in each case can be determined from the initial, conditions for i and didt. Again, we should keep in mind that Eq. (8.49), only applies for finding the inductor current i. But once the inductor, current iL i is known, we can find v L didt, which is the same, voltage across inductor, capacitor, and resistor. Hence, the current, through the resistor is iR vR, while the capacitor current is, iC C dvdt. Alternatively, the complete response for any variable x(t), may be found directly, using, x(t) xss(t) xt(t), , (8.50), , where xss and xt are its final value and transient response, respectively., , Example 8.8, , In the circuit of Fig. 8.23, find i(t) and iR(t) for t 7 0., 20 Ω, , t=0, i, 4A, , 20 H, , iR, 20 Ω, , 8 mF, , +, v, −, , +, −, , 30u(–t) V, , Figure 8.23, For Example 8.8., , Solution:, For t 6 0, the switch is open, and the circuit is partitioned into two independent subcircuits. The 4-A current flows through the inductor, so that, i(0) 4 A
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ale29559_ch08.qxd, , 338, , 07/08/2008, , 11:16 AM, , Page 338, , Chapter 8, , Second-Order Circuits, , Since 30u(t) 30 when t 6 0 and 0 when t 7 0, the voltage source, is operative for t 6 0. The capacitor acts like an open circuit and the, voltage across it is the same as the voltage across the 20- resistor, connected in parallel with it. By voltage division, the initial capacitor, voltage is, v(0) , , 20, (30) 15 V, 20 20, , For t 7 0, the switch is closed, and we have a parallel RLC circuit, with a current source. The voltage source is zero which means it acts, like a short-circuit. The two 20- resistors are now in parallel. They, are combined to give R 20 20 10 . The characteristic roots are, determined as follows:, 1, 1, , 6.25, 2RC, 2 10 8 103, 1, 1, 0 , , 2.5, 2LC, 220 8 103, , a, , s1,2 a 2a2 20 6.25 239.0625 6.25, 6.25 5.7282, or, s1 11.978,, , s2 0.5218, , Since a 7 0, we have the overdamped case. Hence,, i(t) Is A1e11.978t A2e0.5218t, , (8.8.1), , where Is 4 is the final value of i(t). We now use the initial conditions, to determine A1 and A2. At t 0,, i(0) 4 4 A1 A2, , 1, , A2 A1, , (8.8.2), , Taking the derivative of i(t) in Eq. (8.8.1),, di, 11.978A1e11.978t 0.5218A2e0.5218t, dt, so that at t 0,, di(0), 11.978A1 0.5218A2, dt, , (8.8.3), , But, L, , di(0), v(0) 15, dt, , 1, , di(0), 15, 15, , , 0.75, dt, L, 20, , Substituting this into Eq. (8.8.3) and incorporating Eq. (8.8.2), we get, 0.75 (11.978 0.5218)A2, , 1, , A2 0.0655, , Thus, A1 0.0655 and A2 0.0655. Inserting A1 and A2 in Eq. (8.8.1), gives the complete solution as, i(t) 4 0.0655(e0.5218t e11.978t) A, From i(t), we obtain v(t) L didt and, iR(t) , , v(t), L di, , 0.785e11.978t 0.0342e0.5218t A, 20, 20 dt
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 339, , 8.7, , General Second-Order Circuits, , 339, , Practice Problem 8.8, , Find i(t) and v(t) for t 7 0 in the circuit of Fig. 8.24., Answer: 12(1 cos t) A, 60 sin t V., , i, +, v, −, , 12u(t) A, , 8.7, , General Second-Order Circuits, , Figure 8.24, , Now that we have mastered series and parallel RLC circuits, we are, prepared to apply the ideas to any second-order circuit having one or, more independent sources with constant values. Although the series and, parallel RLC circuits are the second-order circuits of greatest interest,, other second-order circuits including op amps are also useful. Given a, second-order circuit, we determine its step response x(t) (which may, be voltage or current) by taking the following four steps:, 1. We first determine the initial conditions x(0) and dx(0)dt and the, final value x(), as discussed in Section 8.2., 2. We turn off the independent sources and find the form of the transient response xt(t) by applying KCL and KVL. Once a second-order, differential equation is obtained, we determine its characteristic, roots. Depending on whether the response is overdamped, critically, damped, or underdamped, we obtain xt(t) with two unknown constants as we did in the previous sections., 3. We obtain the steady-state response as, xss (t) x(), , 5H, , 0.2 F, , For Practice Prob. 8.8., A circuit may look complicated at first., But once the sources are turned off in, an attempt to find the form of the transient response, it may be reducible to, a first-order circuit, when the storage, elements can be combined, or to a, parallel/series RLC circuit. If it is reducible to a first-order circuit, the solution becomes simply what we had in, Chapter 7. If it is reducible to a parallel, or series RLC circuit, we apply the techniques of previous sections in this, chapter., , (8.51), , where x() is the final value of x, obtained in step 1., 4. The total response is now found as the sum of the transient, response and steady-state response, x(t) xt(t) xss(t), , (8.52), , We finally determine the constants associated with the transient, response by imposing the initial conditions x(0) and dx(0)dt,, determined in step 1., We can apply this general procedure to find the step response of, any second-order circuit, including those with op amps. The following, examples illustrate the four steps., , Problems in this chapter can also be, solved by using Laplace transforms,, which are covered in Chapters 15, and 16., , Example 8.9, , Find the complete response v and then i for t 7 0 in the circuit of, Fig. 8.25., Solution:, We first find the initial and final values. At t 0, the circuit is at steady, state. The switch is open; the equivalent circuit is shown in Fig. 8.26(a)., It is evident from the figure that, , , v(0 ) 12 V,, , , , i(0 ) 0, , i(0) i(0) 0, , (8.9.1), , i, , 1H, 2Ω, , 12 V, , At t 0, the switch is closed; the equivalent circuit is in Fig. 8.26(b)., By the continuity of capacitor voltage and inductor current, we know that, v(0) v(0) 12 V,, , 4Ω, , +, −, , 1, 2, , t=0, , Figure 8.25, For Example 8.9., , F, , +, v, −
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 340, , Chapter 8, , 340, , To get dv(0)dt, we use C dvdt iC or dvdt iCC. Applying, KCL at node a in Fig. 8.26(b),, , i, , 4Ω, , Second-Order Circuits, , +, 12 V, , +, −, , i(0) iC (0) , , v, −, , 0 iC (0) , , 12, 2, , v(0), 2, iC (0) 6 A, , 1, , (a), , Hence,, 4Ω, , 1H, , i, , dv(0), 6, , 12 V/s, dt, 0.5, , a, iC, , 12 V, , +, −, , 2Ω, , +, v, −, , 0.5 F, , The final values are obtained when the inductor is replaced by a short, circuit and the capacitor by an open circuit in Fig. 8.26(b), giving, i() , , (b), , Figure 8.26, Equivalent circuit of the circuit in Fig. 8.25, for: (a) t 6 0, (b) t 7 0., , 12, 2 A,, 42, , i, , 1H, , v, , (8.9.3), , v, 1 dv, , 2, 2 dt, , (8.9.4), , Applying KVL to the left mesh results in, , a, 2Ω, , v() 2i() 4 V, , Next, we obtain the form of the transient response for t 7 0. By, turning off the 12-V voltage source, we have the circuit in Fig. 8.27., Applying KCL at node a in Fig. 8.27 gives, i, , 4Ω, , (8.9.2), , +, v, −, , Figure 8.27, , 1, 2, , 4i 1, , F, , di, v0, dt, , (8.9.5), , Since we are interested in v for the moment, we substitute i from, Eq. (8.9.4) into Eq. (8.9.5). We obtain, , Obtaining the form of the transient, response for Example 8.9., , 2v 2, , dv, 1 dv, 1 d 2v, , , v0, dt, 2 dt, 2 dt 2, , or, dv, d 2v, 5, 6v 0, 2, dt, dt, From this, we obtain the characteristic equation as, s2 5s 6 0, with roots s 2 and s 3. Thus, the natural response is, vn(t) Ae2t Be3t, , (8.9.6), , where A and B are unknown constants to be determined later. The, steady-state response is, vss (t) v() 4, , (8.9.7), , v(t) vt vss 4 Ae2t Be3t, , (8.9.8), , The complete response is, , We now determine A and B using the initial values. From Eq. (8.9.1),, v(0) 12. Substituting this into Eq. (8.9.8) at t 0 gives, 12 4 A B, , 1, , AB8, , (8.9.9)
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 341, , 8.7, , General Second-Order Circuits, , 341, , Taking the derivative of v in Eq. (8.9.8),, dv, 2Ae2t 3Be3t, dt, , (8.9.10), , Substituting Eq. (8.9.2) into Eq. (8.9.10) at t 0 gives, 12 2A 3B, , 1, , 2A 3B 12, , (8.9.11), , From Eqs. (8.9.9) and (8.9.11), we obtain, A 12,, , B 4, , so that Eq. (8.9.8) becomes, v(t) 4 12e2t 4e3t V,, , t 7 0, , (8.9.12), , From v, we can obtain other quantities of interest by referring to, Fig. 8.26(b). To obtain i, for example,, i, , v, 1 dv, , 2 6e2t 2e3t 12e2t 6e3t, 2, 2 dt, t 7 0, 2 6e2t 4e3t A,, , (8.9.13), , Notice that i(0) 0, in agreement with Eq. (8.9.1)., , Practice Problem 8.9, , Determine v and i for t 7 0 in the circuit of Fig. 8.28. (See comments, about current sources in Practice Prob. 7.5.), , 10 Ω, , Answer: 12(1 e5t) V, 3(1 e5t) A., , 4Ω, , 3A, , i, 1, 20, , +, v, −, , F, , 2H, , t=0, , Figure 8.28, For Practice Prob. 8.9., , Example 8.10, , Find vo(t) for t 7 0 in the circuit of Fig. 8.29., Solution:, This is an example of a second-order circuit with two inductors. We, first obtain the mesh currents i1 and i2, which happen to be the currents, through the inductors. We need to obtain the initial and final values of, these currents., For t 6 0, 7u(t) 0, so that i1(0) 0 i2(0). For t 7 0,, 7u(t) 7, so that the equivalent circuit is as shown in Fig. 8.30(a). Due, to the continuity of inductor current,, i1(0) i1(0) 0,, , i2(0) i2(0) 0, , (8.10.1), , vL 2(0) vo(0) 1[(i1(0) i2(0)] 0, , (8.10.2), , , , Applying KVL to the left loop in Fig. 8.30(a) at t 0 ,, 7 3i1(0) vL1(0) vo(0), , 3Ω, , 7u(t) V, , +, −, , 1, 2, , H, , 1Ω, i1, , Figure 8.29, For Example 8.10., , +, vo, −, , i2, 1, 5, , H
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 343, , 8.7, , General Second-Order Circuits, , 343, , where A and B are constants. The steady-state response is, 7, A, 3, , i1ss i1() , , (8.10.10), , From Eqs. (8.10.9) and (8.10.10), we obtain the complete response as, i1(t) , , 7, Ae3t Be10t, 3, , (8.10.11), , We finally obtain A and B from the initial values. From Eqs. (8.10.1), and (8.10.11),, 0, , 7, AB, 3, , (8.10.12), , Taking the derivative of Eq. (8.10.11), setting t 0 in the derivative,, and enforcing Eq. (8.10.3), we obtain, 14 3A 10B, , (8.10.13), , From Eqs. (8.10.12) and (8.10.13), A 43 and B 1. Thus,, i1(t) , , 7, 4, e3t e10t, 3, 3, , (8.10.14), , We now obtain i2 from i1. Applying KVL to the left loop in, Fig. 8.30(a) gives, 7 4i1 i2 , , 1 di1, 2 dt, , 1, , i2 7 4i1 , , 1 di1, 2 dt, , Substituting for i1 in Eq. (8.10.14) gives, 28, 16, e3t 4e10t 2e3t 5e10t, 3, 3, (8.10.15), 7, 10 3t, 10t, e, e, 3, 3, , i2(t) 7 , , From Fig. 8.29,, vo(t) 1[i1(t) i2(t)], , (8.10.16), , Substituting Eqs. (8.10.14) and (8.10.15) into Eq. (8.10.16) yields, vo(t) 2(e3t e10t), , (8.10.17), , Note that vo(0) 0, as expected from Eq. (8.10.2)., , For t 7 0, obtain vo(t) in the circuit of Fig. 8.32. (Hint: First find v1, and v2.), t, , Answer: 8(e, , Practice Problem 8.10, 1Ω, , v1, , 6t, , e, , 1Ω, , v2, , + vo −, , ) V, t 7 0., 20u(t) V, , +, −, , 1, 2, , F, , Figure 8.32, For Practice Prob. 8.10., , 1, 3, , F
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 346, , Chapter 8, , 346, , Second-Order Circuits, , Taking the derivative of Eq. (8.11.10),, dvo, et(A cos 2t B sin 2t 2A sin 2t 2B cos 2t), dt, Setting t 0 and incorporating Eq. (8.11.12), we obtain, 0 A 2B, , (8.11.14), , From Eqs. (8.11.13) and (8.11.14), A 10 and B 5. Thus, the, step response becomes, vo(t) 10 et(10 cos 2t 5 sin 2t) mV,, , Practice Problem 8.11, R1, , vs, , +, −, , In the op amp circuit shown in Fig. 8.34, vs 10u(t) V, find vo(t) for, t 7 0. Assume that R1 R2 10 k, C1 20 mF, and C2 100 mF., , R2, , +, –, , +, C2, , C1, , t 7 0, , Answer: (10 12.5et 2.5e5t) V, t 7 0., , vo, −, , 8.9, , Figure 8.34, For Practice Prob. 8.11., , PSpice Analysis of RLC Circuits, , RLC circuits can be analyzed with great ease using PSpice, just like, the RC or RL circuits of Chapter 7. The following two examples will, illustrate this. The reader may review Section D.4 in Appendix D on, PSpice for transient analysis., , Example 8.12, , The input voltage in Fig. 8.35(a) is applied to the circuit in Fig. 8.35(b)., Use PSpice to plot v(t) for 0 6 t 6 4 s., , vs, , Solution:, , 12, , 0, , t (s), , 2, (a), , 60 Ω, , vs, , +, −, , 3H, , 1, 27, , 60 Ω, , (b), , Figure 8.35, For Example 8.12., , F, , +, v, −, , 1. Define. As true with most textbook problems, the problem is, clearly defined., 2. Present. The input is equal to a single square wave of, amplitude 12 V with a period of 2 s. We are asked to plot the, output, using PSpice., 3. Alternative. Since we are required to use PSpice, that is the, only alternative for a solution. However, we can check it using, the technique illustrated in Section 8.5 (a step response for a, series RLC circuit)., 4. Attempt. The given circuit is drawn using Schematics as in, Fig. 8.36. The pulse is specified using VPWL voltage source,, but VPULSE could be used instead. Using the piecewise linear, function, we set the attributes of VPWL as T1 0, V1 0,, T2 0.001, V2 12, and so forth, as shown in Fig. 8.36., Two voltage markers are inserted to plot the input and output, voltages. Once the circuit is drawn and the attributes are set,, we select Analysis/Setup/Transient to open up the Transient, Analysis dialog box. As a parallel RLC circuit, the roots of the, characteristic equation are 1 and 9. Thus, we may set Final, Time as 4 s (four times the magnitude of the lower root). When
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 347, , PSPice Analysis of RLC Circuits, , 8.9, , 347, , the schematic is saved, we select Analysis/Simulate and obtain, the plots for the input and output voltages under the PSpice A/D, window as shown in Fig. 8.37., 12 V, 10 V, V, , T1=0, T2=0.0001, T3=2, T4=2.0001, , V1=0, V2=12, V3=12, V4=0, , V, R1, , L1, , 60, , 3H, , 8 V, 6 V, 4 V, , +, −, , V1, , R2, , 60, , 0.03703, , C1, , 2 V, 0 V, 0s, 2.0s, 1.0s, V(L1:2) V(R1:1), Time, , Figure 8.36, , Figure 8.37, , Schematic for the circuit in Fig. 8.35(b)., , For Example 8.12: input and output., , Now we check using the technique from Section 8.5. We, can start by realizing the Thevenin equivalent for the resistorsource combination is VTh 122 (the open circuit voltage, divides equally across both resistors) 6 V. The equivalent, resistance is 30 (60 60). Thus, we can now solve for the, response using R 30 , L 3 H, and C (127) F., We first need to solve for a and 0:, a R(2L) 306 5, , and, , 0 , , 1, 1, 3, B 27, , 3, , Since 5 is greater than 3, we have the overdamped case, s1,2 5 252 9 1, 9,, i(t) C, where, , v(0) 0,, v() 6 V,, , i(0) 0, , dv(t), ,, dt, , v(t) A1et A2e9t 6, v(0) 0 A1 A2 6, i(0) 0 C(A1 9A2), , which yields A1 9A2. Substituting this into the above, we get, 0 9A2 A2 6, or A2 0.75 and A1 6.75., v(t) (6.75e t 0.75e 9t 6) u(t) V for all 0 6 t 6 2 s., At t 1 s, v(1) 6.75e1 0.75e9 2.483 0.0001 , 6 3.552 V. At t 2 s, v(2) 6.75e2 0 6 5.086 V., Note that from 2 6 t 6 4 s, VTh 0, which implies that, v() 0. Therefore, v(t) (A3e(t2) A4e9(t2))u(t 2) V., At t 2 s, A3 A4 5.086., i(t) , , (A3e(t2) 9A4e9(t2)), 27, , 3.0s, , 4.0s
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 348, , Chapter 8, , 348, , Second-Order Circuits, , and, i(2) , , (6.75e2 6.75e18), 33.83 mA, 27, , Therefore, A3 9A4 0.9135., Combining the two equations, we get A3 9(5.086 A3) , 0.9135, which leads to A3 5.835 and A4 0.749., v(t) (5.835e (t2) 0.749e 9(t2)) u (t 2) V, At t 3 s, v(3) (2.147 0) 2.147 V. At t 4 s, v(4) , 0.7897 V., 5. Evaluate. A check between the values calculated above and the, plot shown in Figure 8.37 shows good agreement within the, obvious level of accuracy., 6. Satisfactory? Yes, we have agreement and the results can be, presented as a solution to the problem., , Practice Problem 8.12, 5Ω, i, vs, , +, −, , Figure 8.38, , 1 mF, , 2H, , Find i(t) using PSpice for 0 6 t 6 4 s if the pulse voltage in Fig. 8.35(a), is applied to the circuit in Fig. 8.38., Answer: See Fig. 8.39., 3.0 A, , 2.0 A, , For Practice Prob. 8.12., 1.0 A, , 0 A, 0 s, , 1.0 s, I(L1), , 2.0 s, , 3.0 s, , 4.0 s, , Time, , Figure 8.39, Plot of i(t) for Practice Prob. 8.12., , Example 8.13, , For the circuit in Fig. 8.40, use PSpice to obtain i(t) for 0 6 t 6 3 s., a, t=0, i(t), , b, 4A, , 5Ω, , 6Ω, , 1, 42, , F, , 7H, , Figure 8.40, For Example 8.13., , Solution:, When the switch is in position a, the 6- resistor is redundant. The, schematic for this case is shown in Fig. 8.41(a). To ensure that current
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 349, , 8.9, 0.0000, , PSPice Analysis of RLC Circuits, , 349, , 4.000E+00, I, , 4A, , R1, , IDC, , 5, , 23.81m, , 7H, , C1, , L1, , R2, , 6, , IC = 0, C1, , 23.81m, , IC = 4A, 7H, , L1, , 0, , 0, , (b), , (a), , Figure 8.41, For Example 8.13: (a) for dc analysis, (b) for transient analysis., , i(t) enters pin 1, the inductor is rotated three times before it is placed in, the circuit. The same applies for the capacitor. We insert pseudocomponents VIEWPOINT and IPROBE to determine the initial capacitor, voltage and initial inductor current. We carry out a dc PSpice analysis, by selecting Analysis/Simulate. As shown in Fig. 8.41(a), we obtain, the initial capacitor voltage as 0 V and the initial inductor current i(0), as 4 A from the dc analysis. These initial values will be used in the, transient analysis., When the switch is moved to position b, the circuit becomes a sourcefree parallel RLC circuit with the schematic in Fig. 8.41(b). We set the, initial condition IC 0 for the capacitor and IC 4 A for the inductor., A current marker is inserted at pin 1 of the inductor. We select Analysis/, Setup/Transient to open up the Transient Analysis dialog box and set, Final Time to 3 s. After saving the schematic, we select Analysis/, Transient. Figure 8.42 shows the plot of i(t). The plot agrees with, i(t) 4.8et 0.8e6t A, which is the solution by hand calculation., , Refer to the circuit in Fig. 8.21 (see Practice Prob. 8.7). Use PSpice to, obtain v(t) for 0 6 t 6 2., Answer: See Fig. 8.43., , 11 V, , 10 V, , 9 V, , 8 V, 0 s, , 0.5 s, V(C1:1), , 1.0 s, , 1.5 s, , Time, , Figure 8.43, Plot of v(t) for Practice Prob. 8.13., , 2.0 s, , 4.00 A, , 3.96 A, , 3.92 A, , 3.88 A, 0 s, , 1.0 s, 2.0 s, I(L1), Time, , 3.0 s, , Figure 8.42, Plot of i(t) for Example 8.13., , Practice Problem 8.13
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ale29559_ch08.qxd, , 07/17/2008, , 11:26 AM, , Page 350, , Chapter 8, , 350, , 8.10, , Second-Order Circuits, , Duality, , The concept of duality is a time-saving, effort-effective measure of, solving circuit problems. Consider the similarity between Eq. (8.4) and, Eq. (8.29). The two equations are the same, except that we must interchange the following quantities: (1) voltage and current, (2) resistance, and conductance, (3) capacitance and inductance. Thus, it sometimes, occurs in circuit analysis that two different circuits have the same equations and solutions, except that the roles of certain complementary elements are interchanged. This interchangeability is known as the, principle of duality., The duality principle asserts a parallelism between pairs of characterizing equations and theorems of electric circuits., , TABLE 8.1, , Dual pairs., Resistance R, Inductance L, Voltage v, Voltage source, Node, Series path, Open circuit, KVL, Thevenin, , Conductance G, Capacitance C, Current i, Current source, Mesh, Parallel path, Short circuit, KCL, Norton, , Even when the principle of linearity, applies, a circuit element or variable, may not have a dual. For example,, mutual inductance (to be covered in, Chapter 13) has no dual., , Dual pairs are shown in Table 8.1. Note that power does not appear in, Table 8.1, because power has no dual. The reason for this is the principle of linearity; since power is not linear, duality does not apply. Also, notice from Table 8.1 that the principle of duality extends to circuit, elements, configurations, and theorems., Two circuits that are described by equations of the same form, but, in which the variables are interchanged, are said to be dual to each, other., Two circuits are said to be duals of one another if they are described, by the same characterizing equations with dual quantities interchanged., , The usefulness of the duality principle is self-evident. Once we, know the solution to one circuit, we automatically have the solution, for the dual circuit. It is obvious that the circuits in Figs. 8.8 and 8.13, are dual. Consequently, the result in Eq. (8.32) is the dual of that in, Eq. (8.11). We must keep in mind that the principle of duality is limited to planar circuits. Nonplanar circuits have no duals, as they cannot be described by a system of mesh equations., To find the dual of a given circuit, we do not need to write down, the mesh or node equations. We can use a graphical technique. Given, a planar circuit, we construct the dual circuit by taking the following, three steps:, 1. Place a node at the center of each mesh of the given circuit. Place, the reference node (the ground) of the dual circuit outside the, given circuit., 2. Draw lines between the nodes such that each line crosses an element. Replace that element by its dual (see Table 8.1)., 3. To determine the polarity of voltage sources and direction of current sources, follow this rule: A voltage source that produces a positive (clockwise) mesh current has as its dual a current source whose, reference direction is from the ground to the nonreference node., In case of doubt, one may verify the dual circuit by writing the nodal, or mesh equations. The mesh (or nodal) equations of the original circuit, are similar to the nodal (or mesh) equations of the dual circuit. The, duality principle is illustrated with the following two examples.
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 351, , 8.10, , Duality, , 351, , Example 8.14, , Construct the dual of the circuit in Fig. 8.44., Solution:, As shown in Fig. 8.45(a), we first locate nodes 1 and 2 in the two, meshes and also the ground node 0 for the dual circuit. We draw a line, between one node and another crossing an element. We replace the line, joining the nodes by the duals of the elements which it crosses. For, example, a line between nodes 1 and 2 crosses a 2-H inductor, and we, place a 2-F capacitor (an inductor’s dual) on the line. A line between, nodes 1 and 0 crossing the 6-V voltage source will contain a 6-A, current source. By drawing lines crossing all the elements, we construct, the dual circuit on the given circuit as in Fig. 8.45(a). The dual circuit, is redrawn in Fig. 8.45(b) for clarity., , 2Ω, , 6V, , +, −, , +, −, , For Example 8.14., , 2H, , 2, , 10 mF, , 2, , 2F, , 10 mH, 6A, , 2F, , 1, 1, , 10 mF, , Figure 8.44, , 2Ω, 6V, , 2H, , t=0, , t=0, , 0.5 Ω, , t=0, , 6A, , 0.5 Ω, , t=0, , 0, , 10 mH, , 0, , (b), , (a), , Figure 8.45, (a) Construction of the dual circuit of Fig. 8.44, (b) dual circuit redrawn., , Practice Problem 8.14, , Draw the dual circuit of the one in Fig. 8.46., Answer: See Fig. 8.47., , 3H, 3F, 50 mA, , 10 Ω, , 4H, , 50 mV, , +, −, , 0.1 Ω, , Figure 8.46, , Figure 8.47, , For Practice Prob. 8.14., , Dual of the circuit in Fig. 8.46., , Obtain the dual of the circuit in Fig. 8.48., Solution:, The dual circuit is constructed on the original circuit as in Fig. 8.49(a)., We first locate nodes 1 to 3 and the reference node 0. Joining nodes, 1 and 2, we cross the 2-F capacitor, which is replaced by a 2-H, inductor., , 4F, , Example 8.15
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 352, , Chapter 8, , 352, , Second-Order Circuits, 5H, , 10 V, , +, −, , i1, , 20 Ω, , i2, , 2F, , i3, , 3A, , Figure 8.48, For Example 8.15., , Joining nodes 2 and 3, we cross the 20- resistor, which is replaced, by a 201 -Æ resistor. We keep doing this until all the elements are crossed., The result is in Fig. 8.49(a). The dual circuit is redrawn in Fig. 8.49(b)., 5F, 5H, 2H, , 1, 10 V, , +, −, , 1, , 2, , 2F, , 20 Ω, , 1, 20, , 2H, , 3, , 2, , Ω, , 3, , 3A, 10 A, , Ω, −, +, , 0, , 1, 20, , 5F, , −, +, , 3V, , 0, , 3V, , 10 A, (a), , (b), , Figure 8.49, For Example 8.15: (a) construction of the dual circuit of Fig. 8.48, (b) dual circuit redrawn., , To verify the polarity of the voltage source and the direction of, the current source, we may apply mesh currents i1, i2, and i3 (all in the, clockwise direction) in the original circuit in Fig. 8.48. The 10-V, voltage source produces positive mesh current i1, so that its dual is a, 10-A current source directed from 0 to 1. Also, i3 3 A in Fig. 8.48, has as its dual v3 3 V in Fig. 8.49(b)., , Practice Problem 8.15, , For the circuit in Fig. 8.50, obtain the dual circuit., Answer: See Fig. 8.51., 1, 3, , 5Ω, 0.2 F, , 2A, , 4H, , 3Ω, , Ω, 4F, , 0.2 H, , +, −, , 20 V, , 2V, , +, −, , 1, 5, , Ω, , Figure 8.50, , Figure 8.51, , For Practice Prob. 8.15., , Dual of the circuit in Fig. 8.50., , 20 A
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 353, , 8.11, , 8.11, , Applications, , 353, , Applications, , Practical applications of RLC circuits are found in control and communications circuits such as ringing circuits, peaking circuits, resonant, circuits, smoothing circuits, and filters. Most of these circuits cannot, be covered until we treat ac sources. For now, we will limit ourselves, to two simple applications: automobile ignition and smoothing circuits., , 8.11.1 Automobile Ignition System, In Section 7.9.4, we considered the automobile ignition system as a, charging system. That was only a part of the system. Here, we consider another part—the voltage generating system. The system is modeled by the circuit shown in Fig. 8.52. The 12-V source is due to the, battery and alternator. The 4- resistor represents the resistance of the, wiring. The ignition coil is modeled by the 8-mH inductor. The 1-mF, capacitor (known as the condenser to automechanics) is in parallel with, the switch (known as the breaking points or electronic ignition). In the, following example, we determine how the RLC circuit in Fig. 8.52 is, used in generating high voltage., , t=0, 4Ω, , 1 F, + v −, C, , i, +, vL, −, , 12 V, , 8 mH, , Spark plug, Ignition coil, , Figure 8.52, Automobile ignition circuit., , Assuming that the switch in Fig. 8.52 is closed prior to t 0, find, the inductor voltage vL for t 7 0., Solution:, If the switch is closed prior to t 0 and the circuit is in steady state,, then, i(0) , , 12, 3 A,, 4, , vC (0) 0, , At t 0, the switch is opened. The continuity conditions require that, i(0) 3 A,, , , vC (0) 0, , (8.16.1), , , , We obtain di(0 )dt from vL(0 ). Applying KVL to the mesh at t 0, yields, 12 4i(0) vL(0) vC (0) 0, 1, vL(0) 0, 12 4 3 vL(0) 0 0, , Example 8.16
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ale29559_ch08.qxd, , 354, , 07/08/2008, , 11:16 AM, , Page 354, , Chapter 8, , Second-Order Circuits, , Hence,, vL(0), di(0), , 0, dt, L, , (8.16.2), , As t S , the system reaches steady state, so that the capacitor acts, like an open circuit. Then, i() 0, , (8.16.3), , If we apply KVL to the mesh for t 7 0, we obtain, 12 Ri L, , di, 1, , dt, C, , , , t, , i dt vC (0), , 0, , Taking the derivative of each term yields, d 2i, R di, i, , , 0, 2, L dt, LC, dt, , (8.16.4), , We obtain the form of the transient response by following the procedure, in Section 8.3. Substituting R 4 , L 8 mH, and C 1 mF, we get, a, , R, 250,, 2L, , 0 , , 1, 2LC, , 1.118 104, , Since a 6 0, the response is underdamped. The damped natural, frequency is, d 220 a2 0 1.118 104, The form of the transient response is, it(t) ea(A cos d t B sin d t), , (8.16.5), , where A and B are constants. The steady-state response is, iss (t) i() 0, , (8.16.6), , so that the complete response is, i(t) it(t) iss (t) e250t(A cos 11,180t B sin 11,180t) (8.16.7), We now determine A and B., i(0) 3 A 0, , 1, , A3, , Taking the derivative of Eq. (8.16.7),, di, 250e250t(A cos 11,180t B sin 11,180t), dt, e250t(11,180A sin 11,180t 11,180B cos 11,180t), Setting t 0 and incorporating Eq. (8.16.2),, 0 250A 11,180B, , 1, , B 0.0671, , Thus,, i(t) e250t(3 cos 11,180t 0.0671 sin 11,180t), , (8.16.8), , The voltage across the inductor is then, vL(t) L, , di, 268e250t sin 11,180t, dt, , (8.16.9)
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 355, , 8.11, , Applications, , 355, , This has a maximum value when sine is unity, that is, at 11,180t0 , p2 or t0 140.5 ms. At time t0, the inductor voltage reaches its, peak, which is, vL(t0) 268e250t0 259 V, , (8.16.10), , Although this is far less than the voltage range of 6000 to 10,000 V, required to fire the spark plug in a typical automobile, a device known, as a transformer (to be discussed in Chapter 13) is used to step up the, inductor voltage to the required level., , Practice Problem 8.16, , In Fig. 8.52, find the capacitor voltage vC for t 7 0., Answer: 12 12e250t cos 11,180t 267.7e250t sin 11,180t V., , 8.11.2 Smoothing Circuits, In a typical digital communication system, the signal to be transmitted, is first sampled. Sampling refers to the procedure of selecting samples, of a signal for processing, as opposed to processing the entire signal., Each sample is converted into a binary number represented by a series, of pulses. The pulses are transmitted by a transmission line such as a, coaxial cable, twisted pair, or optical fiber. At the receiving end, the, signal is applied to a digital-to-analog (D/A) converter whose output is, a “staircase” function, that is, constant at each time interval. In order, to recover the transmitted analog signal, the output is smoothed by letting it pass through a “smoothing” circuit, as illustrated in Fig. 8.53., An RLC circuit may be used as the smoothing circuit., , The output of a D/A converter is shown in Fig. 8.54(a). If the RLC, circuit in Fig. 8.54(b) is used as the smoothing circuit, determine the, output voltage vo(t)., vs, 10, 1, , 1Ω, , 1H, , 3, , 2, , 4, vs, 0, –2, , +, −, , +, v0, −, , 1F, , t (s), 0, (a), , 0, (b), , Figure 8.54, For Example 8.17: (a) output of a D/A converter, (b) an RLC, smoothing circuit., , Solution:, This problem is best solved using PSpice. The schematic is shown in, Fig. 8.55(a). The pulse in Fig. 8.54(a) is specified using the piecewise, , vs (t), , p(t), D/A, , Smoothing, circuit, , v0(t), , Figure 8.53, A series of pulses is applied to the digitalto-analog (D/A) converter, whose output, is applied to the smoothing circuit., , Example 8.17
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 356, , Chapter 8, , 356, , Second-Order Circuits, , V, T1=0, T2=0.001, T3=1, T4=1.001, T5=2, T6=2.001, T7=3, T8=3.001, , V1=0, V2=4, V3=4, V4=10, V5=10, V6=−2, V7=−2, V8=0, , +, −, , V, R1, , L1, , 1, , 1H, , V1, , 10 V, , 5 V, , 1, , C1, , 0 V, , −5 V, 0 s, 0, , 2.0 s, 4.0 s, 6.0 s, V(V1:+), V(C1:1), Time, (b), , (a), , Figure 8.55, For Example 8.17: (a) schematic, (b) input and output voltages., , linear function. The attributes of V1 are set as T1 0, V1 0,, T2 0.001, V2 4, T3 1, V3 4, and so on. To be able to plot both, input and output voltages, we insert two voltage markers as shown. We, select Analysis/Setup/Transient to open up the Transient Analysis dialog, box and set Final Time as 6 s. Once the schematic is saved, we select, Analysis/Simulate to run and obtain the plots shown in Fig. 8.55(b)., , Practice Problem 8.17, , Rework Example 8.17 if the output of the D/A converter is as shown, in Fig. 8.56., Answer: See Fig. 8.57., vs, , 8.0 V, , 8, 7, 4.0 V, , 0 V, , 0, –1, –3, , 1, , 2 3, , 4, , t (s), , −4.0 V, 0 s, , 2.0 s, 4.0 s, 6.0 s, V(V1:+), V(C1:1), Time, , Figure 8.56, , Figure 8.57, , For Practice Prob. 8.17., , Result of Practice Prob. 8.17., , 8.12, , Summary, , 1. The determination of the initial values x(0) and dx(0)dt and final, value x() is crucial to analyzing second-order circuits., 2. The RLC circuit is second-order because it is described by a, second-order differential equation. Its characteristic equation is
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 357, , Review Questions, , 3., , 4., , 5., , 6., 7., , 8., , 357, , s2 2a s 20 0, where a is the damping factor and 0 is the, undamped natural frequency. For a series circuit, a R2L, for a, parallel circuit a 12RC, and for both cases 0 101LC., If there are no independent sources in the circuit after switching, (or sudden change), we regard the circuit as source-free. The complete solution is the natural response., The natural response of an RLC circuit is overdamped, underdamped, or critically damped, depending on the roots of the characteristic equation. The response is critically damped when the, roots are equal (s1 s2 or a 0), overdamped when the roots, are real and unequal (s1 s2 or a 7 0), or underdamped when, the roots are complex conjugate (s1 s*2 or a 6 0)., If independent sources are present in the circuit after switching,, the complete response is the sum of the transient response and the, steady-state response., PSpice is used to analyze RLC circuits in the same way as for RC, or RL circuits., Two circuits are dual if the mesh equations that describe one circuit, have the same form as the nodal equations that describe the other., The analysis of one circuit gives the analysis of its dual circuit., The automobile ignition circuit and the smoothing circuit are typical applications of the material covered in this chapter., , Review Questions, 8.1, , For the circuit in Fig. 8.58, the capacitor voltage at, t 0 ( just before the switch is closed) is:, (a) 0 V, , (b) 4 V, , (c) 8 V, , 8.4, , (a) (A cos 2t B sin 2t)e3t, , (d) 12 V, , (b) (A 2Bt)e3t, , t=0, 2Ω, , If the roots of the characteristic equation of an RLC, circuit are 2 and 3, the response is:, , (c) Ae2t Bte3t, (d) Ae2t Be3t, , 4Ω, , where A and B are constants., 12 V +, −, , 1H, , 2F, , 8.5, , In a series RLC circuit, setting R 0 will produce:, (a) an overdamped response, (b) a critically damped response, , Figure 8.58, , (c) an underdamped response, , For Review Questions 8.1 and 8.2., , (d) an undamped response, 8.2, , For the circuit in Fig. 8.58, the initial inductor, current (at t 0) is:, (a) 0 A, , 8.3, , (b) 2 A, , (c) 6 A, , (d) 12 A, , When a step input is applied to a second-order, circuit, the final values of the circuit variables are, found by:, (a) Replacing capacitors with closed circuits and, inductors with open circuits., , (e) none of the above, 8.6, , A parallel RLC circuit has L 2 H and C 0.25 F., The value of R that will produce unity damping factor is:, (a) 0.5 (b) 1 , , 8.7, , (c) 2 , , (d) 4 , , Refer to the series RLC circuit in Fig. 8.59. What, kind of response will it produce?, (a) overdamped, , (b) Replacing capacitors with open circuits and, inductors with closed circuits., , (b) underdamped, , (c) Doing neither of the above., , (d) none of the above, , (c) critically damped
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 358, , Chapter 8, , 358, 1Ω, , Second-Order Circuits, R, , 1H, , vs, , 1F, , L, , +, −, , C, , Figure 8.59, , L, , is, , R, , C, , (a), , (b), , For Review Question 8.7., C1, , R, , 8.8, , Consider the parallel RLC circuit in Fig. 8.60. What, type of response will it produce?, (a) overdamped, , R2, , R1, , vs, , is, C2, , C1, , (b) underdamped, , +, −, , L, , (c), , (d), , (c) critically damped, , R1, , (d) none of the above, vs, , +, −, , C, , L, , R2, , L, 1H, , R2, , C, , R1, is, , 1Ω, , C2, , 1F, (e), , (f), , Figure 8.61, For Review Question 8.9., , Figure 8.60, For Review Question 8.8., , 8.10 In an electric circuit, the dual of resistance is:, 8.9, , Match the circuits in Fig. 8.61 with the following, items:, , (a) conductance, , (b) inductance, , (c) capacitance, , (d) open circuit, , (i) first-order circuit, , (e) short circuit, , (ii) second-order series circuit, (iii) second-order parallel circuit, , Answers: 8.1a, 8.2c, 8.3b, 8.4d, 8.5d, 8.6c, 8.7b, 8.8b,, 8.9 (i)-c, (ii)-b, e, (iii)-a, (iv)-d, f, 8.10a., , (iv) none of the above, , Problems, Section 8.2 Finding Initial and Final Values, 8.1, , For the circuit in Fig. 8.62, find:, (a) i(0) and v(0),, , 8.2, , Using Fig. 8.63, design a problem to help other, students better understand finding initial and final, values., , (b) di(0)dt and dv(0)dt,, (c) i() and v()., t=0, , iR, , +, −, , R3, , i, 2H, , R2, , 4Ω, , 6Ω, 12 V, , R1, , 0.4 F, , +, v, −, , v +, −, , C, t=0, , Figure 8.62, , Figure 8.63, , For Prob. 8.1., , For Prob. 8.2., , iC, , iL, L
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 359, , Problems, , 8.3, , 359, , Refer to the circuit shown in Fig. 8.64. Calculate:, , , , , R, , Rs, , , , + vR −, , (a) iL(0 ), vC (0 ), and vR(0 ),, (b) diL(0)dt, d vC (0)dt, and dvR(0)dt,, , Vs u(t), , +, −, , C, , (c) iL(), vC (), and vR()., , +, vL, −, , L, , Figure 8.67, For Prob. 8.6., , 40 Ω, , +, vR, −, , 10 Ω, , 2u(t) A, , +, vC, −, , 1, 4, , +, −, , IL, F, 1, 8, , Section 8.3 Source-Free Series RLC Circuit, H, , 10 V, , Figure 8.64, For Prob. 8.3., , 8.4, , In the circuit of Fig. 8.65, find:, , 8.7, , A series RLC circuit has R 10 k, L 0.1 mH,, and C 10 mF. What type of damping is exhibited, by the circuit?, , 8.8, , Design a problem to help other students better, understand source-free RLC circuits., , 8.9, , The current in an RLC circuit is described by, di, d 2i, 10 25i 0, dt, dt 2, , (a) v(0) and i(0),, (b) dv(0)dt and di(0)dt,, , If i(0) 2 A and di(0)dt 0, find i(t) for t 7 0., , (c) v() and i()., 3Ω, , 0.25 H, i, , 10u(–t) V, , 8.10 The differential equation that describes the voltage, in an RLC network is, , +, −, , 0.1 F, , +, v, −, , d 2v, dv, 5 4v 0, 2, dt, dt, , 5Ω, , 1u(t) A, , Given that v(0) 0, dv(0)dt 5 V/s, obtain v(t)., 8.11 The natural response of an RLC circuit is described, by the differential equation, , Figure 8.65, For Prob. 8.4., , 8.5, , dv, d 2v, 2 v0, dt, dt 2, , Refer to the circuit in Fig. 8.66. Determine:, , for which the initial conditions are v(0) 20 V and, dv(0)dt 0. Solve for v(t)., , (a) i(0) and v(0),, (b) di(0)dt and dv(0)dt,, , 8.12 If R 20 , L 0.6 H, what value of C will make, an RLC series circuit:, , (c) i() and v()., , (a) overdamped,, (b) critically damped,, , 1H, , (c) underdamped?, , i, 4u(t) A, , 4Ω, , 1, 4, , F, , 6Ω, , +, v, −, , 8.13 For the circuit in Fig. 8.68, calculate the value of R, needed to have a critically damped response., , Figure 8.66, For Prob. 8.5., , 8.6, , 60 Ω, R, , In the circuit of Fig. 8.67, find:, (a) vR(0) and vL(0),, (b) dvR(0)dt and dvL(0)dt,, , Figure 8.68, , (c) vR() and vL()., , For Prob. 8.13., , 0.01 F, , 4H
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 360, , Chapter 8, , 360, , Second-Order Circuits, 5Ω, , 8.14 The switch in Fig. 8.69 moves from position A to, position B at t 0 (please note that the switch must, connect to point B before it breaks the connection at, A, a make-before-break switch). Find v(t) for t 7 0., , t=0, 20 V, , 30 Ω, , A, , t=0, , +, −, , 1Ω, , 1F, , 0.25 H, , 4H, , Figure 8.72, B, +, −, , 20 V, , For Prob. 8.18., , +, v(t), −, , 0.25 F, , 10 Ω, , 8.19 Obtain v(t) for t 7 0 in the circuit of Fig. 8.73., , Figure 8.69, For Prob. 8.14., , +, v, , 10 Ω, , 8.15 The responses of a series RLC circuit are, , t=0, , vC (t) 30 10e20t 30e10t V, , 90 V, , iL(t) 40e20t 60e10t mA, where vC and iL are the capacitor voltage and, inductor current, respectively. Determine the values, of R, L, and C., 8.16 Find i(t) for t 7 0 in the circuit of Fig. 8.70., , 10 Ω, , t=0, , +, −, , 4H, , Figure 8.73, For Prob. 8.19., 8.20 The switch in the circuit of Fig. 8.74 has been closed, for a long time but is opened at t 0. Determine i(t), for t 7 0., , 60 Ω, i(t), , +, −, , 1, 2, , i(t), , 1 mF, 20 V, , 1F, , −, , H, , 2Ω, , 40 Ω, 2.5 H, , 12 V, +−, , Figure 8.70, , 1, 4, , For Prob. 8.16., 8.17 In the circuit of Fig. 8.71, the switch instantaneously, moves from position A to B at t 0. Find v(t) for all, t 0., , t=0, F, , Figure 8.74, For Prob. 8.20., *8.21 Calculate v(t) for t 7 0 in the circuit of Fig. 8.75., , t=0, , A, , 0.25 H, , 15 Ω, , B, 15 A, , 4Ω, , 10 Ω, , 0.04 F, , 6Ω, , 12 Ω, , +, v (t), –, , t=0, 24 V, , +, −, , Figure 8.71, For Prob. 8.17., , 60 Ω, , 3H, +, v, −, , Figure 8.75, 8.18 Find the voltage across the capacitor as a function of, time for t 7 0 for the circuit in Fig. 8.72. Assume, steady-state conditions exist at t 0., , For Prob. 8.21., * An asterisk indicates a challenging problem., , 1, 27, , F, , 25 Ω
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 361, , Problems, , Section 8.4 Source-Free Parallel RLC Circuit, 8.22 Assuming R 2 k, design a parallel RLC circuit, that has the characteristic equation, , 361, , If the initial conditions are v(0) 0 dv(0)dt,, find v(t)., 8.28 A series RLC circuit is described by, , s2 100s 106 0., , L, , 8.23 For the network in Fig. 8.76, what value of C is, needed to make the response underdamped with, unity damping factor (a 1)?, , d 2i, di, i, R 2, 2, dt, C, dt, , Find the response when L 0.5 H, R 4 ,, and C 0.2 F. Let i(0) 1, di(0)dt 0., 8.29 Solve the following differential equations subject to, the specified initial conditions, , 10 Ω, , 0.5 H, , C, , (a) d 2vdt 2 4v 12, v(0) 0, dv(0)dt 2, , 10 mF, , (b) d 2idt 2 5 didt 4i 8, i(0) 1,, di(0)dt 0, , Figure 8.76, , (c) d 2vdt 2 2 dvdt v 3, v(0) 5,, dv(0)dt 1, , For Prob. 8.23., 8.24 The switch in Fig. 8.77 moves from position A to, position B at t 0 (please note that the switch must, connect to point B before it breaks the connection at, A, a make-before-break switch). Determine i(t) for, t 7 0., , (d) d 2idt 2 2 didt 5i 10, i(0) 4,, di(0)dt 2, 8.30 The step responses of a series RLC circuit are, vC 40 10e2000t 10e4000t V,, 2000t, , iL(t) 3e, , A, t =0, i(t), , B, 12 A, , 20 Ω, , 10 mF, , 10 Ω, , 0.25 H, , 4000t, , 6e, , t 7 0, , mA,, , t 7 0, , (a) Find C. (b) Determine what type of damping is, exhibited by the circuit., 8.31 Consider the circuit in Fig. 8.79. Find vL(0) and, vC (0)., , Figure 8.77, For Prob. 8.24., , 40 Ω, , 8.25 Using Fig. 8.78, design a problem to help other, students better understand source-free RLC circuits., 2u(t), , R1, , io(t), , L, , 0.5 H, , +, vL, −, , 10 Ω, , 1F, , +, vC, −, , +, −, , 50 V, , Figure 8.79, For Prob. 8.31., , t=0, v +, −, , R2, , C, , +, vo(t), −, , 8.32 For the circuit in Fig. 8.80, find v(t) for t 7 0., , Figure 8.78, For Prob. 8.25., 4u(–t) A, , Section 8.5 Step Response of a Series RLC Circuit, 8.26 The step response of an RLC circuit is described by, d 2i, di, 2 5i 10, 2, dt, dt, Given that i(0) 6 A and di(0)dt 12 A/s, solve, for i(t)., , dv, d v, 4, 8v 48, 2, dt, dt, , + v −, , 4Ω, +−, , 100u(t) V, , 8.27 A branch voltage in an RLC circuit is described by, 2, , 0.04 F, , 1H, , Figure 8.80, For Prob. 8.32., , 2Ω
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 362, , Chapter 8, , 362, , Second-Order Circuits, , 8.33 Find v(t) for t 7 0 in the circuit of Fig. 8.81., , *8.37 For the network in Fig. 8.85, solve for i(t) for t 7 0., , 1H, , t=0, , 6Ω, , 6Ω, 6Ω, , 3A, , +, v, −, , 10 Ω, , i(t), , 5Ω, , 4F, , 1, 8, , 4u(t) A, 1, 2, , t=0, +, −, , 30 V, , Figure 8.81, , 10 V, , For Prob. 8.33., , F, , H, , +, −, , Figure 8.85, 8.34 Calculate i(t) for t 7 0 in the circuit of Fig. 8.82., , 8.38 Refer to the circuit in Fig. 8.86. Calculate i(t) for, t 7 0., , + v −, 1, 16, , 20u(−t) V, , +, −, , For Prob. 8.37., , 6(1 − u(t )) A, , i, , F, , i(t), 1, 4, , H, , 3, 4, , 5Ω, 1, 3, , 10 Ω, F, , Figure 8.82, , 5Ω, , For Prob. 8.34., , 10 Ω, , 8.35 Using Fig. 8.83, design a problem to help other, students better understand the step response of series, RLC circuits., , Figure 8.86, For Prob. 8.38., 8.39 Determine v(t) for t 7 0 in the circuit of Fig. 8.87., , R, , +, −, , +, −, , 0.5 F, , 30 Ω, , t=0, V1, , V2, , C, , +, v, −, , 0.25 H, , + v −, 60u(t) V, , L, , +, −, , +, −, , 20 Ω, , 30u(t) V, , Figure 8.87, , Figure 8.83, , For Prob. 8.39., , For Prob. 8.35., , 8.40 The switch in the circuit of Fig. 8.88 is moved from, position a to b at t 0. Determine i(t) for t 7 0., , 8.36 Obtain v(t) and i(t) for t 7 0 in the circuit of, Fig. 8.84., , i(t), , 3u(t) A, , H, , 5H, , 5Ω, , 2Ω, , i(t), , 1Ω, , 0.2 F, , 0.02 F 14 Ω, , b, 2H, , a, t=0, , +, v (t), −, , 6Ω, 4A, , +−, 20 V, , Figure 8.84, , Figure 8.88, , For Prob. 8.36., , For Prob. 8.40., , 2Ω, , +, −, , 12 V
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 363, , Problems, , *8.41 For the network in Fig. 8.89, find i(t) for t 7 0., 5Ω, 20 Ω, , 363, , 8.46 Using Fig. 8.93, design a problem to help other, students better understand the step response of a, parallel RLC circuit., i(t), , 1H, i, , t=0, 50 V +, −, , L, , 5Ω, , 1, 25, , v +, −, , F, , Figure 8.89, , Figure 8.93, , For Prob. 8.41., , For Prob. 8.46., , C, , R, , *8.42 Given the network in Fig. 8.90, find v(t) for t 7 0., 8.47 Find the output voltage vo(t) in the circuit of, Fig. 8.94., , 2A, , 1H, t=0, , 6Ω, 1Ω, , 4A, , 1, 25, , t=0, , +, v, −, , F, , 10 Ω, 5Ω, , 3A, , +, vo, −, , 10 mF, , 1H, , Figure 8.90, For Prob. 8.42., 8.43 The switch in Fig. 8.91 is opened at t 0 after the, circuit has reached steady state. Choose R and C, such that a 8 Np/s and d 30 rad/s., 10 Ω, , t=0, , R, , +, −, , 0.5 H, , Figure 8.94, For Prob. 8.47., 8.48 Given the circuit in Fig. 8.95, find i(t) and v(t) for, t 7 0., , i(t), 40 V, 1H, , C, 1Ω, , Figure 8.91, , 1, 4, , 2Ω, , For Prob. 8.43., , F, , +, v (t), −, , t=0, , 8.44 A series RLC circuit has the following parameters:, R 1 k, L 1 H, and C 10 nF. What type of, damping does this circuit exhibit?, , 12 V, , +, −, , Figure 8.95, For Prob. 8.48., , Section 8.6 Step Response of a Parallel, RLC Circuit, , 8.49 Determine i(t) for t 7 0 in the circuit of Fig. 8.96., , 8.45 In the circuit of Fig. 8.92, find v(t) and i(t) for t 7 0., Assume v(0) 0 V and i(0) 1 A., , 4Ω, t=0, , i, 4u(t) A, , 2Ω, , +, v, −, , 0.5 F, , 1H, , 12 V +, −, , Figure 8.92, , Figure 8.96, , For Prob. 8.45., , For Prob. 8.49., , 5H, , i(t), 1, 20, , F, , 5Ω, , 3A
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 364, , Chapter 8, , 364, , Second-Order Circuits, , 8.55 For the circuit in Fig. 8.101, find v(t) for t 7 0., Assume that v(0) 4 V and i(0) 2 A., , 8.50 For the circuit in Fig. 8.97, find i(t) for t 7 0., 10 Ω, , 2Ω, , i(t), 30 V +, −, , 40 Ω, , 10 mF, , 6u(t) A, , 4H, , Figure 8.97, , Figure 8.101, , For Prob. 8.50., , For Prob. 8.55., , 8.51 Find v(t) for t 7 0 in the circuit of Fig. 8.98., , i, , +, v, −, , 0.1 F, , i, 4, , 0.5 F, , 8.56 In the circuit of Fig. 8.102, find i(t) for t 7 0., 4Ω, , t=0, io, , R, , i, , +, v, −, , L, , C, , t=0, , 6Ω, , 1, 25, , 50 V +, −, , F, , 1, 4, , H, , Figure 8.98, For Prob. 8.51., , Figure 8.102, For Prob. 8.56., , 8.52 The step response of a parallel RLC circuit is, v 10 20e 300t(cos 400t 2 sin 400t) V,, , t, , when the inductor is 50 mH. Find R and C., , 0, , 8.57 If the switch in Fig. 8.103 has been closed for a long, time before t 0 but is opened at t 0, determine:, (a) the characteristic equation of the circuit,, , Section 8.7 General Second-Order Circuits, , (b) ix and vR for t 7 0., , 8.53 After being open for a day, the switch in the circuit, of Fig. 8.99 is closed at t 0. Find the differential, equation describing i(t), t 7 0., , t=0, ix, , t=0, , 80 Ω, , 16 V, i, , 120 V +, −, , 10 mF, , 12 Ω, , +, −, , 1, 36, , +, vR, −, , 8Ω, 1H, , F, , 0.25 H, , Figure 8.103, For Prob. 8.57., , Figure 8.99, For Prob. 8.53., 8.54 Using Fig. 8.100, design a problem to help other, students better understand general second-order, circuits., , 8.58 In the circuit of Fig. 8.104, the switch has been in, position 1 for a long time but moved to position 2 at, t 0. Find:, (a) v(0), dv(0)dt, (b) v(t) for t, , A t=0, , R1, , R3, , 2, i, , B, I, , 0., , R2, , +, v, −, , C, , L, , 1, , 8Ω, , t=0, 0.25 H, , Figure 8.100, , Figure 8.104, , For Prob. 8.54., , For Prob. 8.58., , 0.5 Ω, , v, , +, –, , 1F, , +, 4V −
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 365, , Problems, , 8.59 The make before break switch in Fig. 8.105 has been, in position 1 for t 6 0. At t 0, it is moved, instantaneously to position 2. Determine v(t)., 4Ω, , 1, , t=0, , 365, , 8.64 Using Fig. 8.109, design a problem to help other, students better understand second-order op amp, circuits., , 4H, C1, , 40 V, , +, −, , v, , +, , 1, 16, , –, , 16 Ω, , F, , R1, , R2, +, −, , Figure 8.105, , vs +, −, , For Prob. 8.59., , +, vo, , C2, , −, , 8.60 Obtain i1 and i2 for t 7 0 in the circuit of Fig. 8.106., , For Prob. 8.64., , 3Ω, , 2Ω, , 4u(t) A, , Figure 8.109, , i1, , i2, , 1H, , 1H, , 8.65 Determine the differential equation for the op amp, circuit in Fig. 8.110. If v1(0) 2 V and, v2(0) 0 V, find vo for t 7 0. Let R 100 k, and C 1 mF., , Figure 8.106, For Prob. 8.60., R, , 8.61 For the circuit in Prob. 8.5, find i and v for t 7 0., , C, , 8.62 Find the response vR(t) for t 7 0 in the circuit of, Fig. 8.107. Let R 3 , L 2 H, and C 118 F., , +, , v1, , −, +, , R, , C, , −, +, R, , + vR −, 10u(t) V, , +, −, , C, , v2, , −, +, , +, vo, −, , L, , Figure 8.107, , Figure 8.110, , For Prob. 8.62., , For Prob. 8.65., , −, , Section 8.8 Second-Order Op Amp Circuits, 8.63 For the op amp circuit in Fig. 8.108, find the, differential equation for i(t)., , 8.66 Obtain the differential equations for vo(t) in the op, amp circuit of Fig. 8.111., , C, , R, , vs +, −, , 10 pF, −, +, , i, , 60 kΩ, , 60 kΩ, , −, +, , L, vs +, −, , Figure 8.108, , Figure 8.111, , For Prob. 8.63., , For Prob. 8.66., , 20 pF, , +, vo, −
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 366, , Chapter 8, , 366, , Second-Order Circuits, , *8.67 In the op amp circuit of Fig. 8.112, determine vo(t), for t 7 0. Let vin u(t) V, R1 R2 10 k,, C1 C2 100 mF., , 8.71 Obtain v(t) for 0 6 t 6 4 s in the circuit of Fig. 8.116, using PSpice., 0.4 F, , 1H, , 6Ω, , C1, 20 Ω, , + 39u(t) V, −, , C2, , R1, , v in, , +, v (t), −, , 6Ω, , 13u(t) A, , R2, −, +, , Figure 8.116, , vo, , For Prob. 8.71., , Figure 8.112, For Prob. 8.67., , 8.72 The switch in Fig. 8.117 has been in position 1 for a, long time. At t 0, it is switched to position 2. Use, PSpice to find i(t) for 0 6 t 6 0.2 s., , Section 8.9 PSpice Analysis of RLC Circuit, , 4 kΩ, , 8.68 For the step function vs u(t), use PSpice to find, the response v(t) for 0 6 t 6 6 s in the circuit of, Fig. 8.113., 2Ω, , 1, , t=0, 10 V, , +, −, , 100 mH, , 1 kΩ, , 2, i, , 2 kΩ, , 100 F, , 1H, , Figure 8.117, +, vs, , +, −, , 1F, , For Prob. 8.72., v(t), , −, , 8.73 Design a problem, to be solved using PSpice, to help, other students better understand source-free RLC, circuits., , Figure 8.113, For Prob. 8.68., 8.69 Given the source-free circuit in Fig. 8.114, use PSpice, to get i(t) for 0 6 t 6 20 s. Take v(0) 30 V and, i(0) 2 A., , Section 8.10 Duality, 8.74 Draw the dual of the circuit shown in Fig. 8.118., , 1Ω, , 10 H, , 2.5 F, , +, v, −, , 9V +, −, , Figure 8.118, , For Prob. 8.69., , For Prob. 8.74., , 8.70 For the circuit in Fig. 8.115, use PSpice to obtain, v(t) for 0 6 t 6 4 s. Assume that the capacitor, voltage and inductor current at t 0 are both zero., , 24 V, , +, −, , 2H, , +, −, , 10 Ω, 0.5 F, +, −, , +, v, , 0.4 F, –, , 3A, , 8.75 Obtain the dual of the circuit in Fig. 8.119., , 12 V, , 3Ω, , 1Ω, , 6Ω, , Figure 8.114, , 6Ω, , 4Ω, , 2Ω, , i, , 4Ω, , Figure 8.115, , Figure 8.119, , For Prob. 8.70., , For Prob. 8.75., , 2H, , 24 V
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ale29559_ch08.qxd, , 07/08/2008, , 11:16 AM, , Page 367, , Comprehensive Problems, , Section 8.11 Applications, , 8.76 Find the dual of the circuit in Fig. 8.120., , 20 Ω, , 10 Ω, , 120 V, , +−, , −+, , 4H, , 8.78 An automobile airbag igniter is modeled by the, circuit in Fig. 8.122. Determine the time it takes the, voltage across the igniter to reach its first peak after, switching from A to B. Let R 3 , C 130 F,, and L 60 mH., , 30 Ω, , 60 V, , 1F, , 367, , 2A, , A, , B, t=0, Airbag igniter, , Figure 8.120, , 12 V, , For Prob. 8.76., 8.77 Draw the dual of the circuit in Fig. 8.121., , R, , For Prob. 8.78., , 3Ω, , 2Ω, 0.25 H, , L, , C, , Figure 8.122, , 5A, , 1F, , +, −, , 1Ω, +, −, , 12 V, , 8.79 A load is modeled as a 250-mH inductor in parallel, with a 12- resistor. A capacitor is needed to be, connected to the load so that the network is, critically damped at 60 Hz. Calculate the size of, the capacitor., , Figure 8.121, For Prob. 8.77., , Comprehensive Problems, 8.80 A mechanical system is modeled by a series RLC, circuit. It is desired to produce an overdamped response, with time constants 0.1 ms and 0.5 ms. If a series, 50-k resistor is used, find the values of L and C., 8.81 An oscillogram can be adequately modeled by a, second-order system in the form of a parallel RLC, circuit. It is desired to give an underdamped voltage, across a 200- resistor. If the damping frequency is, 4 kHz and the time constant of the envelope is 0.25 s,, find the necessary values of L and C., 8.82 The circuit in Fig. 8.123 is the electrical analog of, body functions used in medical schools to study, convulsions. The analog is as follows:, C1 Volume of fluid in a drug, C2 Volume of blood stream in a specified region, , +, vo, −, , C1, , v0 Initial concentration of the drug dosage, , +, v (t), −, , For Prob. 8.82., , 8.83 Figure 8.124 shows a typical tunnel-diode oscillator, circuit. The diode is modeled as a nonlinear, resistor with iD f(vD), i.e., the diode current is a, nonlinear function of the voltage across the diode., Derive the differential equation for the circuit in, terms of v and iD., , R, , vs, , L, , +, −, , v(t) Percentage of the drug in the blood stream, Find v(t) for t 7 0 given that C1 0.5 mF, C2 , 5 mF, R1 5 M, R2 2.5 M, and v0 60u(t) V., , C2, , R2, , Figure 8.123, , R1 Resistance in the passage of the drug from, the input to the blood stream, R2 Resistance of the excretion mechanism,, such as kidney, etc., , R1, , t=0, , Figure 8.124, For Prob. 8.83., , i, , +, v, −, , C, , ID, +, vD, −
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 368, , P A R T, , T W O, , AC Circuits, OUTLINE, 9, , Sinusoids and Phasors, , 10, , Sinusoidal Steady-State Analysis, , 11, , AC Power Analysis, , 12, , Three-Phase Circuits, , 13, , Magnetically Coupled Circuits, , 14, , Frequency Response
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 369, , c h a p t e r, , 9, , Sinusoids and, Phasors, He who knows not, and knows not that he knows not, is a fool—, shun him. He who knows not, and knows that he knows not, is a child—, teach him. He who knows, and knows not that he knows, is asleep—wake, him up. He who knows, and knows that he knows, is wise—follow him., —Persian Proverb, , Enhancing Your Skills and Your Career, ABET EC 2000 criteria (3.d), “an ability to function on, multi-disciplinary teams.”, The “ability to function on multidisciplinary teams” is inherently critical for the working engineer. Engineers rarely, if ever, work by themselves. Engineers will always be part of some team. One of the things, I like to remind students is that you do not have to like everyone on a, team; you just have to be a successful part of that team., Most frequently, these teams include individuals from of a variety, of engineering disciplines, as well as individuals from nonengineering, disciplines such as marketing and finance., Students can easily develop and enhance this skill by working in, study groups in every course they take. Clearly, working in study, groups in nonengineering courses as well as engineering courses outside your discipline will also give you experience with multidisciplinary teams., , Photo by Charles Alexander, , 369
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 370, , Chapter 9, , 370, , Sinusoids and Phasors, , Historical, Nikola Tesla (1856–1943) and George Westinghouse (1846–1914), , George Westinghouse. Photo, © Bettmann/Corbis, , helped establish alternating current as the primary mode of electricity, transmission and distribution., Today it is obvious that ac generation is well established as the form, of electric power that makes widespread distribution of electric power, efficient and economical. However, at the end of the 19th century, which, was the better—ac or dc—was hotly debated and had extremely outspoken supporters on both sides. The dc side was lead by Thomas, Edison, who had earned a lot of respect for his many contributions., Power generation using ac really began to build after the successful contributions of Tesla. The real commercial success in ac came from George, Westinghouse and the outstanding team, including Tesla, he assembled., In addition, two other big names were C. F. Scott and B. G. Lamme., The most significant contribution to the early success of ac was, the patenting of the polyphase ac motor by Tesla in 1888. The induction motor and polyphase generation and distribution systems doomed, the use of dc as the prime energy source., , 9.1, , Introduction, , Thus far our analysis has been limited for the most part to dc circuits:, those circuits excited by constant or time-invariant sources. We have, restricted the forcing function to dc sources for the sake of simplicity,, for pedagogic reasons, and also for historic reasons. Historically, dc, sources were the main means of providing electric power up until the, late 1800s. At the end of that century, the battle of direct current versus alternating current began. Both had their advocates among the electrical engineers of the time. Because ac is more efficient and economical, to transmit over long distances, ac systems ended up the winner. Thus,, it is in keeping with the historical sequence of events that we considered dc sources first., We now begin the analysis of circuits in which the source voltage or, current is time-varying. In this chapter, we are particularly interested in, sinusoidally time-varying excitation, or simply, excitation by a sinusoid., A sinusoid is a signal that has the form of the sine or cosine function., , A sinusoidal current is usually referred to as alternating current (ac)., Such a current reverses at regular time intervals and has alternately positive and negative values. Circuits driven by sinusoidal current or voltage sources are called ac circuits., We are interested in sinusoids for a number of reasons. First, nature, itself is characteristically sinusoidal. We experience sinusoidal variation in the motion of a pendulum, the vibration of a string, the ripples, on the ocean surface, and the natural response of underdamped secondorder systems, to mention but a few. Second, a sinusoidal signal is easy, to generate and transmit. It is the form of voltage generated throughout
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 371, , 9.2, , Sinusoids, , 371, , the world and supplied to homes, factories, laboratories, and so on. It, is the dominant form of signal in the communications and electric, power industries. Third, through Fourier analysis, any practical periodic signal can be represented by a sum of sinusoids. Sinusoids,, therefore, play an important role in the analysis of periodic signals., Lastly, a sinusoid is easy to handle mathematically. The derivative, and integral of a sinusoid are themselves sinusoids. For these and, other reasons, the sinusoid is an extremely important function in, circuit analysis., A sinusoidal forcing function produces both a transient response, and a steady-state response, much like the step function, which we studied in Chapters 7 and 8. The transient response dies out with time so, that only the steady-state response remains. When the transient response, has become negligibly small compared with the steady-state response,, we say that the circuit is operating at sinusoidal steady state. It is this, sinusoidal steady-state response that is of main interest to us in this, chapter., We begin with a basic discussion of sinusoids and phasors. We, then introduce the concepts of impedance and admittance. The basic, circuit laws, Kirchhoff’s and Ohm’s, introduced for dc circuits, will be, applied to ac circuits. Finally, we consider applications of ac circuits, in phase-shifters and bridges., , 9.2, , Sinusoids, , Consider the sinusoidal voltage, v(t) Vm sin t, , (9.1), , where, Vm the amplitude of the sinusoid, the angular frequency in radians/s, t the argument of the sinusoid, The sinusoid is shown in Fig. 9.1(a) as a function of its argument and, in Fig. 9.1(b) as a function of time. It is evident that the sinusoid, repeats itself every T seconds; thus, T is called the period of the sinusoid. From the two plots in Fig. 9.1, we observe that T 2 p,, T, , 2p, , , (9.2), , v(t), , v(t), , Vm, , Vm, , 0, –Vm, , π, , 3π, , 2π, , 4π, , t, , 0, –Vm, , (a), , Figure 9.1, , A sketch of Vm sin t: (a) as a function of t, (b) as a function of t., , T, 2, , T, , 3T, 2, (b), , 2T, , t
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ale29559_ch09.qxd, , 07/16/2008, , 12:31 PM, , Page 372, , Chapter 9, , 372, , Sinusoids and Phasors, , Historical, , The Burndy Library Collection, at The Huntington Library,, San Marino, California., , Heinrich Rudorf Hertz (1857–1894), a German experimental physicist, demonstrated that electromagnetic waves obey the same fundamental laws as light. His work confirmed James Clerk Maxwell’s, celebrated 1864 theory and prediction that such waves existed., Hertz was born into a prosperous family in Hamburg, Germany., He attended the University of Berlin and did his doctorate under the, prominent physicist Hermann von Helmholtz. He became a professor, at Karlsruhe, where he began his quest for electromagnetic waves., Hertz successfully generated and detected electromagnetic waves; he, was the first to show that light is electromagnetic energy. In 1887, Hertz, noted for the first time the photoelectric effect of electrons in a molecular structure. Although Hertz only lived to the age of 37, his discovery of electromagnetic waves paved the way for the practical use of, such waves in radio, television, and other communication systems. The, unit of frequency, the hertz, bears his name., , The fact that v(t) repeats itself every T seconds is shown by replacing, t by t T in Eq. (9.1). We get, v(t T) Vm sin (t T) Vm sin at , , 2p, b, , , Vm sin(t 2p) Vm sin t v(t), , (9.3), , Hence,, v(t T) v(t), , (9.4), , that is, v has the same value at t T as it does at t and v(t) is said to, be periodic. In general,, A periodic function is one that satisfies f (t ) f (t nT ), for all t and, for all integers n., , As mentioned, the period T of the periodic function is the time of one, complete cycle or the number of seconds per cycle. The reciprocal of, this quantity is the number of cycles per second, known as the cyclic, frequency f of the sinusoid. Thus,, f, , 1, T, , (9.5), , From Eqs. (9.2) and (9.5), it is clear that, The unit of f is named after the German, physicist Heinrich R. Hertz (1857–1894)., , 2pf, While is in radians per second (rad/s), f is in hertz (Hz)., , (9.6)
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 373, , 9.2, , Sinusoids, , Let us now consider a more general expression for the sinusoid,, v(t) Vm sin(t f), , (9.7), , where (t f) is the argument and f is the phase. Both argument and, phase can be in radians or degrees., Let us examine the two sinusoids, v1(t) Vm sin t, , and, , v2 (t) Vm sin(t f), , (9.8), , shown in Fig. 9.2. The starting point of v2 in Fig. 9.2 occurs first in, time. Therefore, we say that v2 leads v1 by f or that v1 lags v2 by f., If f 0, we also say that v1 and v2 are out of phase. If f 0, then, v1 and v2 are said to be in phase; they reach their minima and maxima at exactly the same time. We can compare v1 and v2 in this manner because they operate at the same frequency; they do not need to, have the same amplitude., , v1 = Vm sin t, Vm, , π, , , , –Vm, , 2π, , t, , v2 = Vm sin(t + ), , Figure 9.2, Two sinusoids with different phases., , A sinusoid can be expressed in either sine or cosine form. When, comparing two sinusoids, it is expedient to express both as either sine, or cosine with positive amplitudes. This is achieved by using the following trigonometric identities:, sin(A B) sin A cos B cos A sin B, cos(A B) cos A cos B sin A sin B, , (9.9), , With these identities, it is easy to show that, sin(t 180) sin t, cos(t 180) cos t, sin(t 90) cos t, cos(t 90) sin t, , (9.10), , Using these relationships, we can transform a sinusoid from sine form, to cosine form or vice versa., , 373
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 374, , Chapter 9, , 374, , + cos t, –90°, , + sin t, (a), , 180°, + cos t, , Sinusoids and Phasors, , A graphical approach may be used to relate or compare sinusoids, as an alternative to using the trigonometric identities in Eqs. (9.9) and, (9.10). Consider the set of axes shown in Fig. 9.3(a). The horizontal axis, represents the magnitude of cosine, while the vertical axis (pointing, down) denotes the magnitude of sine. Angles are measured positively, counterclockwise from the horizontal, as usual in polar coordinates., This graphical technique can be used to relate two sinusoids. For example, we see in Fig. 9.3(a) that subtracting 90 from the argument of, cos t gives sin t, or cos(t 90) sin t. Similarly, adding 180 to, the argument of sin t gives sin t, or sin(t 180) sin t, as, shown in Fig. 9.3(b)., The graphical technique can also be used to add two sinusoids of, the same frequency when one is in sine form and the other is in cosine, form. To add A cos t and B sin t, we note that A is the magnitude of, cos t while B is the magnitude of sin t, as shown in Fig. 9.4(a). The, magnitude and argument of the resultant sinusoid in cosine form is, readily obtained from the triangle. Thus,, A cos t B sin t C cos(t u), , + sin t, , (9.11), , where, (b), , Figure 9.3, A graphical means of relating cosine, and sine: (a) cos(t 90) sin t,, (b) sin(t 180) sin t., , C 2A 2 B 2,, , u tan1, , B, A, , (9.12), , For example, we may add 3 cos t and 4 sin t as shown in Fig. 9.4(b), and obtain, 3 cos t 4 sin t 5 cos(t 53.1), , (9.13), , Compared with the trigonometric identities in Eqs. (9.9) and, (9.10), the graphical approach eliminates memorization. However, we, must not confuse the sine and cosine axes with the axes for complex, numbers to be discussed in the next section. Something else to note in, Figs. 9.3 and 9.4 is that although the natural tendency is to have the, vertical axis point up, the positive direction of the sine function is down, in the present case., , –4, A, , cos t, , 5, , –, , 53.1°, , C, B, , 0, , +3, , cos t, , sin t, , sin t, (a), , Figure 9.4, , (b), , (a) Adding A cos t and B sin t, (b) adding 3 cos t and 4 sin t.
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 375, , 9.2, , Sinusoids, , Find the amplitude, phase, period, and frequency of the sinusoid, , 375, , Example 9.1, , v(t) 12 cos(50 t 10), Solution:, The amplitude is Vm 12 V., The phase is f 10., The angular frequency is 50 rad/s., 2p, 2p, , 0.1257 s., The period T , , 50, 1, The frequency is f 7.958 Hz., T, , Given the sinusoid 5 sin(4 p t 60), calculate its amplitude, phase,, angular frequency, period, and frequency., , Practice Problem 9.1, , Answer: 5, 60, 12.57 rad/s, 0.5 s, 2 Hz., , Calculate the phase angle between v1 10 cos(t 50) and v2 , 12 sin(t 10). State which sinusoid is leading., Solution:, Let us calculate the phase in three ways. The first two methods use, trigonometric identities, while the third method uses the graphical, approach., , ■ METHOD 1 In order to compare v1 and v2, we must express, them in the same form. If we express them in cosine form with positive amplitudes,, v1 10 cos(t 50) 10 cos(t 50 180), v1 10 cos(t 130) or v1 10 cos(t 230), , (9.2.1), , and, v2 12 sin(t 10) 12 cos(t 10 90), v2 12 cos(t 100), , (9.2.2), , It can be deduced from Eqs. (9.2.1) and (9.2.2) that the phase difference between v1 and v2 is 30. We can write v2 as, v2 12 cos(t 130 30), , or, , v2 12 cos(t 260) (9.2.3), , Comparing Eqs. (9.2.1) and (9.2.3) shows clearly that v2 leads v1 by 30., , ■ METHOD 2 Alternatively, we may express v1 in sine form:, v1 10 cos(t 50) 10 sin(t 50 90), 10 sin(t 40) 10 sin(t 10 30), , Example 9.2
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 376, , Chapter 9, , 376, , cos t, 50°, , v1, , 10°, , Sinusoids and Phasors, , But v2 12 sin(t 10). Comparing the two shows that v1 lags v2, by 30. This is the same as saying that v2 leads v1 by 30., , ■ METHOD 3 We may regard v1 as simply 10 cos t with a, phase shift of 50. Hence, v1 is as shown in Fig. 9.5. Similarly, v2, is 12 sin t with a phase shift of 10, as shown in Fig. 9.5. It is easy, to see from Fig. 9.5 that v2 leads v1 by 30, that is, 90 50 10., , v2, sin t, , Figure 9.5, For Example 9.2., , Practice Problem 9.2, , Find the phase angle between, i1 4 sin(377t 25), , and, , i2 5 cos(377t 40), , Does i1 lead or lag i2?, Answer: 155, i1 leads i2., , 9.3, , Phasors, , Sinusoids are easily expressed in terms of phasors, which are more convenient to work with than sine and cosine functions., A phasor is a complex number that represents the amplitude and, phase of a sinusoid., , Charles Proteus Steinmetz (1865–1923), was a German-Austrian mathematician, and electrical engineer., Appendix B presents a short tutorial on, complex numbers., , Phasors provide a simple means of analyzing linear circuits excited by, sinusoidal sources; solutions of such circuits would be intractable otherwise. The notion of solving ac circuits using phasors was first introduced by Charles Steinmetz in 1893. Before we completely define, phasors and apply them to circuit analysis, we need to be thoroughly, familiar with complex numbers., A complex number z can be written in rectangular form as, z x jy, , (9.14a), , where j 11; x is the real part of z; y is the imaginary part of z., In this context, the variables x and y do not represent a location as in, two-dimensional vector analysis but rather the real and imaginary parts, of z in the complex plane. Nevertheless, we note that there are some, resemblances between manipulating complex numbers and manipulating two-dimensional vectors., The complex number z can also be written in polar or exponential, form as, z r lf re jf, , (9.14b)
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 377, , 9.3, , Phasors, , 377, , Historical, Charles Proteus Steinmetz (1865–1923), a German-Austrian, mathematician and engineer, introduced the phasor method (covered in, this chapter) in ac circuit analysis. He is also noted for his work on the, theory of hysteresis., Steinmetz was born in Breslau, Germany, and lost his mother at the, age of one. As a youth, he was forced to leave Germany because of, his political activities just as he was about to complete his doctoral dissertation in mathematics at the University of Breslau. He migrated to, Switzerland and later to the United States, where he was employed by, General Electric in 1893. That same year, he published a paper in, which complex numbers were used to analyze ac circuits for the first, time. This led to one of his many textbooks, Theory and Calculation, of ac Phenomena, published by McGraw-Hill in 1897. In 1901, he, became the president of the American Institute of Electrical Engineers,, which later became the IEEE., , where r is the magnitude of z, and f is the phase of z. We notice that, z can be represented in three ways:, z x jy, z r lf, z re j f, , Rectangular form, Polar form, Exponential form, , (9.15), , The relationship between the rectangular form and the polar form, is shown in Fig. 9.6, where the x axis represents the real part and the, y axis represents the imaginary part of a complex number. Given x and, y, we can get r and f as, r 2x y ,, 2, , 2, , f tan, , 1, , y, x, , y r sin f, , 2j, , (9.16b), , Thus, z may be written as, z x jy r lf r ( cos f j sin f), , z, , (9.16a), , On the other hand, if we know r and f, we can obtain x and y as, x r cos f,, , Imaginary axis, , y, , j, , 0, , x, , Real axis, , –j, , (9.17), , –2j, , Figure 9.6, Addition and subtraction of complex numbers are better performed, in rectangular form; multiplication and division are better done in polar, form. Given the complex numbers, z x jy r lf,, z1 x1 jy1 r1 lf1, z2 x2 jy2 r2 lf2, the following operations are important., Addition:, z1 z2 (x1 x2) j( y1 y2), , r, , (9.18a), , Representation of a complex number z , x jy r lf.
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 379, , 9.3, , Phasors, , 379, , V is thus the phasor representation of the sinusoid v(t), as we said earlier. In other words, a phasor is a complex representation of the magnitude and phase of a sinusoid. Either Eq. (9.20a) or Eq. (9.20b) can, be used to develop the phasor, but the standard convention is to use, Eq. (9.20a)., One way of looking at Eqs. (9.23) and (9.24) is to consider the plot, of the sinor Ve jt Vm e j(tf) on the complex plane. As time increases,, the sinor rotates on a circle of radius Vm at an angular velocity in the, counterclockwise direction, as shown in Fig. 9.7(a). We may regard v(t), as the projection of the sinor Ve jt on the real axis, as shown in, Fig. 9.7(b). The value of the sinor at time t 0 is the phasor V of the, sinusoid v(t). The sinor may be regarded as a rotating phasor. Thus, whenever a sinusoid is expressed as a phasor, the term e jt is implicitly present. It is therefore important, when dealing with phasors, to keep in mind, the frequency of the phasor; otherwise we can make serious mistakes., , A phasor may be regarded as a mathematical equivalent of a sinusoid with, the time dependence dropped., , Rotation at rad ⁄s, , v(t) = Re(Ve jt ), , Re, , Vm, , If we use sine for the phasor instead of, cosine, then v (t ) V m sin(t f) , Im( V m e j(t f)) and the corresponding, phasor is the same as that in Eq. (9.24)., , Vm, , , t0, t, , Im, , at t = t0, , –Vm, (a), , (b), , Figure 9.7, Representation of Ve jt: (a) sinor rotating counterclockwise, (b) its projection, on the real axis, as a function of time., , Equation (9.23) states that to obtain the sinusoid corresponding to, a given phasor V, multiply the phasor by the time factor e jt and take, the real part. As a complex quantity, a phasor may be expressed in rectangular form, polar form, or exponential form. Since a phasor has, magnitude and phase (“direction”), it behaves as a vector and is printed, in boldface. For example, phasors V Vm lf and I Im lu are, graphically represented in Fig. 9.8. Such a graphical representation of, phasors is known as a phasor diagram., Equations (9.21) through (9.23) reveal that to get the phasor corresponding to a sinusoid, we first express the sinusoid in the cosine, form so that the sinusoid can be written as the real part of a complex, number. Then we take out the time factor e jt, and whatever is left is, the phasor corresponding to the sinusoid. By suppressing the time factor, we transform the sinusoid from the time domain to the phasor, domain. This transformation is summarized as follows:, v(t) Vm cos(t f), (Time-domain, representation), , 3, , V Vmlf, (Phasor-domain, representation), , (9.25), , We use lightface italic letters such as, z to represent complex numbers but, boldface letters such as V to represent, phasors, because phasors are vectorlike quantities.
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ale29559_ch09.qxd, , 380, , 07/17/2008, , 11:46 AM, , Page 380, , Chapter 9, , Sinusoids and Phasors, Imaginary axis, V, , , , Vm, Leading direction, , Real axis, –, Lagging direction, , Im, , I, , , Figure 9.8, , A phasor diagram showing V Vm lf and I Im lu ., , Given a sinusoid v(t) Vm cos(t f), we obtain the corresponding phasor as V Vm lf. Equation (9.25) is also demonstrated, in Table 9.1, where the sine function is considered in addition to the, cosine function. From Eq. (9.25), we see that to get the phasor representation of a sinusoid, we express it in cosine form and take the, magnitude and phase. Given a phasor, we obtain the time domain, representation as the cosine function with the same magnitude as the, phasor and the argument as t plus the phase of the phasor. The idea, of expressing information in alternate domains is fundamental to all, areas of engineering., TABLE 9.1, , Sinusoid-phasor transformation., Time domain representation, , Phasor domain representation, , Vm cos(t f), , Vm lf, , Vm sin(t f), , Vm lf 90, , Im cos(t u), , Im lu, , Im sin(t u), , Im lu 90, , Note that in Eq. (9.25) the frequency (or time) factor e jt is suppressed, and the frequency is not explicitly shown in the phasor domain, representation because is constant. However, the response depends, on . For this reason, the phasor domain is also known as the frequency, domain., From Eqs. (9.23) and (9.24), v(t) Re(Ve jt) Vm cos(t f),, so that, dv, Vm sin(t f) Vm cos(t f 90), dt, Re(Vme jte jfe j 90) Re( jVe jt), , (9.26)
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 381, , 9.3, , Phasors, , 381, , This shows that the derivative v(t) is transformed to the phasor domain, as jV, , Differentiating a sinusoid is equivalent, to multiplying its corresponding phasor, by j., , dv, dt, , jV, , 3, , (Time domain), , (9.27), , (Phasor domain), , Similarly, the integral of v(t) is transformed to the phasor domain, as Vj, , v dt, (Time domain), , V, j, , 3, , (9.28), , Integrating a sinusoid is equivalent to, dividing its corresponding phasor, by j., , (Phasor domain), , Equation (9.27) allows the replacement of a derivative with respect, to time with multiplication of j in the phasor domain, whereas, Eq. (9.28) allows the replacement of an integral with respect to time, with division by j in the phasor domain. Equations (9.27) and (9.28), are useful in finding the steady-state solution, which does not require, knowing the initial values of the variable involved. This is one of the, important applications of phasors., Besides time differentiation and integration, another important, use of phasors is found in summing sinusoids of the same frequency. This is best illustrated with an example, and Example 9.6, provides one., The differences between v(t) and V should be emphasized:, , Adding sinusoids of the same frequency is equivalent to adding their, corresponding phasors., , 1. v(t) is the instantaneous or time domain representation, while V is, the frequency or phasor domain representation., 2. v(t) is time dependent, while V is not. (This fact is often forgotten by students.), 3. v(t) is always real with no complex term, while V is generally, complex., Finally, we should bear in mind that phasor analysis applies only when, frequency is constant; it applies in manipulating two or more sinusoidal, signals only if they are of the same frequency., , Evaluate these complex numbers:, (a) (40l50 20l30), , 12, , (b), , 10l30 (3 j4), (2 j4)(3 j5)*, , Solution:, (a) Using polar to rectangular transformation,, 40l50 40(cos 50 j sin 50) 25.71 j30.64, 20l30 20[cos(30) j sin(30)] 17.32 j10, Adding them up gives, 40l50 20l30 43.03 j20.64 47.72l25.63, , Example 9.3
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 383, , 9.3, , Phasors, , Find the sinusoids represented by these phasors:, , 383, , Example 9.5, , (a) I 3 j4 A, (b) V j8ej20 V, Solution:, (a) I 3 j 4 5l126.87. Transforming this to the time domain, gives, i(t) 5 cos(t 126.87) A, (b) Since j 1l90,, V j8l20 (1l90)(8l20), 8l90 20 8l70 V, Converting this to the time domain gives, v(t) 8 cos(t 70) V, , Find the sinusoids corresponding to these phasors:, , Practice Problem 9.5, , (a) V 10l30 V, (b) I j(5 j12) A, Answer: (a) v(t) 10 cos(t 210) V or 10 cos(t 150) V,, (b) i(t) 13 cos(t 22.62) A., , Given i1(t) 4 cos(t 30) A and i2(t) 5 sin(t 20) A, find, their sum., Solution:, Here is an important use of phasors—for summing sinusoids of the, same frequency. Current i1(t) is in the standard form. Its phasor is, I1 4l30, We need to express i2(t) in cosine form. The rule for converting sine, to cosine is to subtract 90. Hence,, i2 5 cos(t 20 90) 5 cos(t 110), and its phasor is, I2 5l110, If we let i i1 i2, then, I I1 I2 4l30 5l110, 3.464 j2 1.71 j4.698 1.754 j2.698, 3.218l56.97 A, , Example 9.6
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , 384, , Page 384, , Chapter 9, , Sinusoids and Phasors, , Transforming this to the time domain, we get, i(t) 3.218 cos(t 56.97) A, Of course, we can find i1 i2 using Eq. (9.9), but that is the hard way., , Practice Problem 9.6, , If v1 10 sin(t 30) V and v2 20 cos(t 45) V, find v , v1 v2., Answer: v(t) 12.158 cos(t 55.95) V., , Example 9.7, , Using the phasor approach, determine the current i(t) in a circuit, described by the integrodifferential equation, , i dt 3 dt 50 cos(2t 75), di, , 4i 8, , Solution:, We transform each term in the equation from time domain to phasor, domain. Keeping Eqs. (9.27) and (9.28) in mind, we obtain the phasor, form of the given equation as, 4I , , 8I, 3jI 50l75, j, , But 2, so, I(4 j4 j6) 50l75, I, , 50l75, 4 j10, , , , 50l75, 10.77l68.2, , 4.642l143.2 A, , Converting this to the time domain,, i(t) 4.642 cos(2t 143.2) A, Keep in mind that this is only the steady-state solution, and it does not, require knowing the initial values., , Practice Problem 9.7, , Find the voltage v(t) in a circuit described by the integrodifferential, equation, 2, , dv, 5v 10, dt, , v dt 50 cos(5t 30), , using the phasor approach., Answer: v(t) 5.3 cos(5t 88) V.
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 385, , 9.4, , 9.4, , Phasor Relationships for Circuit Elements, , 385, i, , Phasor Relationships, for Circuit Elements, , I, , +, , Now that we know how to represent a voltage or current in the phasor or frequency domain, one may legitimately ask how we apply this, to circuits involving the passive elements R, L, and C. What we need, to do is to transform the voltage-current relationship from the time, domain to the frequency domain for each element. Again, we will, assume the passive sign convention., We begin with the resistor. If the current through a resistor R is, i Im cos(t f), the voltage across it is given by Ohm’s law as, v iR RIm cos(t f), , (9.29), , +, R, , v, −, , −, v = iR, (a), , V = IR, (b), , Figure 9.9, Voltage-current relations for a resistor in, the: (a) time domain, (b) frequency domain., Im, , The phasor form of this voltage is, , V, , V RIm lf, , (9.30), , But the phasor representation of the current is I Im lf. Hence,, V RI, , I, , (9.31), , showing that the voltage-current relation for the resistor in the phasor, domain continues to be Ohm’s law, as in the time domain. Figure 9.9, illustrates the voltage-current relations of a resistor. We should note, from Eq. (9.31) that voltage and current are in phase, as illustrated in, the phasor diagram in Fig. 9.10., For the inductor L, assume the current through it is i , Im cos(t f). The voltage across the inductor is, vL, , R, , V, , di, LIm sin(t f), dt, , , 0, , Re, , Figure 9.10, Phasor diagram for the resistor., i, , I, , +, , +, , (9.32), , v, , Recall from Eq. (9.10) that sin A cos(A 90). We can write the, voltage as, , −, , −, , v = L di, dt, (a), , V = jLI, , v LIm cos(t f 90), , (9.33), , L, , V, , (b), , Figure 9.11, , which transforms to the phasor, V LIm e j(f90) LIme jf e j90 LIm lf 90, , (9.34), , Voltage-current relations for an inductor in, the: (a) time domain, (b) frequency domain., , But Im lf I, and from Eq. (9.19), e j90 j. Thus,, V jLI, , Im, , (9.35), , showing that the voltage has a magnitude of LIm and a phase of, f 90. The voltage and current are 90 out of phase. Specifically, the, current lags the voltage by 90. Figure 9.11 shows the voltage-current, relations for the inductor. Figure 9.12 shows the phasor diagram., For the capacitor C, assume the voltage across it is v , Vm cos(t f). The current through the capacitor is, iC, , dv, dt, , (9.36), , By following the same steps as we took for the inductor or by applying Eq. (9.27) on Eq. (9.36), we obtain, I jC V, , L, , 1, , V, , I, jC, , (9.37), , , V, I, , 0, , Re, , Figure 9.12, Phasor diagram for the inductor;, I lags V., Although it is equally correct to say, that the inductor voltage leads the current by 90, convention gives the current, phase relative to the voltage.
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ale29559_ch09.qxd, , 07/17/2008, , 11:46 AM, , 386, , Page 386, , Chapter 9, , Sinusoids and Phasors, , i, , Im, , I, , +, , , , +, I, C, , v, , C, , V, , −, , V, , , −, , dv, i = C dt, (a), , I = jC V, , 0, , Re, , Figure 9.14, , (b), , Phasor diagram for the capacitor; I, leads V., , Figure 9.13, Voltage-current relations for, a capacitor in the: (a) time, domain, (b) frequency, domain., , showing that the current and voltage are 90 out of phase. To be specific, the current leads the voltage by 90. Figure 9.13 shows the voltagecurrent relations for the capacitor; Fig. 9.14 gives the phasor diagram., Table 9.2 summarizes the time domain and phasor domain representations of the circuit elements., TABLE 9.2, , Summary of voltage-current relationships., Element, , Time domain, , R, , v Ri, di, vL, dt, dv, iC, dt, , L, C, , Example 9.8, , Frequency domain, V RI, V jLI, V, , I, jC, , The voltage v 12 cos(60t 45) is applied to a 0.1-H inductor. Find, the steady-state current through the inductor., Solution:, For the inductor, V jLI, where 60 rad/s and V 12l45 V., Hence,, I, , 12l45, 12l45, V, 2l45 A, , , jL, j60 0.1, 6l90, , Converting this to the time domain,, i(t) 2 cos(60t 45) A, , Practice Problem 9.8, , If voltage v 10 cos(100t 30) is applied to a 50 mF capacitor, calculate the current through the capacitor., Answer: 50 cos(100t 120) mA.
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ale29559_ch09.qxd, , 07/17/2008, , 11:46 AM, , Page 387, , 9.5, , 9.5, , Impedance and Admittance, , 387, , Impedance and Admittance, , In the preceding section, we obtained the voltage-current relations for, the three passive elements as, V RI,, , V jLI,, , V, , I, jC, , (9.38), , These equations may be written in terms of the ratio of the phasor voltage to the phasor current as, V, R,, I, , V, jL,, I, , V, 1, , I, jC, , (9.39), , From these three expressions, we obtain Ohm’s law in phasor form for, any type of element as, Z, , V, I, , or, , V ZI, , (9.40), , where Z is a frequency-dependent quantity known as impedance, measured in ohms., TABLE 9.3, The impedance Z of a circuit is the ratio of the phasor voltage V to the, phasor current I, measured in ohms ()., , The impedance represents the opposition that the circuit exhibits to, the flow of sinusoidal current. Although the impedance is the ratio of, two phasors, it is not a phasor, because it does not correspond to a sinusoidally varying quantity., The impedances of resistors, inductors, and capacitors can be, readily obtained from Eq. (9.39). Table 9.3 summarizes their impedances. From the table we notice that ZL jL and ZC jC., Consider two extreme cases of angular frequency. When 0 (i.e.,, for dc sources), ZL 0 and ZC S , confirming what we already, know—that the inductor acts like a short circuit, while the capacitor, acts like an open circuit. When S (i.e., for high frequencies),, ZL S and ZC 0, indicating that the inductor is an open circuit, to high frequencies, while the capacitor is a short circuit. Figure 9.15, illustrates this., As a complex quantity, the impedence may be expressed in rectangular form as, Z R jX, , Element Impedance, R, , ZR, , L, , Z jL, , C, , Z, , (9.42), , 1, jC, , Admittance, 1, R, 1, Y, jL, , Y, , Y jC, , Short circuit at dc, , L, , Open circuit at, high frequencies, , (9.41), , where R Re Z is the resistance and X Im Z is the reactance. The, reactance X may be positive or negative. We say that the impedance is, inductive when X is positive or capacitive when X is negative. Thus,, impedance Z R jX is said to be inductive or lagging since current, lags voltage, while impedance Z R jX is capacitive or leading, because current leads voltage. The impedance, resistance, and reactance, are all measured in ohms. The impedance may also be expressed in, polar form as, Z 0Z 0 lu, , Impedances and admittances, of passive elements., , (a), Open circuit at dc, , C, , Short circuit at, high frequencies, (b), , Figure 9.15, Equivalent circuits at dc and high, frequencies: (a) inductor, (b) capacitor.
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ale29559_ch09.qxd, , 388, , 07/08/2008, , 11:54 AM, , Page 388, , Chapter 9, , Sinusoids and Phasors, , Comparing Eqs. (9.41) and (9.42), we infer that, Z R jX 0Z 0 lu, , (9.43), , where, 0Z 0 2R 2 X 2,, , u tan1, , X, R, , (9.44), , and, X 0Z 0 sin u, , R 0Z 0 cos u,, , (9.45), , It is sometimes convenient to work with the reciprocal of impedance, known as admittance., , The admittance Y is the reciprocal of impedance, measured in, siemens (S)., , The admittance Y of an element (or a circuit) is the ratio of the phasor current through it to the phasor voltage across it, or, , Y, , 1, I, , Z, V, , (9.46), , The admittances of resistors, inductors, and capacitors can be obtained, from Eq. (9.39). They are also summarized in Table 9.3., As a complex quantity, we may write Y as, Y G jB, , (9.47), , where G Re Y is called the conductance and B Im Y is called, the susceptance. Admittance, conductance, and susceptance are, all expressed in the unit of siemens (or mhos). From Eqs. (9.41), and (9.47),, 1, R jX, , (9.48), , R jX, R jX, 1, , 2, R jX R jX, R X2, , (9.49), , G jB , By rationalization,, G jB , , Equating the real and imaginary parts gives, G, , R, ,, R X2, 2, , B, , X, R X2, 2, , (9.50), , showing that G 1R as it is in resistive circuits. Of course, if X 0,, then G 1R.
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 389, , 9.6, , Kirchhoff’s Laws in the Frequency Domain, , 389, , Find v(t) and i(t) in the circuit shown in Fig. 9.16., , Example 9.9, , Solution:, From the voltage source 10 cos 4t, 4,, , i, , Vs 10 l0 V, , vs = 10 cos 4t, , 5Ω, , +, −, , 0.1 F, , +, v, −, , The impedance is, Z5, , Figure 9.16, , 1, 1, 5, 5 j2.5, jC, j4 0.1, , For Example 9.9., , Hence the current, I, , 10l0, 10(5 j2.5), Vs, , 2, Z, 5 j2.5, 5 2.52, 1.6 j0.8 1.789l26.57 A, , (9.9.1), , The voltage across the capacitor is, V IZC , , 1.789l26.57, I, , jC, j4 0.1, , , 1.789l26.57, 0.4l90, , (9.9.2), 4.47l63.43 V, , Converting I and V in Eqs. (9.9.1) and (9.9.2) to the time domain, we get, i(t) 1.789 cos(4t 26.57) A, v(t) 4.47 cos(4t 63.43) V, Notice that i(t) leads v(t) by 90 as expected., , Practice Problem 9.9, , Refer to Fig. 9.17. Determine v(t) and i(t)., , i, , Answer: 8.944 sin(10t 93.43) V, 4.472 sin(10t 3.43) A., vs = 20 sin(10t + 30°) V +, −, , 9.6, , Kirchhoff’s Laws in the, Frequency Domain, , Figure 9.17, For Practice Prob. 9.9., , We cannot do circuit analysis in the frequency domain without Kirchhoff’s current and voltage laws. Therefore, we need to express them in, the frequency domain., For KVL, let v1, v2, p , vn be the voltages around a closed loop., Then, v1 v2 p vn 0, , (9.51), , In the sinusoidal steady state, each voltage may be written in cosine, form, so that Eq. (9.51) becomes, Vm1 cos(t u1) Vm2 cos(t u2), p Vmn cos(t un) 0, , (9.52), , 4Ω, , 0.2 H, , +, v, −
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 393, , 9.7, , Impedance Combinations, , 393, , A delta or wye circuit is said to be balanced if it has equal impedances in all three branches., , When a ¢-Y circuit is balanced, Eqs. (9.67) and (9.68) become, , Z¢ 3ZY, , or, , ZY , , 1, Z¢, 3, , (9.69), , where ZY Z1 Z2 Z3 and Z¢ Za Zb Zc., As you see in this section, the principles of voltage division, current division, circuit reduction, impedance equivalence, and Y-¢ transformation all apply to ac circuits. Chapter 10 will show that other, circuit techniques—such as superposition, nodal analysis, mesh analysis,, source transformation, the Thevenin theorem, and the Norton theorem—, are all applied to ac circuits in a manner similar to their application in, dc circuits., , Example 9.10, , Find the input impedance of the circuit in Fig. 9.23. Assume that the, circuit operates at 50 rad/s., 2 mF, , Solution:, Let, , Zin, , Z1 Impedance of the 2-mF capacitor, Z2 Impedance of the 3- resistor in series with the10-mF, capacitor, Z3 Impedance of the 0.2-H inductor in series with the 8resistor, Then, 1, 1, j10, , jC, j50 2 103, 1, 1, Z2 3 , 3, (3 j2), jC, j50 10 103, Z3 8 jL 8 j50 0.2 (8 j10), Z1 , , The input impedance is, Zin Z1 Z2 Z3 j10 , j10 , , (44 j14)(11 j8), 112 82, , (3 j2)(8 j10), 11 j8, , j10 3.22 j1.07, , Thus,, Zin 3.22 j11.07, , 0.2 H, , 3Ω, 10 mF, , Figure 9.23, For Example 9.10., , 8Ω
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 396, , Chapter 9, , 396, , Practice Problem 9.12, , Find I in the circuit of Fig. 9.30., Answer: 6.364l3.8 A., , I, −j3 Ω, , j4 Ω, , 30 0° V, , Sinusoids and Phasors, , j5 Ω, , 8Ω, , +, −, , 9.8, 5Ω, , 10 Ω, , −j2 Ω, , Figure 9.30, For Practice Prob. 9.12., , Applications, , In Chapters 7 and 8, we saw certain uses of RC, RL, and RLC circuits, in dc applications. These circuits also have ac applications; among them, are coupling circuits, phase-shifting circuits, filters, resonant circuits, ac, bridge circuits, and transformers. This list of applications is inexhaustive. We will consider some of them later. It will suffice here to observe, two simple ones: RC phase-shifting circuits, and ac bridge circuits., , 9.8.1 Phase-Shifters, I, , C, , +, R, , Vi, −, , +, Vo, −, , (a), I, , R, , +, , +, Vo, −, , C, , Vi, −, (b), , Figure 9.31, Series RC shift circuits: (a) leading, output, (b) lagging output., , vo, , A phase-shifting circuit is often employed to correct an undesirable, phase shift already present in a circuit or to produce special desired, effects. An RC circuit is suitable for this purpose because its capacitor, causes the circuit current to lead the applied voltage. Two commonly, used RC circuits are shown in Fig. 9.31. (RL circuits or any reactive, circuits could also serve the same purpose.), In Fig. 9.31(a), the circuit current I leads the applied voltage Vi, by some phase angle u, where 0 6 u 6 90, depending on the values, of R and C. If XC 1C, then the total impedance is Z R jXC,, and the phase shift is given by, u tan 1, , XC, R, , (9.70), , This shows that the amount of phase shift depends on the values of R,, C, and the operating frequency. Since the output voltage Vo across the, resistor is in phase with the current, Vo leads (positive phase shift) Vi, as shown in Fig. 9.32(a)., In Fig. 9.31(b), the output is taken across the capacitor. The current I leads the input voltage Vi by u, but the output voltage vo(t) across, the capacitor lags (negative phase shift) the input voltage vi(t) as illustrated in Fig. 9.32(b)., vi, , vi, , vo, , t, , t, , , Phase shift, , , Phase shift, (a), , Figure 9.32, Phase shift in RC circuits: (a) leading output, (b) lagging output., , (b)
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 397, , 9.8, , Applications, , 397, , We should keep in mind that the simple RC circuits in Fig. 9.31, also act as voltage dividers. Therefore, as the phase shift u approaches, 90, the output voltage Vo approaches zero. For this reason, these, simple RC circuits are used only when small amounts of phase shift, are required. If it is desired to have phase shifts greater than 60,, simple RC networks are cascaded, thereby providing a total phase, shift equal to the sum of the individual phase shifts. In practice, the, phase shifts due to the stages are not equal, because the succeeding, stages load down the earlier stages unless op amps are used to separate the stages., , Example 9.13, , Design an RC circuit to provide a phase of 90 leading., Solution:, If we select circuit components of equal ohmic value, say R 0XC 0 , 20 , at a particular frequency, according to Eq. (9.70), the phase shift, is exactly 45. By cascading two similar RC circuits in Fig. 9.31(a), we, obtain the circuit in Fig. 9.33, providing a positive or leading phase, shift of 90, as we shall soon show. Using the series-parallel combination, technique, Z in Fig. 9.33 is obtained as, Z 20 (20 j20) , , 20(20 j20), 12 j4, 40 j20, , −j20 Ω, , V1, , −j20 Ω, , +, , +, , Vi, , 20 Ω, , 20 Ω, , −, , Vo, −, , Z, , Figure 9.33, (9.13.1), , An RC phase shift circuit with 90 leading, phase shift; for Example 9.13., , Using voltage division,, V1 , , 12 j4, Z, 12, l45 Vi, Vi , Vi , Z j20, 12 j24, 3, , (9.13.2), , 20, 12, l45 V1, V1 , 20 j20, 2, , (9.13.3), , and, Vo , , Substituting Eq. (9.13.2) into Eq. (9.13.3) yields, Vo a, , 12, l45b a 12 l45 Vi b 1 l90 Vi, 3, 3, 2, , Thus, the output leads the input by 90 but its magnitude is only about, 33 percent of the input., , Practice Problem 9.13, , Design an RC circuit to provide a 90 lagging phase shift of the output voltage relative to the input voltage. If an ac voltage of 10 V rms, is applied, what is the output voltage?, , 10 Ω, , 10 Ω, , +, , Answer: Figure 9.34 shows a typical design; 3.33 V rms., , Vi, , −j10 Ω, , −, , Figure 9.34, For Practice Prob. 9.13., , −j10 Ω, , +, Vo, −
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 398, , Chapter 9, , 398, , Example 9.14, 150 Ω, , For the RL circuit shown in Fig. 9.35(a), calculate the amount of phase, shift produced at 2 kHz., , 100 Ω, , 10 mH, , Solution:, At 2 kHz, we transform the 10-mH and 5-mH inductances to the, corresponding impedances., , 5 mH, , (a), 150 Ω, , 100 Ω, , V1, , Sinusoids and Phasors, , +, , 10 mH, , 1, , 5 mH, , 1, , XL L 2p 2 103 10 103, 40p 125.7, XL L 2p 2 103 5 103, 20p 62.83, , +, j125.7 Ω, , Vi, , j62.83 Ω, , Vo, , −, , −, , Consider the circuit in Fig. 9.35(b). The impedance Z is the parallel, combination of j 125.7 and 100 j 62.83 . Hence,, Z j125.7 7 (100 j62.83), , Z, (b), , Figure 9.35, , , , For Example 9.14., , j125.7(100 j62.83), 69.56l60.1, 100 j188.5, , (9.14.1), , Using voltage division,, 69.56 l60.1, Z, Vi , Vi, Z 150, 184.7 j60.3, 0.3582 l42.02 Vi, , (9.14.2), , j62.832, V1 0.532l57.86 V1, 100 j62.832, , (9.14.3), , V1 , , and, Vo , , Combining Eqs. (9.14.2) and (9.14.3),, Vo (0.532 l57.86)(0.3582 l42.02) Vi 0.1906l100 Vi, showing that the output is about 19 percent of the input in magnitude, but leading the input by 100. If the circuit is terminated by a load, the, load will affect the phase shift., , Practice Problem 9.14, 1 mH, , 2 mH, , +, Vi, , +, 10 Ω, , −, , 50 Ω, , Vo, , Refer to the RL circuit in Fig. 9.36. If 1 V is applied, find the magnitude and the phase shift produced at 5 kHz. Specify whether the phase, shift is leading or lagging., Answer: 0.172, 120.4, lagging., , −, , Figure 9.36, For Practice Prob. 9.14., , 9.8.2 AC Bridges, An ac bridge circuit is used in measuring the inductance L of an, inductor or the capacitance C of a capacitor. It is similar in form to, the Wheatstone bridge for measuring an unknown resistance (discussed in Section 4.10) and follows the same principle. To measure, L and C, however, an ac source is needed as well as an ac meter
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 399, , 9.8, , Applications, , instead of the galvanometer. The ac meter may be a sensitive ac, ammeter or voltmeter., Consider the general ac bridge circuit displayed in Fig. 9.37. The, bridge is balanced when no current flows through the meter. This, means that V1 V2. Applying the voltage division principle,, V1 , , Zx, Z2, Vs V2 , Vs, Z1 Z2, Z3 Zx, , 399, , Z1, Vs, , AC, meter, , ≈, , (9.71), , Z2, , Thus,, Zx, Z2, , Z1 Z2, Z3 Zx, , Figure 9.37, , Z2Z3 Z1Zx, , 1, , (9.72), , or, Zx , , Z3, Z2, Z1, , (9.73), , This is the balanced equation for the ac bridge and is similar to Eq. (4.30), for the resistance bridge except that the R’s are replaced by Z’s., Specific ac bridges for measuring L and C are shown in Fig. 9.38,, where Lx and Cx are the unknown inductance and capacitance to be, measured while Ls and Cs are a standard inductance and capacitance, (the values of which are known to great precision). In each case, two, resistors, R1 and R2, are varied until the ac meter reads zero. Then the, bridge is balanced. From Eq. (9.73), we obtain, Lx , , R2, Ls, R1, , (9.74), , Cx , , R1, Cs, R2, , (9.75), , and, , Notice that the balancing of the ac bridges in Fig. 9.38 does not depend, on the frequency f of the ac source, since f does not appear in the relationships in Eqs. (9.74) and (9.75)., , R1, , R2, , R1, , AC, meter, Ls, , R2, AC, meter, , Lx, , Cs, , Cx, , ≈, , ≈, , (a), , (b), , Figure 9.38, Specific ac bridges: (a) for measuring L, (b) for measuring C., , Z3, , A general ac bridge., , +, V1, −, , +, V2, −, , Zx
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ale29559_ch09.qxd, , 07/08/2008, , 400, , Example 9.15, , 11:54 AM, , Page 400, , Chapter 9, , Sinusoids and Phasors, , The ac bridge circuit of Fig. 9.37 balances when Z1 is a 1-k resistor,, Z2 is a 4.2-k resistor, Z3 is a parallel combination of a 1.5-M, resistor and a 12-pF capacitor, and f 2 kHz. Find: (a) the series components that make up Zx, and (b) the parallel components that make, up Zx., Solution:, 1. Define. The problem is clearly stated., 2. Present. We are to determine the unknown components subject, to the fact that they balance the given quantities. Since a parallel, and series equivalent exists for this circuit, we need to find both., 3. Alternative. Although there are alternative techniques that can, be used to find the unknown values, a straightforward equality, works best. Once we have answers, we can check them by using, hand techniques such as nodal analysis or just using PSpice., 4. Attempt. From Eq. (9.73),, Zx , , Z3, Z2, Z1, , (9.15.1), , where Zx Rx jXx ,, Z1 1000, , ,, , Z2 4200, , (9.15.2), , and, R3, jC3, R3, 1, Z3 R3 , , , jC3, R3 1jC3, 1 jR3C3, Since R3 1.5 M, Z3 , , and C3 12 pF,, , 1.5 106, 1.5 106, , 1 j2p 2 103 1.5 106 12 1012 1 j0.2262, , or, Z3 1.427 j 0.3228 M, , (9.15.3), , (a) Assuming that Zx is made up of series components, we, substitute Eqs. (9.15.2) and (9.15.3) in Eq. (9.15.1) and obtain, Rx jXx , , 4200, (1.427 j 0.3228) 106, 1000, , (5.993 j1.356) M, , (9.15.4), , Equating the real and imaginary parts yields Rx 5.993 M, a capacitive reactance, Xx , , 1, 1.356 106, C, , or, C, , 1, 1, , 58.69 pF, 3, Xx, 2p 2 10 1.356 106, , and
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 401, , 9.8, , Applications, , (b) Zx remains the same as in Eq. (9.15.4) but Rx and Xx are in, parallel. Assuming an RC parallel combination,, Zx (5.993 j1.356) M, Rx, 1, , Rx , jCx, 1 jRxCx, By equating the real and imaginary parts, we obtain, Rx , , Real(Zx)2 Imag(Zx)2, 5.9932 1.3562, , 6.3 M, Real(Zx), 5.993, , Cx , , , Imag(Zx), [Real(Zx)2 Imag(Zx)2], 1.356, 2.852 mF, 2 p (2000)(5.9172 1.3562), , We have assumed a parallel RC combination which works in, this case., 5. Evaluate. Let us now use PSpice to see if we indeed have the, correct equalities. Running PSpice with the equivalent circuits,, an open circuit between the “bridge” portion of the circuit,, and a 10-volt input voltage yields the following voltages at the, ends of the “bridge” relative to a reference at the bottom of, the circuit:, FREQ, VM($N_0002) VP($N_0002), 2.000E+03 9.993E+00 -8.634E-03, 2.000E+03 9.993E+00 -8.637E-03, Since the voltages are essentially the same, then no measurable, current can flow through the “bridge” portion of the circuit for, any element that connects the two points together and we have a, balanced bridge, which is to be expected. This indicates we have, properly determined the unknowns., There is a very important problem with what we have done!, Do you know what that is? We have what can be called an, ideal, “theoretical” answer, but one that really is not very good, in the real world. The difference between the magnitudes of the, upper impedances and the lower impedances is much too large, and would never be accepted in a real bridge circuit. For, greatest accuracy, the overall magnitude of the impedances must, at least be within the same relative order. To increase the, accuracy of the solution of this problem, I would recommend, increasing the magnitude of the top impedances to be in the, range of 500 k to 1.5 M . One additional real-world comment:, the size of these impedances also creates serious problems in, making actual measurements, so the appropriate instruments, must be used in order to minimize their loading (which would, change the actual voltage readings) on the circuit., 6. Satisfactory? Since we solved for the unknown terms and then, tested to see if they woked, we validated the results. They can, now be presented as a solution to the problem., , 401
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , 402, , Practice Problem 9.15, , Page 402, , Chapter 9, , Sinusoids and Phasors, , In the ac bridge circuit of Fig. 9.37, suppose that balance is achieved, when Z1 is a 4.8-k resistor, Z2 is a 10- resistor in series with a, 0.25-mH inductor, Z3 is a 12-k resistor, and f 6 MHz. Determine, the series components that make up Zx., Answer: A 25-, , 9.9, , resistor in series with a 0.625-mH inductor., , Summary, , 1. A sinusoid is a signal in the form of the sine or cosine function., It has the general form, v(t) Vm cos(t f), where Vm is the amplitude, 2 p f is the angular frequency,, (t f) is the argument, and f is the phase., 2. A phasor is a complex quantity that represents both the magnitude and the phase of a sinusoid. Given the sinusoid v(t) , Vm cos(t f), its phasor V is, V Vmlf, 3. In ac circuits, voltage and current phasors always have a fixed, relation to one another at any moment of time. If v(t) , Vm cos(t fv) represents the voltage through an element and, i(t) Im cos(t fi) represents the current through the element,, then fi fv if the element is a resistor, fi leads fv by 90 if the, element is a capacitor, and fi lags fv by 90 if the element is an, inductor., 4. The impedance Z of a circuit is the ratio of the phasor voltage, across it to the phasor current through it:, Z, , V, R() jX(), I, , The admittance Y is the reciprocal of impedance:, Y, , 1, G() jB(), Z, , Impedances are combined in series or in parallel the same way as, resistances in series or parallel; that is, impedances in series add, while admittances in parallel add., 5. For a resistor Z R, for an inductor Z j X jL, and for a, capacitor Z jX 1jC., 6. Basic circuit laws (Ohm’s and Kirchhoff’s) apply to ac circuits in, the same manner as they do for dc circuits; that is,, V ZI, Ik 0 (KCL), Vk 0 (KVL)
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ale29559_ch09.qxd, , 07/08/2008, , 11:54 AM, , Page 403, , Problems, , 403, , 7. The techniques of voltage/current division, series/parallel combination of impedance/admittance, circuit reduction, and Y-¢ transformation all apply to ac circuit analysis., 8. AC circuits are applied in phase-shifters and bridges., , Review Questions, 9.1, , 9.2, , 9.3, , Which of the following is not a right way to express, the sinusoid A cos t ?, (b) A cos(2 p tT ), , (a) 0 rad/s, , (b) 1 rad/s, , (c) A cos (t T ), , (d) A sin(t 90), , (d), , (e) none of the above, , (a) a phasor, , (b) harmonic, , (c) periodic, , (d) reactive, , 1Ω, , Which of these frequencies has the shorter period?, , v(t), , (b) 1 kHz, , If v1 30 sin(t 10) and v2 20 sin(t 50),, which of these statements are true?, (a) v1 leads v2, , (b) v2 leads v1, , (c) v2 lags v1, , (d) v1 lags v2, , The voltage across an inductor leads the current, through it by 90., (a) True, , 9.6, , (a) resistance, , (b) admittance, , (c) susceptance, , (d) conductance, , The impedance of a capacitor increases with, increasing frequency., (a) True, , (b) False, , 1, 4, , H, , +, vo(t), −, , For Review Question 9.8., , 9.9, , A series RC circuit has 0 VR 0 12 V and 0VC 0 5 V., The magnitude of the supply voltage is:, (a) 7 V, , (b) False, , The imaginary part of impedance is called:, , +, −, , Figure 9.39, , (b) 7 V, , (c) 13 V, , (d) 17 V, , 9.10 A series RCL circuit has R 30 , XC 50, XL 90 . The impedance of the circuit is:, , (e) reactance, 9.7, , rad/s, , (c) 4 rad/s, , A function that repeats itself after fixed intervals is, said to be:, , (e) v1 and v2 are in phase, 9.5, , At what frequency will the output voltage vo(t) in, Fig. 9.39 be equal to the input voltage v(t) ?, , (a) A cos 2 p ft, , (a) 1 krad/s, 9.4, , 9.8, , (a) 30 j140, , (b) 30 j40, , (c) 30 j40, , (d) 30 j40, , (e) 30 j40, , Answers: 9.1d, 9.2c, 9.3b, 9.4b,d, 9.5a, 9.6e, 9.7b, 9.8d,, 9.9c, 9.10b., , Problems, Section 9.2 Sinusoids, 9.1, , 9.2, , Given the sinusoidal voltage v(t) , 50 cos(30t 10) V, find: (a) the amplitude, Vm, (b) the period T, (c) the frequency f, and, (d) v(t) at t 10 ms., A current source in a linear circuit has, is 8 cos(500p t 25) A, , , and, , (a) What is the amplitude of the current?, (b) What is the angular frequency?, (c) Find the frequency of the current., (d) Calculate is at t 2 ms., 9.3, , Express the following functions in cosine form:, (a) 4 sin(t 30), (c) 10 sin(t 20), , (b) 2 sin 6t
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ale29559_ch09.qxd, , 07/08/2008, , 11:55 AM, , Page 404, , Chapter 9, , 404, , Sinusoids and Phasors, , 9.4, , Design a problem to help other students better, understand sinusoids., , 9.5, , Given v1 20 sin(t 60) and v2 , 60 cos(t 10), determine the phase angle, between the two sinusoids and which one lags, the other., , (a), , For the following pairs of sinusoids, determine, which one leads and by how much., , (c) 2, , 9.6, , (a) v(t) 10 cos(4t 60) and, i(t) 4 sin(4t 50), , (c) x(t) 13 cos 2t 5 sin 2t and, y(t) 15 cos(2t 11.8), , (b), , If f (f) cos f j sin f, show that f (f) e jf., , 9.8, , Calculate these complex numbers and express your, results in rectangular form:, , (b), , 3 j4, , (2 j)(3 j4), , , , 10, 5 j12, , (c) 10 (8l50)(5 j12), Evaluate the following complex numbers and leave, your results in polar form:, (a) 5l30 a6 j8 , (b), , 3l60, 2j, , b, , (10l60)(35l50), (2 j6) (5 j), , 9.10 Design a problem to help other students better, understand phasors., 9.11 Find the phasors corresponding to the following, signals:, (a) v(t) 21 cos(4t 15) V, (b) i(t) 8 sin(10t 70) mA, (c) v(t) 120 sin(10t 50) V, (d) i(t) 60 cos(30t 10) mA, 9.12 Let X 8l40 and Y 10l30. Evaluate the, following quantities and express your results in, polar form:, (a) (X Y)X*, (b) (X Y)*, (c) (X Y)X, , j2, 2, 8 j5, , 2 j3, j2, , (5 j6) (2 j8), (3 j4)(5 j) (4 j6), (240l75 160l30)(60 j80), (67 j84)(20l32), 10 j20 2, b 1(10 j5)(16 j20), 3 j4, , 9.15 Evaluate these determinants:, (a) 2, (b) 2, , j2, , 8l20, , (4l80)(6l50), , (c) a, , 9.7, , 15 l45, , (5l10)(10l40), , 9.14 Simplify the following expressions:, (a), , (a), , 2 j3, 7 j8, , 1 j6, 5 j11, , (b), , (b) v1(t) 4 cos(377t 10) and, v2(t) 20 cos 377t, , Section 9.3 Phasors, , 9.9, , 9.13 Evaluate the following complex numbers:, , 10 j6, 5, , 2 j3, 2, 1 j, , 20l30, , 4l10, , 16l0, , 3l45, , 1j, (c) 3 j, 1, , j, 1, j, , 2, , 0, j 3, 1j, , 9.16 Transform the following sinusoids to phasors:, (a) 10 cos(4t 75), , (b) 5 sin(20t 10), , (c) 4 cos 2t 3 sin 2t, 9.17 Two voltages v1 and v2 appear in series so that their, sum is v v1 v2. If v1 10 cos(50t p3) V, and v2 12 cos(50t 30) V, find v., 9.18 Obtain the sinusoids corresponding to each of the, following phasors:, (a) V1 60l15 V, 1, (b) V2 6 j8 V, 40, (c) I1 2.8ejp3 A, 377, (d) I2 0.5 j1.2 A, 10 3, 9.19 Using phasors, find:, (a) 3 cos(20t 10) 5 cos(20t 30), (b) 40 sin 50t 30 cos(50t 45), (c) 20 sin 400t 10 cos(400t 60), 5 sin(400t 20), 9.20 A linear network has a current input, 4 cos(t 20) A and a voltage output, 10 cos(t 110) V. Determine the associated, impedance.
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ale29559_ch09.qxd, , 07/08/2008, , 11:55 AM, , Page 405, , Problems, , 9.30 A voltage v(t) 100 cos(60t 20) V is applied to, a parallel combination of a 40-k resistor and a, 50-mF capacitor. Find the steady-state currents, through the resistor and the capacitor., , 9.21 Simplify the following:, (a) f (t) 5 cos(2t 15) 4 sin(2t 30), (b) g(t) 8 sin t 4 cos(t 50), t, , (c) h(t) , , (10 cos 40t 50 sin 40t) dt, , 9.31 A series RLC circuit has R 80 , L 240 mH,, and C 5 mF. If the input voltage is v(t) , 10 cos 2t, find the currrent flowing through the circuit., , 0, , 9.22 An alternating voltage is given by v(t) , 20 cos(5t 30) V. Use phasors to find, dv, 10v(t) 4 2, dt, , , , 405, , 9.32 Using Fig. 9.40, design a problem to help other, students better understand phasor relationships for, circuit elements., , t, , v(t) dt, , , , Assume that the value of the integral is zero at, t ., , IL, , 9.23 Apply phasor analysis to evaluate the following., , Load, (R + jL), , v +, −, , (a) v 50 cos(t 30) 30 cos(t 90) V, (b) i 15 cos(t 45) 10 sin(t 45) A, 9.24 Find v(t) in the following integrodifferential, equations using the phasor approach:, , Figure 9.40, For Prob. 9.32., , v dt 5 cos (t 45) V, dv, 5v(t) 4 v dt 20 sin(4t 10) V, (b), dt, (a) v(t) , , 9.33 A series RL circuit is connected to a 110-V ac, source. If the voltage across the resistor is 85 V, find, the voltage across the inductor., , 9.25 Using phasors, determine i(t) in the following, equations:, , 9.34 What value of will cause the forced response vo in, Fig. 9.41 to be zero?, , di, (a) 2 3i(t) 4 cos(2t 45), dt, (b) 10, , i dt dt 6i(t) 5 cos(5t 22) A, di, , 2Ω, +, , 9.26 The loop equation for a series RLC circuit gives, di, 2i , dt, , , , 5 mF, 100 cos(t + 45°) V +, −, , t, , i dt cos 2t A, , , , Assuming that the value of the integral at t , zero, find i(t) using the phasor method., , −, , is, , 9.27 A parallel RLC circuit has the node equation, dv, 50v 100, dt, , vo, 20 mH, , v dt 110 cos(377t 10) V, , Determine v(t) using the phasor method. You may, assume that the value of the integral at t is zero., , Figure 9.41, For Prob. 9.34., , Section 9.5 Impedance and Admittance, 9.35 Find current i in the circuit of Fig. 9.42, when, vs(t) 50 cos 200t V., , Section 9.4 Phasor Relationships for Circuit, Elements, 9.28 Determine the current that flows through an 8resistor connected to a voltage source, vs 110 cos 377t V., 9.29 What is the instantaneous voltage across a 2-mF, capacitor when the current through it is, i 4 sin(106 t 25) A?, , i, , vs, , Figure 9.42, For Prob. 9.35., , +, −, , 10 Ω, , 5 mF, , 20 mH
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ale29559_ch09.qxd, , 07/08/2008, , 11:55 AM, , Page 406, , Chapter 9, , 406, , Sinusoids and Phasors, , 9.36 Using Fig. 9.43, design a problem to help other, students better understand impedance., i, , R1, , 9.40 In the circuit of Fig. 9.47, find io when:, (a) 1 rad/s, , (b) 5 rad/s, , (c) 10 rad/s, , L, , io, , vs, , +, −, , C, , R2, , R3, , 1H, , 10 cos t V +, −, , 2Ω, , 0.05 F, , Figure 9.47, , Figure 9.43, , For Prob. 9.40., , For Prob. 9.36., , 9.41 Find v(t) in the RLC circuit of Fig. 9.48., 9.37 Determine the admittance Y for the circuit in Fig. 9.44., , 1Ω, 1Ω, , Y, , 2Ω, , j4 Ω, , −j5 Ω, , 25 cos t V +, −, , +, v (t), −, , 1F, 1H, , Figure 9.44, , Figure 9.48, , For Prob. 9.37., , For Prob. 9.41., , 9.38 Using Fig. 9.45, design a problem to help other, students better understand admittance., , 9.42 Calculate vo (t) in the circuit of Fig. 9.49., 30 Ω, , i, is(t), , R, , +, v, −, , C, , 50 Ω, , 50 F, , 100 sin 200t V +, −, , 0.1 H, , +, vo(t), −, , (a), , Figure 9.49, , i, R2, R1, , vs(t) +, −, , C, , +, v, −, , For Prob. 9.42., 9.43 Find current Io in the circuit shown in Fig. 9.50., , L, , Io, , 50 Ω, , 100 Ω, , (b), , Figure 9.45, , 60 0° V +, −, , For Prob. 9.38., 9.39 For the circuit shown in Fig. 9.46, find Z eq and use, that to find current I. Let 10 rad/s., I, , 4Ω, , j20 Ω, , j80 Ω, , Figure 9.50, For Prob. 9.43., 9.44 Calculate i(t) in the circuit of Fig. 9.51., , −j14 Ω, , i, 12 0° V +, −, , 16 Ω, , −j40 Ω, , j25 Ω, , 50 cos 200t V +, −, , Figure 9.46, , Figure 9.51, , For Prob. 9.39., , For prob. 9.44., , 5Ω, 4Ω, , 5 mF, , 10 mH, , 3Ω
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ale29559_ch09.qxd, , 07/08/2008, , 11:55 AM, , Page 407, , Problems, , 9.50 Determine vx in the circuit of Fig. 9.57. Let is(t) , 5 cos(100t 40) A., , 9.45 Find current Io in the network of Fig. 9.52., j4 Ω, , 2Ω, , 407, , 0.1 H, , Io, −j2 Ω, , 5 0° A, , 2Ω, , −j2 Ω, , is (t), , Figure 9.52, , 20 Ω, , 1 mF, , +, vx, –, , Figure 9.57, , For Prob. 9.45., , For Prob. 9.50., , 9.46 If is 20 cos(10t 15) A in the circuit of, Fig. 9.53, find io., 4Ω, , 9.51 If the voltage vo across the 2- resistor in the circuit, of Fig. 9.58 is 5 cos 2t V, obtain is., , 3Ω, , 0.1 F, , 0.5 H, , io, is, , 0.2 H, , 0.1 F, , Figure 9.53, , +, vo, −, , 1Ω, , is, , 2Ω, , Figure 9.58, , For Prob. 9.46., , For Prob. 9.51., , 9.47 In the circuit of Fig. 9.54, determine the value, of is(t)., is (t), , 20 cos 2000t V, , 2Ω, , 9.52 If Vo 20l45 V in the circuit of Fig. 9.59, find Is., −j5 Ω, , 2 mH, , +, −, , 50 F, , 20 Ω, , 10 Ω, , Is, , 5Ω, , j5 Ω, , +, Vo, −, , Figure 9.59, , Figure 9.54, , For Prob. 9.52., , For Prob. 9.47., 9.48 Given that vs(t) 20 sin(100t 40) in Fig. 9.55,, determine ix(t)., 10 Ω, , 9.53 Find Io in the circuit of Fig. 9.60., 4Ω, , 30 Ω, , Io, ix, , +, vs (t) −, , 0.2 H, , 0.5 mF, , 2Ω, , –j2 Ω, , +, 60 –30° V −, , j6 Ω, , 8Ω, , 10 Ω, , Figure 9.55, For Prob. 9.48., , Figure 9.60, For Prob. 9.53., , 9.49 Find vs(t) in the circuit of Fig. 9.56 if the current ix, through the 1- resistor is 0.5 sin 200t A., 2Ω, , vs, , +, −, , ix, , 9.54 In the circuit of Fig. 9.61, find Vs if Io 2l0 A., −j2 Ω, , 1Ω, , j2 Ω, , Vs, +−, , −j1 Ω, , 2Ω, , Figure 9.56, , Figure 9.61, , For Prob. 9.49., , For Prob. 9.54., , j4 Ω, , j2 Ω, , −j1 Ω, Io, 1Ω
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ale29559_ch09.qxd, , 07/08/2008, , 11:55 AM, , Page 408, , Chapter 9, , 408, , Sinusoids and Phasors, , *9.55 Find Z in the network of Fig. 9.62, given that, Vo 8l0 V., , 9.59 For the network in Fig. 9.66, find Zin. Let , 10 rad/s., , 12 Ω, , 1, F, 4, , Z, 40 −90° V +, −, , +, Vo, −, , j8 Ω, , −j4 Ω, , Figure 9.62, , Zin, , 5Ω, , 0.5 H, , Figure 9.66, , For Prob. 9.55., , For Prob. 9.59., , Section 9.7 Impedance Combinations, , 9.60 Obtain Zin for the circuit in Fig. 9.67., , 9.56 At 377 rad/s, find the input impedance of the, circuit shown in Fig. 9.63., 12 Ω, , 50 Ω, , 50 F, , j30 Ω, –j100 Ω, , 60 Ω, , Zin, 60 mH, , 40 Ω, , 40 Ω, , j20 Ω, , Figure 9.67, , Figure 9.63, , For Prob. 9.60., , For Prob. 9.56., 9.57 At 1 rad/s, obtain the input admittance in the, circuit of Fig. 9.64., 1Ω, , 2Ω, , Yin, , 2H, , 9.61 Find Zeq in the circuit of Fig. 9.68., , Zeq, , 1F, , 1−jΩ, , 1 + j3 Ω, , 1 + j2 Ω, j5 Ω, , Figure 9.64, For Prob. 9.57., , Figure 9.68, 9.58 Using Fig. 9.65, design a problem to help other, students better understand impedance combinations., , For Prob. 9.61., 9.62 For the circuit in Fig. 9.69, find the input impedance, Zin at 10 krad/s., , R1, , L, , C, , R2, , 50 Ω, , 2 mH, , + v −, 1 F, , Figure 9.65, For Prob. 9.58., , Zin, , Figure 9.69, * An asterisk indicates a challenging problem., , For Prob. 9.62., , +, −, , 2v
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ale29559_ch09.qxd, , 07/08/2008, , 11:55 AM, , Page 410, , Chapter 9, , 410, , Sinusoids and Phasors, , 9.71 Obtain the equivalent impedance of the circuit in, Fig. 9.78., , (a) Calculate the phase shift at 2 MHz., (b) Find the frequency where the phase shift is 45., , j4 Ω, , 5Ω, , −j Ω, , 2Ω, , +, −, , Figure 9.78, , Figure 9.81, , For Prob. 9.71., , For Prob. 9.77., , 9.72 Calculate the value of Zab in the network of, Fig. 9.79., −j9 Ω, , j6 Ω, a, , −j9 Ω, , j6 Ω, j6 Ω, , −j9 Ω, 20 Ω, 20 Ω, , 9.78 A coil with impedance 8 j6 is connected in, series with a capacitive reactance X. The series, combination is connected in parallel with a resistor, R. Given that the equivalent impedance of the, resulting circuit is 5l0 , find the value of R and X., 9.79 (a) Calculate the phase shift of the circuit in Fig. 9.82., (b) State whether the phase shift is leading or, lagging (output with respect to input)., (c) Determine the magnitude of the output when the, input is 120 V., 20 Ω, , 10 Ω, , Figure 9.79, , Vi, , For Prob. 9.72., , −, , 9.73 Determine the equivalent impedance of the circuit in, Fig. 9.80., , j10 Ω, , j30 Ω, , j60 Ω, , +, Vo, −, , Figure 9.82, For Prob. 9.79., , (a) Vo when R is maximum, , −j6 Ω, , (b) Vo when R is minimum, , 4Ω, , (c) the value of R that will produce a phase shift of 45, , a, j8 Ω, , j8 Ω, , 0 < R < 100 Ω, , j12 Ω, , b, , 50 Ω, , +, , Figure 9.80, , vi, , For Prob. 9.73., , −, , Section 9.8 Applications, , 200 mH, , +, vo, −, , Figure 9.83, , 9.74 Design an RL circuit to provide a 90 leading phase, shift., 9.75 Design a circuit that will transform a sinusoidal, voltage input to a cosinusoidal voltage output., 9.76 For the following pairs of signals, determine if v1, leads or lags v2 and by how much., (a) v1 10 cos(5t 20),, , v2 8 sin 5t, , (b) v1 19 cos(2t 90),, , v2 6 sin 2t, , (c) v1 4 cos 10t,, , 30 Ω, , 9.80 Consider the phase-shifting circuit in Fig. 9.83. Let, Vi 120 V operating at 60 Hz. Find:, , −j4 Ω, , j6 Ω, , 40 Ω, , +, , b, , 2Ω, , +, Vo, −, , 20 nF, , Vi, , Zeq, , −j2 Ω, , j2 Ω, , 1Ω, , 9.77 Refer to the RC circuit in Fig. 9.81., , v2 15 sin 10t, , For Prob. 9.80., 9.81 The ac bridge in Fig. 9.37 is balanced when R1 , 400 , R2 600 , R3 1.2 k , and C2 0.3 mF., Find Rx and Cx. Assume R2 and C2 are in series., 9.82 A capacitance bridge balances when R1 100, R2 2 k , and Cs 40 mF. What is Cx, the, capacitance of the capacitor under test?, , ,, , 9.83 An inductive bridge balances when R1 1.2 k ,, R2 500 , and Ls 250 mH. What is the value of, Lx, the inductance of the inductor under test?
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ale29559_ch09.qxd, , 07/08/2008, , 11:55 AM, , Page 411, , Comprehensive Problems, , 9.84 The ac bridge shown in Fig. 9.84 is known as a, Maxwell bridge and is used for accurate measurement, of inductance and resistance of a coil in terms of a, standard capacitance Cs. Show that when the bridge, is balanced,, Lx R2R3Cs, , and, , Rx , , 411, , 9.85 The ac bridge circuit of Fig. 9.85 is called a Wien, bridge. It is used for measuring the frequency of a, source. Show that when the bridge is balanced,, f, , R2, R3, R1, , 1, 2p 2R2R4C2C4, , Find Lx and Rx for R1 40 k , R2 1.6 k ,, R3 4 k , and Cs 0.45 mF., R1, , R1, , R3, Cs, , R3, AC, meter, , AC, meter, R2, , Lx, , R2, , R4, C2, , Rx, , C4, , Figure 9.84, , Figure 9.85, , Maxwell bridge; For Prob. 9.84., , Wein bridge; For Prob. 9.85., , Comprehensive Problems, −j20 Ω, , 9.86 The circuit shown in Fig. 9.86 is used in a television, receiver. What is the total impedance of this circuit?, 250 Hz, 240 Ω, , j95 Ω, , j30 Ω, , 120 Ω, −j20 Ω, , ≈, , −j84 Ω, , Figure 9.88, For Prob. 9.88., , Figure 9.86, For Prob. 9.86., 9.87 The network in Fig. 9.87 is part of the schematic, describing an industrial electronic sensing device., What is the total impedance of the circuit at 2 kHz?, , 50 Ω, , 10 mH, , 2 F, , 80 Ω, , 9.89 An industrial load is modeled as a series combination, of a capacitance and a resistance as shown in Fig. 9.89., Calculate the value of an inductance L across the, series combination so that the net impedance is, resistive at a frequency of 50 kHz., , 100 Ω, , 200 Ω, L, 200 nF, , Figure 9.87, For Prob. 9.87., , Figure 9.89, For Prob. 9.89., , 9.88 A series audio circuit is shown in Fig. 9.88., (a) What is the impedance of the circuit?, (b) If the frequency were halved, what would be the, impedance of the circuit?, , 9.90 An industrial coil is modeled as a series combination, of an inductance L and resistance R, as shown in, Fig. 9.90. Since an ac voltmeter measures only, the magnitude of a sinusoid, the following
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ale29559_ch09.qxd, , 07/08/2008, , 11:55 AM, , Page 412, , Chapter 9, , 412, , Sinusoids and Phasors, , measurements are taken at 60 Hz when the circuit, operates in the steady state:, 0 Vs 0 145 V,, , 0V1 0 50 V,, , 0Vo 0 110 V, , Use these measurements to determine the values of L, and R., 80 Ω, , Coil, , + V −, 1, , +, R, , Vs, , +, −, , Vo, L, −, , 9.92 A transmission line has a series impedance of, Z 100l75 and a shunt admittance of Y , 450l48 mS. Find: (a) the characteristic impedance, Zo 1ZY , (b) the propagation constant g , 1ZY., 9.93 A power transmission system is modeled as shown in, Fig. 9.92. Given the following;, Source voltage, Source impedance, Line impedance, Load impedance, Find the load current, , Vs 115l0 V,, Zs (2 j) ,, Z/ (0.8 j0.6) ,, ZL (46.4 j37.8) ,, IL., , Figure 9.90, For Prob. 9.90., , Zs, , 9.91 Figure 9.91 shows a parallel combination of an, inductance and a resistance. If it is desired to connect, a capacitor in series with the parallel combination, such that the net impedance is resistive at 10 MHz,, what is the required value of C?, , IL, vs, , +, −, , Figure 9.92, For Prob. 9.93., 300 Ω, , Figure 9.91, For Prob. 9.91., , 20 H, , ZL, Zᐉ, , Source, , C, , Zᐉ, , Transmission line, , Load
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ale29559_ch10.qxd, , 07/08/2008, , 11:58 AM, , Page 413, , c h a p t e r, , Sinusoidal SteadyState Analysis, , 10, , Three men are my friends—he that loves me, he that hates me, he that, is indifferent to me. Who loves me, teaches me tenderness; who hates, me, teaches me caution; who is indifferent to me, teaches me selfreliance., —J. E. Dinger, , Enhancing Your Career, Career in Software Engineering, Software engineering is that aspect of engineering that deals with the, practical application of scientific knowledge in the design, construction,, and validation of computer programs and the associated documentation, required to develop, operate, and maintain them. It is a branch of electrical engineering that is becoming increasingly important as more and, more disciplines require one form of software package or another to perform routine tasks and as programmable microelectronic systems are, used in more and more applications., The role of a software engineer should not be confused with, that of a computer scientist; the software engineer is a practitioner,, not a theoretician. A software engineer should have good computerprogramming skills and be familiar with programming languages, in, particular C, which is becoming increasingly popular. Because hardware and software are closely interlinked, it is essential that a software, engineer have a thorough understanding of hardware design. Most, important, the software engineer should have some specialized knowledge of the area in which the software development skill is to be, applied., All in all, the field of software engineering offers a great career to, those who enjoy programming and developing software packages. The, higher rewards will go to those having the best preparation, with the, most interesting and challenging opportunities going to those with, graduate education., , Output of a modeling software., Courtesy Ansoft, , 413
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ale29559_ch10.qxd, , 07/08/2008, , 11:58 AM, , 414, , Page 414, , Chapter 10, , 10.1, , Sinusoidal Steady-State Analysis, , Introduction, , In Chapter 9, we learned that the forced or steady-state response of circuits to sinusoidal inputs can be obtained by using phasors. We also, know that Ohm’s and Kirchhoff’s laws are applicable to ac circuits. In, this chapter, we want to see how nodal analysis, mesh analysis,, Thevenin’s theorem, Norton’s theorem, superposition, and source transformations are applied in analyzing ac circuits. Since these techniques, were already introduced for dc circuits, our major effort here will be, to illustrate with examples., Analyzing ac circuits usually requires three steps., , Steps to Analyze AC Circuits:, 1. Transform the circuit to the phasor or frequency domain., 2. Solve the problem using circuit techniques (nodal analysis,, mesh analysis, superposition, etc.)., 3. Transform the resulting phasor to the time domain., , Frequency domain analysis of an ac, circuit via phasors is much easier, than analysis of the circuit in the, time domain., , Step 1 is not necessary if the problem is specified in the frequency, domain. In step 2, the analysis is performed in the same manner as dc, circuit analysis except that complex numbers are involved. Having read, Chapter 9, we are adept at handling step 3., Toward the end of the chapter, we learn how to apply PSpice in, solving ac circuit problems. We finally apply ac circuit analysis to two, practical ac circuits: oscillators and ac transistor circuits., , 10.2, , Nodal Analysis, , The basis of nodal analysis is Kirchhoff’s current law. Since KCL is, valid for phasors, as demonstrated in Section 9.6, we can analyze ac, circuits by nodal analysis. The following examples illustrate this., , Example 10.1, , Find ix in the circuit of Fig. 10.1 using nodal analysis., 10 Ω, , 1H, ix, , 20 cos 4t V, , +, −, , Figure 10.1, For Example 10.1., , 0.1 F, , 2ix, , 0.5 H
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ale29559_ch10.qxd, , 420, , 07/08/2008, , 11:58 AM, , Page 420, , Chapter 10, , Sinusoidal Steady-State Analysis, , From Eqs. (10.4.5) and (10.4.6), we obtain the matrix equation, c, , 8 j2, 8, I1, 10 j6, d B RB, R, 8, 14 j I3, 24 j35, , We obtain the following determinants, 8 j2, 8, ` 112 j8 j28 2 64 50 j20, 8, 14 j, 10 j6, 8, ¢1 `, ` 140 j10 j84 6 192 j280, 24 j35 14 j, 58 j186, ¢ `, , Current I1 is obtained as, I1 , , 58 j186, ¢1, , 3.618l274.5 A, ¢, 50 j20, , The required voltage V0 is, Vo j2(I1 I2) j2(3.618l274.5 3), 7.2134 j6.568 9.756l222.32 V, , ■ METHOD 2 We can use MATLAB to solve Eqs. (10.4.1) to, (10.4.4). We first cast the equations as, 8 j2 j2, 8, 0, I1, 10, 0, 1, 0, 0, I2, 3, D, TD TD T, 8, j5 8 j4 6 j5 I3, 0, 0, 0, 1, 1, I4, 4, , (10.4.7a), , or, AI B, By inverting A, we can obtain I as, I A1B, , (10.4.7b), , We now apply MATLAB as follows:, >> A = [(8-j*2), 0, -8, 0, >> B = [10 -3 0, >> I = inv(A)*B, , j*2, 1, -j*5, 0, 4]’;, , -8, 0, (8-j*4), -1, , I =, 0.2828 - 3.6069i, -3.0000, -1.8690 - 4.4276i, 2.1310 - 4.4276i, >> Vo = -2*j*(I(1) - I(2)), Vo =, -7.2138 - 6.5655i, as obtained previously., , 0;, 0;, (6+j*5);, 1];
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ale29559_ch10.qxd, , 07/08/2008, , 11:58 AM, , Page 421, , 10.4, , Superposition Theorem, , 421, , Practice Problem 10.4, , Calculate current Io in the circuit of Fig. 10.11., , 10 Ω, , Answer: 2.538l5.943 A., , Io, , –j4 Ω, , j8 Ω, 1 0° A, , 10.4, , 25 0° V +, −, , Superposition Theorem, , Since ac circuits are linear, the superposition theorem applies to ac, circuits the same way it applies to dc circuits. The theorem becomes, important if the circuit has sources operating at different frequencies., In this case, since the impedances depend on frequency, we must have, a different frequency domain circuit for each frequency. The total, response must be obtained by adding the individual responses in the, time domain. It is incorrect to try to add the responses in the phasor, or frequency domain. Why? Because the exponential factor e jt is, implicit in sinusoidal analysis, and that factor would change for every, angular frequency . It would therefore not make sense to add responses, at different frequencies in the phasor domain. Thus, when a circuit has, sources operating at different frequencies, one must add the responses, due to the individual frequencies in the time domain., , 5Ω, , – j6 Ω, , Figure 10.11, For Practice Prob. 10.4., , Example 10.5, , Use the superposition theorem to find Io in the circuit in Fig. 10.7., Solution:, Let, Io I¿o I–o, , (10.5.1), , where I¿o and I–o are due to the voltage and current sources, respectively., To find I¿o, consider the circuit in Fig. 10.12(a). If we let Z be the, parallel combination of j2 and 8 j10, then, Z, , 4Ω, , j2(8 j10), 0.25 j2.25, 2j 8 j10, , j10 Ω, , +, −, , j20 V, , –j2 Ω, , 8Ω, , and current I¿o is, I¿o , , I'o, , –j 2 Ω, , j20, j20, , 4 j2 Z, 4.25 j4.25, , (a), , or, , 4Ω, , I¿o 2.353 j2.353, , (10.5.2), , To get I–o, consider the circuit in Fig. 10.12(b). For mesh 1,, (8 j8)I1 j10I3 j2I2 0, , I3, j10 Ω, 8Ω, , (4 j4)I2 j2I1 j2I3 0, , I2, , (10.5.3), , For mesh 2,, , –j2 Ω, , I1, , (10.5.4), (b), , For mesh 3,, I3 5, , Figure 10.12, (10.5.5), , I''o, , –j 2 Ω, , 5A, , Solution of Example 10.5.
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ale29559_ch10.qxd, , 07/08/2008, , 11:58 AM, , 422, , Page 422, , Chapter 10, , Sinusoidal Steady-State Analysis, , From Eqs. (10.5.4) and (10.5.5),, (4 j4)I2 j2I1 j10 0, Expressing I1 in terms of I2 gives, I1 (2 j2)I2 5, , (10.5.6), , Substituting Eqs. (10.5.5) and (10.5.6) into Eq. (10.5.3), we get, (8 j8)[(2 j2)I2 5] j50 j2I2 0, or, I2 , , 90 j40, 2.647 j1.176, 34, , Current I–o is obtained as, I–o I2 2.647 j1.176, , (10.5.7), , From Eqs. (10.5.2) and (10.5.7), we write, Io I¿o I–o 5 j3.529 6.12l144.78 A, which agrees with what we got in Example 10.3. It should be noted, that applying the superposition theorem is not the best way to solve, this problem. It seems that we have made the problem twice as hard, as the original one by using superposition. However, in Example 10.6,, superposition is clearly the easiest approach., , Practice Problem 10.5, , Find current Io in the circuit of Fig. 10.8 using the superposition, theorem., Answer: 3.582l65.45 A., , Example 10.6, , Find vo of the circuit of Fig. 10.13 using the superposition theorem., 2H, , 1Ω, , 4Ω, , + v −, o, 10 cos 2t V, , +, −, , 2 sin 5t A, , 0.1 F, , +, −, , 5V, , Figure 10.13, For Example 10.6., , Solution:, Since the circuit operates at three different frequencies ( 0 for the, dc voltage source), one way to obtain a solution is to use superposition,, which breaks the problem into single-frequency problems. So we let, vo v1 v2 v3, , (10.6.1)
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ale29559_ch10.qxd, , 07/08/2008, , 11:58 AM, , Page 423, , 10.4, , Superposition Theorem, , 423, , where v1 is due to the 5-V dc voltage source, v2 is due to the 10 cos 2t V, voltage source, and v3 is due to the 2 sin 5t A current source., To find v1, we set to zero all sources except the 5-V dc source., We recall that at steady state, a capacitor is an open circuit to, dc while an inductor is a short circuit to dc. There is an alternative, way of looking at this. Since 0, jL 0, 1jC . Either, way, the equivalent circuit is as shown in Fig. 10.14(a). By voltage, division,, v1 , , 1, (5) 1 V, 14, , (10.6.2), , To find v2, we set to zero both the 5-V source and the 2 sin 5t current, source and transform the circuit to the frequency domain., 10 cos 2t, 2H, , 1, 1, , 0.1 F, , 1, , 10l0,, 2 rad/s, jL j4 , 1, j5 , jC, , The equivalent circuit is now as shown in Fig. 10.14(b). Let, Z j5 4 , , 1Ω, , j5 4, 2.439 j1.951, 4 j5, , j4 Ω, , 4Ω, , + v −, 1, +, −, , 10 0° V, , 5V, , (a), , +, −, , 1Ω, , I1, , 4Ω, , + V −, 2, – j5 Ω, , j10 Ω, , (b), , 1Ω, + V −, 3, 2 –90° A, , –j2 Ω, , 4Ω, , (c), , Figure 10.14, Solution of Example 10.6: (a) setting all sources to zero except the 5-V dc source, (b) setting all sources to zero except the ac, voltage source, (c) setting all sources to zero except the ac current source., , By voltage division,, V2 , , 1, 10, (10l0) , 2.498l30.79, 1 j4 Z, 3.439 j2.049, , In the time domain,, v2 2.498 cos(2t 30.79), , (10.6.3), , To obtain v3, we set the voltage sources to zero and transform what, is left to the frequency domain., 2 sin 5t, 2H, , 1, 1, , 0.1 F, , 1, , 2l90,, 5 rad/s, jL j10 , 1, j2 , jC
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ale29559_ch10.qxd, , 07/08/2008, , 11:58 AM, , 424, , Page 424, , Chapter 10, , Sinusoidal Steady-State Analysis, , The equivalent circuit is in Fig. 10.14(c). Let, Z1 j2 4 , , j2 4, 0.8 j1.6 , 4 j2, , By current division,, I1 , V3 I1 1 , , j10, (2l90) A, j10 1 Z1, j10, (j2) 2.328l80 V, 1.8 j8.4, , In the time domain,, v3 2.33 cos(5t 80) 2.33 sin(5t 10) V, , (10.6.4), , Substituting Eqs. (10.6.2) to (10.6.4) into Eq. (10.6.1), we have, vo(t) 1 2.498 cos(2t 30.79) 2.33 sin(5t 10) V, , Practice Problem 10.6, , Calculate vo in the circuit of Fig. 10.15 using the superposition, theorem., 8Ω, , 50 sin 5t V, , +, vo, −, , +, −, , 0.2 F, , 1H, , 4 cos 10t A, , Figure 10.15, For Practice Prob. 10.6., , Answer: 7.718 sin(5t 81.12) 2.102 cos(10t 86.24) V., , 10.5, , Source Transformation, , As Fig. 10.16 shows, source transformation in the frequency domain, involves transforming a voltage source in series with an impedance to, a current source in parallel with an impedance, or vice versa. As we, go from one source type to another, we must keep the following relationship in mind:, , Vs Zs Is, , 3, , Is , , Vs, Zs, , (10.1)
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ale29559_ch10.qxd, , 07/08/2008, , 11:58 AM, , Page 426, , Chapter 10, , 426, , Practice Problem 10.7, , Sinusoidal Steady-State Analysis, , Find Io in the circuit of Fig. 10.19 using the concept of source, transformation., j1 Ω, , 2Ω, , Io, 4Ω, , 8 90° Α, , j5 Ω, , 1Ω, , – j3 Ω, , – j2 Ω, , Figure 10.19, For Practice Prob. 10.7., , Answer: 6.576l99.46 A., , 10.6, ZTh, a, Linear, circuit, , a, VTh, , +, −, , b, , b, , Figure 10.20, Thevenin equivalent., , Thevenin and Norton, Equivalent Circuits, , Thevenin’s and Norton’s theorems are applied to ac circuits in the same, way as they are to dc circuits. The only additional effort arises from, the need to manipulate complex numbers. The frequency domain version of a Thevenin equivalent circuit is depicted in Fig. 10.20, where, a linear circuit is replaced by a voltage source in series with an impedance. The Norton equivalent circuit is illustrated in Fig. 10.21, where, a linear circuit is replaced by a current source in parallel with an impedance. Keep in mind that the two equivalent circuits are related as, VTh ZNIN,, , a, Linear, circuit, , ZTh ZN, , (10.2), , a, IN, , b, , Figure 10.21, Norton equivalent., , Example 10.8, , ZN, b, , just as in source transformation. VTh is the open-circuit voltage while IN, is the short-circuit current., If the circuit has sources operating at different frequencies (see, Example 10.6, for example), the Thevenin or Norton equivalent circuit, must be determined at each frequency. This leads to entirely different, equivalent circuits, one for each frequency, not one equivalent circuit, with equivalent sources and equivalent impedances., , Obtain the Thevenin equivalent at terminals a-b of the circuit in Fig. 10.22., d, – j6 Ω, , 120 75° V +, −, , 4Ω, , a, , e, , b, j12 Ω, , 8Ω, f, , Figure 10.22, For Example 10.8., , c
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ale29559_ch10.qxd, , 07/08/2008, , 11:58 AM, , Page 429, , 10.6, , Thevenin and Norton Equivalent Circuits, , 429, , To obtain ZTh, we remove the independent source. Due to the, presence of the dependent current source, we connect a 3-A current, source (3 is an arbitrary value chosen for convenience here, a number, divisible by the sum of currents leaving the node) to terminals a-b as, shown in Fig. 10.26(b). At the node, KCL gives, 3 Io 0.5Io, , Io 2 A, , 1, , Applying KVL to the outer loop in Fig. 10.26(b) gives, Vs Io(4 j3 2 j4) 2(6 j), The Thevenin impedance is, ZTh , , 2(6 j), Vs, , 4 j0.6667 , Is, 3, , Determine the Thevenin equivalent of the circuit in Fig. 10.27 as seen, from the terminals a-b., , Practice Problem 10.9, j4 Ω, , 8Ω, , Answer: ZTh 4.473l7.64 , VTh 29.4l72.9 V., , +, , Vo, , −, a, , – j2 Ω, 4Ω, , 20 0° A, , 0.2Vo, b, , Figure 10.27, For Practice Prob. 10.9., , Example 10.10, , Obtain current Io in Fig. 10.28 using Norton’s theorem., a, 5Ω, , 8Ω, , 40 90° V +, −, , Io, , 3 0° A, , –j2 Ω, , 20 Ω, 10 Ω, , j15 Ω, , j4 Ω, b, , Figure 10.28, For Example 10.10., , Solution:, Our first objective is to find the Norton equivalent at terminals a-b. ZN, is found in the same way as ZTh. We set the sources to zero as shown, in Fig. 10.29(a). As evident from the figure, the (8 j2) and (10 j4), impedances are short-circuited, so that, ZN 5 , To get IN, we short-circuit terminals a-b as in Fig. 10.29(b) and, apply mesh analysis. Notice that meshes 2 and 3 form a supermesh, because of the current source linking them. For mesh 1,, j40 (18 j2)I1 (8 j2)I2 (10 j4)I3 0, , (10.10.1)
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ale29559_ch10.qxd, , 07/08/2008, , 11:58 AM, , Page 431, , 10.7, , 10.7, , Op Amp AC Circuits, , 431, , Op Amp AC Circuits, , The three steps stated in Section 10.1 also apply to op amp circuits, as, long as the op amp is operating in the linear region. As usual, we will, assume ideal op amps. (See Section 5.2.) As discussed in Chapter 5,, the key to analyzing op amp circuits is to keep two important properties of an ideal op amp in mind:, 1. No current enters either of its input terminals., 2. The voltage across its input terminals is zero., The following examples will illustrate these ideas., , Example 10.11, , Determine vo(t) for the op amp circuit in Fig. 10.31(a) if vs , 3 cos 1000t V., 20 kΩ, , 20 kΩ, , 10 kΩ, vs, , +, −, , – j 10 kΩ, , 0.1 F, , 10 kΩ, , −, +, , 0.2 F, , 10 kΩ, vo, 3 0° V +, −, , (a), , V1, , 10 kΩ, , – j5 kΩ, , (b), , For Example 10.11: (a) the original circuit in the time domain, (b) its frequency domain equivalent., , Solution:, We first transform the circuit to the frequency domain, as shown in, Fig. 10.31(b), where Vs 3l0, 1000 rad/s. Applying KCL at, node 1, we obtain, , 10, , , , V1, V1 0 V1 Vo, , , j5, 10, 20, , or, 6 (5 j4)V1 Vo, , (10.11.1), , At node 2, KCL gives, 0 Vo, V1 0, , 10, j10, which leads to, V1 jVo, Substituting Eq. (10.11.2) into Eq. (10.11.1) yields, 6 j(5 j4)Vo Vo (3 j5)Vo, Vo , , 6, 1.029l59.04, 3 j5, , Hence,, vo(t) 1.029 cos(1000t 59.04) V, , (10.11.2), , 0V, 2, , 1, , Figure 10.31, , 3l0 V1, , Vo, , −, +, , Vo
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ale29559_ch10.qxd, , 07/08/2008, , 11:58 AM, , Page 432, , Chapter 10, , 432, , Practice Problem 10.11, , Sinusoidal Steady-State Analysis, , Find vo and io in the op amp circuit of Fig. 10.32. Let vs , 4 cos 5000t V., , 10 kΩ, , 10 nF, +, , vs, , 20 kΩ, , +, −, , io, , −, , vo, , 20 nF, , Figure 10.32, For Practice Prob. 10.11., , Answer: 1.3333 sin 5000t V, 133.33 sin 5000t mA., , Example 10.12, , C1, , R1, vs, , C2, , Compute the closed-loop gain and phase shift for the circuit in Fig. 10.33., Assume that R1 R2 10 k, C1 2 mF, C2 1 mF, and , 200 rad/s., , R2, , Solution:, The feedback and input impedances are calculated as, , −, +, , +, −, , R2, 1, , jC2, 1 jR2C2, 1 jR1C1, 1, Zi R1 , , jC1, jC1, Zf R2 2 2, , +, vo, −, , Since the circuit in Fig. 10.33 is an inverting amplifier, the closed-loop, gain is given by, , Figure 10.33, For Example 10.12., , G, , Zf, jC1R2, Vo, , Vs, Zi, (1 jR1C1)(1 jR2C2), , Substituting the given values of R1, R2, C1, C2, and , we obtain, G, , j4, 0.434l130.6, (1 j4)(1 j2), , Thus, the closed-loop gain is 0.434 and the phase shift is 130.6., , Practice Problem 10.12, +, −, vs, , C, , +, −, R, , Figure 10.34, For Practice Prob. 10.12., , R, , vo, , Obtain the closed-loop gain and phase shift for the circuit in Fig. 10.34., Let R 10 k, C 1 mF, and 1000 rad/s., Answer: 1.015, 5.6.
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ale29559_ch10.qxd, , 07/08/2008, , 11:58 AM, , Page 433, , 10.8, , 10.8, , AC Analysis Using PSpice, , 433, , AC Analysis Using PSpice, , PSpice affords a big relief from the tedious task of manipulating complex numbers in ac circuit analysis. The procedure for using PSpice for, ac analysis is quite similar to that required for dc analysis. The reader, should read Section D.5 in Appendix D for a review of PSpice concepts for ac analysis. AC circuit analysis is done in the phasor or frequency domain, and all sources must have the same frequency., Although ac analysis with PSpice involves using AC Sweep, our, analysis in this chapter requires a single frequency f 2p. The output file of PSpice contains voltage and current phasors. If necessary,, the impedances can be calculated using the voltages and currents in the, output file., , Example 10.13, , Obtain vo and io in the circuit of Fig. 10.35 using PSpice., 50 mH, , 4 kΩ, io, 8 sin(1000t + 50°) V, , +, −, , 2 F, , 0.5io, , 2 kΩ, , +, vo, −, , Figure 10.35, For Example 10.13., , Solution:, We first convert the sine function to cosine., 8 sin(1000t 50) 8 cos(1000t 50 90), 8 cos(1000t 40), The frequency f is obtained from as, f, , , 1000, , 159.155 Hz, 2p, 2p, , The schematic for the circuit is shown in Fig. 10.36. Notice that the, current-controlled current source F1 is connected such that its current, flows from node 0 to node 3 in conformity with the original circuit in, Fig. 10.35. Since we only want the magnitude and phase of vo and io, we, set the attributes of IPRINT and VPRINT1 each to AC yes, MAG yes,, PHASE yes. As a single-frequency analysis, we select Analysis/, Setup/AC Sweep and enter Total Pts 1, Start Freq 159.155, and, Final Freq 159.155. After saving the schematic, we simulate it by, selecting Analysis/Simulate. The output file includes the source, frequency in addition to the attributes checked for the pseudocomponents, IPRINT and VPRINT1,, FREQ, 1.592E+02, , IM(V_PRINT3), 3.264E–03, , IP(V_PRINT3), –3.743E+01, , FREQ, 1.592E+02, , VM(3), 1.550E+00, , VP(3), –9.518E+01
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ale29559_ch10.qxd, , 07/08/2008, , 11:58 AM, , Page 434, , Chapter 10, , 434, , R1, , Sinusoidal Steady-State Analysis, , L1, , 2, , AC=ok, MAG=ok, PHASE=ok, , 3, , 50mH, , 4k, IPRINT, ACMAG=8, +, ACPHASE=-40 −, , V, , F1, , AC=yes, MAG=yes, PHASE=ok, , R2, , GAIN=0.5, C1, , 2k, , 2u, , 0, , Figure 10.36, The schematic of the circuit in Fig. 10.35., , From this output file, we obtain, Vo 1.55l95.18 V,, , Io 3.264l37.43 mA, , which are the phasors for, vo 1.55 cos(1000t 95.18) 1.55 sin(1000t 5.18) V, and, io 3.264 cos(1000t 37.43) mA, , Practice Problem 10.13, , Use PSpice to obtain vo and io in the circuit of Fig. 10.37., io, , 2 kΩ, , 20 cos 3000t A, , +, −, , 1 F, , 3 kΩ, , 2H, +, vo, −, , +, −, , 2vo, , 1 kΩ, , Figure 10.37, For Practice Prob. 10.13., , Answer: 536.4 cos(3000t 154.6) mV, 1.088 cos(3000t 55.12) mA., , Example 10.14, , Find V1 and V2 in the circuit of Fig. 10.38., Solution:, 1. Define. In its present form, the problem is clearly stated. Again,, we must emphasize that time spent here will save lots of time, and expense later on! One thing that might have created a, problem for you is that, if the reference was missing for this, problem, you would then need to ask the individual assigning
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ale29559_ch10.qxd, , 07/08/2008, , 11:58 AM, , Page 435, , 10.8, , AC Analysis Using PSpice, , 435, , –j2, , 0.2Vx, , 3 0° A, , 1Ω, , V1, , 2Ω, , +, Vx, −, , – j1 Ω, , j2 Ω, , V2, , j2 Ω, , 2Ω, + 18 30° V, −, , – j1 Ω, , Figure 10.38, For Example 10.14., , the problem where it is to be located. If you could not do that,, then you would need to assume where it should be and then, clearly state what you did and why you did it., 2. Present. The given circuit is a frequency domain circuit and the, unknown node voltages V1 and V2 are also frequency domain, values. Clearly, we need a process to solve for these unknowns, in the frequency domain., 3. Alternative. We have two direct alternative solution techniques, that we can easily use. We can do a straightforward nodal, analysis approach or use PSpice. Since this example is in a, section dedicated to using PSpice to solve problems, we will, use PSpice to find V1 and V2. We can then use nodal analysis, to check the answer., 4. Attempt. The circuit in Fig. 10.35 is in the time domain, whereas, the one in Fig. 10.38 is in the frequency domain. Since we are not, given a particular frequency and PSpice requires one, we select any, frequency consistent with the given impedances. For example, if, we select 1 rad/s, the corresponding frequency is f 2p , 0.15916 Hz. We obtain the values of the capacitance (C , 1XC) and inductances (L XL ). Making these changes, results in the schematic in Fig. 10.39. To ease wiring, we have, , AC=ok, MAG=ok, PHASE=yes, , C1, AC=ok, MAG=ok, PHASE=yes, 1, , −, ACMAG=3A −, , 0.5C, R2, , L1, , L2, , R3, , 2, , 2H, , 2H, , 2, , GAIN=0.2, , I1, , R1, , 1 C2, , ACPHASE=0, , Figure 10.39, Schematic for circuit in the Fig. 10.38., , 1C G1 + − G, , C3, , 1C ACMAG=18V, ACPHASE=30, , +, −, , V1
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ale29559_ch10.qxd, , 436, , 07/08/2008, , 11:58 AM, , Page 436, , Chapter 10, , Sinusoidal Steady-State Analysis, , exchanged the positions of the voltage-controlled current source G1, and the 2 j2 impedance. Notice that the current of G1 flows, from node 1 to node 3, while the controlling voltage is across the, capacitor C2, as required in Fig. 10.38. The attributes of pseudocomponents VPRINT1 are set as shown. As a single-frequency, analysis, we select Analysis/Setup/AC Sweep and enter, Total Pts 1, Start Freq 0.15916, and Final Freq 0.15916., After saving the schematic, we select Analysis/Simulate, to simulate the circuit. When this is done, the output file, includes, FREQ, 1.592E–01, , VM(1), 2.708E+00, , VP(1), –5.673E+01, , FREQ, 1.592E-01, , VM(3), 4.468E+00, , VP(3), –1.026E+02, , from which we obtain,, V1 2.708l56.74 V, , and, , V2 6.911l80.72 V, , 5. Evaluate. One of the most important lessons to be learned is, that when using programs such as PSpice you still need to, validate the answer. There are many opportunities for making a, mistake, including coming across an unknown “bug” in PSpice, that yields incorrect results., So, how can we validate this solution? Obviously, we can, rework the entire problem with nodal analysis, and perhaps, using MATLAB, to see if we obtain the same results. There is, another way we will use here: write the nodal equations and, substitute the answers obtained in the PSpice solution, and see, if the nodal equations are satisfied., The nodal equations for this circuit are given below. Note, we have substituted V1 Vx into the dependent source., 3 , , V1 0 V1 0 V1 V2, V1 V2, , , 0.2V1 , 0, 1, j1, 2 j2, j2, (1 j 0.25 j0.25 0.2 j0.5)V1, (0.25 j0.25 j0.5)V2 3, (1.45 j1.25)V1 (0.25 j0.25)V2 3, 1.9144l40.76 V1 0.3536l45 V2 3, , Now, to check the answer, we substitute the PSpice answers into, this., 1.9144l40.76 2.708l56.74 0.3536l45 6.911l80.72, 5.184l15.98 2.444l35.72, 4.984 j1.4272 1.9842 j1.4269, 3 j0.0003, [Answer checks], 6. Satisfactory? Although we used only the equation from node 1, to check the answer, this is more than satisfactory to validate the, answer from the PSpice solution. We can now present our work, as a solution to the problem.
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ale29559_ch10.qxd, , 07/08/2008, , 11:58 AM, , Page 437, , 10.9, , Applications, , Practice Problem 10.14, , Obtain Vx and Ix in the circuit depicted in Fig. 10.40., 12 0° V, +−, 1Ω, , j2 Ω, , j2 Ω, , Vx, , – j 0.25, , 1Ω, , Ix, 2Ω, , 4 60° A, , +, −, , – j1 Ω, , 4Ix, , Figure 10.40, For Practice Prob. 10.14., , Answer: 9.842l44.78 V, 2.584l158 A., , 10.9, , Applications, , The concepts learned in this chapter will be applied in later chapters, to calculate electric power and determine frequency response. The concepts are also used in analyzing coupled circuits, three-phase circuits,, ac transistor circuits, filters, oscillators, and other ac circuits. In this, section, we apply the concepts to develop two practical ac circuits: the, capacitance multiplier and the sine wave oscillators., , 10.9.1 Capacitance Multiplier, The op amp circuit in Fig. 10.41 is known as a capacitance multiplier,, for reasons that will become obvious. Such a circuit is used in integratedcircuit technology to produce a multiple of a small physical capacitance C when a large capacitance is needed. The circuit in Fig. 10.41, can be used to multiply capacitance values by a factor up to 1000., For example, a 10-pF capacitor can be made to behave like a 100-nF, capacitor., , Vi, Ii, , 1, , +, Zi, , −, +, , R1, , R2, , 0V, 2, −, , A1, , Vi, −, , Figure 10.41, Capacitance multiplier., , +, C, , A2, , 437, , Vo
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ale29559_ch10.qxd, , 07/08/2008, , 438, , 11:58 AM, , Page 438, , Chapter 10, , Sinusoidal Steady-State Analysis, , In Fig. 10.41, the first op amp operates as a voltage follower, while, the second one is an inverting amplifier. The voltage follower isolates, the capacitance formed by the circuit from the loading imposed by the, inverting amplifier. Since no current enters the input terminals of the, op amp, the input current Ii flows through the feedback capacitor., Hence, at node 1,, Ii , , Vi Vo, jC(Vi Vo), 1jC, , (10.3), , Applying KCL at node 2 gives, Vi 0, 0 Vo, , R1, R2, or, Vo , , R2, Vi, R1, , (10.4), , Substituting Eq. (10.4) into (10.3) gives, Ii jC a1 , , R2, b Vi, R1, , or, Ii, R2, j a1 b C, Vi, R1, , (10.5), , The input impedance is, Zi , , Vi, 1, , Ii, jCeq, , (10.6), , where, Ceq a1 , , R2, bC, R1, , (10.7), , Thus, by a proper selection of the values of R1 and R2, the op amp circuit in Fig. 10.41 can be made to produce an effective capacitance, between the input terminal and ground, which is a multiple of the physical capacitance C. The size of the effective capacitance is practically, limited by the inverted output voltage limitation. Thus, the larger the, capacitance multiplication, the smaller is the allowable input voltage to, prevent the op amps from reaching saturation., A similar op amp circuit can be designed to simulate inductance., (See Prob. 10.89.) There is also an op amp circuit configuration to create a resistance multiplier., , Example 10.15, , Calculate Ceq in Fig. 10.41 when R1 10 k, R2 1 M, and C 1 nF., Solution:, From Eq. (10.7), Ceq a1 , , R2, 1 106, b C a1 , b 1 nF 101 nF, R1, 10 103
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ale29559_ch10.qxd, , 07/08/2008, , 11:58 AM, , Page 439, , 10.9, , Applications, , Determine the equivalent capacitance of the op amp circuit in Fig. 10.41, if R1 10 k, R2 10 M, and C 10 nF., , 439, , Practice Problem 10.15, , Answer: 10 mF., , 10.9.2 Oscillators, We know that dc is produced by batteries. But how do we produce ac?, One way is using oscillators, which are circuits that convert dc to ac., An oscillator is a circuit that produces an ac waveform as output when, powered by a dc input., , The only external source an oscillator needs is the dc power supply. Ironically, the dc power supply is usually obtained by converting, the ac supplied by the electric utility company to dc. Having gone, through the trouble of conversion, one may wonder why we need to, use the oscillator to convert the dc to ac again. The problem is that the, ac supplied by the utility company operates at a preset frequency of, 60 Hz in the United States (50 Hz in some other nations), whereas, many applications such as electronic circuits, communication systems,, and microwave devices require internally generated frequencies that, range from 0 to 10 GHz or higher. Oscillators are used for generating, these frequencies., In order for sine wave oscillators to sustain oscillations, they must, meet the Barkhausen criteria:, , This corresponds to 2pf , 377 rad/s., , 1. The overall gain of the oscillator must be unity or greater. Therefore, losses must be compensated for by an amplifying device., 2. The overall phase shift (from input to output and back to the input), must be zero., Three common types of sine wave oscillators are phase-shift, twin T, and, Wien-bridge oscillators. Here we consider only the Wien-bridge oscillator., The Wien-bridge oscillator is widely used for generating sinusoids, in the frequency range below 1 MHz. It is an RC op amp circuit with, only a few components, easily tunable and easy to design. As shown, in Fig. 10.42, the oscillator essentially consists of a noninverting amplifier with two feedback paths: the positive feedback path to the noninverting input creates oscillations, while the negative feedback path to, the inverting input controls the gain. If we define the impedances of, the RC series and parallel combinations as Zs and Zp, then, j, 1, Zs R1 , R1 , jC1, C1, , (10.8), , R2, 1, , jC2, 1 jR2C2, , (10.9), , Zp R2 , , Negative feedback, path to control gain, Rf, Rg, , −, +, , +, v 2 R2, −, , (10.10), , +, vo, −, , C2, , The feedback ratio is, Zp, V2, , Vo, Zs Zp, , C1, , R1, , Positive feedback path, to create oscillations, , Figure 10.42, Wien-bridge oscillator.
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ale29559_ch10.qxd, , 07/08/2008, , 440, , 11:59 AM, , Page 440, , Chapter 10, , Sinusoidal Steady-State Analysis, , Substituting Eqs. (10.8) and (10.9) into Eq. (10.10) gives, V2, , Vo, , , R2, j, R2 aR1 , b (1 jR2C2), C1, R2C1, , (10.11), , (R2C1 R1C1 R2C2) j(2R1C1R2C2 1), , To satisfy the second Barkhausen criterion, V2 must be in phase with, Vo, which implies that the ratio in Eq. (10.11) must be purely real., Hence, the imaginary part must be zero. Setting the imaginary part, equal to zero gives the oscillation frequency o as, 2o R1C1R2C2 1 0, or, o , , 1, 1R1R2C1C2, , (10.12), , In most practical applications, R1 R2 R and C1 C2 C, so that, o , , 1, 2pfo, RC, , (10.13), , 1, 2pRC, , (10.14), , or, fo , , Substituting Eq. (10.13) and R1 R2 R, C1 C2 C into Eq. (10.11), yields, V2, 1, , (10.15), Vo, 3, Thus, in order to satisfy the first Barkhausen criterion, the op amp must, compensate by providing a gain of 3 or greater so that the overall gain, is at least 1 or unity. We recall that for a noninverting amplifier,, Rf, Vo, 1, 3, V2, Rg, , (10.16), , Rf 2Rg, , (10.17), , or, Due to the inherent delay caused by the op amp, Wien-bridge oscillators are limited to operating in the frequency range of 1 MHz or less., , Example 10.16, , Design a Wien-bridge circuit to oscillate at 100 kHz., Solution:, Using Eq. (10.14), we obtain the time constant of the circuit as, RC , , 1, 1, , 1.59 106, 2 p fo 2 p 100 103, , (10.16.1), , If we select R 10 k, then we can select C 159 pF to satisfy, Eq. (10.16.1). Since the gain must be 3, Rf Rg 2. We could select, Rf 20 k while Rg 10 k.
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ale29559_ch10.qxd, , 07/08/2008, , 11:59 AM, , Page 441, , Review Questions, , In the Wien-bridge oscillator circuit in Fig. 10.42, let R1 R2 , 2.5 k, C1 C2 1 nF. Determine the frequency fo of the oscillator., , 441, , Practice Problem 10.16, , Answer: 63.66 kHz., , 10.10, , Summary, , 1. We apply nodal and mesh analysis to ac circuits by applying KCL, and KVL to the phasor form of the circuits., 2. In solving for the steady-state response of a circuit that has independent sources with different frequencies, each independent, source must be considered separately. The most natural approach, to analyzing such circuits is to apply the superposition theorem. A, separate phasor circuit for each frequency must be solved independently, and the corresponding response should be obtained in, the time domain. The overall response is the sum of the time, domain responses of all the individual phasor circuits., 3. The concept of source transformation is also applicable in the frequency domain., 4. The Thevenin equivalent of an ac circuit consists of a voltage, source VTh in series with the Thevenin impedance ZTh., 5. The Norton equivalent of an ac circuit consists of a current source, IN in parallel with the Norton impedance ZN (ZTh)., 6. PSpice is a simple and powerful tool for solving ac circuit problems. It relieves us of the tedious task of working with the complex numbers involved in steady-state analysis., 7. The capacitance multiplier and the ac oscillator provide two typical applications for the concepts presented in this chapter. A capacitance multiplier is an op amp circuit used in producing a multiple, of a physical capacitance. An oscillator is a device that uses a dc, input to generate an ac output., , Review Questions, 10.1, , The voltage Vo across the capacitor in Fig. 10.43 is:, (a) 5l0 V, (c) 7.071l45 V, , (b) 7.071l45 V, (d) 5l45 V, , 10.2, , The value of the current Io in the circuit of Fig. 10.44 is:, (a) 4l0 A, , (b) 2.4l90 A, , (c) 0.6l0 A, , (d) 1 A, , 1Ω, , 10 0° V, , +, −, , – j1 Ω, , +, Vo, −, , Io, 3 0° A, , Figure 10.43, , Figure 10.44, , For Review Question 10.1., , For Review Question 10.2., , j8 Ω, , – j2 Ω
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ale29559_ch10.qxd, , 07/08/2008, , 11:59 AM, , Page 442, , Chapter 10, , 442, , 10.3, , Sinusoidal Steady-State Analysis, , Using nodal analysis, the value of Vo in the circuit of, Fig. 10.45 is:, , 10.6, , For the circuit in Fig. 10.48, the Thevenin, impedance at terminals a-b is:, , (a) 24 V, , (b) 8 V, , (a) 1 , , (b) 0.5 j0.5 , , (c) 8 V, , (d) 24 V, , (c) 0.5 j0.5 , , (d) 1 j2 , , (e) 1 j2 , , Vo, 1Ω, j6 Ω, , 1H, a, , –j 3 Ω, , 4 90° A, , 5 cos t V, , +, −, , 1F, b, , Figure 10.48, , Figure 10.45, , For Review Questions 10.6 and 10.7., , For Review Question 10.3., , 10.4, , In the circuit of Fig. 10.46, current i(t) is:, (a) 10 cos t A, , (b) 10 sin t A, , (d) 5 sin t A, , (e) 4.472 cos(t 63.43) A, , 10.7, , (c) 5 cos t A, , 10.8, 1F, , 1H, , 10 cos t V, , +, −, , In the circuit of Fig. 10.48, the Thevenin voltage at, terminals a-b is:, (a) 3.535l45 V, , (b) 3.535l45 V, , (c) 7.071l45 V, , (d) 7.071l45 V, , Refer to the circuit in Fig. 10.49. The Norton, equivalent impedance at terminals a-b is:, (a) j4 , , (b) j2 , , (c) j2 , , (d) j4 , , 1Ω, , i(t), , –j2 Ω, , Figure 10.46, , a, , For Review Question 10.4., 6 0° V +, −, , 10.5, , Refer to the circuit in Fig. 10.47 and observe that the, two sources do not have the same frequency. The, current ix(t) can be obtained by:, , j4 Ω, b, , Figure 10.49, For Review Questions 10.8 and 10.9., , (a) source transformation, (b) the superposition theorem, (c) PSpice, , 10.9, , 1Ω, , 1H, , +, −, , Figure 10.47, For Review Question 10.5., , 1F, , +, −, , (a) 1l0 A, , (b) 1.5 l90 A, , (c) 1.5l90 A, , (d) 3l90 A, , 10.10 PSpice can handle a circuit with two independent, sources of different frequencies., , ix, sin 2t V, , The Norton current at terminals a-b in the circuit of, Fig. 10.49 is:, , sin 10t V, , (a) True, , (b) False, , Answers: 10.1c, 10.2a, 10.3d, 10.4a, 10.5b, 10.6c,, 10.7a, 10.8a, 10.9d, 10.10b.
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ale29559_ch10.qxd, , 07/08/2008, , 11:59 AM, , Page 443, , Problems, , 443, , Problems, Section 10.2 Nodal Analysis, 10.1, , 10.6, , Determine Vx in Fig. 10.55., , Determine i in the circuit of Fig. 10.50., , 20 Ω, , 1Ω, , i, , +, –, , 4Vx, 2 cos 10t V, , +, −, , 1F, , 1Ω, , 1H, , j10 Ω, 4.5 0° A 20 Ω, , +, Vx, –, , Figure 10.55, For Prob. 10.6., , Figure 10.50, For Prob. 10.1., 10.2, , 10.7, , Using Fig. 10.51, design a problem to help other, students better understand nodal analysis., , Use nodal analysis to find V in the circuit of, Fig. 10.56., , 2Ω, , +, −, , 4 0° V, , j20 Ω, , 40 Ω, –j5 Ω, , j4 Ω, , +, Vo, −, , 120, , –15° V, , +, −, , 6, , V, , –j30 Ω, , 30° A, , 50 Ω, , Figure 10.51, For Prob. 10.2., 10.3 Determine vo in the circuit of Fig. 10.52., 1, 12, , 4Ω, , 32 sin 4t V, , Figure 10.56, For Prob. 10.7., , F, , 2H, , +, −, , vo, −, , Use nodal analysis to find current io in the circuit of, Fig. 10.57. Let is 6 cos(200t 15) A., , 10.8, , +, 1Ω, , 6Ω, , 4 cos 4t A, , 0.1 vo, , Figure 10.52, io, , For Prob. 10.3., 10.4, , Determine i1 in the circuit of Fig. 10.53., i1, , 2 F, , 2 kΩ, , +, 100 cos 103t V −, , is, , +, −, , 0.5 H, , 30i1, , 20 Ω, , 25 cos(4 , , 103t), , V +, −, , –, , 20 Ω, 2 F, , 0.25 H, , 50 F, , 100 mH, , Use nodal analysis to find vo in the circuit of Fig. 10.58., , Find io in the circuit of Fig. 10.54., 2 kΩ, , +, , Figure 10.57, , 10.9, , For Prob. 10.4., , io, , vo, , For Prob. 10.8., , Figure 10.53, 10.5, , 40 Ω, , 50 F, , 10 mH, , io, +, −, , 10 cos 103t V, 10io, , +, −, , Figure 10.54, , Figure 10.58, , For Prob. 10.5., , For Prob. 10.9., , 20 Ω, , 4io, , 30 Ω, , +, vo, −
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ale29559_ch10.qxd, , 07/08/2008, , 11:59 AM, , Page 444, , Chapter 10, , 444, , Sinusoidal Steady-State Analysis, , 10.10 Use nodal analysis to find vo in the circuit of, Fig. 10.59. Let 2 krad/s., , 10.14 Calculate the voltage at nodes 1 and 2 in the circuit, of Fig. 10.63 using nodal analysis., j4 Ω, , 2 F, , 20 30° A, , +, 10 sin t A, , vx, , 2 kΩ, , +, 0.1 vx 4 kΩ, , 50 mH, –, , 1, , vo, –, , 2, , –j2 Ω, , 10 Ω, , –j5 Ω, , j2 Ω, , Figure 10.59, For Prob. 10.10., , Figure 10.63, 10.11 Apply nodal analysis to the circuit in Fig. 10.60 and, determine Io., Io, , 10 0° V, , 10.15 Solve for the current I in the circuit of Fig. 10.64, using nodal analysis., , j5 Ω, 2Ω, , For Prob. 10.14., , 10 0° A, , 2Ω, , +, −, , 2I o, , j1 Ω, , 2Ω, , j8 Ω, , I, 40 –90° V, , Figure 10.60, , +, −, , –j2 Ω, , 4Ω, , 2I, , For Prob. 10.11., 10.12 Using Fig. 10.61, design a problem to help other, students better understand nodal analysis., , Figure 10.64, For Prob. 10.15., 10.16 Use nodal analysis to find Vx in the circuit shown in, Fig. 10.65., , 2io, , j4 Ω, , R2, io, R1, , is, , C, , + Vx −, , L, , 10.13 Determine Vx in the circuit of Fig. 10.62 using any, method of your choice., –j2 Ω, , 30° V, , 3 45° A, , For Prob. 10.16., , For Prob. 10.12., , 40, , –j3 Ω, , Figure 10.65, , Figure 10.61, , +, −, , 5Ω, , 2 0° A, , 8Ω, , j6 Ω, , j4 Ω, 150 20° V, , +, Vx, –, , 10.17 By nodal analysis, obtain current Io in the circuit of, Fig. 10.66., , 3Ω, , 10 Ω, , 5 0° A, , +, −, 3Ω, , Figure 10.62, , Figure 10.66, , For Prob. 10.13., , For Prob. 10.17., , Io, , 2Ω, , 1Ω, , –j 2 Ω
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ale29559_ch10.qxd, , 07/08/2008, , 11:59 AM, , Page 445, , Problems, , 445, , 10.18 Use nodal analysis to obtain Vo in the circuit of Fig. 10.67 below., 8Ω, +, Vx, −, , 8 45° A, , 2Ω, , j6 Ω, , 4Ω, , j5 Ω, , 2Vx, , –j1 Ω, , –j2 Ω, , +, Vo, −, , Figure 10.67, For Prob. 10.18., 10.22 For the circuit in Fig. 10.71, determine VoVs., , 10.19 Obtain Vo in Fig. 10.68 using nodal analysis., , R1, , j2 Ω, 12 0° V, +−, , Vs, , +, −, , C, L, , +, Vo, −, , 2Ω, , R2, , 4Ω, , –j4 Ω, , +, Vo, −, , 0.2Vo, , Figure 10.71, For Prob. 10.22., , Figure 10.68, For Prob. 10.19., 10.23 Using nodal analysis obtain V in the circuit of, Fig. 10.72., 10.20 Refer to Fig. 10.69. If vs(t) Vm sin t and, vo(t) A sin(t f), derive the expressions for, A and f., , jL, , +, Vs −, , R, , +, −, , vs(t), , R, , L, , +, –, , V, , 1, jC, , +, vo(t), −, , C, , 1, jC, , Figure 10.72, For Prob. 10.23., , Figure 10.69, For Prob. 10.20., , Section 10.3 Mesh Analysis, 10.21 For each of the circuits in Fig. 10.70, find VoVi for, 0, S , and 2 1LC., , 10.24 Design a problem to help other students better, understand mesh analysis., 10.25 Solve for io in Fig. 10.73 using mesh analysis., , R, , L, , +, Vi, , C, , −, (a), , R, +, , +, , Vo, , Vi, , −, , −, , C, 4Ω, , +, L, , io, , Vo, −, , 2H, , 20 cos 2t V +, −, , (b), , Figure 10.70, , Figure 10.73, , For Prob. 10.21., , For Prob. 10.25., , 0.25 F, , + 12 sin 2t V, −
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ale29559_ch10.qxd, , 07/08/2008, , 11:59 AM, , Page 447, , Problems, , 10.32 Determine Vo and Io in the circuit of Fig. 10.80, using mesh analysis., , 447, , 10.38 Using mesh analysis, obtain Io in the circuit shown, in Fig. 10.83., Io, , j4 Ω, , 2Ω, , 4 –30° A, , Io, , +, Vo, −, , 3Vo, , −, +, , 2 0° A, , j2 Ω, , –j2 Ω, , 2Ω, , –j4 Ω, , 1Ω, , Figure 10.80, , +, −, , 10 90° V, , 1Ω, , 4 0° A, , For Prob. 10.32., , Figure 10.83, For Prob. 10.38., 10.33 Compute I in Prob. 10.15 using mesh analysis., 10.39 Find I1, I2, I3, and Ix in the circuit of Fig. 10.84., 10 Ω, , 10.34 Use mesh analysis to find Io in Fig. 10.28 (for, Example 10.10)., 20 Ω, , 10.35 Calculate Io in Fig. 10.30 (for Practice Prob. 10.10), using mesh analysis., , –j 15 Ω, , I3, , j 16 Ω, , Ix, I1, , 10.36 Compute Vo in the circuit of Fig. 10.81 using mesh, analysis., , 8 90° A, 2Ω, , 2Ω, , +, Vo, −, , 2Ω, , I2, , +, −, , –j25 Ω, , 8Ω, , Figure 10.84, , –j3 Ω, , j4 Ω, , 20 64° V, , For Prob. 10.39., + 24 0° V, −, , Section 10.4 Superposition Theorem, 10.40 Find io in the circuit shown in Fig. 10.85 using, superposition., 4Ω, , 4 0° A, , Figure 10.81, , 2Ω, io, , For Prob. 10.36., 20 cos 4t V, , 10.37 Use mesh analysis to find currents I1, I2, and I3 in, the circuit of Fig. 10.82., , +, −, , + 16 V, −, , 1H, , Figure 10.85, For Prob. 10.40., , I1, , 120, , –90° V, , +, −, , I2, , 10.41 Find vo for the circuit in Fig. 10.86, assuming that, vs 3 cos 2t 8 sin 4t V., Z, 0.25 F, , Z = 80 – j 35 Ω, 120 –30° V, , −, +, , vs +, −, , Z, I3, , 2Ω, , +, vo, –, , Figure 10.82, , Figure 10.86, , For Prob. 10.37., , For Prob. 10.41.
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ale29559_ch10.qxd, , 07/08/2008, , 11:59 AM, , Page 448, , Chapter 10, , 448, , Sinusoidal Steady-State Analysis, , 10.42 Using Fig. 10.87, design a problem to help other, students better understand the superposition theorem., Io, , j XL, , V1 +, −, , 10.46 Solve for vo(t) in the circuit of Fig. 10.91 using the, superposition principle., , R2, , 6Ω, + V, −, 2, , −j XC, , R1, , 12 cos 3t V, , +, −, , 1, 12, , 2H, +, vo, −, , F, , +, −, , 4 sin 2t A, , 20 V, , Figure 10.91, , Figure 10.87, , For Prob. 10.46., , For Prob. 10.42., , 10.43 Using the superposition principle, find ix in the, circuit of Fig. 10.88., , 1, 8, , F, , 3Ω, , 1Ω, , ix, +, −, , 4H, , 5 cos(2t + 10°) A, , 10.47 Determine io in the circuit of Fig. 10.92, using the, superposition principle., , 10 cos(2t – 60°) V, , 10 sin(t – 30°) V, , +, −, , 1, 6, , F, , 24 V, , 2H, , −+, , io, , 2Ω, , 4Ω, , 2 cos 3t, , Figure 10.92, , Figure 10.88, For Prob. 10.43., , For Prob. 10.47., , 10.44 Use the superposition principle to obtain vx in the, circuit of Fig. 10.89. Let vs 25 sin 2t V and, is 6 cos(6t 10) A., , 10.48 Find io in the circuit of Fig. 10.93 using superposition., , 20 Ω, , is, , 16 Ω, , +, vx, –, , 20 F, io, , 5H, , 50 cos 2000t V, , +, −, , vs, , +, −, , 80 Ω, , 60 Ω, , 2 sin 4000t A, , +, −, , 24 V, , Figure 10.93, , Figure 10.89, , For Prob. 10.48., , For Prob. 10.44., , 10.45 Use superposition to find i(t) in the circuit of, Fig. 10.90., , i, , Section 10.5 Source Transformation, 10.49 Using source transformation, find i in the circuit of, Fig. 10.94., , 20 Ω, 3Ω, , 20 cos(10t + 30°) V, , 100 Ω, , 40 mH, , +, −, , + 10 sin 4t V, −, , i, , 5 mH, 5Ω, , 16 sin(200t + 30°) A, 1 mF, , 300 mH, , Figure 10.90, , Figure 10.94, , For Prob. 10.45., , For Prob. 10.49.
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ale29559_ch10.qxd, , 07/08/2008, , 11:59 AM, , Page 449, , Problems, , 449, –j5 Ω, , 10.50 Using Fig. 10.95, design a problem to help other, students understand source transformation., 4 0° A, , L, , R1, , vs(t) +, , R2, , C, , a, 8Ω, , j10 Ω, b, , +, vo, , (b), , Figure 10.98, For Prob. 10.55., , Figure 10.95, , 10.56 For each of the circuits in Fig. 10.99, obtain Thevenin, and Norton equivalent circuits at terminals a-b., , For Prob. 10.50., 10.51 Use source transformation to find Io in the circuit of, Prob. 10.42., , 6Ω, , a, , 10.52 Use the method of source transformation to find Ix in, the circuit of Fig. 10.96., 2Ω, , –j2 Ω, 2 0° A, , –j2 Ω, , j4 Ω, , j4 Ω, , Ix, 30 0° V +, −, , b, , 4Ω, , 6Ω, , (a), 2.5 90° A, 30 Ω, , –j3 Ω, , j10 Ω, , Figure 10.96, , 120 45° V +, −, , For Prob. 10.52., , 60 Ω, , a, –j5 Ω, , 10.53 Use the concept of source transformation to find Vo, in the circuit of Fig. 10.97., , b, (b), , Figure 10.99, –j3 Ω, , 4Ω, , 20 0° V +, −, , 2Ω, , j2 Ω, , For Prob. 10.56., , j4 Ω, , –j2 Ω, , +, Vo, −, , 10.57 Using Fig. 10.100, design a problem to help other, students better understand Thevenin and Norton, equivalent circuits., R1, , – j XC, , R2, , Figure 10.97, For Prob. 10.53., , Vs +, −, , j XL, , 10.54 Rework Prob. 10.7 using source transformation., , Figure 10.100, , Section 10.6 Thevenin and Norton, Equivalent Circuits, , For Prob. 10.57., , 10.55 Find the Thevenin and Norton equivalent circuits at, terminals a-b for each of the circuits in Fig. 10.98., , 10.58 For the circuit depicted in Fig. 10.101, find the, Thevenin equivalent circuit at terminals a-b., a, , j20 Ω, , 10 Ω, , 8Ω, , a, , j10 Ω, , 2 30° A, –j6 Ω, , –j10 Ω, , 50 30° V +, −, , b, b, (a), , Figure 10.101, For Prob. 10.58.
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ale29559_ch10.qxd, , 07/08/2008, , 11:59 AM, , Page 450, , Chapter 10, , 450, , Sinusoidal Steady-State Analysis, , 10.59 Calculate the output impedance of the circuit shown, in Fig. 10.102., –j 2 Ω, , 10.63 Obtain the Norton equivalent of the circuit depicted, in Fig. 10.106 at terminals a-b., , 10 Ω, , 5 F, a, , + Vo −, j 40 Ω, , 0.2Vo, , 10 H, , 4 cos(200t + 30°) A, , 2 kΩ, b, , Figure 10.106, , Figure 10.102, For Prob. 10.59., , For Prob. 10.63., , 10.60 Find the Thevenin equivalent of the circuit in, Fig. 10.103 as seen from:, , 10.64 For the circuit shown in Fig. 10.107, find the Norton, equivalent circuit at terminals a-b., , (a) terminals a-b, , (b) terminals c-d, , c, 10 Ω, , d, –j4 Ω, 3 60° A, , 40 0° V, , +, −, , 8 0° A, , j5 Ω, , 40 Ω, , 60 Ω, , a, , a, , 4Ω, , b, , j80 Ω, b, , Figure 10.103, , –j30 Ω, , Figure 10.107, For Prob. 10.64., , For Prob. 10.60., 10.61 Find the Thevenin equivalent at terminals a-b of the, circuit in Fig. 10.104., , 10.65 Using Fig. 10.108, design a problem to help other, students better understand Norton’s theorem., , 4Ω, a, Ix, , +−, , io, , –j3 Ω, , 2 0° A, , vs(t), , R, , 1.5Ix, , L, , C1, , C2, , b, , Figure 10.104, , Figure 10.108, , For Prob. 10.61., , For Prob. 10.65., , 10.62 Using Thevenin’s theorem, find vo in the circuit of, Fig. 10.105., 3io, , io, , 20 cos(t + 30°) V +, −, , 4Ω, , 2H, , 1, 4, , 1, 8, , F, , 10.66 At terminals a-b, obtain Thevenin and Norton, equivalent circuits for the network depicted in, Fig. 10.109. Take 10 rad/s., 10 mF 12 cos t V, −+, , F, , 2Ω, , +, vo, −, , 2 sin t A, , +, vo, −, , 10 Ω, , 1, 2, , H, , a, 2vo, b, , Figure 10.105, , Figure 10.109, , For Prob. 10.62., , For Prob. 10.66.
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ale29559_ch10.qxd, , 07/08/2008, , 11:59 AM, , Page 451, , Problems, , 451, , 10.67 Find the Thevenin and Norton equivalent circuits at, terminals a-b in the circuit of Fig. 10.110., , R2, C, R1, , –j5 Ω, 12 Ω, , 13 Ω, +, −, , 60 45° V, , a, , −, +, , +, vo, −, , +, −, , vs, , b, j6 Ω, , 10 Ω, , Figure 10.113, For Prob. 10.70., , 8Ω, , 10.71 Find vo in the op amp circuit of Fig. 10.114., , Figure 10.110, For Prob. 10.67., , +, −, , 10.68 Find the Thevenin equivalent at terminals a-b in the, circuit of Fig. 10.111., , +, , 0.5 F, 16 cos(2t + 45°) V +, −, , vo, , 10 kΩ, 2 kΩ, , io, , 4Ω, , –, , a, +, 6 sin10t V, , vo, 3, , +, −, , +, −, , 1, F, 20, , 4io, , 1 H vo, , Figure 10.114, For Prob. 10.71., , −, , Figure 10.111, , b, , 10.72 Compute io(t) in the op amp circuit in Fig. 10.115 if, vs 10 cos(104t 30) V., , For Prob. 10.68., , 50 kΩ, +, −, vs, , Section 10.7 Op Amp AC Circuits, 10.69 For the differentiator shown in Fig. 10.112, obtain, VoVs. Find vo(t) when vs(t) Vm sin t and, 1RC., , +, −, , io, , 1 nF, , 100 kΩ, , Figure 10.115, For Prob. 10.72., 10.73 If the input impedance is defined as Zin VsIs,, find the input impedance of the op amp circuit in, Fig. 10.116 when R1 10 k, R2 20 k, C1 , 10 nF, C2 20 nF, and 5000 rad/s., , R, C, −, +, vs, , +, −, , C1, , +, vo, −, , Figure 10.112, For Prob. 10.69., , Is, , R1, , +, −, Vs +, −, , C2, , Zin, , 10.70 Using Fig. 10.113, design a problem to help other, students better understand op amps in AC circuits., , R2, , Figure 10.116, For Prob. 10.73., , Vo
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ale29559_ch10.qxd, , 07/08/2008, , 11:59 AM, , Page 455, , Problems, , 10.92 The oscillator circuit in Fig. 10.134 uses an ideal, op amp., , 455, , 10.95 Figure 10.136 shows a Hartley oscillator. Show that, the frequency of oscillation is, , (a) Calculate the minimum value of Ro that will, cause oscillation to occur., , fo , , 1, 2p1C(L1 L2), , (b) Find the frequency of oscillation., Rf, , 1 MΩ, 100 kΩ, , −, +, , Ri, , −, +, , Vo, , Ro, C, , 10 H, , 10 kΩ, , 2 nF, , L2, , L1, , Figure 10.134, , Figure 10.136, , For Prob. 10.92., , A Hartley oscillator; for Prob. 10.95., , 10.93 Figure 10.135 shows a Colpitts oscillator. Show that, the oscillation frequency is, , 10.96 Refer to the oscillator in Fig. 10.137., , fo , , (a) Show that, , 1, 2p1LCT, , V2, 1, , Vo, 3 j(LR RL), , where CT C1C2(C1 C2). Assume Ri W XC2., , (b) Determine the oscillation frequency fo., Rf, Ri, , −, +, , (c) Obtain the relationship between R1 and R2 in, order for oscillation to occur., Vo, R2, R1, , L, C2, , −, +, , C1, , Vo, L, , R, , Figure 10.135, , V2, , A Colpitts oscillator; for Prob. 10.93., , L, , (Hint: Set the imaginary part of the impedance in the, feedback circuit equal to zero.), 10.94 Design a Colpitts oscillator that will operate at 50 kHz., , Figure 10.137, For Prob. 10.96., , R
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ale29559_ch11.qxd, , 07/08/2008, , 12:01 PM, , Page 457, , c h a p t e r, , AC Power Analysis, , 11, , Four things come not back: the spoken word; the sped arrow; time, past; the neglected opportunity., —Al Halif Omar Ibn, , Enhancing Your Career, Career in Power Systems, The discovery of the principle of an ac generator by Michael Faraday, in 1831 was a major breakthrough in engineering; it provided a convenient way of generating the electric power that is needed in every, electronic, electrical, or electromechanical device we use now., Electric power is obtained by converting energy from sources such, as fossil fuels (gas, oil, and coal), nuclear fuel (uranium), hydro energy, (water falling through a head), geothermal energy (hot water, steam),, wind energy, tidal energy, and biomass energy (wastes). These various, ways of generating electric power are studied in detail in the field of, power engineering, which has become an indispensable subdiscipline, of electrical engineering. An electrical engineer should be familiar with, the analysis, generation, transmission, distribution, and cost of electric, power., The electric power industry is a very large employer of electrical, engineers. The industry includes thousands of electric utility systems, ranging from large, interconnected systems serving large regional areas, to small power companies serving individual communities or factories., Due to the complexity of the power industry, there are numerous electrical engineering jobs in different areas of the industry: power plant, (generation), transmission and distribution, maintenance, research, data, acquisition and flow control, and management. Since electric power is, used everywhere, electric utility companies are everywhere, offering, exciting training and steady employment for men and women in thousands of communities throughout the world., , A pole-type transformer with a lowvoltage, three-wire distribution system., © Vol. 129 PhotoDisc/Getty, , 457
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ale29559_ch11.qxd, , 07/08/2008, , 12:01 PM, , Page 458, , Chapter 11, , 458, , 11.1, , AC Power Analysis, , Introduction, , Our effort in ac circuit analysis so far has been focused mainly on calculating voltage and current. Our major concern in this chapter is, power analysis., Power analysis is of paramount importance. Power is the most, important quantity in electric utilities, electronic, and communication, systems, because such systems involve transmission of power from one, point to another. Also, every industrial and household electrical, device—every fan, motor, lamp, pressing iron, TV, personal computer—, has a power rating that indicates how much power the equipment, requires; exceeding the power rating can do permanent damage to an, appliance. The most common form of electric power is 50- or 60-Hz, ac power. The choice of ac over dc allowed high-voltage power transmission from the power generating plant to the consumer., We will begin by defining and deriving instantaneous power and, average power. We will then introduce other power concepts. As practical applications of these concepts, we will discuss how power is, measured and reconsider how electric utility companies charge their, customers., , 11.2, , Instantaneous and Average Power, , As mentioned in Chapter 2, the instantaneous power p(t) absorbed by, an element is the product of the instantaneous voltage v(t) across the, element and the instantaneous current i(t) through it. Assuming the passive sign convention,, p(t) v(t)i(t), , We can also think of the instantaneous, power as the power absorbed by the, element at a specific instant of time., Instantaneous quantities are denoted, by lowercase letters., , +, v (t), −, , The instantaneous power (in watts) is the power at any instant of time., , It is the rate at which an element absorbs energy., Consider the general case of instantaneous power absorbed by an, arbitrary combination of circuit elements under sinusoidal excitation,, as shown in Fig. 11.1. Let the voltage and current at the terminals of, the circuit be, v(t) Vm cos(t uv), , (11.2a), , i(t) Im cos(t ui ), , (11.2b), , where Vm and Im are the amplitudes (or peak values), and uv and ui are, the phase angles of the voltage and current, respectively. The instantaneous power absorbed by the circuit is, , i(t), Sinusoidal, source, , (11.1), , Passive, linear, network, , Figure 11.1, Sinusoidal source and passive linear circuit., , p(t) v(t)i(t) Vm Im cos(t uv) cos(t ui), , (11.3), , We apply the trigonometric identity, 1, cos A cos B [cos(A B) cos(A B)], 2, , (11.4)
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ale29559_ch11.qxd, , 07/08/2008, , 12:01 PM, , Page 459, , 11.2, , Instantaneous and Average Power, , and express Eq. (11.3) as, 1, 1, p(t) Vm Im cos(uv ui) Vm Im cos(2t uv ui), 2, 2, , (11.5), , This shows us that the instantaneous power has two parts. The first part, is constant or time independent. Its value depends on the phase difference between the voltage and the current. The second part is a sinusoidal function whose frequency is 2, which is twice the angular, frequency of the voltage or current., A sketch of p(t) in Eq. (11.5) is shown in Fig. 11.2, where T , 2p is the period of voltage or current. We observe that p(t) is periodic, p(t) p(t T0), and has a period of T0 T2, since its frequency is twice that of voltage or current. We also observe that p(t), is positive for some part of each cycle and negative for the rest of, the cycle. When p(t) is positive, power is absorbed by the circuit., When p(t) is negative, power is absorbed by the source; that is,, power is transferred from the circuit to the source. This is possible, because of the storage elements (capacitors and inductors) in the, circuit., , p(t), , 1, V I, 2 m m, , 1, V I, 2 m m, , 0, , T, , T, 2, , cos(v − i ), t, , Figure 11.2, The instantaneous power p(t) entering a circuit., , The instantaneous power changes with time and is therefore difficult to measure. The average power is more convenient to measure. In, fact, the wattmeter, the instrument for measuring power, responds to, average power., , The average power, in watts, is the average of the instantaneous power, over one period., , Thus, the average power is given by, P, , 1, T, , , , T, , p(t) dt, , (11.6), , 0, , Although Eq. (11.6) shows the averaging done over T, we would get, the same result if we performed the integration over the actual period, of p(t) which is T0 T2., , 459
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ale29559_ch11.qxd, , 460, , 07/08/2008, , 12:01 PM, , Page 460, , Chapter 11, , AC Power Analysis, , Substituting p(t) in Eq. (11.5) into Eq. (11.6) gives, P, , 1, T, , T, , , , 0, , 1, , T, , 1, Vm Im cos(uv ui) dt, 2, , , , T, , 0, , 1, Vm Im cos(2t uv ui) dt, 2, , 1, 1, Vm Im cos(uv ui), 2, T, 1, 1, Vm Im, 2, T, , , , , , T, , dt, , 0, , T, , cos(2t uv ui) dt, , (11.7), , 0, , The first integrand is constant, and the average of a constant is the same, constant. The second integrand is a sinusoid. We know that the average of a sinusoid over its period is zero because the area under the, sinusoid during a positive half-cycle is canceled by the area under it, during the following negative half-cycle. Thus, the second term in, Eq. (11.7) vanishes and the average power becomes, 1, P Vm Im cos(uv ui), 2, , (11.8), , Since cos(uv ui) cos(ui uv), what is important is the difference, in the phases of the voltage and current., Note that p(t) is time-varying while P does not depend on time. To, find the instantaneous power, we must necessarily have v(t) and i(t) in, the time domain. But we can find the average power when voltage and, current are expressed in the time domain, as in Eq. (11.8), or when they, are expressed in the frequency domain. The phasor forms of v(t) and i(t), in Eq. (11.2) are V Vmluv and I Imlui, respectively. P is calculated, using Eq. (11.8) or using phasors V and I. To use phasors, we notice that, 1, 1, VI* Vm Imluv ui, 2, 2, 1, Vm Im[cos(uv ui) j sin(uv ui)], 2, , (11.9), , We recognize the real part of this expression as the average power P, according to Eq. (11.8). Thus,, 1, 1, P Re[VI*] Vm Im cos(uv ui), 2, 2, , (11.10), , Consider two special cases of Eq. (11.10). When uv ui, the voltage and current are in phase. This implies a purely resistive circuit or, resistive load R, and, 1, 1, 1, P Vm Im I 2m R 0I 0 2 R, 2, 2, 2, , (11.11), , where 0 I 0 2 I I*. Equation (11.11) shows that a purely resistive circuit absorbs power at all times. When uv ui 90, we have a purely, reactive circuit, and, 1, P Vm Im cos 90 0, (11.12), 2
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ale29559_ch11.qxd, , 07/08/2008, , 12:01 PM, , Page 461, , 11.2, , Instantaneous and Average Power, , 461, , showing that a purely reactive circuit absorbs no average power. In, summary,, A resistive load (R ) absorbs power at all times, while a reactive load, (L or C ) absorbs zero average power., , Example 11.1, , Given that, v(t) 120 cos(377t 45) V, , and, , i(t) 10 cos(377t 10) A, , find the instantaneous power and the average power absorbed by the, passive linear network of Fig. 11.1., Solution:, The instantaneous power is given by, p vi 1200 cos(377t 45) cos(377t 10), Applying the trigonometric identity, 1, cos A cos B [cos(A B) cos(A B)], 2, gives, p 600[cos(754t 35) cos 55], or, p(t) 344.2 600 cos(754t 35) W, The average power is, 1, 1, P Vm Im cos(uv ui) 120(10) cos[45 (10)], 2, 2, 600 cos 55 344.2 W, which is the constant part of p(t) above., , Calculate the instantaneous power and average power absorbed by the, passive linear network of Fig. 11.1 if, v(t) 165 cos(10t 20) V, , and, , Practice Problem 11.1, , i(t) 20 sin(10t 60) A, , Answer: 1.0606 1.65 cos(20t 10) kW, 1.0606 kW., , Calculate the average power absorbed by an impedance Z 30 j70 , when a voltage V 120l0 is applied across it., Solution:, The current through the impedance is, I, , 120l0, 120l0, V, , , 1.576l66.8 A, Z, 30 j 70, 76.16l66.8, , Example 11.2
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ale29559_ch11.qxd, , 07/08/2008, , 12:01 PM, , Page 462, , Chapter 11, , 462, , AC Power Analysis, , The average power is, 1, 1, P Vm Im cos(uv ui) (120)(1.576) cos(0 66.8) 37.24 W, 2, 2, , Practice Problem 11.2, , A current I 20l30 A flows through an impedance Z 40l22 ., Find the average power delivered to the impedance., Answer: 3.709 kW., , Example 11.3, I, , 5 30° V, , For the circuit shown in Fig. 11.3, find the average power supplied by, the source and the average power absorbed by the resistor., , 4Ω, , +, −, , − j2 Ω, , Solution:, The current I is given by, I, , Figure 11.3, For Example 11.3., , 5l30, 4 j2, , , , 5l30, 4.472l26.57, , 1.118l56.57 A, , The average power supplied by the voltage source is, 1, P (5)(1.118) cos(30 56.57) 2.5 W, 2, The current through the resistor is, IR I 1.118l56.57 A, and the voltage across it is, VR 4IR 4.472l56.57 V, The average power absorbed by the resistor is, P, , 1, (4.472)(1.118) 2.5 W, 2, , which is the same as the average power supplied. Zero average power, is absorbed by the capacitor., , Practice Problem 11.3, 3Ω, , 160 45° V, , +, −, , Figure 11.4, For Practice Prob. 11.3., , j1 Ω, , In the circuit of Fig. 11.4, calculate the average power absorbed by the, resistor and inductor. Find the average power supplied by the voltage, source., Answer: 3.84 kW, 0 W, 3.84 kW.
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ale29559_ch11.qxd, , 07/08/2008, , 12:01 PM, , Page 463, , 11.2, , Instantaneous and Average Power, , 463, , Example 11.4, , Determine the average power generated by each source and the average, power absorbed by each passive element in the circuit of Fig. 11.5(a)., , 4 0° Α, , 20 Ω, , − j5 Ω, , 2, , 4, , 1, , j10 Ω, , 3, , +, 5, , +, −, , 4 0° Α, , 60 30° V, , +, V1, −, , (a), , V2, I1, , −, j10 Ω, , (b), , Figure 11.5, For Example 11.4., , Solution:, We apply mesh analysis as shown in Fig. 11.5(b). For mesh 1,, I1 4 A, For mesh 2,, ( j10 j5)I2 j10I1 60l30 0,, , I1 4 A, , or, j5I2 60l30 j40, , − j5 Ω, , 20 Ω, , 1, , I2 12l60 8, 10.58l79.1 A, , For the voltage source, the current flowing from it is I2 10.58l79.1 A, and the voltage across it is 60l30 V, so that the average power is, 1, P5 (60)(10.58) cos(30 79.1) 207.8 W, 2, Following the passive sign convention (see Fig. 1.8), this average power, is absorbed by the source, in view of the direction of I2 and the polarity, of the voltage source. That is, the circuit is delivering average power to, the voltage source., For the current source, the current through it is I1 4l0 and the, voltage across it is, V1 20I1 j10(I1 I2) 80 j10(4 2 j10.39), 183.9 j20 184.984l6.21 V, The average power supplied by the current source is, 1, P1 (184.984)(4) cos(6.21 0) 367.8 W, 2, It is negative according to the passive sign convention, meaning that, the current source is supplying power to the circuit., For the resistor, the current through it is I1 4l0 and the voltage, across it is 20I1 80l0, so that the power absorbed by the resistor is, 1, P2 (80)(4) 160 W, 2, , I2, , +, −, , 60 30° V
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ale29559_ch11.qxd, , 07/08/2008, , 12:01 PM, , 464, , Page 464, , Chapter 11, , AC Power Analysis, , For the capacitor, the current through it is I2 10.58l79.1 and the voltage across it is j5I2 (5l90)(10.58l79.1) 52.9l79.1 90., The average power absorbed by the capacitor is, 1, P4 (52.9)(10.58) cos(90) 0, 2, For the inductor, the current through it is I1 I2 , 2 j10.39 10.58l79.1. The voltage across it is j10(I1 I2), 10.58l79.190. Hence, the average power absorbed by the, inductor is, 1, P3 (105.8)(10.58) cos 90 0, 2, Notice that the inductor and the capacitor absorb zero average power, and that the total power supplied by the current source equals the power, absorbed by the resistor and the voltage source, or, P1 P2 P3 P4 P5 367.8 160 0 0 207.8 0, indicating that power is conserved., , Practice Problem 11.4, , Calculate the average power absorbed by each of the five elements in, the circuit of Fig. 11.6., , 8Ω, 40 0° V, , +, −, , j4 Ω, − j2 Ω, , +, −, , 20 90° V, , Figure 11.6, For Practice Prob. 11.4., , Answer: 40-V Voltage source: 60 W; j20-V Voltage source: 40 W;, resistor: 100 W; others: 0 W., , 11.3, , Maximum Average Power Transfer, , In Section 4.8 we solved the problem of maximizing the power delivered by a power-supplying resistive network to a load RL. Representing the circuit by its Thevenin equivalent, we proved that the, maximum power would be delivered to the load if the load resistance, is equal to the Thevenin resistance RL RTh. We now extend that, result to ac circuits., Consider the circuit in Fig. 11.7, where an ac circuit is connected, to a load ZL and is represented by its Thevenin equivalent. The load is, usually represented by an impedance, which may model an electric
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ale29559_ch11.qxd, , 07/08/2008, , 12:01 PM, , Page 466, , Chapter 11, , 466, , AC Power Analysis, , This means that for maximum average power transfer to a purely resistive load, the load impedance (or resistance) is equal to the magnitude, of the Thevenin impedance., , Example 11.5, 4Ω, , 10 0° V, , j5 Ω, 8Ω, , +, −, , Determine the load impedance ZL that maximizes the average power, drawn from the circuit of Fig. 11.8. What is the maximum average, power?, ZL, , − j6 Ω, , Figure 11.8, , Solution:, First we obtain the Thevenin equivalent at the load terminals. To get, ZTh, consider the circuit shown in Fig. 11.9(a). We find, ZTh j5 4 (8 j6) j5 , , For Example 11.5., , 4(8 j6), 2.933 j4.467 , 4 8 j6, , j5 Ω, , 4Ω, , j5 Ω, , 4Ω, , +, 8Ω, , Z Th, , 8Ω, , 10 V +, −, , − j6 Ω, , − j6 Ω, , (a), , VTh, −, , (b), , Figure 11.9, Finding the Thevenin equivalent of the circuit in Fig. 11.8., , To find VTh, consider the circuit in Fig. 11.8(b). By voltage division,, VTh , , 8 j6, (10) 7.454l10.3 V, 4 8 j6, , The load impedance draws the maximum power from the circuit when, ZL Z*Th 2.933 j4.467 , According to Eq. (11.20), the maximum average power is, Pmax , , Practice Problem 11.5, − j4 Ω, , 8Ω, , j10 Ω, , 6A, , Figure 11.10, For Practice Prob. 11.5., , 5Ω, , ZL, , 0 VTh 0 2, (7.454)2, , 2.368 W, 8RTh, 8(2.933), , For the circuit shown in Fig. 11.10, find the load impedance ZL that, absorbs the maximum average power. Calculate that maximum average power., Answer: 3.415 j0.7317 , 12.861 W.
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ale29559_ch11.qxd, , 07/08/2008, , 12:01 PM, , Page 467, , 11.4, , Effective or RMS Value, , 467, , Example 11.6, , In the circuit in Fig. 11.11, find the value of RL that will absorb the, maximum average power. Calculate that power., , 40 Ω − j30 Ω, , Solution:, We first find the Thevenin equivalent at the terminals of RL., ZTh, , j20(40 j30), (40 j30) j20 , 9.412 j22.35 , j20 40 j30, , 150 30° V, , +, −, , j20 Ω, , Figure 11.11, , By voltage division,, , For Example 11.6., , j20, VTh , (150l30) 72.76l134 V, j20 40 j30, The value of RL that will absorb the maximum average power is, RL 0ZTh 0 29.4122 22.352 24.25 , , The current through the load is, I, , 72.76l134, VTh, , 1.8l100.42 A, ZTh RL, 33.66 j22.35, , The maximum average power absorbed by RL is, Pmax , , 1 2, 1, 0 I 0 RL (1.8)2(24.25) 39.29 W, 2, 2, , In Fig. 11.12, the resistor RL is adjusted until it absorbs the maximum, average power. Calculate RL and the maximum average power, absorbed by it., 80 Ω, , 120 60° V, , +, −, , j60 Ω, , 90 Ω, , − j30 Ω, , RL, , Figure 11.12, For Practice Prob. 11.6., , Answer: 30 , 6.863 W., , 11.4, , Effective or RMS Value, , The idea of effective value arises from the need to measure the effectiveness of a voltage or current source in delivering power to a resistive load., The effective value of a periodic current is the dc current that delivers the same average power to a resistor as the periodic current., , Practice Problem 11.6, , RL
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ale29559_ch11.qxd, , 07/08/2008, , 12:01 PM, , Page 468, , Chapter 11, , 468, , In Fig. 11.13, the circuit in (a) is ac while that of (b) is dc. Our objective, is to find Ieff that will transfer the same power to resistor R as the sinusoid i. The average power absorbed by the resistor in the ac circuit is, , i(t), , v(t), , +, −, , AC Power Analysis, , R, , P, (a), , 1, T, , , , T, , 0, , , , T, , i2 dt, , (11.22), , 0, , while the power absorbed by the resistor in the dc circuit is, , I eff, +, V eff, −, , R, T, , i2R dt , , P I 2eff R, R, , (b), , Figure 11.13, Finding the effective current: (a) ac circuit,, (b) dc circuit., , (11.23), , Equating the expressions in Eqs. (11.22) and (11.23) and solving for, Ieff , we obtain, 1 T 2, Ieff , i dt, (11.24), BT 0, , , , The effective value of the voltage is found in the same way as current;, that is,, 1 T 2, Veff , v dt, (11.25), BT 0, , , , This indicates that the effective value is the (square) root of the mean, (or average) of the square of the periodic signal. Thus, the effective, value is often known as the root-mean-square value, or rms value for, short; and we write, Ieff Irms,, , Veff Vrms, , (11.26), , For any periodic function x(t) in general, the rms value is given by, 1, BT, , Xrms , , , , T, , x2 dt, , (11.27), , 0, , The effective value of a periodic signal is its root mean square (rms) value., , Equation 11.27 states that to find the rms value of x(t), we first, find its square x2 and then find the mean of that, or, 1, T, , , , T, , x2 dt, , 0, , and then the square root ( 1, ) of that mean. The rms value of a, constant is the constant itself. For the sinusoid i(t) Im cos t, the, effective or rms value is, Irms , I 2m, , BT, , 1, BT, , , , T, , 0, , , , T, , I 2m cos2 t dt, , 0, , (11.28), , Im, 1, (1 cos 2t) dt , 2, 12, , Similarly, for v(t) Vm cos t,, Vrms , , Vm, 12, , (11.29), , Keep in mind that Eqs. (11.28) and (11.29) are only valid for sinusoidal, signals.
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ale29559_ch11.qxd, , 07/08/2008, , 12:01 PM, , Page 469, , 11.4, , Effective or RMS Value, , 469, , The average power in Eq. (11.8) can be written in terms of the rms, values., Vm Im, 1, P Vm Im cos(uv ui) , cos(uv ui), 2, 12 12, (11.30), Vrms Irms cos(uv ui), Similarly, the average power absorbed by a resistor R in Eq. (11.11), can be written as, V 2rms, P I 2rms R , (11.31), R, When a sinusoidal voltage or current is specified, it is often in terms, of its maximum (or peak) value or its rms value, since its average value, is zero. The power industries specify phasor magnitudes in terms of their, rms values rather than peak values. For instance, the 110 V available at, every household is the rms value of the voltage from the power company. It is convenient in power analysis to express voltage and current, in their rms values. Also, analog voltmeters and ammeters are designed, to read directly the rms value of voltage and current, respectively., , Determine the rms value of the current waveform in Fig. 11.14. If the, current is passed through a 2- resistor, find the average power absorbed, by the resistor., Solution:, The period of the waveform is T 4. Over a period, we can write the, current waveform as, 5t, 0 6 t 6 2, i(t) b, 10, 2 6 t 6 4, , Irms, , , , , , i(t), 10, , 0, , 2, , 4, , 6, , 8, , 10, , t, , −10, , The rms value is, 1, , BT, , Example 11.7, , Figure 11.14, T, , 0, , 1, i2 dt , c, B4, 3 2, , , , 2, , (5t)2 dt , , 0, , , , 4, , 2, , (10)2 dt d, , For Example 11.7., , 4, , 1, t, 1 200, c 25 ` 100t ` d , 200b 8.165 A, a, B4, 3 0, 4, 3, B, 2, , The power absorbed by a 2- resistor is, P I 2rms R (8.165)2(2) 133.3 W, , Find the rms value of the current waveform of Fig. 11.15. If the current, flows through a 9- resistor, calculate the average power absorbed by, the resistor., Answer: 4.318 A, 192 W., , Practice Problem 11.7, i(t), 8, , 0, , 1, , 2, , 3, , Figure 11.15, For Practice Prob. 11.7., , 4, , 5, , 6, , t
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ale29559_ch11.qxd, , 07/08/2008, , 12:01 PM, , Page 470, , Chapter 11, , 470, , Example 11.8, , AC Power Analysis, , The waveform shown in Fig. 11.16 is a half-wave rectified sine wave., Find the rms value and the amount of average power dissipated in a, 10- resistor., , v (t), 10, , Solution:, The period of the voltage waveform is T 2 p, and, 0, , , , 2, , 3, , 10 sin t, 0 6 t 6 p, v(t) b, 0,, p 6 t 6 2p, , t, , Figure 11.16, For Example 11.8., , The rms value is obtained as, V 2rms , But sin t , 2, , 1, T, 1, 2 (1, , V 2rms , , , , T, , v2(t) dt , , 0, , 1, c, 2p, , , , p, , (10 sin t)2 dt , , , , 2p, , p, , 0, , 02 dt d, , cos 2t). Hence,, , 1, 2p, , , , p, , 0, , 100, 50, sin 2t p, (1 cos 2t) dt , at , b`, 2, 2p, 2, 0, , 50, 1, , ap sin 2 p 0b 25,, 2p, 2, , Vrms 5 V, , The average power absorbed is, P, , Practice Problem 11.8, , V 2rms, 52, , 2.5 W, R, 10, , Find the rms value of the full-wave rectified sine wave in Fig. 11.17., Calculate the average power dissipated in a 6- resistor., , v (t), , Answer: 7.071 V, 8.333 W., , 10, , 0, , , , Figure 11.17, For Practice Prob. 11.8., , 2, , 3, , t, , 11.5, , Apparent Power and Power Factor, , In Section 11.2 we saw that if the voltage and current at the terminals, of a circuit are, v(t) Vm cos(t uv), , and, , i(t) Im cos(t ui) (11.32), , or, in phasor form, V Vmluv and I Imlui, the average power is, 1, P Vm Im cos(uv ui), 2, , (11.33), , In Section 11.4, we saw that, P Vrms Irms cos(uv ui) S cos(uv ui), , (11.34), , We have added a new term to the equation:, S Vrms Irms, , (11.35)
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ale29559_ch11.qxd, , 07/08/2008, , 12:01 PM, , Page 471, , 11.5, , Apparent Power and Power Factor, , 471, , The average power is a product of two terms. The product Vrms Irms is, known as the apparent power S. The factor cos(uv ui) is called the, power factor (pf)., The apparent power (in VA) is the product of the rms values of voltage and current., , The apparent power is so called because it seems apparent that the power, should be the voltage-current product, by analogy with dc resistive circuits. It is measured in volt-amperes or VA to distinguish it from the, average or real power, which is measured in watts. The power factor is, dimensionless, since it is the ratio of the average power to the apparent, power,, , pf , , P, cos(uv ui), S, , (11.36), , The angle uv ui is called the power factor angle, since it is the, angle whose cosine is the power factor. The power factor angle is equal, to the angle of the load impedance if V is the voltage across the load, and I is the current through it. This is evident from the fact that, Vmluv, Vm, V, luv ui, , , I, Im, Imlui, , Z, , (11.37), , Alternatively, since, Vrms , , V, Vrmsluv, 12, , (11.38a), , Irms , , I, Irmslui, 12, , (11.38b), , and, , the impedance is, Z, , Vrms, Vrms, V, luv ui, , , I, Irms, Irms, , (11.39), , The power factor is the cosine of the phase difference between voltage and current. It is also the cosine of the angle of the load impedance., , From Eq. (11.36), the power factor may be seen as that factor by which, the apparent power must be multiplied to obtain the real or average, power. The value of pf ranges between zero and unity. For a purely, resistive load, the voltage and current are in phase, so that uv ui 0, and pf 1. This implies that the apparent power is equal to the average power. For a purely reactive load, uv ui 90 and pf 0. In, this case the average power is zero. In between these two extreme, cases, pf is said to be leading or lagging. Leading power factor means, that current leads voltage, which implies a capacitive load. Lagging, power factor means that current lags voltage, implying an inductive, , From Eq. (11.36), the power factor, may also be regarded as the ratio of, the real power dissipated in the load, to the apparent power of the load.
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ale29559_ch11.qxd, , 07/08/2008, , 12:01 PM, , 472, , Page 472, , Chapter 11, , AC Power Analysis, , load. Power factor affects the electric bills consumers pay the electric, utility companies, as we will see in Section 11.9.2., , Example 11.9, , A series-connected load draws a current i(t) 4 cos(100pt 10) A, when the applied voltage is v(t) 120 cos(100pt 20) V. Find the, apparent power and the power factor of the load. Determine the element values that form the series-connected load., Solution:, The apparent power is, S Vrms Irms , , 120 4, 240 VA, 12 12, , The power factor is, pf cos(uv ui) cos(20 10) 0.866, , (leading), , The pf is leading because the current leads the voltage. The pf may, also be obtained from the load impedance., Z, , 120l20, V, , 30l30 25.98 j15 , I, 4l10, pf cos(30) 0.866, , (leading), , The load impedance Z can be modeled by a 25.98- resistor in series, with a capacitor with, XC 15 , , 1, C, , or, C, , Practice Problem 11.9, , 1, 1, , 212.2 mF, 15, 15 100p, , Obtain the power factor and the apparent power of a load whose, impedance is Z 60 j40 when the applied voltage is v(t) , 160 cos(377t 10) V., Answer: 0.8321 lagging, 177.5l33.69 VA., , Example 11.10, , Determine the power factor of the entire circuit of Fig. 11.18 as seen, by the source. Calculate the average power delivered by the source., Solution:, The total impedance is, Z 6 4 (j2) 6 , , j2 4, 6.8 j1.6 7l13.24 , 4 j2
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ale29559_ch11.qxd, , 07/08/2008, , 12:01 PM, , Page 473, , 11.6, , Complex Power, , 473, 6Ω, , The power factor is, pf cos(13.24) 0.9734, , (leading), , since the impedance is capacitive. The rms value of the current is, Irms , , 30l0, Vrms, , 4.286l13.24 A, Z, 7l13.24, , 30 0° V rms, , +, −, , −j2 Ω, , 4Ω, , Figure 11.18, For Example 11.10., , The average power supplied by the source is, P Vrms Irms pf (30)(4.286)0.9734 125 W, or, P I 2rms R (4.286)2(6.8) 125 W, where R is the resistive part of Z., , Calculate the power factor of the entire circuit of Fig. 11.19 as seen by, the source. What is the average power supplied by the source?, , Practice Problem 11.10, 10 Ω, , 8Ω, , Answer: 0.936 lagging, 1.062 kW., 120 0° V rms, , +, −, , j4 Ω, , −j6 Ω, , Figure 11.19, For Practice Prob. 11.10., , 11.6, , Complex Power, , Considerable effort has been expended over the years to express, power relations as simply as possible. Power engineers have coined, the term complex power, which they use to find the total effect of, parallel loads. Complex power is important in power analysis because, it contains all the information pertaining to the power absorbed by a, given load., Consider the ac load in Fig. 11.20. Given the phasor form V , Vmluv and I Imlui of voltage v(t) and current i(t), the complex, power S absorbed by the ac load is the product of the voltage and the, complex conjugate of the current, or, 1, S VI*, 2, , (11.40), , assuming the passive sign convention (see Fig. 11.20). In terms of the, rms values,, S Vrms I*rms, , (11.41), , where, Vrms , , V, Vrmsluv, 12, , (11.42), , Irms , , I, Irmslui, 12, , (11.43), , and, , I, +, V, , Load, Z, , −, , Figure 11.20, The voltage and current phasors associated, with a load.
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ale29559_ch11.qxd, , 07/08/2008, , 12:01 PM, , 474, , When working with the rms values of, currents or voltages, we may drop the, subscript rms if no confusion will be, caused by doing so., , Page 474, , Chapter 11, , AC Power Analysis, , Thus we may write Eq. (11.41) as, S Vrms Irmsluv ui, Vrms Irms cos(uv ui) jVrms Irms sin(uv ui), , (11.44), , This equation can also be obtained from Eq. (11.9). We notice from, Eq. (11.44) that the magnitude of the complex power is the apparent, power; hence, the complex power is measured in volt-amperes (VA). Also,, we notice that the angle of the complex power is the power factor angle., The complex power may be expressed in terms of the load impedance Z. From Eq. (11.37), the load impedance Z may be written as, Z, , Vrms, Vrms, V, luv ui, , , I, Irms, Irms, , (11.45), , Thus, Vrms ZIrms. Substituting this into Eq. (11.41) gives, S I 2rms Z , , 2, V rms, Vrms I*rms, Z*, , (11.46), , Since Z R jX, Eq. (11.46) becomes, S I 2rms(R jX) P jQ, , (11.47), , where P and Q are the real and imaginary parts of the complex power;, that is,, P Re(S) I 2rms R, (11.48), Q Im(S) I 2rms X, , (11.49), , P is the average or real power and it depends on the load’s resistance, R. Q depends on the load’s reactance X and is called the reactive (or, quadrature) power., Comparing Eq. (11.44) with Eq. (11.47), we notice that, P Vrms Irms cos(uv ui),, , Q Vrms Irms sin(uv ui) (11.50), , The real power P is the average power in watts delivered to a load; it, is the only useful power. It is the actual power dissipated by the load., The reactive power Q is a measure of the energy exchange between the, source and the reactive part of the load. The unit of Q is the volt-ampere, reactive (VAR) to distinguish it from the real power, whose unit is the, watt. We know from Chapter 6 that energy storage elements neither dissipate nor supply power, but exchange power back and forth with the, rest of the network. In the same way, the reactive power is being transferred back and forth between the load and the source. It represents a, lossless interchange between the load and the source. Notice that:, 1. Q 0 for resistive loads (unity pf)., 2. Q 6 0 for capacitive loads (leading pf)., 3. Q 7 0 for inductive loads (lagging pf)., Thus,, Complex power (in VA) is the product of the rms voltage phasor and, the complex conjugate of the rms current phasor. As a complex quantity,, its real part is real power P and its imaginary part is reactive power Q.
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ale29559_ch11.qxd, , 07/22/2008, , 01:25 PM, , Page 475, , 11.6, , Complex Power, , 475, , Introducing the complex power enables us to obtain the real and reactive powers directly from voltage and current phasors., , Complex Power S P jQ Vrms(Irms)*, Vrms Irmsluv ui, , Apparent Power S 0S 0 Vrms Irms 2P 2 Q 2, Real Power P Re(S) S cos(uv ui), , (11.51), , Reactive Power Q Im(S) S sin(uv ui), Power Factor , , P, cos(uv ui), S, , This shows how the complex power contains all the relevant power, information in a given load., It is a standard practice to represent S, P, and Q in the form of a, triangle, known as the power triangle, shown in Fig. 11.21(a). This is, similar to the impedance triangle showing the relationship between Z,, R, and X, illustrated in Fig. 11.21(b). The power triangle has four, items—the apparent/complex power, real power, reactive power, and, the power factor angle. Given two of these items, the other two can, easily be obtained from the triangle. As shown in Fig. 11.22, when S, lies in the first quadrant, we have an inductive load and a lagging pf., When S lies in the fourth quadrant, the load is capacitive and the pf is, leading. It is also possible for the complex power to lie in the second, or third quadrant. This requires that the load impedance have a negative resistance, which is possible with active circuits., , S contains all power information of, a load. The real part of S is the real, power P ; its imaginary part is the reactive power Q ; its magnitude is the apparent power S; and the cosine of its, phase angle is the power factor pf., , Im, , S, , Q, , |Z |, , X, , +Q (lagging pf ), , S, , , , , , P, , R, , v − i, , (a), , (b), , v − i, , P, , Re, , Figure 11.21, (a) Power triangle, (b) impedance triangle., , S, , −Q (leading pf ), , Figure 11.22, Power triangle., , The voltage across a load is v(t) 60 cos(t 10) V and the current, through the element in the direction of the voltage drop is i(t) , 1.5 cos(t 50) A. Find: (a) the complex and apparent powers,, (b) the real and reactive powers, and (c) the power factor and the load, impedance., , Example 11.11
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 476, , 476, , Chapter 11, , AC Power Analysis, , Solution:, (a) For the rms values of the voltage and current, we write, Vrms , , 60, 22, , l10,, , Irms , , 1.5, 22, , l50, , The complex power is, S Vrms I*rms a, , 60, 22, , l10b a 1.5 l50b 45l60 VA, 22, , The apparent power is, S 0 S 0 45 VA, (b) We can express the complex power in rectangular form as, S 45l60 45[cos(60) j sin(60)] 22.5 j38.97, Since S P jQ, the real power is, P 22.5 W, while the reactive power is, Q 38.97 VAR, (c) The power factor is, pf cos(60) 0.5 (leading), It is leading, because the reactive power is negative. The load impedance is, Z, , 60l10, V, , 40l60 , I, 1.5l50, , which is a capacitive impedance., , Practice Problem 11.11, , For a load, Vrms 110l85 V, Irms 0.4l15 A. Determine: (a) the, complex and apparent powers, (b) the real and reactive powers, and, (c) the power factor and the load impedance., Answer: (a) 44l70 VA, 44 VA, (b) 15.05 W, 41.35 VAR, (c) 0.342, lagging, 94.06 j258.4 ., , Example 11.12, , A load Z draws 12 kVA at a power factor of 0.856 lagging from a, 120-V rms sinusoidal source. Calculate: (a) the average and reactive powers delivered to the load, (b) the peak current, and (c) the load impedance., Solution:, (a) Given that pf cos u 0.856, we obtain the power angle as, u cos1 0.856 31.13. If the apparent power is S 12,000 VA,, then the average or real power is, P S cos u 12,000 0.856 10.272 kW
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 477, , 11.7, , Conservation of AC Power, , 477, , while the reactive power is, Q S sin u 12,000 0.517 6.204 kVA, (b) Since the pf is lagging, the complex power is, S P jQ 10.272 j6.204 kVA, From S Vrms I*rms, we obtain, I*rms , , 10,272 j6204, S, , 85.6 j51.7 A 100l31.13 A, Vrms, 120l0, , Thus Irms 100l31.13 and the peak current is, Im 22Irms 22(100) 141.4 A, (c) The load impedance, Z, , 120l0, Vrms, , 1.2l31.13 , Irms, 100l31.13, , which is an inductive impedance., A sinusoidal source supplies 20 kVAR reactive power to load Z , 250l75 . Determine: (a) the power factor, (b) the apparent power, delivered to the load, and (c) the rms voltage., , Practice Problem 11.12, , Answer: (a) 0.2588 leading, (b) 20.71 kVA, (c) 2.275 kV., , 11.7, , Conservation of AC Power, , The principle of conservation of power applies to ac circuits as well as, to dc circuits (see Section 1.5)., To see this, consider the circuit in Fig. 11.23(a), where two load, impedances Z1 and Z2 are connected in parallel across an ac source V., KCL gives, I I1 I2, , (11.52), , The complex power supplied by the source is (from now on, unless, otherwise specified, all values of voltages and currents will be assumed, to be rms values), S VI* V(I1* I2*) VI*1 VI*2 S1 S2 (11.53), I, I, V +, −, , I1, , I2, , Z1, , Z2, , (a), , V +, −, , Z1, , Z2, , +V −, 1, , +V −, 2, , (b), , Figure 11.23, An ac voltage source supplied loads connected in: (a) parallel, (b) series., , In fact, we already saw in Examples, 11.3 and 11.4 that average power is, conserved in ac circuits.
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , 478, , Page 478, , Chapter 11, , AC Power Analysis, , where S1 and S2 denote the complex powers delivered to loads Z1 and, Z2, respectively., If the loads are connected in series with the voltage source, as, shown in Fig. 11.23(b), KVL yields, V V1 V2, , (11.54), , The complex power supplied by the source is, S VI* (V1 V2)I* V1I* V2I* S1 S2 (11.55), where S1 and S2 denote the complex powers delivered to loads Z1 and, Z2, respectively., We conclude from Eqs. (11.53) and (11.55) that whether the loads, are connected in series or in parallel (or in general), the total power, supplied by the source equals the total power delivered to the load., Thus, in general, for a source connected to N loads,, S S1 S2 p SN, , In fact, all forms of ac power are conserved: instantaneous, real, reactive,, and complex., , (11.56), , This means that the total complex power in a network is the sum of, the complex powers of the individual components. (This is also true of, real power and reactive power, but not true of apparent power.) This, expresses the principle of conservation of ac power:, The complex, real, and reactive powers of the sources equal the, respective sums of the complex, real, and reactive powers of the individual loads., , From this we imply that the real (or reactive) power flow from sources, in a network equals the real (or reactive) power flow into the other elements in the network., , Example 11.13, , Figure 11.24 shows a load being fed by a voltage source through a, transmission line. The impedance of the line is represented by the, (4 j2) impedance and a return path. Find the real power and reactive power absorbed by: (a) the source, (b) the line, and (c) the load., I, , 220 0° V rms, , 4Ω, , 15 Ω, , +, −, , Source, , j2 Ω, , − j10 Ω, Line, , Load, , Figure 11.24, For Example 11.13., , Solution:, The total impedance is, Z (4 j2) (15 j10) 19 j8 20.62l22.83
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 479, , 11.7, , Conservation of AC Power, , 479, , The current through the circuit is, I, , 220l0, Vs, , 10.67l22.83 A rms, Z, 20.62l22.83, , (a) For the source, the complex power is, Ss Vs I* (220l0)(10.67l22.83), 2347.4l22.83 (2163.5 j910.8) VA, From this, we obtain the real power as 2163.5 W and the reactive, power as 910.8 VAR (leading)., (b) For the line, the voltage is, Vline (4 j2)I (4.472l26.57)(10.67l22.83), 47.72l49.4 V rms, The complex power absorbed by the line is, Sline Vline I* (47.72l49.4)(10.67l22.83), 509.2l26.57 455.4 j227.7 VA, or, Sline 0I 0 2Zline (10.67)2(4 j2) 455.4 j227.7 VA, That is, the real power is 455.4 W and the reactive power is 227.76, VAR (lagging)., (c) For the load, the voltage is, VL (15 j10)I (18.03l33.7)(10.67l22.83), 192.38l10.87 V rms, The complex power absorbed by the load is, SL VL I* (192.38l10.87)(10.67l22.83), 2053l33.7 (1708 j1139) VA, The real power is 1708 W and the reactive power is 1139 VAR, (leading). Note that Ss Sline SL, as expected. We have used the rms, values of voltages and currents., , In the circuit in Fig. 11.25, the 60- resistor absorbs an average power, of 240 W. Find V and the complex power of each branch of the circuit. What is the overall complex power of the circuit? (Assume the, current through the 60- resistor has no phase shift.), , Practice Problem 11.13, 20 Ω, 30 Ω, , Answer: 240.67l21.45 V (rms); the 20- resistor: 656 VA; the, (30 j10) impedance: 480 j160 VA; the (60 j20) impedance: 240 j80 VA; overall: 1376 j80 VA., , V, , +, −, , − j10 Ω, , Figure 11.25, For Practice Prob. 11.13., , j20 Ω, 60 Ω
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 480, , Chapter 11, , 480, , Example 11.14, , In the circuit of Fig. 11.26, Z1 60l30 and Z2 40l45 ., Calculate the total: (a) apparent power, (b) real power, (c) reactive, power, and (d) pf, supplied by the source and seen by the source., , It, , 120 10° V rms, , +, −, , AC Power Analysis, , I1, , I2, , Z1, , Z2, , Solution:, The current through Z1 is, I1 , , Figure 11.26, For Example 11.14., , 120l10, V, , 2l40 A rms, Z1, 60l30, , while the current through Z2 is, I2 , , 120l10, V, , 3l35 A rms, Z2, 40l45, , The complex powers absorbed by the impedances are, S1 , , 2, V rms, (120)2, , 240l30 207.85 j120 VA, Z1*, 60l30, , S2 , , 2, V rms, (120)2, , 360l45 254.6 j254.6 VA, Z*2, 40l45, , The total complex power is, St S1 S2 462.4 j134.6 VA, (a) The total apparent power is, 0 St 0 2462.42 134.62 481.6 VA., (b) The total real power is, Pt Re(St) 462.4 W or Pt P1 P2., (c) The total reactive power is, Qt Im(St) 134.6 VAR or Qt Q1 Q2., , (d) The pf Pt 0St 0 462.4481.6 0.96 (lagging)., We may cross check the result by finding the complex power Ss supplied, by the source., It I1 I2 (1.532 j1.286) (2.457 j1.721), 4 j0.435 4.024l6.21 A rms, Ss VI*t (120l10)(4.024l6.21), 482.88l16.21 463 j135 VA, which is the same as before., , Practice Problem 11.14, , Two loads connected in parallel are respectively 2 kW at a pf of 0.75, leading and 4 kW at a pf of 0.95 lagging. Calculate the pf of the two, loads. Find the complex power supplied by the source., Answer: 0.9972 (leading), 6 j0.4495 kVA.
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 481, , 11.8, , 11.8, , Power Factor Correction, , 481, , Power Factor Correction, , Most domestic loads (such as washing machines, air conditioners, and, refrigerators) and industrial loads (such as induction motors) are inductive and operate at a low lagging power factor. Although the inductive, nature of the load cannot be changed, we can increase its power factor., The process of increasing the power factor without altering the voltage, or current to the original load is known as power factor correction., , Since most loads are inductive, as shown in Fig. 11.27(a), a load’s, power factor is improved or corrected by deliberately installing a, capacitor in parallel with the load, as shown in Fig. 11.27(b). The effect, of adding the capacitor can be illustrated using either the power triangle or the phasor diagram of the currents involved. Figure 11.28 shows, the latter, where it is assumed that the circuit in Fig. 11.27(a) has a, power factor of cos u1, while the one in Fig. 11.27(b) has a power factor of cos u2. It is evident from Fig. 11.28 that adding the capacitor has, caused the phase angle between the supplied voltage and current to, reduce from u1 to u2, thereby increasing the power factor. We also, notice from the magnitudes of the vectors in Fig. 11.28 that with the, same supplied voltage, the circuit in Fig. 11.27(a) draws larger current, IL than the current I drawn by the circuit in Fig. 11.27(b). Power companies charge more for larger currents, because they result in increased, power losses (by a squared factor, since P I 2L R). Therefore, it is beneficial to both the power company and the consumer that every effort, is made to minimize current level or keep the power factor as close to, unity as possible. By choosing a suitable size for the capacitor, the current can be made to be completely in phase with the voltage, implying unity power factor., , I, +, , IL, , +, , V, , Inductive, load, , V, , Alternatively, power factor correction, may be viewed as the addition of a reactive element (usually a capacitor) in, parallel with the load in order to make, the power factor closer to unity., An inductive load is modeled as a, series combination of an inductor and, a resistor., , IC, IL, , Inductive, load, , IC, 1, , C, , 2, , V, IC, , I, , −, , −, (a), , (b), , Figure 11.27, Power factor correction: (a) original inductive load,, (b) inductive load with improved power factor., , IL, , Figure 11.28, Phasor diagram showing the effect of, adding a capacitor in parallel with the, inductive load., , We can look at the power factor correction from another perspective. Consider the power triangle in Fig. 11.29. If the original inductive, load has apparent power S1, then, P S1 cos u1,, , Q1 S1 sin u1 P tan u1, , (11.57)
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 482, , Chapter 11, , 482, , AC Power Analysis, , If we desire to increase the power factor from cos u1 to cos u2 without, altering the real power (i.e., P S2 cos u2), then the new reactive power is, , QC, , Q2 P tan u2, S1, Q1, , S2, Q2, , (11.58), , The reduction in the reactive power is caused by the shunt capacitor;, that is,, QC Q1 Q2 P(tan u1 tan u2), , (11.59), , But from Eq. (11.46), QC V 2rmsXC CV 2rms. The value of the, required shunt capacitance C is determined as, , 1 2, P, , Figure 11.29, Power triangle illustrating power factor, correction., , C, , QC, 2, V rms, , , , P(tan u1 tan u2), 2, V rms, , (11.60), , Note that the real power P dissipated by the load is not affected by the, power factor correction because the average power due to the capacitance is zero., Although the most common situation in practice is that of an, inductive load, it is also possible that the load is capacitive; that is, the, load is operating at a leading power factor. In this case, an inductor, should be connected across the load for power factor correction. The, required shunt inductance L can be calculated from, QL , , V 2rms, V 2rms, , XL, L, , 1, , L, , V 2rms, QL, , (11.61), , where QL Q1 Q2, the difference between the new and old reactive powers., , Example 11.15, , When connected to a 120-V (rms), 60-Hz power line, a load absorbs, 4 kW at a lagging power factor of 0.8. Find the value of capacitance, necessary to raise the pf to 0.95., Solution:, If the pf 0.8, then, cos u1 0.8, , 1, , u1 36.87, , where u1 is the phase difference between voltage and current. We, obtain the apparent power from the real power and the pf as, S1 , , P, 4000, , 5000 VA, cos u1, 0.8, , The reactive power is, Q1 S1 sin u 5000 sin 36.87 3000 VAR, When the pf is raised to 0.95,, cos u2 0.95, , 1, , u2 18.19
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 483, , 11.9, , Applications, , 483, , The real power P has not changed. But the apparent power has, changed; its new value is, S2 , , P, 4000, , 4210.5 VA, cos u2, 0.95, , The new reactive power is, Q2 S2 sin u2 1314.4 VAR, The difference between the new and old reactive powers is due to the, parallel addition of the capacitor to the load. The reactive power due, to the capacitor is, QC Q1 Q2 3000 1314.4 1685.6 VAR, and, C, , QC, 2, V rms, , , , 1685.6, 310.5 mF, 2p 60 1202, , Note: Capacitors are normally purchased for voltages they expect to, see. In this case, the maximum voltage this capacitor will see is about, 170 V peak. We would suggest purchasing a capacitor with a voltage, rating equal to, say, 200 V., , Find the value of parallel capacitance needed to correct a load of, 140 kVAR at 0.85 lagging pf to unity pf. Assume that the load is supplied by a 110-V (rms), 60-Hz line., , Practice Problem 11.15, , Answer: 30.69 mF., , 11.9, , Applications, , In this section, we consider two important application areas: how power, is measured and how electric utility companies determine the cost of, electricity consumption., , 11.9.1 Power Measurement, The average power absorbed by a load is measured by an instrument, called the wattmeter., , Reactive power is measured by an, instrument called the varmeter. The, varmeter is often connected to the, load in the same way as the wattmeter., , The wattmeter is the instrument for measuring the average power., , Figure 11.30 shows a wattmeter that consists essentially of two, coils: the current coil and the voltage coil. A current coil with very low, impedance (ideally zero) is connected in series with the load (Fig. 11.31), and responds to the load current. The voltage coil with very high impedance (ideally infinite) is connected in parallel with the load as shown, in Fig. 11.31 and responds to the load voltage. The current coil acts, like a short circuit because of its low impedance; the voltage coil, , Some wattmeters do not have coils;, the wattmeter considered here is the, electromagnetic type.
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 484, , Chapter 11, , 484, , AC Power Analysis, , i, , i, , ±, Current coil, ±, , ±, , +, v, −, , R, Voltage coil, , +, v, −, , ZL, , Figure 11.31, The wattmeter connected to the load., i, ±, , Figure 11.30, A wattmeter., , behaves like an open circuit because of its high impedance. As a result,, the presence of the wattmeter does not disturb the circuit or have an, effect on the power measurement., When the two coils are energized, the mechanical inertia of the moving system produces a deflection angle that is proportional to the average value of the product v(t)i(t). If the current and voltage of the load, are v(t) Vm cos(t uv) and i(t) Im cos(t ui), their corresponding rms phasors are, Vrms , , Vm, 22, , luv, , Irms , , and, , Im, 22, , lui, , (11.62), , and the wattmeter measures the average power given by, P |Vrms ||Irms | cos(uv ui) Vrms Irms cos(uv ui) (11.63), As shown in Fig. 11.31, each wattmeter coil has two terminals with, one marked . To ensure upscale deflection, the terminal of the current coil is toward the source, while the terminal of the voltage coil, is connected to the same line as the current coil. Reversing both coil, connections still results in upscale deflection. However, reversing one, coil and not the other results in downscale deflection and no wattmeter, reading., , Example 11.16, , Find the wattmeter reading of the circuit in Fig. 11.32., 12 Ω, , j10 Ω, , ±, ±, , 150 0° V rms, , +, −, , 8Ω, − j6 Ω, , Figure 11.32, For Example 11.16., , Solution:, 1. Define. The problem is clearly defined. Interestingly, this is a, problem where the student could actually validate the results by, doing the problem in the laboratory with a real wattmeter.
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 485, , 11.9, , Applications, , 2. Present. This problem consists of finding the average power, delivered to a load by an external source with a series impedance., 3. Alternative. This is a straightforward circuit problem where all, we need to do is find the magnitude and phase of the current, through the load and the magnitude and the phase of the voltage, across the load. These quantities could also be found by using, PSpice, which we will use as a check., 4. Attempt. In Fig. 11.32, the wattmeter reads the average power, absorbed by the (8 j6) impedance because the current coil, is in series with the impedance while the voltage coil is in, parallel with it. The current through the circuit is, Irms , , 150l0, (12 j10) (8 j6), , 150, A, 20 j4, , , , The voltage across the (8 j6) impedance is, Vrms Irms(8 j6) , , 150(8 j6), V, 20 j4, , The complex power is, S Vrms I*rms , , 150(8 j6), 1502(8 j6), 150, , , 20 j4, 20 j4, 202 42, 423.7 j324.6 VA, , The wattmeter reads, P Re(S) 432.7 W, 5. Evaluate. We can check our results by using PSpice., AC=ok, MAG=ok, PHASE=yes, , AC=ok, MAG=ok, IPRINT, PHASE=yes, R1, , L1, , 12, , 10, R2, , ACMAG=150V, ACPHASE=0, , +, −, , 8, , V1, C2, , 0.16667, , Simulation yields:, FREQ, 1.592E-01, , IM(V_PRINT2), 7.354E+00, , IP(V_PRINT2), -1.131E+01, , FREQ, 1.592E-01, , VM($N_0004), 7.354E+01, , VP($N_0004), -4.818E+01, , and, , 485
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 486, , 486, , Chapter 11, , AC Power Analysis, , To check our answer, all we need is the magnitude of the current, (7.354 A) flowing through the load resistor:, P (IL)2R (7.354)28 432.7 W, As expected, the answer does check!, 6. Satisfactory? We have satisfactorily solved the problem and the, results can now be presented as a solution to the problem., , Practice Problem 11.16, , For the circuit in Fig. 11.33, find the wattmeter reading., 4Ω, , ±, , − j2 Ω, , ±, 120 30° V rms, , +, −, , j9 Ω, , 12 Ω, , Figure 11.33, For Practice Prob. 11.16., , Answer: 1437 W., , 11.9.2 Electricity Consumption Cost, In Section 1.7, we considered a simplified model of the way the cost of, electricity consumption is determined. But the concept of power factor, was not included in the calculations. Now we consider the importance, of power factor in electricity consumption cost., Loads with low power factors are costly to serve because they, require large currents, as explained in Section 11.8. The ideal situation would be to draw minimum current from a supply so that S , P, Q 0, and pf 1. A load with nonzero Q means that energy flows, back and forth between the load and the source, giving rise to additional power losses. In view of this, power companies often encourage their customers to have power factors as close to unity as possible, and penalize some customers who do not improve their load power, factors., Utility companies divide their customers into categories: as residential (domestic), commercial, and industrial, or as small power,, medium power, and large power. They have different rate structures for, each category. The amount of energy consumed in units of kilowatthours (kWh) is measured using a kilowatt-hour meter installed at the, customer’s premises., Although utility companies use different methods for charging customers, the tariff or charge to a consumer is often two-part. The first, part is fixed and corresponds to the cost of generation, transmission,, and distribution of electricity to meet the load requirements of the consumers. This part of the tariff is generally expressed as a certain price
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 487, , 11.9, , Applications, , 487, , per kW of maximum demand. Or it may be based on kVA of maximum demand, to account for the power factor (pf) of the consumer. A, pf penalty charge may be imposed on the consumer whereby a certain, percentage of kW or kVA maximum demand is charged for every 0.01, fall in pf below a prescribed value, say 0.85 or 0.9. On the other hand,, a pf credit may be given for every 0.01 that the pf exceeds the prescribed value., The second part is proportional to the energy consumed in kWh;, it may be in graded form, for example, the first 100 kWh at 16 cents/kWh,, the next 200 kWh at 10 cents/kWh and so forth. Thus, the bill is determined based on the following equation:, Total Cost Fixed Cost Cost of Energy, , (11.64), , A manufacturing industry consumes 200 MWh in one month. If the, maximum demand is 1600 kW, calculate the electricity bill based on, the following two-part rate:, , Example 11.17, , Demand charge: $5.00 per month per kW of billing demand., Energy charge: 8 cents per kWh for the first 50,000 kWh,, 5 cents per kWh for the remaining energy., Solution:, The demand charge is, $5.00 1600 $8000, , (11.17.1), , The energy charge for the first 50,000 kWh is, $0.08 50,000 $4000, , (11.17.2), , The remaining energy is 200,000 kWh 50,000 kWh 150,000 kWh,, and the corresponding energy charge is, $0.05 150,000 $7500, , (11.17.3), , Adding the results of Eqs. (11.17.1) to (11.17.3) gives, Total bill for the month $8000 $4000 $7500 $19,500, It may appear that the cost of electricity is too high. But this is often a, small fraction of the overall cost of production of the goods manufactured, or the selling price of the finished product., , The monthly reading of a paper mill’s meter is as follows:, Maximum demand: 32,000 kW, Energy consumed: 500 MWh, Using the two-part rate in Example 11.17, calculate the monthly bill, for the paper mill., Answer: $186,500., , Practice Problem 11.17
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 488, , 488, , Example 11.18, , Chapter 11, , AC Power Analysis, , A 300-kW load supplied at 13 kV (rms) operates 520 hours a month, at 80 percent power factor. Calculate the average cost per month based, on this simplified tariff:, Energy charge: 6 cents per kWh, Power-factor penalty: 0.1 percent of energy charge for every 0.01, that pf falls below 0.85., Power-factor credit: 0.1 percent of energy charge for every 0.01, that pf exceeds 0.85., Solution:, The energy consumed is, W 300 kW 520 h 156,000 kWh, The operating power factor pf 80% 0.8 is 5 0.01 below the, prescribed power factor of 0.85. Since there is 0.1 percent energy charge, for every 0.01, there is a power-factor penalty charge of 0.5 percent., This amounts to an energy charge of, ¢W 156,000 , , 5 0.1, 780 kWh, 100, , The total energy is, Wt W ¢W 156,000 780 156,780 kWh, The cost per month is given by, Cost 6 cents Wt $0.06 156,780 $9,406.80, , Practice Problem 11.18, , An 800-kW induction furnace at 0.88 power factor operates 20 hours, per day for 26 days in a month. Determine the electricity bill per month, based on the tariff in Example 11.18., Answer: $24,885.12., , 11.10, , Summary, , 1. The instantaneous power absorbed by an element is the product of, the element’s terminal voltage and the current through the element:, p vi., 2. Average or real power P (in watts) is the average of instantaneous, power p:, 1, P, T, , , , T, , p dt, , 0, , If v(t) Vm cos(t uv) and i(t) Im cos(t ui), then Vrms , Vm12, Irms Im12, and, P, , 1, Vm Im cos(uv ui) Vrms Irms cos(uv ui), 2
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 489, , 11.10, , Summary, , Inductors and capacitors absorb no average power, while the average power absorbed by a resistor is (12)I 2m R I 2rms R., 3. Maximum average power is transferred to a load when the load, impedance is the complex conjugate of the Thevenin impedance, as seen from the load terminals, ZL Z*Th., 4. The effective value of a periodic signal x(t) is its root-mean-square, (rms) value., Xeff Xrms , , 1, BT, , , , T, , x2 dt, , 0, , For a sinusoid, the effective or rms value is its amplitude divided, by 12., 5. The power factor is the cosine of the phase difference between, voltage and current:, pf cos(uv ui), It is also the cosine of the angle of the load impedance or the ratio, of real power to apparent power. The pf is lagging if the current, lags voltage (inductive load) and is leading when the current leads, voltage (capacitive load)., 6. Apparent power S (in VA) is the product of the rms values of voltage and current:, S Vrms Irms, It is also given by S 0S 0 2P2 Q2, where P is the real, power and Q is reactive power., 7. Reactive power (in VAR) is:, 1, Q Vm Im sin(uv ui) Vrms Irms sin(uv ui), 2, 8. Complex power S (in VA) is the product of the rms voltage phasor and the complex conjugate of the rms current phasor. It is also, the complex sum of real power P and reactive power Q., S Vrms I*rms Vrms Irmsluv ui P jQ, Also,, S I 2rms Z , , 2, V rms, Z*, , 9. The total complex power in a network is the sum of the complex, powers of the individual components. Total real power and reactive power are also, respectively, the sums of the individual real, powers and the reactive powers, but the total apparent power is, not calculated by the process., 10. Power factor correction is necessary for economic reasons; it is the, process of improving the power factor of a load by reducing the, overall reactive power., 11. The wattmeter is the instrument for measuring the average power., Energy consumed is measured with a kilowatt-hour meter., , 489
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 490, , Chapter 11, , 490, , AC Power Analysis, , Review Questions, 11.1, , The average power absorbed by an inductor is zero., (a) True, , 11.2, , (b) False, , The Thevenin impedance of a network seen from the, load terminals is 80 j55 . For maximum power, transfer, the load impedance must be:, (a) 80 j55 , (c) 80 j55 , , 11.3, , 11.4, , 11.5, , 11.6, , If the load impedance is 20 j20, the power factor is, (a) l45, (b) 0, (c) 1, (d) 0.7071, (e) none of these, A quantity that contains all the power information in, a given load is the, , 11.7, , (a), , 11.8, , For the power triangle in Fig. 11.34(b), the apparent, power is:, (a) 2000 VA, (c) 866 VAR, , 11.9, , (b) 1000 VAR, (d) 500 VAR, , A source is connected to three loads Z1, Z2, and Z3, in parallel. Which of these is not true?, (a) P P1 P2 P3, (c) S S1 S2 S3, , (b) Q Q1 Q2 Q3, (d) S S1 S2 S3, , 11.10 The instrument for measuring average power is the:, , (b) VA, (d) none of these, , (a) voltmeter, (c) wattmeter, (e) kilowatt-hour meter, , In the power triangle shown in Fig. 11.34(a), the, reactive power is:, (a) 1000 VAR leading, (c) 866 VAR leading, , (b), , Figure 11.34, , (b) apparent power, (d) reactive power, , Reactive power is measured in:, (a) watts, (c) VAR, , 500 W, , For Review Questions 11.7 and 11.8., , (b) 120 V, (d) 210 V, , (a) power factor, (c) average power, (e) complex power, , 1000 VAR, , 60°, , (b) 80 j55 , (d) 80 j55 , , The amplitude of the voltage available in the 60-Hz,, 120-V power outlet in your home is:, (a) 110 V, (c) 170 V, , 30°, , (b) 1000 VAR lagging, (d) 866 VAR lagging, , (b) ammeter, (d) varmeter, , Answers: 11.1a, 11.2c, 11.3c, 11.4d, 11.5e, 11.6c, 11.7d,, 11.8a, 11.9c, 11.10c., , Problems1, Section 11.2 Instantaneous and Average Power, 11.1, , If v(t) 160 cos 50t V and i(t) , 20 sin(50t 30) A, calculate the instantaneous, power and the average power., , 11.2, , Given the circuit in Fig. 11.35, find the average, power supplied or absorbed by each element., , 11.3, , A load consists of a 60- resistor in parallel with a, 90-mF capacitor. If the load is connected to a voltage, source vs(t) 40 cos 2000t, find the average power, delivered to the load., , 11.4, , Using Fig. 11.36, design a problem to help other, students better understand instantaneous and average, power., , j1 Ω, , j4 Ω, , R1, 6 0° Α, , 5Ω, , Vs +, −, , Figure 11.35, , Figure 11.36, , For Prob. 11.2., , For Prob. 11.4., , 1, , jXL, , Starting with problem 11.22, unless otherwise specified, assume that all values of currents and voltages are rms., , R2, − jXC
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 491, , Problems, , 11.5, , Assuming that vs 16 cos(2t 40) V in the, circuit of Fig. 11.37, find the average power, delivered to each of the passive elements., , 1Ω, , 491, , For the op amp circuit in Fig. 11.41, Vs 2l30 V., Find the average power absorbed by the 20-k, resistor., , 11.9, , 2Ω, , vs +, −, , 3H, , +, –, Vs, , 0.25 F, , +, −, , 10 kΩ, , j6 kΩ, , 2 kΩ, , Figure 11.37, , 20 kΩ, j12 kΩ, , j4 kΩ, , For Prob. 11.5., , Figure 11.41, 11.6, , For the circuit in Fig. 11.38, is 3 cos 10 t A. Find, the average power absorbed by the 50- resistor., 3, , For Prob. 11.9., 11.10 In the op amp circuit in Fig. 11.42, find the total, average power absorbed by the resistors., , 20i x, ix, is, , + −, , R, , 50 Ω, , 20 mH, , 40 F, , 10 Ω, , +, −, , +, −, R, +, −, , Vo cos t V, , R, , Figure 11.38, For Prob. 11.6., , Figure 11.42, For Prob. 11.10., , 11.7, , Given the circuit of Fig. 11.39, find the average, power absorbed by the 10- resistor., , 4Ω, , 8 20° V, , +, −, , 0.1Vo, , 11.11 For the network in Fig. 11.43, assume that the port, impedance is, , − j5 Ω, , Io, , +, −, , 8Io, , j5 Ω, , 10 Ω, , Zab , +, Vo, −, , R, 21 R C, , 2, , ltan1 RC, , Find the average power consumed by the network, when R 10 k, C 200 nF, and i , 2 sin(377t 22) mA., , Figure 11.39, , i, , For Prob. 11.7., , 11.8, , 2 2, , a, Linear, network, , In the circuit of Fig. 11.40, determine the average, power absorbed by the 40- resistor., , +, v, −, , b, , Figure 11.43, Io, , 5 0° A, , Figure 11.40, For Prob. 11.8., , − j20 Ω, , j10 Ω, , For Prob. 11.11., , 0.5Io, , 40 Ω, , Section 11.3 Maximum Average Power Transfer, 11.12 For the circuit shown in Fig. 11.44, determine the, load impedance Z for maximum power transfer, (to Z). Calculate the maximum power absorbed by, the load.
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 492, , Chapter 11, , 492, , AC Power Analysis, , j2 Ω, j3 Ω, , 4Ω, , +, −, , 40 0° V, , 11.17 Calculate the value of ZL in the circuit of Fig. 11.48, in order for ZL to receive maximum average power., What is the maximum average power received by ZL?, − j10 Ω, , 5Ω, , ZL, , 30 Ω, ZL, 2 90° A, , Figure 11.44, For Prob. 11.12., , j20 Ω, , 40 Ω, , 11.13 The Thevenin impedance of a source is ZTh , 120 j60 , while the peak Thevenin voltage is, VTh 110 j0 V. Determine the maximum, available average power from the source., , Figure 11.48, For Prob. 11.17., , 11.14 Using Fig. 11.45, design a problem to help other, students better understand maximum average power, transfer., C, , R2, , 11.18 Find the value of ZL in the circuit of Fig. 11.49 for, maximum power transfer., , −, +, , 40 Ω, 40 Ω, , is, , L, , R1, , Figure 11.45, , 5 0° A, , ZL, , Figure 11.49, , For Prob. 11.14., , For Prob. 11.18., , 11.15 In the circuit of Fig. 11.46, find the value of ZL that, will absorb the maximum power and the value of the, maximum power., − j1 Ω, , 1Ω, , 11.19 The variable resistor R in the circuit of Fig. 11.50 is, adjusted until it absorbs the maximum average, power. Find R and the maximum average power, absorbed., − j2 Ω, , 3Ω, , +, 120 0° V, , 80 Ω, , Z, j20 Ω, , +, −, , − j10 Ω, , 60 0° V, , j1 Ω, , Vo, −, , 2Vo, , ZL, j1 Ω, , 6Ω, , 4 0° A, , R, , Figure 11.46, For Prob. 11.15., , Figure 11.50, For Prob. 11.19., , 11.16 For the circuit of Fig. 11.47, find the maximum, power delivered to the load ZL., , 11.20 The load resistance RL in Fig. 11.51 is adjusted until, it absorbs the maximum average power. Calculate, the value of RL and the maximum average power., , 0.5 v o, , 2Ω, , 160 cos 4t V +, −, , +, vo, –, , 1, 20, , F, , 4I o, , I o 40 Ω, , 4Ω, , 1H, , ZL, , 120 0° V, , +, −, , Figure 11.47, , Figure 11.51, , For Prob. 11.16., , For Prob. 11.20., , j20 Ω, , +−, − j10 Ω, , − j10 Ω, , RL
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 493, , Problems, , 11.21 Assuming that the load impedance is to be purely, resistive, what load should be connected to terminals, a-b of the circuits in Fig. 11.52 so that the maximum, power is transferred to the load?, , 493, , 11.25 Find the rms value of the signal shown in Fig. 11.56., , f (t), 100 Ω, , 10, , − j10 Ω, a, , 0, , 1, , 2, , 3, , 4, , 5, , t, , –10, , 40 Ω, +, −, , 120 60° V, , –1, , 50 Ω, , Figure 11.56, , 2 90° A, , For Prob. 11.25., , j30 Ω, b, , Figure 11.52, , 11.26 Find the effective value of the voltage waveform in, Fig. 11.57., , For Prob. 11.21., , Section 11.4 Effective or RMS Value, v(t), , 11.22 Find the rms value of the offset sine wave shown in, Fig. 11.53., i(t), 4, , 10, 5, , 0, , , 0, , 2, , 3, , t, , 2, , 4, , 6, , 8, , 10, , t, , Figure 11.57, For Prob. 11.26., , Figure 11.53, For Prob. 11.22., 11.23 Using Fig. 11.54, design a problem to help other, students better understand how to find the rms value, of a waveshape., , 11.27 Calculate the rms value of the current waveform of, Fig. 11.58., , i(t), , v (t), Vp, , 5, , 0, 0, , T/3, , 2T/3, , T, , 4T/3, , 5, , 10, , 15, , 20, , 25, , t, , Figure 11.58, , t, , For Prob. 11.27., , Figure 11.54, For Prob. 11.23., 11.24 Determine the rms value of the waveform in, Fig. 11.55., , 11.28 Find the rms value of the voltage waveform of, Fig. 11.59 as well as the average power absorbed by, a 2- resistor when the voltage is applied across the, resistor., , v(t), v(t), 10, 10, 0, 1, −10, , 2, , 3, , 4, , t, 0, , Figure 11.55, , Figure 11.59, , For Prob. 11.24., , For Prob. 11.28., , 2, , 5, , 7, , 10, , 12, , t
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 494, , Chapter 11, , 494, , AC Power Analysis, , 11.29 Calculate the effective value of the current waveform, in Fig. 11.60 and the average power delivered to a, 12- resistor when the current runs through the, resistor., , 11.33 Determine the rms value for the waveform in, Fig. 11.64., i(t), 10, , i(t), 10, , 0, 0, , 5, , 10, , 15, , 20, , 25, , 30, , t, , −10, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , t, , 9 10, , Figure 11.64, For Prob. 11.33., 11.34 Find the effective value of f(t) defined in Fig. 11.65., , Figure 11.60, For Prob. 11.29., f (t), 10, , 11.30 Compute the rms value of the waveform depicted in, Fig. 11.61., , –1, , 0, , 1, , 2, , 3, , 4, , t, , 5, , Figure 11.65, , v (t), , For Prob. 11.34., 2, 0, −1, , 2, , 4, , 6, , 8, , 10, , t, , 11.35 One cycle of a periodic voltage waveform is depicted, in Fig. 11.66. Find the effective value of the voltage., Note that the cycle starts at t 0 and ends at t 6 s., , Figure 11.61, v(t), , For Prob. 11.30., , 30, , 11.31 Find the rms value of the signal shown in Fig. 11.62., 20, , v (t), 2, , 10, 0, , 1, , 2, , 3, , 4, , 5, , t, 0, , –4, , 1, , 2, , 3, , 4, , 5, , 6, , t, , Figure 11.66, For Prob. 11.35., , Figure 11.62, For Prob. 11.31., 11.32 Obtain the rms value of the current waveform shown, in Fig. 11.63., , 11.36 Calculate the rms value for each of the following, functions:, (a) i(t) 10 A, , (b) v(t) 4 3 cos 5t V, , (c) i(t) 8 6 sin 2t A (d) v(t) 5 sin t 4 cos t V, i(t), , 11.37 Design a problem to help other students better, understand how to determine the rms value of the, sum of multiple currents., , 10t 2, 10, , Section 11.5 Apparent Power and Power Factor, 0, , Figure 11.63, For Prob. 11.32., , 1, , 2, , 3, , 4, , 5, , t, , 11.38 For the power system in Fig. 11.67, find: (a) the, average power, (b) the reactive power, (c) the power, factor. Note that 220 V is an rms value.
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 495, , Problems, , 495, , 11.43 Design a problem to help other students understand, complex power., , +, 220 V, 60 Hz, −, , 124 0° Ω, , 11.44 Find the complex power delivered by vs to the, network in Fig. 11.69. Let vs 160 cos 2000t V., , 20 − j25 Ω, 90 + j80 Ω, , 30 Ω, , Figure 11.67, For Prob. 11.38., , 40 F, , 20 Ω, ix, , vs, , 11.39 An ac motor with impedance ZL 4.2 j3.6 is, supplied by a 220-V, 60-Hz source. (a) Find pf, P,, and Q. (b) Determine the capacitor required to be, connected in parallel with the motor so that the, power factor is corrected to unity., 11.40 Design a problem to help other students better, understand apparent power and power factor., , +, −, , 60 mH, , +, −, , 4ix, , Figure 11.69, For Prob. 11.44., , 11.45 The voltage across a load and the current through it, are given by, , 11.41 Obtain the power factor for each of the circuits in, Fig. 11.68. Specify each power factor as leading or, lagging., , v(t) 20 60 cos 100t V, i(t) 1 0.5 sin 100t A, Find:, (a) the rms values of the voltage and of the current, , 4Ω, , (b) the average power dissipated in the load, , j5 Ω, − j2 Ω, , − j2 Ω, , (a) V 220l30 V rms, I 0.5l60 A rms, , (a), − j1 Ω, , 11.46 For the following voltage and current phasors,, calculate the complex power, apparent power, real, power, and reactive power. Specify whether the pf is, leading or lagging., (b) V 250l10 V rms,, I 6.2l25 A rms, , 4Ω, , (c) V 120l0 V rms, I 2.4l15 A rms, 1Ω, , j2 Ω, , j1 Ω, , (b), , Figure 11.68, For Prob. 11.41., , (d) V 160l45 V rms, I 8.5l90 A rms, 11.47 For each of the following cases, find the complex, power, the average power, and the reactive power:, (a) v(t) 112 cos(t 10) V,, i(t) 4 cos(t 50) A, (b) v(t) 160 cos 377t V,, i(t) 4 cos(377t 45) A, , Section 11.6 Complex Power, 11.42 A 110-V rms, 60-Hz source is applied to a load, impedance Z. The apparent power entering the load, is 120 VA at a power factor of 0.707 lagging., , (c) V 80l60 V rms, Z 50l30 , (d) I 10l60 A rms, Z 100l45 , 11.48 Determine the complex power for the following, cases:, , (a) Calculate the complex power., , (a) P 269 W, Q 150 VAR (capacitive), , (b) Find the rms current supplied to the load., , (b) Q 2000 VAR, pf 0.9 (leading), , (c) Determine Z., , (c) S 600 VA, Q 450 VAR (inductive), , (d) Assuming that Z R jL, find the values of, R and L., , (d) Vrms 220 V, P 1 kW,, 0Z 0 40 (inductive)
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 496, , Chapter 11, , 496, , AC Power Analysis, , 11.49 Find the complex power for the following cases:, , I, , (a) P 4 kW, pf 0.86 (lagging), , A, , +, , (b) S 2 kVA, P 1.6 kW (capacitive), (c) Vrms 208l20 V, Irms 6.5l50 A, , 120 30° V, , C, , –, , (d) Vrms 120l30 V, Z 40 j60 , 11.50 Obtain the overall impedance for the following cases:, , B, , Figure 11.72, For Prob. 11.53., , (a) P 1000 W, pf 0.8 (leading),, Vrms 220 V, (b) P 1500 W, Q 2000 VAR (inductive),, Irms 12 A, (c) S 4500l60 VA, V 120l45 V, , Section 11.7 Conservation of AC Power, 11.54 For the network in Fig. 11.73, find the complex, power absorbed by each element., , 11.51 For the entire circuit in Fig. 11.70, calculate:, (a) the power factor, (b) the average power delivered by the source, , − j3 Ω, +, −, , 120 −20° V, , j5 Ω, , 4Ω, , (c) the reactive power, (d) the apparent power, , Figure 11.73, , (e) the complex power, , For Prob. 11.54., 11.55 Using Fig. 11.74, design a problem to help other, students better understand the conservation of AC, power., , 2Ω, − j5 Ω, 120 45° V, , +, −, , − jXC, , j6 Ω, , 10 Ω, , 8Ω, , Figure 11.70, , jXL, , V1 +, −, , + V, − 2, , R, , Figure 11.74, , For Prob. 11.51., , For Prob. 11.55., , 11.52 In the circuit of Fig. 11.71, device A receives 2 kW, at 0.8 pf lagging, device B receives 3 kVA at 0.4 pf, leading, while device C is inductive and consumes, 1 kW and receives 500 VAR., (a) Determine the power factor of the entire system., (b) Find I given that Vs 240l45 V rms., I, , A, , 11.56 Obtain the complex power delivered by the source in, the circuit of Fig. 11.75., 3Ω, , j4 Ω, , 5Ω, , 10 30° Α, , − j2 Ω, , 6Ω, , Figure 11.75, , +, , For Prob. 11.56., Vs, , B, , C, , –, , Figure 11.71, , 11.57 For the circuit in Fig. 11.76, find the average, reactive,, and complex power delivered by the dependent, current source., , For Prob. 11.52., 11.53 In the circuit of Fig. 11.72, load A receives 4 kVA, at 0.8 pf leading. Load B receives 2.4 kVA at 0.6 pf, lagging. Box C is an inductive load that consumes, 1 kW and receives 500 VAR., , 4Ω, , 24 0° V, , +, −, , (a) Determine I., , Figure 11.76, , (b) Calculate the power factor of the combination., , For Prob. 11.57., , 1Ω, , − j1 Ω, +, Vo, −, , 2Ω, , j2 Ω, , 2Vo
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 497, , Problems, , 497, , 11.58 Obtain the complex power delivered to the 10-k, resistor in Fig. 11.77 below., , − j3 kΩ, , 500 Ω Io, , 0.6 0° V rms, , +, −, , 20Io, , j1 kΩ, , 4 kΩ, , 10 kΩ, , Figure 11.77, For Prob. 11.58., , 11.59 Calculate the reactive power in the inductor and, capacitor in the circuit of Fig. 11.78., , 11.61 Given the circuit in Fig. 11.80, find Io and the overall, complex power supplied., , Io, j30 Ω, , 50 Ω, , 200 90° V, 240 0° V, , +, −, , 1.2 kW, 0.8 kVAR (cap), , − j20 Ω, , 4 0° A, , +, −, , 4 kW, 0.9 pf lagging, , 2 kVA, 0.707 pf leading, , 40 Ω, , Figure 11.80, For Prob. 11.61., , Figure 11.78, For Prob. 11.59., , 11.62 For the circuit in Fig. 11.81, find Vs., 11.60 For the circuit in Fig. 11.79, find Vo and the input, power factor., , 0.2 Ω, , j0.04 Ω, , 0.3 Ω, , j0.15 Ω, +, , Vs, +, 12 0° A rms, , 20 kW, 0.8 pf lagging, , 16 kW, 0.9 pf lagging, , Vo, −, , +, −, , 10 W, 0.9 pf lagging, , 15 W, 0.8 pf leading, , −, , Figure 11.81, For Prob. 11.62., , Figure 11.79, For Prob. 11.60., 11.63 Find Io in the circuit of Fig. 11.82., , Io, , 110 0° V, , +, −, , Figure 11.82, For Prob. 11.63., , 12 kW, 0.866 pf leading, , 120 V rms, , 16 kW, 0.85 pf lagging, , 20 kVAR, 0.6 pf lagging
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 498, , Chapter 11, , 498, , AC Power Analysis, , 11.64 Determine Is in the circuit of Fig. 11.83, if the, voltage source supplies 2.5 kW and 0.4 kVAR, (leading)., , 11.68 Compute the complex power supplied by the current, source in the series RLC circuit in Fig. 11.87., R, , 8Ω, +, −, , Is, , L, , Io cos t, , C, , 120 0° V, , Figure 11.87, , j12 Ω, , For Prob. 11.68., , Figure 11.83, Section 11.8 Power Factor Correction, , For Prob. 11.64., 11.65 In the op amp circuit of Fig. 11.84, vs 4 cos 104t V., Find the average power delivered to the 50-k, resistor., 100 kΩ, , vs, , +, −, , 11.69 Refer to the circuit shown in Fig. 11.88., (a) What is the power factor?, (b) What is the average power dissipated?, (c) What is the value of the capacitance that will give, a unity power factor when connected to the load?, , +, −, , 50 kΩ, , 1 nF, , 120 V rms, 60 Hz, , +, −, , C, , Z = 10 + j12 Ω, , Figure 11.84, Figure 11.88, , For Prob. 11.65., , For Prob. 11.69., 11.66 Obtain the average power absorbed by the 6-k, resistor in the op amp circuit in Fig. 11.85., 2 kΩ, 4 kΩ, , 11.70 Design a problem to help other students better, understand power factor correction., , j4 kΩ, , j3 kΩ, −, +, , 2 45° V +, −, , 6 kΩ, − j2 kΩ, , Figure 11.85, For Prob. 11.66., 11.67 For the op amp circuit in Fig. 11.86, calculate:, (a) the complex power delivered by the voltage source, (b) the average power dissipated in the 12- resistor, 10 Ω, , 0.6 sin(2t + 20°) V +, −, , Figure 11.86, For Prob. 11.67., , 1.5 H, , 11.72 Two loads connected in parallel draw a total of, 2.4 kW at 0.8 pf lagging from a 120-V rms, 60-Hz, line. One load absorbs 1.5 kW at a 0.707 pf lagging., Determine: (a) the pf of the second load, (b) the, parallel element required to correct the pf to, 0.9 lagging for the two loads., 11.73 A 240-V rms 60-Hz supply serves a load that is, 10 kW (resistive), 15 kVAR (capacitive), and 22 kVAR, (inductive). Find:, , 0.1 F, 4Ω, , 11.71 Three loads are connected in parallel to a, 120l0 V rms source. Load 1 absorbs 60 kVAR, at pf 0.85 lagging, load 2 absorbs 90 kW and, 50 kVAR leading, and load 3 absorbs 100 kW at, pf 1. (a) Find the equivalent impedance., (b) Calculate the power factor of the parallel, combination. (c) Determine the current supplied, by the source., , −, +, , (a) the apparent power, 12 Ω, , (b) the current drawn from the supply, (c) the kVAR rating and capacitance required to, improve the power factor to 0.96 lagging, (d) the current drawn from the supply under the new, power-factor conditions
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 499, , Problems, , 11.74 A 120-V rms 60-Hz source supplies two loads, connected in parallel, as shown in Fig. 11.89., , 499, , 11.77 What is the reading of the wattmeter in the network, of Fig. 11.92?, , (a) Find the power factor of the parallel combination., 6Ω, , (b) Calculate the value of the capacitance connected, in parallel that will raise the power factor to unity., , 4H, , ±, ±, , 120 cos 2t V +, −, , Load 1, 24 kW, pf = 0.8, lagging, , Load 2, 40 kW, pf = 0.95, lagging, , 0.1 F, , 15 Ω, , Figure 11.92, For Prob. 11.77., 11.78 Find the wattmeter reading of the circuit shown in, Fig. 11.93., , Figure 11.89, For Prob. 11.74., , 10 Ω, , 11.75 Consider the power system shown in Fig. 11.90., Calculate:, , 5Ω, , ±, , 1H, , ±, 20 cos 4t V +, −, , 1, 12, , 4Ω, , F, , (a) the total complex power, , Figure 11.93, , (b) the power factor, , For Prob. 11.78., 11.79 Determine the wattmeter reading of the circuit in, Fig. 11.94., , +, 240 V rms, 50 Hz, −, , 20 Ω i, 80 − j50 Ω, , 40 Ω, , 120 + j70 Ω, , 10 mH, , ±, ±, , 60 + j0, , 10 cos100t, , Figure 11.90, , +, −, , 2 i, , For Prob. 11.75., , Figure 11.94, For Prob. 11.79., , Section 11.9 Applications, , 11.80 The circuit of Fig. 11.95 portrays a wattmeter, connected into an ac network., , 11.76 Obtain the wattmeter reading of the circuit in, Fig. 11.91., , (a) Find the load current., (b) Calculate the wattmeter reading., , 4 Ω − j3 Ω, , ±, , WM, , ±, 12 0° V +, −, , j2 Ω, , 8Ω, , 3 30° A, , 110 V, , Figure 11.91, , Figure 11.95, , For Prob. 11.76., , For Prob. 11.80., , +, −, , Z L = 6.4 Ω, pf = 0.825, , 500 F
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 500, , Chapter 11, , 500, , AC Power Analysis, , (b) Calculate the charge per kWh with a flat-rate, tariff if the revenue to the utility company is to, remain the same as for the two-part tariff., , 11.81 Design a problem to help other students better understand, how to correct power factor to values other than unity., 11.82 A 240-V rms 60-Hz source supplies a parallel combination of a 5-kW heater and a 30-kVA induction, motor whose power factor is 0.82. Determine:, (a) the system apparent power, (b) the system reactive power, , 11.85 A regular household system of a single-phase threewire circuit allows the operation of both 120-V and, 240-V, 60-Hz appliances. The household circuit is, modeled as shown in Fig. 11.96. Calculate:, (a) the currents I1, I2, and In, , (c) the kVA rating of a capacitor required to adjust, the system power factor to 0.9 lagging, , (b) the total complex power supplied, (c) the overall power factor of the circuit, , (d) the value of the capacitor required, 11.83 Oscilloscope measurements indicate that the peak, voltage across a load and the peak current through, it are, respectively, 210l60 V and 8l25 A., Determine:, (a) the real power, , I1, , 120, , 0° V, , +, −, , 10 Ω, , Lamp, , In, , (b) the apparent power, , 30 Ω, , (c) the reactive power, 10 Ω, , (d) the power factor, 11.84 A consumer has an annual consumption of 1200, MWh with a maximum demand of 2.4 MVA. The, maximum demand charge is $30 per kVA per annum,, and the energy charge per kWh is 4 cents., (a) Determine the annual cost of energy., , 120, , 0° V, , Kitchen ramp, , Refrigerator, , +, −, I2, , 15 mH, , Figure 11.96, For Prob. 11.85., , Comprehensive Problems, 11.86 A transmitter delivers maximum power to an antenna, when the antenna is adjusted to represent a load of, 75- resistance in series with an inductance of, 4 mH. If the transmitter operates at 4.12 MHz, find, its internal impedance., 11.87 In a TV transmitter, a series circuit has an impedance, of 3 k and a total current of 50 mA. If the voltage, across the resistor is 80 V, what is the power factor, of the circuit?, 11.88 A certain electronic circuit is connected to a 110-V, ac line. The root-mean-square value of the current, drawn is 2 A, with a phase angle of 55., (a) Find the true power drawn by the circuit., (b) Calculate the apparent power., 11.89 An industrial heater has a nameplate that reads:, 210 V 60 Hz 12 kVA 0.78 pf lagging, Determine:, , of capacitors is required to operate the turbinegenerator but keep it from being overloaded?, 11.91 The nameplate of an electric motor has the following, information:, Line voltage: 220 V rms, Line current: 15 A rms, Line frequency: 60 Hz, Power: 2700 W, Determine the power factor (lagging) of the motor., Find the value of the capacitance C that must be, connected across the motor to raise the pf to unity., 11.92 As shown in Fig. 11.97, a 550-V feeder line supplies, an industrial plant consisting of a motor drawing, 60 kW at 0.75 pf (inductive), a capacitor with a, rating of 20 kVAR, and lighting drawing 20 kW., (a) Calculate the total reactive power and apparent, power absorbed by the plant., , (a) the apparent and the complex power, , (b) Determine the overall pf., , (b) the impedance of the heater, , (c) Find the current in the feeder line., , *11.90 A 2000-kW turbine-generator of 0.85 power factor, operates at the rated load. An additional load of, 300 kW at 0.8 power factor is added.What kVAR, , * An asterisk indicates a challenging problem.
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ale29559_ch11.qxd, , 07/08/2008, , 12:02 PM, , Page 501, , Comprehensive Problems, , 501, , Amplifier, 550 V, , +, −, , 60 kW, pf = 0.75, , Coupling capacitor, 20 kVAR, , 20 kW, , Speaker, , Vin, , Figure 11.97, (a), , For Prob. 11.92., 10 Ω, , 11.93 A factory has the following four major loads:, • A motor rated at 5 hp, 0.8 pf lagging, (1 hp 0.7457 kW)., • A heater rated at 1.2 kW, 1.0 pf., • Ten 120-W lightbulbs., • A synchronous motor rated at 1.6 kVAR, 0.6 pf, leading., (a) Calculate the total real and reactive power., (b) Find the overall power factor., , 11.94 A 1-MVA substation operates at full load at 0.7 power, factor. It is desired to improve the power factor to 0.95, by installing capacitors. Assume that new substation, and distribution facilities cost $120 per kVA installed,, and capacitors cost $30 per kVA installed., (a) Calculate the cost of capacitors needed., (b) Find the savings in substation capacity released., (c) Are capacitors economical for releasing the, amount of substation capacity?, , 40 nF, , 4Ω, vs, 80 mH, , Amplifier, , Speaker, (b), , Figure 11.98, For Prob. 11.95., 11.96 A power amplifier has an output impedance of, 40 j8 . It produces a no-load output voltage of, 146 V at 300 Hz., (a) Determine the impedance of the load that, achieves maximum power transfer., (b) Calculate the load power under this matching, condition., 11.97 A power transmission system is modeled as shown in, Fig. 11.99. If Vs 240l0 rms, find the average, power absorbed by the load., j1 Ω, , 0.1 Ω, , 11.95 A coupling capacitor is used to block dc current from, an amplifier as shown in Fig. 11.98(a). The amplifier, and the capacitor act as the source, while the speaker, is the load as in Fig. 11.98(b)., (a) At what frequency is maximum power, transferred to the speaker?, (b) If Vs 4.6 V rms, how much power is delivered, to the speaker at that frequency?, , 100 Ω, Vs, , +, −, , Source, , Figure 11.99, For Prob. 11.97., , j20 Ω, , j1 Ω, , 0.1 Ω, Line, , Load
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 503, , c h a p t e r, , Three-Phase Circuits, , 12, , He who cannot forgive others breaks the bridge over which he must, pass himself., —G. Herbert, , Enhancing Your Skills and Your Career, ABET EC 2000 criteria (3.e), “an ability to identify, formulate,, and solve engineering problems.”, Developing and enhancing your “ability to identify, formulate, and, solve engineering problems” is a primary focus of textbook. Following our six step problem-solving process is the best way to practice, this skill. Our recommendation is that you use this process whenever, possible. You may be pleased to learn that this process works well for, nonengineering courses., , ABET EC 2000 criteria (f), “an understanding of professional, and ethical responsibility.”, “An understanding of professional and ethical responsibility” is required, of every engineer. To some extent, this understanding is very personal, for each of us. Let us identify some pointers to help you develop this, understanding. One of my favorite examples is that an engineer has the, responsibility to answer what I call the “unasked question.” For, instance, assume that you own a car that has a problem with the transmission. In the process of selling that car, the prospective buyer asks, you if there is a problem in the right-front wheel bearing. You answer, no. However, as an engineer, you are required to inform the buyer that, there is a problem with the transmission without being asked., Your responsibility both professionally and ethically is to perform, in a manner that does not harm those around you and to whom you are, responsible. Clearly, developing this capability will take time and maturity on your part. I recommend practicing this by looking for professional and ethical components in your day-to-day activities., , Photo by Charles Alexander, , 503
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 504, , Chapter 12, , 504, , Three-Phase Circuits, , 12.1, , Historical note: Thomas Edison invented, a three-wire system, using three wires, instead of four., , Introduction, , So far in this text, we have dealt with single-phase circuits. A single-phase, ac power system consists of a generator connected through a pair of wires, (a transmission line) to a load. Figure 12.1(a) depicts a single-phase twowire system, where Vp is the rms magnitude of the source voltage and f, is the phase. What is more common in practice is a single-phase threewire system, shown in Fig. 12.1(b). It contains two identical sources, (equal magnitude and the same phase) that are connected to two loads by, two outer wires and the neutral. For example, the normal household system is a single-phase three-wire system because the terminal voltages, have the same magnitude and the same phase. Such a system allows the, connection of both 120-V and 240-V appliances., , Vp +, −, Vp +, −, , ZL, , Vp +, −, , a, , A, , n, , N, , b, , B, , ZL1, , ZL2, , (b), , (a), , Figure 12.1, Single-phase systems: (a) two-wire type, (b) three-wire type., , +, −, , Vp 0°, , Vp −90° +, −, , a, , A, , n, , N, , b, , B, , Figure 12.2, Two-phase three-wire system., , Vp, , 0°, , −+, Vp −120°, −+, Vp +120°, −+, , a, , A, , ZL1, , b, , B, , ZL2, , c, , C, , ZL3, , n, , N, , Figure 12.3, Three-phase four-wire system., , ZL1, , ZL2, , Circuits or systems in which the ac sources operate at the same frequency but different phases are known as polyphase. Figure 12.2 shows, a two-phase three-wire system, and Fig. 12.3 shows a three-phase fourwire system. As distinct from a single-phase system, a two-phase system, is produced by a generator consisting of two coils placed perpendicular, to each other so that the voltage generated by one lags the other by 90., By the same token, a three-phase system is produced by a generator consisting of three sources having the same amplitude and frequency but out, of phase with each other by 120. Since the three-phase system is by far, the most prevalent and most economical polyphase system, discussion in, this chapter is mainly on three-phase systems., Three-phase systems are important for at least three reasons. First,, nearly all electric power is generated and distributed in three-phase, at, the operating frequency of 60 Hz (or 377 rad/s) in the United, States or 50 Hz (or 314 rad/s) in some other parts of the world., When one-phase or two-phase inputs are required, they are taken from, the three-phase system rather than generated independently. Even when, more than three phases are needed—such as in the aluminum industry,, where 48 phases are required for melting purposes—they can be provided, by manipulating the three phases supplied. Second, the instantaneous, power in a three-phase system can be constant (not pulsating), as we, will see in Section 12.7. This results in uniform power transmission, and less vibration of three-phase machines. Third, for the same amount, of power, the three-phase system is more economical than the singlephase. The amount of wire required for a three-phase system is less, than that required for an equivalent single-phase system.
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 505, , 12.2, , Balanced Three-Phase Voltages, , 505, , Historical, Nikola Tesla (1856–1943) was a Croatian-American engineer whose, inventions—among them the induction motor and the first polyphase ac, power system—greatly influenced the settlement of the ac versus dc debate in favor of ac. He was also responsible for the adoption of 60 Hz as, the standard for ac power systems in the United States., Born in Austria-Hungary (now Croatia), to a clergyman, Tesla had, an incredible memory and a keen affinity for mathematics. He moved, to the United States in 1884 and first worked for Thomas Edison. At, that time, the country was in the “battle of the currents” with George, Westinghouse (1846–1914) promoting ac and Thomas Edison rigidly, leading the dc forces. Tesla left Edison and joined Westinghouse, because of his interest in ac. Through Westinghouse, Tesla gained the, reputation and acceptance of his polyphase ac generation, transmission,, and distribution system. He held 700 patents in his lifetime. His other, inventions include high-voltage apparatus (the tesla coil) and a wireless transmission system. The unit of magnetic flux density, the tesla,, was named in honor of him., , We begin with a discussion of balanced three-phase voltages. Then, we analyze each of the four possible configurations of balanced threephase systems. We also discuss the analysis of unbalanced three-phase, systems. We learn how to use PSpice for Windows to analyze a balanced or unbalanced three-phase system. Finally, we apply the concepts, developed in this chapter to three-phase power measurement and residential electrical wiring., , 12.2, , Balanced Three-Phase Voltages, , Three-phase voltages are often produced with a three-phase ac generator (or alternator) whose cross-sectional view is shown in Fig. 12.4., The generator basically consists of a rotating magnet (called the rotor), surrounded by a stationary winding (called the stator). Three separate, a, Threephase b, output, , c, b′, , N, , a′, , Stator, , c, Rotor, a, , S, c′, , n, , Figure 12.4, A three-phase generator., , b, , Courtesy Smithsonian, Institution
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 506, , Chapter 12, , 506, Van, , 0, , 120°, , Vbn, , Vcn, , windings or coils with terminals a-a¿, b-b¿, and c-c¿ are physically, placed 120 apart around the stator. Terminals a and a¿, for example,, stand for one of the ends of coils going into and the other end coming, out of the page. As the rotor rotates, its magnetic field “cuts” the flux, from the three coils and induces voltages in the coils. Because the coils, are placed 120 apart, the induced voltages in the coils are equal in, magnitude but out of phase by 120 (Fig. 12.5). Since each coil can be, regarded as a single-phase generator by itself, the three-phase generator can supply power to both single-phase and three-phase loads., A typical three-phase system consists of three voltage sources connected to loads by three or four wires (or transmission lines). (Threephase current sources are very scarce.) A three-phase system is, equivalent to three single-phase circuits. The voltage sources can be, either wye-connected as shown in Fig. 12.6(a) or delta-connected as in, Fig. 12.6(b)., , t, , 240°, , Three-Phase Circuits, , Figure 12.5, The generated voltages are 120 apart, from each other., , a, n, +, −, , +, −, , Vcn, , a, , Van, Vbn, b, , Vca, , +, −, , +, −, , + Vab, −, −+, , b, , Vbc, c, (a), , c, (b), , Figure 12.6, Three-phase voltage sources: (a) Y-connected source, (b) ¢-connected, source., Vcn, , , 120°, , 120°, Van, , −120°, , Let us consider the wye-connected voltages in Fig. 12.6(a) for, now. The voltages Van, Vbn, and Vcn are respectively between lines a, b,, and c, and the neutral line n. These voltages are called phase voltages., If the voltage sources have the same amplitude and frequency and, are out of phase with each other by 120, the voltages are said to be, balanced. This implies that, Van Vbn Vcn 0, 0Van 0 0Vbn 0 0Vcn 0, , Vbn, (a), , (12.1), (12.2), , Thus,, Vbn, , , , Balanced phase voltages are equal in magnitude and are out of phase, with each other by 120., , 120°, 120°, −120°, , Van, , Since the three-phase voltages are 120 out of phase with each, other, there are two possible combinations. One possibility is shown in, Fig. 12.7(a) and expressed mathematically as, Van Vpl0, , Vcn, (b), , Figure 12.7, Phase sequences: (a) abc or positive, sequence, (b) acb or negative sequence., , Vbn Vpl120, Vcn Vpl240 Vpl120, , (12.3)
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 507, , 12.2, , Balanced Three-Phase Voltages, , where Vp is the effective or rms value of the phase voltages. This is, known as the abc sequence or positive sequence. In this phase sequence,, Van leads Vbn, which in turn leads Vcn. This sequence is produced when, the rotor in Fig. 12.4 rotates counterclockwise. The other possibility is, shown in Fig. 12.7(b) and is given by, , Van Vpl0, Vcn Vpl120, Vbn Vpl240 Vpl120, , 507, , As a common tradition in power, systems, voltage and current in this, chapter are in rms values unless, otherwise stated., , (12.4), , This is called the acb sequence or negative sequence. For this phase, sequence, Van leads Vcn, which in turn leads Vbn. The acb sequence is, produced when the rotor in Fig. 12.4 rotates in the clockwise direction., It is easy to show that the voltages in Eqs. (12.3) or (12.4) satisfy, Eqs. (12.1) and (12.2). For example, from Eq. (12.3),, Van Vbn Vcn Vpl0 Vpl120 Vpl120, Vp(1.0 0.5 j0.866 0.5 j0.866) (12.5), 0, The phase sequence is the time order in which the voltages pass, through their respective maximum values., , The phase sequence is determined by the order in which the phasors, pass through a fixed point in the phase diagram., In Fig. 12.7(a), as the phasors rotate in the counterclockwise, direction with frequency , they pass through the horizontal axis in a, sequence abcabca . . . . Thus, the sequence is abc or bca or cab. Similarly, for the phasors in Fig. 12.7(b), as they rotate in the counterclockwise direction, they pass the horizontal axis in a sequence, acbacba . . . . This describes the acb sequence. The phase sequence is, important in three-phase power distribution. It determines the direction of the rotation of a motor connected to the power source, for, example., Like the generator connections, a three-phase load can be either, wye-connected or delta-connected, depending on the end application., Figure 12.8(a) shows a wye-connected load, and Fig. 12.8(b) shows a, delta-connected load. The neutral line in Fig. 12.8(a) may or may not, be there, depending on whether the system is four- or three-wire. (And,, of course, a neutral connection is topologically impossible for a delta, connection.) A wye- or delta-connected load is said to be unbalanced, if the phase impedances are not equal in magnitude or phase., , The phase sequence may also be regarded as the order in which the phase, voltages reach their peak (or maximum), values with respect to time., , Reminder: As time increases, each, phasor (or sinor) rotates at an angular, velocity ., , a, b, Z2, Z1, , n, Z3, c, (a), a, , Zb, , Zc, b, Za, , A balanced load is one in which the phase impedances are equal in, magnitude and in phase., , c, (b), , Figure 12.8, For a balanced wye-connected load,, Z1 Z2 Z3 ZY, , (12.6), , Two possible three-phase load configurations: (a) a Y-connected load, (b) a, ¢ -connected load.
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 508, , 508, , Reminder: A Y-connected load consists, of three impedances connected to a, neutral node, while a ¢ -connected, load consists of three impedances, connected around a loop. The load is, balanced when the three impedances, are equal in either case., , Chapter 12, , Three-Phase Circuits, , where ZY is the load impedance per phase. For a balanced deltaconnected load,, Za Zb Zc Z¢, , (12.7), , where Z¢ is the load impedance per phase in this case. We recall from, Eq. (9.69) that, Z¢ 3ZY, , or, , 1, ZY Z¢, 3, , (12.8), , so we know that a wye-connected load can be transformed into a deltaconnected load, or vice versa, using Eq. (12.8)., Since both the three-phase source and the three-phase load can be, either wye- or delta-connected, we have four possible connections:, • Y-Y connection (i.e., Y-connected source with a Y-connected, load)., • Y-¢ connection., • ¢-¢ connection., • ¢ -Y connection., In subsequent sections, we will consider each of these possible configurations., It is appropriate to mention here that a balanced delta-connected, load is more common than a balanced wye-connected load. This is due, to the ease with which loads may be added or removed from each phase, of a delta-connected load. This is very difficult with a wye-connected, load because the neutral may not be accessible. On the other hand,, delta-connected sources are not common in practice because of the circulating current that will result in the delta-mesh if the three-phase voltages are slightly unbalanced., , Example 12.1, , Determine the phase sequence of the set of voltages, vbn, , van 200 cos(t 10), 200 cos(t 230),, vcn 200 cos(t 110), , Solution:, The voltages can be expressed in phasor form as, Van 200l10 V,, , Vbn 200l230 V,, , Vcn 200l110 V, , We notice that Van leads Vcn by 120 and Vcn in turn leads Vbn by 120., Hence, we have an acb sequence., , Practice Problem 12.1, , Given that Vbn 110l30 V, find Van and Vcn, assuming a positive, (abc) sequence., Answer: 110l150 V, 110l90 V.
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 509, , 12.3, , 12.3, , Balanced Wye-Wye Connection, , 509, , Balanced Wye-Wye Connection, , We begin with the Y-Y system, because any balanced three-phase system can be reduced to an equivalent Y-Y system. Therefore, analysis, of this system should be regarded as the key to solving all balanced, three-phase systems., A balanced Y-Y system is a three-phase system with a balanced, Y-connected source and a balanced Y-connected load., , Consider the balanced four-wire Y-Y system of Fig. 12.9, where a, Y-connected load is connected to a Y-connected source. We assume a, balanced load so that load impedances are equal. Although the impedance ZY is the total load impedance per phase, it may also be regarded, as the sum of the source impedance Zs, line impedance Z/, and load, impedance ZL for each phase, since these impedances are in series. As, illustrated in Fig. 12.9, Zs denotes the internal impedance of the phase, winding of the generator; Z/ is the impedance of the line joining a, phase of the source with a phase of the load; ZL is the impedance of, each phase of the load; and Zn is the impedance of the neutral line., Thus, in general, ZY Zs Z/ ZL, Zl, , a, , (12.9), , A, , Zs, ZL, +, −, , Van, , Zn, , n, , N, −, +, , Vcn +−, , Vbn, , ZL, , ZL, , Zs, , Zs, , b, , c, , C, , B, Zl, , Zl, , Figure 12.9, A balanced Y-Y system, showing the source, line,, and load impedances., , Ia, a, , Zs and Z/ are often very small compared with ZL, so one can assume, that ZY ZL if no source or line impedance is given. In any event, by, lumping the impedances together, the Y-Y system in Fig. 12.9 can be, simplified to that shown in Fig. 12.10., Assuming the positive sequence, the phase voltages (or line-toneutral voltages) are, Van Vpl0, Vbn Vpl120,, , Vcn Vpl120, , (12.10), , A, , +, −, , Van, , ZY, , In, , n, Vcn, , N, , −, +, , −, +, , ZY, , Vbn, , c, , Ib, Ic, , C, , b, , Figure 12.10, Balanced Y-Y connection., , ZY, B
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 510, , Chapter 12, , 510, , Three-Phase Circuits, , The line-to-line voltages or simply line voltages Vab, Vbc, and Vca are, related to the phase voltages. For example,, Vab Van Vnb Van Vbn Vpl0 Vpl120, (12.11a), 13, 1, Vp a1 j, ) 13Vpl30, 2, 2, Similarly, we can obtain, Vbc Vbn Vcn 13Vpl90, , (12.11b), , Vca Vcn Van 13Vpl210, , (12.11c), , Thus, the magnitude of the line voltages VL is 13 times the magnitude, of the phase voltages Vp, or, , where, and, , Vab = Van + Vnb, , Vnb, Vcn, , VL 13Vp, , (12.12), , Vp 0 Van 0 0Vbn 0 0Vcn 0, , (12.13), , VL 0 Vab 0 0Vbc 0 0Vca 0, , (12.14), , Also the line voltages lead their corresponding phase voltages by 30., Figure 12.11(a) illustrates this. Figure 12.11(a) also shows how to, determine Vab from the phase voltages, while Fig. 12.11(b) shows the, same for the three line voltages. Notice that Vab leads Vbc by 120, and, Vbc leads Vca by 120, so that the line voltages sum up to zero as do, the phase voltages., Applying KVL to each phase in Fig. 12.10, we obtain the line currents as, , 30°, , Ia , , Van, , Van, ,, ZY, , Ib , , Vanl120, Vbn, , Ial120, ZY, ZY, , Vanl240, Vcn, Ic , , Ial240, ZY, ZY, , Vbn, , We can readily infer that the line currents add up to zero,, , (a), Vca, , (12.15), , Vcn, , Vab, , Ia Ib Ic 0, , (12.16), , In (Ia Ib Ic) 0, , (12.17a), , VnN ZnIn 0, , (12.17b), , so that, Van, Vbn, , Vbc, (b), , Figure 12.11, Phasor diagrams illustrating the relationship between line voltages and phase, voltages., , or, that is, the voltage across the neutral wire is zero. The neutral line can, thus be removed without affecting the system. In fact, in long distance, power transmission, conductors in multiples of three are used with the, earth itself acting as the neutral conductor. Power systems designed in, this way are well grounded at all critical points to ensure safety., While the line current is the current in each line, the phase current, is the current in each phase of the source or load. In the Y-Y system, the, line current is the same as the phase current. We will use single subscripts
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 511, , 12.3, , Balanced Wye-Wye Connection, , for line currents because it is natural and conventional to assume that line, currents flow from the source to the load., An alternative way of analyzing a balanced Y-Y system is to do, so on a “per phase” basis. We look at one phase, say phase a, and analyze the single-phase equivalent circuit in Fig. 12.12. The single-phase, analysis yields the line current Ia as, , 511, Ia, , a, , A, , Van +, −, , ZY, n, , N, , Figure 12.12, A single-phase equivalent circuit., , Van, Ia , ZY, , (12.18), , From Ia, we use the phase sequence to obtain other line currents. Thus,, as long as the system is balanced, we need only analyze one phase., We may do this even if the neutral line is absent, as in the three-wire, system., , Calculate the line currents in the three-wire Y-Y system of Fig. 12.13., 5 – j2 Ω, , a, , A, , + 110 0° V, −, 10 + j8 Ω, 110 −240° V +−, c, , − 110 −120° V, +, 5 – j2 Ω, b, , B, , 5 – j2 Ω, , C, , 10 + j8 Ω, 10 + j8 Ω, , Figure 12.13, Three-wire Y-Y system; for Example 12.2., , Solution:, The three-phase circuit in Fig. 12.13 is balanced; we may replace it, with its single-phase equivalent circuit such as in Fig. 12.12. We obtain, Ia from the single-phase analysis as, Ia , , Van, ZY, , where ZY (5 j2) (10 j8) 15 j6 16.155l21.8. Hence,, Ia , , 110l0, 16.155l21.8, , 6.81l21.8 A, , Since the source voltages in Fig. 12.13 are in positive sequence, the, line currents are also in positive sequence:, Ib Ial120 6.81l141.8 A, Ic Ial240 6.81l261.8 A 6.81l98.2 A, , Example 12.2
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , 512, , Practice Problem 12.2, , Page 512, , Chapter 12, , Three-Phase Circuits, , A Y-connected balanced three-phase generator with an impedance of, 0.4 j0.3 per phase is connected to a Y-connected balanced load, with an impedance of 24 j19 per phase. The line joining the generator and the load has an impedance of 0.6 j0.7 per phase., Assuming a positive sequence for the source voltages and that Van , 120l30 V, find: (a) the line voltages, (b) the line currents., Answer: (a) 207.85l60 V, 207.85l60 V, 207.85l180 V,, (b) 3.75l8.66 A, 3.75l128.66 A, 3.75l111.34 A., , 12.4, , Balanced Wye-Delta Connection, , A balanced Y- ¢ system consists of a balanced Y-connected source, feeding a balanced ¢ -connected load., This is perhaps the most practical, three-phase system, as the three-phase, sources are usually Y-connected while, the three-phase loads are usually, ¢ -connected., , The balanced Y-delta system is shown in Fig. 12.14, where the, source is Y-connected and the load is ¢ -connected. There is, of course,, no neutral connection from source to load for this case. Assuming the, positive sequence, the phase voltages are again, Van Vpl0, Vpl120,, Vcn Vpl120, , Vbn, , (12.19), , As shown in Section 12.3, the line voltages are, Vab 13Vpl30 VAB,, Vbc 13Vpl90 VBC, l, Vca 13Vp 150 VCA, , (12.20), , showing that the line voltages are equal to the voltages across the load, impedances for this system configuration. From these voltages, we can, obtain the phase currents as, IAB , , VAB, ,, Z¢, , IBC , , VBC, ,, Z¢, , ICA , , VCA, Z¢, , (12.21), , These currents have the same magnitude but are out of phase with each, other by 120., Ia, , a, Van, , +, −, n, , I AB, , A, Z∆, , Vcn, , −, +, , − Vbn, +, , c, , b, , Z∆, Ib, Ic, , Figure 12.14, Balanced Y- ¢ connection., , ICA, Z∆, , B, , C, I BC
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 513, , 12.4, , Balanced Wye-Delta Connection, , 513, , Another way to get these phase currents is to apply KVL. For, example, applying KVL around loop aABbna gives, Van Z¢IAB Vbn 0, or, IAB , , Van Vbn, Vab, VAB, , , Z¢, Z¢, Z¢, , (12.22), , which is the same as Eq. (12.21). This is the more general way of finding the phase currents., The line currents are obtained from the phase currents by applying KCL at nodes A, B, and C. Thus,, Ia IAB ICA,, , Ib IBC IAB,, , Ic ICA IBC, , (12.23), , Ia IAB ICA IAB(1 1l240), IAB(1 0.5 j0.866) IAB 13l30, , (12.24), , Since ICA IABl240,, , showing that the magnitude IL of the line current is 13 times the magnitude Ip of the phase current, or, IL 13Ip, , (12.25), , IL 0Ia 0 0 Ib 0 0Ic 0, , (12.26), , Ip 0IAB 0 0IBC 0 0ICA 0, , (12.27), , where, Ic, , and, , Also, the line currents lag the corresponding phase currents by 30,, assuming the positive sequence. Figure 12.15 is a phasor diagram illustrating the relationship between the phase and line currents., An alternative way of analyzing the Y-¢ circuit is to transform, the -connected load to an equivalent Y-connected load. Using the ¢-Y, transformation formula in Eq. (12.8),, , I CA, , 30°, I AB, 30°, Ia, , 30°, Ib, , I BC, , Figure 12.15, Phasor diagram illustrating the relationship, between phase and line currents., , ZY , , Z¢, 3, , (12.28), Ia, , After this transformation, we now have a Y-Y system as in Fig. 12.10., The three-phase Y-¢ system in Fig. 12.14 can be replaced by the singlephase equivalent circuit in Fig. 12.16. This allows us to calculate only, the line currents. The phase currents are obtained using Eq. (12.25) and, utilizing the fact that each of the phase currents leads the corresponding line current by 30., , A balanced abc-sequence Y-connected source with Van 100l10 V, is connected to a ¢ -connected balanced load (8 j4) per phase. Calculate the phase and line currents., , Z∆, 3, , Van +, −, , Figure 12.16, A single-phase equivalent circuit of a balanced Y-¢ circuit., , Example 12.3
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ale29559_ch12.qxd, , 07/22/2008, , 01:38 PM, , 514, , Page 514, , Chapter 12, , Three-Phase Circuits, , Solution:, This can be solved in two ways., , ■ METHOD 1 The load impedance is, Z¢ 8 j4 8.944l26.57 , If the phase voltage Van 100l10, then the line voltage is, Vab Van 13l30 10013l10 30 VAB, or, VAB 173.2l40 V, The phase currents are, 173.2l40, VAB, , 19.36l13.43 A, Z¢, 8.944l26.57, IBC IABl120 19.36l106.57 A, ICA IABl120 19.36l133.43 A, , IAB , , The line currents are, Ia IAB 13l30 13(19.36)l13.43 30, 33.53l16.57 A, Ib Ial120 33.53l136.57 A, Ic Ial120 33.53l103.43 A, , ■ METHOD 2 Alternatively, using single-phase analysis,, 100l10, Van, , 33.54l16.57 A, Ia , Z¢3, 2.981l26.57, as above. Other line currents are obtained using the abc phase sequence., , Practice Problem 12.3, , One line voltage of a balanced Y-connected source is VAB , 240l20 V. If the source is connected to a ¢ -connected load of, 20l40 , find the phase and line currents. Assume the abc sequence., Answer: 12l60 A, 12l180 A, 12l60 A, 20.79l90 A,, 20.79l150 A, 20.79l30 A., , 12.5, , Balanced Delta-Delta Connection, , A balanced ¢-¢ system is one in which both the balanced source, and balanced load are ¢ -connected., , The source as well as the load may be delta-connected as shown, in Fig. 12.17. Our goal is to obtain the phase and line currents as usual.
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 515, , 12.5, , Balanced Delta-Delta Connection, , Ia, , a, , A, IAB, , Vca, , −, +, , + Vab, −, , Z∆, , Z∆, , ICA, , Ib, c, , −+, , b, , Vbc, , Ic, , 515, , B, , C, IBC, , Z∆, , Figure 12.17, A balanced ¢-¢ connection., , Assuming a positive sequence, the phase voltages for a delta-connected, source are, Vab Vpl0, Vbc Vpl120,, Vca Vpl120, , (12.29), , The line voltages are the same as the phase voltages. From Fig. 12.17,, assuming there is no line impedances, the phase voltages of the deltaconnected source are equal to the voltages across the impedances; that is,, Vab VAB,, , Vbc VBC,, , Vca VCA, , (12.30), , Hence, the phase currents are, IAB , , VBC, Vab, Vbc, VAB, , ,, IBC , , Z¢, Z¢, Z¢, Z¢, VCA, Vca, ICA , , Z¢, Z¢, , (12.31), , Since the load is delta-connected just as in the previous section, some, of the formulas derived there apply here. The line currents are obtained, from the phase currents by applying KCL at nodes A, B, and C, as we, did in the previous section:, Ia IAB ICA,, , Ib IBC IAB,, , Ic ICA IBC, , (12.32), , Also, as shown in the last section, each line current lags the corresponding phase current by 30; the magnitude IL of the line current is 23 times, the magnitude Ip of the phase current,, IL 13Ip, , (12.33), , An alternative way of analyzing the ¢-¢ circuit is to convert both, the source and the load to their Y equivalents. We already know that, ZY Z¢3. To convert a ¢ -connected source to a Y-connected source,, see the next section., , A balanced ¢ -connected load having an impedance 20 j15 is, connected to a ¢ -connected, positive-sequence generator having, Vab 330l0 V. Calculate the phase currents of the load and the line, currents., , Example 12.4
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 517, , 12.6, , Balanced Delta-Wye Connection, , Ia, , a, , 517, , A, ZY, , Vca, , −, +, , + Vab, −, , N, ZY, , Ib, c, , −+, Vbc, , C, , B, , b, , ZY, , Ic, , Figure 12.18, A balanced ¢-Y connection., , But Ib lags Ia by 120, since we assumed the abc sequence; that is,, Ib Ial120. Hence,, Ia Ib Ia(1 1l120), Ia a1 , , 1, 13, j, b Ia 13l30, 2, 2, , (12.36), , Substituting Eq. (12.36) into Eq. (12.35) gives, Ia , , Vp13l30, , (12.37), , ZY, , From this, we obtain the other line currents Ib and Ic using the positive phase sequence, i.e., Ib Ial120, Ic Ial120. The phase, currents are equal to the line currents., Another way to obtain the line currents is to replace the deltaconnected source with its equivalent wye-connected source, as shown, in Fig. 12.19. In Section 12.3, we found that the line-to-line voltages, of a wye-connected source lead their corresponding phase voltages by, 30. Therefore, we obtain each phase voltage of the equivalent wyeconnected source by dividing the corresponding line voltage of the, delta-connected source by 13 and shifting its phase by 30. Thus,, the equivalent wye-connected source has the phase voltages, Van , Vbn , , Vp, 13, , Vp, 13, , l150,, , Vp, 13, , (12.38), , l90, , Ia , , which is the same as Eq. (12.37)., , ZY, , n, , −, Vbn +, , +−, Vcn, , + V, − ab, , +−, , b, , Vbc, , Figure 12.19, Transforming a ¢ -connected source to an, equivalent Y-connected source., , If the delta-connected source has source impedance Zs per phase, the, equivalent wye-connected source will have a source impedance of Zs3, per phase, according to Eq. (9.69)., Once the source is transformed to wye, the circuit becomes a wyewye system. Therefore, we can use the equivalent single-phase circuit, shown in Fig. 12.20, from which the line current for phase a is, Vp13l30, , Vca, , + V, an, −, , −, +, , c, , l30, Vcn , , a, , (12.39), , Ia, Vp −30°, √3, , +, −, , Figure 12.20, The single-phase equivalent circuit., , ZY
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 519, , 12.7, , Power in a Balanced System, , 519, , obtained by directly applying KCL and KVL to the appropriate threephase circuits., , A balanced Y-connected load with a phase impedance of 40 j25 is, supplied by a balanced, positive sequence ¢ -connected source with a line, voltage of 210 V. Calculate the phase currents. Use Vab as reference., , Example 12.5, , Solution:, The load impedance is, ZY 40 j25 47.17l32 , and the source voltage is, Vab 210l0 V, When the ¢ -connected source is transformed to a Y-connected source,, Van , , Vab, l30 121.2l30 V, 13, , The line currents are, 121.2l30, Van, , 2.57l62 A, ZY, 47.12l32, Ib Ial120 2.57l178 A, , Ia , , Ic Ial120 2.57l58 A, which are the same as the phase currents., , In a balanced ¢-Y circuit, Vab 240l15 and ZY (12 j15) ., Calculate the line currents., Answer: 7.21l66.34 A, 7.21l173.66 A, 7.21l53.66 A., , 12.7, , Power in a Balanced System, , Let us now consider the power in a balanced three-phase system. We, begin by examining the instantaneous power absorbed by the load. This, requires that the analysis be done in the time domain. For a Y-connected, load, the phase voltages are, vAN 12Vp cos t,, vBN 12Vp cos(t 120), vCN 12Vp cos(t 120), , (12.41), , where the factor 12 is necessary because Vp has been defined as the rms, value of the phase voltage. If ZY Zlu, the phase currents lag behind, their corresponding phase voltages by u. Thus,, ia 12Ip cos(t u),, ib 12Ip cos(t u 120) (12.42), ic 12Ip cos(t u 120), , Practice Problem 12.5
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ale29559_ch12.qxd, , 520, , 07/08/2008, , 12:05 PM, , Page 520, , Chapter 12, , Three-Phase Circuits, , where Ip is the rms value of the phase current. The total instantaneous, power in the load is the sum of the instantaneous powers in the three, phases; that is,, p pa pb pc vAN ia vBN ib vCN ic, 2Vp Ip[cos t cos(t u), cos(t 120) cos(t u 120), cos(t 120) cos(t u 120)], , (12.43), , Applying the trigonometric identity, 1, cos A cos B [cos(A B) cos(A B)], 2, , (12.44), , gives, p Vp Ip[3 cos u cos(2t u) cos(2t u 240), cos(2t u 240)], Vp Ip[3 cos u cos a cos a cos 240 sin a sin 240, (12.45), cos a cos 240 sin a sin 240], where a 2t u, 1, Vp Ip c 3 cos u cos a 2a b cos a d 3Vp Ip cos u, 2, Thus the total instantaneous power in a balanced three-phase system is, constant—it does not change with time as the instantaneous power of, each phase does. This result is true whether the load is Y- or ¢ -connected., This is one important reason for using a three-phase system to generate, and distribute power. We will look into another reason a little later., Since the total instantaneous power is independent of time, the, average power per phase Pp for either the ¢ -connected load or the, Y-connected load is p3, or, Pp Vp Ip cos u, , (12.46), , and the reactive power per phase is, Qp Vp Ip sin u, , (12.47), , The apparent power per phase is, Sp Vp Ip, , (12.48), , The complex power per phase is, Sp Pp jQp Vp I*p, , (12.49), , where Vp and Ip are the phase voltage and phase current with magnitudes Vp and Ip, respectively. The total average power is the sum of the, average powers in the phases:, P Pa Pb Pc 3Pp 3Vp Ip cos u 13VL IL cos u, , (12.50), , For a Y-connected load, IL Ip but VL 13Vp, whereas for a, ¢ -connected load, IL 13Ip but VL Vp. Thus, Eq. (12.50) applies, for both Y-connected and ¢ -connected loads. Similarly, the total reactive power is, Q 3Vp Ip sin u 3Qp 13VL IL sin u, , (12.51)
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 521, , 12.7, , Power in a Balanced System, , 521, , and the total complex power is, , S 3Sp 3Vp I*p 3I 2p Zp , , 3Vp2, Z *p, , (12.52), , where Zp Zplu is the load impedance per phase. (Zp could be ZY or, Z¢.) Alternatively, we may write Eq. (12.52) as, S P jQ 13VL ILlu, , (12.53), , Remember that Vp, Ip, VL, and IL are all rms values and that u is the, angle of the load impedance or the angle between the phase voltage, and the phase current., A second major advantage of three-phase systems for power distribution is that the three-phase system uses a lesser amount of wire, than the single-phase system for the same line voltage VL and the same, absorbed power PL. We will compare these cases and assume in both, that the wires are of the same material (e.g., copper with resistivity r),, of the same length /, and that the loads are resistive (i.e., unity power, factor). For the two-wire single-phase system in Fig. 12.21(a),, IL PLVL, so the power loss in the two wires is, Ploss 2I 2L R 2R, , P 2L, , (12.54), , V 2L, , R′, , IL, , R, , +, Threephase, balanced, source, , +, Singlephase, source, , PL, R, , Ia, , Load, , VL, −, , R′, , Ib, , R′, , Ic, , VL 0°, −, +, VL −120°, −, , Transmission lines, , Transmission lines, , (a), , (b), , Figure 12.21, Comparing the power loss in (a) a single-phase system, and (b) a three-phase system., , For the three-wire three-phase system in Fig. 12.21(b), I¿L 0 Ia 0 0Ib 0 , 0Ic 0 PL 13VL from Eq. (12.50). The power loss in the three wires is, P¿loss 3(I¿L)2R¿ 3R¿, , P 2L, , 3V L2, , R¿, , PL2, , V 2L, , (12.55), , Equations (12.54) and (12.55) show that for the same total power delivered PL and same line voltage VL,, Ploss, 2R, , P¿loss, R¿, , (12.56), , Threephase, balanced, load
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ale29559_ch12.qxd, , 07/08/2008, , 522, , 12:05 PM, , Page 522, , Chapter 12, , Three-Phase Circuits, , But from Chapter 2, R r/p r 2 and R¿ r/p r¿ 2, where r and r¿, are the radii of the wires. Thus,, Ploss, 2r¿ 2, 2, P¿loss, r, , (12.57), , If the same power loss is tolerated in both systems, then r 2 2r¿ 2. The, ratio of material required is determined by the number of wires and, their volumes, so, Material for single-phase, 2(pr 2/), 2r 2, , , Material for three-phase, 3(pr¿ 2/), 3r¿ 2, 2, (2) 1.333, 3, , (12.58), , since r 2 2r¿ 2. Equation (12.58) shows that the single-phase system, uses 33 percent more material than the three-phase system or that the, three-phase system uses only 75 percent of the material used in the, equivalent single-phase system. In other words, considerably less material is needed to deliver the same power with a three-phase system than, is required for a single-phase system., , Example 12.6, , Refer to the circuit in Fig. 12.13 (in Example 12.2). Determine the total, average power, reactive power, and complex power at the source and, at the load., Solution:, It is sufficient to consider one phase, as the system is balanced. For, phase a,, Vp 110l0 V, , and, , Ip 6.81l21.8 A, , Thus, at the source, the complex power absorbed is, Ss 3Vp I*p 3(110l0 )(6.81l21.8 ), 2247l21.8 (2087 j834.6) VA, The real or average power absorbed is 2087 W and the reactive, power is 834.6 VAR., At the load, the complex power absorbed is, SL 3 0 Ip 0 2Zp, , where Zp 10 j8 12.81l38.66 and Ip Ia 6.81l21.8. Hence,, SL 3(6.81)212.81l38.66 1782l38.66, (1392 j1113) VA, The real power absorbed is 1391.7 W and the reactive power absorbed, is 1113.3 VAR. The difference between the two complex powers is, absorbed by the line impedance (5 j2) . To show that this is the, case, we find the complex power absorbed by the line as, S/ 3 0 Ip 0 2 Z/ 3(6.81)2(5 j2) 695.6 j278.3 VA, , which is the difference between Ss and SL; that is, Ss S/ SL 0,, as expected.
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 523, , 12.7, , Power in a Balanced System, , For the Y-Y circuit in Practice Prob. 12.2, calculate the complex power, at the source and at the load., , 523, , Practice Problem 12.6, , Answer: (1054 j843.3) VA, (1012 j801.6) VA., , A three-phase motor can be regarded as a balanced Y-load. A threephase motor draws 5.6 kW when the line voltage is 220 V and the line, current is 18.2 A. Determine the power factor of the motor., , Example 12.7, , Solution:, The apparent power is, S 13VL IL 13(220)(18.2) 6935.13 VA, Since the real power is, P S cos u 5600 W, the power factor is, pf cos u , , P, 5600, , 0.8075, S, 6935.13, , Calculate the line current required for a 30-kW three-phase motor having a power factor of 0.85 lagging if it is connected to a balanced, source with a line voltage of 440 V., , Practice Problem 12.7, , Answer: 46.31 A., , Two balanced loads are connected to a 240-kV rms 60-Hz line, as, shown in Fig. 12.22(a). Load 1 draws 30 kW at a power factor of 0.6, lagging, while load 2 draws 45 kVAR at a power factor of 0.8 lagging., Assuming the abc sequence, determine: (a) the complex, real, and reactive powers absorbed by the combined load, (b) the line currents, and, (c) the kVAR rating of the three capacitors ¢ -connected in parallel with, the load that will raise the power factor to 0.9 lagging and the capacitance of each capacitor., Solution:, (a) For load 1, given that P1 30 kW and cos u1 0.6, then sin u1 0.8., Hence,, S1 , , P1, 30 kW, , 50 kVA, cos u1, 0.6, , and Q1 S1 sin u1 50(0.8) 40 kVAR. Thus, the complex power, due to load 1 is, S1 P1 jQ1 30 j40 kVA, , (12.8.1), , Example 12.8
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 524, , Chapter 12, , 524, , Three-Phase Circuits, , For load 2, if Q2 45 kVAR and cos u2 0.8, then sin u2 0.6. We find, Q2, 45 kVA, , 75 kVA, sin u2, 0.6, , S2 , , Balanced, load 1, , and P2 S2 cos u2 75(0.8) 60 kW. Therefore the complex power due, to load 2 is, S2 P2 jQ2 60 j45 kVA, (12.8.2), , Balanced, load 2, , From Eqs. (12.8.1) and (12.8.2), the total complex power absorbed by, the load is, , (a), , S S1 S2 90 j85 kVA 123.8l43.36 kVA, C, C, , C, , (12.8.3), , which has a power factor of cos 43.36 0.727 lagging. The real, power is then 90 kW, while the reactive power is 85 kVAR., (b) Since S 13VL IL, the line current is, IL , , S, 13VL, , (12.8.4), , We apply this to each load, keeping in mind that for both loads, VL , 240 kV. For load 1,, , Combined, load, , IL1 , , (b), , Figure 12.22, For Example 12.8: (a) The original, balanced loads, (b) the combined load, with improved power factor., , 50,000, 120.28 mA, 13 240,000, , Since the power factor is lagging, the line current lags the line voltage, by u1 cos1 0.6 53.13. Thus,, Ia1 120.28l53.13, For load 2,, IL2 , , 75,000, 180.42 mA, 13 240,000, , and the line current lags the line voltage by u2 cos1 0.8 36.87., Hence,, Ia2 180.42l36.87, The total line current is, Ia Ia1 Ia2 120.28l53.13 180.42l36.87, (72.168 j96.224) (144.336 j108.252), 216.5 j204.472 297.8l43.36 mA, Alternatively, we could obtain the current from the total complex, power using Eq. (12.8.4) as, IL , , 123,800, 297.82 mA, 13 240,000, , and, Ia 297.82l43.36 mA, which is the same as before. The other line currents, Ib2 and Ica, can be, obtained according to the abc sequence (i.e., Ib 297.82l163.36 mA, and Ic 297.82l76.64 mA)., (c) We can find the reactive power needed to bring the power factor to, 0.9 lagging using Eq. (11.59),, QC P(tan uold tan unew)
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 525, , 12.8, , Unbalanced Three-Phase Systems, , 525, , where P 90 kW, uold 43.36, and unew cos1 0.9 25.84., Hence,, QC 90,000(tan 43.36 tan 25.84) 41.4 kVAR, This reactive power is for the three capacitors. For each capacitor, the, rating Q¿C 13.8 kVAR. From Eq. (11.60), the required capacitance is, C, , Q¿C, V 2rms, , Since the capacitors are ¢ -connected as shown in Fig. 12.22(b), Vrms, in the above formula is the line-to-line or line voltage, which is 240 kV., Thus,, 13,800, 635.5 pF, C, (2 p 60)(240,000)2, , Assume that the two balanced loads in Fig. 12.22(a) are supplied by, an 840-V rms 60-Hz line. Load 1 is Y-connected with 30 j40 per, phase, while load 2 is a balanced three-phase motor drawing 48 kW at, a power factor of 0.8 lagging. Assuming the abc sequence, calculate:, (a) the complex power absorbed by the combined load, (b) the kVAR, rating of each of the three capacitors ¢ -connected in parallel with the, load to raise the power factor to unity, and (c) the current drawn from, the supply at unity power factor condition., , Practice Problem 12.8, , Answer: (a) 56.47 j47.29 kVA, (b) 15.7 kVAR, (c) 38.813 A., , 12.8, , Unbalanced Three-Phase Systems, , This chapter would be incomplete without mentioning unbalanced, three-phase systems. An unbalanced system is caused by two possible, situations: (1) the source voltages are not equal in magnitude and/or, differ in phase by angles that are unequal, or (2) load impedances are, unequal. Thus,, , Ia, , A, , ZA, , VAN, , An unbalanced system is due to unbalanced voltage sources or an, unbalanced load., , To simplify analysis, we will assume balanced source voltages, but an, unbalanced load., Unbalanced three-phase systems are solved by direct application, of mesh and nodal analysis. Figure 12.23 shows an example of an, unbalanced three-phase system that consists of balanced source voltages (not shown in the figure) and an unbalanced Y-connected load, (shown in the figure). Since the load is unbalanced, ZA, ZB, and ZC are, not equal. The line currents are determined by Ohm’s law as, Ia , , VAN, ,, ZA, , Ib , , VBN, ,, ZB, , Ic , , VCN, ZC, , (12.59), , In, N, Ib, , VBN, , Ic, , B, VCN, , ZB, , ZC, C, , Figure 12.23, Unbalanced three-phase Y-connected, load., A special technique for handling unbalanced three-phase systems is the, method of symmetrical components,, which is beyond the scope of this text.
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 526, , Chapter 12, , 526, , Three-Phase Circuits, , This set of unbalanced line currents produces current in the neutral line,, which is not zero as in a balanced system. Applying KCL at node N, gives the neutral line current as, In (Ia Ib Ic), , (12.60), , In a three-wire system where the neutral line is absent, we can still, find the line currents Ia, Ib, and Ic using mesh analysis. At node N,, KCL must be satisfied so that Ia Ib Ic 0 in this case. The same, could be done for an unbalanced ¢ -Y, Y- ¢, or ¢-¢ three-wire system., As mentioned earlier, in long distance power transmission, conductors, in multiples of three (multiple three-wire systems) are used, with the, earth itself acting as the neutral conductor., To calculate power in an unbalanced three-phase system requires, that we find the power in each phase using Eqs. (12.46) to (12.49). The, total power is not simply three times the power in one phase but the, sum of the powers in the three phases., , Example 12.9, , The unbalanced Y-load of Fig. 12.23 has balanced voltages of 100 V, and the acb sequence. Calculate the line currents and the neutral current. Take ZA 15 , ZB 10 j5 , ZC 6 j8 ., Solution:, Using Eq. (12.59), the line currents are, Ia , Ib , , 100l120, , 100l0, , 6.67l0 A, , 100l120, , 8.94l93.44 A, 11.18l26.56, 100l120, 100l120, Ic , , 10l66.87 A, 6 j8, 10l53.13, 10 j5, , , , 15, , Using Eq. (12.60), the current in the neutral line is, In (Ia Ib Ic) (6.67 0.54 j8.92 3.93 j9.2), 10.06 j0.28 10.06l178.4 A, , Practice Problem 12.9, Ia, , A, , – j5 Ω, , j6 Ω, , Ib, Ic, , B, , Answer: 21.66l41.06 A, 34.98l139.8 A, 38.24l74.27 A., , 8Ω, , 10 Ω, , 16 Ω, , The unbalanced ¢ -load of Fig. 12.24 is supplied by balanced line-to-line, voltages of 240 V in the positive sequence. Find the line currents. Take, Vab as reference., , C, , Figure 12.24, Unbalanced ¢ -load; for Practice Prob. 12.9.
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , 528, , Page 528, , Chapter 12, , Three-Phase Circuits, , The line currents are, Ia I1 56.78 A,, Ic I2 42.75l155.1 A, Ib I2 I1 38.78 j18 56.78 25.46l135 A, (b) We can now calculate the complex power absorbed by the load. For, phase A,, SA 0 Ia 0 2ZA (56.78)2( j5) j16,120 VA, , For phase B,, , SB 0 Ib 0 2ZB (25.46)2(10) 6480 VA, , For phase C,, SC 0 Ic 0 2 ZC (42.75)2(j10) j18,276 VA, The total complex power absorbed by the load is, SL SA SB SC 6480 j2156 VA, (c) We check the result above by finding the power absorbed by the, source. For the voltage source in phase a,, Sa Van I*a (120l0)(56.78) 6813.6 VA, For the source in phase b,, Sb Vbn I*b (120l120)(25.46l135), 3055.2l105 790 j2951.1 VA, For the source in phase c,, Sc VbnI*c (120l120)(42.75l155.1), 5130l275.1 456.03 j5109.7 VA, The total complex power absorbed by the three-phase source is, Ss Sa Sb Sc 6480 j2156 VA, showing that Ss SL 0 and confirming the conservation principle of, ac power., , Practice Problem 12.10, , Find the line currents in the unbalanced three-phase circuit of Fig. 12.26, and the real power absorbed by the load., a, , 220 −120° rms V +−, , A, , + 220 0° rms V, −, −+, , c, , 220 120° rms V, , b, , − j5 Ω, , B, , 10 Ω, , C, j10 Ω, , Figure 12.26, For Practice Prob. 12.10., , Answer: 64l80.1 A, 38.1l60 A, 42.5l225 A, 4.84 kW.
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ale29559_ch12.qxd, , 07/22/2008, , 01:38 PM, , Page 529, , PSpice for Three-Phase Circuits, , 12.9, , 529, , PSpice for Three-Phase Circuits, , 12.9, , PSpice can be used to analyze three-phase balanced or unbalanced circuits in the same way it is used to analyze single-phase ac circuits., However, a delta-connected source presents two major problems to, PSpice. First, a delta-connected source is a loop of voltage sources—, which PSpice does not like. To avoid this problem, we insert a resistor of negligible resistance (say, 1 m per phase) into each phase of, the delta-connected source. Second, the delta-connected source does, not provide a convenient node for the ground node, which is necessary, to run PSpice. This problem can be eliminated by inserting balanced, wye-connected large resistors (say, 1 M per phase) in the deltaconnected source so that the neutral node of the wye-connected resistors serves as the ground node 0. Example 12.12 will illustrate this., , For the balanced Y-¢ circuit in Fig. 12.27, use PSpice to find the line, current IaA, the phase voltage VAB, and the phase current IAC. Assume, that the source frequency is 60 Hz., 100 0° V, −+, , a, , 1Ω, , A, 100 Ω, , 100 −120° V, −+, , n, , b, , 100 Ω, , 0.2 H, , 1Ω, , B, 100 Ω, , 100 120° V, −+, , c, , 1Ω, , 0.2 H, , 0.2 H, C, , Figure 12.27, For Example 12.11., , Solution:, The schematic is shown in Fig. 12.28. The pseudocomponents IPRINT, are inserted in the appropriate lines to obtain IaA and IAC, while VPRINT2, is inserted between nodes A and B to print differential voltage VAB., We set the attributes of IPRINT and VPRINT2 each to AC yes,, MAG yes, PHASE yes, to print only the magnitude and phase of, the currents and voltages. As a single-frequency analysis, we select, Analysis/Setup/AC Sweep and enter Total Pts 1, Start Freq 60,, and Final Freq 60. Once the circuit is saved, it is simulated by, selecting Analysis/Simulate. The output file includes the following:, FREQ, 6.000E+01, , V(A,B), 1.699E+02, , VP(A,B), 3.081E+01, , FREQ, 6.000E+01, , IM(V_PRINT2), 2.350E+00, , IP(V_PRINT2), -3.620E+01, , FREQ, 6.000E+01, , IM(V_PRINT3), 1.357E+00, , IP(V_PRINT3), -6.620E+01, , Example 12.11
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 534, , 534, , Chapter 12, , Three-Phase Circuits, , V=, 1.0e+002*, 2.0480+0.3612i, -0.7114-1.9546i, 0, 0, >>I=inv(Z)*V, I=, 8.9309+2.6973i, 0.0093+4.5159i, 5.4623+3.7954i, 2.9801+2.4044i, Iab I1 I4 (8.931 j2.697) (2.98 j2.404), 5.951 j0.293 5.958l177.18 A., Answer checks., 6. Satisfactory? We have a satisfactory solution and an adequate, check for the solution. We can now present the results as a, solution to the problem., , Practice Problem 12.12, , For the unbalanced circuit in Fig. 12.32, use PSpice to find the generator, current Ica, the line current IcC, and the phase current IAB., a, , A, 10 Ω, , + 220 −30° V, −, j10 Ω, 220 90° V, , −, +, , b, , B, , 10 Ω, , 10 Ω, , + 220 −150° V, −, , − j10 Ω, c, , C, , Figure 12.32, For Practice Prob. 12.12., , Answer: 24.68l90 A, 37.25l83.8 A, 15.556l75 A., , 12.10, , Applications, , Both wye and delta source connections have important practical applications. The wye source connection is used for long distance transmission of electric power, where resistive losses (I 2R) should be
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 535, , 12.10, , Applications, , 535, , minimal. This is due to the fact that the wye connection gives a line, voltage that is 13 greater than the delta connection; hence, for the, same power, the line current is 13 smaller. The delta source connection is used when three single-phase circuits are desired from a, three-phase source. This conversion from three-phase to single-phase, is required in residential wiring, because household lighting and, appliances use single-phase power. Three-phase power is used in, industrial wiring where a large power is required. In some applications, it is immaterial whether the load is wye- or delta-connected., For example, both connections are satisfactory with induction motors., In fact, some manufacturers connect a motor in delta for 220 V and, in wye for 440 V so that one line of motors can be readily adapted, to two different voltages., Here we consider two practical applications of those concepts covered in this chapter: power measurement in three-phase circuits and, residential wiring., , 12.10.1 Three-Phase Power Measurement, Section 11.9 presented the wattmeter as the instrument for measuring, the average (or real) power in single-phase circuits. A single wattmeter, can also measure the average power in a three-phase system that is balanced, so that P1 P2 P3; the total power is three times the reading of that one wattmeter. However, two or three single-phase wattmeters, are necessary to measure power if the system is unbalanced. The threewattmeter method of power measurement, shown in Fig. 12.33, will, work regardless of whether the load is balanced or unbalanced, wyeor delta-connected. The three-wattmeter method is well suited for, power measurement in a three-phase system where the power factor is, constantly changing. The total average power is the algebraic sum of, the three wattmeter readings,, PT P1 P2 P3, , ±, , a, , W1, ±, , ±, , b, , W2, ±, , o, , c, , ±, , W3, , Three-phase, load (wye, or delta,, balanced or, unbalanced), , ±, , Figure 12.33, Three-wattmeter method for measuring, three-phase power., , (12.61), , where P1, P2, and P3 correspond to the readings of wattmeters W1, W2,, and W3, respectively. Notice that the common or reference point o in, Fig. 12.33 is selected arbitrarily. If the load is wye-connected, point o, can be connected to the neutral point n. For a delta-connected load,, point o can be connected to any point. If point o is connected to point, b, for example, the voltage coil in wattmeter W2 reads zero and P2 0,, indicating that wattmeter W2 is not necessary. Thus, two wattmeters are, sufficient to measure the total power., The two-wattmeter method is the most commonly used method for, three-phase power measurement. The two wattmeters must be properly, connected to any two phases, as shown typically in Fig. 12.34. Notice, that the current coil of each wattmeter measures the line current, while, the respective voltage coil is connected between the line and the third, line and measures the line voltage. Also notice that the terminal of, the voltage coil is connected to the line to which the corresponding, current coil is connected. Although the individual wattmeters no longer, read the power taken by any particular phase, the algebraic sum of the, two wattmeter readings equals the total average power absorbed by the, load, regardless of whether it is wye- or delta-connected, balanced or, , a, , ±, , W1, , ±, b, , c, , ±, , W2, , Three-phase, load (wye, or delta,, balanced or, unbalanced), , ±, , Figure 12.34, Two-wattmeter method for measuring, three-phase power.
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ale29559_ch12.qxd, , 536, , 07/08/2008, , 12:05 PM, , Page 536, , Chapter 12, , Three-Phase Circuits, , unbalanced. The total real power is equal to the algebraic sum of the, two wattmeter readings,, PT P1 P2, , (12.62), , We will show here that the method works for a balanced three-phase, system., Consider the balanced, wye-connected load in Fig. 12.35. Our, objective is to apply the two-wattmeter method to find the average, power absorbed by the load. Assume the source is in the abc sequence, and the load impedance ZY ZYlu. Due to the load impedance, each, voltage coil leads its current coil by u, so that the power factor is cos u., We recall that each line voltage leads the corresponding phase voltage, by 30. Thus, the total phase difference between the phase current Ia, and line voltage Vab is u 30, and the average power read by, wattmeter W1 is, P1 Re[Vab I*a] Vab Ia cos(u 30) VL IL cos(u 30) (12.63), W1, a, +, Vab, b, , Ia, , ±, , ±, , Ib, , −, , ZY, , −, , ZY, ZY, , Vcb, +, c, , ±, , W2, , ±, , Ic, , Figure 12.35, Two-wattmeter method applied to a balanced wye load., , Similarly, we can show that the average power read by wattmeter 2 is, P2 Re[Vcb I*c] Vcb Ic cos(u 30) VL IL cos(u 30) (12.64), We now use the trigonometric identities, cos(A B) cos A cos B sin A sin B, cos(A B) cos A cos B sin A sin B, , (12.65), , to find the sum and the difference of the two wattmeter readings in, Eqs. (12.63) and (12.64):, P1 P2 VL IL[cos(u 30) cos(u 30)], VL IL(cos u cos 30 sin u sin 30, cos u cos 30 sin u sin 30), VL IL2 cos 30 cos u 13VL IL cos u, , (12.66), , since 2 cos 30 13. Comparing Eq. (12.66) with Eq. (12.50) shows, that the sum of the wattmeter readings gives the total average power,, PT P1 P2, , (12.67)
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 537, , 12.10, , Applications, , 537, , Similarly,, P1 P2 VL IL[cos(u 30) cos(u 30)], Vl IL(cos u cos 30 sin u sin 30, cos u cos 30 sin u sin 30), , (12.68), , VL IL 2 sin 30 sin u, P2 P1 VL IL sin u, since 2 sin 30 1. Comparing Eq. (12.68) with Eq. (12.51) shows, that the difference of the wattmeter readings is proportional to the total, reactive power, or, QT 13(P2 P1), , (12.69), , From Eqs. (12.67) and (12.69), the total apparent power can be, obtained as, ST 2PT2 Q T2, , (12.70), , Dividing Eq. (12.69) by Eq. (12.67) gives the tangent of the power factor angle as, tan u , , QT, P2 P1, 13, PT, P2 P1, , (12.71), , from which we can obtain the power factor as pf cos u. Thus, the, two-wattmeter method not only provides the total real and reactive powers, it can also be used to compute the power factor. From Eqs. (12.67),, (12.69), and (12.71), we conclude that:, 1. If P2 P1, the load is resistive., 2. If P2 7 P1, the load is inductive., 3. If P2 6 P1, the load is capacitive., Although these results are derived from a balanced wye-connected, load, they are equally valid for a balanced delta-connected load. However, the two-wattmeter method cannot be used for power measurement, in a three-phase four-wire system unless the current through the neutral line is zero. We use the three-wattmeter method to measure the real, power in a three-phase four-wire system., , Three wattmeters W1, W2, and W3 are connected, respectively, to phases, a, b, and c to measure the total power absorbed by the unbalanced wyeconnected load in Example 12.9 (see Fig. 12.23). (a) Predict the, wattmeter readings. (b) Find the total power absorbed., Solution:, Part of the problem is already solved in Example 12.9. Assume that, the wattmeters are properly connected as in Fig. 12.36., , Example 12.13
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 538, , 538, , Chapter 12, , Three-Phase Circuits, Ia, , A, , W1 +, VAN, −, , In, −, , −, , 15 Ω, N, 6Ω, , 10 Ω, , VBN, VCN, , Ib, , +, , Ic, , W2, , − j8 Ω, j5 Ω, , +, , C, , B, , W3, , Figure 12.36, For Example 12.13., , (a) From Example 12.9,, VAN 100l0,, , VBN 100l120,, , VCN 100l120 V, , while, Ia 6.67l0,, , Ib 8.94l93.44,, , Ic 10l66.87 A, , We calculate the wattmeter readings as follows:, P1 Re(VAN I*a) VAN Ia cos(uVAN uIa), 100 6.67 cos(0 0) 667 W, P2 Re(VBN I*b) VBN Ib cos(uVBN uIb), 100 8.94 cos(120 93.44) 800 W, *, P3 Re(VCN I c ) VCN Ic cos(uVCN uIc), 100 10 cos(120 66.87) 600 W, (b) The total power absorbed is, PT P1 P2 P3 667 800 600 2067 W, We can find the power absorbed by the resistors in Fig. 12.36 and use, that to check or confirm this result., PT 0 Ia 0 2(15) 0Ib 0 2(10) 0Ic 0 2(6), 6.672(15) 8.942(10) 102(6), 667 800 600 2067 W, which is exactly the same thing., , Practice Problem 12.13, , Repeat Example 12.13 for the network in Fig. 12.24 (see Practice, Prob. 12.9). Hint: Connect the reference point o in Fig. 12.33 to point B., Answer: (a) 3.92 kW, 0 W, 8.895 kW, (b) 12.815 kW., , Example 12.14, , The two-wattmeter method produces wattmeter readings P1 1560 W, and P2 2100 W when connected to a delta-connected load. If the line, voltage is 220 V, calculate: (a) the per-phase average power, (b) the perphase reactive power, (c) the power factor, and (d) the phase impedance.
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 539, , 12.10, , Applications, , 539, , Solution:, We can apply the given results to the delta-connected load. (a) The total, real or average power is, PT P1 P2 1560 2100 3660 W, The per-phase average power is then, 1, Pp PT 1220 W, 3, (b) The total reactive power is, QT 13(P2 P1) 13(2100 1560) 935.3 VAR, so that the per-phase reactive power is, 1, Qp QT 311.77 VAR, 3, (c) The power angle is, u tan1, , QT, 935.3, tan1, 14.33, PT, 3660, , Hence, the power factor is, cos u 0.9689 (lagging), It is a lagging pf because QT is positive or P2 7 P1., (c) The phase impedance is Zp Zplu. We know that u is the same as, the pf angle; that is, u 14.33., Zp , , Vp, Ip, , We recall that for a delta-connected load, Vp VL 220 V. From, Eq. (12.46),, Pp Vp Ip cos u, , Ip , , 1, , 1220, 5.723 A, 220 0.9689, , Hence,, Zp , , Vp, Ip, , , , 220, 38.44 , 5.723, , and, Zp 38.44l14.33 , , Let the line voltage VL 208 V and the wattmeter readings of the, balanced system in Fig. 12.35 be P1 560 W and P2 800 W., Determine:, (a) the total average power, (b) the total reactive power, (c) the power factor, (d) the phase impedance, Is the impedance inductive or capacitive?, Answer: (a) 240 W, (b) 2355.6 VAR, (c) 0.1014, (d) 18.25l84.18 ,, inductive., , Practice Problem 12.14
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 540, , 540, , Example 12.15, , Chapter 12, , Three-Phase Circuits, , The three-phase balanced load in Fig. 12.35 has impedance per phase, of ZY 8 j6 . If the load is connected to 208-V lines, predict the, readings of the wattmeters W1 and W2. Find PT and QT., Solution:, The impedance per phase is, ZY 8 j6 10l36.87 , so that the pf angle is 36.87. Since the line voltage VL 208 V, the, line current is, IL , , Vp, , 0 ZY 0, , , , 20813, 12 A, 10, , Then, P1 VL IL cos(u 30) 208 12 cos(36.87 30), 980.48 W, P2 VL IL cos(u 30) 208 12 cos(36.87 30), 2478.1 W, Thus, wattmeter 1 reads 980.48 W, while wattmeter 2 reads 2478.1 W., Since P2 7 P1, the load is inductive. This is evident from the load ZY, itself. Next,, PT P1 P2 3.459 kW, and, QT 13(P2 P1) 13(1497.6) VAR 2.594 kVAR, , Practice Problem 12.15, , If the load in Fig. 12.35 is delta-connected with impedance per phase, of Zp 30 j40 and VL 440 V, predict the readings of the, wattmeters W1 and W2. Calculate PT and QT., Answer: 6.166 kW, 0.8021 kW, 6.968 kW, 9.291 kVAR., , 12.10.2 Residential Wiring, In the United States, most household lighting and appliances operate on, 120-V, 60-Hz, single-phase alternating current. (The electricity may also, be supplied at 110, 115, or 117 V, depending on the location.) The local, power company supplies the house with a three-wire ac system. Typically, as in Fig. 12.37, the line voltage of, say, 12,000 V is stepped down, to 120/240 V with a transformer (more details on transformers in the next, chapter). The three wires coming from the transformer are typically colored red (hot), black (hot), and white (neutral). As shown in Fig. 12.38,, the two 120-V voltages are opposite in phase and hence add up to zero., That is, VW 0l0, VB 120l0, VR 120l180 VB., VBR VB VR VB (VB) 2VB 240l0, , (12.72)
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 541, , 12.10, , Applications, , 541, , Step-down, transformer, Circuit, #1, 120 V, , Wall of, house, , Circuit, #2, 120 V, , Fuse, , Circuit, #3, 240 V, , Fuses, , Switch, Fuse, Light, pole, , Watt-hour meter, , Grounded metal, stake, , Ground, , Figure 12.37, A 120/240 household power system., A. Marcus and C. M. Thomson, Electricity for Technicians, 2nd ed. [Englewood Cliffs,, NJ: Prentice Hall, 1975], p. 324., , To other houses, , Black, (hot) B, White, (neutral), , W, , Ground, R, , +, 120 V, −, −, 120 V, +, , Red (hot), , 120 V, lights, , 120 V, appliance, 240 V, appliance, , 120 V, lights, , 120 V, appliance, , Transformer, House, , Figure 12.38, Single-phase three-wire residential wiring., , Since most appliances are designed to operate with 120 V, the lighting, and appliances are connected to the 120-V lines, as illustrated in, Fig. 12.39 for a room. Notice in Fig. 12.37 that all appliances are conLamp sockets, nected in parallel. Heavy appliances that consume large currents, such, as air conditioners, dishwashers, ovens, and laundry machines, are connected to the 240-V power line., Switch, Because of the dangers of electricity, house wiring is carefully regBase outlets, ulated by a code drawn by local ordinances and by the National ElecNeutral, trical Code (NEC). To avoid trouble, insulation, grounding, fuses, and, circuit breakers are used. Modern wiring codes require a third wire for 120 volts, a separate ground. The ground wire does not carry power like the neuUngrounded conductor, tral wire but enables appliances to have a separate ground connection., Figure 12.40 shows the connection of the receptacle to a 120-V rms, line and to the ground. As shown in the figure, the neutral line is con- Figure 12.39, A typical wiring diagram of a room., nected to the ground (the earth) at many critical locations. Although A. Marcus and C. M. Thomson, Electricity for, the ground line seems redundant, grounding is important for many Technicians, 2nd ed. [Englewood Cliffs, NJ:, reasons. First, it is required by NEC. Second, grounding provides a Prentice Hall, 1975], p. 325.
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ale29559_ch12.qxd, , 542, , 07/08/2008, , 12:05 PM, , Page 542, , Chapter 12, , Three-Phase Circuits, Fuse or circuit breaker, , Hot wire, Receptacle, , 120 V rms, , +, −, , To other appliances, Neutral wire, , Power system, ground, , Service, panel ground, , Ground wire, , Figure 12.40, Connection of a receptacle to the hot line and to the ground., , convenient path to ground for lightning that strikes the power line., Third, grounds minimize the risk of electric shock. What causes shock, is the passage of current from one part of the body to another. The, human body is like a big resistor R. If V is the potential difference, between the body and the ground, the current through the body is determined by Ohm’s law as, I, , V, R, , (12.73), , The value of R varies from person to person and depends on whether, the body is wet or dry. How great or how deadly the shock is depends, on the amount of current, the pathway of the current through the, body, and the length of time the body is exposed to the current. Currents less than 1 mA may not be harmful to the body, but currents, greater than 10 mA can cause severe shock. A modern safety device, is the ground-fault circuit interrupter (GFCI), used in outdoor circuits, and in bathrooms, where the risk of electric shock is greatest. It is, essentially a circuit breaker that opens when the sum of the currents, iR, iW, and iB through the red, white, and the black lines is not equal, to zero, or iR iW iB 0., The best way to avoid electric shock is to follow safety guidelines concerning electrical systems and appliances. Here are some of, them:, • Never assume that an electrical circuit is dead. Always check to, be sure., • Use safety devices when necessary, and wear suitable clothing, (insulated shoes, gloves, etc.)., • Never use two hands when testing high-voltage circuits, since the, current through one hand to the other hand has a direct path, through your chest and heart., • Do not touch an electrical appliance when you are wet. Remember that water conducts electricity., • Be extremely careful when working with electronic appliances, such as radio and TV because these appliances have large capacitors in them. The capacitors take time to discharge after the power, is disconnected., • Always have another person present when working on a wiring, system, just in case of an accident.
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 543, , Review Questions, , 12.11, , 543, , Summary, , 1. The phase sequence is the order in which the phase voltages of a, three-phase generator occur with respect to time. In an abc, sequence of balanced source voltages, Van leads Vbn by 120,, which in turn leads Vcn by 120. In an acb sequence of balanced, voltages, Van leads Vcn by 120, which in turn leads Vbn by 120., 2. A balanced wye- or delta-connected load is one in which the threephase impedances are equal., 3. The easiest way to analyze a balanced three-phase circuit is to, transform both the source and the load to a Y-Y system and then, analyze the single-phase equivalent circuit. Table 12.1 presents a, summary of the formulas for phase currents and voltages and line, currents and voltages for the four possible configurations., 4. The line current IL is the current flowing from the generator to the, load in each transmission line in a three-phase system. The line, voltage VL is the voltage between each pair of lines, excluding the, neutral line if it exists. The phase current Ip is the current flowing, through each phase in a three-phase load. The phase voltage Vp is, the voltage of each phase. For a wye-connected load,, VL 13Vp, , and, , IL Ip, , For a delta-connected load,, VL Vp, , and, , IL 13Ip, , 5. The total instantaneous power in a balanced three-phase system is, constant and equal to the average power., 6. The total complex power absorbed by a balanced three-phase, Y-connected or ¢ -connected load is, S P jQ 13VL ILlu, where u is the angle of the load impedances., 7. An unbalanced three-phase system can be analyzed using nodal or, mesh analysis., 8. PSpice is used to analyze three-phase circuits in the same way as, it is used for analyzing single-phase circuits., 9. The total real power is measured in three-phase systems using, either the three-wattmeter method or the two-wattmeter method., 10. Residential wiring uses a 120/240-V, single-phase, three-wire system., , Review Questions, 12.1, , What is the phase sequence of a three-phase motor, for which VAN 220l100 V and, VBN 220l140 V?, (a) abc, , 12.2, , (b) acb, , If in an acb phase sequence, Van 100l20, then, Vcn is:, (a) 100l140, (c) 100l50, , (b) 100l100, (d) 100l10, , 12.3, , Which of these is not a required condition for a, balanced system:, (a) 0Van 0 0Vbn 0 0Vcn 0, (b) Ia Ib Ic 0, , (c) Van Vbn Vcn 0, (d) Source voltages are 120 out of phase with each, other., (e) Load impedances for the three phases are equal.
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Chapter 12, , 544, , 12.4, , 12.7, , 12.8, , (b) False, , In a ¢ -connected load, the line current and phase, current are equal., (a) True, , 12.6, , Three-Phase Circuits, , In a Y-connected load, the line current and phase, current are equal., (a) True, , 12.5, , Page 544, , (a) True, 12.9, , (b) False, , In a Y-Y system, a line voltage of 220 V produces a, phase voltage of:, (a) 381 V, , (b) 311 V, , (d) 156 V, , (e) 127 V, , (c) 220 V, , (b) 71 V, , (d) 173 V, , (e) 141 V, , (b) False, , In a balanced three-phase circuit, the total, instantaneous power is equal to the average power., (a) True, , (b) False, , 12.10 The total power supplied to a balanced ¢ -load is, found in the same way as for a balanced Y-load., (a) True, , In a ¢-¢ system, a phase voltage of 100 V produces, a line voltage of:, (a) 58 V, , When a Y-connected load is supplied by voltages in, abc phase sequence, the line voltages lag the, corresponding phase voltages by 30., , (c) 100 V, , (b) False, , Answers: 12.1a, 12.2a, 12.3c, 12.4a, 12.5b, 12.6e, 12.7c,, 12.8b, 12.9a, 12.10a., , Problems1, , 12.1, , −+, , If Vab 400 V in a balanced Y-connected threephase generator, find the phase voltages, assuming, the phase sequence is:, , VP, n, , (a) abc, 12.2, , (b) acb, , What is the phase sequence of a balanced threephase circuit for which Van 160l30 V and, Vcn 160l90 V? Find Vbn., , 12.3, , Determine the phase sequence of a balanced threephase circuit in which Vbn 208l130 V and, Vcn 208l10 V. Obtain Van., , 12.4, , A three-phase system with abc sequence and, VL 200 V feeds a Y-connected load with, ZL 40l30 . Find the line currents., , 12.5, , For a Y-connected load, the time-domain expressions, for three line-to-neutral voltages at the terminals are:, , −120° V, −+, , VP 120° V, −+, , jXL, , b, , B, , R, , jXL, N, , c, , C, , R, , jXL, , Obtain the line currents in the three-phase circuit of, Fig. 12.42 on the next page., , 12.8, , In a balanced three-phase Y-Y system, the source is an, abc sequence of voltages and Van 220l20 V rms., The line impedance per phase is 0.6 j1.2 ,, while the per-phase impedance of the load is, 10 j14 . Calculate the line currents and the, load voltages., , 12.9, , A balanced Y-Y four-wire system has phase voltages, Van 120l0,, , Vbn 120l120, , Vcn 120l120 V, , Section 12.3 Balanced Wye-Wye Connection, , 1, , R, , 12.7, , vBN 150 cos(t 88) V, , Using Fig. 12.41, design a problem to help other, students better understand balanced wye-wye, connected circuits., , A, , For Prob. 12.6., , vCN 150 cos(t 152) V, , 12.6, , a, , Figure 12.41, , vAN 150 cos(t 32) V, , Write the time-domain expressions for the line-toline voltages vAB, vBC, and vCA., , 0° V, , VP, , Section 12.2 Balanced Three-Phase Voltages, , The load impedance per phase is 19 j13 , and the, line impedance per phase is 1 j2 . Solve for, the line currents and neutral current., , Remember that unless stated otherwise, all given voltages and currents are rms values.
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 545, , Problems, , 545, , Ia, , a, , A, , +, −, , 440 0° V, , 6 − j8 Ω, , n, , N, 6 − j8 Ω, , 6 − j8 Ω, 440 120° V +−, , − 440 −120° V, +, Ib, , Ic, , Figure 12.42, For Prob. 12.7., , 12.10 For the circuit in Fig. 12.43, determine the current in, the neutral line., , 12.12 Using Fig. 12.45, design a problem to help other, students better understand wye-delta connected, circuits., Ia, , 2Ω, , 220 0° V, , +, −, , 2Ω, , 10 + j5 Ω, , −, +, , VP 0° V, Z∆, , n, VP 120° V +−, , 20 Ω, , 220 −120° V, 220 120° V, , +, −, , 25 − j10 Ω, −+, , A, , a, , −, +, , Z∆, , VP −120° V, Ib, , Z∆, , c, b, , 2Ω, , Ic, , C, , B, , Figure 12.43, , Figure 12.45, , For Prob. 12.10., , For Prob. 12.12., , Section 12.4 Balanced Wye-Delta Connection, , 12.13 In the balanced three-phase Y- ¢ system in Fig. 12.46,, find the line current IL and the average power delivered, to the load., , 12.11 In the Y- ¢ system shown in Fig. 12.44, the source is, a positive sequence with Van 120l0 V and phase, impedance Zp 2 j3 . Calculate the line, voltage VL and the line current IL., , 220 0° V rms, −+, 220 ⫺120° V rms, , Van, −+, , −+, , a, , 220 120° V rms, Vbn, n, , −+, Vcn, −+, , Figure 12.44, For Prob. 12.11., , Zp, b, , −+, Zp, , Zp, , 2Ω, , 2Ω, , 2Ω, , 9⫺j6 Ω, 9⫺j6 Ω, 9⫺j6 Ω, , Figure 12.46, For Prob. 12.13., , c, , 12.14 Obtain the line currents in the three-phase circuit of, Fig. 12.47 on the next page.
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 546, , Chapter 12, , 546, , Three-Phase Circuits, 1 + j2 Ω, , A, , a, ZL, , + 120 0° V, −, n, , ZL, , C, , 120 120° V +−, , + 120 –120° V, −, b, , c, , B, ZL = 12 + j2 Ω, 1 + j2 Ω, , 1 + j2 Ω, , Figure 12.47, For Prob. 12.14., 12.15 The circuit in Fig. 12.48 is excited by a balanced, three-phase source with a line voltage of 210 V. If, Zl 1 j1 , Z¢ 24 j30 , and, ZY 12 j5 , determine the magnitude of the, line current of the combined loads., Zl, , Section 12.5 Balanced Delta-Delta Connection, 12.19 For the ¢-¢ circuit of Fig. 12.50, calculate the phase, and line currents., , ZY, , a, , a, Z∆, , A, 30 Ω, , Z∆, , Zl, , ZY, , + 173 0° V, −, , b, , j10 Ω, Z∆, , Zl, , ZY, , B, j10 Ω, , 30 Ω, , c, + 173 −120° V, −, , Figure 12.48, For Prob. 12.15., , j10 Ω, , 12.16 A balanced delta-connected load has a phase current, IAC 10l30 A., (a) Determine the three line currents assuming that, the circuit operates in the positive phase sequence., (b) Calculate the load impedance if the line voltage, is VAB 110l0 V., 12.17 A balanced delta-connected load has line current, Ia 10l25 A. Find the phase currents IAB, IBC,, and ICA., , c, , C, , Figure 12.50, For Prob. 12.19., , 12.20 Using Fig. 12.51, design a problem to help other, students better understand balanced delta-delta, connected circuits., , 12.18 If Van 440l60 V in the network of Fig. 12.49,, find the load phase currents IAB, IBC, and ICA., , 12 Ω, , Three-phase,, Y-connected, generator, , A, I AB, , j9 Ω, , j9 Ω, B, , VL 120° V +−, , +, −, , 12 Ω, , b, c, , Ia, , A, , a, , (+) phase, sequence, , 30 Ω, , b, , −, 173 120° V +, , 12 Ω, , j9 Ω, , VL, , 0° V, , ZL, , ZL, I CA, , Ib, −+, , C, , VL −120° V, , Figure 12.49, , Figure 12.51, , For Prob. 12.18., , For Prob. 12.20., , Ic, , B, , C, I BC, , ZL
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 547, , Problems, , 547, , Section 12.6 Balanced Delta-Wye Connection, , 12.21 Three 440-V generators form a delta-connected source, that is connected to a balanced delta-connected load of, ZL 10 j8 per phase as shown in Fig. 12.52., , 12.25 In the circuit of Fig. 12.54, if Vab 220l10,, Vbc 220l110, Vca 220l130 V, find the line, currents., , (a) Determine the value of IAC., (b) What is the value of IbB?, , a, a, 440 120° +−, −+, , c, , 3 + j2 Ω, , Ia, , 3 + j2 Ω, , Ib, , 3 + j2 Ω, , Ic, , A, + 440 0°, −, b B, , ZL, , +, −, , ZL, −, Vca +, , C, , 10 − j8 Ω, , b, , ZL, , 440 –120°, , 10 − j8 Ω, , Vab, , + V, −, bc, , Figure 12.52, For Prob. 12.21., , 10 − j8 Ω, , c, , Figure 12.54, For Prob. 12.25., 12.22 Find the line currents Ia, Ib, and Ic in the three-phase, network of Fig. 12.53 below. Take Z¢ 12 j15 ,, ZY 4 j6 , and Zl 2 ., , 12.26 Using Fig. 12.55, design a problem to help other, students better understand balanced delta connected, sources delivering power to balanced wye connected, loads., , 12.23 A three-phase balanced system with a line voltage of, 202 V rms feeds a delta-connected load with, Zp 25l60 ., , I aA, , a, , A, , (a) Find the line current., , R, , (b) Determine the total power supplied to the load, using two wattmeters connected to the A and C, lines., , Three-phase,, ∆-connected, generator, (+) phase, sequence, , 12.24 A balanced delta-connected source has phase voltage, Vab 440l30 V and a positive phase sequence. If, this is connected to a balanced delta-connected load,, find the line and phase currents. Take the load, impedance per phase as 60l30 and line, impedance per phase as 1 j1 ., , Zl, , 208 120° V +−, , −+, 208 −120° V, , Figure 12.53, For Prob. 12.22., , − jXC, R, , I cC, , B, , Figure 12.55, For Prob. 12.26., , Ia, , A, , ZY, , Z∆, , Ib, Ic, , Z∆, , ZY, , ZY, B, , Zl, , I bB, , b, c, , + 208 0° V, −, Zl, , − jXC, N, , Z∆, , C, , − jXC, R, C
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ale29559_ch12.qxd, , 07/09/2008, , 11:41 AM, , Page 548, , Chapter 12, , 548, , Three-Phase Circuits, , 12.32 Design a problem to help other students better, understand power in a balanced three-phase, system., , 12.27 A ¢ -connected source supplies power to a Yconnected load in a three-phase balanced system., Given that the line impedance is 2 j1 per phase, while the load impedance is 6 j4 per phase, find, the magnitude of the line voltage at the load. Assume, the source phase voltage Vab 208l0 V rms., , 12.33 A three-phase source delivers 9.6 kVA to a wyeconnected load with a phase voltage of 208 V and a, power factor of 0.9 lagging. Calculate the source line, current and the source line voltage., , 12.28 The line-to-line voltages in a Y-load have a magnitude, of 220 V and are in the positive sequence at 60 Hz. If, the loads are balanced with Z1 Z2 Z3 25l30,, find all line currents and phase voltages., , 12.34 A balanced wye-connected load with a phase, impedance of 10 j16 is connected to a balanced, three-phase generator with a line voltage of 220 V., Determine the line current and the complex power, absorbed by the load., , Section 12.7 Power in a Balanced System, 12.29 A balanced three-phase Y- ¢ system has Van 120l0, V rms and Z¢ 51 j45 . If the line impedance, per phase is 0.4 j1.2 , find the total complex, power delivered to the load., , 12.35 Three equal impedances, 60 j30 each, are deltaconnected to a 230-V rms, three-phase circuit., Another three equal impedances, 40 j10 each,, are wye-connected across the same circuit at the, same points. Determine:, , 12.30 In Fig. 12.56, the rms value of the line voltage is, 208 V. Find the average power delivered to the load., , (a) the line current, a, , A, + V V, a, −, b, b, n, −+, −, +, , c, , B, , ZL, , (b) the total complex power supplied to the two, loads, , ZL, , N, , (c) the power factor of the two loads combined, , Z L = 30 45°, Vc, , 12.36 A 4200-V, three-phase transmission line has an, impedance of 4 j per phase. If it supplies a load, of 1 MVA at 0.75 power factor (lagging), find:, , C, , Figure 12.56, For Prob. 12.30., , (a) the complex power, (b) the power loss in the line, (c) the voltage at the sending end, , 12.31 A balanced delta-connected load is supplied by a, 60-Hz three-phase source with a line voltage of 240 V., Each load phase draws 6 kW at a lagging power, factor of 0.8. Find:, , 12.37 The total power measured in a three-phase system, feeding a balanced wye-connected load is 12 kW, at a power factor of 0.6 leading. If the line voltage, is 208 V, calculate the line current IL and the load, impedance ZY., , (a) the load impedance per phase, (b) the line current, (c) the value of capacitance needed to be connected, in parallel with each load phase to minimize the, current from the source, , 12.38 Given the circuit in Fig. 12.57 below, find the total, complex power absorbed by the load., , 1Ω, 208 0° V +−, , j2 Ω, 9Ω, , 208 240° V, −+, , −, + 208 120° V, , Figure 12.57, For Prob. 12.38., , 1Ω, , j2 Ω, , 9Ω, , j12 Ω, , j12 Ω, j12 Ω, , 1Ω, , j2 Ω, , 1Ω, , j2 Ω, , 9Ω
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 549, , Problems, , 12.39 Find the real power absorbed by the load in Fig. 12.58., 5Ω, , a, , A, − j6 Ω, , 100 120° V +−, , + 100 0° V, −, 5Ω, , −+, , c, , 100 −120° V, , 8Ω, , b, , 4Ω, j3 Ω, , 10 Ω, , C, , B, 5Ω, , Figure 12.58, For Prob. 12.39., 12.40 For the three-phase circuit in Fig. 12.59, find the, average power absorbed by the delta-connected load, with Z¢ 21 j24 ., 100 0° V rms, −+, 100 −120° V rms, −+, , 100 120° V rms, −+, , 1Ω, , j0.5 Ω, , 1Ω, , j0.5 Ω, , 12.45 A balanced wye-connected load is connected to the, generator by a balanced transmission line with an, impedance of 0.5 j2 per phase. If the load is, rated at 450 kW, 0.708 power factor lagging, 440-V, line voltage, find the line voltage at the generator., 12.46 A three-phase load consists of three 100- resistors, that can be wye- or delta-connected. Determine, which connection will absorb the most average, power from a three-phase source with a line voltage, of 110 V. Assume zero line impedance., 12.47 The following three parallel-connected three-phase, loads are fed by a balanced three-phase source:, Load 1: 250 kVA, 0.8 pf lagging, Load 2: 300 kVA, 0.95 pf leading, Load 3: 450 kVA, unity pf, If the line voltage is 13.8 kV, calculate the line, current and the power factor of the source. Assume, that the line impedance is zero., , Z∆, Z∆, , 1Ω, , 549, , Z∆, , j0.5 Ω, , 12.48 A balanced, positive-sequence wye-connected source, has Van 240l0 V rms and supplies an unbalanced, delta-connected load via a transmission line with, impedance 2 j3 per phase., (a) Calculate the line currents if ZAB 40 j15 ,, ZBC 60 , ZCA 18 j12 ., (b) Find the complex power supplied by the source., , Figure 12.59, For Prob. 12.40., 12.41 A balanced delta-connected load draws 5 kW at a, power factor of 0.8 lagging. If the three-phase system, has an effective line voltage of 400 V, find the line, current., 12.42 A balanced three-phase generator delivers 9.6 kW to a, wye-connected load with impedance 30 j40 per, phase. Find the line current IL and the line voltage VL., 12.43 Refer to Fig. 12.48. Obtain the complex power, absorbed by the combined loads., 12.44 A three-phase line has an impedance of 1 j3 per, phase. The line feeds a balanced delta-connected load,, which absorbs a total complex power of 12 j5 kVA., If the line voltage at the load end has a magnitude of, 240 V, calculate the magnitude of the line voltage at, the source end and the source power factor., , 12.49 Each phase load consists of a 20- resistor and a 10-, inductive reactance. With a line voltage of 220 V rms,, calculate the average power taken by the load if:, (a) the three-phase loads are delta-connected, (b) the loads are wye-connected, 12.50 A balanced three-phase source with VL 240 V rms, is supplying 8 kVA at 0.6 power factor lagging to, two wye-connected parallel loads. If one load draws, 3 kW at unity power factor, calculate the impedance, per phase of the second load., , Section 12.8 Unbalanced Three-Phase Systems, , 12.51 Consider the ¢-¢ system shown in Fig. 12.60. Take, Z1 8 j6 , Z2 4.2 j2.2 , Z3 10 j0 ., , a, , 240 0° V, , b, , A, , −, +, , +, −, +−, , 240 120° V, , Figure 12.60, For Prob. 12.51., , 240 −120° V, c, , Z3, C, , Z1, , Z2, B
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ale29559_ch12.qxd, , 07/08/2008, , 12:05 PM, , Page 550, , Chapter 12, , 550, , Three-Phase Circuits, , (a) Find the phase current IAB, IBC, and ICA., , 12.56 Using Fig. 12.63, design a problem to help other, students to better understand unbalanced three-phase, systems., , (b) Calculate line currents IaA, IbB, and IcC., 12.52 A four-wire wye-wye circuit has, Van 120l120,, , Vbn 120l0, a, , Vcn 120l120 V, ZAN 20l60,, , −+, , ZBN 30l0, VP 120° V + −, , Zcn 40l30 , , 12.53 Using Fig. 12.61, design a problem that will help, other students better understand unbalanced threephase systems., , Vp 0°, , −, +, , B, , VP –120° V, , − jXC, R, C, , Figure 12.63, For Prob. 12.56., , 12.57 Determine the line currents for the three-phase, circuit of Fig. 12.64. Let Va 110l0, Vb , 110l120, Vc 110l120 V., , Ia, , Vp 120°, , b, , c, , find the current in the neutral line., , +, −, , 0° V +−, , VP, , If the impedances are, , A, jXL, , jXL, Ia, , Vp −120°, −, +, , R2, Ib, , Va, , R1, , +, −, , 80 + j50 Ω, , 60 – j40 Ω, , 20 + j30 Ω, , Ic, −, +, , Vc +−, , Figure 12.61, For Prob. 12.53., , Ib, Ic, , 12.54 A balanced three-phase Y-source with VP 210 V rms, drives a Y-connected three-phase load with phase, impedance ZA 80 , ZB 60 j90 , and, ZC j80 . Calculate the line currents and total, complex power delivered to the load. Assume that, the neutrals are connected., 12.55 A three-phase supply, with the line voltage 240 V rms, positively phased, has an unbalanced delta-connected, load as shown in Fig. 12.62. Find the phase currents, and the total complex power., , Figure 12.64, For Prob. 12.57., , Section 12.9 PSpice for Three-Phase Circuits, , 12.58 Solve Prob. 12.10 using PSpice., 12.59 The source in Fig. 12.65 is balanced and exhibits a, positive phase sequence. If f 60 Hz, use PSpice to, find VAN, VBN, and VCN., , A, a, j25 Ω, , 40 Ω, , A, , 100 0° V +−, n, , B, , +−, , C, 30 30° Ω, , c, , Figure 12.62, , Figure 12.65, , For Prob. 12.55., , For Prob. 12.59., , −+, , b, , B 40 Ω, , 0.2 mF, N, 10 mF, C
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ale29559_ch12.qxd, , 07/08/2008, , 12:06 PM, , Page 551, , Problems, , 12.60 Use PSpice to determine Io in the single-phase,, three-wire circuit of Fig. 12.66. Let Z1 15 j10 ,, Z2 30 j20 , and Z3 12 j5 ., , 551, , 12.63 Use PSpice to find currents IaA and IAC in the unbalanced three-phase system shown in Fig. 12.69. Let, Zl 2 j,, , Z1 40 j20 ,, , Z2 50 j30 ,, 4Ω, , 220 0° V, −+, , Io, 220 0° V, , +, −, , A, , a, , Z1, 4Ω, , 220 –120° V, , Z3, 220 0° V, , Z1, , Z3 25 , , +, −, , 4Ω, , −+, , Z1, Z1, B, , b, , Z2, , Z3, , Z2, 220 120° V, , Figure 12.66, , −+, , For Prob. 12.60., , Z1, , C, , c, , Figure 12.69, For Prob. 12.63., , 12.61 Given the circuit in Fig. 12.67, use PSpice to, determine currents IaA and voltage VBN., , 240 0° V, −+, , a, , 4Ω, , j3 Ω, , A 10 Ω, , 12.64 For the circuit in Fig. 12.58, use PSpice to find the, line currents and the phase currents., 12.65 A balanced three-phase circuit is shown in Fig. 12.70, on the next page. Use PSpice to find the line currents, IaA, IbB, and IcC., , j15 Ω, , Section 12.10 Applications, − j36 Ω, 240 −120° V, n, , −+, , b, , 4Ω, , j3 Ω, , − j36 Ω, 10 Ω, , B, , j15 Ω, N, , − j36 Ω, 240 120° V, −+, , (a) the voltage to neutral, c, , 4Ω, , 10 Ω, , j3 Ω, , j15 Ω, , (b) the currents I1, I2, I3, and In, , C, , (c) the readings of the wattmeters, , Figure 12.67, , (d) the total power absorbed by the load, , For Prob. 12.61., , *12.67 As shown in Fig. 12.72, a three-phase four-wire line, with a phase voltage of 120 V rms and positive, phase sequence supplies a balanced motor load at, 260 kVA at 0.85 pf lagging. The motor load is, connected to the three main lines marked a, b, and c., In addition, incandescent lamps (unity pf) are, connected as follows: 24 kW from line c to the, neutral, 15 kW from line b to the neutral, and 9 kW, from line c to the neutral., , 12.62 Using Fig. 12.68, design a problem to help other, students better understand how to use PSpice to, analyze three-phase circuits., , a, +, −, VL 120° V, , −, +, , 12.66 A three-phase, four-wire system operating with a, 208-V line voltage is shown in Fig. 12.71. The, source voltages are balanced. The power absorbed, by the resistive wye-connected load is measured by, the three-wattmeter method. Calculate:, , Rline, , Lline, , VL 0° V, Lline, Rline, , A, , R, , B, , C, , b, , N, +, −, c, , (b) Find the magnitude of the current in the neutral, line., , VL −120° V, Rline, , Lline, , (a) If three wattmeters are arranged to measure the, power in each line, calculate the reading of each, meter., , C, , L, , 12.68 Meter readings for a three-phase wye-connected, alternator supplying power to a motor indicate that, , Figure 12.68, For Prob. 12.62., , * An asterisk indicates a challenging problem.
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ale29559_ch12.qxd, , 07/08/2008, , 12:06 PM, , Page 552, , Chapter 12, , 552, , Three-Phase Circuits, 0.6 Ω, , a, , j0.5 Ω, , A, , 0.2 Ω, , 30 Ω, , j1 Ω, , 0.2 Ω, , j1 Ω, , 0.6 Ω, , −, +, , 30 Ω, , j0.5 Ω, B, , b, +, −, , 240 130° V, , −j20 Ω, , +, −, , 240 10° V, , 240 −110° V, , −j20 Ω, , 30 Ω, , j1 Ω, −j20 Ω, , 0.2 Ω, , 0.6 Ω, , j0.5 Ω, , c, , C, , Figure 12.70, For Prob. 12.65., , a, I1, , b, , W1, , c, I2, , Motor load, 260 kVA,, 0.85 pf, lagging, , d, , W2, , 40 Ω, In, , 48 Ω, n, , I3, , 60 Ω, , 24 kW 15 kW 9 kW, Lighting loads, , Figure 12.72, For Prob. 12.67., , W3, , Figure 12.71, For Prob. 12.66., , the line voltages are 330 V, the line currents are 8.4 A,, and the total line power is 4.5 kW. Find:, (a) the load in VA, (b) the load pf, , 240-V line. Assume that the motor load is wyeconnected and that it draws a line current of 6 A., Calculate the pf of the motor and its phase, impedance., 12.71 In Fig. 12.73, two wattmeters are properly connected, to the unbalanced load supplied by a balanced source, such that Vab 208l0 V with positive phase, sequence., , (c) the phase current, , (a) Determine the reading of each wattmeter., , (d) the phase voltage, , (b) Calculate the total apparent power absorbed by, the load., , 12.69 A certain store contains three balanced three-phase, loads. The three loads are:, Load 1: 16 kVA at 0.85 pf lagging, Load 2: 12 kVA at 0.6 pf lagging, Load 3: 8 kW at unity pf, The line voltage at the load is 208 V rms at 60 Hz,, and the line impedance is 0.4 j0.8 . Determine, the line current and the complex power delivered to, the loads., 12.70 The two-wattmeter method gives P1 1200 W and, P2 400 W for a three-phase motor running on a, , a, , A, , W1, 20 Ω, , b, , 0, , B, , 12 Ω, , 10 Ω, c, , Figure 12.73, For Prob. 12.71., , W2, , j5 Ω, , − j10 Ω, C
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ale29559_ch12.qxd, , 07/08/2008, , 12:06 PM, , Page 553, , Comprehensive Problems, , 12.72 If wattmeters W1 and W2 are properly connected, respectively between lines a and b and lines b and c, to measure the power absorbed by the deltaconnected load in Fig. 12.44, predict their readings., , 208 0° V, , +, −, Z, W2, , −, +, ±, , ±, 240 − 60° V +, −, , Z, , ±, , 208 −60° V, , W1, , W1, , ±, , 12.73 For the circuit displayed in Fig. 12.74, find the, wattmeter readings., ±, , 553, , ±, , Z = 60 − j30 Ω, , Figure 12.75, , Z, , For Prob. 12.74., Z = 10 + j30 Ω, , W2, , −, 240 −120° V +, , Z, ±, , ±, , Figure 12.74, For Prob. 12.73., 12.74 Predict the wattmeter readings for the circuit in, Fig. 12.75., , 12.75 A man has a body resistance of 600 . How much, current flows through his ungrounded body:, (a) when he touches the terminals of a 12-V, autobattery?, (b) when he sticks his finger into a 120-V light socket?, 12.76 Show that the I 2R losses will be higher for a 120-V, appliance than for a 240-V appliance if both have the, same power rating., , Comprehensive Problems, 12.77 A three-phase generator supplied 3.6 kVA at a power, factor of 0.85 lagging. If 2500 W are delivered to the, load and line losses are 80 W per phase, what are the, losses in the generator?, 12.78 A three-phase 440-V, 51-kW, 60-kVA inductive load, operates at 60 Hz and is wye-connected. It is desired, to correct the power factor to 0.95 lagging. What, value of capacitor should be placed in parallel with, each load impedance?, 12.79 A balanced three-phase generator has an abc phase, sequence with phase voltage Van 255l0 V. The, generator feeds an induction motor which may be, represented by a balanced Y-connected load with an, impedance of 12 j5 per phase. Find the line, currents and the load voltages. Assume a line, impedance of 2 per phase., 12.80 A balanced three-phase source furnishes power to, the following three loads:, Load 1: 6 kVA at 0.83 pf lagging, Load 2: unknown, Load 3: 8 kW at 0.7071 pf leading, If the line current is 84.6 A rms, the line voltage at, the load is 208 V rms, and the combined load has a, 0.8 pf lagging, determine the unknown load., , 12.81 A professional center is supplied by a balanced, three-phase source. The center has four balanced, three-phase loads as follows:, Load 1: 150 kVA at 0.8 pf leading, Load 2: 100 kW at unity pf, Load 3: 200 kVA at 0.6 pf lagging, Load 4: 80 kW and 95 kVAR (inductive), If the line impedance is 0.02 j0.05 per phase, and the line voltage at the loads is 480 V, find the, magnitude of the line voltage at the source., 12.82 A balanced three-phase system has a distribution, wire with impedance 2 j6 per phase. The, system supplies two three-phase loads that are, connected in parallel. The first is a balanced wyeconnected load that absorbs 400 kVA at a power, factor of 0.8 lagging. The second load is a balanced, delta-connected load with impedance of 10 j8 , per phase. If the magnitude of the line voltage at the, loads is 2400 V rms, calculate the magnitude of the, line voltage at the source and the total complex, power supplied to the two loads., 12.83 A commercially available three-phase inductive, motor operates at a full load of 120 hp (1 hp 746 W), at 95 percent efficiency at a lagging power factor of
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ale29559_ch12.qxd, , 07/08/2008, , 12:06 PM, , Page 554, , Chapter 12, , 554, , Three-Phase Circuits, , 0.707. The motor is connected in parallel to a, 80-kW balanced three-phase heater at unity power, factor. If the magnitude of the line voltage is, 480 V rms, calculate the line current., *12.84 Figure 12.76 displays a three-phase delta-connected, motor load which is connected to a line voltage of, 440 V and draws 4 kVA at a power factor of 72 percent lagging. In addition, a single 1.8 kVAR capacitor, is connected between lines a and b, while a 800-W, lighting load is connected between line c and neutral., Assuming the abc sequence and taking Van Vpl0,, find the magnitude and phase angle of currents, Ia, Ib, Ic, and In., , 12.86 For the single-phase three-wire system in Fig. 12.77,, find currents IaA, IbB, and InN., 1Ω, , a, +, −, , 240 0° V rms, , A, 24 − j2 Ω, , 1Ω, , n, 240 0° V rms +, −, , N, 15 + j4 Ω, , 1Ω, , b, , B, , Figure 12.77, For Prob. 12.86., , Ia, a, Ib, , 1.8 kVAR, , b, Ic, c, In, d, , Motor load, 4 kVA,, pf = 72%, lagging, , 12.87 Consider the single-phase three-wire system shown, in Fig. 12.78. Find the current in the neutral wire and, the complex power supplied by each source. Take Vs, as a 115l0-V, 60-Hz source., 1Ω, , 800 W lighting load, , Figure 12.76, , Vs +, −, , 2Ω, , 20 Ω, , 15 Ω, , For Prob. 12.84., 12.85 Design a three-phase heater with suitable symmetric, loads using wye-connected pure resistance. Assume, that the heater is supplied by a 240-V line voltage, and is to give 27 kW of heat., , Vs +, −, , Figure 12.78, For Prob. 12.87., , 1Ω, , 30 Ω, , 50 mH
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ale29559_ch13.qxd, , 07/14/2008, , 01:00 PM, , Page 555, , c h a p t e r, , Magnetically, Coupled Circuits, , 13, , If you would increase your happiness and prolong your life, forget your, neighbor’s faults . . . . Forget the peculiarities of your friends, and only, remember the good points which make you fond of them . . . . Obliterate, everything disagreeable from yesterday; write upon today’s clean sheet, those things lovely and lovable., —Anonymous, , Enhancing Your Career, Career in Electromagnetics, Electromagnetics is the branch of electrical engineering (or physics), that deals with the analysis and application of electric and magnetic, fields. In electromagnetics, electric circuit analysis is applied at low, frequencies., The principles of electromagnetics (EM) are applied in various allied, disciplines, such as electric machines, electromechanical energy conversion, radar meteorology, remote sensing, satellite communications, bioelectromagnetics, electromagnetic interference and compatibility, plasmas,, and fiber optics. EM devices include electric motors and generators, transformers, electromagnets, magnetic levitation, antennas, radars, microwave, ovens, microwave dishes, superconductors, and electrocardiograms. The, design of these devices requires a thorough knowledge of the laws and, principles of EM., EM is regarded as one of the more difficult disciplines in electrical engineering. One reason is that EM phenomena are rather abstract., But if one enjoys working with mathematics and can visualize the, invisible, one should consider being a specialist in EM, since few electrical engineers specialize in this area. Electrical engineers who specialize in EM are needed in microwave industries, radio/TV broadcasting, stations, electromagnetic research laboratories, and several communications industries., , Telemetry receiving station for space, satellites. © DV169/Getty Images, , 555
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ale29559_ch13.qxd, , 556, , 07/10/2008, , 03:59 PM, , Page 556, , Chapter 13, , Magnetically Coupled Circuits, , Historical, James Clerk Maxwell (1831–1879), a graduate in mathematics from, Cambridge University, in 1865 wrote a most remarkable paper in which, he mathematically unified the laws of Faraday and Ampere. This relationship between the electric field and magnetic field served as the basis, for what was later called electromagnetic fields and waves, a major, field of study in electrical engineering. The Institute of Electrical and, Electronics Engineers (IEEE) uses a graphical representation of this, principle in its logo, in which a straight arrow represents current and, a curved arrow represents the electromagnetic field. This relationship, is commonly known as the right-hand rule. Maxwell was a very active, theoretician and scientist. He is best known for the “Maxwell equations.” The maxwell, a unit of magnetic flux, was named after him., , © Bettmann/Corbis, , 13.1, , © Bettmann/Corbis, , Introduction, , The circuits we have considered so far may be regarded as conductively coupled, because one loop affects the neighboring loop through, current conduction. When two loops with or without contacts between, them affect each other through the magnetic field generated by one of, them, they are said to be magnetically coupled., The transformer is an electrical device designed on the basis of the, concept of magnetic coupling. It uses magnetically coupled coils to, transfer energy from one circuit to another. Transformers are key circuit elements. They are used in power systems for stepping up or stepping down ac voltages or currents. They are used in electronic circuits, such as radio and television receivers for such purposes as impedance, matching, isolating one part of a circuit from another, and again for, stepping up or down ac voltages and currents., We will begin with the concept of mutual inductance and introduce the dot convention used for determining the voltage polarities of, inductively coupled components. Based on the notion of mutual inductance, we then introduce the circuit element known as the transformer., We will consider the linear transformer, the ideal transformer, the ideal, autotransformer, and the three-phase transformer. Finally, among their
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 557, , 13.2, , Mutual Inductance, , 557, , important applications, we look at transformers as isolating and matching devices and their use in power distribution., , 13.2, , Mutual Inductance, , When two inductors (or coils) are in a close proximity to each other,, the magnetic flux caused by current in one coil links with the other, coil, thereby inducing voltage in the latter. This phenomenon is known, as mutual inductance., Let us first consider a single inductor, a coil with N turns. When, current i flows through the coil, a magnetic flux f is produced around, it (Fig. 13.1). According to Faraday’s law, the voltage v induced in the, coil is proportional to the number of turns N and the time rate of change, of the magnetic flux f; that is,, vN, , df, dt, , i(t), , v, −, , (13.1), Figure 13.1, , But the flux f is produced by current i so that any change in f is, caused by a change in the current. Hence, Eq. (13.1) can be written as, vN, , , , +, , df di, di dt, , (13.2), , di, dt, , (13.3), , Magnetic flux produced by a single coil, with N turns., , or, vL, , which is the voltage-current relationship for the inductor. From Eqs. (13.2), and (13.3), the inductance L of the inductor is thus given by, LN, , df, di, , (13.4), , This inductance is commonly called self-inductance, because it relates, the voltage induced in a coil by a time-varying current in the same coil., Now consider two coils with self-inductances L 1 and L 2 that are, in close proximity with each other (Fig. 13.2). Coil 1 has N1 turns,, while coil 2 has N2 turns. For the sake of simplicity, assume that the, second inductor carries no current. The magnetic flux f1 emanating, from coil 1 has two components: one component f11 links only coil 1,, and another component f12 links both coils. Hence,, f1 f11 f12, , df1, dt, , (13.6), , Only flux f12 links coil 2, so the voltage induced in coil 2 is, v2 N2, , df12, dt, , +, i1(t), , (13.5), , Although the two coils are physically separated, they are said to be, magnetically coupled. Since the entire flux f1 links coil 1, the voltage, induced in coil 1 is, v1 N1, , L1, , (13.7), , 11, , L2, , 12, , +, , v1, , v2, , −, , −, N1 turns, , N2 turns, , Figure 13.2, Mutual inductance M21 of coil 2 with, respect to coil 1.
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 559, , 13.2, , Mutual Inductance, , and we refer to M as the mutual inductance between the two coils. Like, self-inductance L, mutual inductance M is measured in henrys (H)., Keep in mind that mutual coupling only exists when the inductors or, coils are in close proximity, and the circuits are driven by time-varying, sources. We recall that inductors act like short circuits to dc., From the two cases in Figs. 13.2 and 13.3, we conclude that, mutual inductance results if a voltage is induced by a time-varying, current in another circuit. It is the property of an inductor to produce, a voltage in reaction to a time-varying current in another inductor near, it. Thus,, Mutual inductance is the ability of one inductor to induce a voltage, across a neighboring inductor, measured in henrys (H)., , Although mutual inductance M is always a positive quantity, the, mutual voltage M didt may be negative or positive, just like the selfinduced voltage L didt. However, unlike the self-induced L didt,, whose polarity is determined by the reference direction of the current, and the reference polarity of the voltage (according to the passive sign, convention), the polarity of mutual voltage M didt is not easy to determine, because four terminals are involved. The choice of the correct, polarity for M didt is made by examining the orientation or particular way in which both coils are physically wound and applying Lenz’s, law in conjunction with the right-hand rule. Since it is inconvenient to, show the construction details of coils on a circuit schematic, we apply, the dot convention in circuit analysis. By this convention, a dot is, placed in the circuit at one end of each of the two magnetically coupled coils to indicate the direction of the magnetic flux if current enters, that dotted terminal of the coil. This is illustrated in Fig. 13.4. Given, a circuit, the dots are already placed beside the coils so that we need, not bother about how to place them. The dots are used along with the, dot convention to determine the polarity of the mutual voltage. The dot, convention is stated as follows:, If a current enters the dotted terminal of one coil, the reference polarity, of the mutual voltage in the second coil is positive at the dotted, terminal of the second coil., , 12, 21, , i1, +, v1, −, , 11, , Coil 1, , Figure 13.4, Illustration of the dot convention., , i2, +, v2, −, , 22, , Coil 2, , 559
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 560, , Chapter 13, , 560, , Magnetically Coupled Circuits, , Alternatively,, , M, i1, , If a current leaves the dotted terminal of one coil, the reference polarity of the mutual voltage in the second coil is negative at the dotted, terminal of the second coil., , +, v2 = M, , di1, dt, , −, , Thus, the reference polarity of the mutual voltage depends on the reference direction of the inducing current and the dots on the coupled, coils. Application of the dot convention is illustrated in the four pairs, of mutually coupled coils in Fig. 13.5. For the coupled coils in, Fig. 13.5(a), the sign of the mutual voltage v2 is determined by the reference polarity for v2 and the direction of i1. Since i1 enters the dotted, terminal of coil 1 and v2 is positive at the dotted terminal of coil 2, the, mutual voltage is M di1dt. For the coils in Fig. 13.5(b), the current, i1 enters the dotted terminal of coil 1 and v2 is negative at the dotted, terminal of coil 2. Hence, the mutual voltage is M di1dt. The same, reasoning applies to the coils in Fig. 13.5(c) and 13.5(d)., Figure 13.6 shows the dot convention for coupled coils in series., For the coils in Fig. 13.6(a), the total inductance is, , (a), M, i1, +, v2 = – M, , di1, dt, , −, (b), M, i2, , L L 1 L 2 2M, , +, di2, v1 = – M, dt, , (Series-aiding connection), , (13.18), , (Series-opposing connection), , (13.19), , For the coils in Fig. 13.6(b),, , −, , L L 1 L 2 2M, , (c), M, , Now that we know how to determine the polarity of the mutual, voltage, we are prepared to analyze circuits involving mutual inductance. As the first example, consider the circuit in Fig. 13.7. Applying, KVL to coil 1 gives, , i2, +, v1 = M, , di2, dt, , v1 i1R1 L1, , −, (d), , di1, di2, M, dt, dt, , (13.20a), , di2, di1, M, dt, dt, , (13.20b), , For coil 2, KVL gives, , Figure 13.5, Examples illustrating how to apply the, dot convention., , v2 i2R2 L2, , We can write Eq. (13.20) in the frequency domain as, V1 (R1 jL1)I1 jM I2, V2 jM I1 (R2 jL2)I2, M, i, , M, , i, L1, , (+), (a), , i, L2, , i, L1, , (−), , L2, , (b), , Figure 13.6, Dot convention for coils in series; the sign indicates the polarity of the mutual voltage:, (a) series-aiding connection, (b) series-opposing connection., , (13.21a), (13.21b)
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , 562, , Page 562, , Chapter 13, , Magnetically Coupled Circuits, , Substituting this in Eq. (13.1.1), we get, ( j2 4 j3)I2 (4 j)I2 12, or, I2 , , 12, 2.91l14.04 A, 4j, , (13.1.3), , From Eqs. (13.1.2) and (13.1.3),, I1 (2 j4)I2 (4.472l63.43)(2.91l14.04), 13.01l49.39 A, , Practice Problem 13.1, , Determine the voltage Vo in the circuit of Fig. 13.10., j1 Ω, , 4Ω, 100 45° V, , +, −, , j8 Ω, , I1, , j5 Ω, , I2, , +, 10 Ω Vo, −, , Figure 13.10, For Practice Prob. 13.1., , Answer: 10l135 V., , Example 13.2, , Calculate the mesh currents in the circuit of Fig. 13.11., 4Ω, , − j3 Ω, , j8 Ω, j2 Ω, , 100 0° V, , +, −, , I1, , j6 Ω, , I2, , 5Ω, , Figure 13.11, For Example 13.2., , Solution:, The key to analyzing a magnetically coupled circuit is knowing, the polarity of the mutual voltage. We need to apply the dot rule. In, Fig. 13.11, suppose coil 1 is the one whose reactance is 6 , and coil, 2 is the one whose reactance is 8 . To figure out the polarity of the, mutual voltage in coil 1 due to current I2, we observe that I2 leaves the, dotted terminal of coil 2. Since we are applying KVL in the clockwise, direction, it implies that the mutual voltage is negative, that is, j2I2., Alternatively, it might be best to figure out the mutual voltage by, redrawing the relevant portion of the circuit, as shown in Fig. 13.12(a),, where it becomes clear that the mutual voltage is V1 2 j I2.
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 564, , Chapter 13, , 564, , 13.3, , Magnetically Coupled Circuits, , Energy in a Coupled Circuit, , In Chapter 6, we saw that the energy stored in an inductor is given by, w, , M, i1, , i2, , +, v1, , +, L1, , −, , L2, , (13.23), , We now want to determine the energy stored in magnetically coupled, coils., Consider the circuit in Fig. 13.14. We assume that currents i1 and i2, are zero initially, so that the energy stored in the coils is zero. If we let i1, increase from zero to I1 while maintaining i2 0, the power in coil 1 is, p1(t) v1i1 i1L1, , v2, −, , 1 2, Li, 2, , di1, dt, , (13.24), , and the energy stored in the circuit is, , Figure 13.14, , w1 , , The circuit for deriving energy stored, in a coupled circuit., , , , p1 dt L1, , , , I1, , 0, , 1, i1 di1 L1I 21, 2, , (13.25), , If we now maintain i1 I1 and increase i2 from zero to I2, the mutual, voltage induced in coil 1 is M12 di2dt, while the mutual voltage induced, in coil 2 is zero, since i1 does not change. The power in the coils is now, p2(t) i1M12, , di2, di2, di2, i2v2 I1M12, i2 L 2, dt, dt, dt, , (13.26), , and the energy stored in the circuit is, w2 , , p dt M I , 2, , I2, , 12 1, , di2 L2, , 0, , M12 I1I2 , , , , I2, , i2 di2, , 0, , 1, L 2 I 22, 2, , (13.27), , The total energy stored in the coils when both i1 and i2 have reached, constant values is, 1, 1, w w1 w2 L1 I 21 L 2 I 22 M12 I1I2, 2, 2, , (13.28), , If we reverse the order by which the currents reach their final values,, that is, if we first increase i2 from zero to I2 and later increase i1 from, zero to I1, the total energy stored in the coils is, 1, 1, w L1 I 21 L 2 I 22 M21I1I2, 2, 2, , (13.29), , Since the total energy stored should be the same regardless of how we, reach the final conditions, comparing Eqs. (13.28) and (13.29) leads us, to conclude that, M12 M21 M, , (13.30a), , and, w, , 1, 1, L1I 21 L 2 I 22 MI1I2, 2, 2, , (13.30b), , This equation was derived based on the assumption that the coil, currents both entered the dotted terminals. If one current enters one
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 565, , 13.3, , Energy in a Coupled Circuit, , dotted terminal while the other current leaves the other dotted terminal, the mutual voltage is negative, so that the mutual energy MI1I2 is, also negative. In that case,, 1, 1, w L1I 21 L 2I 22 MI1I2, 2, 2, , (13.31), , Also, since I1 and I2 are arbitrary values, they may be replaced by i1 and, i2, which gives the instantaneous energy stored in the circuit the general, expression, w, , 1 2, 1, L1i 1 L 2 i 22, 2, 2, , Mi1i 2, , (13.32), , The positive sign is selected for the mutual term if both currents enter, or leave the dotted terminals of the coils; the negative sign is selected, otherwise., We will now establish an upper limit for the mutual inductance M., The energy stored in the circuit cannot be negative because the circuit, is passive. This means that the quantity 12L1i 21 12L 2i 22 Mi1i2, must be greater than or equal to zero:, 1, 1, L1i 21 L 2i 22 Mi1i2 0, 2, 2, , (13.33), , To complete the square, we both add and subtract the term i1i2 1L1L 2, on the right-hand side of Eq. (13.33) and obtain, 1, (i1 1L1 i2 1L 2)2 i1i2(1L1L 2 M) 0, 2, , (13.34), , The squared term is never negative; at its least it is zero. Therefore,, the second term on the right-hand side of Eq. (13.34) must be greater, than zero; that is,, 1L1L 2 M 0, or, M 1L1L 2, , (13.35), , Thus, the mutual inductance cannot be greater than the geometric mean, of the self-inductances of the coils. The extent to which the mutual, inductance M approaches the upper limit is specified by the coefficient, of coupling k, given by, k, , M, 1L1L 2, , (13.36), , or, M k1L1L 2, , (13.37), , where 0 k 1 or equivalently 0 M 1L1L2. The coupling coefficient is the fraction of the total flux emanating from one coil that links, the other coil. For example, in Fig. 13.2,, k, , f12, f12, , f1, f11 f12, , (13.38), , 565
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 566, , Chapter 13, , 566, , Magnetically Coupled Circuits, , and in Fig. 13.3,, , Air or ferrite core, , k, , f21, f21, , f2, f21 f22, , (13.39), , If the entire flux produced by one coil links another coil, then k 1, and we have 100 percent coupling, or the coils are said to be perfectly, coupled. For k 6 0.5, coils are said to be loosely coupled; and for, k 7 0.5, they are said to be tightly coupled. Thus,, The coupling coefficient k is a measure of the magnetic coupling, between two coils; 0 k 1., (a), , (b), , Figure 13.15, Windings: (a) loosely coupled, (b) tightly, coupled; cutaway view demonstrates both, windings., , Example 13.3, , Consider the circuit in Fig. 13.16. Determine the coupling coefficient., Calculate the energy stored in the coupled inductors at time t 1 s if, v 60 cos(4t 30) V., , 2.5 H, 10 Ω, , v, , +, −, , 5H, , 4H, , We expect k to depend on the closeness of the two coils, their, core, their orientation, and their windings. Figure 13.15 shows loosely, coupled windings and tightly coupled windings. The air-core transformers used in radio frequency circuits are loosely coupled, whereas, iron-core transformers used in power systems are tightly coupled. The, linear transformers discussed in Section 3.4 are mostly air-core; the, ideal transformers discussed in Sections 13.5 and 13.6 are principally, iron-core., , 1, 16, , F, , Solution:, The coupling coefficient is, k, , Figure 13.16, For Example 13.3., , 2.5, M, , 0.56, 1L1L 2, 120, , indicating that the inductors are tightly coupled. To find the energy, stored, we need to calculate the current. To find the current, we need, to obtain the frequency-domain equivalent of the circuit., 60 cos(4t 30), 5H, 2.5 H, 4H, , 1, 1, 1, 1, , 60l30, 4 rad/s, jL1 j 20 , jM j10 , jL2 j16 , , 1, F, 16, , 1, , 1, j4 , jC, , The frequency-domain equivalent is shown in Fig. 13.17. We now apply, mesh analysis. For mesh 1,, (10 j20)I1 j10I2 60l30, , (13.3.1), , For mesh 2,, j10I1 ( j16 j4)I2 0, or, I1 1.2I2, , (13.3.2)
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 567, , 13.4, , Linear Transformers, , 567, , Substituting this into Eq. (13.3.1) yields, I2(12 j14) 60l30, , I2 3.254l160.6 A, , 1, , and, I1 1.2I2 3.905l19.4 A, In the time-domain,, i1 3.905 cos(4t 19.4),, , i2 3.254 cos(4t 160.6), , At time t 1 s, 4t 4 rad 229.2, and, i1 3.905 cos(229.2 19.4) 3.389 A, i2 3.254 cos(229.2 160.6) 2.824 A, The total energy stored in the coupled inductors is, 1 2, 1, L1i 1 L 2i 22 Mi1i 2, 2, 2, 1, 1, (5)(3.389)2 (4)(2.824)2 2.5(3.389)(2.824) 20.73 J, 2, 2, , w, , j10 Ω, , 10 Ω, 60 30° V, , +, −, , I1, , j20 Ω, , j16 Ω, , I2, , − j4 Ω, , Figure 13.17, Frequency-domain equivalent of the circuit in Fig. 13.16., , For the circuit in Fig. 13.18, determine the coupling coefficient and the, energy stored in the coupled inductors at t 1.5 s., , 4Ω, , 40 cos 2t V, , +, −, , 1, 8, , F, , 1H, , 2H, , 1H, , 2Ω, , Figure 13.18, For Practice Prob. 13.3., , Answer: 0.7071, 39.4 J., , 13.4, , Linear Transformers, , Here we introduce the transformer as a new circuit element. A transformer is a magnetic device that takes advantage of the phenomenon, of mutual inductance., , Practice Problem 13.3
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 568, , Chapter 13, , 568, , Magnetically Coupled Circuits, , A transformer is generally a four-terminal device comprising two (or, more) magnetically coupled coils., , A linear transformer may also be regarded as one whose flux is proportional to the currents in its windings., , As shown in Fig. 13.19, the coil that is directly connected to the voltage source is called the primary winding. The coil connected to the, load is called the secondary winding. The resistances R1 and R2 are, included to account for the losses (power dissipation) in the coils. The, transformer is said to be linear if the coils are wound on a magnetically linear material—a material for which the magnetic permeability, is constant. Such materials include air, plastic, Bakelite, and wood. In, fact, most materials are magnetically linear. Linear transformers are, sometimes called air-core transformers, although not all of them are, necessarily air-core. They are used in radio and TV sets. Figure 13.20, portrays different types of transformers., , M, R1, , V, , +, −, , I1, , R2, , L1, , Primary coil, , I2, , L2, , Secondary coil, , Figure 13.19, A linear transformer., , (a), , (b), , Figure 13.20, Different types of transformers: (a) copper wound dry power transformer, (b) audio transformers., Courtesy of: (a) Electric Service Co., (b) Jensen Transformers., , ZL
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 569, , 13.4, , Linear Transformers, , 569, , We would like to obtain the input impedance Zin as seen from the, source, because Zin governs the behavior of the primary circuit. Applying KVL to the two meshes in Fig. 13.19 gives, V (R1 jL1)I1 jMI2, 0 jMI1 (R2 jL 2 ZL)I2, , (13.40a), (13.40b), , In Eq. (13.40b), we express I2 in terms of I1 and substitute it into, Eq. (13.40a). We get the input impedance as, Zin , , V, 2M 2, R1 jL1 , I1, R2 jL 2 ZL, , (13.41), , Notice that the input impedance comprises two terms. The first term,, (R1 jL1), is the primary impedance. The second term is due to the, coupling between the primary and secondary windings. It is as though, this impedance is reflected to the primary. Thus, it is known as the, reflected impedance ZR, and, ZR , , 2M 2, R2 jL 2 ZL, , Some authors call this the coupled, impedance., , (13.42), , It should be noted that the result in Eq. (13.41) or (13.42) is not, affected by the location of the dots on the transformer, because the, same result is produced when M is replaced by M., The little bit of experience gained in Sections 13.2 and 13.3 in analyzing magnetically coupled circuits is enough to convince anyone that, analyzing these circuits is not as easy as circuits in previous chapters., For this reason, it is sometimes convenient to replace a magnetically, coupled circuit by an equivalent circuit with no magnetic coupling. We, want to replace the linear transformer in Fig. 13.21 by an equivalent T, or ß circuit, a circuit that would have no mutual inductance., The voltage-current relationships for the primary and secondary, coils give the matrix equation, c, , V1, jL1 jM, I1, d c, d c d, V2, jM jL 2 I2, , M, I1, , I2, , +, V1, , +, L1, , L2, , −, , V2, −, , Figure 13.21, (13.43), , Determining the equivalent circuit of a, linear transformer., , By matrix inversion, this can be written as, L2, c, , I1, j(L1L2 M ), d ≥, M, I2, j(L1L 2 M 2), 2, , M, V1, j(L1L 2 M 2), ¥ c d, L1, V2, 2, j(L1L 2 M ), , (13.44), , Our goal is to match Eqs. (13.43) and (13.44) with the corresponding, equations for the T and ß networks., For the T (or Y) network of Fig. 13.22, mesh analysis provides the, terminal equations as, j(La Lc), I1, jLc, V1, d c d, c d c, V2, jLc, j(Lb Lc) I2, , (13.45), , I1, , La, , Lb, , +, V1, , I2, +, , Lc, , −, , Figure 13.22, An equivalent T circuit., , V2, −
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 570, , Chapter 13, , 570, , Magnetically Coupled Circuits, , If the circuits in Figs. 13.21 and 13.22 are equivalents, Eqs. (13.43), and (13.45) must be identical. Equating terms in the impedance matrices of Eqs. (13.43) and (13.45) leads to, La L1 M,, LC, , I1, , I2, , +, V1, , +, LA, , LB, , −, , Figure 13.23, An equivalent ß circuit., , V2, −, , Lb L 2 M,, , Lc M, , (13.46), , For the ß (or ¢ ) network in Fig. 13.23, nodal analysis gives the, terminal equations as, 1, 1, , I1, jLA, jLC, c d ≥, I2, 1, , jLC, , 1, jLC, V1, ¥ c d, 1, 1, V2, , jLB, jLC, , , (13.47), , Equating terms in admittance matrices of Eqs. (13.44) and (13.47), we, obtain, LA , , L1L 2 M 2, L1L 2 M 2, ,, LB , L2 M, L1 M, 2, L1L 2 M, LC , M, , (13.48), , Note that in Figs. 13.22 and 13.23, the inductors are not magnetically, coupled. Also note that changing the locations of the dots in Fig. 13.21, can cause M to become M . As Example 13.6 illustrates, a negative, value of M is physically unrealizable but the equivalent model is still, mathematically valid., , Example 13.4, , In the circuit of Fig. 13.24, calculate the input impedance and current, I1. Take Z1 60 j100 , Z2 30 j40 , and ZL 80 j60 ., j5 Ω, , Z1, , 50 60° V, , +, −, , I1, , j20 Ω, , Z2, , j40 Ω, , I2, , Figure 13.24, For Example 13.4., , Solution:, From Eq. (13.41),, (5)2, j40 Z2 ZL, 25, 60 j100 j20 , 110 j140, , Zin Z1 j20 , , 60 j80 0.14l51.84, 60.09 j80.11 100.14l53.1 , , ZL
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 571, , 13.4, , Linear Transformers, , 571, , Thus,, I1 , , 50l60, V, , 0.5l113.1 A, Zin, 100.14l53.1, , Find the input impedance of the circuit in Fig. 13.25 and the current, from the voltage source., j3 Ω, , 4Ω, , Practice Problem 13.4, , − j6 Ω, 6Ω, , 20 0° V, , +, −, , j8 Ω, , j10 Ω, j4 Ω, , Figure 13.25, For Practice Prob. 13.4., , Answer: 8.58l58.05 , 2.331l58.05 A., , Determine the T-equivalent circuit of the linear transformer in, Fig. 13.26(a)., , 2H, I1, , I2, , a, , 8H, c, , 10 H, , 4H, , b, , c, 2H, , d, (a), , 2H, , a, , b, , d, (b), , Figure 13.26, For Example 13.5: (a) a linear transformer, (b) its, T-equivalent circuit., , Solution:, Given that L1 10, L 2 4, and M 2, the T-equivalent network has, the following parameters:, La L1 M 10 2 8 H, L b L 2 M 4 2 2 H,, Lc M 2 H, The T-equivalent circuit is shown in Fig. 13.26(b). We have assumed that, reference directions for currents and voltage polarities in the primary and, secondary windings conform to those in Fig. 13.21. Otherwise, we may, need to replace M with M, as Example 13.6 illustrates., , Example 13.5
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 572, , Chapter 13, , 572, , Practice Problem 13.5, , Magnetically Coupled Circuits, , For the linear transformer in Fig. 13.26(a), find the ß equivalent, network., Answer: LA 18 H, LB 4.5 H, LC 18 H., , Example 13.6, , Solve for I1, I2, and Vo in Fig. 13.27 (the same circuit as for Practice, Prob. 13.1) using the T-equivalent circuit for the linear transformer., j1 Ω, , 4Ω, , 60 90° V, , +, −, , I1, , j8 Ω, , j5 Ω, , I2, , +, Vo, −, , 10 Ω, , Figure 13.27, For Example 13.6., , j1 Ω, I1, , I2, , +, V1, , +, j8 Ω, , j5 Ω, , −, , V2, −, , (a), j9 Ω, , j6 Ω, , Solution:, Notice that the circuit in Fig. 13.27 is the same as that in Fig. 13.10, except that the reference direction for current I2 has been reversed, just, to make the reference directions for the currents for the magnetically, coupled coils conform with those in Fig. 13.21., We need to replace the magnetically coupled coils with the Tequivalent circuit. The relevant portion of the circuit in Fig. 13.27 is, shown in Fig. 13.28(a). Comparing Fig. 13.28(a) with Fig. 13.21 shows, that there are two differences. First, due to the current reference, directions and voltage polarities, we need to replace M by M to make, Fig. 13.28(a) conform with Fig. 13.21. Second, the circuit in Fig. 13.21, is in the time-domain, whereas the circuit in Fig. 13.28(a) is in, the frequency-domain. The difference is the factor j; that is, L in, Fig. 13.21 has been replaced with jL and M with jM. Since is, not specified, we can assume 1 rad/s or any other value; it really, does not matter. With these two differences in mind,, La L1 (M ) 8 1 9 H, Lb L 2 (M ) 5 1 6 H,, Lc M 1 H, , − j1 Ω, , (b), , Figure 13.28, For Example 13.6: (a) circuit for coupled, coils of Fig. 13.27, (b) T-equivalent, circuit., , Thus, the T-equivalent circuit for the coupled coils is as shown in, Fig. 13.28(b)., Inserting the T-equivalent circuit in Fig. 13.28(b) to replace the two, coils in Fig. 13.27 gives the equivalent circuit in Fig. 13.29, which can be, solved using nodal or mesh analysis. Applying mesh analysis, we obtain, j6 I1(4 j9 j1) I2(j1), , (13.6.1), , 0 I1(j1) I2(10 j6 j1), , (13.6.2), , (10 j5), I2 (5 j10)I2, j, , (13.6.3), , and, , From Eq. (13.6.2),, I1
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 573, , 13.5, I1, , j6 V, , 4Ω, , +, −, , I1, , j9 Ω, , − j1 Ω, , j6 Ω, , I2, , I2, , +, Vo, −, , Ideal Transformers, , 573, , 10 Ω, , Figure 13.29, For Example 13.6., , Substituting Eq. (13.6.3) into Eq. (13.6.1) gives, j6 (4 j8)(5 j10)I2 j I2 (100 j)I2 100I2, Since 100 is very large compared with 1, the imaginary part of, (100 j) can be ignored so that 100 j 100. Hence,, I2 , , j6, j0.06 0.06l90 A, 100, , From Eq. (13.6.3),, I1 (5 j10) j0.06 0.6 j0.3 A, and, Vo 10I2 j0.6 0.6l90 V, This agrees with the answer to Practice Prob. 13.1. Of course, the, direction of I2 in Fig. 13.10 is opposite to that in Fig. 13.27. This will, not affect Vo, but the value of I2 in this example is the negative of that, of I2 in Practice Prob. 13.1. The advantage of using the T-equivalent, model for the magnetically coupled coils is that in Fig. 13.29 we do, not need to bother with the dot on the coupled coils., , Solve the problem in Example 13.1 (see Fig. 13.9) using the T-equivalent, model for the magnetically coupled coils., Answer: 13l49.4 A, 2.91l14.04 A., , 13.5, , Ideal Transformers, , An ideal transformer is one with perfect coupling (k 1). It consists, of two (or more) coils with a large number of turns wound on a common core of high permeability. Because of this high permeability of, the core, the flux links all the turns of both coils, thereby resulting in, a perfect coupling., To see how an ideal transformer is the limiting case of two coupled, inductors where the inductances approach infinity and the coupling is perfect, let us reexamine the circuit in Fig. 13.14. In the frequency domain,, V1 jL1I1 jM I2, V2 jM I1 jL 2 I2, , (13.49a), (13.49b), , Practice Problem 13.6
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 574, , Chapter 13, , 574, , Magnetically Coupled Circuits, , From Eq. (13.49a), I1 (V1 jM I2)jL1 (we could have also use, this equation to develop the current ratios instead of using the conservation of power which we will do shortly). Substituting this in, Eq. (13.49b) gives, V2 jL 2I2 , , jM 2 I2, M V1, , L1, L1, , But M 1L1L 2 for perfect coupling (k 1). Hence,, V2 jL 2I2 , , jL1L 2I2, 2L1L 2V1, L2, , , V1 nV1, L1, L1, B L1, , where n 1L 2L1 and is called the turns ratio. As L1, L2, M S such, that n remains the same, the coupled coils become an ideal transformer., A transformer is said to be ideal if it has the following properties:, 1. Coils have very large reactances (L1, L2, M S )., 2. Coupling coefficient is equal to unity (k 1)., 3. Primary and secondary coils are lossless (R1 0 R2)., An ideal transformer is a unity-coupled, lossless transformer in which, the primary and secondary coils have infinite self-inductances., , N1, , Iron-core transformers are close approximations to ideal transformers., These are used in power systems and electronics., Figure 13.30(a) shows a typical ideal transformer; the circuit symbol is in Fig. 13.30(b). The vertical lines between the coils indicate an, iron core as distinct from the air core used in linear transformers. The, primary winding has N1 turns; the secondary winding has N2 turns., When a sinusoidal voltage is applied to the primary winding as, shown in Fig. 13.31, the same magnetic flux f goes through both, windings. According to Faraday’s law, the voltage across the primary, winding is, , N2, , (a), , N1, , v1 N1, , N2, , df, dt, , (13.50a), , while that across the secondary winding is, (b), , v2 N2, , Figure 13.30, (a) Ideal transformer, (b) circuit symbol, for ideal transformers., , df, dt, , Dividing Eq. (13.50b) by Eq. (13.50a), we get, N2, v2, , n, v1, N1, , I1, +, V, , +, −, , I2, , 1:n, V1, −, , +, V2, −, , (13.50b), , ZL, , Figure 13.31, Relating primary and secondary quantities, in an ideal transformer., , (13.51), , where n is, again, the turns ratio or transformation ratio. We can use, the phasor voltages V1 and V2 rather than the instantaneous values v1, and v2. Thus, Eq. (13.51) may be written as, V2, N2, , n, V1, N1, , (13.52)
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 575, , 13.5, , Ideal Transformers, , 575, , For the reason of power conservation, the energy supplied to the primary must equal the energy absorbed by the secondary, since there are, no losses in an ideal transformer. This implies that, v1i1 v2i2, , (13.53), , In phasor form, Eq. (13.53) in conjunction with Eq. (13.52) becomes, V2, I1, , n, I2, V1, , (13.54), , showing that the primary and secondary currents are related to the turns, ratio in the inverse manner as the voltages. Thus,, N1, I2, 1, , , n, I1, N2, , I1, , When n 1, we generally call the transformer an isolation transformer. The reason will become obvious in Section 13.9.1. If n 7 1,, we have a step-up transformer, as the voltage is increased from primary to secondary (V2 7 V1). On the other hand, if n 6 1, the transformer is a step-down transformer, since the voltage is decreased from, primary to secondary (V2 6 V1)., , +, V1, −, , The ratings of transformers are usually specified as V1V2. A transformer with rating 2400/120 V should have 2400 V on the primary and, 120 in the secondary (i.e., a step-down transformer). Keep in mind that, the voltage ratings are in rms., Power companies often generate at some convenient voltage and, use a step-up transformer to increase the voltage so that the power can, be transmitted at very high voltage and low current over transmission, lines, resulting in significant cost savings. Near residential consumer, premises, step-down transformers are used to bring the voltage down, to 120 V. Section 13.9.3 will elaborate on this., It is important that we know how to get the proper polarity of the, voltages and the direction of the currents for the transformer in Fig. 13.31., If the polarity of V1 or V2 or the direction of I1 or I2 is changed, n in, Eqs. (13.51) to (13.55) may need to be replaced by n. The two simple rules to follow are:, 1. If V1 and V2 are both positive or both negative at the dotted terminals, use n in Eq. (13.52). Otherwise, use n., 2. If I1 and I2 both enter into or both leave the dotted terminals, use, n in Eq. (13.55). Otherwise, use n., The rules are demonstrated with the four circuits in Fig. 13.32., , +, V2, −, I2, N, = 1, I1, N2, , V2 N2, =, V1 N1, (a), I1, , I2, , N1:N2, +, V1, −, , A step-down transformer is one whose secondary voltage is less than, its primary voltage., , A step-up transformer is one whose secondary voltage is greater than, its primary voltage., , I2, , N1:N2, , (13.55), , +, V2, −, I2, N, =− 1, I1, N2, , V2 N2, =, V1 N1, (b), I1, , I2, , N1:N2, +, V1, −, , +, V2, −, I2, N, = 1, I1, N2, , V2, N, =− 2, V1, N1, (c), I1, , I2, , N1:N2, +, V1, −, , +, V2, −, , V2, N, =− 2, V1, N1, , I2, N, =− 1, I1, N2, (d), , Figure 13.32, Typical circuits illustrating proper voltage, polarities and current directions in an, ideal transformer.
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 576, , 576, , Chapter 13, , Magnetically Coupled Circuits, , Using Eqs. (13.52) and (13.55), we can always express V1 in terms, of V2 and I1 in terms of I2, or vice versa:, V1 , , V2, n, , V2 nV1, , or, , I1 nI2, , I2 , , or, , (13.56), , I1, n, , (13.57), , The complex power in the primary winding is, S1 V1I *1 , , V2, (nI2)* V2I *2 S2, n, , (13.58), , showing that the complex power supplied to the primary is delivered, to the secondary without loss. The transformer absorbs no power. Of, course, we should expect this, since the ideal transformer is lossless., The input impedance as seen by the source in Fig. 13.31 is found from, Eqs. (13.56) and (13.57) as, Zin , , V1, 1 V2, 2, I1, n I2, , (13.59), , It is evident from Fig. 13.31 that V2 I2 ZL, so that, Zin , , Notice that an ideal transformer reflects, an impedance as the square of the, turns ratio., , ZL, , (13.60), , n2, , The input impedance is also called the reflected impedance, since it, appears as if the load impedance is reflected to the primary side. This, ability of the transformer to transform a given impedance into another, impedance provides us a means of impedance matching to ensure, maximum power transfer. The idea of impedance matching is very useful in practice and will be discussed more in Section 13.9.2., In analyzing a circuit containing an ideal transformer, it is common practice to eliminate the transformer by reflecting impedances, and sources from one side of the transformer to the other. In the circuit of Fig. 13.33, suppose we want to reflect the secondary side of, the circuit to the primary side. We find the Thevenin equivalent of, the circuit to the right of the terminals a-b. We obtain VTh as the, open-circuit voltage at terminals a-b, as shown in Fig. 13.34(a)., , Z1, , Vs1, , +, −, , I1, , a, +, V1, −, b, , 1:n, , I2, , c, , +, V2, −, , Z2, , +, −, d, , Figure 13.33, Ideal transformer circuit whose equivalent circuits are to, be found., , Vs2
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 577, , 13.5, , a, +, VTh, −, , I1, , 1:n, +, V1, −, , I2, , Ideal Transformers, , Z2, , a, , +, V2, −, , +, −, , Vs2, , 1 0° V, , 577, , I1, +, V1, −, , +, −, , I2, , 1:n, +, V2, −, , Z2, , b, , b, , (b), , (a), , Figure 13.34, (a) Obtaining VTh for the circuit in Fig. 13.33, (b) obtaining ZTh for the circuit in Fig. 13.33., , Since terminals a-b are open, I1 0 I2 so that V2 Vs2. Hence,, From Eq. (13.56),, VTh V1 , , Vs2, V2, , n, n, , (13.61), , To get ZTh, we remove the voltage source in the secondary winding and, insert a unit source at terminals a-b, as in Fig. 13.34(b). From Eqs. (13.56), and (13.57), I1 nI2 and V1 V2 n, so that, ZTh , , V2 n, V1, Z2, , 2,, I1, nI2, n, , V2 Z2I2, , (13.62), , which is what we should have expected from Eq. (13.60). Once we, have VTh and ZTh, we add the Thevenin equivalent to the part of the, circuit in Fig. 13.33 to the left of terminals a-b. Figure 13.35 shows, the result., , Z2, Z1, , a, , n2, , +, , The general rule for eliminating the transformer and reflecting the secondary circuit to the primary side is: divide the secondary impedance, by n 2, divide the secondary voltage by n, and multiply the secondary, current by n., , Vs1, , +, −, , + Vs2, − n, , V1, −, b, , Figure 13.35, We can also reflect the primary side of the circuit in Fig. 13.33 to, the secondary side. Figure 13.36 shows the equivalent circuit., The rule for eliminating the transformer and reflecting the primary, circuit to the secondary side is: multiply the primary impedance by, n 2, multiply the primary voltage by n, and divide the primary current, by n., , According to Eq. (13.58), the power remains the same, whether calculated on the primary or the secondary side. But realize that this reflection approach only applies if there are no external connections between, the primary and secondary windings. When we have external connections between the primary and secondary windings, we simply use regular mesh and nodal analysis. Examples of circuits where there are, external connections between the primary and secondary windings are, in Figs. 13.39 and 13.40. Also note that if the locations of the dots in, Fig. 13.33 are changed, we might have to replace n by n in order to, obey the dot rule, illustrated in Fig. 13.32., , Equivalent circuit for Fig. 13.33 obtained, by reflecting the secondary circuit to the, primary side., , n 2Z1, , c, , Z2, , +, nVs1, , +, −, , V2, −, , +, −, , Vs2, , d, , Figure 13.36, Equivalent circuit for Fig. 13.33 obtained, by reflecting the primary circuit to the, secondary side.
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , 578, , Example 13.7, , Page 578, , Chapter 13, , Magnetically Coupled Circuits, , An ideal transformer is rated at 2400/120 V, 9.6 kVA, and has 50 turns, on the secondary side. Calculate: (a) the turns ratio, (b) the number of, turns on the primary side, and (c) the current ratings for the primary, and secondary windings., Solution:, (a) This is a step-down transformer, since V1 2,400 V 7 V2 120 V., V2, 120, , 0.05, V1, 2,400, , n, (b), n, , N2, N1, , 0.05 , , 1, , 50, N1, , or, N1 , , 50, 1,000 turns, 0.05, , (c) S V1I1 V2 I2 9.6 kVA. Hence,, I1 , I2 , , Practice Problem 13.7, , 9,600, 9,600, , 4A, V1, 2,400, , 9,600, 9,600, , 80 A, V2, 120, , or, , I2 , , I1, 4, , 80 A, n, 0.05, , The primary current to an ideal transformer rated at 3300/110 V is 5 A., Calculate: (a) the turns ratio, (b) the kVA rating, (c) the secondary current., Answer: (a) 130, (b) 16.5 kVA, (c) 150 A., , Example 13.8, , For the ideal transformer circuit of Fig. 13.37, find: (a) the source current I1, (b) the output voltage Vo, and (c) the complex power supplied, by the source., I1, , 120 0° V rms, , 4Ω, , − j6 Ω, , I2, 1:2, , +, V1, −, , +, −, , +, V2, −, , 20 Ω, , +, Vo, −, , Figure 13.37, For Example 13.8., , Solution:, (a) The 20- impedance can be reflected to the primary side and we get, ZR , , 20, 20, 5, , 2, 4, n
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , 580, , Page 580, , Chapter 13, , Magnetically Coupled Circuits, , secondary sides due to the 30- resistor. We apply mesh analysis. For, mesh 1,, 120 (20 30)I1 30I2 V1 0, or, 50I1 30I2 V1 120, , (13.9.1), , For mesh 2,, V2 (10 30)I2 30I1 0, or, 30I1 40I2 V2 0, , (13.9.2), , At the transformer terminals,, 1, V2 V1, 2, , (13.9.3), , I2 2I1, , (13.9.4), , (Note that n 12.) We now have four equations and four unknowns,, but our goal is to get I2. So we substitute for V1 and I1 in terms of V2, and I2 in Eqs. (13.9.1) and (13.9.2). Equation (13.9.1) becomes, 55I2 2V2 120, , (13.9.5), , and Eq. (13.9.2) becomes, 15I2 40I2 V2 0, , V2 55I2, , 1, , Substituting Eq. (13.9.6) in Eq. (13.9.5),, 165I2 120, , 1, , I2 , , 120, 0.7272 A, 165, , The power absorbed by the 10- resistor is, P (0.7272)2(10) 5.3 W, , Practice Problem 13.9, , Find Vo in the circuit of Fig. 13.40., 8Ω, + V −, o, 4Ω, , 240 0° V, , +, −, , Figure 13.40, For Practice Prob. 13.9., , Answer: 96 V., , 1:2, , 2Ω, 8Ω, , (13.9.6)
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 581, , 13.6, , 13.6, , Ideal Autotransformers, , 581, , Ideal Autotransformers, , Unlike the conventional two-winding transformer we have considered, so far, an autotransformer has a single continuous winding with a connection point called a tap between the primary and secondary sides., The tap is often adjustable so as to provide the desired turns ratio for, stepping up or stepping down the voltage. This way, a variable voltage, is provided to the load connected to the autotransformer., Figure 13.41, An autotransformer is a transformer in which both the primary and the, secondary are in a single winding., , Figure 13.41 shows a typical autotransformer. As shown in, Fig. 13.42, the autotransformer can operate in the step-down or stepup mode. The autotransformer is a type of power transformer. Its, major advantage over the two-winding transformer is its ability to, transfer larger apparent power. Example 13.10 will demonstrate this., Another advantage is that an autotransformer is smaller and lighter, than an equivalent two-winding transformer. However, since both the, primary and secondary windings are one winding, electrical isolation, (no direct electrical connection) is lost. (We will see how the property, of electrical isolation in the conventional transformer is practically, employed in Section 13.9.1.) The lack of electrical isolation between, the primary and secondary windings is a major disadvantage of the, autotransformer., Some of the formulas we derived for ideal transformers apply to, ideal autotransformers as well. For the step-down autotransformer circuit of Fig. 13.42(a), Eq. (13.52) gives, , A typical autotransformer., Courtesy of Todd Systems, Inc., , I1, +, , V, , +, −, , V1, , I2, , N1, N2, , +, V2, −, , −, (a), I2, , +, I1, , N2, N1, , +, , V1, N1 N2, N1, , 1, V2, N2, N2, , V +, −, , (13.63), , ZL, , V2, , ZL, , V1, −, , −, (b), , As an ideal autotransformer, there are no losses, so the complex power, remains the same in the primary and secondary windings:, S1 V1I *1 S 2 V2 I *2, , (13.64), , Equation (13.64) can also be expressed as, V1I1 V2 I2, or, I1, V2, , V1, I2, , (13.65), , Thus, the current relationship is, I1, N2, , I2, N1 N2, For the step-up autotransformer circuit of Fig. 13.42(b),, V2, V1, , N1, N1 N2, , (13.66), , Figure 13.42, (a) Step-down autotransformer, (b) step-up, autotransformer.
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ale29559_ch13.qxd, , 07/10/2008, , 582, , 03:59 PM, , Page 582, , Chapter 13, , Magnetically Coupled Circuits, , or, N1, V1, , V2, N1 N2, , (13.67), , The complex power given by Eq. (13.64) also applies to the step-up, autotransformer so that Eq. (13.65) again applies. Hence, the current, relationship is, N1 N2, N2, I1, , 1, I2, N1, N1, , (13.68), , A major difference between conventional transformers and autotransformers is that the primary and secondary sides of the autotransformer are not only coupled magnetically but also coupled conductively., The autotransformer can be used in place of a conventional transformer, when electrical isolation is not required., , Example 13.10, , Compare the power ratings of the two-winding transformer in, Fig. 13.43(a) and the autotransformer in Fig. 13.43(b)., , 4A, , 0.2 A, , +, Vs = 12 V, −, , 4.2 A, , +, , 4.2 A, +, , +, , +, Vs, −, , 240 V Vp, −, −, , +, , + 0.2 A, , 12 V, , 240 V, , −, , −, , 252 V, +, Vp = 240 V, −, , −, , (b), , (a), , Figure 13.43, For Example 13.10., , Solution:, Although the primary and secondary windings of the autotransformer, are together as a continuous winding, they are separated in Fig. 13.43(b), for clarity. We note that the current and voltage of each winding of the, autotransformer in Fig. 13.43(b) are the same as those for the twowinding transformer in Fig. 13.43(a). This is the basis of comparing, their power ratings., For the two-winding transformer, the power rating is, S1 0.2(240) 48 VA, , or, , S2 4(12) 48 VA, , For the autotransformer, the power rating is, S1 4.2(240) 1,008 VA, , or, , S2 4(252) 1,008 VA, , which is 21 times the power rating of the two-winding transformer.
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 584, , Chapter 13, , 584, , Practice Problem 13.11, I1, , Magnetically Coupled Circuits, , In the autotransformer circuit of Fig. 13.45, find currents I1, I2, and Io., Take V1 1,250 V, V2 500 V., Answer: 12.8 A, 32 A, 19.2 A., , +, I2, V1, , +, V2, Io, , −, , 16 kW load, , −, , 13.7, , Three-Phase Transformers, , To meet the demand for three-phase power transmission, transformer, connections compatible with three-phase operations are needed. We can, achieve the transformer connections in two ways: by connecting three, single-phase transformers, thereby forming a so-called transformer, bank, or by using a special three-phase transformer. For the same kVA, rating, a three-phase transformer is always smaller and cheaper than, three single-phase transformers. When single-phase transformers are, used, one must ensure that they have the same turns ratio n to achieve, a balanced three-phase system. There are four standard ways of connecting three single-phase transformers or a three-phase transformer for, three-phase operations: Y-Y, ¢-¢, Y-¢, and ¢-Y., For any of the four connections, the total apparent power ST, real, power PT, and reactive power QT are obtained as, , Figure 13.45, For Practice Prob. 13.11., , ST 13VL IL, PT ST cos u 13VL IL cos u, QT ST sin u 13VL IL sin u, , (13.69a), (13.69b), (13.69c), , where VL and IL are, respectively, equal to the line voltage VLp and the, line current ILp for the primary side, or the line voltage VLs and the, line current ILs for the secondary side. Notice From Eq. (13.69) that, for each of the four connections, VLs ILs VLp ILp, since power must, be conserved in an ideal transformer., For the Y-Y connection (Fig. 13.46), the line voltage VLp at the primary side, the line voltage VLs on the secondary side, the line current ILp, on the primary side, and the line current ILs on the secondary side are, related to the transformer per phase turns ratio n according to Eqs. (13.52), and (13.55) as, VLs nVLp, (13.70a), ILp, ILs , (13.70b), n, For the ¢ - ¢ connection (Fig. 13.47), Eq. (13.70) also applies for, the line voltages and line currents. This connection is unique in the, ILp, ILs = n, , ILp, 1:n, , ILp, ILs = n, , +, , +, , VL p, , VLs = nVLp, , +, VLp, , +, VLs = nVLp, , −, , −, , −, , −, , ILp, 1:n, , Figure 13.46, , Figure 13.47, , Y-Y three-phase transformer connection., , ¢ - ¢ three-phase transformer connection.
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 585, , 13.7, , Three-Phase Transformers, , 585, , sense that if one of the transformers is removed for repair or maintenance, the other two form an open delta, which can provide threephase voltages at a reduced level of the original three-phase transformer., For the Y- ¢ connection (Fig. 13.48), there is a factor of 13 arising from the line-phase values in addition to the transformer per phase, turns ratio n. Thus,, VLs , ILs , , nVLp, , (13.71a), , 13, , 13ILp, n, , (13.71b), , Similarly, for the ¢ -Y connection (Fig. 13.49),, VLs n13VLp, ILs , , ILp, , (13.72b), , n13, , ILs =, , ILp, , (13.72a), , 3 ILp, n, ILp, , 1:n, ILs =, , +, , +, , VLs =, −, , nVLp, 3, , ILp, , n 3, , 1:n, +, , +, VLp, , VLp, , −, , VLs = n 3 VLp, , −, , −, , Figure 13.48, , Figure 13.49, , Y- ¢ three-phase transformer connection., , ¢ -Y three-phase transformer connection., , The 42-kVA balanced load depicted in Fig. 13.50 is supplied by a threephase transformer. (a) Determine the type of transformer connections., (b) Find the line voltage and current on the primary side. (c) Determine the kVA rating of each transformer used in the transformer bank., Assume that the transformers are ideal., , a, , 1:5, , A, 240 V, , b, , 42 kVA, B Three-phase, load, C, , c, , Figure 13.50, For Example 13.12., , Example 13.12
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 586, , 586, , Chapter 13, , Magnetically Coupled Circuits, , Solution:, (a) A careful observation of Fig. 13.50 shows that the primary side is, Y-connected, while the secondary side is ¢ -connected. Thus, the threephase transformer is Y- ¢, similar to the one shown in Fig. 13.48., (b) Given a load with total apparent power ST 42 kVA, the turns ratio, n 5, and the secondary line voltage VLs 240 V, we can find the, secondary line current using Eq. (13.69a), by, IL s , , ST, 42,000, 101 A, , 13VL s, 13(240), , From Eq. (13.71),, ILp , VLp , , n, 5 101, ILs , 292 A, 13, 13, 13, 13 240, V , 83.14 V, n Ls, 5, , (c) Because the load is balanced, each transformer equally shares the, total load and since there are no losses (assuming ideal transformers),, the kVA rating of each transformer is S ST3 14 kVA. Alternatively, the transformer rating can be determined by the product of the, phase current and phase voltage of the primary or secondary side. For, the primary side, for example, we have a delta connection, so that the, phase voltage is the same as the line voltage of 240 V, while the phase, current is ILp13 58.34 A. Hence, S 240 58.34 14 kVA., , Practice Problem 13.12, , A three-phase ¢ - ¢ transformer is used to step down a line voltage of, 625 kV, to supply a plant operating at a line voltage of 12.5 kV. The, plant draws 40 MW with a lagging power factor of 85 percent. Find:, (a) the current drawn by the plant, (b) the turns ratio, (c) the current, on the primary side of the transformer, and (d) the load carried by each, transformer., Answer: (a) 2.1736 kA, (b) 0.02, (c) 43.47 A, (d) 15.69 MVA., , 13.8, , PSpice Analysis of Magnetically, Coupled Circuits, , PSpice analyzes magnetically coupled circuits just like inductor circuits, except that the dot convention must be followed. In PSpice Schematic,, the dot (not shown) is always next to pin 1, which is the left-hand terminal of the inductor when the inductor with part name L is placed, (horizontally) without rotation on a schematic. Thus, the dot or pin 1, will be at the bottom after one 90 counterclockwise rotation, since, rotation is always about pin 1. Once the magnetically coupled inductors are arranged with the dot convention in mind and their value
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , 13.8, , Page 587, , PSpice Analysis of Magnetically Coupled Circuits, , 587, , attributes are set in henries, we use the coupling symbol K_LINEAR, to define the coupling. For each pair of coupled inductors, take the following steps:, 1. Select Draw/Get New Part and type K_LINEAR., 2. Hit <enter> or click OK and place the K_LINEAR symbol on the, schematic, as shown in Fig. 13.51. (Notice that K_LINEAR is not, a component and therefore has no pins.), 3. DCLICKL on COUPLING and set the value of the coupling coefficient k., 4. DCLICKL on the boxed K (the coupling symbol) and enter the, reference designator names for the coupled inductors as values, of Li, i 1, 2, . . . , 6. For example, if inductors L20 and L23, are coupled, we set L1 L20 and L2 L23. L1 and at least, one other Li must be assigned values; other Li’s may be left, blank., In step 4, up to six coupled inductors with equal coupling can be, specified., For the air-core transformer, the partname is XFRM_LINEAR., It can be inserted in a circuit by selecting Draw/Get Part Name and, then typing in the part name or by selecting the part name from the, analog.slb library. As shown typically in Fig. 13.52(a), the main, attributes of the linear transformer are the coupling coefficient k and, the inductance values L1 and L2 in henries. If the mutual inductance, M is specified, its value must be used along with L1 and L2 to, calculate k. Keep in mind that the value of k should lie between 0, and 1., For the ideal transformer, the part name is XFRM NONLINEAR, and is located in the breakout.slb library. Select it by clicking, Draw/Get Part Name and then typing in the part name. Its attributes, are the coupling coefficient and the numbers of turns associated with, L1 and L2, as illustrated typically in Fig. 13.52(b). The value of the, coefficient of mutual coupling k 1., PSpice has some additional transformer configurations that we will, not discuss here., , Use PSpice to find i1, i2, and i3 in the circuit displayed in Fig. 13.53., i2, , 70 Ω, , 2H, 1H, , 100 Ω, , 3H, , 2H, , i1, 60 cos (12t – 10°) V, , +, −, , 3H, , 1.5 H, , 4H, 270 F, , Figure 13.53, For Example 13.13., , i3, +, −, , 40 cos 12t V, , K K1, K_Linear, COUPLING = 1, , Figure 13.51, K_Linear for defining coupling., , TX2, , COUPLING = 0.5, L1_VALUE = 1mH, L2_VALUE = 25mH, (a), TX4, , kbreak, COUPLING = 0.5, L1_TURNS = 500, L2_TURNS = 1000, (b), , Figure 13.52, (a) Linear transformer, XFRM_LINEAR, (b) ideal transformer, XFRM_NONLINEAR., , Example 13.13
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ale29559_ch13.qxd, , 07/21/2008, , 11:15 AM, , 590, , Page 590, , Chapter 13, , Magnetically Coupled Circuits, , Solution:, 1. Define. The problem is clearly defined and we can proceed to, the next step., 2. Present. We have an ideal transformer and we are to find the, input and the output voltages for that transformer. In addition,, we are to use PSpice to solve for the voltages., 3. Alternative. We are required to use PSpice. We can use mesh, analysis to perform a check., 4. Attempt. As usual, we assume 1 and find the corresponding, values of capacitance and inductance of the elements:, j10 jL, 1, j40 , jC, Reminder: For an ideal transformer, the, inductances of both the primary and, secondary windings are infinitely large., , 1, , L 10 H, , 1, , C 25 mF, , Figure 13.57 shows the schematic. For the ideal transformer, we set the, coupling factor to 0.99999 and the numbers of turns to 400,000 and, 100,000. The two VPRINT2 pseudocomponents are connected across, the transformer terminals to obtain V1 and V2. As a single-frequency, analysis, we select Analysis/Setup/AC Sweep and enter Total Pts 1,, Start Freq 0.1592, and Final Freq 0.1592. After saving the, schematic, we select Analysis/Simulate to simulate it. The output file, includes:, FREQ, VM($N_0003,$N_0006) VP($N_0003,$N_0006), 1.592E-01 9.112E+01, 3.792E+01, FREQ, VM($N_0006,$N_0005) VP($N_0006,$N_0005), 1.592E-01 2.278E+01, -1.421E+02, This can be written as, V1 91.12l37.92 V, , and, , V2 22.78l142.1 V, , 5. Evaluate. We can check the answer by using mesh analysis as, follows:, Loop 1, , 120l30 (80 j40)I1 V1 20(I1 I2 ) 0, , Loop 2, , 20(I1 I2) V2 (6 j10)I2 0, , C1, , R1, 80, , 0.025, AC = yes, MAG = yes, PHASE = yes, , COUPLING = 0.99999, L1_TURNS = 400000, L2_TURNS = 100000 AC = yes, MAG = yes, TX2, PHASE = yes, , R3, , 6, , L1, , 10, , kbreak, , V1, + ACMAG = 120V, − ACPHASE = 30, R2, , 20, , Figure 13.57, The schematic for the circuit in Fig. 13.56.
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 592, , Chapter 13, , 592, , For more information on the many, kinds of transformers, a good text is, W. M. Flanagan, Handbook of Transformer Design and Applications,, 2nd ed. (New York: McGraw-Hill, 1993)., , Magnetically Coupled Circuits, , Because of these diverse uses, there are many special designs for, transformers (only some of which are discussed in this chapter): voltage transformers, current transformers, power transformers, distribution, transformers, impedance-matching transformers, audio transformers,, single-phase transformers, three-phase transformers, rectifier transformers, inverter transformers, and more. In this section, we consider, three important applications: transformer as an isolation device, transformer as a matching device, and power distribution system., , 13.9.1 Transformer as an Isolation Device, , Fuse, 1:n, va, , +, −, , Rectifier, , Isolation transformer, , Figure 13.59, A transformer used to isolate an ac supply, from a rectifier., , Electrical isolation is said to exist between two devices when there is, no physical connection between them. In a transformer, energy is transferred by magnetic coupling, without electrical connection between the, primary circuit and secondary circuit. We now consider three simple, practical examples of how we take advantage of this property., First, consider the circuit in Fig. 13.59. A rectifier is an electronic, circuit that converts an ac supply to a dc supply. A transformer is often, used to couple the ac supply to the rectifier. The transformer serves two, purposes. First, it steps up or steps down the voltage. Second, it provides, electrical isolation between the ac power supply and the rectifier, thereby, reducing the risk of shock hazard in handling the electronic device., As a second example, a transformer is often used to couple two, stages of an amplifier, to prevent any dc voltage in one stage from, affecting the dc bias of the next stage. Biasing is the application of a, dc voltage to a transistor amplifier or any other electronic device in, order to produce a desired mode of operation. Each amplifier stage is, biased separately to operate in a particular mode; the desired mode of, operation will be compromised without a transformer providing dc isolation. As shown in Fig. 13.60, only the ac signal is coupled through, the transformer from one stage to the next. We recall that magnetic, coupling does not exist with a dc voltage source. Transformers are used, in radio and TV receivers to couple stages of high-frequency amplifiers. When the sole purpose of a transformer is to provide isolation,, its turns ratio n is made unity. Thus, an isolation transformer has n 1., As a third example, consider measuring the voltage across 13.2-kV, lines. It is obviously not safe to connect a voltmeter directly to, such high-voltage lines. A transformer can be used both to electrically, isolate the line power from the voltmeter and to step down the voltage, to a safe level, as shown in Fig. 13.61. Once the voltmeter is used to, , 1:1, Power lines, Amplifier, stage 1, , ac + dc, , ac only, , Amplifier, stage 2, , +, 13,200 V, –, n:1, +, , Isolation transformer, , 120 V, , Figure 13.60, A transformer providing dc isolation between two, amplifier stages., , V Voltmeter, , –, , Figure 13.61, A transformer providing isolation between the, power lines and the voltmeter.
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 593, , 13.9, , Applications, , 593, , measure the secondary voltage, the turns ratio is used to determine the, line voltage on the primary side., , Example 13.15, , Determine the voltage across the load in Fig. 13.62., Solution:, We can apply the superposition principle to find the load voltage. Let, vL vL1 vL2, where vL1 is due to the dc source and vL2 is due to, the ac source. We consider the dc and ac sources separately, as shown, in Fig. 13.63. The load voltage due to the dc source is zero, because a, time-varying voltage is necessary in the primary circuit to induce a, voltage in the secondary circuit. Thus, vL1 0. For the ac source and, a value of Rs so small it can be neglected,, V2, V2, 1, , , V1, 120, 3, , or, , V2 , , Rs, , 3:1, , 120 V +, −, ac, , RL = 5 kΩ, , 12 V +, dc −, , Figure 13.62, For Example 13.15., , 120, 40 V, 3, , Hence, VL2 40 V ac or vL2 40 cos t; that is, only the ac voltage, is passed to the load by the transformer. This example shows how the, transformer provides dc isolation., Rs, , 3:1, , 3:1, +, V2 = 0, −, , 12 V +, dc −, , RL, , 120 V +, −, ac, , (a), , +, V1, −, , +, V2, −, , RL, , (b), , Figure 13.63, For Example 13.15: (a) dc source, (b) ac source., , Refer to Fig. 13.61. Calculate the turns ratio required to step down the, 13.2-kV line voltage to a safe level of 120 V., , Practice Problem 13.15, , Answer: 110., , 13.9.2 Transformer as a Matching Device, We recall that for maximum power transfer, the load resistance RL, must be matched with the source resistance Rs. In most cases, the, two resistances are not matched; both are fixed and cannot be altered., However, an iron-core transformer can be used to match the load, resistance to the source resistance. This is called impedance matching. For example, to connect a loudspeaker to an audio power amplifier requires a transformer, because the speaker’s resistance is only, a few ohms while the internal resistance of the amplifier is several, thousand ohms., Consider the circuit shown in Fig. 13.64. We recall from Eq. (13.60), that the ideal transformer reflects its load back to the primary with a, , Rs, , vs, , 1:n, , +, −, Source, , RL, , Matching transformer, , Load, , Figure 13.64, Transformer used as a matching device.
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 594, , Chapter 13, , 594, , Magnetically Coupled Circuits, , scaling factor of n2. To match this reflected load RL n2 with the source, resistance Rs, we set them equal,, Rs , , RL, , (13.73), , n2, , Equation (13.73) can be satisfied by proper selection of the turns ratio n., From Eq. (13.73), we notice that a step-down transformer (n 6 1) is, needed as the matching device when Rs 7 RL, and a step-up (n 7 1), is required when Rs 6 RL., , Example 13.16, , The ideal transformer in Fig. 13.65 is used to match the amplifier circuit, to the loudspeaker to achieve maximum power transfer. The Thevenin, (or output) impedance of the amplifier is 192 , and the internal impedance of the speaker is 12 . Determine the required turns ratio., , 1:n, Amplifier, circuit, Speaker, , Figure 13.65, Using an ideal transformer to match, the speaker to the amplifier; for, Example 13.16., , Solution:, We replace the amplifier circuit with the Thevenin equivalent and, reflect the impedance ZL 12 of the speaker to the primary side of, the ideal transformer. Figure 13.66 shows the result. For maximum, power transfer,, Z Th , , Z Th, , VTh, , +, −, , ZL, n2, , ZL, , n2 , , or, , n2, , ZL, 12, 1, , , Z Th, 192, 16, , Thus, the turns ratio is n 14 0.25., Using P I 2R, we can show that indeed the power delivered to the, speaker is much larger than without the ideal transformer. Without, the ideal transformer, the amplifier is directly connected to the speaker., The power delivered to the speaker is, PL a, , Figure 13.66, Equivalent circuit of the circuit in, Fig. 13.65; for Example 13.16., , 2, VTh, b Z L 288 V 2Th mW, Z Th Z L, , With the transformer in place, the primary and secondary currents are, Ip , , VTh, Z Th Z L n, , 2, , Is , , ,, , Ip, n, , Hence,, VTh n, , 2, , PL I 2s Z L a, , Z Th Z L n, , a, , n Z Th Z L, , nVTh, 2, , b ZL, 2, 2, , b Z L 1,302 V 2Th mW, , confirming what was said earlier., , Practice Problem 13.16, , Calculate the turns ratio of an ideal transformer required to match a, 400- load to a source with internal impedance of 2.5 k. Find the, load voltage when the source voltage is 30 V., Answer: 0.4, 6 V.
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 595, , 13.9, , Applications, , 595, , 13.9.3 Power Distribution, A power system basically consists of three components: generation,, transmission, and distribution. The local electric company operates a, plant that generates several hundreds of megavolt-amperes (MVA), typically at about 18 kV. As Fig. 13.67 illustrates, three-phase step-up, transformers are used to feed the generated power to the transmission, line. Why do we need the transformer? Suppose we need to transmit, 100,000 VA over a distance of 50 km. Since S VI, using a line voltage of 1,000 V implies that the transmission line must carry 100 A and, this requires a transmission line of a large diameter. If, on the other, hand, we use a line voltage of 10,000 V, the current is only 10 A. The, smaller current reduces the required conductor size, producing considerable savings as well as minimizing transmission line I 2R losses. To, minimize losses requires a step-up transformer. Without the transformer, the majority of the power generated would be lost on the transmission line. The ability of the transformer to step up or step down, voltage and distribute power economically is one of the major reasons, for generating ac rather than dc. Thus, for a given power, the larger the, voltage, the better. Today, 1 MV is the largest voltage in use; the level, may increase as a result of research and experiments., , Insulators, , 3, 345,000 V, , Neutral, Tower, , Neutral, , 345,000 V, , Tower, 345,000 V, , Neutral, , 3, Step-up, transformer, , 3 60 Hz ac, 18,000 V, Generator, , Neutral, 3 60 Hz ac, 208 V, , 3, Step-down, transformer, , Figure 13.67, A typical power distribution system., A. Marcus and C. M. Thomson, Electricity for Technicians, 2nd ed. [Englewood Cliffs,, NJ: Prentice Hall, 1975], p. 337., , Beyond the generation plant, the power is transmitted for hundreds, of miles through an electric network called the power grid. The threephase power in the power grid is conveyed by transmission lines hung, overhead from steel towers which come in a variety of sizes and shapes., The (aluminum-conductor, steel-reinforced) lines typically have overall diameters up to about 40 mm and can carry current of up to 1,380 A., At the substations, distribution transformers are used to step down, the voltage. The step-down process is usually carried out in stages., Power may be distributed throughout a locality by means of either, overhead or underground cables. The substations distribute the power, to residential, commercial, and industrial customers. At the receiving, end, a residential customer is eventually supplied with 120/240 V, while, industrial or commercial customers are fed with higher voltages such, , One may ask, How would increasing, the voltage not increase the current,, thereby increasing I 2R losses? Keep in, mind that I V/R, where V/ is the potential difference between the sending, and receiving ends of the line. The, voltage that is stepped up is the sending end voltage V, not V/. If the receiving end is VR , then V/ V VR. Since, V and VR are close to each other, V/ is, small even when V is stepped up.
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 596, , 596, , Chapter 13, , Magnetically Coupled Circuits, , as 460/208 V. Residential customers are usually supplied by distribution transformers often mounted on the poles of the electric utility company. When direct current is needed, the alternating current is converted, to dc electronically., , Example 13.17, , A distribution transformer is used to supply a household as in Fig. 13.68., The load consists of eight 100-W bulbs, a 350-W TV, and a 15-kW, kitchen range. If the secondary side of the transformer has 72 turns,, calculate: (a) the number of turns of the primary winding, and (b) the, current Ip in the primary winding., Ip, , +, 2400 V, −, , +, 120 V, −, –, 120 V, +, , TV, Kitchen, range, , 8 bulbs, , Figure 13.68, For Example 13.17., , Solution:, (a) The dot locations on the winding are not important, since we are, only interested in the magnitudes of the variables involved. Since, Np, Ns, , , , Vp, Vs, , we get, Np Ns, , Vp, Vs, , 72, , 2,400, 720 turns, 240, , (b) The total power absorbed by the load is, S 8 100 350 15,000 16.15 kW, But S Vp I p Vs Is, so that, Ip , , Practice Problem 13.17, , S, 16,150, 6.729 A, , Vp, 2,400, , In Example 13.17, if the eight 100-W bulbs are replaced by twelve, 60-W bulbs and the kitchen range is replaced by a 4.5-kW airconditioner, find: (a) the total power supplied, (b) the current Ip in the, primary winding., Answer: (a) 5.57 kW, (b) 2.321 A.
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 597, , 13.10, , 13.10, , Summary, , Summary, , 1. Two coils are said to be mutually coupled if the magnetic flux f, emanating from one passes through the other. The mutual inductance between the two coils is given by, M k1L1L2, where k is the coupling coefficient, 0 6 k 6 1., 2. If v1 and i1 are the voltage and current in coil 1, while v2 and i2, are the voltage and current in coil 2, then, v1 L 1, , di1, di2, M, dt, dt, , v2 L 2, , and, , di2, di1, M, dt, dt, , Thus, the voltage induced in a coupled coil consists of self-induced, voltage and mutual voltage., 3. The polarity of the mutually-induced voltage is expressed in the, schematic by the dot convention., 4. The energy stored in two coupled coils is, 1, 1 2, L1i1 L2i22, 2, 2, , Mi1i2, , 5. A transformer is a four-terminal device containing two or more, magnetically coupled coils. It is used in changing the current, voltage, or impedance level in a circuit., 6. A linear (or loosely coupled) transformer has its coils wound on a, magnetically linear material. It can be replaced by an equivalent T, or ß network for the purposes of analysis., 7. An ideal (or iron-core) transformer is a lossless (R1 R2 0), transformer with unity coupling coefficient (k 1) and infinite, inductances (L1, L2, M S )., 8. For an ideal transformer,, V2 nV1,, , 9., 10., 11., , 12., , I2 , , I1, ,, n, , S1 S2,, , ZR , , ZL, n2, , where n N2 N1 is the turns ratio. N1 is the number of turns of, the primary winding and N2 is the number of turns of the secondary winding. The transformer steps up the primary voltage when, n 7 1, steps it down when n 6 1, or serves as a matching device, when n 1., An autotransformer is a transformer with a single winding common to both the primary and the secondary circuits., PSpice is a useful tool for analyzing magnetically coupled, circuits., Transformers are necessary in all stages of power distribution systems. Three-phase voltages may be stepped up or down by threephase transformers., Important uses of transformers in electronics applications are as, electrical isolation devices and impedance-matching devices., , 597
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 598, , Chapter 13, , 598, , Magnetically Coupled Circuits, , Review Questions, 13.1, , Refer to the two magnetically coupled coils of, Fig. 13.69(a). The polarity of the mutual voltage is:, (a) Positive, , 13.6, , (b) Negative, , (a) 10, , 13.7, , M, , M, i2, , i1, , For the ideal transformer in Fig. 13.70(b),, N2 N1 10. The ratio I2 I1 is:, , i2, , i1, , 2V, , For Review Questions 13.1 and 13.2., , +, 50 V, −, , For the two magnetically coupled coils of, Fig. 13.69(b), the polarity of the mutual voltage is:, (a) Positive, , (d) 10, , (b) 0.75, (d) 5.333, , 8V, , (b), , (a), , 13.8, , If the three-winding transformer is connected as in, Fig. 13.71(b), the value of the output voltage Vo is:, (a) 10, , A transformer is used in stepping down or stepping up:, (a) dc voltages, , +, Vo, −, , For Review Questions 13.7 and 13.8., , The coefficient of coupling for two coils having, L1 2 H, L2 8 H, M 3 H is:, (c) 1.333, , −, , 2V, +, 50 V, −, , Figure 13.71, , (b) Negative, , (a) 0.1875, , +, Vo, , 8V, , 13.4, , (c) 6, , (b) 6, , (b), , (a), , Figure 13.69, , 13.3, , (d) 10, , A three-winding transformer is connected as, portrayed in Fig. 13.71(a). The value of the output, voltage Vo is:, (a) 10, , 13.2, , (c) 0.1, , (b) 0.1, , (b) ac voltages, , (c) both dc and ac voltages, , 13.9, , (b) 6, , (c) 6, , (d) 10, , In order to match a source with internal impedance, of 500 to a 15- load, what is needed is:, (a) step-up linear transformer, (b) step-down linear transformer, , 13.5, , (c) step-up ideal transformer, , The ideal transformer in Fig. 13.70(a) has, N2N1 10. The ratio V2V1 is:, (a) 10, , I1, , (b) 0.1, , I2, , N1 : N2, +, , (c) 0.1, , I1, , (a), , Figure 13.70, For Review Questions 13.5 and 13.6., , (e) autotransformer, , (d) 10, , N1 : N2, , +, V2, −, , V1, −, , (d) step-down ideal transformer, , I2, , 13.10 Which of these transformers can be used as an, isolation device?, (a) linear transformer, , (b) ideal transformer, , (c) autotransformer, , (d) all of the above, , (b), , Answers: 13.1b, 13.2a, 13.3b, 13.4b, 13.5d, 13.6b,, 13.7c, 13.8a, 13.9d, 13.10b.
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 599, , Problems, , 599, , Problems1, Section 13.2 Mutual Inductance, 13.1, , 13.5, , For the three coupled coils in Fig. 13.72, calculate, the total inductance., , (a) the two coils are connected in series, , 2H, 4H, , 6H, , Two coils are mutually coupled, with L1 25 mH,, L2 60 mH, and k 0.5. Calculate the maximum, possible equivalent inductance if:, (b) the coils are connected in parallel, , 5H, , 13.6, , 10 H, , 8H, , The coils in Fig. 13.75 have L1 40 mH,, L2 5 mH, and coupling coefficient k 0.6. Find, i1(t) and v2(t), given that v1(t) 10 cos t and i2(t) , 2 sin t, 2000 rad/s., , Figure 13.72, , M, , For Prob. 13.1., 13.2, , i1, , Using Fig. 13.73, design a problem to help other, students better understand mutual inductance., , +, , L1, , +, L1, , v1, , Figure 13.75, , M23, , L2, , For Prob. 13.6., 13.7 For the circuit in Fig. 13.76, find Vo., L3, , j1 Ω, , 2Ω, , For Prob. 13.2., , 13.4, , v2, −, , Figure 13.73, 13.3, , L2, , −, , M13, M12, , i2, , Two coils connected in series-aiding fashion have a, total inductance of 250 mH. When connected in a, series-opposing configuration, the coils have a total, inductance of 150 mH. If the inductance of one coil, (L 1) is three times the other, find L 1, L 2, and M., What is the coupling coefficient?, , 12 0°, , +, −, , 1Ω, , j6 Ω, , j4 Ω, , j1 Ω, , 1Ω, , +, Vo, −, , 2Ω, , +, v(t), −, , Figure 13.76, For Prob. 13.7., 13.8, , Find v(t) for the circuit in Fig. 13.77., , (a) For the coupled coils in Fig. 13.74(a), show that, 1H, , Leq L 1 L 2 2M, , 4Ω, , (b) For the coupled coils in Fig. 13.74(b), show that, L 1L 2 M 2, Leq , L1 L 2 2M, , +, −, , 100 cos 4t, , 2H, , 1H, , Figure 13.77, L1, , For Prob. 13.8., 13.9, , M, M, , L2, , L1, , Find Vx in the network shown in Fig. 13.78., , L2, , 2Ω, , j1 Ω, , 2Ω, + V −, x, , Leq, , Leq, (a), , 16 30° V, , +, −, , j4 Ω, , (b), , Figure 13.74, For Prob. 13.4., , Figure 13.78, For Prob. 13.9., , 1, , Remember, unless otherwise specified, assume all values of currents and voltages are rms., , j4 Ω, , − j1 Ω, , 4 0° A
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 602, , Chapter 13, , 602, , Magnetically Coupled Circuits, , Section 13.4 Linear Transformers, , 13.25 For the network in Fig. 13.94, find Zab and Io., , io, , 13.29 In the circuit of Fig. 13.98, find the value of the, coupling coefficient k that will make the 10-, resistor dissipate 320 W. For this value of k, find the, energy stored in the coupled coils at t 1.5 s., , k = 0.5, 4Ω, , 1Ω, , a, , 3Ω, , 0.5 F, 12 sin 2t V, , +, −, , 2Ω, , 1H, , 1H, , 2H, , k, 10 Ω, , b, , +, −, , 165 cos 10 3t V, , Figure 13.94, , 30 mH, , 20 Ω, , 50 mH, , For Prob. 13.25., , Figure 13.98, For Prob. 13.29., , 13.26 Find Io in the circuit of Fig. 13.95. Switch the dot on, the winding on the right and calculate Io again., − j30 Ω, , 10 60° A, , 50 Ω, , 13.30 (a) Find the input impedance of the circuit in, Fig. 13.99 using the concept of reflected, impedance., , k = 0.601, Io, , j20 Ω, , j40 Ω, , (b) Obtain the input impedance by replacing the, linear transformer by its T equivalent., , 10 Ω, , j40 Ω, , Figure 13.95, , j10 Ω, , 25 Ω, , 8Ω, , For Prob. 13.26., j20 Ω, , j30 Ω, , − j6 Ω, , 13.27 Find the average power delivered to the 50-, resistor in the circuit of Fig. 13.96., Zin, 10 Ω, , Figure 13.99, For Prob. 13.30., , 0.5 H, 8Ω, , 120 cos 20t V, , +, −, , 1H, , 13.31 Using Fig. 13.100, design a problem to help other, students better understand linear transformers and, 50 Ω, how to find T-equivalent and ß -equivalent circuits., , 2H, , Figure 13.96, M, , For Prob. 13.27., *13.28 In the circuit of Fig. 13.97, find the value of X, that will give maximum power transfer to the, 20- load., , 8Ω, , Vs +, −, , Figure 13.97, For Prob. 13.28., , – jX, , j12 Ω, , L1, , L2, , Figure 13.100, , j10 Ω, , For Prob. 13.31., , j15 Ω, , 20 Ω, , *13.32 Two linear transformers are cascaded as shown in, Fig. 13.101. Show that, , Zin , , 2R(L 2a L a L b M 2a ), j3(L 2a L b L a L 2b L a M 2b L b M 2a), 2(L a L b L 2b M 2b) jR(L a L b)
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 604, , Chapter 13, , 604, , Magnetically Coupled Circuits, , 13.40 The primary of an ideal transformer with a turns, ratio of 5 is connected to a voltage source with, Thevenin parameters vTh 10 cos 2000t V and, RTh 100 . Determine the average power, delivered to a 200- load connected across the, secondary winding., , *13.44 In the ideal transformer circuit of Fig. 13.109, find, i1(t) and i2(t)., , R, , 13.41 Determine I1 and I2 in the circuit of Fig. 13.106., , I1, , 10 Ω, , 3:1, , I2, , 2Ω, , i1(t), , i2(t), 1:n, , Vo, dc, , + V cos t, m, −, , Figure 13.109, For Prob. 13.44., , 220 0° V, , +, −, , 13.45 For the circuit shown in Fig. 13.110, find the, value of the average power absorbed by the 8-, resistor., , Figure 13.106, For Prob. 13.41., , 13.42 For the circuit in Fig. 13.107, determine the power, absorbed by the 2- resistor. Assume the 80 V is an, rms value., , 48 Ω, , 160 sin (30t) V, 50 Ω, , – j1 Ω, , j20, , 1:2, , 1, 120, , 3:1, , F, +, 8Ω, –, , +, −, , Figure 13.110, For Prob. 13.45., , 120 0° V +, −, , 2Ω, , 13.46 (a) Find I1 and I2 in the circuit of Fig. 13.111 below., Ideal, , (b) Switch the dot on one of the windings. Find I1, and I2 again., , Figure 13.107, For Prob. 13.42., , 13.47 Find v(t) for the circuit in Fig. 13.112., , 13.43 Obtain V1 and V2 in the ideal transformer circuit of, Fig. 13.108., 1, 3, , 2Ω, , 1:4, +, 2 0° A, , 10 Ω, , V1, −, , +, V2 12 Ω, −, , 1 0° A, , 120 cos 3t, , +, −, , Figure 13.108, , Figure 13.112, , For Prob. 13.43., , For Prob. 13.47., , I1, , 160 60° V, , +, −, , Figure 13.111, For Prob. 13.46., , j16 Ω, , 10 Ω, , 1:2, , 12 Ω, , – j8 Ω, , F, 1: 4, , 1Ω, 5Ω, , I2, , + 100 30°, −, , +, v(t), −
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 605, , Problems, , 13.48 Using Fig. 13.113, design a problem to help other, students better understand how ideal transformers, work., R1, , 605, , 13.52 For the circuit in Fig. 13.117, determine the turns, ratio n that will cause maximum average power, transfer to the load. Calculate that maximum, average power., , R2, , n:1, , 40 Ω, Vs, , 1:n, , jXL, , +, −, Ix, , 120 0° V rms, , –jXC, , +, −, , 10 Ω, , Figure 13.113, Figure 13.117, , For Prob. 13.48., , For Prob. 13.52., 13.49 Find current ix in the ideal transformer circuit shown, in Fig. 13.114., 13.53 Refer to the network in Fig. 13.118., ix, 2Ω, , 12 cos 2t V, , 1, 20 F, , (a) Find n for maximum power supplied to the, 200- load., (b) Determine the power in the 200- load if, n 10., , 1:3, , +, −, , 6Ω, 3Ω, , Figure 13.114, , 4 0° A rms, , For Prob. 13.49., 13.50 Calculate the input impedance for the network in, Fig. 13.115., 8Ω, , j12 Ω, , 1:5, , a, , 1:n, 200 Ω, , 5Ω, , Figure 13.118, For Prob. 13.53., 24 Ω, , 4:1, , 6Ω, , − j10 Ω, b, , Z in, , Figure 13.115, For Prob. 13.50., 13.51 Use the concept of reflected impedance to find, the input impedance and current I1 in, Fig. 13.116., I1, , 240 0° V, , +, −, , Figure 13.116, For Prob. 13.51., , 5Ω, , – j2 Ω, 1:2, , 8Ω, , 1:3, , 36 Ω, , j18 Ω
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 606, , Chapter 13, , 606, , Magnetically Coupled Circuits, , 13.54 A transformer is used to match an amplifier with an, 8- load as shown in Fig. 13.119. The Thevenin, equivalent of the amplifier is: VTh 10 V,, ZTh 128 ., (a) Find the required turns ratio for maximum energy, power transfer., , (a) I1 and I2,, (b) V1, V2, and Vo,, (c) the complex power supplied by the source., 13.58 Determine the average power absorbed by each, resistor in the circuit of Fig. 13.123., , (b) Determine the primary and secondary currents., (c) Calculate the primary and secondary voltages., 20 Ω, , 1:n, Amplifier, circuit, , 20 Ω, , 8Ω, +, −, , 80 cos 4t V, , Figure 13.119, For Prob. 13.54., , 1:5, 100 Ω, , Figure 13.123, For Prob. 13.58., , 13.55 For the circuit in Fig. 13.120, calculate the, equivalent resistance., 1:4, , 20 Ω, , 1: 3, , 13.59 In the circuit of Fig. 13.124, let vs 160 cos 1000t., Find the average power delivered to each resistor., , 60 Ω, , Req, , 10 Ω, , Figure 13.120, , 1:4, , For Prob. 13.55., 13.56 Find the power absorbed by the 10- resistor in the, ideal transformer circuit of Fig. 13.121., 2Ω, , vs, , +, −, , 20 Ω, 12 Ω, , 1:2, , Figure 13.124, For Prob. 13.59., 230 0° V +, −, , 10 Ω, 5Ω, , 13.60 Refer to the circuit in Fig. 13.125 on the following, page., , Figure 13.121, For Prob. 13.56., , (a) Find currents I1, I2, and I3., 13.57 For the ideal transformer circuit of Fig. 13.122, below, find:, I1, , 60 90° V rms, , +, −, , Figure 13.122, For Prob. 13.57., , (b) Find the power dissipated in the 40- resistor., , I2, , 2Ω, , 1:2, +, V1, −, , +, V2, −, , − j6 Ω, 12 Ω, , j3 Ω, , +, Vo, −
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 607, , Problems, I1, , I2, , 4Ω, , 5Ω, , 607, I3, , 1:2, , 1:4, +, −, , 120 0° V, , 10 Ω, , 40 Ω, , Figure 13.125, For Prob. 13.60., *13.61 For the circuit in Fig. 13.126, find I1, I2, and Vo., I1, , 2Ω, , 14 Ω, , 1:5, +, Vo, −, , +, −, , 24 0° V, , I2, , 3:4, , 60 Ω, , 160 Ω, , Figure 13.126, For Prob. 13.61., 13.62 For the network in Fig. 13.127, find, (a) the complex power supplied by the source,, (b) the average power delivered to the 18- resistor., j4 Ω, , 6Ω, , – j20 Ω, , 8Ω, , 2:5, , 1:3, 18 Ω, , 240 0° V, , +, −, , j45 Ω, , Figure 13.127, For Prob. 13.62., 13.63 Find the mesh currents in the circuit of Fig. 13.128, 1Ω, , 12 0° V, , +, −, , – j6 Ω, , 7Ω, , 1:2, , 1:3, , I2, , I1, , 9Ω, , I3, , j18 Ω, , Figure 13.128, For Prob. 13.63., 13.64 For the circuit in Fig. 13.129, find the turns ratio so, that the maximum power is delivered to the 30-k, resistor., , 8 kΩ, , 12 0° V, , +, −, , *13.65 Calculate the average power dissipated by the 20-, resistor in Fig. 13.130., 40 Ω, 10 Ω, , 1:n, 30 kΩ, , +, −, , 200 V, rms, , Figure 13.129, , Figure 13.130, , For Prob. 13.64., , For Prob. 13.65., , 1:2, , 50 Ω, , 1:3, , 20 Ω
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 608, , Chapter 13, , 608, , Magnetically Coupled Circuits, , Section 13.6 Ideal Autotransformers, , 13.70 In the ideal transformer circuit shown in Fig. 13.133,, determine the average power delivered to the load., , 13.66 Design a problem to help other students better, understand how the ideal autotransformer, works., , 30 + j12 Ω, , 13.67 An autotransformer with a 40 percent tap is supplied, by a 400-V, 60-Hz source and is used for stepdown operation. A 5-kVA load operating at unity, power factor is connected to the secondary, terminals. Find:, , 1000 turns, 120 0° V rms, , +, −, 20 – j40 Ω, , 200 turns, , (a) the secondary voltage, (b) the secondary current, , Figure 13.133, , (c) the primary current, , For Prob. 13.70., , 13.68 In the ideal autotransformer of Fig. 13.131, calculate, I1, I2, and Io. Find the average power delivered to, the load., , 13.71 In the autotransformer circuit in Fig. 13.134, show, that, Z in a1 , , N1 2, b ZL, N2, , I2, , 2 – j6 Ω, , 200 turns, I1, , 120 30° V +, −, , ZL, , 10 + j40 Ω, , 80 turns, Io, , Z in, , Figure 13.134, , Figure 13.131, , For Prob. 13.71., , For Prob. 13.68., , Section 13.7 Three-Phase Transformers, *13.69 In the circuit of Fig. 13.132, ZL is adjusted until, maximum average power is delivered to ZL. Find ZL, and the maximum average power transferred to it., Take N1 600 turns and N2 200 turns., , 13.72 In order to meet an emergency, three single-phase, transformers with 12,470/7,200 V rms are connected, in ¢ -Y to form a three-phase transformer which is, fed by a 12,470-V transmission line. If the, transformer supplies 60 MVA to a load, find:, (a) the turns ratio for each transformer,, (b) the currents in the primary and secondary, windings of the transformer,, , N1, 75 Ω, , j125 Ω, ZL, , 120 0° V rms, , +, −, , N2, , (c) the incoming and outgoing transmission line, currents., 13.73 Figure 13.135 on the following page shows a threephase transformer that supplies a Y-connected load., (a) Identify the transformer connection., , Figure 13.132, , (b) Calculate currents I2 and Ic., , For Prob. 13.69., , (c) Find the average power absorbed by the load.
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 609, , Problems, I1, 450 0° V, , 3:1, , 609, , Ia, , I2, , 450 –120° V, , Ib, I3, , 450 120° V, , Ic, 8Ω, , 8Ω, , − j6 Ω, , 8Ω, , − j6 Ω, , − j6 Ω, , Figure 13.135, For Prob. 13.73., 13.74 Consider the three-phase transformer shown in, Fig. 13.136. The primary is fed by a three-phase, source with line voltage of 2.4 kV rms, while the, secondary supplies a three-phase 120-kW balanced, load at pf of 0.8. Determine:, (a) the type of transformer connections,, , (c) the values of ILP and IPP,, (d) the kVA rating of each phase of the transformer., 13.75 A balanced three-phase transformer bank with the, ¢ -Y connection depicted in Fig. 13.137 is used to, step down line voltages from 4,500 V rms to 900 V, rms. If the transformer feeds a 120-kVA load, find:, , (b) the values of ILS and IPS,, , (a) the turns ratio for the transformer,, (b) the line currents at the primary and secondary, sides., , 2.4 kV, ILP, 4:1, , IPS, 1:n, 4500 V, , 900 V, , 42 kVA, Three-phase, load, , Figure 13.137, For Prob. 13.75., ILS, , IPP, Load, 120 kW pf = 0.8, , Figure 13.136, For Prob. 13.74., 1:n, , 13.76 Using Fig. 13.138, design a problem to help other, students better understand a Y- ¢ , three-phase, transformer and how they work., Rline, , jXL, , Rline, , jXL, , Rline, , jXL, , Vs, , Figure 13.138, For Prob. 13.76., , Vline, Balanced, load
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ale29559_ch13.qxd, , 07/10/2008, , 03:59 PM, , Page 611, , Comprehensive Problems, , 13.84 Determine I1, I2, and I3 in the ideal transformer, circuit of Fig. 13.145 using PSpice., , 50 Ω, , j80 Ω, , I1, , I2, 1:2, 40 Ω, − j30 Ω, , 440 0° V, , +, −, , 1:3, , 60 Ω, , I3, , j50 Ω, , Figure 13.145, , 611, , impedance mismatch occurs. By inserting an, impedance-matching transformer ahead of the, receiver, maximum power can be realized. Calculate, the required turns ratio., 13.88 A step-down power transformer with a turns ratio of, n 0.1 supplies 12.6 V rms to a resistive load. If the, primary current is 2.5 A rms, how much power is, delivered to the load?, 13.89 A 240/120-V rms power transformer is rated at, 10 kVA. Determine the turns ratio, the primary, current, and the secondary current., 13.90 A 4-kVA, 2,400/240-V rms transformer has 250, turns on the primary side. Calculate:, (a) the turns ratio,, , For Prob. 13.84., , (b) the number of turns on the secondary side,, , Section 13.9 Applications, , (c) the primary and secondary currents., , 13.85 A stereo amplifier circuit with an output impedance, of 7.2 k is to be matched to a speaker with an input, impedance of 8 by a transformer whose primary, side has 3,000 turns. Calculate the number of turns, required on the secondary side., 13.86 A transformer having 2,400 turns on the primary and, 48 turns on the secondary is used as an impedancematching device. What is the reflected value of a, 3- load connected to the secondary?, 13.87 A radio receiver has an input resistance of 300., When it is connected directly to an antenna system, with a characteristic impedance of 75 , an, , 13.91 A 25,000/240-V rms distribution transformer has a, primary current rating of 75 A., (a) Find the transformer kVA rating., (b) Calculate the secondary current., 13.92 A 4,800-V rms transmission line feeds a distribution, transformer with 1,200 turns on the primary and, 28 turns on the secondary. When a 10- load is, connected across the secondary, find:, (a) the secondary voltage,, (b) the primary and secondary currents,, (c) the power supplied to the load., , Comprehensive Problems, 13.93 A four-winding transformer (Fig. 13.146) is often, used in equipment (e.g., PCs, VCRs) that may be, operated from either 110 V or 220 V. This makes the, equipment suitable for both domestic and foreign use., Show which connections are necessary to provide:, (a) an output of 14 V with an input of 110 V,, (b) an output of 50 V with an input of 220 V., a, , e, , 110 V, , 32 V, b, , f, , c, , g, , d, , h, , 110 V, , are four possible connections, two of which are, wrong. Find the output voltage of:, (a) a wrong connection,, (b) the right connection., 13.95 Ten bulbs in parallel are supplied by a 7,200/120-V, transformer as shown in Fig. 13.147, where the bulbs, are modeled by the 144- resistors. Find:, (a) the turns ratio n,, (b) the current through the primary winding., , 18 V, , Figure 13.146, For Prob. 13.93., *13.94 A 440/110-V ideal transformer can be connected to, become a 550/440-V ideal autotransformer. There, , 1:n, 7200 V, , Figure 13.147, For Prob. 13.95., , 120 V, , 144 Ω, , 144 Ω
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ale29559_ch13.qxd, , 612, , 07/10/2008, , 03:59 PM, , Page 612, , Chapter 13, , Magnetically Coupled Circuits, , *13.96 Some modern power transmission systems now have, major, high voltage DC transmission segments., There are a lot of good reasons for doing this but we, will not go into them here. To go from the AC to DC,, power electronics are used. We start with three-phase, AC and then rectify it (using a full-wave rectifier). It, was found that using a delta to wye and delta, combination connected secondary would give us a, much smaller ripple after the full-wave rectifier., How is this accomplished? Remember that these are, real devices and are wound on common cores., , Hint: using Figs. 13.47 and 13.49, and the fact that, each coil of the wye connected secondary and each, coil of the delta connected secondary are wound, around the same core of each coil of the delta connected primary so the voltage of each of the, corresponding coils are in phase. When the output, leads of both secondaries are connected through fullwave rectifiers with the same load, you will see that, the ripple is now greatly reduced. Please consult the, instructor for more help if necessary.
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ale29559_ch14.qxd, , 07/14/2008, , 04:19 PM, , Page 613, , c h a p t e r, , 14, , Frequency, Response, Dost thou love Life? Then do not squander Time; for that is the stuff, Life is made., —Benjamin Franklin, , Enhancing Your Career, Career in Control Systems, Control systems are another area of electrical engineering where circuit, analysis is used. A control system is designed to regulate the behavior, of one or more variables in some desired manner. Control systems play, major roles in our everyday life. Household appliances such as heating and air-conditioning systems, switch-controlled thermostats, washers and dryers, cruise controllers in automobiles, elevators, traffic, lights, manufacturing plants, navigation systems—all utilize control, systems. In the aerospace field, precision guidance of space probes, the, wide range of operational modes of the space shuttle, and the ability, to maneuver space vehicles remotely from earth all require knowledge, of control systems. In the manufacturing sector, repetitive production, line operations are increasingly performed by robots, which are programmable control systems designed to operate for many hours without fatigue., Control engineering integrates circuit theory and communication, theory. It is not limited to any specific engineering discipline but may, involve environmental, chemical, aeronautical, mechanical, civil, and, electrical engineering. For example, a typical task for a control system, engineer might be to design a speed regulator for a disk drive head., A thorough understanding of control systems techniques is essential to the electrical engineer and is of great value for designing control systems to perform the desired task., , A welding robot. © Vol. 1 PhotoDisc/, Getty Images, , 613
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ale29559_ch14.qxd, , 07/10/2008, , 03:49 PM, , Page 614, , Chapter 14, , 614, , 14.1, , Frequency Response, , Introduction, , In our sinusoidal circuit analysis, we have learned how to find voltages and currents in a circuit with a constant frequency source. If, we let the amplitude of the sinusoidal source remain constant and, vary the frequency, we obtain the circuit’s frequency response. The, frequency response may be regarded as a complete description of, the sinusoidal steady-state behavior of a circuit as a function of, frequency., The frequency response of a circuit is the variation in its behavior with, change in signal frequency., The frequency response of a circuit, may also be considered as the, variation of the gain and phase, with frequency., , The sinusoidal steady-state frequency responses of circuits are of, significance in many applications, especially in communications and, control systems. A specific application is in electric filters that block, out or eliminate signals with unwanted frequencies and pass signals of, the desired frequencies. Filters are used in radio, TV, and telephone, systems to separate one broadcast frequency from another., We begin this chapter by considering the frequency response of, simple circuits using their transfer functions. We then consider Bode, plots, which are the industry-standard way of presenting frequency, response. We also consider series and parallel resonant circuits and, encounter important concepts such as resonance, quality factor, cutoff, frequency, and bandwidth. We discuss different kinds of filters and network scaling. In the last section, we consider one practical application, of resonant circuits and two applications of filters., , 14.2, , X(), , Linear network, , Y(), , Input, , H(), , Output, , Figure 14.1, A block diagram representation of a linear, network., , In this context, X () and Y () denote, the input and output phasors of a network; they should not be confused, with the same symbolism used for reactance and admittance. The multiple, usage of symbols is conventionally, permissible due to lack of enough letters in the English language to express, all circuit variables distinctly., , Transfer Function, , The transfer function H() (also called the network function) is a, useful analytical tool for finding the frequency response of a circuit., In fact, the frequency response of a circuit is the plot of the circuit’s, transfer function H() versus , with varying from 0 to, ., A transfer function is the frequency-dependent ratio of a forced, function to a forcing function (or of an output to an input). The idea, of a transfer function was implicit when we used the concepts of, impedance and admittance to relate voltage and current. In general,, a linear network can be represented by the block diagram shown in, Fig. 14.1., The transfer function H () of a circuit is the frequency-dependent, ratio of a phasor output Y() (an element voltage or current) to a phasor, input X() (source voltage or current)., , Thus,, H() , , Y(), X(), , (14.1)
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ale29559_ch14.qxd, , 07/10/2008, , 03:49 PM, , Page 615, , 14.2, , Transfer Function, , assuming zero initial conditions. Since the input and output can be either, voltage or current at any place in the circuit, there are four possible transfer functions:, Vo(), H() Voltage gain , (14.2a), Vi (), Io(), H() Current gain , (14.2b), Ii (), Vo(), Ii (), Io(), H() Transfer Admittance , Vi (), , H() Transfer Impedance , , 615, , Some authors use H ( j) for transfer, instead of H (), since and j are an, inseparable pair., , (14.2c), (14.2d), , where subscripts i and o denote input and output values. Being a complex, quantity, H() has a magnitude H() and a phase f; that is, H() , H()lf., To obtain the transfer function using Eq. (14.2), we first obtain the, frequency-domain equivalent of the circuit by replacing resistors,, inductors, and capacitors with their impedances R, jL, and 1jC. We, then use any circuit technique(s) to obtain the appropriate quantity in, Eq. (14.2). We can obtain the frequency response of the circuit by plotting the magnitude and phase of the transfer function as the frequency, varies. A computer is a real time-saver for plotting the transfer function., The transfer function H() can be expressed in terms of its numerator polynomial N() and denominator polynomial D() as, H() , , N(), D(), , (14.3), , where N() and D() are not necessarily the same expressions for the, input and output functions, respectively. The representation of H() in, Eq. (14.3) assumes that common numerator and denominator factors in, H() have canceled, reducing the ratio to lowest terms. The roots of, N() 0 are called the zeros of H() and are usually represented as, j z1, z2, p . Similarly, the roots of D() 0 are the poles of H(), and are represented as j p1, p2, p ., A zero, as a root of the numerator polynomial, is a value that results in, a zero value of the function. A pole, as a root of the denominator polynomial, is a value for which the function is infinite., , A zero may also be regarded as the, value of s j that makes H (s) zero,, and a pole as the value of s j that, makes H(s) infinite., , To avoid complex algebra, it is expedient to replace j temporarily, with s when working with H() and replace s with j at the end., For the RC circuit in Fig. 14.2(a), obtain the transfer function VoVs, and its frequency response. Let vs Vm cos t., Solution:, The frequency-domain equivalent of the circuit is in Fig. 14.2(b). By, voltage division, the transfer function is given by, H() , , 1jC, Vo, 1, , , Vs, R 1jC, 1 jRC, , Example 14.1
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ale29559_ch14.qxd, , 07/17/2008, , 01:07 PM, , Page 616, , Chapter 14, , 616, , Frequency Response, R, , v s (t), , +, −, , R, +, v o (t), −, , C, , H, , 1, jC, , Vs +, −, , +, Vo, −, , 1, (a), , (b), , Figure 14.2, , 0.707, , For Example 14.1: (a) time-domain RC circuit,, (b) frequency-domain RC circuit., , , , 0 = 1, 0 RC, (a), = 1, 0 0 RC, , , , −45°, , Comparing this with Eq. (9.18e), we obtain the magnitude and phase of, H() as, 1, , H, ,, f tan 1, 0, 21 (0)2, where 0 1RC. To plot H and f for 0 6 6 , we obtain their, values at some critical points and then sketch., At 0, H 1 and f 0. At , H 0 and f 90., Also, at 0, H 112 and f 45. With these and a few, more points as shown in Table 14.1, we find that the frequency response, is as shown in Fig. 14.3. Additional features of the frequency response, in Fig. 14.3 will be explained in Section 14.6.1 on lowpass filters., TABLE 14.1, , −90°, , , For Example 14.1., (b), , Figure 14.3, Frequency response of the RC circuit:, (a) amplitude response, (b) phase, response., , Practice Problem 14.1, R, , 0, , H, , F, , 0, , H, , F, , 0, 1, 2, 3, , 1, 0.71, 0.45, 0.32, , 0, 45, 63, 72, , 10, 20, 100, , , 0.1, 0.05, 0.01, 0, , 84, 87, 89, 90, , Obtain the transfer function VoVs of the RL circuit in Fig. 14.4,, assuming vs Vm cos t. Sketch its frequency response., Answer: jL(R jL); see Fig. 14.5 for the response., , +, vo, −, , H, 1, , RL circuit for Practice Prob. 14.1., , 0.707, , vs +, −, , L, , , 90°, , Figure 14.4, 45°, , , , 0 =R, 0 L, , , , 0 =R, 0 L, , (a), , Figure 14.5, Frequency response of the RL circuit in Fig. 14.4., , (b)
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ale29559_ch14.qxd, , 07/10/2008, , 03:49 PM, , Page 617, , 14.3, , The Decibel Scale, , 617, , Example 14.2, , For the circuit in Fig. 14.6, calculate the gain Io()Ii () and its poles, and zeros., , 4Ω, , Solution:, By current division,, , 0.5 F, , i i (t), , Io() , , io (t), , 2H, , 4 j2, Ii (), 4 j2 1j0.5, , Figure 14.6, For Example 14.2., , or, j0.5(4 j2), Io(), s(s 2), ,, , 2, 2, Ii (), 1 j2 ( j), s 2s 1, , s j, , The zeros are at, s(s 2) 0, , 1, , z1 0, z2 2, , The poles are at, s2 2s 1 (s 1)2 0, Thus, there is a repeated pole (or double pole) at p 1., , Find the transfer function Vo()Ii () for the circuit in Fig. 14.7., Obtain its zeros and poles., Answer:, 7.317., , 10(s 1)(s 3), , s j; zeros: 1, 3; poles: 0.683,, s2 8s 5, , Practice Problem 14.2, ii (t), , v o (t), , +, −, , 10 Ω, 0.1 F, , 6Ω, 2H, , Figure 14.7, For Practice Prob. 14.2., , 14.3, , The Decibel Scale, , It is not always easy to get a quick plot of the magnitude and phase of, the transfer function as we did above. A more systematic way of obtaining the frequency response is to use Bode plots. Before we begin to, construct Bode plots, we should take care of two important issues: the, use of logarithms and decibels in expressing gain., Since Bode plots are based on logarithms, it is important that we, keep the following properties of logarithms in mind:, 1., 2., 3., 4., , log P1P2 log P1 log P2, log P1P2 log P1 log P2, log P n n log P, log 1 0, , In communications systems, gain is measured in bels. Historically,, the bel is used to measure the ratio of two levels of power or power, gain G; that is,, G Number of bels log10, , P2, P1, , (14.4), , Historical note: The bel is named after, Alexander Graham Bell, the inventor of, the telephone.
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ale29559_ch14.qxd, , 07/10/2008, , 03:49 PM, , Page 618, , Chapter 14, , 618, , Frequency Response, , Historical, Alexander Graham Bell (1847–1922) inventor of the telephone,, was a Scottish-American scientist., Bell was born in Edinburgh, Scotland, a son of Alexander Melville, Bell, a well-known speech teacher. Alexander the younger also became, a speech teacher after graduating from the University of Edinburgh and, the University of London. In 1866 he became interested in transmitting speech electrically. After his older brother died of tuberculosis, his, father decided to move to Canada. Alexander was asked to come to, Boston to work at the School for the Deaf. There he met Thomas A., Watson, who became his assistant in his electromagnetic transmitter, experiment. On March 10, 1876, Alexander sent the famous first telephone message: “Watson, come here I want you.” The bel, the logarithmic unit introduced in Chapter 14, is named in his honor., The decibel (dB) provides us with a unit of less magnitude. It is 110th of, a bel and is given by, GdB 10 log10, , P2, P1, , (14.5), , When P1 P2, there is no change in power and the gain is 0 dB. If, P2 2P1, the gain is, GdB 10 log10 2 3 dB, , (14.6), , and when P2 0.5P1, the gain is, GdB 10 log10 0.5 3 dB, , I1, , I2, , +, V1, , R1, , R2, , Network, , −, , +, V2, −, , (14.7), , Equations (14.6) and (14.7) show another reason why logarithms are, greatly used: The logarithm of the reciprocal of a quantity is simply, negative the logarithm of that quantity., Alternatively, the gain G can be expressed in terms of voltage, or current ratio. To do so, consider the network shown in Fig. 14.8. If, P1 is the input power, P2 is the output (load) power, R1 is the input, resistance, and R2 is the load resistance, then P1 0.5V 21R1 and, P2 0.5V 22R2, and Eq. (14.5) becomes, V 22R2, P2, 10 log10 2, P1, V 1R1, V2 2, R1, 10 log10 a b 10 log10, V1, R2, , GdB 10 log10, P1, , P2, , Figure 14.8, Voltage-current relationships for a fourterminal network., , GdB 20 log10, , V2, R2, 10 log10, V1, R1, , (14.8), , (14.9), , For the case when R2 R1, a condition that is often assumed when, comparing voltage levels, Eq. (14.9) becomes, GdB 20 log10, , V2, V1, , (14.10)
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ale29559_ch14.qxd, , 07/10/2008, , 03:49 PM, , Page 619, , 14.4, , Bode Plots, , 619, , Instead, if P1 I 21 R1 and P2 I 22 R2, for R1 R2, we obtain, GdB 20 log10, , I2, I1, , (14.11), , Three things are important to note from Eqs. (14.5), (14.10), and (14.11):, 1. That 10 log10 is used for power, while 20 log10 is used for voltage, or current, because of the square relationship between them, (P V 2R I 2R)., 2. That the dB value is a logarithmic measurement of the ratio of one, variable to another of the same type. Therefore, it applies in, expressing the transfer function H in Eqs. (14.2a) and (14.2b),, which are dimensionless quantities, but not in expressing H in, Eqs. (14.2c) and (14.2d)., 3. It is important to note that we only use voltage and current magnitudes in Eqs. (14.10) and (14.11). Negative signs and angles will, be handled independently as we will see in Section 14.4., With this in mind, we now apply the concepts of logarithms and decibels to construct Bode plots., , 14.4, , Bode Plots, , Obtaining the frequency response from the transfer function as we, did in Section 14.2 is an uphill task. The frequency range required in, frequency response is often so wide that it is inconvenient to use a, linear scale for the frequency axis. Also, there is a more systematic, way of locating the important features of the magnitude and phase, plots of the transfer function. For these reasons, it has become standard practice to plot the transfer function on a pair of semilogarithmic plots: the magnitude in decibels is plotted against the logarithm, of the frequency; on a separate plot, the phase in degrees is plotted, against the logarithm of the frequency. Such semilogarithmic plots of, the transfer function—known as Bode plots—have become the industry standard., Bode plots are semilog plots of the magnitude (in decibels) and phase, (in degrees) of a transfer function versus frequency., , Bode plots contain the same information as the nonlogarithmic plots, discussed in the previous section, but they are much easier to construct,, as we shall see shortly., The transfer function can be written as, H Hlf He jf, , (14.12), , Taking the natural logarithm of both sides,, ln H ln H ln e jf ln H jf, , (14.13), , Thus, the real part of ln H is a function of the magnitude while the, imaginary part is the phase. In a Bode magnitude plot, the gain, HdB 20 log10 H, , (14.14), , Historical note: Named after Hendrik, W. Bode (1905–1982), an engineer, with the Bell Telephone Laboratories,, for his pioneering work in the 1930s, and 1940s.
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ale29559_ch14.qxd, , 07/17/2008, , 01:07 PM, , Page 620, , Chapter 14, , 620, , is plotted in decibels (dB) versus frequency. Table 14.2 provides a few, values of H with the corresponding values in decibels. In a Bode phase, plot, f is plotted in degrees versus frequency. Both magnitude and, phase plots are made on semilog graph paper., A transfer function in the form of Eq. (14.3) may be written in terms, of factors that have real and imaginary parts. One such representation, might be, K( j) 1 (1 jz1)[1 j2z1k ( jk)2] p, H() , (14.15), (1 jp1)[1 j2z2n ( jn)2] p, , TABLE 14.2, , Specific gain and their decibel, values.*, Magnitude H, , 20 log10 H (dB), , 0.001, 0.01, 0.1, 0.5, 112, 1, 12, 2, 10, 20, 100, 1000, , 60, 40, 20, 6, 3, 0, 3, 6, 20, 26, 40, 60, , Frequency Response, , which is obtained by dividing out the poles and zeros in H(). The, representation of H() as in Eq. (14.15) is called the standard form., H() may include up to seven types of different factors that can appear, in various combinations in a transfer function. These are:, 1., 2., 3., 4., , * Some of these values are approximate., , A gain K, A pole ( j)1 or zero ( j) at the origin, A simple pole 1(1 jp1) or zero (1 jz1), A quadratic pole 1[1 j2z2n ( jn)2], [1 j2z1k ( jk)2], , or, , zero, , In constructing a Bode plot, we plot each factor separately and then, add them graphically. The factors can be considered one at a time and, then combined additively because of the logarithms involved. It is this, mathematical convenience of the logarithm that makes Bode plots a, powerful engineering tool., We will now make straight-line plots of the factors listed above., We shall find that these straight-line plots known as Bode plots approximate the actual plots to a reasonable degree of accuracy., , The origin is where 1 or log 0, and the gain is zero., , Constant term: For the gain K, the magnitude is 20 log10 K and the, phase is 0; both are constant with frequency. Thus, the magnitude and, phase plots of the gain are shown in Fig. 14.9. If K is negative, the, magnitude remains 20 log10 0K 0 but the phase is 180., A decade is an interval between two, frequencies with a ratio of 10; e.g.,, between 0 and 100, or between, 10 and 100 Hz. Thus, 20 dB/decade, means that the magnitude changes, 20 dB whenever the frequency, changes tenfold or one decade., , Pole/zero at the origin: For the zero ( j) at the origin, the magnitude is 20 log10 and the phase is 90. These are plotted in Fig. 14.10,, where we notice that the slope of the magnitude plot is 20 dB/decade,, while the phase is constant with frequency., The Bode plots for the pole ( j)1 are similar except that the slope, of the magnitude plot is 20 dB/decade while the phase is 90. In, general, for ( j)N, where N is an integer, the magnitude plot will have, a slope of 20N dB/decade, while the phase is 90N degrees., , H, 20 log 10 K, , , 0, , 0.1, , 1, , 10, , 100 , , 0.1, , (a), , Figure 14.9, Bode plots for gain K: (a) magnitude plot, (b) phase plot., , 1, , 10, (b), , 100
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ale29559_ch14.qxd, , 07/10/2008, , 03:49 PM, , Page 621, , 14.4, , Bode Plots, , 621, , Simple pole/zero: For the simple zero (1 jz1), the magnitude is, 20 log10 01 jz1 0 and the phase is tan1 z1. We notice that, , The special case of dc ( 0) does, not appear on Bode plots because log, 0 , implying that zero frequency, is infinitely far to the left of the origin of, Bode plots., , HdB 20 log10 ` 1 , , j, `, z1, , 1, , 20 log10 1 0, , (14.16), , as S 0, HdB 20 log10 ` 1 , , j, `, z1, , 1, , , z1, as S , 20 log10, , (14.17), , showing that we can approximate the magnitude as zero (a straight line, with zero slope) for small values of and by a straight line with slope, 20 dB/decade for large values of . The frequency z1 where the two, asymptotic lines meet is called the corner frequency or break frequency., Thus the approximate magnitude plot is shown in Fig. 14.11(a), where, the actual plot is also shown. Notice that the approximate plot is close, to the actual plot except at the break frequency, where z1 and the, deviation is 20 log10 0(1 j1) 0 20 log10 12 3 dB., The phase tan1(z1) can be expressed as, , 0.1, , 1.0, , 10, , , , –20, (a), , (14.18), , , As a straight-line approximation, we let f 0 for z110, f 45, for z1, and f 90 for 10z1. As shown in Fig. 14.11(b), along with the actual plot, the straight-line plot has a slope of 45 per, decade., The Bode plots for the pole 1(1 jp1) are similar to those in, Fig. 14.11 except that the corner frequency is at p1, the magnitude, has a slope of 20 dB/decade, and the phase has a slope of 45 per, decade., Quadratic pole/zero: The magnitude of the quadratic pole 1[1 , j2z2 n ( jn)2] is 20 log10 01 j2z2 n ( jn)2 0 and, the phase is tan1(2z2 n)(1 22n). But, j2z2 , j 2, a b `, n, n, , 0, , Slope = 20 dB/decade, , 0, 0, , f tan1 a b • 45, z1, z1, 90, S , , HdB 20 log10 ` 1 , , H, 20, , 90°, , 0°, , 0.1, , 1.0, , 10, , , , (b), , Figure 14.10, Bode plot for a zero ( j) at the origin:, (a) magnitude plot, (b) phase plot., , 0, , 1, , as S 0, (14.19), , 90°, Exact, , H, , Approximate, , Approximate, 20, , 45°, , 45°/decade, , Exact, , 0.1z1, , z1, , 3 dB 10z1, , , , 0°, , 0.1z1, , (a), , Figure 14.11, , Bode plots of zero (1 jz1): (a) magnitude plot, (b) phase plot., , z1, (b), , 10z1, ,
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ale29559_ch14.qxd, , 07/10/2008, , 03:49 PM, , Page 622, , Chapter 14, , 622, , Frequency Response, , and, HdB 20 log10 ` 1 , , j2z2, j 2, a b `, n, n, , 1, , , n, as S , (14.20), , 40 log10, , Thus, the amplitude plot consists of two straight asymptotic lines: one, with zero slope for 6 n and the other with slope 40 dB/decade, for 7 n, with n as the corner frequency. Figure 14.12(a) shows, the approximate and actual amplitude plots. Note that the actual plot, depends on the damping factor z2 as well as the corner frequency n., The significant peaking in the neighborhood of the corner frequency, should be added to the straight-line approximation if a high level of, accuracy is desired. However, we will use the straight-line approximation for the sake of simplicity., H, 20, , , , 2 = 0.05, 2 = 0.2, 2 = 0.4, , 0, , 0°, 2 = 1.5, , 2 = 0.707, 2 = 1.5, , –20, , –90°, – 40 dB/dec, , –40, 0.01n, , 0.1n, , n, , 10n, , 100n , , –180°, 0.01n, , 2 = 0.707, –90°/dec, , 2 = 0.4, 2 = 0.2, 2 = 0.05, 0.1n, , (a), , n, , 10n, , 100n , , (b), , Figure 14.12, , Bode plots of quadratic pole [1 j 2zn 22n]1: (a) magnitude plot, (b) phase plot., , The phase can be expressed as, There is another procedure for obtaining Bode plots that is faster and perhaps, more efficient than the one we have just, discussed. It consists in realizing that, zeros cause an increase in slope, while, poles cause a decrease. By starting with, the low-frequency asymptote of the, Bode plot, moving along the frequency, axis, and increasing or decreasing the, slope at each corner frequency, one can, sketch the Bode plot immediately from, the transfer function without the effort, of making individual plots and adding, them. This procedure can be used once, you become proficient in the one, discussed here., Digital computers have rendered, the procedure discussed here almost, obsolete. Several software packages, such as PSpice, MATLAB, Mathcad,, and Micro-Cap can be used to generate frequency response plots. We will, discuss PSpice later in the chapter., , 1, , f tan, , 0, 0, • 90, n, 1 22n, 180, S , 2z2n, , (14.21), , The phase plot is a straight line with a slope of 90 per decade starting, at n10 and ending at 10n, as shown in Fig. 14.12(b). We see again, that the difference between the actual plot and the straight-line plot is due, to the damping factor. Notice that the straight-line approximations for both, magnitude and phase plots for the quadratic pole are the same as those, for a double pole, i.e. (1 jn)2. We should expect this because, the double pole (1 jn)2 equals the quadratic pole 1[1 , j2z2n ( jn)2] when z2 1. Thus, the quadratic pole can be, treated as a double pole as far as straight-line approximation is concerned., For the quadratic zero [1 j2z1k ( jk)2], the plots in, Fig. 14.12 are inverted because the magnitude plot has a slope of, 40 dB/decade while the phase plot has a slope of 90 per decade., Table 14.3 presents a summary of Bode plots for the seven factors., Of course, not every transfer function has all seven factors. To sketch, the Bode plots for a function H() in the form of Eq. (14.15), for example, we first record the corner frequencies on the semilog graph paper,, sketch the factors one at a time as discussed above, and then combine
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ale29559_ch14.qxd, , 07/17/2008, , 01:07 PM, , Page 623, , 14.4, , Bode Plots, , 623, , TABLE 14.3, , Summary of Bode straight-line magnitude and phase plots., Factor, , Magnitude, , Phase, , 20 log10 K, , K, 0°, , , , 90N°, , 20N dB ⁄decade, , ( j) N, , 1, ( j)N, , 1, , , , , , 1, , , , , , −20N dB ⁄decade, , a1 , , −90N°, 90N°, , 20N dB ⁄decade, , N, , j, b, z, , 0°, z, 10, , , , z, , p, 10, , p, , 1, (1 jp)N, , , , z, , 10z, , p, , 10p, , , 0°, , −20N dB ⁄decade, , −90N°, 180N°, , 40N dB ⁄decade, , B1 , , , , 2 jz, j 2 N, a bR, n, n, 0°, n, , , , k, , , n, 10, k, 10, 0°, , 1, , n, , 10n, , k, , 10k, , , , , , [1 2 jzk ( jk)2]N, −40N dB ⁄decade, −180N°
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ale29559_ch14.qxd, , 07/10/2008, , 03:49 PM, , Page 624, , Chapter 14, , 624, , Frequency Response, , additively the graphs of the factors. The combined graph is often drawn, from left to right, changing slopes appropriately each time a corner frequency is encountered. The following examples illustrate this procedure., , Example 14.3, , Construct the Bode plots for the transfer function, H() , , 200 j, ( j 2)( j 10), , Solution:, We first put H() in the standard form by dividing out the poles and, zeros. Thus,, H() , , , 10 j, (1 j2)(1 j10), 10 0 j 0, 01 j2 0 01 j10 0, , l90 tan1 2 tan1 10, , Hence, the magnitude and phase are, HdB 20 log10 10 20 log10 0 j 0 20 log10 ` 1 , 20 log10 ` 1 , f 90 tan1, , j, `, 2, , j, `, 10, , , , tan1, 2, 10, , We notice that there are two corner frequencies at 2, 10. For both, the magnitude and phase plots, we sketch each term as shown by the, dotted lines in Fig. 14.13. We add them up graphically to obtain the, overall plots shown by the solid curves., H (dB), , 20 log1010, , 20, 20 log10 j, 0, 0.1, , 0.2, , 1, , 2, , 20 log10, , 10, , 20, , 1, 1 + j/2 , , 100 200, , , , 20 log10, , 1, 1 + j/10 , , 100 200, , , , (a), , , 90°, , 90°, , 0°, 0.1, , 0.2, , –90°, , 1, –tan–1 , 2, , 2, , 10, –tan–1 , 10, (b), , Figure 14.13, For Example 14.3: (a) magnitude plot, (b) phase plot., , 20
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ale29559_ch14.qxd, , 07/10/2008, , 03:49 PM, , Page 625, , 14.4, , Bode Plots, , Practice Problem 14.3, , Draw the Bode plots for the transfer function, H() , , 625, , 5( j 2), j( j 10), , Answer: See Fig. 14.14., H (dB), 20, 20 log10 1 +, 1 2, 0, 0.1, , 10, , 100, , , 20 log101, , 20 log10, , 1, j , , 20 log10, , –20, , j, 2, , 1, 1+ j/10 , , (a), , 90°, , , tan–1 2, , –tan–110, , 0°, 0.1 0.2, , 10 20, , 1 2, , –90°, , 100, , , , −90°, (b), , Figure 14.14, For Practice Prob. 14.3: (a) magnitude plot, (b) phase plot., , Example 14.4, , Obtain the Bode plots for, H() , , j 10, j( j 5)2, , Solution:, Putting H() in the standard form, we get, H() , , 0.4(1 j10), j(1 j5)2, , From this, we obtain the magnitude and phase as, HdB 20 log10 0.4 20 log10 ` 1 , 40 log10 ` 1 , f 0 tan1, , j, ` 20 log10 0 j 0, 10, , j, `, 5, , , , 90 2 tan1, 10, 5, , There are two corner frequencies at 5, 10 rad/s. For the pole with corner frequency at 5, the slope of the magnitude plot is 40 dB/decade, and that of the phase plot is 90 per decade due to the power of 2. The
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ale29559_ch14.qxd, , 07/10/2008, , 03:49 PM, , Page 626, , Chapter 14, , 626, , Frequency Response, , magnitude and the phase plots for the individual terms (in dotted lines), and the entire H( j) (in solid lines) are in Fig. 14.15., H (dB), 20, , 20 log10, , 20 log10 1 +, , 1, j , , j, 10, , , 90°, , 20 log10 0.4, 0, –80.1, –20, , 0.5, , 1, , 5, , 10, , –20 dB/decade, , –40, , , 50 100, 1, 40 log10, 1 + j/5 , , , tan–110, , 0°, 0.1, , 0.5, , 1, , 5, , 10, , 50 100, –90°, , –90°, – 60 dB/decade, , , , , –2 tan–1 5, , –90°/decade, –180°, – 45°/decade, , – 40 dB/decade, , 45°/decade, , (b), , (a), , Figure 14.15, Bode plots for Example 14.4: (a) magnitude plot, (b) phase plot., , Practice Problem 14.4, , Sketch the Bode plots for, H() , , 50 j, ( j 4) ( j 10)2, , Answer: See Fig. 14.16., H (dB), 20, 0.1, 0, , 1, , 4, , , , 20 log10 j , 10, 40, 100, , 90°, , 90°, , , 0.1, 0°, , 0.4, , 1, , 10, , 4, , –20 log10 8, 40 log10, , (a), , –90°, , 1, 1 + j/10 , 20 log10, , 100, , , – tan–1 4, , –20, –40, , 40, , 1, 1 + j/4 , , –180°, , , –2 tan –1 10, (b), , Figure 14.16, For Practice Prob. 14.4: (a) magnitude plot, (b) phase plot., , Example 14.5, , Draw the Bode plots for, H(s) , , s1, s2 12s 100, , Solution:, 1. Define. The problem is clearly stated and we follow the, technique outlined in the chapter., 2. Present. We are to develop the approximate bode plot for the, given function, H(s)., 3. Alternative. The two most effective choices would be the, approximation technique outlined in the chapter, which we will, ,
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ale29559_ch14.qxd, , 07/10/2008, , 03:49 PM, , Page 627, , 14.4, , Bode Plots, , 627, , use here, and MATLAB, which can actually give us the exact, Bode plots., 4. Attempt. We express H(s) as, H() , , 1100(1 j), 1 j1.210 ( j10)2, , For the quadratic pole, n 10 rad/s, which serves as the, corner frequency. The magnitude and phase are, HdB 20 log10 100 20 log10 01 j 0, j1.2, 2, 20 log10 ` 1 , , `, 10, 100, 1.210, d, f 0 tan1 tan1 c, 1 2100, Figure 14.17 shows the Bode plots. Notice that the quadratic, pole is treated as a repeated pole at k, that is, (1 jk)2,, which is an approximation., , H (dB), , , , 20 log10 1 + j , , 20, , 90°, , tan–1 , 0, 0.1, , 10, , 1, , 100, , 1, 1 + j6/10 – 2/100 , , 20 log10, –20, , 0°, 0.1, , , , –90°, – tan–1, , –20 log10 100, –40, , –180°, (a), , 6/10, 1 – 2/100, (b), , Figure 14.17, Bode plots for Example 14.5: (a) magnitude plot, (b) phase plot., , 5. Evaluate. Although we could use MATLAB to validate the, solution, we will use a more straightforward approach. First, we, must realize that the denominator assumes that z 0 for the, approximation, so we will use the following equation to check, our answer:, H(s) , , s1, s 102, 2, , We also note that we need to actually solve for HdB and the, corresponding phase angle f. First, let 0., HdB 20 log10 (1100) 40, , and, , f 0, , Now try 1., HdB 20 log10 (1.414299) 36.9 dB, which is the expected 3 dB up from the corner frequency., f 45, , 10, , 1, , from, , H( j) , , j1, 1 100, , 100, ,
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ale29559_ch14.qxd, , 07/10/2008, , 03:49 PM, , Page 628, , Chapter 14, , 628, , Frequency Response, , Now try 100., HdB 20 log10 (100) 20 log10 (9900) 39.91 dB, f is 90 from the numerator minus 180, which gives 90. We, now have checked three different points and got close agreement,, and, since this is an approximation, we can feel confident that, we have worked the problem successfully., You can reasonably ask why did we not check at 10?, If we just use the approximate value we used above, we end up, with an infinite value, which is to be expected from z 0 (see, Fig. 14.12a). If we used the actual value of H( j10) we will still, end up being far from the approximate values, since z 0.6 and, Fig. 14.12a shows a significant deviation from the approximation., We could have reworked the problem with z 0.707, which, would have gotten us closer to the approximation. However, we, really have enough points without doing this., 6. Satisfactory? We are satisfied the problem has been worked, successfully and we can present the results as a solution to the, problem., , Practice Problem 14.5, , Construct the Bode plots for, H(s) , , 10, s(s 80s 400), 2, , Answer: See Fig. 14.18., , , H (dB), 20, , 1, 20 log10, j , , 0, 0.1, , 1, 20 log10, 1 + j0.2 – 2/400 , 100 200 , , 10 20, , 12, , 0°, 0.1, , 100 200 , , 10 20, , 12, , –90°, –90°, , –20, , –20 log10 40, , –32, –40, , –20 dB/decade, , –180°, –tan–1, –270°, , , 1 – 2/400 , , – 60 dB/decade, (a), , (b), , Figure 14.18, For Practice Prob. 14.5: (a) magnitude plot, (b) phase plot., , Example 14.6, , Given the Bode plot in Fig. 14.19, obtain the transfer function H()., Solution:, To obtain H() from the Bode plot, we keep in mind that a zero always, causes an upward turn at a corner frequency, while a pole causes a
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ale29559_ch14.qxd, , 07/10/2008, , 03:49 PM, , Page 629, , 14.5, , Series Resonance, , downward turn. We notice from Fig. 14.19 that there is a zero j at the, origin which should have intersected the frequency axis at 1. This is, indicated by the straight line with slope 20 dB/decade. The fact that this, straight line is shifted by 40 dB indicates that there is a 40-dB gain; that is,, 40 20 log10 K, , 1, , 629, H, 40 dB, , –20 dB/decade, , +20 dB/decade, , log10 K 2, , or, , – 40 dB/decade, , K 102 100, In addition to the zero j at the origin, we notice that there are three, factors with corner frequencies at 1, 5, and 20 rad/s. Thus, we have:, 1. A pole at p 1 with slope 20 dB/decade to cause a downward turn and counteract the zero at the origin. The pole at, p 1 is determined as 1(1 j1)., 2. Another pole at p 5 with slope 20 dB/decade causing a, downward turn. The pole is 1(1 j5)., 3. A third pole at p 20 with slope 20 dB/decade causing a, further downward turn. The pole is 1(1 j20)., , 0, 0.1, , 1, , 5 10, , 20, , 100 , , Figure 14.19, For Example 14.6., , Putting all these together gives the corresponding transfer function as, 100 j, (1 j1) (1 j5) (1 j20), j10 4, , ( j 1) ( j 5) ( j 20), , H() , , or, H(s) , , 104s, ,, (s 1) (s 5) (s 20), , s j, , Obtain the transfer function H() corresponding to the Bode plot in, Fig. 14.20., , Practice Problem 14.6, H, , Answer: H() , , 4,000(s 5), ., (s 10) (s 100)2, , To see how to use MATLAB to produce Bode plots, refer to Section 14.11., , +20 dB/decade, –40 dB/decade, 0 dB, , 1, , 5, , 10, , Figure 14.20, For Practice Prob. 14.6., , 14.5, , Series Resonance, , The most prominent feature of the frequency response of a circuit may, be the sharp peak (or resonant peak) exhibited in its amplitude characteristic. The concept of resonance applies in several areas of science, and engineering. Resonance occurs in any system that has a complex, conjugate pair of poles; it is the cause of oscillations of stored energy, from one form to another. It is the phenomenon that allows frequency, , 100, , 1000
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ale29559_ch14.qxd, , 07/10/2008, , 03:49 PM, , Page 630, , Chapter 14, , 630, , Frequency Response, , discrimination in communications networks. Resonance occurs in any, circuit that has at least one inductor and one capacitor., Resonance is a condition in an RLC circuit in which the capacitive and, inductive reactances are equal in magnitude, thereby resulting in a, purely resistive impedance., , jL, , R, , Vs = Vm , , +, −, , Resonant circuits (series or parallel) are useful for constructing filters,, as their transfer functions can be highly frequency selective. They are, used in many applications such as selecting the desired stations in radio, and TV receivers., Consider the series RLC circuit shown in Fig. 14.21 in the frequency domain. The input impedance is, , I, , 1, j C, , Z H() , , Vs, 1, R jL , I, jC, , (14.22), , or, Figure 14.21, , Z R j aL , , The series resonant circuit., , 1, b, C, , (14.23), , Resonance results when the imaginary part of the transfer function is, zero, or, 1, Im(Z) L , 0, (14.24), C, The value of that satisfies this condition is called the resonant frequency 0. Thus, the resonance condition is, 1, 0C, , (14.25), , 1, rad/s, 1LC, , (14.26), , 1, Hz, 2 p 1LC, , (14.27), , 0 L , or, 0 , Since 0 2 p f 0,, f0 , Note that at resonance:, Note No. 4 becomes evident from the, fact that, 0 VL 0 , , Vm, 0 L QVm, R, , 0 VC 0 , , Vm 1, QVm, R 0C, , where Q is the quality factor, defined, in Eq. (14.38)., , 1. The impedance is purely resistive, thus, Z R. In other words,, the LC series combination acts like a short circuit, and the entire, voltage is across R., 2. The voltage Vs and the current I are in phase, so that the power, factor is unity., 3. The magnitude of the transfer function H() Z() is minimum., 4. The inductor voltage and capacitor voltage can be much more, than the source voltage., The frequency response of the circuit’s current magnitude, I 0I 0 , , Vm, 2R (L 1C )2, 2, , (14.28)
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ale29559_ch14.qxd, , 07/10/2008, , 03:49 PM, , Page 631, , 14.5, , Series Resonance, , is shown in Fig. 14.22; the plot only shows the symmetry illustrated, in this graph when the frequency axis is a logarithm. The average, power dissipated by the RLC circuit is, , 631, , I, Vm /R, 0.707Vm /R, , 1, P() I 2R, 2, , (14.29), , The highest power dissipated occurs at resonance, when I Vm R, so, that, P(0) , , 1 V 2m, 2 R, , (Vm 12 )2, V 2m, , 2R, 4R, , (14.31), , Hence, 1 and 2 are called the half-power frequencies., The half-power frequencies are obtained by setting Z equal to 12R,, and writing, B, , R2 aL , , 1 2, b 12 R, C, , (14.32), , Solving for , we obtain, , 1 , , R, R 2, 1, a b , 2L B 2L, LC, , (14.33), , R, R 2, 1, 2 , a b , 2L B 2L, LC, We can relate the half-power frequencies with the resonant frequency., From Eqs. (14.26) and (14.33),, 0 112, , (14.34), , showing that the resonant frequency is the geometric mean of the halfpower frequencies. Notice that 1 and 2 are in general not symmetrical around the resonant frequency 0, because the frequency response, is not generally symmetrical. However, as will be explained shortly,, symmetry of the half-power frequencies around the resonant frequency, is often a reasonable approximation., Although the height of the curve in Fig. 14.22 is determined by R,, the width of the curve depends on other factors. The width of the, response curve depends on the bandwidth B, which is defined as the, difference between the two half-power frequencies,, B 2 1, , 1 0 2, Bandwidth B, , (14.30), , At certain frequencies 1, 2, the dissipated power is half the, maximum value; that is,, P(1) P(2) , , 0, , (14.35), , This definition of bandwidth is just one of several that are commonly, used. Strictly speaking, B in Eq. (14.35) is a half-power bandwidth,, because it is the width of the frequency band between the half-power, frequencies., The “sharpness” of the resonance in a resonant circuit is measured, quantitatively by the quality factor Q. At resonance, the reactive energy, , Figure 14.22, The current amplitude versus frequency, for the series resonant circuit of, Fig. 14.21., ,
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ale29559_ch14.qxd, , 07/10/2008, , 03:49 PM, , Page 632, , Chapter 14, , 632, , Although the same symbol Q is used, for the reactive power, the two are not, equal and should not be confused. Q, here is dimensionless, whereas reactive, power Q is in VAR. This may help distinguish between the two., , Frequency Response, , in the circuit oscillates between the inductor and the capacitor. The, quality factor relates the maximum or peak energy stored to the energy, dissipated in the circuit per cycle of oscillation:, Q 2p, , Peak energy stored in the circuit, Energy dissipated by the circuit, in one period at resonance, , (14.36), , It is also regarded as a measure of the energy storage property of a circuit in relation to its energy dissipation property. In the series RLC, circuit, the peak energy stored is 12 LI 2, while the energy dissipated in, one period is 12 (I 2R)(1f0). Hence,, Q 2p 1, , 2I, , 1, 2, 2 LI, 2, , R(1f0), , , , 2 p f0 L, R, , (14.37), , or, , Amplitude, Q1 (least selectivity), Q2 (medium selectivity), Q3 (greatest selectivity), , Q, , 0 L, 1, , R, 0CR, , (14.38), , Notice that the quality factor is dimensionless. The relationship, between the bandwidth B and the quality factor Q is obtained by substituting Eq. (14.33) into Eq. (14.35) and utilizing Eq. (14.38)., B, , , , 0, R, , L, Q, , (14.39), , or B 20CR. Thus, , B3, B2, , The quality factor of a resonant circuit is the ratio of its resonant, frequency to its bandwidth., , B1, , Figure 14.23, The higher the circuit Q, the smaller the, bandwidth., The quality factor is a measure of, the selectivity (or “sharpness” of, resonance) of the circuit., , Keep in mind that Eqs. (14.33), (14.38), and (14.39) only apply to a, series RLC circuit., As illustrated in Fig. 14.23, the higher the value of Q, the more, selective the circuit is but the smaller the bandwidth. The selectivity of, an RLC circuit is the ability of the circuit to respond to a certain frequency and discriminate against all other frequencies. If the band of, frequencies to be selected or rejected is narrow, the quality factor of, the resonant circuit must be high. If the band of frequencies is wide,, the quality factor must be low., A resonant circuit is designed to operate at or near its resonant frequency. It is said to be a high-Q circuit when its quality factor is equal, to or greater than 10. For high-Q circuits (Q 10), the half-power, frequencies are, for all practical purposes, symmetrical around the resonant frequency and can be approximated as, 1 0 , , B, ,, 2, , 2 0 , , B, 2, , High-Q circuits are used often in communications networks., , (14.40)
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ale29559_ch14.qxd, , 07/10/2008, , 03:49 PM, , Page 633, , 14.5, , Series Resonance, , 633, , We see that a resonant circuit is characterized by five related, parameters: the two half-power frequencies 1 and 2, the resonant frequency 0, the bandwidth B, and the quality factor Q., , Example 14.7, , In the circuit of Fig. 14.24, R 2 , L 1 mH, and C 0.4 mF., (a) Find the resonant frequency and the half-power frequencies. (b) Calculate the quality factor and bandwidth. (c) Determine the amplitude, of the current at 0, 1, and 2., , R, , 20 sin t +, −, , Solution:, (a) The resonant frequency is, 0 , , 1, 2LC, , , , 1, 3, , 210, , 0.4, , 50 krad/s, , 6, , 10, , ■ METHOD 1 The lower half-power frequency is, R, R 2, 1, a b , 2L B 2L, LC, 2, , 2(103)2 (50 103)2, 2 103, 1 11 2500 krad/s 49 krad/s, , 1 , , Similarly, the upper half-power frequency is, 2 1 11 2500 krad/s 51 krad/s, (b) The bandwidth is, B 2 1 2 krad/s, or, B, , R, 2, 3 2 krad/s, L, 10, , The quality factor is, Q, , 0, 50, , 25, B, 2, , ■ METHOD 2 Alternatively, we could find, Q, , 0 L, 50, , R, , 103, 2, , 103, , 25, , From Q, we find, B, , 0, 50 103, , 2 krad/s, Q, 25, , Since Q 7 10, this is a high-Q circuit and we can obtain the halfpower frequencies as, B, 50 1 49 krad/s, 2, B, 2 0 50 1 51 krad/s, 2, , 1 0 , , Figure 14.24, For Example 14.7., , L, , C
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 634, , Chapter 14, , 634, , Frequency Response, , as obtained earlier., (c) At 0,, I, , Vm, 20, , 10 A, R, 2, , At 1, 2,, I, , Practice Problem 14.7, , Vm, 10, , 7.071 A, 12R, 12, , A series-connected circuit has R 4 and L 25 mH. (a) Calculate, the value of C that will produce a quality factor of 50. (b) Find 1, 2,, and B. (c) Determine the average power dissipated at 0, 1, 2., Take Vm 100 V., Answer: (a) 0.625 mF, (b) 7920 rad/s, 8080 rad/s, 160 rad/s, (c) 1.25 kW,, 0.625 kW, 0.625 kW., , I = Im, , +, V, −, , , , R, , jL, , 1, jC, , 14.6, , Parallel Resonance, , The parallel RLC circuit in Fig. 14.25 is the dual of the series RLC circuit. So we will avoid needless repetition. The admittance is, , Figure 14.25, The parallel resonant circuit., , Y H() , , I, 1, 1, jC , V, R, jL, , (14.41), , or, V , , Y, , Im R, 0.707 Im R, , 1, 1, j aC , b, R, L, , (14.42), , Resonance occurs when the imaginary part of Y is zero,, 1, 0, L, , (14.43), , 1, rad/s, 1LC, , (14.44), , C , 1 0, , 0, , 2, , , , or, , Bandwidth B, , Figure 14.26, The current amplitude versus frequency for, the series resonant circuit of Fig. 14.25., , We can see this from the fact that, 0 IL 0 , , Im R, QIm, 0 L, , 0 IC 0 0CIm R QIm, where Q is the quality factor, defined, in Eq. (14.47)., , 0 , , which is the same as Eq. (14.26) for the series resonant circuit. The, voltage |V| is sketched in Fig. 14.26 as a function of frequency., Notice that at resonance, the parallel LC combination acts like an, open circuit, so that the entire current flows through R. Also, the, inductor and capacitor current can be much more than the source, current at resonance., We exploit the duality between Figs. 14.21 and 14.25 by comparing Eq. (14.42) with Eq. (14.23). By replacing R, L, and C in the
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ale29559_ch14.qxd, , 07/17/2008, , 01:08 PM, , Page 635, , 14.6, , Parallel Resonance, , expressions for the series circuit with 1R, C, and L respectively, we, obtain for the parallel circuit, 1, 1 2, 1, a, b , 2RC B 2RC, LC, 1, 1 2, 1, 2 , a, b , 2RC B 2RC, LC, , 1 , , B 2 1 , , Q, , (14.45), , 1, RC, , (14.46), , 0, R, 0 RC , B, 0L, , (14.47), , It should be noted that Eqs. (14.45) to (14.47) apply only to a parallel, RLC circuit. Using Eqs. (14.45) and (14.47), we can express the halfpower frequencies in terms of the quality factor. The result is, 1 0, , B, , 1a, , 0, 1 2, b , ,, 2Q, 2Q, , 2 0, , B, , 1a, , 0, 1 2, b , 2Q, 2Q, (14.48), , Again, for high-Q circuits (Q 10), 1 0 , , B, ,, 2, , 2 0 , , B, 2, , (14.49), , Table 14.4 presents a summary of the characteristics of the series and parallel resonant circuits. Besides the series and parallel RLC considered, here, other resonant circuits exist. Example 14.9 treats a typical example., TABLE 14.4, , Summary of the characteristics of resonant RLC circuits., Characteristic, Resonant frequency, 0, Quality factor, Q, Bandwidth, B, Half-power frequencies, 1, 2, For Q 10, 1, 2, , Series circuit, , Parallel circuit, , 1, 1LC, , 1, 1LC, , 1, 0 L, or, 0 RC, R, 0, Q, , R, or 0 RC, 0 L, 0, Q, , 1 2 0, 1 2 0, 0 1 a b , 0 1 a b , B, 2Q, 2Q, B, 2Q, 2Q, B, B, 0 , 0 , 2, 2, , 635
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 636, , Chapter 14, , 636, , Example 14.8, , In the parallel RLC circuit of Fig. 14.27, let R 8 k , L 0.2 mH,, and C 8 mF. (a) Calculate 0, Q, and B. (b) Find 1 and 2., (c) Determine the power dissipated at 0, 1, and 2., , io, , 10 sin t +, −, , R, , Frequency Response, , L, , Solution:, C, , (a), , Figure 14.27, , 0 , , For Example 14.8., , 1, , 1LC, 20.2, Q, , 1, 103, , R, , 0 L, 25, B, , 8, , 8, 103, , 106, , 103, 0.2, , , , 103, , 105, 25 krad/s, 4, 1,600, , 0, 15.625 rad/s, Q, , (b) Due to the high value of Q, we can regard this as a high-Q circuit,, Hence,, B, 25,000 7.812 24,992 rad/s, 2, B, 2 0 25,000 7.812 25,008 rad/s, 2, 1 0 , , (c) At 0, Y 1R or Z R 8 k . Then, Io , , 10l90, V, , 1.25l90 mA, Z, 8,000, , Since the entire current flows through R at resonance, the average, power dissipated at 0 is, P, , 1, 1 2, 0 Io 0 R (1.25, 2, 2, , 103)2 (8, , 103 ) 6.25 mW, , or, P, , V 2m, , 2R, 2, , 100, 6.25 mW, 8 103, , At 1, 2,, P, , Practice Problem 14.8, , V 2m, 3.125 mW, 4R, , A parallel resonant circuit has R 100 k , L 20 mH, and C 5 nF., Calculate 0, 1, 2, Q, and B., Answer: 100 krad/s, 99 krad/s, 101 krad/s, 50, 2 krad/s.
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 637, , 14.7, , Passive Filters, , 637, , Example 14.9, , Determine the resonant frequency of the circuit in Fig. 14.28., Solution:, The input admittance is, Y j0.1 , , 2H, , 2 j2, 1, 1, , 0.1 j0.1 , 10, 2 j2, 4 42, , At resonance, Im(Y) 0 and, 00.1 , , 20, 4 420, , Im cos t, , 0.1 F, , 2Ω, , Figure 14.28, For Example 14.9., , 0, , 1, , 0 2 rad/s, , Calculate the resonant frequency of the circuit in Fig. 14.29., , Practice Problem 14.9, 100 mH, , Answer: 100 rad/s., Vm cos t +, −, , 14.7, , 10 Ω, , Passive Filters, , The concept of filters has been an integral part of the evolution of electrical engineering from the beginning. Several technological achievements would not have been possible without electrical filters. Because, of this prominent role of filters, much effort has been expended on the, theory, design, and construction of filters and many articles and books, have been written on them. Our discussion in this chapter should be, considered introductory., A filter is a circuit that is designed to pass signals with desired, frequencies and reject or attenuate others., , As a frequency-selective device, a filter can be used to limit the frequency spectrum of a signal to some specified band of frequencies., Filters are the circuits used in radio and TV receivers to allow us to, select one desired signal out of a multitude of broadcast signals in the, environment., A filter is a passive filter if it consists of only passive elements R,, L, and C. It is said to be an active filter if it consists of active elements, (such as transistors and op amps) in addition to passive elements R, L,, and C. We consider passive filters in this section and active filters in, the next section. LC filters have been used in practical applications for, more than eight decades. LC filter technology feeds related areas such, as equalizers, impedance-matching networks, transformers, shaping, networks, power dividers, attenuators, and directional couplers, and is, continuously providing practicing engineers with oppurtunities to innovate and experiment. Besides the LC filters we study in these sections,, there are other kinds of filters—such as digital filters, electromechanical, filters, and microwave filters—which are beyond the level of this text., , 0.5 mF, , Figure 14.29, For Practice Prob. 14.9, , 20 Ω
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ale29559_ch14.qxd, , 07/17/2008, , 01:08 PM, , Page 638, , Chapter 14, , 638, H(), , As shown in Fig. 14.30, there are four types of filters whether passive or active:, , Passband, , 1, , , , 1. A lowpass filter passes low frequencies and stops high frequencies,, as shown ideally in Fig. 14.30(a)., 2. A highpass filter passes high frequencies and rejects low frequencies,, as shown ideally in Fig. 14.30(b)., 3. A bandpass filter passes frequencies within a frequency band and, blocks or attenuates frequencies outside the band, as shown ideally in Fig. 14.30(c)., 4. A bandstop filter passes frequencies outside a frequency band and, blocks or attenuates frequencies within the band, as shown ideally, in Fig. 14.30(d)., , Stopband, , Table 14.5 presents a summary of the characteristics of these filters. Be, aware that the characteristics in Table 14.5 are only valid for first- or, second-order filters—but one should not have the impression that only, these kinds of filter exist. We now consider typical circuits for realizing the filters shown in Table 14.5., , Stopband, , , , c, , 0, , (a), H(), Passband, , 1, Stopband, , c, , 0, , Frequency Response, , (b), H(), Passband, , 1, Stopband, , , , 2, , 1, , 0, , (c), , TABLE 14.5, , H(), 1, , Summary of the characteristics of ideal filters., Passband, , Passband, , Type of Filter, Stopband, , 1, , 0, , 2, , , , (d), , Figure 14.30, Ideal frequency response of four types of, filter: (a) lowpass filter, (b) highpass filter,, (c) bandpass filter, (d) bandstop filter., , Lowpass, Highpass, Bandpass, Bandstop, , H(0), , H(), , H(c) or H(0), , 1, 0, 0, 1, , 0, 1, 0, 1, , 112, 112, 1, 0, , c is the cutoff frequency for lowpass and highpass filters; 0 is the center frequency for, bandpass and bandstop filters., , 14.7.1 Lowpass Filter, R, vi (t) +, −, , C, , +, vo(t), −, , A typical lowpass filter is formed when the output of an RC circuit is, taken off the capacitor as shown in Fig. 14.31. The transfer function, (see also Example 14.1) is, H() , , Figure 14.31, A lowpass filter., , 1jC, Vo, , Vi, R 1jC, , H() , H(), 1, Ideal, 0.707, , Actual, , c, , , , Figure 14.32, Ideal and actual frequency response of a, lowpass filter., , (14.50), , Note that H(0) 1, H() 0. Figure 14.32 shows the plot of 0H() 0,, along with the ideal characteristic. The half-power frequency, which is, equivalent to the corner frequency on the Bode plots but in the context of filters is usually known as the cutoff frequency c, is obtained, by setting the magnitude of H() equal to 112, thus,, H(c) , , 0, , 1, 1 jRC, , 1, 21 , , 2c R2C2, , , , 1, 12, , or, c , , 1, RC, , (14.51)
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 639, , 14.7, , Passive Filters, , The cutoff frequency is also called the rolloff frequency., A lowpass filter is designed to pass only frequencies from dc up to, the cutoff frequency c., , A lowpass filter can also be formed when the output of an RL circuit is taken off the resistor. Of course, there are many other circuits, for lowpass filters., , 639, , The cutoff frequency is the frequency, at which the transfer function H drops, in magnitude to 70.71% of its maximum, value. It is also regarded as the frequency at which the power dissipated, in a circuit is half of its maximum value., , 14.7.2. Highpass Filter, C, , A highpass filter is formed when the output of an RC circuit is taken, off the resistor as shown in Fig. 14.33. The transfer function is, Vo, R, H() , , Vi, R 1jC, H() , , jRC, 1 jRC, , v i (t) +, −, , (14.52), , A highpass filter., H(), , Ideal, , 1, , (14.53), , 0.707, Actual, , A highpass filter is designed to pass all frequencies above its cutoff, frequency c., , A highpass filter can also be formed when the output of an RL circuit is taken off the inductor., , c, , 0, , , , Figure 14.34, Ideal and actual frequency response of a, highpass filter., , 14.7.3 Bandpass Filter, The RLC series resonant circuit provides a bandpass filter when the, output is taken off the resistor as shown in Fig. 14.35. The transfer, function is, Vo, R, H() , , Vi, R j(L 1C), , 1, 1LC, , L, , C, , vi (t) +, −, , (14.54), , We observe that H(0) 0, H() 0. Figure 14.36 shows the plot of, 0 H() 0 . The bandpass filter passes a band of frequencies (1 6 6 2), centered on 0, the center frequency, which is given by, 0 , , +, v o(t), −, , Figure 14.33, , Note that H(0) 0, H() 1. Figure 14.34 shows the plot of 0 H() 0 ., Again, the corner or cutoff frequency is, 1, c , RC, , R, , R, , +, vo (t), −, , Figure 14.35, A bandpass filter., H(), , (14.55), , Ideal, , 1, 0.707, , Actual, , A bandpass filter is designed to pass all frequencies within a band of, frequencies, 1 6 6 2., , Since the bandpass filter in Fig. 14.35 is a series resonant circuit, the, half-power frequencies, the bandwidth, and the quality factor are determined as in Section 14.5. A bandpass filter can also be formed by, cascading the lowpass filter (where 2 c) in Fig. 14.31 with the, , 0, , 1, , 0, , 2, , , , Figure 14.36, Ideal and actual frequency response of a, bandpass filter.
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 640, , Chapter 14, , 640, , Frequency Response, , highpass filter (where 1 c) in Fig. 14.33. However, the result, would not be the same as just adding the output of the lowpass filter, to the input of the highpass filter, because one circuit loads the other, and alters the desired transfer function., R, , v i (t) +, −, , 14.7.4 Bandstop Filter, , +, , C, , A filter that prevents a band of frequencies between two designated, values (1 and 2) from passing is variably known as a bandstop, bandreject, or notch filter. A bandstop filter is formed when the output RLC, series resonant circuit is taken off the LC series combination as shown, in Fig. 14.37. The transfer function is, , v o(t), , L, , –, , Figure 14.37, A bandstop filter., , H() , , j(L 1C), Vo, , Vi, R j(L 1C), , (14.56), , Notice that H(0) 1, H() 1. Figure 14.38 shows the plot of, 0H() 0 . Again, the center frequency is given by, , H(), 1, 0.707, , 0 , , Actual, Ideal, , 0, , 1, , 0, , 2, , Figure 14.38, Ideal and actual frequency response of a, bandstop filter., , , , 1, 1LC, , (14.57), , while the half-power frequencies, the bandwidth, and the quality factor are calculated using the formulas in Section 14.5 for a series resonant circuit. Here, 0 is called the frequency of rejection, while the, corresponding bandwidth (B 2 1) is known as the bandwidth, of rejection. Thus,, A bandstop filter is designed to stop or eliminate all frequencies, within a band of frequencies, 1 6 6 2., , Notice that adding the transfer functions of the bandpass and the, bandstop gives unity at any frequency for the same values of R, L, and, C. Of course, this is not true in general but true for the circuits treated, here. This is due to the fact that the characteristic of one is the inverse, of the other., In concluding this section, we should note that:, 1. From Eqs. (14.50), (14.52), (14.54), and (14.56), the maximum gain, of a passive filter is unity. To generate a gain greater than unity,, one should use an active filter as the next section shows., 2. There are other ways to get the types of filters treated in this section., 3. The filters treated here are the simple types. Many other filters, have sharper and complex frequency responses., , Example 14.10, , Determine what type of filter is shown in Fig. 14.39. Calculate the corner or cutoff frequency. Take R 2 k , L 2 H, and C 2 mF., Solution:, The transfer function is, H(s) , , R 1sC, Vo, ,, , Vi, sL R 1sC, , s j, , (14.10.1)
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 641, , 14.7, , Passive Filters, , 641, , But, , L, , R g, , RsC, 1, R, , , sC, R 1sC, 1 sRC, , v i (t) +, −, , R, , C, , Substituting this into Eq. (14.10.1) gives, H(s) , , R(1 sRC), R, ,, 2, sL R(1 sRC), s RLC sL R, , s j, , +, v o (t), −, , Figure 14.39, For Example 14.10., , or, H() , , R, RLC jL R, 2, , (14.10.2), , Since H(0) 1 and H() 0, we conclude from Table 14.5 that the, circuit in Fig. 14.39 is a second-order lowpass filter. The magnitude, of H is, H, , R, 2(R RLC )2 2L2, 2, , (14.10.3), , The corner frequency is the same as the half-power frequency, i.e.,, where H is reduced by a factor of 112. Since the dc value of H(), is 1, at the corner frequency, Eq. (14.10.3) becomes after squaring, H2 , , 1, R2, , 2, 2, (R c RLC)2 2c L2, , or, 2 (1 2c LC)2 a, , c L 2, b, R, , Substituting the values of R, L, and C, we obtain, 2 (1 2c 4, , 10 6 )2 (c 103)2, , Assuming that c is in krad/s,, 2 (1 42c )2 2c, , or, , 164c 72c 1 0, , Solving the quadratic equation in 2c , we get 2c 0.5509 and 0.1134., Since c is real,, c 0.742 krad/s 742 rad/s, , For the circuit in Fig. 14.40, obtain the transfer function Vo ()Vi ()., Identify the type of filter the circuit represents and determine the corner frequency. Take R1 100 R2, L 2 mH., j, R2, Answer:, a, b, highpass filter, R1 R2 j c, R1R2, c , 25 krad/s., (R1 R2)L, , Practice Problem 14.10, R1, v i (t) +, −, , L, , Figure 14.40, For Practice Prob. 14.10., , R2, , +, vo (t), −
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 642, , 642, , Example 14.11, , Chapter 14, , Frequency Response, , If the bandstop filter in Fig. 14.37 is to reject a 200-Hz sinusoid while, passing other frequencies, calculate the values of L and C. Take, R 150 and the bandwidth as 100 Hz., Solution:, We use the formulas for a series resonant circuit in Section 14.5., B 2 p(100) 200 p rad/s, But, B, , R, L, , 1, , L, , R, 150, , 0.2387 H, B, 200 p, , Rejection of the 200-Hz sinusoid means that f0 is 200 Hz, so that 0, in Fig. 14.38 is, 0 2 p f0 2 p (200) 400 p, Since 0 11LC,, C, , Practice Problem 14.11, , 1, 1, , 2.653 mF, 2, 2, 0 L, (400 p) (0.2387), , Design a bandpass filter of the form in Fig. 14.35 with a lower cutoff, frequency of 20.1 kHz and an upper cutoff frequency of 20.3 kHz. Take, R 20 k . Calculate L, C, and Q., Answer: 15.92 H, 3.9 pF, 101., , 14.8, , Active Filters, , There are three major limitations to the passive filters considered in the, previous section. First, they cannot generate gain greater than 1; passive elements cannot add energy to the network. Second, they may, require bulky and expensive inductors. Third, they perform poorly at, frequencies below the audio frequency range (300 Hz 6 f 6 3,000 Hz)., Nevertheless, passive filters are useful at high frequencies., Active filters consist of combinations of resistors, capacitors, and, op amps. They offer some advantages over passive RLC filters. First,, they are often smaller and less expensive, because they do not require, inductors. This makes feasible the integrated circuit realizations of filters. Second, they can provide amplifier gain in addition to providing, the same frequency response as RLC filters. Third, active filters can be, combined with buffer amplifiers (voltage followers) to isolate each, stage of the filter from source and load impedance effects. This isolation allows designing the stages independently and then cascading them, to realize the desired transfer function. (Bode plots, being logarithmic,, may be added when transfer functions are cascaded.) However, active, filters are less reliable and less stable. The practical limit of most active
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 643, , 14.8, , Active Filters, , 643, , filters is about 100 kHz—most active filters operate well below that, frequency., Filters are often classified according to their order (or number of, poles) or their specific design type., , Zf, Zi, , –, , One type of first-order filter is shown in Fig. 14.41. The components, selected for Zi and Zf determine whether the filter is lowpass or highpass, but one of the components must be reactive., Figure 14.42 shows a typical active lowpass filter. For this filter,, the transfer function is, Zf, Vo, , Vi, Zi, , (14.58), , Rf jCf, Rf, 1, , , jCf, Rf 1jCf, 1 jCf Rf, , (14.59), , Figure 14.41, A general first-order active filter., , Rf, Cf, , where Zi Ri and, Zf Rf g, , +, Vo, −, , Vi, , 14.8.1 First-Order Lowpass Filter, , H() , , −, +, , +, , Ri, , −, +, , +, , +, Vo, –, , Vi, , Therefore,, H() , , Rf, , 1, Ri 1 jCf Rf, , –, , (14.60), Figure 14.42, , We notice that Eq. (14.60) is similar to Eq. (14.50), except that there, is a low frequency ( S 0) gain or dc gain of RfRi. Also, the corner frequency is, 1, c , (14.61), Rf Cf, , Active first-order lowpass filter., , which does not depend on Ri. This means that several inputs with different Ri could be summed if required, and the corner frequency would, remain the same for each input., , 14.8.2 First-Order Highpass Filter, Rf, , Figure 14.43 shows a typical highpass filter. As before,, H() , , Zf, Vo, , Vi, Zi, , Rf, Ri 1jCi, , , , jCiRf, 1 jCiRi, , 1, RiCi, , −, +, , Vi, , +, Vo, , –, , –, , (14.63), , This is similar to Eq. (14.52), except that at very high frequencies, ( S ), the gain tends to RfRi. The corner frequency is, c , , Ci, , +, , where Zi Ri 1jCi and Zf Rf so that, H() , , Ri, , (14.62), , Figure 14.43, Active first-order highpass filter., , (14.64), , 14.8.3 Bandpass Filter, The circuit in Fig. 14.42 may be combined with that in Fig. 14.43 to, form a bandpass filter that will have a gain K over the required range, of frequencies. By cascading a unity-gain lowpass filter, a unity-gain, , This way of creating a bandpass filter,, not necessarily the best, is perhaps the, easiest to understand.
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 644, , Chapter 14, , 644, , Frequency Response, , highpass filter, and an inverter with gain Rf Ri, as shown in the block, diagram of Fig. 14.44(a), we can construct a bandpass filter whose frequency response is that in Fig. 14.44(b). The actual construction of the, bandpass filter is shown in Fig. 14.45., , H, K, 0.707 K, B, vi, , Low-pass, filter, , High-pass, filter, , vo, , Inverter, , 1, , 0, , 0, , (a), , 2, , , , (b), , Figure 14.44, Active bandpass filter: (a) block diagram, (b) frequency response., , R, C1, R, R, , −, +, , +, , R, , C2, −, +, , Rf, Ri, , −, +, , vi, –, , Stage 1, Low-pass filter, sets 2 value, , Stage 2, High-pass filter, sets 1 value, , +, vo, –, , Stage 3, An inverter, provides gain, , Figure 14.45, Active bandpass filter., , The analysis of the bandpass filter is relatively simple. Its transfer, function is obtained by multiplying Eqs. (14.60) and (14.63) with the, gain of the inverter; that is,, H() , , Rf, jC2R, Vo, 1, a, b a, b a b, Vi, 1 jC1R, 1 jC2R, Ri, Rf, , jC2R, 1, , Ri 1 jC1R 1 jC2R, , (14.65), , The lowpass section sets the upper corner frequency as, 2 , , 1, RC1, , (14.66), , while the highpass section sets the lower corner frequency as, 1 , , 1, RC2, , (14.67)
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 645, , 14.8, , Active Filters, , 645, , With these values of 1 and 2, the center frequency, bandwidth, and, quality factor are found as follows:, 0 112, , (14.68), , B 2 1, , (14.69), , Q, , 0, B, , (14.70), , To find the passband gain K, we write Eq. (14.65) in the standard, form of Eq. (14.15),, H() , , Rf, , Rf, j1, j2, , Ri (1 j1)(1 j2), Ri (1 j)(2 j), (14.71), , At the center frequency 0 112, the magnitude of the transfer, function is, 0H(0) 0 `, , Rf, j02, 2, ` , Ri (1 j0)(2 j0), Ri 1 2, Rf, , (14.72), , Thus, the passband gain is, K, , 2, Ri 1 2, Rf, , (14.73), , 14.8.4 Bandreject (or Notch) Filter, A bandreject filter may be constructed by parallel combination of a, lowpass filter and a highpass filter and a summing amplifier, as shown, in the block diagram of Fig. 14.46(a). The circuit is designed such that, the lower cutoff frequency 1 is set by the lowpass filter while the, upper cutoff frequency 2 is set by the highpass filter. The gap between, 1 and 2 is the bandwidth of the filter. As shown in Fig. 14.46(b), the, filter passes frequencies below 1 and above 2. The block diagram in, Fig. 14.46(a) is actually constructed as shown in Fig. 14.47. The transfer function is, H() , , Rf, jC2R, Vo, 1, a, , b, Vi, Ri, 1 jC1R, 1 jC2R, , (14.74), , H, K, 0.707 K, , vi, , Low-pass, filter sets, 1, High-pass, filter sets, 2 > 1, , v1, Summing, amplifier, v2, , vo = v 1 + v 2, 0, , 1, , 0, B, , (a), , Figure 14.46, Active bandreject filter: (a) block diagram, (b) frequency response., , (b), , 2, ,
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ale29559_ch14.qxd, , 07/10/2008, , 646, , 03:50 PM, , Page 646, , Chapter 14, , Frequency Response, R, C1, R, , Ri, , −, +, , vi, , Ri, , R, , +, C2, , R, , Rf, −, +, , −, +, , +, vo, –, , –, , Figure 14.47, Active bandreject filter., , The formulas for calculating the values of 1, 2, the center frequency,, bandwidth, and quality factor are the same as in Eqs. (14.66) to (14.70)., To determine the passband gain K of the filter, we can write, Eq. (14.74) in terms of the upper and lower corner frequencies as, H() , , Rf, , a, , j1, 1, , b, 1 j2, 1 j1, , Ri, Rf (1 j21 ( j)211), , Ri (1 j2)(1 j1), , (14.75), , Comparing this with the standard form in Eq. (14.15) indicates that in, the two passbands ( S 0 and S ) the gain is, K, , Rf, , (14.76), , Ri, , We can also find the gain at the center frequency by finding the magnitude of the transfer function at 0 112, writing, Rf (1 j20 1 ( j0)211), `, Ri (1 j0 2)(1 j0 1), Rf 21, , H(0) `, , , (14.77), , Ri 1 2, , Again, the filters treated in this section are only typical. There are, many other active filters that are more complex., , Example 14.12, , Design a lowpass active filter with a dc gain of 4 and a corner frequency, of 500 Hz., Solution:, From Eq. (14.61), we find, c 2 p fc 2 p (500) , , 1, Rf Cf, , (14.12.1)
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 647, , 14.8, , Active Filters, , 647, , The dc gain is, H(0) , , Rf, Ri, , 4, , (14.12.2), , We have two equations and three unknowns. If we select Cf 0.2 mF,, then, Rf , , 1, 2 p (500)0.2, , 106, , 1.59 k, , and, Ri , , Rf, 4, , 397.5, , We use a 1.6-k resistor for Rf and a 400shows the filter., , resistor for Ri. Figure 14.42, , Design a highpass filter with a high-frequency gain of 5 and a corner, frequency of 2 kHz. Use a 0.1-mF capacitor in your design., Answer: Ri 800, , and Rf 4 k ., , Design a bandpass filter in the form of Fig. 14.45 to pass frequencies, between 250 Hz and 3,000 Hz and with K 10. Select R 20 k ., Solution:, 1. Define. The problem is clearly stated and the circuit to be used, in the design is specified., 2. Present. We are asked to use the op amp circuit specified in, Fig. 14.45 to design a bandpass filter. We are given the value of, R to use (20 k ). In addition, the frequency range of the signals, to be passed is 250 Hz to 3 kHz., 3. Alternative. We will use the equations developed in Section 14.8.3, to obtain a solution. We will then use the resulting transfer, function to validate the answer., 4. Attempt. Since 1 1RC2, we obtain, C2 , , 1, 1, , , R1, 2 p f1R, 2p, , 1, 250, , 20, , 103, , 31.83 nF, , Similarly, since 2 1RC1,, C1 , , 1, 1, , , R2, 2 p f2 R, 2p, , 1, 3,000, , 20, , 103, , 2.65 nF, , From Eq. (14.73),, Rf, Ri, , K, , Practice Problem 14.12, , f1 f2, 1 2, 10(3,250), K, , 10.83, 2, f2, 3,000, , Example 14.13
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 648, , 648, , Chapter 14, , Frequency Response, , If we select Ri 10 k , then Rf 10.83Ri 108.3 k ., 5. Evaluate. The output of the first op amp is given by, 109(V1 0), 1, , Vi 0, V1 0, s2.65, , , 20 k, 20 k, 0 S V1 , , Vi, 1 5.3, , 105s, , The output of the second op amp is given by, V1 0, 20 k, V2 , , , , , 1, s31.83 nF, 6.366 104sV1, , , V2 0, 0S, 20 k, , 1 6.366 104s, 6.366 104sVi, , (1 6.366, , 104s)(1 5.3, , 105s), , The output of the third op amp is given by, Vo 0, V2 0, , 0 S Vo 10.83V2 S j2 p, 10 k, 108.3 k, Vo , Let j2 p, , 6.894, (1 6.366, , 25, , 103sVi, , 104s)(1 5.3, , 105s), , 25 and solve for the magnitude of VoVi., j10.829, Vo, , Vi, (1 j1)(1), , 0 Vo Vi 0 (0.7071)10.829, which is the lower corner frequency point., Let s j2 p 3000 j18.849 k . We then get, j129.94, Vo, , Vi, (1 j12)(1 j1), 129.94l90, (0.7071)10.791l18.61, , (12.042l85.24)(1.4142l45), Clearly this is the upper corner frequency and the answer checks., 6. Satisfactory? We have satisfactorily designed the circuit and can, present the results as a solution to the problem., , Practice Problem 14.13, , Design a notch filter based on Fig. 14.47 for 0 20 krad/s, K 5,, and Q 10. Use R Ri 10 k ., Answer: C1 4.762 nF, C2 5.263 nF, and Rf 50 k ., , 14.9, , Scaling, , In designing and analyzing filters and resonant circuits or in circuit, analysis in general, it is sometimes convenient to work with element, values of 1 , 1 H, or 1 F, and then transform the values to realistic
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 649, , 14.9, , Scaling, , values by scaling. We have taken advantage of this idea by not using, realistic element values in most of our examples and problems; mastering circuit analysis is made easy by using convenient component values. We have thus eased calculations, knowing that we could use, scaling to then make the values realistic., There are two ways of scaling a circuit: magnitude or impedance, scaling, and frequency scaling. Both are useful in scaling responses, and circuit elements to values within the practical ranges. While magnitude scaling leaves the frequency response of a circuit unaltered,, frequency scaling shifts the frequency response up or down the frequency spectrum., , 14.9.1 Magnitude Scaling, Magnitude scaling is the process of increasing all impedances in a network by a factor, the frequency response remaining unchanged., , Recall that impedances of individual elements R, L, and C are, given by, ZR R,, , ZL jL,, , ZC , , 1, jC, , (14.78), , In magnitude scaling, we multiply the impedance of each circuit element by a factor Km and let the frequency remain constant. This gives, the new impedances as, Z¿R Km ZR Km R,, , Z¿L Km Z L jKm L, 1, Z¿C Km Z C , jCKm, , (14.79), , Comparing Eq. (14.79) with Eq. (14.78), we notice the following changes, in the element values: R S Km R, L S Km L, and C S CKm. Thus, in, magnitude scaling, the new values of the elements and frequency are, R¿ Km R,, C, C¿ , ,, Km, , L¿ Km L, ¿ , , (14.80), , The primed variables are the new values and the unprimed variables, are the old values. Consider the series or parallel RLC circuit. We, now have, ¿0 , , 1, 2L¿C¿, , , , 1, 2Km LCKm, , , , 1, 2LC, , 0, , (14.81), , showing that the resonant frequency, as expected, has not changed., Similarly, the quality factor and the bandwidth are not affected by magnitude scaling. Also, magnitude scaling does not affect transfer functions in the forms of Eqs. (14.2a) and (14.2b), which are dimensionless, quantities., , 649
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 650, , 650, , Chapter 14, , Frequency Response, , 14.9.2 Frequency Scaling, Frequency scaling is equivalent to relabeling the frequency axis of a frequency response plot. It is needed, when translating frequencies such as a, resonant frequency, a corner frequency,, a bandwidth, etc., to a realistic level. It, can be used to bring capacitance and, inductance values into a range that is, convenient to work with., , Frequency scaling is the process of shifting the frequency response of, a network up or down the frequency axis while leaving the impedance, the same., , We achieve frequency scaling by multiplying the frequency by a factor Kf while keeping the impedance the same., From Eq. (14.78), we see that the impedances of L and C are, frequency-dependent. If we apply frequency scaling to ZL() and ZC (), in Eq. (14.78), we obtain, L, Kf, , ZL j(Kf)L¿ jL, , 1, , L¿ , , 1, 1, , j(Kf)C¿, jC, , 1, , C¿ , , ZC , , C, Kf, , (14.82a), (14.82b), , since the impedance of the inductor and capacitor must remain the, same after frequency scaling. We notice the following changes in the, element values: L S LKf and C S CKf . The value of R is not, affected, since its impedance does not depend on frequency. Thus, in, frequency scaling, the new values of the elements and frequency are, R¿ R,, C¿ , , L¿ , , C, ,, Kf, , L, Kf, , (14.83), , ¿ Kf , , Again, if we consider the series or parallel RLC circuit, for the resonant frequency, ¿0 , , 1, 2L¿C¿, , , , 1, 2(LKf )(CKf ), , , , Kf, 2LC, , Kf 0 (14.84), , and for the bandwidth, B¿ Kf B, , (14.85), , but the quality factor remains the same (Q¿ Q)., , 14.9.3 Magnitude and Frequency Scaling, If a circuit is scaled in magnitude and frequency at the same time, then, R¿ Km R,, 1, C¿ , C,, Km Kf, , L¿ , , Km, L, Kf, , (14.86), , ¿ Kf , , These are more general formulas than those in Eqs. (14.80) and (14.83)., We set Km 1 in Eq. (14.86) when there is no magnitude scaling or, Kf 1 when there is no frequency scaling.
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 651, , 14.9, , Scaling, , 651, , Example 14.14, , A fourth-order Butterworth lowpass filter is shown in Fig. 14.48(a). The, filter is designed such that the cutoff frequency c 1 rad/s. Scale the, circuit for a cutoff frequency of 50 kHz using 10- k resistors., 1Ω, , 0.765 H, , 1.848 H, , 10 kΩ, , 58.82 mH, , 24.35 H, , +, vs, , +, −, , 0.765 F, , 1.848 F, , 1Ω, , +, , vo, , vs, , +, −, , 243.5 pF, , 588.2 pF, , 10 kΩ vo, , −, , −, , (a), , (b), , Figure 14.48, For Example 14.14: (a) Normalized Butterworth lowpass filter, (b) scaled version of the same lowpass filter., , Solution:, If the cutoff frequency is to shift from c 1 rad/s to ¿c 2 p (50), krad/s, then the frequency scale factor is, Kf , , ¿c, 100 p 103, , p, c, 1, , 105, , Also, if each 1- resistor is to be replaced by a 10- k, the magnitude scale factor must be, Km , , resistor, then, , R¿, 10 103, , 104, R, 1, , Using Eq. (14.86),, Km, 10 4, L1 , (1.848) 58.82 mH, Kf, p 105, Km, 104, L¿2 , L2 , (0.765) 24.35 mH, Kf, p 105, C1, 0.765, C¿1 , , 243.5 pF, Km Kf, p 109, L¿1 , , C¿2 , , C2, 1.848, , 588.2 pF, Km Kf, p 109, , The scaled circuit is shown in Fig. 14.48(b). This circuit uses practical, values and will provide the same transfer function as the prototype in, Fig. 14.48(a), but shifted in frequency., , A third-order Butterworth filter normalized to c 1 rad/s is shown, in Fig. 14.49. Scale the circuit to a cutoff frequency of 10 kHz. Use, 15-nF capacitors., , Practice Problem 14.14, , Answer: R¿1 R¿2 1.061 k , C¿1 C¿2 15 nF, L¿ 33.77 mH., , vs +, −, , 1Ω, , 2H, +, 1F, , 1F, , 1Ω, , vo, −, , Figure 14.49, For Practice Prob. 14.14.
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 652, , Chapter 14, , 652, , 14.10, , Frequency Response, , Frequency Response Using PSpice, , PSpice is a useful tool in the hands of the modern circuit designer for, obtaining the frequency response of circuits. The frequency response, is obtained using the AC Sweep as discussed in Section D.5 (Appendix D). This requires that we specify in the AC Sweep dialog box Total, Pts, Start Freq, End Freq, and the sweep type. Total Pts is the number of points in the frequency sweep, and Start Freq and End Freq are,, respectively, the starting and final frequencies, in hertz. In order to, know what frequencies to select for Start Freq and End Freq, one must, have an idea of the frequency range of interest by making a rough, sketch of the frequency response. In a complex circuit where this may, not be possible, one may use a trial-and-error approach., There are three types of sweeps:, Linear: The frequency is varied linearly from Start Freq to End, Freq with Total Pts equally spaced points (or responses)., Octave: The frequency is swept logarithmically by octaves from, Start Freq to End Freq with Total Pts per octave. An octave is, a factor of 2 (e.g., 2 to 4, 4 to 8, 8 to 16)., Decade: The frequency is varied logarithmically by decades from, Start Freq to End Freq with Total Pts per decade. A decade, is a factor of 10 (e.g., from 2 Hz to 20 Hz, 20 Hz to 200 Hz,, 200 Hz to 2 kHz)., It is best to use a linear sweep when displaying a narrow frequency, range of interest, as a linear sweep displays the frequency range well, in a narrow range. Conversely, it is best to use a logarithmic (octave, or decade) sweep for displaying a wide frequency range of interest—, if a linear sweep is used for a wide range, all the data will be crowded, at the high- or low-frequency end and insufficient data at the other end., With the above specifications, PSpice performs a steady-state sinusoidal analysis of the circuit as the frequency of all the independent, sources is varied (or swept) from Start Freq to End Freq., The PSpice A/D program produces a graphical output. The output, data type may be specified in the Trace Command Box by adding one, of the following suffixes to V or I:, M Amplitude of the sinusoid., P, Phase of the sinusoid., dB Amplitude of the sinusoid in decibels, i.e., 20 log10 (amplitude)., , Example 14.15, , Determine the frequency response of the circuit shown in Fig. 14.50., , 8 kΩ, +, vs, , +, 1 kΩ, , −, , Figure 14.50, For Example 14.15., , 1 F, , vo, −, , Solution:, We let the input voltage vs be a sinusoid of amplitude 1 V and phase 0., Figure 14.51 is the schematic for the circuit. The capacitor is rotated 270, counterclockwise to ensure that pin 1 (the positive terminal) is on, top. The voltage marker is inserted to the output voltage across the, capacitor. To perform a linear sweep for 1 6 f 6 1000 Hz with 50, points, we select Analysis/Setup/AC Sweep, DCLICK Linear, type, 50 in the Total Pts box, type 1 in the Start Freq box, and type 1000 in, the End Freq box. After saving the file, we select Analysis/Simulate to, simulate the circuit. If there are no errors, the PSpice A/D window will
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 653, , Frequency Response Using PSpice, , 14.10, , 653, , V, , R1, 8k, ACMAG =1V, ACPHASE =0, , V1, -, , 1k, , R2, , 1u, , C1, , 0, , Figure 14.51, The schematic for the circuit in Fig. 14.50., , display the plot of V(C1:1), which is the same as Vo or H() Vo1,, as shown in Fig. 14.52(a). This is the magnitude plot, since V(C1:1) is, the same as VM(C1:1). To obtain the phase plot, select Trace/Add in, the PSpice A/D menu and type VP(C1:1) in the Trace Command box., Figure 14.52(b) shows the result. By hand, the transfer function is, H() , , Vo, 1,000, , Vs, 9,000 j8, , H() , , 1, 9 j16 p, , or, 10 3, , showing that the circuit is a lowpass filter as demonstrated in Fig. 14.52., Notice that the plots in Fig. 14.52 are similar to those in Fig. 14.3 (note, that the horizontal axis in Fig. 14.52 is logrithic while the horizontal, axis in Fig. 14.3 is linear.), 0 d, 120 mV, –20 d, 80 mV, – 40 d, 40 mV, , – 60 d, , 0 V, 1.0 Hz, , 10 Hz, , 100 Hz, , V(C1:1), Frequency, , 1.0 KHz, , –80 d, 1.0 Hz, , 10 Hz, , 100 Hz 1.0 KHz, , VP(C1:1), Frequency, (b), , (a), , Figure 14.52, For Example 14.15: (a) magnitude plot, (b) phase plot of the frequency response., , Obtain the frequency response of the circuit in Fig. 14.53 using PSpice., Use a linear frequency sweep and consider 1 6 f 6 1000 Hz with, 100 points., Answer: See Fig. 14.54., , Practice Problem 14.15, 1 F, , +, , 6 kΩ, , vs, −, , Figure 14.53, For Practice Prob. 14.15., , +, 2 kΩ, , vo, −
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 654, , Chapter 14, , 654, , Frequency Response, , 1.0 V, 40 d, , 0.5 V, , 20 d, , 0 V, 1.0 Hz, , 10 Hz, , 100 Hz 1.0 KHz, , 0 d, 1.0 Hz, , 100 Hz 1.0 KHz, , 10 Hz, , VP(R2:2), Frequency, , V(R2:2), Frequency, , (b), , (a), , Figure 14.54, For Practice Problem 14.15: (a) magnitude plot, (b) phase plot of the frequency response., , Example 14.16, , ACMAG = 10V, ACPHASE = 0, , Use PSpice to generate the gain and phase Bode plots of V in the circuit of Fig. 14.55., V, , R1, , L1, , 2, , 10mH, , +, V1, −, , 4u, , C1, , Solution:, The circuit treated in Example 14.15 is first-order while the one in this, example is second-order. Since we are interested in Bode plots, we use, decade frequency sweep for 300 6 f 6 3,000 Hz with 50 points per, decade. We select this range because we know that the resonant, frequency of the circuit is within the range. Recall that, , 0, , Figure 14.55, , 0 , , For Example 14.16., , 1, 5 krad/s, 1LC, , or, , f0 , , , 795.8 Hz, 2p, , After drawing the circuit as in Fig. 14.55, we select Analysis/Setup/AC, Sweep, DCLICK Decade, enter 50 in the Total Pts box, 300 as the, Start Freq, and 3,000 in the End Freq box. Upon saving the file, we, simulate it by selecting Analysis/Simulate. This will automatically, bring up the PSpice A/D window and display V(C1:1) if there are no, errors. Since we are interested in the Bode plot, we select Trace/Add, in the PSpice A/D menu and type dB(V(C1:1)) in the Trace Command, box. The result is the Bode magnitude plot in Fig. 14.56(a). For the, 0 d, 50, , –50 d, , –100 d, 0, –150 d, , –50, 100 Hz, 1.0 KHz, dB(V(C1:1)), Frequency, , 10 KHz, , –200 d, 100 Hz, 1.0 KHz, VP(C1:1), Frequency, , (a), , Figure 14.56, For Example 14.16: (a) Bode plot, (b) phase plot of the response., , (b), , 10 KHz
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 655, , 14.11, , Computation Using MATLAB, , 655, , phase plot, we select Trace/Add in the PSpice A/D menu and type, VP(C1:1) in the Trace Command box. The result is the Bode phase, plot of Fig. 14.56(b). Notice that the plots confirm the resonant, frequency of 795.8 Hz., , Consider the network in Fig. 14.57. Use PSpice to obtain the Bode, plots for Vo over a frequency from 1 kHz to 100 kHz using 20 points, per decade., , 1 0° A, , 0.4 mH, , 1 F, , 1 kΩ, , Practice Problem 14.16, , +, Vo, −, , Figure 14.57, For Practice Prob. 14.16., , Answer: See Fig. 14.58., 60, , 0 d, , 40, , –100 d, , 20, , –200 d, , 0, 1.0 KHz, 10 KHz, dB(V(R1:1)), Frequency, , 100 KHz, , –300 d, 1.0 KHz, 10 KHz, VP(R1:1), Frequency, , (a), , Figure 14.58, For Practice Prob. 14.16: Bode (a) magnitude plot, (b) phase plot., , 14.11, , Computation Using MATLAB, , MATLAB is a software package that is widely used for engineering, computation and simulation. A review of MATLAB is provided in, Appendix E for the beginner. This section shows how to use the software to numerically perform most of the operations presented in this, chapter and Chapter 15. The key to describing a system in MATLAB is, to specify the numerator (num) and denominator (den) of the transfer, function of the system. Once this is done, we can use several MATLAB, commands to obtain the system’s Bode plots (frequency response) and, the system’s response to a given input., The command bode produces the Bode plots (both magnitude and, phase) of a given transfer function H(s). The format of the command, is bode (num, den), where num is the numerator of H(s) and den is its, denominator. The frequency range and number of points are automatically selected. For example, consider the transfer function in Example 14.3. It is better to first write the numerator and denominator in, polynomial forms., , (b), , 100 KHz
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ale29559_ch14.qxd, , 07/17/2008, , 01:08 PM, , Page 656, , Chapter 14, , 656, , Frequency Response, , Thus,, 200 j, 200s, ,, 2, s j, ( j 2)( j 10), s 12s 20, Using the following commands, the Bode plots are generated as shown, in Fig. 14.59. If necessary, the command logspace can be included to, generate a logarithmically spaced frequency and the command semilogx, can be used to produce a semilog scale., H(s) , , >> num = [200 0];, % specify the numerator of H(s), >> den = [1 12 20]; % specify the denominator of H(s), >> bode(num, den); % determine and draw Bode plots, The step response y(t) of a system is the output when the input x(t) is, the unit step function. The command step plots the step response of a system given the numerator and denominator of the transfer function of that, system. The time range and number of points are automatically selected., For example, consider a second-order system with the transfer function, H(s) , , 12, s 3s 12, 2, , We obtain the step response of the system shown in Fig. 14.60 by using, the following commands., >> n = 12;, >> d = [1 3 12];, >> step(n,d);, We can verify the plot in Fig. 14.60 by obtaining y (t) x(t) * u(t) or, Y(s) X(s) H(s)., Step response, 1.2, , Phase (deg), , 1, Amplitude, , Magnitude (dB), , Bode diagrams, 20, 10, 0, –10, –20, 50, , 0.8, 0.6, 0.4, 0.2, , 0, 0, , –50, , 0, 10–2, , 101, 10–1, 100, Frequency (rad/s), , Figure 14.59, Magnitude and phase plots., , 0.5, , 1, , 102, , 1.5 2 2.5, Time (s), , 3, , 3.5, , 4, , Figure 14.60, The Step response of, H(s) 12(s2 3s 12)., , The command lsim is a more general command than step. It calculates the time response of a system to any arbitrary input signal. The, format of the command is y lsim (num, den, x, t), where x(t) is the, input signal, t is the time vector, and y(t) is the output generated. For, example, assume a system is described by the transfer function, H(s) , , s4, s 2s2 5s 10, 3
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ale29559_ch14.qxd, , 07/17/2008, , 01:54 PM, , Page 657, , 14.12, , Applications, , To find the response y(t) of the system to input x(t) 10etu(t), we, use the following MATLAB commands. Both the response y(t) and the, input x(t) are plotted in Fig. 14.61., >> t = 0:0.02:5; % time vector 0 < t < 5 with increment, 0.02, >> x = 10*exp(-t);, >> num = [1 4];, >> den = [1 2 5 10];, >> y = lsim(num,den,x,t);, >> plot(t,x,t,y), , y(t), , x(t), , 10, 8, 6, 4, 2, 0, –2, –4, 0, , 0.5, , 1, , 1.5, , 2, , 2.5, , 3, , 3.5, , 4, , 4.5, , 5, , Figure 14.61, , The response of the system described by H(s) , (s 4)(s2 2s2 5s 10) to an exponential, input., , 14.12, , Applications, , Resonant circuits and filters are widely used, particularly in electronics, power systems, and communications systems. For example, a, Notch filter with a cutoff frequency at 60 Hz may be used to eliminate, the 60-Hz power line noise in various communications electronics. Filtering of signals in communications systems is necessary in order to, select the desired signal from a host of others in the same range (as in, the case of radio receivers discussed next) and also to minimize the, effects of noise and interference on the desired signal. In this section,, we consider one practical application of resonant circuits and two, applications of filters. The focus of each application is not to understand the details of how each device works but to see how the circuits, considered in this chapter are applied in the practical devices., , 14.12.1 Radio Receiver, Series and parallel resonant circuits are commonly used in radio and TV, receivers to tune in stations and to separate the audio signal from the, radio-frequency carrier wave. As an example, consider the block diagram, , 657
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 658, , Chapter 14, , 658, , Frequency Response, , of an AM radio receiver shown in Fig. 14.62. Incoming amplitudemodulated radio waves (thousands of them at different frequencies, from different broadcasting stations) are received by the antenna. A, resonant circuit (or a bandpass filter) is needed to select just one of the, incoming waves. The selected signal is very weak and is amplified in, stages in order to generate an audible audio-frequency wave. Thus, we, have the radio frequency (RF) amplifier to amplify the selected broadcast signal, the intermediate frequency (IF) amplifier to amplify an, internally generated signal based on the RF signal, and the audio amplifier to amplify the audio signal just before it reaches the loudspeaker., It is much easier to amplify the signal at three stages than to build an, amplifier to provide the same amplification for the entire band., , Carrier, frequency, Audio frequency, , Amplitudemodulated, radio waves, , 800 kHz, RF, amplifier, , Mixer, , 455 kHz, , IF, amplifier, stages, , 455 kHz, , Detector, , Audio, amplifier, Loudspeaker, , 1255, kHz, Ganged tuning, , Audio to, 5 kHz, , Local, oscillator, , Figure 14.62, A simplified block diagram of a superheterodyne AM radio receiver., , The type of AM receiver shown in Fig. 14.62 is known as the superheterodyne receiver. In the early development of radio, each amplification stage had to be tuned to the frequency of the incoming signal. This, way, each stage must have several tuned circuits to cover the entire AM, band (540 to 1600 kHz). To avoid the problem of having several resonant circuits, modern receivers use a frequency mixer or heterodyne circuit, which always produces the same IF signal (445 kHz) but retains, the audio frequencies carried on the incoming signal. To produce the, constant IF frequency, the rotors of two separate variable capacitors are, mechanically coupled with one another so that they can be rotated, simultaneously with a single control; this is called ganged tuning. A, local oscillator ganged with the RF amplifier produces an RF signal that, is combined with the incoming wave by the frequency mixer to produce, an output signal that contains the sum and the difference frequencies of, the two signals. For example, if the resonant circuit is tuned to receive an, 800-kHz incoming signal, the local oscillator must produce a 1,255-kHz, signal, so that the sum (1,255 800 2,055 kHz) and the difference, (1,255 800 455 kHz) of frequencies are available at the output of
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 659, , 14.12, , Applications, , 659, , the mixer. However, only the difference, 455 kHz, is used in practice., This is the only frequency to which all the IF amplifier stages are tuned,, regardless of the station dialed. The original audio signal (containing, the “intelligence”) is extracted in the detector stage. The detector basically removes the IF signal, leaving the audio signal. The audio signal, is amplified to drive the loudspeaker, which acts as a transducer converting the electrical signal to sound., Our major concern here is the tuning circuit for the AM radio, receiver. The operation of the FM radio receiver is different from that, of the AM receiver discussed here, and in a much different range of, frequencies, but the tuning is similar., , Example 14.17, , The resonant or tuner circuit of an AM radio is portrayed in Fig. 14.63., Given that L 1 mH, what must be the range of C to have the resonant frequency adjustable from one end of the AM band to another?, , RF amplifier, Tuner, , Solution:, The frequency range for AM broadcasting is 540 to 1,600 kHz. We, consider the low and high ends of the band. Since the resonant circuit, in Fig. 14.63 is a parallel type, we apply the ideas in Section 14.6., From Eq. (14.44),, , C, , L, , R, , A, , Input resistance, to amplifier, , 1, 0 2 p f0 , 1LC, , Figure 14.63, The tuner circuit for Example 14.17., , or, C, , 1, 4 p 2 f 20 L, , For the high end of the AM band, f0 1,600 kHz, and the corresponding, C is, C1 , , 4p2, , 1, 1,6002 106, , 106, , 9.9 nF, , For the low end of the AM band, f0 540 kHz, and the corresponding, C is, C2 , , 1, 4p, , 2, , 2, , 540, , 106, , 106, , 86.9 nF, , Thus, C must be an adjustable (gang) capacitor varying from 9.9 nF to, 86.9 nF., , For an FM radio receiver, the incoming wave is in the frequency range, from 88 to 108 MHz. The tuner circuit is a parallel RLC circuit with a, 4-mH coil. Calculate the range of the variable capacitor necessary to, cover the entire band., Answer: From 0.543 pF to 0.818 pF., , Practice Problem 14.17
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660, , 07/10/2008, , 03:50 PM, , Page 660, , Chapter 14, , Frequency Response, , 14.12.2 Touch-Tone Telephone, A typical application of filtering is the touch-tone telephone set, shown in Fig. 14.64. The keypad has 12 buttons arranged in four rows, and three columns. The arrangement provides 12 distinct signals by, using seven tones divided into two groups: the low-frequency group, (697 to 941 Hz) and the high-frequency group (1,209 to 1,477 Hz)., Pressing a button generates a sum of two sinusoids corresponding, to its unique pair of frequencies. For example, pressing the number, 6 button generates sinusoidal tones with frequencies 770 Hz and, 1,477 Hz., , 1, , 697 Hz, , Low-band frequencies, , ale29559_ch14.qxd, , 770 Hz, , 852 Hz, , 941 Hz, , 2, , 3, , ABC, , DEF, , 4, , 5, , 6, , GHI, , JKL, , MNO, , 7, , 8, , 9, , PRS, , TUV, , WXY, , *, , O, , #, , OPER, 1209 Hz, , 1336 Hz, , 1477 Hz, , High-band frequencies, , Figure 14.64, Frequency assignments for touch-tone dialing., Adapted from G. Daryanani, Principles of Active Network, Synthesis and Design [New York: John Wiley & Sons],, 1976, p. 79., , When a caller dials a telephone number, a set of signals is transmitted to the telephone office, where the touch-tone signals are, decoded by detecting the frequencies they contain. Figure 14.65, shows the block diagram for the detection scheme. The signals are, first amplified and separated into their respective groups by the lowpass (LP) and highpass (HP) filters. The limiters (L) are used to convert the separated tones into square waves. The individual tones are, identified using seven bandpass (BP) filters, each filter passing one, tone and rejecting other tones. Each filter is followed by a detector, (D), which is energized when its input voltage exceeds a certain, level. The outputs of the detectors provide the required dc signals, needed by the switching system to connect the caller to the party, being called.
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 661, , 14.12, , LP, , L1, , Lowpass, filter, , Limiter, , A, , Applications, , 661, , BP1, , D1, , 697 Hz, , BP2, , D2, , 770 Hz, , BP3, , D3, , 852 Hz, , BP4, , D4, , 941 Hz, , Bandpass, filters, , Detectors, , BP5, , D5, , 1209 Hz, , BP6, , D6, , 1336 Hz, , BP7, , D7, , 1477 Hz, , Bandpass, filters, , Detectors, , Low-group, signals, , Amplifier, , HP, , L2, , Highpass, filter, , Limiter, , High-group, signals, , Figure 14.65, Block diagram of detection scheme., G. Daryanani, Principles of Active Network Synthesis and Design [New York: John Wiley & Sons],, 1976, p. 79., , Using the standard 600- resistor used in telephone circuits and a, series RLC circuit, design the bandpass filter BP2 in Fig. 14.65., , Example 14.18, , Solution:, The bandpass filter is the series RLC circuit in Fig. 14.35. Since BP2, passes frequencies 697 Hz to 852 Hz and is centered at f0 770 Hz,, its bandwidth is, B 2 p ( f2 f1) 2 p (852 697) 973.89 rad/s, From Eq. (14.39),, L, , R, 600, , 0.616 H, B, 973.89, , From Eq. (14.27) or (14.55),, C, , 1, 1, , , 2 2, 2, 0 L, 4p f 0L, 4p2, , 1, 7702, , 0.616, , 69.36 nF, , Repeat Example 14.18 for bandpass filter BP6., Answer: 0.356 H, 39.83 nF., , 14.12.3 Crossover Network, Another typical application of filters is the crossover network that, couples an audio amplifier to woofer and tweeter speakers, as shown, in Fig. 14.66(a). The network basically consists of one highpass RC, , Practice Problem 14.18
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 662, , Chapter 14, , 662, C, , filter and one lowpass RL filter. It routes frequencies higher than a, prescribed crossover frequency fc to the tweeter (high-frequency, loudspeaker) and frequencies below fc into the woofer (low-frequency, loudspeaker). These loudspeakers have been designed to accommodate certain frequency responses. A woofer is a low-frequency loudspeaker designed to reproduce the lower part of the frequency range,, up to about 3 kHz. A tweeter can reproduce audio frequencies from, about 3 kHz to about 20 kHz. The two speaker types can be combined to reproduce the entire audio range of interest and provide the, optimum in frequency response., By replacing the amplifier with a voltage source, the approximate, equivalent circuit of the crossover network is shown in Fig. 14.66(b),, where the loudspeakers are modeled by resistors. As a highpass filter,, the transfer function V1Vs is given by, , Tweeter, S1, , One channel, of a stereo, amplifier, , L, S2, Woofer, (a), , C, , L, , Vs +, −, , +, V1, −, , R1, , +, V2, −, , R2, , S1, , Frequency Response, , H1() , , S2, , (b), , (14.87), , Similarly, the transfer function of the lowpass filter is given by, , Figure 14.66, (a) A crossover network for two, loudspeakers, (b) equivalent model., , H 2 (), , jR1C, V1, , Vs, 1 jR1C, , H2() , , H1(), , c, , Figure 14.67, Frequency responses of the crossover, network in Fig. 14.66., , Example 14.19, , , , V2, R2, , Vs, R2 jL, , (14.88), , The values of R1, R2, L, and C may be selected such that the two filters, have the same cutoff frequency, known as the crossover frequency, as, shown in Fig. 14.67., The principle behind the crossover network is also used in the resonant circuit for a TV receiver, where it is necessary to separate the, video and audio bands of RF carrier frequencies. The lower-frequency, band (picture information in the range from about 30 Hz to about, 4 MHz) is channeled into the receiver’s video amplifier, while the highfrequency band (sound information around 4.5 MHz) is channeled to, the receiver’s sound amplifier., , In the crossover network of Fig. 14.66, suppose each speaker acts as, a 6- resistance. Find C and L if the crossover frequency is 2.5 kHz., Solution:, For the highpass filter,, 1, R1C, , c 2 p fc , or, C, , 1, , 2 p fc R1, 2p, , 1, 2.5, , 10 3, , 6, , 10.61 mF, , For the lowpass filter,, c 2 p fc , , R2, L, , or, L, , R2, , 2 p fc, 2p, , 6, 2.5, , 103, , 382 mH
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 663, , 14.13, , If each speaker in Fig. 14.66 has an 8find L and the crossover frequency., , Summary, , resistance and C 10 mF,, , Answer: 0.64 mH, 1.989 kHz., , 14.13, , Summary, , 1. The transfer function H() is the ratio of the output response Y(), to the input excitation X(); that is, H() Y()X()., 2. The frequency response is the variation of the transfer function, with frequency., 3. Zeros of a transfer function H(s) are the values of s j that make, H(s) 0, while poles are the values of s that make H(s) S ., 4. The decibel is the unit of logarithmic gain. For a voltage or current, gain G, its decibel equivalent is GdB 20 log10 G., 5. Bode plots are semilog plots of the magnitude and phase of the, transfer function as it varies with frequency. The straight-line, approximations of H (in dB) and f (in degrees) are constructed, using the corner frequencies defined by the poles and zeros of H()., 6. The resonant frequency is that frequency at which the imaginary part, of a transfer function vanishes. For series and parallel RLC circuits., 0 , , 1, 1LC, , 7. The half-power frequencies (1, 2) are those frequencies at which, the power dissipated is one-half of that dissipated at the resonant, frequency. The geometric mean between the half-power frequencies is the resonant frequency, or, 0 112, 8. The bandwidth is the frequency band between half-power, frequencies:, B 2 1, 9. The quality factor is a measure of the sharpness of the resonance, peak. It is the ratio of the resonant (angular) frequency to the, bandwidth,, Q, , 0, B, , 10. A filter is a circuit designed to pass a band of frequencies and, reject others. Passive filters are constructed with resistors, capacitors, and inductors. Active filters are constructed with resistors,, capacitors, and an active device, usually an op amp., 11. Four common types of filters are lowpass, highpass, bandpass, and, bandstop. A lowpass filter passes only signals whose frequencies are, below the cutoff frequency c. A highpass filter passes only signals, whose frequencies are above the cutoff frequency c. A bandpass, filter passes only signals whose frequencies are within a prescribed, , 663, , Practice Problem 14.19
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 664, , Chapter 14, , 664, , Frequency Response, , range (1 6 6 2). A bandstop filter passes only signals whose, frequencies are outside a prescribed range (1 7 7 2)., 12. Scaling is the process whereby unrealistic element values are, magnitude-scaled by a factor Km and/or frequency-scaled by a factor Kf to produce realistic values., R¿ Km R,, , L¿ , , Km, L,, Kf, , C¿ , , 1, C, Km Kf, , 13. PSpice can be used to obtain the frequency response of a circuit, if a frequency range for the response and the desired number of, points within the range are specified in the AC Sweep., 14. The radio receiver—one practical application of resonant circuits—, employs a bandpass resonant circuit to tune in one frequency, among all the broadcast signals picked up by the antenna., 15. The touch-tone telephone and the crossover network are two typical applications of filters. The touch-tone telephone system employs, filters to separate tones of different frequencies to activate electronic switches. The crossover network separates signals in different frequency ranges so that they can be delivered to different, devices such as tweeters and woofers in a loudspeaker system., , Review Questions, 14.1, , A zero of the transfer function, H(s) , , 10(s 1), (s 2)(s 3), , 14.7, , is at, (a) 10, 14.2, , (b) 1, , (c) 2, , (a) 20 dB/decade, , 14.8, , (b) 40 dB/decade, , (c) 40 dB/decade (d) 20 dB/decade, 14.3, , 14.4, , 14.5, , 14.6, , On the Bode phase plot for 0.5 6 6 50, the slope, of [1 j10 225]2 is, (a) 45/decade, , (b) 90/decade, , (c) 135/decade, , (d) 180/decade, , How much inductance is needed to resonate at 5 kHz, with a capacitance of 12 nF?, (a) 2,652 H, , (b) 11.844 H, , (c) 3.333 H, , (d) 84.43 mH, , The difference between the half-power frequencies is, called the:, (a) quality factor, , (b) resonant frequency, , (c) bandwidth, , (d) cutoff frequency, , In a series RLC circuit, which of these quality factors, has the steepest magnitude response curve near, resonance?, , (b) Q 12, , (c) Q 8, , (d) Q 4, , In a parallel RLC circuit, the bandwidth B is directly, proportional to R., (a) True, , (d) 3, , On the Bode magnitude plot, the slope of 1(5 j)2, for large values of is, , (a) Q 20, , 14.9, , (b) False, , When the elements of an RLC circuit are both, magnitude-scaled and frequency-scaled, which, quality is unaffected?, (a) resistor, , (b) resonant frequency, , (c) bandwidth, , (d) quality factor, , What kind of filter can be used to select a signal of, one particular radio station?, (a) lowpass, , (b) highpass, , (c) bandpass, , (d) bandstop, , 14.10 A voltage source supplies a signal of constant, amplitude, from 0 to 40 kHz, to an RC lowpass filter., A load resistor, connected in parallel across the, capacitor, experiences the maximum voltage at:, (a) dc, , (b) 10 kHz, , (c) 20 kHz, , (d) 40 kHz, , Answers: 14.1b, 14.2c, 14.3d, 14.4d, 14.5c, 14.6a,, 14.7b, 14.8d, 14.9c, 14.10a.
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 665, , Problems, , 665, , Problems, Section 14.2 Transfer Function, 14.1, , 14.5, , Find the transfer function VoVi of the RC circuit, in Fig. 14.68. Express it using 0 1RC., , For each of the circuits shown in Fig. 14.72, find, H(s) Vo(s)Vs(s)., , C, , v i (t) +, −, , Rs, , R, , +, v o (t), −, , Vs +, −, , L, , +, Vo, −, , C, , +, Vo, −, , R, , Figure 14.68, , (a), , For Prob. 14.1., L, , 14.2, , Using Fig. 14.69, design a problem to help other, students better understand how to determine, transfer functions., , Vs +, −, , R, , R1, +, Vo, −, , R2, , Vi +, −, , (b), , Figure 14.72, For Prob. 14.5., , C, , Figure 14.69, For Prob. 14.2., 14.3, , 14.6, , Given the circuit in Fig. 14.70, R1 2 , R2 5 ,, C1 0.1 F, and C2 0.2 F, determine the transfer, function H(s) Vo(s)Vi(s)., R1, , Vi, , +, −, , R2, , 1H, , C2, , C1, , Io, , +, Vo, −, , Figure 14.70, , +, Is, , 1Ω, , Vo, −, , 1Ω, , 1H, , Figure 14.73, , For Prob. 14.3., 14.4, , For the circuit shown in Fig. 14.73, find H(s) , Io(s)Is(s)., , Find the transfer function H() VoVi of the, circuits shown in Fig. 14.71., , For Prob. 14.6., , L, +, , +, , Vi, , C, , R, , Vo, , Section 14.3 The Decibel Scale, 14.7, , −, , −, , (a) 0.05 dB, 14.8, , (a), C, +, , Calculate |H()| if HdB equals, (b) 6.2 dB, , (c) 104.7 dB, , Design a problem to help other students, calculate the magnitude in dB and phase in, degrees of a variety of transfer functions at a, single value of ., , +, R, , Vi, , Vo, L, −, , −, (b), , Figure 14.71, For Prob. 14.4., , Section 14.4 Bode Plots, 14.9, , A ladder network has a voltage gain of, H() , , 10, (1 j)(10 j), , Sketch the Bode plots for the gain.
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 666, , Chapter 14, , 666, , Frequency Response, , 14.10, , Design a problem to help other students better, understand how to determine the Bode magnitude, and phase plots of a given transfer function in, terms of j., , 14.11, , Sketch the Bode plots for, H() , , 14.12, , 10 j, j(2 j), , 14.20, , Design a more complex problem than given in, Prob. 14.10, to help other students better, understand how to determine the Bode magnitude, and phase plots of a given transfer function in, terms of j. Include at least a second order, repeated root., , 14.21, , Sketch the magnitude Bode plot for, H(s) , , A transfer function is given by, T(s) , , s1, s(s 10), , 14.22, , Sketch the magnitude and phase Bode plots., 14.13, , H (dB), , s j, , 50( j 1), , 20, , j( 10 j 25), 2, , s j, , 0, 10, , 100, , 1000, , (rad/s), , Figure 14.74, For Prob. 14.22., , 10, , s j, s(s s 16), , 14.23, , 2, , The Bode magnitude plot of H() is shown in, Fig. 14.75. Find H()., , Sketch the Bode plots for, s, ,, (s 2)2(s 1), , s j, , H (dB), , A linear network has this transfer function, 7s 2 s 4, ,, H(s) 3, s 8s 2 14s 5, , 0, , 1, , s j, , 10, , 100, , (rad/s), , +20 dB/decade, , Use MATLAB or equivalent to plot the magnitude, and phase (in degrees) of the transfer function., Take 0.1 6 6 10 rad/s., 14.19, , –20 dB/decade, , 40, , Sketch Bode magnitude and phase plots for, , G(s) , 14.18, , s1, ,, s (s 10), 2, , 40(s 1), ,, (s 2)(s 10), , H(s) , 14.17, , s j, , Find the transfer function H() with the Bode, magnitude plot shown in Fig. 14.74., , Construct the Bode magnitude and phase plots for, H(s) , , 14.16, , ,, , Draw the Bode plots for, H() , , 14.15, , (s 1)(s 2 60s 400), , Construct the Bode plots for, G(s) , , 14.14, , s(s 20), , – 40 dB/decade, , Figure 14.75, For Prob. 14.23., , Sketch the asymptotic Bode plots of the magnitude, and phase for, H(s) , , 100s, ,, (s 10)(s 20)(s 40), , s j, , H (dB), 40, , 14.24, , The magnitude plot in Fig. 14.76 represents the, transfer function of a preamplifier. Find H(s)., , 20 dB/decade, , 2,122, 0, 50, , Figure 14.76, For Prob. 14.24., , 500, , , 20 dB/decade
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 667, , Problems, , Section 14.5 Series Resonance, 14.25, , 14.35, , A series RLC network has R 2 k , L 40 mH,, and C 1 mF. Calculate the impedance at, resonance and at one-fourth, one-half, twice, and, four times the resonant frequency., , 14.26, , Design a problem to help other students better, understand 0, Q, and B at resonance in series, RLC circuits., , 14.27, , Design a series RLC resonant circuit with 0 , 40 rad/s and B 10 rad/s., , 14.28, , Design a series RLC circuit with B 20 rad/s and, 0 1,000 rad/s. Find the circuit’s Q. Let, R 10 ., , 14.29, , 667, , A parallel RLC circuit has R 5 k , L 8 mH,, and C 60 mF. Determine:, (a) the resonant frequency, (b) the bandwidth, (c) the quality factor, , 14.36, , It is expected that a parallel RLC resonant circuit, has a midband admittance of 25 103 S, quality, factor of 80, and a resonant frequency of 200 krad/s., Calculate the values of R, L, and C. Find the, bandwidth and the half-power frequencies., , 14.37, , Rework Prob. 14.25 if the elements are connected, in parallel., , 14.38, , Find the resonant frequency of the circuit in, Fig. 14.78., , Let vs 120 cos(at) V in the circuit of Fig. 14.77., Find 0, Q, and B, as seen by the capacitor., , C, 12 kΩ, R, , L, vs, , +, −, , 45 kΩ, , 1 F, , 60 mH, , Figure 14.78, Figure 14.77, , For Prob. 14.38., , For Prob. 14.29., 14.39, 14.30, , A circuit consisting of a coil with inductance 10 mH, and resistance 20 is connected in series with a, capacitor and a generator with an rms voltage of, 120 V. Find:, , For the “tank” circuit in Fig. 14.79, find the, resonant frequency., , 40 mH, , (a) the value of the capacitance that will cause the, circuit to be in resonance at 15 kHz, , Io cos t, , 1 F, 50 Ω, , (b) the current through the coil at resonance, (c) the Q of the circuit, , Figure 14.79, Section 14.6 Parallel Resonance, 14.31, , 14.32, , 14.33, , 14.34, , Design a parallel resonant RLC circuit with, 0 10 rad/s and Q 20. Calculate the, bandwidth of the circuit. Let R 10 ., , For Probs. 14.39 and 14.91., , 14.40, , Design a problem to help other students better, understand the quality factor, the resonant, frequency, and bandwidth of a parallel RLC circuit., , (a) the capacitance, (b) the inductance, (c) the resonant frequency, , A parallel resonant circuit with quality factor 120 has, a resonant frequency of 6 106 rad/s. Calculate the, bandwidth and half-power frequencies., A parallel RLC circuit is resonant at 5.6 MHz, has a, Q of 80, and has a resistive branch of 40 k ., Determine the values of L and C in the other two, branches., , A parallel resonance circuit has a resistance of, 2 k and half-power frequencies of 86 kHz and, 90 kHz. Determine:, , (d) the bandwidth, (e) the quality factor, 14.41, , Using Fig. 14.80, design a problem to help, other students better understand the quality factor,, the resonant frequency, and bandwidth of RLC, circuits.
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 668, , Chapter 14, , 668, , Frequency Response, 30 kΩ, , R2, , C, , R1, , Vs, , L, , Figure 14.80, , 50 F, , 50 kΩ, , 10 mH, , Figure 14.84, , For Prob. 14.41., 14.42, , +, −, , For Prob. 14.45., , For the circuits in Fig. 14.81, find the resonant, frequency 0, the quality factor Q, and the, bandwidth B., , 14.46, , For the network illustrated in Fig. 14.85, find, (a) the transfer function H() Vo()I(),, (b) the magnitude of H at 0 1 rad/s., , 2Ω, , 20 mH, 6Ω, , 1Ω, , 3 F, , 1H, 2 kΩ, , 6 F, , 0.4 F, , I, , 1Ω, , 1H, , 1Ω, , 1F, , +, Vo, −, , (b), , (a), , Figure 14.81, , Figure 14.85, , For Prob. 14.42., , For Probs. 14.46, 14.78, and 14.92., , 14.43, , Section 14.7 Passive Filters, , Calculate the resonant frequency of each of the, circuits in Fig. 14.82., , 14.47, , Show that a series LR circuit is a lowpass filter if, the output is taken across the resistor. Calculate the, corner frequency fc if L 2 mH and R 10 k ., , 14.48, , Find the transfer function VoVs of the circuit in, Fig. 14.86. Show that the circuit is a lowpass filter., , C, , L, , C, , R, , R, , (a), , L, , 1H, , (b), , Figure 14.82, For Prob. 14.43., , vs +, −, , 0.25 Ω, , 1F, , +, vo, −, , *14.44 For the circuit in Fig. 14.83, find:, , Figure 14.86, , (a) the resonant frequency 0, , For Prob. 14.48., , (b) Zin(0), 9 F, Zin, , 1Ω, , 20 mH, , 0.1 Ω, , 14.49, , Design a problem to help other students better, understand lowpass filters described by transfer, functions., , 14.50, , Determine what type of filter is in Fig. 14.87., Calculate the corner frequency fc., , Figure 14.83, , 200 Ω, , For Prob. 14.44., 14.45, , For the circuit shown in Fig. 14.84, find 0, B, and, Q, as seen by the voltage across the inductor., , v i (t) +, −, , Figure 14.87, * An asterisk indicates a challenging problem., , For Prob. 14.50., , 0.1 H, , +, v o(t), −
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 669, , Problems, , 14.51, , Design an RL lowpass filter that uses a 40-mH coil, and has a cutoff frequency of 5 kHz., , 14.52, , Design a problem to help other students better, understand passive highpass filters., , 14.53, , Design a series RLC type bandpass filter with, cutoff frequencies of 10 kHz and 11 kHz., Assuming C 80 pF, find R, L, and Q., , 14.54, , Design a passive bandstop filter with 0 10 rad/s, and Q 20., , 14.55, , Determine the range of frequencies that will be, passed by a series RLC bandpass filter with, R 10 , L 25 mH, and C 0.4 mF. Find the, quality factor., , 14.56, , 6Ω, , sB, ,, s2 sB 20, , s j, , +, , 4 F, Vi, , +, −, , 4Ω, , 1 mH, , Vo, –, , Figure 14.89, For Prob. 14.59., , Section 14.8 Active Filters, 14.60, , Obtain the transfer function of a highpass filter, with a passband gain of 10 and a cutoff frequency, of 50 rad/s., , 14.61, , Find the transfer function for each of the active, filters in Fig. 14.90., , (a) Show that for a bandpass filter,, H(s) , , 669, , where B bandwidth of the filter and 0 is the, center frequency., (b) Similarly, show that for a bandstop filter,, H(s) , 14.57, , s2 20, s2 sB 20, , ,, , −, +, , R, , s j, +, vi, –, , Determine the center frequency and bandwidth of, the bandpass filters in Fig. 14.88., , C, , +, vo, –, , (a), 1F, , 1Ω, Vs +, −, , 1F, , 1Ω, , +, Vo, −, , −, +, , C, +, vi, –, , (a), , R, , +, vo, –, , 1Ω, , 1H, , (b), Vs +, −, , 1Ω, , 1H, , +, Vo, −, , Figure 14.90, For Probs. 14.61 and 14.62., , (b), , Figure 14.88, , 14.62, , For Prob. 14.57., , 14.58, , The circuit parameters for a series RLC bandstop, filter are R 2 k , L 0.1 H, C 40 pF., Calculate:, , (a) 200 Hz, 14.63, , (a) the center frequency, , (c) 10 kHz, , Design an active first-order highpass filter with, 100s, ,, s 10, , s j, , Use a 1-mF capacitor., , (c) the quality factor, Find the bandwidth and center frequency of the, bandstop filter of Fig. 14.89., , (b) 2 kHz, , H(s) , , (b) the half-power frequencies, 14.59, , The filter in Fig. 14.90(b) has a 3-dB cutoff, frequency at 1 kHz. If its input is connected to a, 120-mV variable frequency signal, find the output, voltage at:, , 14.64, , Obtain the transfer function of the active filter in, Fig. 14.91 on the next page. What kind of filter is it?
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 670, , Chapter 14, , 670, , Frequency Response, , Rf, , 14.67, , Design an active lowpass filter with dc gain of 0.25, and a corner frequency of 500 Hz., , Cf, , 14.68, , Design a problem to help other students better, understand the design of active highpass filters, when specifying a high-frequency gain and a, corner frequency., , 14.69, , Design the filter in Fig. 14.94 to meet the following, requirements:, , Ci, , Ri, , −, +, , +, vi, , +, vo, , –, , –, , (a) It must attenuate a signal at 2 kHz by 3 dB, compared with its value at 10 MHz., , Figure 14.91, , (b) It must provide a steady-state output of vo(t) , 10 sin(2 p 10 8t 180) V for an input, vs(t) 4 sin(2 p 10 8t) V., , For Prob. 14.64., 14.65, , A highpass filter is shown in Fig. 14.92. Show that, the transfer function is, , Rf, , Rf, , jRC, b, H() a1 , Ri 1 jRC, , C, , R, , −, +, , +, vo, –, , vs +, −, , C, +, −, , +, , +, Rf, , vi, , R, , Figure 14.94, , vo, , For Prob. 14.69., , Ri, –, , –, , *14.70 A second-order active filter known as a Butterworth, filter is shown in Fig. 14.95., , Figure 14.92, , (a) Find the transfer function VoVi., , For Prob. 14.65., 14.66, , (b) Show that it is a lowpass filter., , A “general” first-order filter is shown in Fig. 14.93., , C1, , (a) Show that the transfer function is, H(s) , , R1, , s (1R1C)[R1R2 R3R4], ,, s 1R2C, , R4, R3 R4, , Vi, , (b) What condition must be satisfied for the circuit, to operate as a highpass filter?, (c) What condition must be satisfied for the circuit, to operate as a lowpass filter?, , C, −, +, R3, , –, , For Prob. 14.66., , +, Vo, –, , Figure 14.95, For Prob. 14.70., , 14.71, , Use magnitude and frequency scaling on the circuit, of Fig. 14.76 to obtain an equivalent circuit in, which the inductor and capacitor have magnitude, 1 H and 1 F respectively., , 14.72, , Design a problem to help other students better, understand magnitude and frequency scaling., , 14.73, , Calculate the values of R, L, and C that will result, in R 12 k , L 40 mH, and C 300 nF, respectively when magnitude-scaled by 800 and, frequency-scaled by 1000., , vo, , R4, , Figure 14.93, , C2, , Section 14.9 Scaling, , R2, , R1, , +, −, , +, , s j, , vs, , R2
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 671, , Problems, , 14.74, , 14.75, , 14.76, , 14.77, , A circuit has R1 3 , R2 10 , L 2H, and, C 110 F. After the circuit is magnitude-scaled, by 100 and frequency-scaled by 106, find the new, values of the circuit elements., , 14.81, , 671, , The circuit shown in Fig. 14.98 has the impedance, Z(s) , , 1,000(s 1), ,, (s 1 j50)(s 1 j50), , Find:, , In an RLC circuit, R 20 , L 4 H, and, C 1 F. The circuit is magnitude-scaled by 10 and, frequency-scaled by 105. Calculate the new values, of the elements., , (a) the values of R, L, C, and G, (b) the element values that will raise the resonant, frequency by a factor of 103 by frequency scaling, , Given a parallel RLC circuit with R 5 k ,, L 10 mH, and C 20 mF, if the circuit is, magnitude-scaled by Km 500 and frequencyscaled by Kf 105, find the resulting values of, R, L, and C., A series RLC circuit has R 10 , 0 40 rad/s,, and B 5 rad/s. Find L and C when the circuit is, scaled:, (a) in magnitude by a factor of 600,, , s j, , R, C, , G, L, , Z(s), , Figure 14.98, For Prob. 14.81., , (b) in frequency by a factor of 1,000,, (c) in magnitude by a factor of 400 and in, frequency by a factor of 105., 14.78, , 14.82, , Scale the lowpass active filter in Fig. 14.99 so, that its corner frequency increases from 1 rad/s to, 200 rad/s. Use a 1-mF capacitor., , Redesign the circuit in Fig. 14.85 so that all, resistive elements are scaled by a factor of 1,000, and all frequency-sensitive elements are frequencyscaled by a factor of 104., , 2Ω, 1F, , *14.79 Refer to the network in Fig. 14.96., , 1Ω, , (a) Find Zin(s)., (b) Scale the elements by Km 10 and Kf 100., Find Zin(s) and 0., 4Ω, 5Ω, , 0.1 F, , +−, , Zin(s), , 3Vo, , 2H, , −, +, , +, , +, , Vi, , Vo, –, , –, , Figure 14.99, +, Vo, −, , For Prob. 14.82., 14.83, , The op amp circuit in Fig. 14.100 is to be, magnitude-scaled by 100 and frequency-scaled by, 105. Find the resulting element values., , Figure 14.96, For Prob. 14.79., 14.80, , 1 F, , (a) For the circuit in Fig. 14.97, draw the new, circuit after it has been scaled by Km 200 and, Kf 104., , 10 kΩ, , (b) Obtain the Thevenin equivalent impedance, at terminals a-b of the scaled circuit at, 104 rad/s., , +, −, , 5 F, , +, −, vo, , Figure 14.100, , 1H, a, , For Prob. 14.83., , Ix, 0.5 F, , vs, , 20 kΩ, , 2Ω, , 0.5I x, , Section 14.10 Frequency Response Using PSpice, , b, , Figure 14.97, For Prob. 14.80., , 14.84, , Using PSpice, obtain the frequency response of the, circuit in Fig. 14.101 on the next page.
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 672, , Chapter 14, , 672, , Frequency Response, , 1 F, , 4 kΩ, , 14.89, , +, , +, , Vi, , Obtain the magnitude plot of the response Vo in the, network of Fig. 14.106 for the frequency interval, 100 6 f 6 1,000 Hz., , Vo, , 1 kΩ, , −, , −, , 50 Ω, , Figure 14.101, , 10 F, , For Prob. 14.84., 14.85, , Use PSpice to obtain the magnitude and phase plots, of VoIs of the circuit in Fig. 14.102., , 10 Ω, , 1 0° A, , 20 Ω, , 4 mH, , +, Vo, −, , 10 nF, , Figure 14.106, For Prob. 14.89., Is, , 200 Ω, , 30 mH 100 Ω, , +, Vo, –, , 14.90, , Obtain the frequency response of the circuit in, Fig. 14.40 (see Practice Problem 14.10). Take R1 , R2 100 , L 2 mH. Use 1 6 f 6 100,000 Hz., , 14.91, , For the “tank” circuit of Fig. 14.79, obtain the, frequency response (voltage across the capacitor), using PSpice. Determine the resonant frequency of, the circuit., , 14.92, , Using PSpice, plot the magnitude of the frequency, response of the circuit in Fig. 14.85., , Figure 14.102, For Prob. 14.85., 14.86, , Using Fig. 14.103, design a problem to help other, students better understand how to use PSpice to, obtain the frequency response (magnitude and, phase of I) in electrical circuits., R2, , R1, , I, , +, Vo, −, , +, Vs −, , R3, , C, , kVo, , L, , Section 14.12 Applications, 14.93, , For the phase shifter circuit shown in Fig. 14.107,, find H VoVs., , Figure 14.103, For Prob. 14.86., 14.87, , R, , In the interval 0.1 6 f 6 100 Hz, plot the, response of the network in Fig. 14.104. Classify, this filter and obtain 0., 1F, , 1F, , 1F, , 1Ω, , 1Ω, , Vo, −, , −, , 14.94, , For an emergency situation, an engineer needs, to make an RC highpass filter. He has one, 10-pF capacitor, one 30-pF capacitor, one, 1.8-k resistor, and one 3.3-k resistor available., Find the greatest cutoff frequency possible using, these elements., , 14.95, , A series-tuned antenna circuit consists of a, variable capacitor (40 pF to 360 pF) and a, 240-mH antenna coil that has a dc resistance, of 12 ., , Figure 14.104, For Prob. 14.87., 14.88, , Use PSpice to generate the magnitude and phase, Bode plots of Vo in the circuit of Fig. 14.105., 1Ω, , 1 0° V +, −, , Figure 14.105, For Prob. 14.88., , 2F, , 2H, , 1F, , R, , For Prob. 14.93., , +, 1Ω, , C, , C, , Figure 14.107, , +, Vi, , Vs +, −, , + Vo −, , 1H, , 1Ω, , +, Vo, −, , (a) Find the frequency range of radio signals to, which the radio is tunable., (b) Determine the value of Q at each end of the, frequency range.
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ale29559_ch14.qxd, , 07/10/2008, , 03:50 PM, , Page 673, , Comprehensive Problems, , 14.96, , The crossover circuit in Fig. 14.108 is a lowpass, filter that is connected to a woofer. Find the transfer, function H() Vo()Vi ()., , 14.97, , 673, , The crossover circuit in Fig. 14.109 is a highpass, filter that is connected to a tweeter. Determine the, transfer function H() Vo()Vi ()., , Tweeter, Woofer, , Amplifier, Ri, , C1, , C2 RL, , C1, , Ri, , +, Vo, −, , Woofer, , Amplifier, , Speakers, , L, , Vi +, −, , Tweeter, , C2, , Vi +, −, , Figure 14.109, , For Prob. 14.96., , For Prob. 14.97., , +, Vo, −, , RL, , L, , Figure 14.108, , Speakers, , Comprehensive Problems, 14.98, , A certain electronic test circuit produced a resonant, curve with half-power points at 432 Hz and 454 Hz., If Q 20, what is the resonant frequency of the, circuit?, , 14.99, , In an electronic device, a series circuit is employed, that has a resistance of 100 , a capacitive, reactance of 5 k , and an inductive reactance of, 300 when used at 2 MHz. Find the resonant, frequency and bandwidth of the circuit., , 14.100 In a certain application, a simple RC lowpass filter, is designed to reduce high frequency noise. If, the desired corner frequency is 20 kHz and, C 0.5 mF, find the value of R., 14.101 In an amplifier circuit, a simple RC highpass filter, is needed to block the dc component while passing, the time-varying component. If the desired rolloff, frequency is 15 Hz and C 10 mF, find the value, of R., , 14.103 The RC circuit in Fig. 14.111 is used for a lead, compensator in a system design. Obtain the transfer, function of the circuit., C, R1, From, photoresistor, output, , Vi, , For Prob. 14.103., 14.104 A low-quality-factor, double-tuned bandpass filter, is shown in Fig. 14.112. Use PSpice to generate the, magnitude plot of Vo()., 0.2 F, , 40 Ω, , +, , 1.24 mH, 2 F, , Vo, 0.124 mH, –, , R, , Figure 14.112, , For Prob. 14.102., , −, , −, , 1 0° V +, −, , (b) Rs 1 k , RL 5 k ., , Figure 14.110, , To, amplifier, input, , Vo, , 4Ω, , (a) Rs 0, RL ,, , Vs +, −, , +, R2, , Figure 14.111, , 14.102 Practical RC filter design should allow for source, and load resistances as shown in Fig. 14.110. Let, R 4 k and C 40-nF. Obtain the cutoff, frequency when:, , Rs, , +, , C, , RL, , For Prob. 14.104.
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ale29559_ch15.qxd, , 07/10/2008, , 04:05 PM, , Page 674, , P A R T, , T H R E E, , Advanced Circuit, Analysis, OUTLINE, 15, , Introduction to the Laplace Transform, , 16, , Applications of the Laplace Transform, , 17, , The Fourier Series, , 18, , Fourier Transform, , 19, , Two-Port Networks
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ale29559_ch15.qxd, , 07/10/2008, , 04:05 PM, , Page 675, , c h a p t e r, , Introduction to the, Laplace Transform, , 15, , The important thing about a problem is not its solution, but the strength, we gain in finding the solution., —Anonymous, , Enhancing Your Skills and Your Career, ABET EC 2000 criteria (3.h), “the broad education necessary, to understand the impact of engineering solutions in a, global and societal context.”, As a student, you must make sure you acquire “the broad education, necessary to understand the impact of engineering solutions in a global, and societal context.” To some extent, if you are already enrolled in an, ABET-accredited engineering program, then some of the courses you, are required to take must meet this criteria. My recommendation is that, even if you are in such a program, you look at all the elective courses, you take to make sure that you expand your awareness of global issues, and societal concerns. The engineers of the future must fully understand that they and their activities affect all of us in one way or another., , Photo by Charles Alexander, , ABET EC 2000 criteria (3.i), “need for, and an ability to, engage in life-long learning.”, You must be fully aware of and recognize the “need for, and an ability to engage in life-long learning.” It almost seems absurd that this, need and ability must be stated. Yet, you would be surprised at how, many engineers do not really understand this concept. The only way to, be really able to keep up with the explosion in technology we are facing now and will be facing in the future is through constant learning., This learning must include nontechnical issues as well as the latest, technology in your field., The best way to keep up with the state of the art in your field is, through your colleagues and association with individuals you meet, through your technical organization or organizations (especially IEEE)., Reading state-of-the-art technical articles is the next best way to stay, current., , 675
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ale29559_ch15.qxd, , 676, , 07/10/2008, , 04:06 PM, , Page 676, , Chapter 15, , Introduction to the Laplace Transform, , Historical, Pierre Simon Laplace (1749–1827), a French astronomer and mathematician, first presented the transform that bears his name and its, applications to differential equations in 1779., Born of humble origins in Beaumont-en-Auge, Normandy, France,, Laplace became a professor of mathematics at the age of 20. His mathematical abilities inspired the famous mathematician Simeon Poisson,, who called Laplace the Isaac Newton of France. He made important, contributions in potential theory, probability theory, astronomy, and, celestial mechanics. He was widely known for his work, Traite de, Mecanique Celeste (Celestial Mechanics), which supplemented the, work of Newton on astronomy. The Laplace transform, the subject of, this chapter, is named after him., , 15.1, , Introduction, , Our goal in this and the following chapters is to develop techniques, for analyzing circuits with a wide variety of inputs and responses. Such, circuits are modeled by differential equations whose solutions describe, the total response behavior of the circuits. Mathematical methods have, been devised to systematically determine the solutions of differential, equations. We now introduce the powerful method of Laplace transformation, which involves turning differential equations into algebraic, equations, thus greatly facilitating the solution process., The idea of transformation should be familiar by now. When using, phasors for the analysis of circuits, we transform the circuit from, the time domain to the frequency or phasor domain. Once we obtain, the phasor result, we transform it back to the time domain. The Laplace, transform method follows the same process: we use the Laplace transformation to transform the circuit from the time domain to the frequency domain, obtain the solution, and apply the inverse Laplace, transform to the result to transform it back to the time domain., The Laplace transform is significant for a number of reasons. First,, it can be applied to a wider variety of inputs than phasor analysis. Second, it provides an easy way to solve circuit problems involving initial conditions, because it allows us to work with algebraic equations, instead of differential equations. Third, the Laplace transform is capable of providing us, in one single operation, the total response of the, circuit comprising both the natural and forced responses., We begin with the definition of the Laplace transform which gives, rise to its most essential properties. By examining these properties, we, shall see how and why the method works. This also helps us to better, appreciate the idea of mathematical transformations. We also consider, some properties of the Laplace transform that are very helpful in circuit, analysis. We then consider the inverse Laplace transform, transfer functions, and convolution. In this chapter, we will focus on the mechanics of the Laplace transformation. In Chapter 16 we will examine how
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 677, , 15.2, , Definition of the Laplace Transform, , 677, , the Laplace transform is applied in circuit analysis, network stability,, and network synthesis., , 15.2, , Definition of the Laplace Transform, , Given a function f(t), its Laplace transform, denoted by F(s) or L[ f (t)],, is defined by, , L[ f (t)] F(s) , , , , , , f (t)est dt, , (15.1), , 0, , where s is a complex variable given by, s s j, , (15.2), , Since the argument st of the exponent e in Eq. (15.1) must be dimensionless, it follows that s has the dimensions of frequency and units of, inverse seconds (s1) or “frequency.” In Eq. (15.1), the lower limit is, specified as 0 to indicate a time just before t 0. We use 0 as the, lower limit to include the origin and capture any discontinuity of f(t) at, t 0; this will accommodate functions—such as singularity functions—, that may be discontinuous at t 0., It should be noted that the integral in Eq. (15.1) is a definite integral with respect to time. Hence, the result of integration is independent of time and only involves the variable “s.”, Equation (15.1) illustrates the general concept of transformation., The function f(t) is transformed into the function F(s). Whereas the, former function involves t as its argument, the latter involves s. We say, the transformation is from t-domain to s-domain. Given the interpretation of s as frequency, we arrive at the following description of the, Laplace transform:, The Laplace transform is an integral transformation of a function f (t ), from the time domain into the complex frequency domain, giving F (s)., , When the Laplace transform is applied to circuit analysis, the differential equations represent the circuit in the time domain. The terms, in the differential equations take the place of f(t). Their Laplace transform, which corresponds to F(s), constitutes algebraic equations representing the circuit in the frequency domain., We assume in Eq. (15.1) that f (t) is ignored for t 6 0. To ensure, that this is the case, a function is often multiplied by the unit step., Thus, f(t) is written as f(t)u(t) or f (t), t 0., The Laplace transform in Eq. (15.1) is known as the one-sided (or, unilateral) Laplace transform. The two-sided (or bilateral) Laplace, transform is given by, F(s) , , , , , , f (t)est dt, , (15.3), , , , The one-sided Laplace transform in Eq. (15.1), being adequate for our, purposes, is the only type of Laplace transform that we will treat in, this book., , For an ordinary function f (t ), the lower, limit can be replaced by 0.
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 678, , Chapter 15, , 678, , 0 e jt 0 2 cos 2 t sin 2 t 1, , Introduction to the Laplace Transform, , A function f (t) may not have a Laplace transform. In order for f(t), to have a Laplace transform, the integral in Eq. (15.1) must converge, to a finite value. Since 0 e jt 0 1 for any value of t, the integral converges when, , , , , , 0, , j, , 0, , c, , 1, , , , est 0 f (t) 0 dt 6 , , (15.4), , for some real value s sc. Thus, the region of convergence for the, Laplace transform is Re(s) s 7 sc, as shown in Fig. 15.1. In this, region, 0F(s) 0 6 and F(s) exists. F(s) is undefined outside the, region of convergence. Fortunately, all functions of interest in circuit, analysis satisfy the convergence criterion in Eq. (15.4) and have, Laplace transforms. Therefore, it is not necessary to specify sc in, what follows., A companion to the direct Laplace transform in Eq. (15.1) is the, inverse Laplace transform given by, L1[F(s)] f (t) , , Figure 15.1, Region of convergence for the Laplace, transform., , , , 1, 2pj, , , , s1 j, , F(s)e st ds, , (15.5), , s1j, , where the integration is performed along a straight line (s1 j, 6, 6 ) in the region of convergence, s1 7 sc. See Fig. 15.1. The, direct application of Eq. (15.5) involves some knowledge about complex analysis beyond the scope of this book. For this reason, we will, not use Eq. (15.5) to find the inverse Laplace transform. We will rather, use a look-up table, to be developed in Section 15.3. The functions f(t), and F(s) are regarded as a Laplace transform pair where, f (t), , 3, , F(s), , (15.6), , meaning that there is one-to-one correspondence between f (t) and, F(s). The following examples derive the Laplace transforms of some, important functions., , Example 15.1, , Determine the Laplace transform of each of the following functions:, (a) u(t), (b) eatu(t), a 0, and (c) d(t)., Solution:, (a) For the unit step function u(t), shown in Fig. 15.2(a), the Laplace, transform is, L[u(t)] , , , , , , 0, , , 1, 1est dt est `, s, , 0, , (15.1.1), , 1, 1, 1, (0) (1) , s, s, s, (b) For the exponential function, shown in Fig. 15.2(b), the Laplace, transform is, L[eat u (t)] , , , , , , eat est dt, , , , 0, , , , , 1, 1, e(sa)t ` , sa, s, , a, 0, , (15.1.2)
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ale29559_ch15.qxd, , 07/22/2008, , 01:19 PM, , Page 679, , 15.3, , Properties of the Laplace Transform, , 679, , (c) For the unit impulse function, shown in Fig. 15.2(c),, , , , L[d(t)] , , , , d(t)est dt e0 1, , (15.1.3), , , , 0, , since the impulse function d(t) is zero everywhere except at t 0. The, sifting property in Eq. (7.33) has been applied in Eq. (15.1.3)., e−atu(t), , u(t), , (t), , 1, , 1, , 1, , 0, , t, , 0, , (a), , 0, , t, (b), , t, (c), , Figure 15.2, For Example 15.1: (a) unit step function, (b) exponential function,, (c) unit impulse function., , Find the Laplace transforms of these functions: r (t) tu(t), that is, the, ramp function; eat u (t); and ejt u (t)., , Practice Problem 15.1, , Answer: 1s2, 1(s a), 1(s j)., , Determine the Laplace transform of f (t) sin t u(t)., , Example 15.2, , Solution:, Using Eq. (B.27) in addition to Eq. (15.1), we obtain the Laplace, transform of the sine function as, F(s) L[sin t] , , , , , , (sin t)est dt , , 0, , 1, , 2j, , , , , , 0, , , , , , a, , e jt ejt st, b e dt, 2j, , (e(sj)t e(sj)t ) dt, , 0, , 1, , 1, 1, , b 2, a, 2 j s j, s j, s 2, Find the Laplace transform of f (t) 10 cos t u(t)., Answer: 10s(s 2 2)., , 15.3, , Properties of the Laplace Transform, , The properties of the Laplace transform help us to obtain transform, pairs without directly using Eq. (15.1) as we did in Examples 15.1 and, 15.2. As we derive each of these properties, we should keep in mind, the definition of the Laplace transform in Eq. (15.1)., , Practice Problem 15.2
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ale29559_ch15.qxd, , 680, , 07/10/2008, , 04:06 PM, , Page 680, , Chapter 15, , Introduction to the Laplace Transform, , Linearity, If F1(s) and F2(s) are, respectively, the Laplace transforms of f1(t) and, f2(t), then, L[a1 f1(t) a2 f2 (t)] a1F1(s) a2F2 (s), , (15.7), , where a1 and a2 are constants. Equation 15.7 expresses the linearity, property of the Laplace transform. The proof of Eq. (15.7) follows, readily from the definition of the Laplace transform in Eq. (15.1)., For example, by the linearity property in Eq. (15.7), we may write, 1, 1, 1, L[cos t u(t)] L c (e jt ejt ) d L[e jt] L[ejt], 2, 2, 2, , (15.8), , But from Example 15.1(b), L[eat] 1(s a). Hence,, 1, 1, 1, s, L[cos t u(t)] a, , b 2, 2 s j, s j, s 2, , (15.9), , Scaling, If F(s) is the Laplace transform of f(t), then, L[ f (at)] , , , , , , f (at)est dt, , (15.10), , , , 0, , where a is a constant and a 7 0. If we let x at, dx a dt, then, L[ f (at)] , , , , , , f (x)ex(sa), , , , 0, , 1, dx, , a, a, , , , , , 0, , f (x)ex(sa) dx, , (15.11), , , , Comparing this integral with the definition of the Laplace transform in, Eq. (15.1) shows that s in Eq. (15.1) must be replaced by sa while, the dummy variable t is replaced by x. Hence, we obtain the scaling, property as, 1, s, L[ f (at)] F a b, a, a, , (15.12), , For example, we know from Example 15.2 that, L[sin t u(t)] , , , s 2, , (15.13), , 2, , Using the scaling property in Eq. (15.12),, L[sin 2t u(t)] , , 1, , 2, , 2 (s2)2 2 s2 42, , (15.14), , which may also be obtained from Eq. (15.13) by replacing with 2., , Time Shift, If F(s) is the Laplace transform of f(t), then, L[ f (t a) u (t a)] , , , , , , f (t a) u (t a)est dt, , , , 0, , a0, , (15.15)
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 681, , 15.3, , Properties of the Laplace Transform, , But u(t a) 0 for t 6 a and u(t a) 1 for t 7 a. Hence,, , , , L[ f (t a) u (t a)] , , , , f (t a)est dt, , (15.16), , a, , If we let x t a, then dx dt and t x a. As t S a, x S 0 and, as t S , x S . Thus,, , , , L[ f (t a) u (t a)] , , , , f (x)es(xa) dx, , , , 0, , eas, , , , , , f (x)esx dx eas F(s), , , , 0, , or, L[ f (t a) u (t a)] eas F(s), , (15.17), , In other words, if a function is delayed in time by a, the result in the, s-domain is found by multiplying the Laplace transform of the function (without the delay) by eas. This is called the time-delay or timeshift property of the Laplace transform., As an example, we know from Eq. (15.9) that, L[cos t u(t)] , , s, s 2, 2, , Using the time-shift property in Eq. (15.17),, L[cos (t a) u (t a)] eas, , s, s 2, 2, , (15.18), , Frequency Shift, If F(s) is the Laplace transform of f (t), then, L[eat f (t) u (t)] , , , , , , , , , , eat f (t)est dt, , 0, , , , f (t)e(sa)t dt F(s a), , 0, , or, L[eat f (t) u (t)] F(s a), , (15.19), , That is, the Laplace transform of eat f (t) can be obtained from the, Laplace transform of f (t) by replacing every s with s a. This is, known as frequency shift or frequency translation., As an example, we know that, cos t u(t), , 3, , s, s 2, , 3, , , 2, s 2, , 2, , and, , (15.20), sin t u(t), , 681
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ale29559_ch15.qxd, , 682, , 07/10/2008, , 04:06 PM, , Page 682, , Chapter 15, , Introduction to the Laplace Transform, , Using the shift property in Eq. (15.19), we obtain the Laplace transform of the damped sine and damped cosine functions as, sa, (s a)2 2, , L[eat sin t u (t)] , (s a)2 2, , L[eat cos t u (t)] , , (15.21a), (15.21b), , Time Differentiation, Given that F(s) is the Laplace transform of f(t), the Laplace transform, of its derivative is, , df, df st, L c u (t) d , e dt, (15.22), dt, dt, 0, , , , To integrate this by parts, we let u est, du sest dt, and dv , (d fdt) dt d f (t), v f (t). Then, Lc, , , df, u (t) d f (t)est ` , dt, 0, , , , 0 f (0) s, , , , f (t)[sest] dt, , , , 0, , , , , , f (t)est dt sF(s) f (0), , , , 0, , or, , L[ f ¿(t)] sF(s) f (0), , (15.23), , The Laplace transform of the second derivative of f (t) is a repeated, application of Eq. (15.23) as, Lc, , d 2f, dt 2, , d sL[ f ¿(t)] f ¿(0) s[sF(s) f (0)] f ¿(0), s2F(s) s f (0) f ¿(0), , or, L[ f –(t)] s2F(s) s f (0) f ¿(0), , (15.24), , Continuing in this manner, we can obtain the Laplace transform of the, nth derivative of f(t) as, Lc, , d nf, d s nF(s) s n1 f (0), dt n, s n2 f ¿(0) p s0 f (n1)(0), , (15.25), , As an example, we can use Eq. (15.23) to obtain the Laplace transform of the sine from that of the cosine. If we let f (t) cos t u(t), then, f (0) 1 and f ¿(t) sin t u (t). Using Eq. (15.23) and the scaling, property,, 1, 1, L[sin t u (t)] L[ f ¿(t)] [sF(s) f (0)], , , , 1, s, 1b 2, as 2, s 2, s 2, as expected., , (15.26)
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 683, , 15.3, , Properties of the Laplace Transform, , Time Integration, If F(s) is the Laplace transform of f(t), the Laplace transform of its, integral is, Lc, , , , t, , 0, , f (t) dt d , , , , t, , f (x) dx d e, , , 0, , c, , st, , dt, , (15.27), , 0, , To integrate this by parts, we let, t, , f (x) dx,, , u, , du f (t) dt, , 0, , and, 1, v est, s, , dv est dt,, Then, Lc, , t, , t, , 0, , 0, , f (t) dt d c f (x) dx d a s e, , , , , 1, , st, , b2, , , 0, , , , 1, a b est f (t)dt, s, , , 0, , For the first term on the right-hand side of the equation, evaluating the, term at t yields zero due to es and evaluating it at t 0 gives, 0, 1, f (x) dx 0. Thus, the first term is zero, and, s, , , , 0, , Lc, , , , t, , 0, , f (t) dt d , , 1, s, , , , , , 1, f (t)est dt F(s), s, , , 0, , or simply,, Lc, , t, , f (t) dt d s F(s), 1, , (15.28), , 0, , As an example, if we let f (t) u(t), from Example 15.1(a),, F(s) 1s. Using Eq. (15.28),, Lc, , t, , f (t) dt d L[t] s a s b, 1 1, , 0, , Thus, the Laplace transform of the ramp function is, L[t] , , 1, s2, , (15.29), , Applying Eq. (15.28), this gives, Lc, , , , t, , 0, , t dt d L c, , t2, 1 1, d , s s2, 2, , or, L[t 2] , , 2, s3, , (15.30), , 683
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 684, , Chapter 15, , 684, , Introduction to the Laplace Transform, , Repeated applications of Eq. (15.28) lead to, L[t n] , , n!, s, , n1, , (15.31), , Similarly, using integration by parts, we can show that, Lc, , , , t, , , , 1, 1, f (t) dt d F(s) f 1(0), s, s, , (15.32), , where, f, , 1, , , , , , (0 ) , , 0, , f (t) dt, , , , Frequency Differentiation, If F(s) is the Laplace transform of f (t), then, F(s) , , , , , , f (t)est dt, , , , 0, , Taking the derivative with respect to s,, dF(s), , ds, , , , , , f (t)(test ) dt , , 0, , , , , , (t f (t))est dt L[t f(t)], , 0, , and the frequency differentiation property becomes, , L[t f (t)] , , dF(s), ds, , (15.33), , Repeated applications of this equation lead to, , f(t), , L[t nf (t)] (1)n, 0, , T, , 2T, , 3T, , t, , Figure 15.3, A periodic function., , (15.34), , For example, we know from Example 15.1(b) that L[eat] , 1(s a). Using the property in Eq. (15.33),, L[teat u(t)] , , f1(t), , d nF(s), ds n, , 1, 1, d, a, b, ds s a, (s a)2, , (15.35), , Note that if a 0, we obtain L[t] 1s 2 as in Eq. (15.29), and, repeated applications of Eq. (15.33) will yield Eq. (15.31)., 0, , T, , t, , Time Periodicity, , f2(t), , 0, , T, , 2T, , t, , f3(t), , If function f(t) is a periodic function such as shown in Fig. 15.3, it can, be represented as the sum of time-shifted functions shown in Fig. 15.4., Thus,, f(t) f1(t) f2 (t) f3 (t) p, f1(t) f1(t T)u(t T), f1(t 2T)u(t 2T) p, , 0, , T, , 2T, , 3T, , t, , Figure 15.4, Decomposition of the periodic function, in Fig. 15.2., , (15.36), , where f1(t) is the same as the function f (t) gated over the interval, 0 6 t 6 T, that is,, f1(t) f (t)[u(t) u(t T)], , (15.37a)
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 685, , 15.3, , Properties of the Laplace Transform, , or, f1(t) b, , f (t),, 0,, , 0 6 t 6 T, otherwise, , (15.37b), , We now transform each term in Eq. (15.36) and apply the time-shift, property in Eq. (15.17). We obtain, F(s) F1(s) F1(s)eTs F1(s)e2Ts F1(s)e3Ts p, F1(s)31 eTs e2Ts e3Ts p 4, , (15.38), , But, 1 x x2 x3 p , if 0x 0 6 1. Hence,, , 1, 1x, , (15.39), , F1(s), 1 eTs, , F(s) , , (15.40), , where F1(s) is the Laplace transform of f1(t); in other words, F1(s) is, the transform f(t) defined over its first period only. Equation (15.40), shows that the Laplace transform of a periodic function is the transform of the first period of the function divided by 1 eTs., , Initial and Final Values, The initial-value and final-value properties allow us to find the initial, value f (0) and the final value f () of f(t) directly from its Laplace, transform F(s). To obtain these properties, we begin with the differentiation property in Eq. (15.23), namely,, sF(s) f (0) L c, , df, d , dt, , , , , , 0, , d f st, e dt, dt, , (15.41), , If we let s S , the integrand in Eq. (15.41) vanishes due to the damping exponential factor, and Eq. (15.41) becomes, lim [sF(s) f (0)] 0, , sS , , Since f (0) is independent of s, we can write, f (0) lim sF(s), , (15.42), , sS , , This is known as the initial-value theorem. For example, we know from, Eq. (15.21a) that, f (t) e2t cos 10t, , 3, , F(s) , , s2, (s 2)2 102, , Using the initial-value theorem,, f (0) lim sF(s) lim, sS , , sS , , lim, , sS , , s 2 2s, s 2 4s 104, 1 2s, 1 4s 104s2, , 1, , which confirms what we would expect from the given f (t)., , (15.43), , 685
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ale29559_ch15.qxd, , 686, , 07/10/2008, , 04:06 PM, , Page 686, , Chapter 15, , Introduction to the Laplace Transform, , In Eq. (15.41), we let s S 0; then, lim [sF(s) f (0)] , , sS0, , , , , , 0, , , , d f 0t, e dt , dt, , , , , , d f f () f (0), , , , 0, , or, f () lim sF(s), , (15.44), , sS0, , This is referred to as the final-value theorem. In order for the finalvalue theorem to hold, all poles of F(s) must be located in the left, half of the s plane (see Fig. 15.1 or Fig. 15.9); that is, the poles must, have negative real parts. The only exception to this requirement is, the case in which F(s) has a simple pole at s 0, because the effect, of 1s will be nullified by sF(s) in Eq. (15.44). For example, from, Eq. (15.21b),, f (t) e2t sin 5t u(t), , F(s) , , 3, , 5, (s 2)2 52, , (15.45), , Applying the final-value theorem,, f () lim s F (s) lim, sS0, , sS0, , 5s, 0, s 2 4s 29, , as expected from the given f (t). As another example,, f (t) sin t u(t), , f (s) , , 3, , 1, s 1, 2, , (15.46), , so that, f () lim s F (s) lim, sS0, , sS0, , s, 0, s2 1, , This is incorrect, because f (t) sin t oscillates between 1 and 1 and, does not have a limit as t S . Thus, the final-value theorem cannot, be used to find the final value of f (t) sin t, because F(s) has poles at, s j, which are not in the left half of the s plane. In general, the finalvalue theorem does not apply in finding the final values of sinusoidal, functions—these functions oscillate forever and do not have final values., The initial-value and final-value theorems depict the relationship, between the origin and infinity in the time domain and the s-domain., They serve as useful checks on Laplace transforms., Table 15.1 provides a list of the properties of the Laplace transform. The last property (on convolution) will be proved in Section 15.5. There are other properties, but these are enough for present, purposes. Table 15.2 summarizes the Laplace transforms of some, common functions. We have omitted the factor u(t) except where it, is necessary., We should mention that many software packages, such as Mathcad, MATLAB, Maple, and Mathematica, offer symbolic math. For, example, Mathcad has symbolic math for the Laplace, Fourier, and Z, transforms as well as the inverse function.
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ale29559_ch15.qxd, , 07/17/2008, , 01:17 PM, , Page 687, , 15.3, , Properties of the Laplace Transform, , 687, , TABLE 15.2, , TABLE 15.1, , Laplace transform pairs.*, , Properties of the Laplace transform., f(t), , Linearity, , a1 f1(t) a2 f2(t), , a1F1(s) a2F2 (s), , d(t), , Scaling, , f (at), , 1, s, Fa b, a, a, , u(t), , Time shift, , f (t a)u(t a), , eas F(s), , eat, , Frequency shift, Time, differentiation, , at, , e f (t), df, dt, d 2f, dt 2, d 3f, , , , Frequency, integration, , sF(s) f (0), s2F(s) s f (0) f ¿(0), , snF(s) sn1 f (0) sn2 f ¿(0), p f (n1) (0), , t, , f (t) dt, , 0, , Frequency, differentiation, , F(s a), , s 3F(s) s2 f (0) sf ¿(0), f –(0), , dt 3, d nf, dt n, Time integration, , F(s), , f(t), , Property, , t f (t), f (t), t, , Time periodicity, , f (t) f (t nT ), , Initial value, , f (0), , Final value, , f (), , Convolution, , f1(t) * f2 (t), , 1, F(s), s, , , , , d, F(s), ds, , , , F(s) ds, , s, , t, tn, teat, t neat, sin t, cos t, sin(t u), cos(t u), , F1(s), 1 esT, lim sF(s), , sS , , lim sF(s), , sS0, , F1(s)F2(s), , Obtain the Laplace transform of f (t) d(t) 2 u (t) 3e2tu(t)., , eat sin t, eat cos t, , F(s), 1, 1, s, 1, sa, 1, s2, n!, n1, s, 1, (s a)2, n!, (s a)n1, , s2 2, s, s2 2, s sin u cos u, s2 2, s cos u sin u, s2 2, , (s a)2 2, sa, (s a)2 2, , *Defined for t 0; f (t) 0, for t 6 0., , Example 15.3, , Solution:, By the linearity property,, F(s) L[d(t)] 2L[ u (t)] 3L[e2tu (t)], 1, 1, s2 s 4, 12 3, , s, s2, s(s 2), , Find the Laplace transform of f (t) (cos (3t) e5t)u(t)., Answer:, , 2s2 5s 9, ., (s 5) (s2 9), , Practice Problem 15.3
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 688, , Chapter 15, , 688, , Example 15.4, , Introduction to the Laplace Transform, , Determine the Laplace transform of f (t) t 2 sin 2t u (t)., Solution:, We know that, L[sin 2t] , , 2, s2 22, , Using frequency differentiation in Eq. (15.34),, d2, 2, b, a 2, 2, ds s 4, 4s, 12s2 16, d, b 2, a 2, 2, ds (s 4), (s 4)3, , F(s) L[t 2 sin 2t] (1)2, , Practice Problem 15.4, , Find the Laplace transform of f (t) t 2 cos 3t u(t)., Answer:, , Example 15.5, , 2s (s2 27), ., (s2 9)3, , Find the Laplace transform of the gate function in Fig. 15.5., , g(t), , Solution:, We can express the gate function in Fig. 15.5 as, , 10, , g(t) 10[u (t 2) u (t 3)], 0, , 1, , 2, , 3, , Since we know the Laplace transform of u(t), we apply the time-shift, property and obtain, , t, , Figure 15.5, , G(s) 10 a, , The gate function; for Example 15.5., , Practice Problem 15.5, , Find the Laplace transform of the function h(t) in Fig. 15.6., , h(t), , Answer:, 10, , 5, , 0, , 4, , Figure 15.6, For Practice Prob. 15.5., , 8, , t, , e3s, 10, e2s, , b (e2s e3s ), s, s, s, , 5, (2 e4s e8s )., s
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 689, , 15.3, , Properties of the Laplace Transform, , 689, , Example 15.6, , Calculate the Laplace transform of the periodic function in Fig. 15.7., f (t), , Solution:, The period of the function is T 2. To apply Eq. (15.40), we first, obtain the transform of the first period of the function., f1(t) 2t[u (t) u (t 1)] 2tu (t) 2tu (t 1), 2tu (t) 2(t 1 1) u (t 1), 2tu (t) 2(t 1) u (t 1) 2 u (t 1), , 2, , 0, , 1, , 2, , 3, , 4, , 5, , t, , Figure 15.7, For Example 15.6., , Using the time-shift property,, F1(s) , , 2, es, 2, 2, , 2, es 2 (1 es ses), 2, 2, s, s, s, s, , Thus, the transform of the periodic function in Fig. 15.7 is, F(s) , , F1(s), 2, (1 es ses), Ts 2, 1e, s (1 e2s), , Determine the Laplace transform of the periodic function in Fig. 15.8., Answer:, , Practice Problem 15.6, , 1 e2s, ., s(1 e5s), , f (t), 1, , 0, , 2, , 5, , 7, , 10, , 12 t, , Figure 15.8, For Practice Prob. 15.6., , Example 15.7, , Find the initial and final values of the function whose Laplace transform is, H(s) , , 20, (s 3) (s 8s 25), 2, , Solution:, Applying the initial-value theorem,, h(0) lim sH(s) lim, sS, , sS, , 20s, (s 3) (s 8s 25), , j, , 2, , 20s2, , 0, , lim, 0, sS (1 3s) (1 8s 25s2), (1 0) (1 0 0), To be sure that the final-value theorem is applicable, we check where, the poles of H(s) are located. The poles of H(s) are s 3, 4 j3,, which all have negative real parts: they are all located on the left half, of the s plane (Fig. 15.9). Hence, the final-value theorem applies and, 20s, h() lim sH(s) lim, sS0, sS0 (s 3) (s2 8s 25), 0, 0, , (0 3) (0 0 25), , ×, , 3, 2, 1, , −4, , ×, −3, , −2, , −1, , 1, , 2, , −1, −2, , ×, , −3, , Figure 15.9, For Example 15.7: Poles of H(s)., , 3, ,
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 690, , 690, , Chapter 15, , Introduction to the Laplace Transform, , Both the initial and final values could be determined from h (t) if we, knew it. See Example 15.11, where h(t) is given., , Practice Problem 15.7, , Obtain the initial and the final values of, G(s) , , 3s3 2s 10, s(s 2)2(s 3), , Answer: 3, 0.8333., , 15.4, , The Inverse Laplace Transform, , Given F(s), how do we transform it back to the time domain and obtain, the corresponding f(t)? By matching entries in Table 15.2, we avoid, using Eq. (15.5) to find f (t)., Suppose F(s) has the general form of, F(s) , , Software packages such as MATLAB,, Mathcad, and Maple are capable of, finding partial fraction expansions, quite easily., , N(s), D(s), , (15.47), , where N(s) is the numerator polynomial and D(s) is the denominator, polynomial. The roots of N(s) 0 are called the zeros of F(s), while, the roots of D(s) 0 are the poles of F(s). Although Eq. (15.47) is, similar in form to Eq. (14.3), here F(s) is the Laplace transform of a, function, which is not necessarily a transfer function. We use partial, fraction expansion to break F(s) down into simple terms whose inverse, transform we obtain from Table 15.2. Thus, finding the inverse Laplace, transform of F(s) involves two steps., , Steps to Find the Inverse Laplace Transform:, 1. Decompose F(s) into simple terms using partial fraction, expansion., 2. Find the inverse of each term by matching entries in Table 15.2., , Let us consider the three possible forms F(s) may take and how to, apply the two steps to each form., , 15.4.1 Simple Poles, Recall from Chapter 14 that a simple pole is a first-order pole. If F(s), has only simple poles, then D(s) becomes a product of factors, so that, Otherwise, we must first apply long, division so that F (s) N (s)D (s) , Q (s) R (s)D (s), where the degree, of R (s), the remainder of the long, division, is less than the degree of D (s)., , F(s) , , N(s), (s p1) (s p2) p (s pn ), , (15.48), , where s p1, p2, p , pn are the simple poles, and pi pj for all, i j (i.e., the poles are distinct). Assuming that the degree of N(s) is
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 691, , 15.4, , The Inverse Laplace Transform, , 691, , less than the degree of D(s), we use partial fraction expansion to, decompose F(s) in Eq. (15.48) as, F(s) , , kn, k1, k2, , p, s p1, s p2, s pn, , (15.49), , The expansion coefficients k1, k2, p , kn are known as the residues of, F(s). There are many ways of finding the expansion coefficients. One, way is using the residue method. If we multiply both sides of Eq. (15.49), by (s p1), we obtain, (s p1)F(s) k1 , , (s p1)k2, (s p1)k n, p, s p2, s pn, , (15.50), , Since pi pj, setting s p1 in Eq. (15.50) leaves only k1 on the, right-hand side of Eq. (15.50). Hence,, (s p1)F(s) 0 sp1 k1, , (15.51), , ki (s pi) F (s) 0 spi, , (15.52), , Thus, in general,, , This is known as Heaviside’s theorem. Once the values of ki are known,, we proceed to find the inverse of F(s) using Eq. (15.49). Since the, inverse transform of each term in Eq. (15.49) is L1[k(s a)] , keat u(t), then, from Table 15.2,, f (t) (k1ep1t k 2 ep2t p k nepnt ) u (t), , (15.53), , 15.4.2 Repeated Poles, Suppose F(s) has n repeated poles at s p. Then we may represent, F(s) as, F(s) , , kn, k n1, k2, p, n , n1, (s p), (s p), (s p)2, k1, , F1(s), sp, , (15.54), , where F1(s) is the remaining part of F(s) that does not have a pole at, s p. We determine the expansion coefficient kn as, k n (s p)n F(s) 0 sp, , (15.55), , as we did above. To determine k n1, we multiply each term in Eq. (15.54), by (s p)n and differentiate to get rid of kn, then evaluate the result at, s p to get rid of the other coefficients except k n1. Thus, we obtain, d, [(s p)n F(s)] 0 sp, ds, , (15.56), , 1 d2, [(s p)n F(s)] 0 sp, 2! ds2, , (15.57), , k n1 , Repeating this gives, k n2 , , Historical note: Named after Oliver, Heaviside (1850–1925), an English, engineer, the pioneer of operational, calculus.
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ale29559_ch15.qxd, , 692, , 07/10/2008, , 04:06 PM, , Page 692, , Chapter 15, , Introduction to the Laplace Transform, , The mth term becomes, k nm , , 1 dm, [(s p)n F(s)] 0 sp, m! ds m, , (15.58), , where m 1, 2, p , n 1. One can expect the differentiation to be difficult to handle as m increases. Once we obtain the values of k1, k2, p , kn, by partial fraction expansion, we apply the inverse transform, L1 c, , 1, t n1eat, u(t), nd , (s a), (n 1)!, , (15.59), , to each term on the right-hand side of Eq. (15.54) and obtain, f (t) ak1ept k 2tept , p, , k 3 2 pt, t e, 2!, , kn, t n1ept b u(t) f1(t), (n 1)!, , (15.60), , 15.4.3 Complex Poles, A pair of complex poles is simple if it is not repeated; it is a double, or multiple pole if repeated. Simple complex poles may be handled the, same way as simple real poles, but because complex algebra is involved, the result is always cumbersome. An easier approach is a method, known as completing the square. The idea is to express each complex, pole pair (or quadratic term) in D(s) as a complete square such as, (s a)2 b2 and then use Table 15.2 to find the inverse of the term., Since N(s) and D(s) always have real coefficients and we know, that the complex roots of polynomials with real coefficients must occur, in conjugate pairs, F(s) may have the general form, F(s) , , A1s A2, s as b, 2, , F1(s), , (15.61), , where F1(s) is the remaining part of F(s) that does not have this pair, of complex poles. If we complete the square by letting, s2 as b s2 2a s a 2 b 2 (s a)2 b 2, , (15.62), , and we also let, A1s A2 A1(s a) B1b, , (15.63), , then Eq. (15.61) becomes, F(s) , , A1(s a), (s a) b, 2, , 2, , , , B1b, (s a)2 b 2, , F1(s), , (15.64), , From Table 15.2, the inverse transform is, f (t) (A1eat cos bt B1eat sin bt) u (t) f1(t), , (15.65), , The sine and cosine terms can be combined using Eq. (9.11)., Whether the pole is simple, repeated, or complex, a general, approach that can always be used in finding the expansion coefficients
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 693, , 15.4, , The Inverse Laplace Transform, , 693, , is the method of algebra, illustrated in Examples 15.9 to 15.11. To, apply the method, we first set F(s) N(s)D(s) equal to an expansion, containing unknown constants. We multiply the result through by a, common denominator. Then we determine the unknown constants by, equating coefficients (i.e., by algebraically solving a set of simultaneous, equations for these coefficients at like powers of s)., Another general approach is to substitute specific, convenient values of s to obtain as many simultaneous equations as the number of, unknown coefficients, and then solve for the unknown coefficients. We, must make sure that each selected value of s is not one of the poles of, F(s). Example 15.11 illustrates this idea., , Example 15.8, , Find the inverse Laplace transform of, F(s) , , 3, 5, 6, , 2, s, s1, s 4, , Solution:, The inverse transform is given by, 3, 5, 6, b, f (t) L1[F(s)] L1 a b L1 a, b L1 a 2, s, s1, s 4, t0, (3 5et 3 sin 2t) u (t),, where Table 15.2 has been consulted for the inverse of each term., , Practice Problem 15.8, , Determine the inverse Laplace transform of, F(s) 1 , , 3, 5s, 2, s4, s 25, , Answer: d(t) (4e4t 5 cos(5t)) u (t)., , Example 15.9, , Find f(t) given that, F(s) , , s2 12, s(s 2) (s 3), , Solution:, Unlike in the previous example where the partial fractions have been, provided, we first need to determine the partial fractions. Since there, are three poles, we let, B, A, C, s2 12, , , s, s(s 2) (s 3), s2, s3, , (15.9.1), , where A, B, and C are the constants to be determined. We can find the, constants using two approaches.
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , 694, , Page 694, , Chapter 15, , Introduction to the Laplace Transform, , ■ METHOD 1 Residue method:, s2 12, 12, `, 2, , (s 2) (s 3) s0 (2) (3), s2 12, 4 12, , B (s 2) F(s) 0 s2 , `, 8, s(s 3) s2 (2) (1), s2 12, 9 12, C (s 3) F(s) 0 s3 , `, , 7, s(s 2) s3 (3) (1), A sF(s) 0 s0 , , ■ METHOD 2 Algebraic method: Multiplying both sides of, Eq. (15.9.1) by s(s 2) (s 3) gives, , s2 12 A(s 2) (s 3) Bs(s 3) Cs(s 2), or, s2 12 A(s2 5s 6) B(s2 3s) C(s2 2s), Equating the coefficients of like powers of s gives, Constant: 12 6A, 1, A2, s:, 0 5A 3B 2C, 1, 3B 2C 10, 2, s:, 1ABC, 1, B C 1, Thus, A 2, B 8, C 7, and Eq. (15.9.1) becomes, F(s) , , 8, 2, 7, , , s, s2, s3, , By finding the inverse transform of each term, we obtain, f (t) (2 8e2t 7e3t ) u (t), , Practice Problem 15.9, , Find f(t) if, F(s) , , 6(s 2), (s 1) (s 3) (s 4), , Answer: f (t) (et 3e3t 4e4t ) u (t)., , Example 15.10, , Calculate v(t) given that, V(s) , , 10s2 4, s(s 1) (s 2)2, , Solution:, While the previous example is on simple roots, this example is on, repeated roots. Let, V(s) , , 10s2 4, s(s 1) (s 2)2, , B, C, D, A, , , , 2, s, s1, s2, (s 2), , (15.10.1)
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 695, , 15.4, , The Inverse Laplace Transform, , 695, , ■ METHOD 1 Residue method:, A sV(s) 0 s0 , , 10s2 4, 4, `, , 1, 2, (s 1) (s 2) s0 (1) (2)2, , 10s2 4, 14, `, , 14, s (s 2)2 s1 (1) (1)2, 10s2 4, 44, C (s 2)2V(s) 0 s2 , `, , 22, s (s 1) s2 (2) (1), , B (s 1)V(s) 0 s1 , , d, d 10s2 4, b`, [(s 2)2V(s)] `, a 2, ds, ds s s, s2, s2, 2, 2, (s s) (20s) (10s 4) (2s 1), 52, , `, , 13, 4, (s2 s)2, s2, , D, , ■ METHOD 2 Algebraic method: Multiplying Eq. (15.10.1) by, s (s 1) (s 2)2, we obtain, , 10s2 4 A(s 1) (s 2)2 Bs (s 2)2, Cs (s 1) Ds (s 1) (s 2), or, 10s2 4 A(s3 5s2 8s 4) B(s3 4s2 4s), C(s2 s) D(s3 3s2 2s), Equating coefficients,, Constant:, s:, s2:, s3:, , 4 4A, 1, A1, 0 8A 4B C 2D, 1, 4B C 2D 8, 10 5A 4B C 3D, 1, 4B C 3D 5, 0ABD, 1, B D 1, , Solving these simultaneous equations gives A 1, B 14, C 22,, D 13, so that, V(s) , , 14, 1, 13, 22, , , , s, s1, s2, (s 2)2, , Taking the inverse transform of each term, we get, v(t) (1 14et 13e2t 22te2t ) u (t), , Practice Problem 15.10, , Obtain g(t) if, s 2s 6, s (s 1)2(s 3), 3, , G(s) , , Answer: (2 3.25et 1.5tet 2.25e3t) u (t)., , Find the inverse transform of the frequency-domain function in, Example 15.7:, H(s) , , 20, (s 3) (s 8s 25), 2, , Example 15.11
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ale29559_ch15.qxd, , 696, , 07/10/2008, , 04:06 PM, , Page 696, , Chapter 15, , Introduction to the Laplace Transform, , Solution:, In this example, H(s) has a pair of complex poles at s2 8s 25 0, or s 4 j3. We let, H(s) , , 20, A, Bs C, , 2, s3, (s 3) (s 8s 25), (s 8s 25), 2, , (15.11.1), , We now determine the expansion coefficients in two ways., , ■ METHOD 1 Combination of methods: We can obtain A using, the method of residue,, A (s 3)H(s) 0 s3 , , 20, 20, `, , 2, s2 8s 25 s3 10, , Although B and C can be obtained using the method of residue, we, will not do so, to avoid complex algebra. Rather, we can substitute two, specific values of s [say s 0, 1, which are not poles of F(s)] into, Eq. (15.11.1). This will give us two simultaneous equations from which, to find B and C. If we let s 0 in Eq. (15.11.1), we obtain, 20, A, C, , 75, 3, 25, or, 20 25A 3C, , (15.11.2), , Since A 2, Eq. (15.11.2) gives C 10. Substituting s 1 into, Eq. (15.11.1) gives, A, BC, 20, , (4) (34), 4, 34, or, 20 34A 4B 4C, , (15.11.3), , But A 2, C 10, so that Eq. (15.11.3) gives B 2., , ■ METHOD 2 Algebraic method: Multiplying both sides of, Eq. (15.11.1) by (s 3)(s2 8s 25) yields, , 20 A(s2 8s 25) (Bs C) (s 3), (15.11.4), A(s2 8s 25) B(s2 3s) C(s 3), Equating coefficients gives, s2:, 0AB, 1, A B, s:, 0 8A 3B C 5A C, Constant: 20 25A 3C 25A 15A, , 1, 1, , C 5A, A2, , That is, B 2, C 10. Thus,, 2(s 4) 2, 2, 2, 2s 10, , 2, , s3, s3, (s 8s 25), (s 4)2 9, 2(s 4), 2, 2, 3, , , , 2, s3, 3 (s 4)2 9, (s 4) 9, , H(s)
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 697, , 15.5, , The Convolution Integral, , 697, , Taking the inverse of each term, we obtain, 2, h(t) a2e3t 2e4t cos 3t e4t sin 3tb u (t) (15.11.5), 3, It is alright to leave the result this way. However, we can combine the, cosine and sine terms as, h(t) (2e3t Re4t cos(3t u)) u (t), , (15.11.6), , To obtain Eq. (15.11.6) from Eq. (15.11.5), we apply Eq. (9.11). Next,, we determine the coefficient R and the phase angle u:, 2, 3, , u tan1 18.43, 2, , R 222 (23)2 2.108,, Thus,, , h(t) (2e3t 2.108e4t cos(3t 18.43)) u (t), , Practice Problem 15.11, , Find g(t) given that, 10, (s 1) (s 4s 13), , G(s) , , 2, , 1, Answer: et e2t cos 3t e2t sin 3t, t 0., 3, , 15.5, , The Convolution Integral, , The term convolution means “folding.” Convolution is an invaluable tool, to the engineer because it provides a means of viewing and characterizing physical systems. For example, it is used in finding the response y(t), of a system to an excitation x(t), knowing the system impulse response, h(t). This is achieved through the convolution integral, defined as, y(t) , , , , , , x(l)h (t l) dl, , (15.66), , , , or simply, y(t) x(t) * h(t), , (15.67), , where l is a dummy variable and the asterisk denotes convolution. Equation (15.66) or (15.67) states that the output is equal to the input convolved, with the unit impulse response. The convolution process is commutative:, y(t) x(t) * h (t) h(t) * x(t), , (15.68a), , or, y(t) , , , , , , , , x(l) h (t l) dl , , , , , , h(l) x (t l) dl, , (15.68b), , , , This implies that the order in which the two functions are convolved, is immaterial. We will see shortly how to take advantage of this commutative property when performing graphical computation of the convolution integral.
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ale29559_ch15.qxd, , 698, , 07/10/2008, , 04:06 PM, , Page 698, , Chapter 15, , Introduction to the Laplace Transform, , The convolution of two signals consists of time-reversing one of the, signals, shifting it, and multiplying it point by point with the second, signal, and integrating the product., , The convolution integral in Eq. (15.66) is the general one; it, applies to any linear system. However, the convolution integral can be, simplified if we assume that a system has two properties. First, if, x(t) 0 for t 6 0, then, y(t) , , , , , , x (l) h (t l) dl , , , , , , , , x (l) h (t l) dl, , (15.69), , 0, , Second, if the system’s impulse response is causal (i.e., h(t) 0 for, t 6 0), then h (t l) 0 for t l 6 0 or l 7 t, so that Eq. (15.69), becomes, t, , x (l) h (t l) dl, , y (t) h (t) * x (t) , , (15.70), , 0, , Here are some properties of the convolution integral., 1. x(t) * h(t) h (t) * x(t) (Commutative), 2. f (t) * [x(t) y(t)] f (t) * x(t) f (t) * y(t) (Distributive), 3. f (t) * [x(t) * y(t)] [ f (t) * x(t)] * y(t) (Associative), , , , , 4. f (t) * d(t) , , f (l) d (t l) dl f (t), , , , 5. f (t) * d(t to) f (t to), 6. f (t) * d¿(t) , , , , , , f (l) d¿(t l) dl f ¿(t), , , , , 7. f (t) * u (t) , , , , f (l) u (t l) dl , , , , , , t, , f (l) dl, , , , Before learning how to evaluate the convolution integral in, Eq. (15.70), let us establish the link between the Laplace transform and, the convolution integral. Given two functions f1(t) and f2(t) with Laplace, transforms F1(s) and F2(s), respectively, their convolution is, t, , f (l) f (t l) dl, , f (t) f1(t) * f2 (t) , , 1, , 2, , (15.71), , 0, , Taking the Laplace transform gives, F(s) L[ f1(t) * f2 (t)] F1(s)F2(s), , (15.72), , To prove that Eq. (15.72) is true, we begin with the fact that F1(s), is defined as, F1(s) , , , , , , f1(l)esl dl, , (15.73), , , , 0, , Multiplying this with F2(s) gives, F1(s)F2(s) , , , , , , 0, , , , f1(l)[F2(s)esl] dl, , (15.74)
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 699, , 15.5, , The Convolution Integral, , We recall from the time shift property in Eq. (15.17) that the term in, brackets can be written as, F2(s)esl L[ f2(t l) u (t l)], , , , , , , f2 (t l) u (t l)esl dt, , (15.75), , 0, , Substituting Eq. (15.75) into Eq. (15.74) gives, F1(s)F2(s) , , , , , , 0, , , , f1(l) c, , , , , , , 0, , f2(t l) u (t l)esl dt d dl (15.76), , Interchanging the order of integration results in, , , F1(s)F2(s) , , , , , 0, , c, , t, , 0, , , , f1(l) f2 (t l) dl d esl dl, , (15.77), , The integral in brackets extends only from 0 to t because the delayed, unit step u(t l) 1 for l 6 t and u (t l) 0 for l 7 t. We notice, that the integral is the convolution of f1(t) and f2(t) as in Eq. (15.71)., Hence,, F1(s)F2(s) L[ f1(t) * f2 (t)], , (15.78), , as desired. This indicates that convolution in the time domain is equivalent to multiplication in the s-domain. For example, if x(t) 4et and, h(t) 5e2t, applying the property in Eq. (15.78), we get, h(t) * x (t) L1[H(s)X(s)] L1 c a, , 5, 4, ba, bd, s2 s1, , 20, 20, , d, s1 s2, 20(et e2t ),, t0, , L1 c, , (15.79), , Although we can find the convolution of two signals using, Eq. (15.78), as we have just done, if the product F1(s)F2(s) is very, complicated, finding the inverse may be tough. Also, there are situations in which f1(t) and f2(t) are available in the form of experimental, data and there are no explicit Laplace transforms. In these cases, one, must do the convolution in the time domain., The process of convolving two signals in the time domain is better appreciated from a graphical point of view. The graphical procedure for evaluating the convolution integral in Eq. (15.70) usually, involves four steps., , Steps to Evaluate the Convolution Integral:, 1. Folding: Take the mirror image of h (l) about the ordinate axis, to obtain h (l)., 2. Displacement: Shift or delay h (l) by t to obtain h (t l)., 3. Multiplication: Find the product of h (t l) and x (l)., 4. Integration: For a given time t, calculate the area under the, product h (t l) x (l) for 0 6 l 6 t to get y(t) at t., , 699
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 700, , Chapter 15, , 700, , Introduction to the Laplace Transform, , The folding operation in step 1 is the reason for the term convolution., The function h (t l) scans or slides over x (l). In view of this, superposition procedure, the convolution integral is also known as the, superposition integral., To apply the four steps, it is necessary to be able to sketch x(l), and h (t l). To get x (l) from the original function x(t) involves, merely replacing every t with l. Sketching h (t l) is the key to the, convolution process. It involves reflecting h (l) about the vertical axis, and shifting it by t. Analytically, we obtain h (t l) by replacing every, t in h(t) by t l. Since convolution is commutative, it may be more, convenient to apply steps 1 and 2 to x(t) instead of h(t). The best way, to illustrate the procedure is with some examples., , Example 15.12, , Find the convolution of the two signals in Fig. 15.10., , x2(t), , x1(t), 2, , 1, 0, , 0, , 1 t, , 1, , 2, , 3 t, , Figure 15.10, For Example 15.12., , Solution:, We follow the four steps to get y(t) x1(t) * x2(t). First, we fold x1(t), as shown in Fig. 15.11(a) and shift it by t as shown in Fig. 15.11(b)., For different values of t, we now multiply the two functions and, integrate to determine the area of the overlapping region., For 0 6 t 6 1, there is no overlap of the two functions, as shown, in Fig. 15.12(a). Hence,, y(t) x1(t) * x2(t) 0,, , −1, , 0, , y(t) , , 2, , 2, , , , (a), , t−1, , (15.12.1), , For 1 6 t 6 2, the two signals overlap between 1 and t, as shown in, Fig. 15.12(b)., , x1(t − ), , x1(−), , 0 6 t 6 1, , 0, , t, , , , (b), , Figure 15.11, (a) Folding x1(l), (b) shifting x1(l) by t., , t, , t, , 1, , 1, , (2)(1) dl 2l `, , 2(t 1),, , 1 6 t 6 2, , (15.12.2), , For 2 6 t 6 3, the two signals completely overlap between (t 1), and t, as shown in Fig. 15.12(c). It is easy to see that the area under, the curve is 2. Or, y(t) , , , , t, , t1, , (2)(1) dl 2l `, , t, , 2,, , 2 6 t 6 3, , (15.12.3), , t1, , For 3 6 t 6 4, the two signals overlap between (t 1) and 3, as, shown in Fig. 15.12(d)., y(t) , , , , 3, , t1, , (2)(1) dl 2l `, , 3, , (15.12.4), , t1, , 2(3 t 1) 8 2t,, , 3 6 t 6 4, , For t 7 4, the two signals do not overlap [Fig. 15.12(e)], and, y(t) 0,, , t 7 4, , (15.12.5), , Combining Eqs. (15.12.1) to (15.12.5), we obtain, 0,, 2t 2,, y(t) e 2,, 8 2t,, 0,, , 0, 1, 2, 3, t, , t, t, t, t, 4, , 1, 2, 3, 4, , (15.12.6)
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 701, , 15.5, x1(t − ), , The Convolution Integral, , 701, , x1(t − ), , 2, , 2, , x2(), 1, , x1(t − ), 2, , x2(), , 1, 0, , 1, , t, , 2, , , , 3, , 1, 0, , t−1 1, , 3 , , t, , (a), , 0, , 1t−1, , t, , (b), , 1, , , , x1(t − ), 2, , x2(), , 3, , (c), , x1(t − ), 2, , x2(), , x2(), , 1, 0, , 1, , t−1 3, , , t 4, , 0, , 1, , (d), , 2, , , , 3 t−14 t, (e), , Figure 15.12, , Overlapping of x1(t l) and x2(l) for: (a) 0 6 t 6 1, (b) 1 6 t 6 2, (c) 2 6 t 6 3,, (d) 3 6 t 6 4, (e) t 7 4., y(t), 2, , which is sketched in Fig. 15.13. Notice that y(t) in this equation is, continuous. This fact can be used to check the results as we move, from one range of t to another. The result in Eq. (15.12.6) can be, obtained without using the graphical procedure—by directly using, Eq. (15.70) and the properties of step functions. This will be illustrated, in Example 15.14., , Graphically convolve the two functions in Fig. 15.14. To show how, powerful working in the s-domain is, verify your answer by performing the equivalent operation in the s-domain., , 0, , 1, , 2, , 3, , 4, , t, , Figure 15.13, Convolution of signals x1(t) and x2(t) in, Fig. 15.10., , Practice Problem 15.12, , Answer: The result of the convolution y(t) is shown in Fig. 15.15,, where, t,, 0 t 2, y(t) c 6 2t, 2 t 3, 0,, otherwise., x 2 (t), y(t), 2, x1(t), , 2, , 1, , 1, , 0, , 1, , t, , 0, , 1, , 2, , t, , 0, , 1, , 2, , 3, , t, , Figure 15.14, , Figure 15.15, , For Practice Prob. 15.12., , Convolution of the signals in Fig. 15.14.
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 702, , Chapter 15, , 702, , Example 15.13, , Introduction to the Laplace Transform, , Graphically convolve g(t) and u(t) shown in Fig. 15.16., , g(t), , Solution:, Let y(t) g(t) * u(t). We can find y(t) in two ways., , 1, , 0, , 1, , ■ METHOD 1 Suppose we fold g(t), as in Fig. 15.17(a), and shift, , t, , it by t, as in Fig. 15.17(b). Since g(t) t, 0 6 t 6 1 originally, we, expect that g(t l) t l, 0 6 t l 6 1 or t 1 6 l 6 t., There is no overlap of the two functions when t 6 0 so that y (0) 0, for this case., , u(t), 1, , 0, , t, , g(−), , Figure 15.16, , u(), , For Example 15.13., , u(), , 1, , 1, , 1, g(t − ), , −1, , t−1 0, , , , 0, , g(t − ), , , , t, , t−1, , 0, , (b), , (a), , t, , , , (c), , Figure 15.17, Convolution of g(t) and u(t) in Fig. 15.16 with g(t) folded., , For 0 6 t 6 1, g(t l) and u (l) overlap from 0 to t, as evident in, Fig. 15.17(b). Therefore,, y(t) , , , , t, , 0, , t, 1, (1)(t l) dl atl l2 b `, 2, 0, , t2, t2, t ,, 2, 2, 2, , 0, , t, , (15.13.1), , 1, , For t 7 1, the two functions overlap completely between (t 1) and, t [see Fig. 15.17(c)]. Hence,, y(t) , , , , t, , (1)(t l) dl, , t1, , (15.13.2), , t, 1, 1, atl l2 b `, ,, 2, 2, t1, , t1, , Thus, from Eqs. (15.13.1) and (15.13.2),, 1 2, t ,, 2, y (t) d, 1, ,, 2, , 0, , t, , 1, , t1, , ■ METHOD 2 Instead of folding g, suppose we fold the unit step, function u(t), as in Fig. 15.18(a), and then shift it by t, as in Fig. 15.18(b)., Since u (t) 1 for t 7 0, u (t l) 1 for t l 7 0 or l 6 t, the, two functions overlap from 0 to t, so that, t, , y (t) , , (1)l dl 2 l, 0, , 1, , 2, , t, , ` , 0, , t2, ,, 2, , 0, , t, , 1, , (15.13.3)
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 703, , 15.5, , The Convolution Integral, , 1, , g() = , , u(t − ) = 1, , u(−), 1, , 1, , 0, , g() = , , , , 0, , t, , (a), , 1, , 703, , u(t − ) = 1, , , , 0, , (b), , 1, , t, , , , (c), , Figure 15.18, Convolution of g(t) and u(t) in Fig. 15.16 with u(t) folded., , For t 7 1, the two functions overlap between 0 and 1, as shown in, Fig. 15.18(c). Hence,, y (t) , , , , 1, , 1 2 1 1, l ` ,, 2, 2, 0, , (1)l dl , , 0, , t1, , (15.13.4), , And, from Eqs. (15.13.3) and (15.13.4),, 1 2, t ,, 2, y (t) d, 1, ,, 2, , 0, , t, , 1, y(t), , t1, , 1, 2, , Although the two methods give the same result, as expected, notice, that it is more convenient to fold the unit step function u(t) than fold, g(t) in this example. Figure 15.19 shows y(t)., , Figure 15.19, , Given g(t) and f(t) in Fig. 15.20, graphically find y(t) g(t) * f (t)., , Practice Problem 15.13, , 0, , 1, , t, , Result of Example 15.13., , f (t), 3, g(t), 3e −t, , 1, , 0, , 1, , t, , 0, , t, , Figure 15.20, For Practice Prob. 15.13., , 3(1 et ), 0 t 1, Answer: y (t) c 3(e 1)et, t 1, 0,, elsewhere., , For the RL circuit in Fig. 15.21(a), use the convolution integral to find, the response io(t) due to the excitation shown in Fig. 15.21(b)., Solution:, 1. Define. The problem is clearly stated and the method of solution, is also specified., , Example 15.14
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 704, , Chapter 15, , 704, , io, 1Ω, , i s(t), , 1H, , (a), i s (t) A, 1, 0, , 2, , t(s), , (b), , Figure 15.21, For Example 15.14., , Introduction to the Laplace Transform, , 2. Present. We are to use the convolution integral to solve for the, response io(t) due to is(t) shown in Fig. 15.21(b)., 3. Alternative. We have learned to do convolution by using the, convolution integral and how to do it graphically. In addition,, we could always work in the s-domain to solve for the current., We will solve for the current using the convolution integral and, then check it using the graphical approach., 4. Attempt. As we stated, this problem can be solved in two, ways: directly using the convolution integral or using the, graphical technique. To use either approach, we first need, the unit impulse response h(t) of the circuit. In the s-domain,, applying the current division principle to the circuit in, Fig. 15.22(a) gives, Io , , Io, 1Ω, , Is, , Hence,, , s, , H(s) , , Io, 1, , Is, s1, , (15.14.1), , and the inverse Laplace transform of this gives, , (a), , h(t) et u (t), , h(t), 1, , 1, Is, s1, , e −t, t, , Figure 15.22(b) shows the impulse response h(t) of the circuit., To use the convolution integral directly, recall that the, response is given in the s-domain as, , (b), , Io(s) H(s) Is(s), , Figure 15.22, For the circuit in Fig. 15.21(a): (a) its, s-domain equivalent, (b) its impulse, response., , (15.14.2), , With the given is(t) in Fig. 15.21(b),, is(t) u (t) u (t 2), so that, t, , io(t) h (t) * is (t) , , i (l)h (t l) dl, s, , 0, , , , , , t, , (15.14.3), , [u (l) u (l 2)]e(tl) dl, , 0, , Since u (l 2) 0 for 0 6 l 6 2, the integrand involving, u (l) is nonzero for all l 7 0, whereas the integrand involving, u (l 2) is nonzero only for l 7 2. The best way to handle, the integral is to do the two parts separately. For 0 6 t 6 2,, io¿ (t) , , , , t, , (1)e(tl) dl et, , 0, t, , t, , (1)e, , l, , dl, , 0, , t, , e (e 1) 1 e ,, t, , (15.14.4), , 0 6 t 6 2, , For t 7 2,, io–(t) , , , , t, , (1)e(tl) dl et, , 2, t, , t, , e, , 2, 2 t, , e (e e ) 1 e e ,, t, , 2, , l, , dl, t 7 2, , (15.14.5)
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 705, , 15.6, , Application to Integrodifferential Equations, , 705, , Substituting Eqs. (15.14.4) and (15.14.5) into Eq. (15.14.3) gives, io (t) io¿ (t) io–(t), (1 et )[u(t 2) u(t)] (1 e2et ) u (t 2), 1 et A,, 0 6 t 6 2, b 2, t, (e 1)e A, t 7 2, , is(t − ), 1, , (15.14.6), 0, , t−2, , , , t, , is(t − ), 1, , 0, , t, , 2 (15.14.7), , For t 7 2, the two functions overlap between (t 2) and t, as, in Fig. 15.23(b). Hence,, io(t) , , , , t, , t2, , (1)el dl el `, , t, , et e(t2), , t2, t, , (e 1) e, 2, , (b), , Figure 15.23, For Example 15.14., , (15.14.8), , From Eqs. (15.14.7) and (15.14.8), the response is, , Excitation is, , 2, , (15.14.9), , which is the same as in Eq. (15.14.6). Thus, the response io(t), along the excitation is(t) is as shown in Fig. 15.24., 6. Satisfactory? We have satisfactorily solved the problem and can, present the results as a solution to the problem., Use convolution to find vo(t) in the circuit of Fig. 15.25(a) when the, excitation is the signal shown in Fig. 15.25(b). To show how powerful, working in the s-domain is, verify your answer by performing the, equivalent operation in the s-domain., vs (V), 1Ω, vs, , +, −, , 0.5 F, , 10, , 10e −t, , +, vo, −, 0, , (a), , t, (b), , Figure 15.25, For Practice Prob. 15.14., , Answer: 20(et e2t ) V., , 15.6, , t , , t0, , A,, , 1 et A,, 0 t, io(t) b 2, t, (e 1) e A, t 2, , h(), 0 t−2, , t, , (1)el dl el ` (1 et ) A,, 0, , 0, , , , t, (a), , 5. Evaluate. To use the graphical technique, we may fold is(t) in, Fig. 15.21(b) and shift by t, as shown in Fig. 15.23(a). For, 0 6 t 6 2, the overlap between is(t l) and h(l) is from 0 to, t, so that, io(t) , , h(), , Application to Integrodifferential, Equations, , The Laplace transform is useful in solving linear integrodifferential, equations. Using the differentiation and integration properties of Laplace, transforms, each term in the integrodifferential equation is transformed., , 1, , Response io, , 0, , 1, , 2, , 3, , 4 t, , Figure 15.24, For Example 15.14; excitation and, response., , Practice Problem 15.14
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ale29559_ch15.qxd, , 07/10/2008, , 706, , 04:06 PM, , Page 706, , Chapter 15, , Introduction to the Laplace Transform, , Initial conditions are automatically taken into account. We solve the, resulting algebraic equation in the s-domain. We then convert the solution, back to the time domain by using the inverse transform. The following examples illustrate the process., , Example 15.15, , Use the Laplace transform to solve the differential equation, d 2v(t), dv(t), 6, 8v(t) 2u(t), 2, dt, dt, subject to v(0) 1, v¿(0) 2., Solution:, We take the Laplace transform of each term in the given differential, equation and obtain, [s2V(s) sv(0) v¿(0)] 6[sV(s) v(0)] 8V(s) , , 2, s, , Substituting v(0) 1, v¿(0) 2,, s 2V(s) s 2 6sV(s) 6 8V(s) , , 2, s, , or, (s2 6s 8)V(s) s 4 , , s 2 4s 2, 2, , s, s, , Hence,, V(s) , , B, A, C, s 2 4s 2, , , s, s(s 2) (s 4), s2 s4, , where, A sV(s) 0 s0 , , s 2 4s 2, 2, 1, , `, , (s 2) (s 4) s0 (2)(4), 4, , B (s 2)V(s) 0 s2 , C (s 4)V(s) 0 s4 , , s 2 4s 2, 2, 1, `, , , s (s 4), (2), (2), 2, s2, , 1, s 2 4s 2, 2, `, , , s (s 2), (4) (2), 4, s4, , Hence,, V(s) , , 1, 4, , s, , , , 1, 2, , s2, , , , 1, 4, , s4, , By the inverse Laplace transform,, 1, v(t) (1 2e2t e4t )u(t), 4
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 707, , 15.6, , Application to Integrodifferential Equations, , Solve the following differential equation using the Laplace transform, method., , 707, , Practice Problem 15.15, , d 2v(t), dv(t), 4, 4v(t) et, dt, dt 2, if v(0) v¿(0) 2., Answer: (2et 4te2t ) u (t)., , Solve for the response y(t) in the following integrodifferential equation., dy, 5y(t) 6, dt, , Example 15.16, , t, , y(t) dt u (t),, , y(0) 2, , 0, , Solution:, Taking the Laplace transform of each term, we get, 1, 6, [sY(s) y(0)] 5Y(s) Y(s) , s, s, Substituting y(0) 2 and multiplying through by s,, Y(s) (s2 5s 6) 1 2s, or, Y(s) , , A, B, 2s 1, , , (s 2) (s 3), s2, s3, , where, A (s 2)Y(s) 0 s2 , , 2s 1, 3, `, , 3, s 3 s2, 1, , B (s 3)Y(s) 0 s3 , , 2s 1, 5, `, , 5, s 2 s3 1, , Thus,, Y(s) , , 3, 5, , s2, s3, , Its inverse transform is, y (t) (3e2t 5e3t ) u (t), , Use the Laplace transform to solve the integrodifferential equation, dy, 3y(t) 2, dt, , t, , y(t) dt 2e, 0, , Answer: (et 4e2t 3e3t ) u (t)., , 3t, , ,, , y(0) 0, , Practice Problem 15.16
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 708, , Chapter 15, , 708, , Introduction to the Laplace Transform, , 15.7, , Summary, , 1. The Laplace transform allows a signal represented by a function, in the time domain to be analyzed in the s-domain (or complex, frequency domain). It is defined as, L[ f (t)] F(s) , , , , , , f (t)est dt, , 0, , 2. Properties of the Laplace transform are listed in Table 15.1, while, the Laplace transforms of basic common functions are listed in, Table 15.2., 3. The inverse Laplace transform can be found using partial fraction, expansions and using the Laplace transform pairs in Table 15.2 as, a look-up table. Real poles lead to exponential functions and complex poles to damped sinusoids., 4. The convolution of two signals consists of time-reversing one of, the signals, shifting it, multiplying it point by point with the second signal, and integrating the product. The convolution integral, relates the convolution of two signals in the time domain to the, inverse of the product of their Laplace transforms:, L1[F1(s)F2 (s)] f1(t) * f2 (t) , , t, , f (l)f (t l) dl, 1, , 2, , 0, , 5. In the time domain, the output y(t) of the network is the convolution of the impulse response with the input x(t),, y(t) h (t) * x (t), Convolution may be regarded as the flip-shift-multiply-time-area, method., 6. The Laplace transform can be used to solve a linear integrodifferential equation., , Review Questions, 15.1, , 15.2, , 15.3, , Every function f (t) has a Laplace transform., , is at, , (a) True, , (a) 4, , (b) 3, , (c) 2, , (d) 1, , (b) False, , The variable s in the Laplace transform H(s) is called, (a) complex frequency, , (b) transfer function, , (c) zero, , (d) pole, , The poles of the function, F(s) , , The Laplace transform of u (t 2) is:, (a), , 1, s2, , (b), , 1, s2, , The zero of the function, F(s) , , s1, (s 2) (s 3) (s 4), , s1, (s 2) (s 3) (s 4), , are at, , e2s, (d), s, , e2s, (c), s, 15.4, , 15.5, , 15.6, , (a) 4, , (b) 3, , (c) 2, , (d) 1, , If F(s) 1(s 2), then f(t) is, (a) e2tu (t), , (b) e2tu (t), , (c) u (t 2), , (d) u (t 2)
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 709, , Problems, , 15.7, , Given that F(s) e2s(s 1), then f(t) is, 2(t1), , (a) e, , is:, (a) et cos 2t, , (t2), , u (t 1), , u (t 2), , (b) e, , t, , 2t, , t, , (c) e u (t 2), , (d) e u (t 1), , (c) e, , (e) e(t2)u (t), 15.8, , 709, , (b) et sin 2t, (d) e2t sin 2t, , cos t, , (e) none of the above, 15.10 The result of u (t) * u (t) is:, , The initial value of f (t) with transform, F(s) , , s1, (s 2)(s 3), , (a) u2(t), , (b) tu(t), , (c) t 2u(t), , (d) d(t), , is:, (a) nonexistent, , Answers: 15.1b, 15.2a, 15.3d, 15.4d, 15.5a,b,c, 15.6b,, 15.7b, 15.8d, 15.9c, 15.10b., , (c) 0, , 1, 6, The inverse Laplace transform of, (d) 1, , 15.9, , (b) , (e), , s2, (s 2)2 1, , Problems, Sections 15.2 and 15.3 Definition and Properties, of the Laplace Transform, 15.1, , 15.6, , 2t,, f (t) • t,, 0,, , Find the Laplace transform of:, (a) cosh at, , (b) sinh at, , 1, [Hint: cosh x (e x ex ),, 2, , 15.7, , 15.3, , (c) h(t) (6 sin(3t) 8 cos(3t)) u (t), (d) x(t) (e2t cosh(4t)) u (t), , (b) sin(t u), , 15.8, , Obtain the Laplace transform of each of the, following functions:, (a) e2t cos 3tu(t), , (b) e2t sin 4 tu(t), , (c) e3t cosh 2tu(t), , (d) e4t sinh tu(t), , 15.5, , Design a problem to help other students better, understand how to find the Laplace transform of, different time varying functions., Find the Laplace transform of each of the following, functions:, (a) t 2 cos(2t 30) u (t), d, (c) 2tu(t) 4 d(t), dt, (e) 5 u (t2), dn, (g) n d(t), dt, , (b) 3t 4e2t u (t), (t1), , (d) 2e, , u (t), , (f) 6et3 u (t), , Find the Laplace transform F(s), given that f (t) is:, (a) 2tu(t 4), (b) 5 cos(t) d(t 2), (c) et u (t t), (d) sin(2t) u (t t), , (e) tet sin 2tu(t), 15.4, , Find the Laplace transform of the following signals:, (b) g(t) (4 3e2t ) u (t), , Determine the Laplace transform of:, (a) cos(t u), , 0 6 t 6 1, 1 6 t 6 2, otherwise, , (a) f (t) (2t 4) u (t), , 1, sinh x (e x ex).], 2, 15.2, , Find F(s) given that, , 15.9, , Determine the Laplace transforms of these, functions:, (a) f (t) (t 4) u (t 2), (b) g(t) 2e4t u (t 1), (c) h(t) 5 cos(2t 1) u (t), (d) p(t) 6[ u (t 2) u (t 4)], , 15.10 In two different ways, find the Laplace transform of, g(t) , , d, (2e4t cos (2t)), dt
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 710, , Chapter 15, , 710, , Introduction to the Laplace Transform, , 15.11 Find F(s) if:, , f (t), t, , (a) f (t) 6e, , 2t, , (b) f (t) 3t e, , cosh 2t, , sinh 4t, , a1, , (c) f (t) 8e3t cosh tu (t 2), 15.12 If g(t) e2t cos 4t, find G(s)., , a2, , 15.13 Find the Laplace transform of the following, functions:, , (c), , 0, , (b) ett sin tu (t), , (a) t cos tu (t), , t(s), , t2, , t1, , Figure 15.29, , sin bt, u (t), t, , For Prob. 15.17., , 15.14 Find the Laplace transform of the signal in Fig. 15.26., 15.18 Obtain the Laplace transforms of the functions in, Fig. 15.30., f (t), 10, g(t), 3, h(t), , 2, 0, , 2, , 4, , t, , 6, , 2, 1, , Figure 15.26, For Prob. 15.14., , 0, , 15.15 Determine the Laplace transform of the function in, Fig. 15.27., , 1, , 2, , 3 t, , 0, , 1, , (a), , 2, , 3, , 4 t, , (b), , Figure 15.30, For Prob. 15.18., , f (t), , 15.19 Calculate the Laplace transform of the train of unit, impulses in Fig. 15.31., , 5, , 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7 t(s), , Figure 15.27, , f (t), , For Prob. 15.15., 1, , 15.16 Obtain the Laplace transform of f (t) in Fig. 15.28., , 0, , 1, , 2, , 3, , 4 t, , Figure 15.31, For Prob. 15.19., f (t), 6, , 15.20 Using Fig. 15.32, design a problem to help other, students better understand the Laplace transform of a, simple, periodic waveshape., , 3, , 0, , 1, , 2, , 3, , 4 t, , Figure 15.28, , g(t), , For Prob. 15.16., , a, , 15.17 Using Fig. 15.29, design a problem to help other, students better understand the Laplace transform of a, simple, non-periodic waveshape., , 0, , Figure 15.32, For Prob. 15.20., , t1, , t2, , t3 t
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 711, , Problems, , 711, , 15.26 Determine the initial and final values of f (t), if they, exist, given that:, , 15.21 Obtain the Laplace transform of the periodic, waveform in Fig. 15.33., , s2 3, s3 4s2 6, s2 2s 1, (b) F(s) , (s 2) (s2 2s 4), (a) F(s) , , f (t), 1, 0, , 2, , 4, , 6, , 8 t, , Section 15.4 The Inverse Laplace Transform, , Figure 15.33, For Prob. 15.21., , 15.27 Determine the inverse Laplace transform of each of, the following functions:, 2, 1, , s, s1, 3s 1, (b) G(s) , s4, 4, (c) H(s) , (s 1) (s 3), 12, (d) J(s) , (s 2)2(s 4), , 15.22 Find the Laplace transforms of the functions in, Fig. 15.34., , (a) F(s) , , h(t), 3, , g(t), 2, , 1, 0, , 1, , 2, , 3 t, , 0, , 1, , 2, , (a), , 3, , 4, , 5 t, , (b), , Figure 15.34, , 15.28 Design a problem to help other students better, understand how to find the inverse Laplace, transform., 15.29 Find the inverse Laplace transform of:, , For Prob. 15.22., V(s) , 15.23 Determine the Laplace transforms of the periodic, functions in Fig. 15.35., , s 13, s (s2 4s 13), , 15.30 Find the inverse Laplace transform of:, 6s2 8s 3, s (s2 2s 5), s2 5s 6, (b) F2(s) , (s 1)2(s 4), 10, (c) F3(s) , (s 1) (s2 4s 8), (a) F1(s) , , f (t), 1, , h(t), , t2, , 4, , 0, , 1, , 2, , 3, , 4 t, , −1, 0, , 2, , (a), , 4, , 6 t, , (b), , Figure 15.35, For Prob. 15.23., , 15.24 Design a problem to help other students better, understand how to find the initial and final value of a, transfer function., 15.25 Let, F(s) , , 5(s 1), (s 2) (s 3), , (a) Use the initial and final value theorems to find, f (0) and f ()., (b) Verify your answer in part (a) by finding f(t),, using partial fractions., , 15.31 Find f (t) for each F(s):, 10s, (s 1) (s 2) (s 3), 2s2 4s 1, (b), (s 1) (s 2)3, s1, (c), (s 2) (s2 2s 5), (a), , 15.32 Determine the inverse Laplace transform of each of, the following functions:, (a), , 8(s 1) (s 3), s (s 2) (s 4), , s2 2s 4, (s 1) (s 2)2, s2 1, (c), (s 3) (s2 4s 5), (b)
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 712, , Chapter 15, , 712, , Introduction to the Laplace Transform, , 15.37 Find the inverse Laplace transform of:, , 15.33 Calculate the inverse Laplace transform of:, (a), (c), , ps, , 6(s 1), , (b), , s4 1, , e, s2 1, , (a) H(s) , , s2 4s 5, (s 3) (s2 2s 2), e4s, (c) F(s) , s2, , 3, s (s 1)3, , (b) G(s) , , 15.34 Find the time functions that have the following, Laplace transforms:, , (d) D(s) , , s 1, s2 4, s, e 4e2s, (b) G(s) 2, s 6s 8, (s 1)e2s, (c) H(s) , s (s 3) (s 4), 2, , (a) F(s) 10 , , 10s, (s 1) (s2 4), 2, , 15.38 Find f (t) given that:, s2 4s, s 10s 26, 5s2 7s 29, (b) F(s) , s (s2 4s 29), (a) F(s) , , 2, , *15.39 Determine f(t) if:, , 15.35 Obtain f(t) for the following transforms:, , 2s3 4s2 1, (s2 2s 17) (s2 4s 20), s2 4, (b) F(s) 2, (s 9) (s2 6s 3), , (s 3)e6s, (a) F(s) , (s 1) (s 2), , (a) F(s) , , 4 e2s, s 5s 4, ses, (c) F(s) , (s 3) (s2 4), (b) F(s) , , s4, s (s 2), , 2, , 15.40 Show that, L1 c, , 4s2 7s 13, d , (s 2) (s2 2s 5), , c 12et cos(2t 45) 3e2t d u (t), , 15.36 Obtain the inverse Laplace transforms of the, following functions:, 1, s2(s 2) (s 3), 1, (b) Y(s) , s(s 1)2, 1, (c) Z(s) , s (s 1) (s2 6s 10), (a) X(s) , , Section 15.5 The Convolution Integral, *15.41 Let x(t) and y(t) be as shown in Fig. 15.36. Find, z(t) x(t) * y(t)., 15.42 Design a problem to help other students better, understand how to convolve two functions together., y(t), 4, , x (t), 2, , 0, 0, , 2, , 4, , 6, , t, , −4, , Figure 15.36, For Prob. 15.41., , * An asterisk indicates a challenging problem., , 2, , 4, , 6, , 8, , t
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 713, , Problems, , 15.43 Find y(t) x (t) * h (t) for each paired x(t) and h(t) in, Fig. 15.37., , 713, , 15.46 Given the following functions, x (t) 2d(t),, , z(t) e2tu (t),, , y (t) 4 u (t),, , evaluate the following convolution operations., x(t), , h(t), , 1, , 1, , (a) x(t) * y(t), (b) x(t) * z(t), (c) y(t) * z(t), , 0, , 0, , t, , 1, , (d) y(t) * [ y(t) z(t)], , t, , 1, , (a), , 15.47 A system has the transfer function, h(t), 2, , H(s) , , x(t), , (a) Find the impulse response of the system., , 2e −t, , 1, , s, (s 1) (s 2), , (b) Determine the output y(t), given that the input is, x (t) u (t)., 0, , 0, , t, , t, , 15.48 Find f (t) using convolution given that:, , (b), , 4, (s 2s 5)2, 2s, (b) F(s) , (s 1) (s2 4), (a) F(s) , , x(t), , h(t), , 1, , 1, , 2, , *15.49 Use the convolution integral to find:, −1, , 0, , 0, , t, , 1, , 2, , 1, , t, , (a) t * e at u (t), , (c), , (b) cos(t) * cos(t) u (t), , Figure 15.37, For Prob. 15.43., , Section 15.6 Application to Integrodifferential, Equations, 15.44 Obtain the convolution of the pairs of signals in, Fig. 15.38., x(t), , h(t), , 1, , 1, , 15.50 Use the Laplace transform to solve the differential, equation, d 2 v(t), dt 2, , 1, , 0, , t, , 15.51 Given that v(0) 2 and dv(0)dt 4, solve, , t, , 1, , d v(t), 10v(t) 3 cos 2t, dt, , subject to v(0) 1, dv(0)dt 2., , 2, 0, , 2, , dv, d 2v, 5, 6v 6et u (t), 2, dt, dt, , −1, (a), , 15.52 Use the Laplace transform to find i(t) for t 7 0 if, f 1(t), , f 2(t), , 1, , 1, , 0, , 1, , t, , d 2i, di, 3 2i d(t) 0,, dt, dt 2, 0, , 1, , 2, , 3, , 4, , 5, , t, , (b), , Figure 15.38, For Prob. 15.44., 15.45 Given h (t) 4e2tu (t) and x(t) d(t) 2e2tu (t),, find y (t) x (t) * h (t)., , i(0) 0,, , i¿(0) 3, , *15.53 Use Laplace transforms to solve for x(t) in, t, , x(t) cos t , , e, , lt, , x(l) dl, , 0, , 15.54 Design a problem to help other students better, understand solving second order differential, equations with a time varying input.
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ale29559_ch15.qxd, , 07/10/2008, , 04:06 PM, , Page 714, , Chapter 15, , 714, , Introduction to the Laplace Transform, , 15.55 Solve for y(t) in the following differential equation if, the initial conditions are zero., d 3y, , 15.58 Given that, dv, 2v 5, dt, , d 2y, , dy, 6 2 8, et cos 2t, 3, dt, dt, dt, , 15.56 Solve for v(t) in the integrodifferential equation, dv, 12, 4, dt, , , , t, , v(l) dl 4 u (t), 0, , with v(0) 1, determine v(t) for t 7 0., 15.59 Solve the integrodifferential equation, , t, , v dt 0, , dy, 4y 3, dt, , , , given that v(0) 2., 15.57 Design a problem to help other students better, understand solving integrodifferential equations with, a periodic input, using Laplace transforms., , t, , y dt 6e, , 2t, , ,, , y (0) 1, , 0, , 15.60 Solve the following integrodifferential equation, 2, , dx, 5x 3, dt, , t, , x dt 4 sin 4t,, 0, , x (0) 1
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 715, , c h a p t e r, , Applications of the, Laplace Transform, , 16, , Communication skills are the most important skills any engineer can, have. A very critical element in this tool set is the ability to ask a question and understand the answer, a very simple thing and yet it may, make the difference between success and failure!, —James A. Watson, , Enhancing Your Skills and Your Career, Asking Questions, In over 30 years of teaching, I have struggled with determining how best, to help students learn. Regardless of how much time students spend in, studying for a course, the most helpful activity for students is learning how, to ask questions in class and then asking those questions. The student, by, asking questions, becomes actively involved in the learning process and no, longer is merely a passive receptor of information. I think this active, involvement contributes so much to the learning process that it is probably, the single most important aspect to the development of a modern engineer., In fact, asking questions is the basis of science. As Charles P. Steinmetz, rightly said, “No man really becomes a fool until he stops asking questions.”, It seems very straightforward and quite easy to ask questions. Have, we not been doing that all our lives? The truth is to ask questions in, an appropriate manner and to maximize the learning process takes, some thought and preparation., I am sure that there are several models one could effectively use. Let, me share what has worked for me. The most important thing to keep in, mind is that you do not have to form a perfect question. Since the questionand-answer format allows the question to be developed in an iterative manner, the original question can easily be refined as you go. I frequently tell, students that they are most welcome to read their questions in class., Here are three things you should keep in mind when asking questions. First, prepare your question. If you are like many students who are, either shy or have not learned to ask questions in class, you may wish to, start with a question you have written down outside of class. Second, wait, for an appropriate time to ask the question. Simply use your judgment on, that. Third, be prepared to clarify your question by paraphrasing it or saying it in a different way in case you are asked to repeat the question., , Photo by Charles Alexander, , 715
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ale29559_ch16.qxd, , 716, , 07/10/2008, , 04:11 PM, , Page 716, , Chapter 16, , Applications of the Laplace Transform, , One last comment: not all professors like students to ask questions, in class even though they may say they do. You need to find out which, professors like classroom questions. Good luck in enhancing one of, your most important skills as an engineer., , 16.1, , Introduction, , Now that we have introduced the Laplace transform, let us see what, we can do with it. Please keep in mind that with the Laplace transform, we actually have one of the most powerful mathematical tools for, analysis, synthesis, and design. Being able to look at circuits and systems in the s-domain can help us to understand how our circuits and, systems really function. In this chapter we will take an in-depth look, at how easy it is to work with circuits in the s-domain. In addition, we, will briefly look at physical systems. We are sure you have studied, some mechanical systems and may have used the same differential, equations to describe them as we use to describe our electric circuits., Actually that is a wonderful thing about the physical universe in which, we live; the same differential equations can be used to describe any, linear circuit, system, or process. The key is the term linear., A system is a mathematical model of a physical process relating the, input to the output., , It is entirely appropriate to consider circuits as systems. Historically, circuits have been discussed as a separate topic from systems, so, we will actually talk about circuits and systems in this chapter realizing that circuits are nothing more than a class of electrical systems., The most important thing to remember is that everything we discussed, in the last chapter and in this chapter applies to any linear system. In the, last chapter, we saw how we can use Laplace transforms to solve linear, differential equations and integral equations. In this chapter, we introduce, the concept of modeling circuits in the s-domain. We can use that principle to help us solve just about any kind of linear circuit. We will take a, quick look at how state variables can be used to analyze systems with multiple inputs and multiple outputs. Finally, we examine how the Laplace, transform is used in network stability analysis and in network synthesis., , 16.2, , Circuit Element Models, , Having mastered how to obtain the Laplace transform and its inverse,, we are now prepared to employ the Laplace transform to analyze circuits. This usually involves three steps., , Steps in Applying the Laplace Transform:, 1. Transform the circuit from the time domain to the s-domain., 2. Solve the circuit using nodal analysis, mesh analysis, source, transformation, superposition, or any circuit analysis technique, with which we are familiar., 3. Take the inverse transform of the solution and thus obtain the, solution in the time domain.
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 717, , 16.2, , Circuit Element Models, , 717, , Only the first step is new and will be discussed here. As we did in phasor analysis, we transform a circuit in the time domain to the frequency, or s-domain by Laplace transforming each term in the circuit., For a resistor, the voltage-current relationship in the time domain is, , As one can infer from step 2, all the circuit analysis techniques applied for dc, circuits are applicable to the s-domain., , v(t) Ri(t), , (16.1), , Taking the Laplace transform, we get, V(s) RI(s), , (16.2), , di(t), dt, , (16.3), , For an inductor,, v(t) L, , Taking the Laplace transform of both sides gives, , i(t), , , , , , V(s) L[sI(s) i(0 )] sLI(s) Li(0 ), , (16.4), , +, , I(s), +, , i(0), , sL, , or, , L, , v (t), , i(0), 1, I(s) V(s) , s, sL, , V(s), , −, , (16.5), , (b), , (a), I(s), , The s-domain equivalents are shown in Fig. 16.1, where the initial condition is modeled as a voltage or current source., For a capacitor,, dv(t), i(t) C, dt, , −, −, + Li(0 ), , −, , +, V(s), , (16.6), , sL, , i(0−), s, , −, , which transforms into the s-domain as, I(s) C[sV(s) v(0)] sCV(s) Cv(0 ), , (c), , (16.7), , Figure 16.1, Representation of an inductor: (a) timedomain, (b,c) s-domain equivalents., , or, V(s) , , v(0), 1, I(s) , s, sC, , (16.8), , The s-domain equivalents are shown in Fig. 16.2. With the s-domain, equivalents, the Laplace transform can be used readily to solve first- and, , +, , +, v (t), , +, v (0), −, , I(s), , I(s), , i(t), , C, , V(s), , (a), , 1, sC, , −, , V(s), +, −, , −, , −, , +, , +, , (b), , v (0), s, , +, 1, sC, −, , Cv (0), , −, (c), , Figure 16.2, Representation of a capacitor: (a) time-domain, (b,c) s-domain equivalents.
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ale29559_ch16.qxd, , 07/17/2008, , 01:22 PM, , Page 718, , Chapter 16, , 718, , The elegance of using the Laplace, transform in circuit analysis lies in the, automatic inclusion of the initial conditions in the transformation process,, thus providing a complete (transient, and steady-state) solution., , i(t), , I(s), , +, , +, R, , v (t), −, , V(s) RI(s), V(s) sLI(s), 1, Capacitor: V(s) , I(s), sC, , −, (a), I(s), , i(t), +, , +, L, , v (t), , sL, , V(s), , −, (b), , v (t), , 1, sC, , V(s), , Z(s) R, Z(s) sL, 1, Capacitor: Z(s) , sC, , Resistor:, Inductor:, , +, , −, , V(s), I(s), , (16.10), , Thus, the impedances of the three circuit elements are, I(s), , i(t), , C, , (16.9), , The s-domain equivalents are shown in Fig. 16.3., We define the impedance in the s-domain as the ratio of the voltage transform to the current transform under zero initial conditions;, that is,, Z(s) , , −, , +, , second-order circuits such as those we considered in Chapters 7 and 8., We should observe from Eqs. (16.3) to (16.8) that the initial conditions, are part of the transformation. This is one advantage of using the, Laplace transform in circuit analysis. Another advantage is that a complete response—transient and steady state—of a network is obtained., We will illustrate this with Examples 16.2 and 16.3. Also, observe the, duality of Eqs. (16.5) and (16.8), confirming what we already know, from Chapter 8 (see Table 8.1), namely, that L and C, I(s) and V(s),, and v(0) and i(0) are dual pairs., If we assume zero initial conditions for the inductor and the capacitor, the above equations reduce to:, Resistor:, Inductor:, , R, , V(s), , Applications of the Laplace Transform, , −, , (16.11), , Table 16.1 summarizes these. The admittance in the s-domain is the, reciprocal of the impedance, or, , (c), , Figure 16.3, Time-domain and s-domain representations, of passive elements under zero initial, conditions., , TABLE 16.1, , Impedance of an element in, the s-domain.*, , Y(s) , , I(s), 1, , Z(s), V(s), , (16.12), , The use of the Laplace transform in circuit analysis facilitates the use, of various signal sources such as impulse, step, ramp, exponential, and, sinusoidal., The models for dependent sources and op amps are easy to develop, drawing from the simple fact that if the Laplace transform of f(t) is, F(s), then the Laplace transform of af(t) is aF(s)—the linearity property. The dependent source model is a little easier in that we deal with, a single value. The dependent source can have only two controlling, values, a constant times either a voltage or a current. Thus,, L[av(t)] aV(s), , (16.13), (16.14), , Element, , Z(s) V(s)I(s), , L[ai(t)] aI(s), , Resistor, Inductor, Capacitor, , R, sL, 1sC, , The ideal op amp can be treated just like a resistor. Nothing within, an op amp, either real or ideal, does anything more than multiply a, voltage by a constant. Thus, we only need to write the equations as we, always do using the constraint that the input voltage to the op amp has, to be zero and the input current has to be zero., , * Assuming zero initial conditions
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 719, , 16.2, , Circuit Element Models, , 719, , Example 16.1, , Find vo(t) in the circuit of Fig. 16.4, assuming zero initial conditions., 1Ω, , Solution:, We first transform the circuit from the time domain to the s-domain., u(t) +, −, , 1, 3, , 5Ω, , u(t), , 1, , 1H, , 1, , sL s, , Figure 16.4, , 1, F, 3, , 1, 3, , s, sC, , For Example 16.1., , 1, , The resulting s-domain circuit is in Fig. 16.5. We now apply mesh, analysis. For mesh 1,, 3, 1, 3, a1 b I1 I2, s, s, s, , (16.1.1), , 1Ω, , 1, s, , 5Ω, , 3, s, , +, −, , s, , I1(s), , For mesh 2,, 3, 3, 0 I1 as 5 b I2, s, s, , +, Vo (s), −, , I2(s), , Figure 16.5, Mesh analysis of the frequency-domain, equivalent of the same circuit., , or, 1, I1 (s2 5s 3)I2, 3, , +, v o(t), −, , 1H, , F, , 1, s, , (16.1.2), , Substituting this into Eq. (16.1.1),, 1, 3 1, 3, a1 b (s2 5s 3) I2 I2, s, s 3, s, Multiplying through by 3s gives, 3 (s3 8s2 18s) I2, Vo(s) sI2 , , 1, , I2 , , 3, s3 8s2 18s, , 3, 3, 12, , 2, 12 (s 4) ( 12)2, s 8s 18, 2, , Taking the inverse transform yields, vo(t) , , 3 4t, e sin 12t V,, 12, , t0, , Determine vo(t) in the circuit of Fig. 16.6, assuming zero initial, conditions., , Practice Problem 16.1, 1H, , 2t, , Answer: 20(1 e, , 2t, , 2te, , ) u (t) V., 1, 4, , F, , 5u(t) V, , Figure 16.6, For Practice Prob. 16.1., , 4Ω, , +, v o(t), −
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ale29559_ch16.qxd, , 07/10/2008, , 720, , Example 16.2, , 04:11 PM, , Page 720, , Chapter 16, , Applications of the Laplace Transform, , Find vo(t) in the circuit of Fig. 16.7. Assume vo(0) 5 V., 10 Ω, , 10e, , −t u(t), , V +, −, , +, v o (t), −, , 10 Ω, , 0.1 F, , 2(t) A, , Figure 16.7, For Example 16.2., , Solution:, We transform the circuit to the s-domain as shown in Fig. 16.8. The initial, condition is included in the form of the current source Cvo(0) 0.1(5) , 0.5 A. [See Fig. 16.2(c).] We apply nodal analysis. At the top node,, 10(s 1) Vo, Vo, Vo, 2 0.5 , , 10, 10, 10s, or, 2Vo, sVo, 1, 1, 2.5 , , , Vo(s 2), s1, 10, 10, 10, 10 Ω, , 10, +, s+1 −, , V o (s), , 10, s, , 10 Ω, , 0.5 A, , 2A, , Figure 16.8, Nodal analysis of the equivalent of the circuit in Fig. 16.7., , Multiplying through by 10,, 10, 25 Vo(s 2), s1, or, Vo , , 25s 35, A, B, , , (s 1) (s 2), s1, s2, , where, 25s 35, 10, `, , 10, (s 2) s1, 1, 25s 35, 15, , `, , 15, (s 1) s2, 1, , A (s 1)Vo(s) 0 s1 , B (s 2)Vo(s) 0 s2, Thus,, Vo(s) , , 10, 15, , s1, s2, , Taking the inverse Laplace transform, we obtain, vo(t) (10et 15e2t ) u (t) V
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 721, , 16.2, , Circuit Element Models, , Find vo(t) in the circuit shown in Fig. 16.9. Note that, since the voltage input is multiplied by u(t), the voltage source is a short for all, t 6 0 and iL(0) 0., Answer: (24e2t 4et3) u (t) V., , 721, , Practice Problem 16.2, 1Ω, 30e−2t u(t) V +, −, , +, v o (t), −, , 2Ω, , 2H, , Figure 16.9, For Practice Prob. 16.2., , Example 16.3, , In the circuit of Fig. 16.10(a), the switch moves from position a to position b at t 0. Find i(t) for t 7 0., Solution:, The initial current through the inductor is i(0) Io. For t 7 0,, Fig. 16.10(b) shows the circuit transformed to the s-domain. The initial, condition is incorporated in the form of a voltage source as, Li(0) LIo. Using mesh analysis,, I(s)(R sL) LIo , , a, , t=0, , R, i(t), , b, Io, , L, , + V, o, −, (a), , Vo, 0, s, , (16.3.1), R, , or, I(s) , , Vo, Io, VoL, LIo, , , , R sL, s(R sL), s RL, s(s RL), , sL, , (16.3.2), , Vo +, s −, , I(s), , Applying partial fraction expansion on the second term on the righthand side of Eq. (16.3.2) yields, VoR, Io, VoR, , I(s) , , s, s RL, (s RL), , (16.3.3), , The inverse Laplace transform of this gives, i(t) aIo , , Vo tt Vo, be, ,, R, R, , t0, , (16.3.4), , where t RL. The term in parentheses is the transient response, while, the second term is the steady-state response. In other words, the final, value is i() VoR, which we could have predicted by applying the, final-value theorem on Eq. (16.3.2) or (16.3.3); that is,, lim sI(s) lim a, , sS0, , sS0, , sIo, VoL, Vo, , b, s RL, s RL, R, , (16.3.5), , Equation (16.3.4) may also be written as, i(t) Io ett , , Vo, (1 ett ),, R, , t0, , (16.3.6), , The first term is the natural response, while the second term is the, forced response. If the initial condition Io 0, Eq. (16.3.6) becomes, i(t) , , Vo, (1 ett ),, R, , t0, , (16.3.7), , which is the step response, since it is due to the step input Vo with no, initial energy., , −, +, , (b), , Figure 16.10, For Example 16.3., , LIo
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 722, , Chapter 16, , 722, , Practice Problem 16.3, a, , t=0, , R, , Vo +, −, , The switch in Fig. 16.11 has been in position b for a long time. It is, moved to position a at t 0. Determine v(t) for t 7 0., Answer: v(t) (Vo Io R)ett Io R, t 7 0, where t RC., , b, Io, , Applications of the Laplace Transform, , C, , +, v (t), −, , Figure 16.11, , 16.3, , For Practice Prob. 16.3., , Circuit Analysis, , Circuit analysis is again relatively easy to do when we are in the, s-domain. We merely need to transform a complicated set of mathematical relationships in the time domain into the s-domain where we, convert operators (derivatives and integrals) into simple multipliers of, s and 1s. This now allows us to use algebra to set up and solve our, circuit equations. The exciting thing about this is that all of the circuit, theorems and relationships we developed for dc circuits are perfectly, valid in the s-domain., Remember, equivalent circuits, with capacitors and inductors, only, exist in the s-domain; they cannot be transformed back into the time, domain., , Example 16.4, 10, 3, , v s (t), , Consider the circuit in Fig. 16.12(a). Find the value of the voltage, across the capacitor assuming that the value of vs(t) 10u(t) V and, assume that at t 0, 1 A flows through the inductor and 5 V is, across the capacitor., , Ω, , +, −, , 5H, , 0.1 F, , Solution:, Figure 16.12(b) represents the entire circuit in the s-domain with the, initial conditions incorporated. We now have a straightforward nodal, analysis problem. Since the value of V1 is also the value of the, capacitor voltage in the time domain and is the only unknown node, voltage, we only need to write one equation., , i(0), s, , 10, s, + v (0), −, s, , V1 10s, V1 [v(0)s], V1 0, i(0), , , , 0, s, 103, 5s, 1(0.1s), , (16.4.1), , 1, 2, 3, 0.1as 3 b V1 0.5, s, s, s, , (16.4.2), , (a), 10, 3, , 10, s, , +, −, , Ω, , V1, , 5s, , or, (b), , Figure 16.12, For Example 16.4., , where v(0) 5 V and i(0) 1 A. Simplifying we get, (s2 3s 2) V1 40 5s, or, V1 , , 35, 30, 40 5s, , , (s 1) (s 2), s1, s2, , (16.4.3), , Taking the inverse Laplace transform yields, v1(t) (35et 30e2t ) u (t) V, , (16.4.4)
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ale29559_ch16.qxd, , 07/22/2008, , 01:15 PM, , Page 723, , 16.3, , Circuit Analysis, , For the circuit shown in Fig. 16.12 with the same initial conditions,, find the current through the inductor for all time t 7 0., , 723, , Practice Problem 16.4, , Answer: i(t) (3 7et 3e2t ) u (t) A., , Example 16.5, , For the circuit shown in Fig. 16.12, and the initial conditions used, in Example 16.4, use superposition to find the value of the capacitor, voltage., Solution:, Since the circuit in the s-domain actually has three independent sources,, we can look at the solution one source at a time. Figure 16.13 presents, the circuits in the s-domain considering one source at a time. We now, have three nodal analysis problems. First, let us solve for the capacitor, voltage in the circuit shown in Fig. 16.13(a)., , 10, 3, , Ω, , V1, 10, s, , 10, s, , +, −, , 5s, , 0, , +, −, , 0, , (a), , V1 10s, V1 0, V1 0, , 0, 0, 103, 5s, 1(0.1s), , 10, 3, , Ω, , V2, 10, s, , or, 0 +, −, , 2, 3, 0.1as 3 b V1 , s, s, , i(0), s, , 5s, , +, −, , 0, , Simplifying we get, (b), , (s2 3s 2) V1 30, , 10, 3, , 30, 30, 30, V1 , , , (s 1) (s 2), s1, s2, 0, , v1(t) (30e, , 2t, , 30e, , ) u (t) V, , V3, 10, s, , or, t, , Ω, , +, −, , 0, , 5s, , (16.5.1), , For Fig. 16.13(b) we get,, V2 0, V2 0, V2 0, 1, , , 0, s, 103, 5s, 1(0.1s), , (c), , Figure 16.13, For Example 16.5., , or, 2, 1, 0.1as 3 b V2 , s, s, This leads to, V2 , , 10, 10, 10, , , (s 1) (s 2), s1, s2, , Taking the inverse Laplace transform, we get, v2(t) (10et 10e2t ) u (t) V, For Fig. 16.13(c),, V3 0, V3 5s, V3 0, , 0, 0, 103, 5s, 1(0.1s), , (16.5.2), , +, −, , v (0)
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 724, , Chapter 16, , 724, , Applications of the Laplace Transform, , or, 2, 0.1as 3 b V3 0.5, s, V3 , , 5s, 5, 10, , , (s 1) (s 2), s1, s2, , This leads to, v3(t) (5et 10e2t ) u (t) V, , (16.5.3), , Now all we need to do is to add Eqs. (16.5.1), (16.5.2), and (16.5.3):, v(t) v1(t) v2(t) v3(t), 5(30 10 5)et (30 10 10)e2t 6 u (t) V, or, v(t) (35et 30e2t ) u (t) V, which agrees with our answer in Example 16.4., , Practice Problem 16.5, , For the circuit shown in Fig. 16.12, and the same initial conditions in, Example 16.4, find the current through the inductor for all time t 7 0, using superposition., Answer: i(t) (3 7et 3e2t ) u (t) A., , Example 16.6, Ix, , is, , +, −, , 2H, , 2ix, 5Ω, , Figure 16.14, For Example 16.6., , Assume that there is no initial energy stored in the circuit of Fig. 16.14, at t 0 and that is 10 u (t) A. (a) Find Vo(s) using Thevenin’s theorem. (b) Apply the initial- and final-value theorems to find vo(0) and, vo(). (c) Determine vo(t)., 5Ω, , +, v o(t), −, , Solution:, Since there is no initial energy stored in the circuit, we assume that the, initial inductor current and initial capacitor voltage are zero at t 0., (a) To find the Thevenin equivalent circuit, we remove the 5-, resistor and then find Voc (VTh) and Isc. To find VTh, we use the Laplacetransformed circuit in Fig. 16.15(a). Since Ix 0, the dependent, voltage source contributes nothing, so, Voc VTh 5 a, , 50, 10, b, s, s, , To find Z Th, we consider the circuit in Fig. 16.15(b), where we first, find Isc. We can use nodal analysis to solve for V1 which then leads to, Isc (Isc Ix V12s)., , , (V1 2Ix) 0, V1 0, 10, , , 0, s, 5, 2s, , along with, Ix , , V1, 2s
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ale29559_ch16.qxd, , 07/14/2008, , 01:04 PM, , Page 725, , 16.3, , Circuit Analysis, , 725, , leads to, , Ix, , 2s, a, +, , 100, V1 , 2s 3, Hence,, 100(2s 3), V1, 50, Isc , , , 2s, 2s, s(2s 3), , 10, s, , +, −, , V Th, , 2I x, 5, , −, b, , and, Z Th , , (a), , Voc, 50s, , 2s 3, Isc, 50[s(2s 3)], , V1 I x, , 2s, a, , The given circuit is replaced by its Thevenin equivalent at terminals, a-b as shown in Fig. 16.16. From Fig. 16.16,, Vo , , 250, 5, 5, 50, 125, VTh , a b, , 5 Z Th, 5 2s 3 s, s(2s 8), s(s 4), , 10, s, , +, −, 5, , (b) Using the initial-value theorem we find, 125s, 125, 0, vo(0) lim sVo(s) lim, lim, 0, sS , sS s 4, sS 1 4s, 1, Using the final-value theorem we find, 125, 125, vo() lim sVo(s) lim, , 31.25 V, sS0, sS0 s 4, 4, , b, (b), , Figure 16.15, For Example 16.6: (a) finding VTh,, (b) determining Z Th., , (c) By partial fraction,, B, 125, A, Vo , , s, s (s 4), s4, 125, A sVo(s) 2, , 2, 31.25, s, 4 s0, s0, 125, 2, B (s 4)Vo(s) 2, , 31.25, s s4, s4, Vo , , 31.25, 31.25, , s, s4, , I sc, , 2Ix, , Z Th, a, V Th, , 5Ω, , +, −, , +, Vo, −, b, , Figure 16.16, The Thevenin equivalent of the circuit in, Fig. 16.14 in the s-domain., , Taking the inverse Laplace transform gives, vo(t) 31.25(1 e4t ) u (t) V, Notice that the values of vo(0) and vo() obtained in part (b) are, confirmed., , The initial energy in the circuit of Fig. 16.17 is zero at t 0. Assume, that vs 15u (t) V. (a) Find Vo(s) using the Thevenin theorem., (b) Apply the initial- and final-value theorems to find vo(0) and vo()., (c) Obtain vo(t)., Answer: (a) Vo(s) 4(s0.25), s(s0.3) , (b) 4, 3.333 V,, (c) (10 2e0.3t ) u (t) V., , Practice Problem 16.6, ix, , vs, , 1F, , 1Ω, , +, −, , +, vo, −, , 2Ω, , Figure 16.17, For Practice Prob. 16.6., , +, −, , 4ix
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , 726, , Page 726, , Chapter 16, , 16.4, For electrical networks, the transfer, function is also known as the network, function., , Applications of the Laplace Transform, , Transfer Functions, , The transfer function is a key concept in signal processing because it, indicates how a signal is processed as it passes through a network. It, is a fitting tool for finding the network response, determining (or, designing for) network stability, and network synthesis. The transfer, function of a network describes how the output behaves with respect, to the input. It specifies the transfer from the input to the output in the, s-domain, assuming no initial energy., The transfer function H (s) is the ratio of the output response Y (s) to, the input excitation X (s), assuming all initial conditions are zero., , Thus,, H(s) , , Some authors would not consider, Eqs. (16.16c) and (16.16d) transfer, functions., , Y(s), X(s), , (16.15), , The transfer function depends on what we define as input and output., Since the input and output can be either current or voltage at any place, in the circuit, there are four possible transfer functions:, H(s) Voltage gain , , Vo(s), Vi (s), , (16.16a), , H(s) Current gain , , Io(s), Ii (s), , (16.16b), , H(s) Impedance , , V(s), I(s), , (16.16c), , H(s) Admittance , , I(s), V(s), , (16.16d), , Thus, a circuit can have many transfer functions. Note that H(s) is, dimensionless in Eqs. (16.16a) and (16.16b)., Each of the transfer functions in Eq. (16.16) can be found in two, ways. One way is to assume any convenient input X(s), use any circuit analysis technique (such as current or voltage division, nodal or, mesh analysis) to find the output Y(s), and then obtain the ratio of, the two. The other approach is to apply the ladder method, which, involves walking our way through the circuit. By this approach, we, assume that the output is 1 V or 1 A as appropriate and use the basic, laws of Ohm and Kirchhoff (KCL only) to obtain the input. The, transfer function becomes unity divided by the input. This approach, may be more convenient to use when the circuit has many loops or, nodes so that applying nodal or mesh analysis becomes cumbersome., In the first method, we assume an input and find the output; in the, second method, we assume the output and find the input. In, both methods, we calculate H(s) as the ratio of output to input transforms. The two methods rely on the linearity property, since we only, deal with linear circuits in this book. Example 16.8 illustrates these, methods.
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 727, , 16.4, , Transfer Functions, , 727, , Equation (16.15) assumes that both X(s) and Y(s) are known., Sometimes, we know the input X(s) and the transfer function H(s). We, find the output Y(s) as, Y(s) H(s)X(s), , (16.17), , and take the inverse transform to get y(t). A special case is when the, input is the unit impulse function, x(t) d(t), so that X(s) 1. For this, case,, Y(s) H(s), , or, , y (t) h (t), , (16.18), , where, h (t) L1[H(s)], , (16.19), , The term h(t) represents the unit impulse response—it is the time-domain, response of the network to a unit impulse. Thus, Eq. (16.19) provides a, new interpretation for the transfer function: H(s) is the Laplace transform, of the unit impulse response of the network. Once we know the impulse, response h(t) of a network, we can obtain the response of the network, to any input signal using Eq. (16.17) in the s-domain or using the convolution integral (section 15.5) in the time domain., , The output of a linear system is y (t) 10et cos 4t u (t) when the input, is x (t) etu (t). Find the transfer function of the system and its, impulse response., , The unit impulse response is the output, response of a circuit when the input is, a unit impulse., , Example 16.7, , Solution:, If x (t) etu (t) and y (t) 10et cos 4t u (t), then, 10(s 1), (s 1)2 42, , X(s) , , 1, s1, , H(s) , , 10(s2 2s 1), 10(s 1)2, Y(s), 2, , 2, X(s), (s 1) 16, s 2s 17, , and, , Y(s) , , Hence,, , To find h(t), we write H(s) as, H(s) 10 40, , 4, (s 1)2 42, , From Table 15.2, we obtain, h(t) 10d(t) 40et sin 4t u (t), , The transfer function of a linear system is, H(s) , , 2s, s6, , Find the output y(t) due to the input 5e3tu (t) and its impulse response., Answer: 10e3t 20e6t, t 0, 2d(t) 12e6tu (t)., , Practice Problem 16.7
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 729, , 16.4, , Transfer Functions, , 729, , Example 16.9, , For the s-domain circuit in Fig. 16.19, find: (a) the transfer function, H(s) VoVi, (b) the impulse response, (c) the response when, vi (t) u (t) V, (d) the response when vi (t) 8 cos 2t V., Solution:, , 1Ω, , Vi +, −, , a, , 1Ω, , 1Ω, , (a) Using voltage division,, Vo , , 1, Vab, s1, , (16.9.1), , For Example 16.9., , But, Vab , , 1 (s 1), (s 1)(s 2), Vi , Vi, 1 (s 1)(s 2), 1 1 (s 1), , or, Vab , , s1, Vi, 2s 3, , (16.9.2), , Substituting Eq. (16.9.2) into Eq. (16.9.1) results in, Vi, 2s 3, , Vo , Thus, the transfer function is, H(s) , , Vo, 1, , Vi, 2s 3, , (b) We may write H(s) as, H(s) , , 1 1, 2 s 32, , Its inverse Laplace transform is the required impulse response:, 1, h (t) e3t2u (t), 2, (c) When vi (t) u (t), Vi (s) 1s, and, Vo(s) H(s)Vi (s) , , B, 1, A, 3 s , 2s(s 2), s 32, , where, A sVo (s) 0 s0 , , 1, 1, `, , 2(s 32) s0 3, , 3, 1, 1, B as b Vo(s) `, , `, , 2, 2s s32, 3, s32, Hence, for vi (t) u (t),, Vo(s) , , 1, 1 1, a , b, 3 s, s 32, , and its inverse Laplace transform is, 1, vo(t) (1 e3t2 ) u (t) V, 3, , b, , Figure 16.19, , s, +, Vo, −
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 730, , Chapter 16, , 730, , Applications of the Laplace Transform, , (d) When vi (t) 8 cos 2t, then Vi (s) , Vo(s) H(s)Vi (s) , , 8s, , and, s2 4, 4s, , (s , , 3, 2, 2 ) (s, , 4), , A, Bs C, , 3 2, s 4, s2, , (16.9.3), , where, 4s, 24, 3, 2, `, , A as b Vo(s) `, 2, 25, s, , 4, s32, s32, To get B and C, we multiply Eq. (16.9.3) by (s 32)(s 2 4). We get, 3, 3, 4s A(s2 4) B as2 sb C as b, 2, 2, Equating coefficients,, 3, 8, Constant: 0 4A C, 1, C A, 2, 3, 3, s:, 4 BC, 2, 2, s :, 0AB, 1, B A, Solving these gives A 2425, B 2425, C 6425. Hence, for, vi (t) 8 cos 2t V,, Vo(s) , , 24, 25, s 32, , , , 32 2, 24, s, , 25 s2 4, 25 s2 4, , and its inverse is, vo(t) , , Practice Problem 16.9, 1Ω, , Vi +, −, , 1Ω, , 2, s, , +, Vo, −, , Figure 16.20, , z1, z2, , Linear, system, , zm, Input signals, , Rework Example 16.9 for the circuit shown in Fig. 16.20., Answer: (a) 2(s 4), (b) 2e4tu (t), (c) 12 (1 e4t) u (t) V,, (d) 3.2(e4t cos 2t 12 sin 2t) u (t) V., , 16.5, , For Practice Prob. 16.9., y1, y2, , yp, Output signals, , Figure 16.21, A linear system with m inputs and p outputs., , 24, 4, ae3t2 cos 2t sin 2tbu (t) V, 25, 3, , State Variables, , Thus far in this book we have considered techniques for analyzing systems with only one input and only one output. Many engineering systems, have many inputs and many outputs, as shown in Fig. 16.21. The state, variable method is a very important tool in analyzing systems and understanding such highly complex systems. Thus, the state variable model is, more general than the single-input, single-output model, such as a transfer function. Although the topic cannot be adequately covered in one chapter, let alone one section of a chapter, we will cover it briefly at this point.
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 731, , 16.5, , State Variables, , In the state variable model, we specify a collection of variables, that describe the internal behavior of the system. These variables are, known as the state variables of the system. They are the variables that, determine the future behavior of a system when the present state of, the system and the input signals are known. In other words, they are, those variables which, if known, allow all other system parameters to, be determined by using only algebraic equations., A state variable is a physical property that characterizes the state of a, system, regardless of how the system got to that state., , Common examples of state variables are the pressure, volume, and, temperature. In an electric circuit, the state variables are the inductor, current and capacitor voltage since they collectively describe the, energy state of the system., The standard way to represent the state equations is to arrange, them as a set of first-order differential equations:, x Ax Bz, #, , (16.20), , where, x1(t), x2(t), #, x(t) D, T state vector representing n state vectors, o, xn(t), and the dot represents the first derivative with respect to time, i.e.,, #, x1(t), #, x2(t), #, T, x(t) D, o, #, xn(t), and, z1(t), z2(t), , T input vector representing m inputs, o, zm(t), , z(t) D, , A and B are respectively n n and n m matrices. In addition to the, state equation in Eq. (16.20), we need the output equation. The complete state model or state space is, #, x Ax Bz, y Cx Dz, , (16.21a), (16.21b), , where, y1(t), y2(t), T the output vector representing p outputs, y(t) D, o, yp(t), , 731
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ale29559_ch16.qxd, , 732, , 07/10/2008, , 04:11 PM, , Page 732, , Chapter 16, , Applications of the Laplace Transform, , C and D are, respectively, p n and p m matrices. For the special, case of single-input single-output, n m p 1., Assuming zero initial conditions, the transfer function of the system is found by taking the Laplace transform of Eq. (16.21a); we obtain, sX(s) AX(s) BZ(s), , S, , (sI A)X(s) BZ(s), , or, X(s) (sI A)1BZ(s), , (16.22), , where I is the identity matrix. Taking the Laplace transform of, Eq. (16.21b) yields, Y(s) CX(s) DZ(s), , (16.23), , Substituting Eq. (16.22) into Eq. (16.23) and dividing by Z(s) gives, the transfer function as, H(s) , , Y(s), C(sI A)1B D, Z(s), , (16.24), , where, A, B, C, D, , system matrix, input coupling matrix, output matrix, feedforward matrix, , In most cases, D 0, so the degree of the numerator of H(s) in, Eq. (16.24) is less than that of the denominator. Thus,, H(s) C(sI A)1B, , (16.25), , Because of the matrix computation involved, MATLAB can be used to, find the transfer function., To apply state variable analysis to a circuit, we follow the following three steps., , Steps to Apply the State Variable Method to, Circuit Analysis:, 1. Select the inductor current i and capacitor voltage v as the state, variables, making sure they are consistent with the passive sign, convention., 2. Apply KCL and KVL to the circuit and obtain circuit variables, (voltages and currents) in terms of the state variables. This, should lead to a set of first-order differential equations necessary and sufficient to determine all state variables., 3. Obtain the output equation and put the final result in state-space, representation., , Steps 1 and 3 are usually straightforward; the major task is in, step 2. We will illustrate this with examples.
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 733, , 16.5, , State Variables, , 733, , Example 16.10, , Find the state-space representation of the circuit in Fig. 16.22. Determine the transfer function of the circuit when vs is the input and ix is, the output. Take R 1, C 0.25 F, and L 0.5 H., Solution:, We select the inductor current i and capacitor voltage v as the state, variables., di, vL L, (16.10.1), dt, iC C, , C, , S, , (16.10.2), , dv, v, i, dt, R, , or, v, i, #, v, , RC, C, , (16.10.3), , since the same voltage v is across both R and C. Applying KVL around, the outer loop yields, di, v vs, dt, #, vs, v, i , L, L, , vs vL v, , L, , S, , (16.10.4), , Equations (16.10.3) and (16.10.4) constitute the state equations. If we, regard ix as the output,, v, ix , (16.10.5), R, Putting Eqs. (16.10.3), (16.10.4), and (16.10.5) in the standard form, leads to, #, 1, 1, v, v, 0, RC, C, (16.10.6a), #, c d c1, d c i d c 1 d vs, i, 0, L, L, ix c, , 1, R, , v, 0d c d, i, , (16.10.6b), , If R 1, C 14, and L 12, we obtain from Eq. (16.10.6) matrices, A, , 1, RC, , 1, C, , L, , 0, , c 1, , d, , c, , 4 4, d,, 2 0, , L, + vL −, , vs, , +, −, , 0, 0, B c 1 d c d,, 2, L, , C c, , 1, 0 d [1 0], R, s 0, 4 4, s 4 4, sI A c, d c, d c, d, 0 s, 2 0, 2, s, Taking the inverse of this gives, 4, s, d, c, adjoint, of, A, 2, s, , 4, 1, 2, (sI A) , determinant of A, s 4s 8, , ic, , 1, , R, , Figure 16.22, For Example 16.10., , dv, dt, , Applying KCL at node 1 gives, i ix iC, , i, , ix, C, , +, v, −
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 734, , Chapter 16, , 734, , Applications of the Laplace Transform, , Thus, the transfer function is given by, s, 4, 0, 8, [1 0] c, [1 0] c, d c d, d, 2, s, , 4, 2, 2s, , 8, H(s) C(sI A)1B , , s2 4s 8, s2 4s 8, 8, 2, s 4s 8, which is the same thing we would get by directly Laplace transforming, the circuit and obtaining H(s) Ix(s)Vs(s). The real advantage of the, state variable approach comes with multiple inputs and multiple outputs., In this case, we have one input vs and one output ix. In the next, example, we will have two inputs and two outputs., , Practice Problem 16.10, L, , R1, , i, vs, , +, −, , Obtain the state variable model for the circuit shown in Fig. 16.23. Let, R1 1, R2 2, C 0.5, and L 0.2 and obtain the transfer function., , +, v, −, , C, , Figure 16.23, , R2, , +, vo, −, , Answer:, #, v, c#d , i, H(s) , , For Practice Prob. 16.10., , Example 16.11, , 1, R1C, , 1, C, , L, , R2, L, , c1, , d, , 1, v, c d c R1C d vs,, i, 0, , v, vo [0 R2 ] c d, i, , 20, s2 12s 30, , Consider the circuit in Fig. 16.24, which may be regarded as a twoinput, two-output system. Determine the state variable model and find, the transfer function of the system., i1, , 1Ω, , 1, , 3Ω, , 2, , io, , + v −, o, , i, vs, , 2Ω, , 1H, 6, , +, −, , +, v, −, , 1, 3F, , +, −, , vi, , Figure 16.24, For Example 16.11., , Solution:, In this case, we have two inputs vs and vi and two outputs vo and io., Again, we select the inductor current i and capacitor voltage v as the, state variables. Applying KVL around the left-hand loop gives, 1#, vs i1 i 0, 6, , S, , #, i 6vs 6i1, , (16.11.1), , We need to eliminate i1. Applying KVL around the loop containing, vs, 1- resistor, 2- resistor, and 13-F capacitor yields, vs i1 vo v, , (16.11.2), , But at node 1, KCL gives, i1 i , , vo, 2, , S, , vo 2(i1 i), , (16.11.3)
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 735, , 16.5, , State Variables, , 735, , Substituting this in Eq. (16.11.2),, vs 3i1 v 2i, , i1 , , S, , 2i v vs, 3, , Substituting this in Eq. (16.11.1) gives, #, i 2v 4i 4vs, , (16.11.4), , (16.11.5), , which is one state equation. To obtain the second one, we apply KCL, at node 2., vo, 1#, 3, #, v io S v vo 3io, (16.11.6), 2, 3, 2, We need to eliminate vo and io. From the right-hand loop, it is evident that, io , , v vi, 3, , (16.11.7), , Substituting Eq. (16.11.4) into Eq. (16.11.3) gives, vo 2 a, , 2i v vs, 2, ib (v i vs), 3, 3, , (16.11.8), , Substituting Eqs. (16.11.7) and (16.11.8) into Eq. (16.11.6) yields the, second state equation as, #, (16.11.9), v 2v i vs vi, The two output equations are already obtained in Eqs. (16.11.7) and, (16.11.8). Putting Eqs. (16.11.5) and (16.11.7) to (16.11.9) together in, the standard form leads to the state model for the circuit, namely,, #, v, 2 1 v, 1 1 vs, (16.11.10a), c#d c, d c d c, d c d, i, 2 4 i, 4 0 vi, c, , 2, 23 23 v, 0 vs, vo, 3, d c 1, c d c, d, dc d, io, 0 i, 0 13 vi, 3, , (16.11.10b), , For the electric circuit in Fig. 16.25, determine the state model. Take, vo and io as the output variables., Answer:, , #, v, 2 2 v, 2, 0 i1, c#d c, d c d c, d c d, i, 4 8, i, 0 8 i2, c, , 1, vo, d c, io, 0, , 0 v, 0, d c d c, 1, i, 0, 1, 4, , vo, , 0 i1, d c d, 1 i2, , H, io, , i1, , 1Ω, , Figure 16.25, For Practice Prob. 16.11., , 1, 2, , F, , 2Ω, , i2, , Practice Problem 16.11
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ale29559_ch16.qxd, , 07/10/2008, , 736, , Example 16.12, , 04:11 PM, , Page 736, , Chapter 16, , Applications of the Laplace Transform, , Assume we have a system where the output is y(t) and the input is z(t)., Let the following differential equation describe the relationship between, the input and the output., d 2y(t), dt, , 2, , 3, , dy(t), 2y(t) 5z(t), dt, , (16.12.1), , Obtain the state model and the transfer function of the system., Solution:, First, we select the state variables. Let x1 y(t), therefore, #, #, x1 y(t), , (16.12.2), , Now let, #, #, x2 x1 y(t), , (16.12.3), , Note that at this time we are looking at a second-order system that, would normally have two first-order terms in the solution., #, #, $, Now we have x2 y(t), where we can find the value x2 from, Eq. (16.12.1), i.e.,, #, $, #, x2 y(t) 2y(t) 3y(t) 5z(t) 2x1 3x2 5z(t) (16.12.4), From Eqs. (16.12.2) to (16.12.4), we can now write the following, matrix equations:, #, x1, 0, 1 x1, 0, (16.12.5), c# d c, d c d c d z(t), x2, 2 3 x2, 5, y(t) [1 0] c, , x1, d, x2, , (16.12.6), , We now obtain the transfer function., sI A s c, , 1 0, 0, 1, s, d c, d c, 0 1, 2 3, 2, , 1, d, s3, , The inverse is, s3 1, d, 2 s, 1, (sI A) , s (s 3) 2, c, , The transfer function is, s3 1 0, 5, da b, (1 0) a b, 2 s 5, 5s, H(s) C(sI A)1B , , s (s 3) 2, s (s 3) 2, 5, , (s 1) (s 2), (1 0) c, , To check this, we directly apply the Laplace transfer to each term in, Eq. (16.12.1). Since initial conditions are zero, we get, [s2 3s 2]Y(s) 5Z(s), , S, , H(s) , , Y(s), 5, 2, Z(s), s 3s 2, , which is in agreement with what we got previously.
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 737, , 16.6, , Applications, , Develop a set of state variable equations that represent the following, differential equation., d 3y, dt 3, , 18, , d 2y, dt 2, , 20, , 737, , Practice Problem 16.12, , dy, 5y z(t), dt, , Answer:, 0, 1, 0, A £ 0, 0, 1 §,, 5 20 18, , 16.6, , 0, B £0§,, 1, , C [1 0, , 0]., , Applications, , So far we have considered three applications of Laplace’s transform:, circuit analysis in general, obtaining transfer functions, and solving linear integrodifferential equations. The Laplace transform also finds, application in other areas in circuit analysis, signal processing, and control systems. Here we will consider two more important applications:, network stability and network synthesis., , 16.6.1 Network Stability, A circuit is stable if its impulse response h (t) is bounded (i.e., h(t) converges to a finite value) as t S ; it is unstable if h (t) grows without, bound as t S . In mathematical terms, a circuit is stable when, lim 0 h (t) 0 finite, , tS , , (16.26), , Since the transfer function H(s) is the Laplace transform of the impulse, response h (t), H(s) must meet certain requirements in order for, Eq. (16.26) to hold. Recall that H(s) may be written as, H(s) , , N(s), D(s), , N(s), N(s), , D(s), (s p1) (s p2) p (s pn), , Zero, Pole, , O, , X, , X, , , O, , (a), , (16.28), , H(s) must meet two requirements for the circuit to be stable. First, the, degree of N(s) must be less than the degree of D(s); otherwise, long, division would produce, R(s), H(s) k n s n kn1s n1 p k1s k0 , D(s), , X, , j, , where the roots of N(s) 0 are called the zeros of H(s) because they, make H(s) 0, while the roots of D(s) 0 are called the poles of H(s), since they cause H(s) S . The zeros and poles of H(s) are often located, in the s plane as shown in Fig. 16.26(a). Recall from Eqs. (15.47) and, (15.48) that H(s) may also be written in terms of its poles as, H(s) , , O, , (16.27), , j, X, , 0, , , , X, , (16.29), (b), , where the degree of R(s), the remainder of the long division, is less, than the degree of D(s). The inverse of H(s) in Eq. (16.29) does not, meet the condition in Eq. (16.26). Second, all the poles of H(s) in, , Figure 16.26, The complex s plane: (a) poles and zeros, plotted, (b) left-half plane.
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 738, , Chapter 16, , 738, , Applications of the Laplace Transform, , Eq. (16.27) (i.e., all the roots of D(s) 0) must have negative real, parts; in other words, all the poles must lie in the left half of the s plane,, as shown typically in Fig. 16.26(b). The reason for this will be apparent if we take the inverse Laplace transform of H(s) in Eq. (16.27)., Since Eq. (16.27) is similar to Eq. (15.48), its partial fraction expansion is similar to the one in Eq. (15.49) so that the inverse of H(s) is, similar to that in Eq. (15.53). Hence,, h(t) (k1ep1t k2ep2t p knepnt ) u (t), , (16.30), , We see from this equation that each pole pi must be positive (i.e., pole, s pi in the left-half plane) in order for epi t to decrease with, increasing t. Thus,, A circuit is stable when all the poles of its transfer function H (s) lie in, the left half of the s plane., , R, Vs +, −, , An unstable circuit never reaches steady state because the transient, response does not decay to zero. Consequently, steady-state analysis is, only applicable to stable circuits., A circuit made up exclusively of passive elements (R, L, and C), and independent sources cannot be unstable, because that would imply, that some branch currents or voltages would grow indefinitely with, sources set to zero. Passive elements cannot generate such indefinite, growth. Passive circuits either are stable or have poles with zero real, parts. To show that this is the case, consider the series RLC circuit in, Fig. 16.27. The transfer function is given by, , sL, 1, sC, , +, Vo, −, , H(s) , , Vo, 1sC, , Vs, R sL 1sC, , or, H(s) , , Figure 16.27, A typical RLC circuit., , 1L, s sRL 1LC, 2, , (16.31), , Notice that D(s) s 2 sRL 1LC 0 is the same as the characteristic equation obtained for the series RLC circuit in Eq. (8.8). The, circuit has poles at, p1,2 a 2a2 20, , (16.32), , where, a, , R, ,, 2L, , 0 , , 1, LC, , For R, L, C 7 0, the two poles always lie in the left half of the s plane,, implying that the circuit is always stable. However, when R 0, a 0, and the circuit becomes unstable. Although ideally this is possible, it, does not really happen, because R is never zero., On the other hand, active circuits or passive circuits with controlled sources can supply energy, and they can be unstable. In fact,, an oscillator is a typical example of a circuit designed to be unstable., An oscillator is designed such that its transfer function is of the form, H(s) , , N(s), N(s), , 2, (s j0) (s j0), s 0, 2, , so that its output is sinusoidal., , (16.33)
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 739, , 16.6, , Applications, , 739, , Example 16.13, , Determine the values of k for which the circuit in Fig. 16.28 is stable., R, , Solution:, Applying mesh analysis to the first-order circuit in Fig. 16.28 gives, I2, 1, Vi aR , b I1 , sC, sC, , (16.13.1), , Vi +, −, , R, 1, sC, , I1, , –, +, , I2, , Figure 16.28, , and, , For Example 16.13., , 0 k I1 aR , , I1, 1, b I2 , sC, sC, , or, 0 ak , , 1, 1, b I1 aR , b I2, sC, sC, , (16.13.2), , We can write Eqs. (16.13.1) and (16.13.2) in matrix form as, aR , , 1, b, Vi, sC, c d ≥, 0, 1, ak , b, sC, , 1, sC, I1, ¥ c d, 1, I2, aR , b, sC, , , The determinant is, ¢ aR , , 1 2, k, 1, sR 2C 2R k, b , 2 2, sC, sC, sC, sC, , (16.13.3), , The characteristic equation ( ¢ 0) gives the single pole as, p, , k 2R, R 2C, , which is negative when k 6 2R. Thus, we conclude the circuit is stable, when k 6 2R and unstable for k 7 2R., , For what value of b is the circuit in Fig. 16.29 stable?, , Practice Problem 16.13, Vo, , Answer: b 7 1R., R, , C, , C, , R, , +, Vo, −, , Figure 16.29, For Practice Prob. 16.13., , An active filter has the transfer function, H(s) , , k, s2 s (4 k) 1, , For what values of k is the filter stable?, , Example 16.14, , kI1
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ale29559_ch16.qxd, , 07/22/2008, , 01:15 PM, , Page 740, , 740, , Chapter 16, , Applications of the Laplace Transform, , Solution:, As a second-order circuit, H(s) may be written as, H(s) , , N(s), s bs c, 2, , where b 4 k, c 1, and N(s) k. This has poles at p2 bp , c 0; that is,, p1,2 , , b 2b2 4c, 2, , For the circuit to be stable, the poles must be located in the left half, of the s plane. This implies that b 7 0., Applying this to the given H(s) means that for the circuit to be, stable, 4 k 7 0 or k 6 4., , Practice Problem 16.14, , A second-order active circuit has the transfer function, H(s) , , 1, s s (25 a) 25, 2, , Find the range of the values of a for which the circuit is stable. What, is the value of a that will cause oscillation?, Answer: a 7 25, a 25., , 16.6.2 Network Synthesis, Network synthesis may be regarded as the process of obtaining an, appropriate network to represent a given transfer function. Network, synthesis is easier in the s-domain than in the time domain., In network analysis, we find the transfer function of a given network. In network synthesis, we reverse the approach: given a transfer, function, we are required to find a suitable network., Network synthesis is finding a network that represents a given transfer, function., , Keep in mind that in synthesis, there may be many different, answers—or possibly no answers—because there are many circuits that, can be used to represent the same transfer function; in network analysis, there is only one answer., Network synthesis is an exciting field of prime engineering importance. Being able to look at a transfer function and come up with the, type of circuit it represents is a great asset to a circuit designer., Although network synthesis constitutes a whole course by itself and, requires some experience, the following examples are meant to whet, your appetite.
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 741, , 16.6, , Applications, , 741, , Example 16.15, , Given the transfer function, H(s) , , Vo(s), 10, 2, Vi (s), s 3s 10, , realize the function using the circuit in Fig. 16.30(a). (a) Select, R 5 , and find L and C. (b) Select R 1 , and find L and C., , L, , v i (t) +, −, , C, , R, , +, vo(t), −, , R, , +, V o(s), −, , Solution:, 1. Define. The problem is clearly and completely defined. This, problem is what we call a synthesis problem: given a transfer, function, synthesize a circuit that produces the given transfer, function. However, to keep the problem more manageable, we, give a circuit that produces the desired transfer function., Had one of the variables, R in this case, not been given a value,, then the problem would have had an infinite number of answers., An open-ended problem of this kind would require some additional, assumptions that would have narrowed the set of solutions., 2. Present. A transfer function of the voltage out versus the voltage, in is equal to 10(s2 3s 10). A circuit, Fig. 16.30, is also, given that should be able to produce the required transfer function., Two different values of R, 5 and 1 , are to be used to calculate, the values of L and C that produce the given transfer function., 3. Alternative. All solution paths involve determining the transfer, function of Fig. 16.30 and then matching the various terms of, the transfer function. Two approaches would be to use mesh, analysis or nodal analysis. Since we are looking for a ratio of, voltages, nodal analysis makes the most sense., 4. Attempt. Using nodal analysis leads to, Vo(s) Vi (s), Vo(s) 0, Vo(s) 0, , , 0, sL, 1(sC), R, Now multiply through by sLR:, RVo(s) RVi (s) s2RLCVo(s) sLVo(s) 0, Collecting terms we get, (s2RLC sL R)Vo(s) RVi (s), or, Vo(s), 1(LC), 2, Vi (s), s [1(RC)]s 1(LC), Matching the two transfer functions produces two equations with, three unknowns., 0.1, LC 0.1, or, L, C, and, RC , , 1, 3, , or, , C, , 1, 3R, , We have a constraint equation, R 5 for (a) and 1 for (b)., (a) C 1(3 5) 66.67 mF and L 1.5 H, (b) C 1(3 1) 333.3 mF and L 300 mH, , (a), sL, , Vi (s) +, −, , i1, , 1, sC, , (b), , Figure 16.30, For Example 16.15., , i2
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 743, , 16.6, , Applications, , 743, , but Vo(s) 1 I2 10(s3 3s2 10s), and the answer checks., 6. Satisfactory? We have clearly identified values of L and C for, each of the conditions. In addition, we have carefully checked, the answers to see if they are correct. The problem has been, adequately solved. The results can now be presented as a, solution to the problem., , Practice Problem 16.15, , Realize the function, G(s) , , Vo(s), 4s, 2, Vi (s), s 4s 20, , C, , using the circuit in Fig. 16.31. Select R 2 , and determine L and C., Answer: 0.5 H, 0.1 F., , L, , v i (t) +, −, , R, , +, v o(t), −, , Figure 16.31, For Practice Prob. 16.15., , Example 16.16, , Synthesize the function, T(s) , , Vo(s), 106, 2, Vs(s), s 100s 106, , using the topology in Fig. 16.32., , Vo, , Y2, Y1, , 1, , Y3, , 2 V2, , −, +, , Vo, , V1, Vs, , +, −, , Y4, , Figure 16.32, For Example 16.16., , Solution:, We apply nodal analysis to nodes 1 and 2. At node 1,, (Vs V1)Y1 (V1 Vo)Y2 (V1 V2)Y3, , (16.16.1), , At node 2,, (V1 V2)Y3 (V2 0)Y4, , (16.16.2)
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 745, , 16.7, , Summary, , 745, , C1 = 2 F, −, +, , R2 = 10 kΩ, , R1 = 10 kΩ, , Vs +, −, , Vo, , C2 = 5 nF, , Figure 16.33, For Example 16.16., , Practice Problem 16.16, , Synthesize the function, Vo(s), 2s, 2, Vin, s 6s 10, using the op amp circuit shown in Fig. 16.34. Select, Y1 , , 1, ,, R1, , Y2 sC1,, , Y3 sC2,, , Y4 , , 1, R2, , Let R1 1 k, and determine C1, C2, and R2., Y3, Y4, Y1, , Y2, , −, +, , Vo, , Vin +, −, , Figure 16.34, For Practice Prob. 16.16., , Answer: 0.1 mF, 0.5 mF, 2 k., , 16.7, , Summary, , 1. The Laplace transform can be used to analyze a circuit. We convert each element from the time domain to the s-domain, solve the, problem using any circuit analysis technique, and convert the result, to the time domain using the inverse transform., 2. In the s-domain, the circuit elements are replaced with the initial, condition at t 0 as follows. (Please note, voltage models are
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 746, , Chapter 16, , 746, , Applications of the Laplace Transform, , given below, but the corresponding current models work equally, well.):, Resistor: vR Ri S VR R I, di, Inductor: vL L, S VL sLI Li(0 ), dt, v(0), 1, , Capactior: vC i dt S VC , s, sC, , , , 3. Using the Laplace transform to analyze a circuit results in a complete (both transient and steady state) response because the initial, conditions are incorporated in the transformation process., 4. The transfer function H(s) of a network is the Laplace transform, of the impulse response h(t)., 5. In the s-domain, the transfer function H(s) relates the output, response Y(s) and an input excitation X(s); that is, H(s) Y(s)X(s)., 6. The state variable model is a useful tool for analyzing complex, systems with several inputs and outputs. State variable analysis is, a powerful technique that is most popularly used in circuit theory, and control. The state of a system is the smallest set of quantities, (known as state variables) that we must know in order to determine its future response to any given input. The state equation in, state variable form is, #, x Ax Bz, while the output equation is, y Cx Dz, 7. For an electric circuit, we first select capacitor voltages and inductor current as state variables. We then apply KCL and KVL to, obtain the state equations., 8. Two other areas of applications of the Laplace transform covered, in this chapter are circuit stability and synthesis. A circuit is stable when all the poles of its transfer function lie in the left half of, the s plane. Network synthesis is the process of obtaining an appropriate network to represent a given transfer function for which, analysis in the s-domain is well suited., , Review Questions, 16.1, , The voltage through a resistor with current i(t) in the, s-domain is sRI(s)., (a) True, , 16.2, , (b) False, , 16.3, , (a) 10s, 16.4, , The current through an RL series circuit with input, voltage v(t) is given in the s-domain as:, 1, (a) V(s) c R d, sL, (c), , V(s), R 1sL, , (b) V(s)(R sL), (d), , V(s), R sL, , The impedance of a 10-F capacitor is:, (c) 110s, , (d) 10s, , We can usually obtain the Thevenin equivalent in the, time domain., (a) True, , 16.5, , (b) s10, , (b) False, , A transfer function is defined only when all initial, conditions are zero., (a) True, , (b) False
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ale29559_ch16.qxd, , 07/10/2008, , 04:11 PM, , Page 747, , Problems, , 16.6, , If the input to a linear system is d(t) and the output is, e2tu (t), the transfer function of the system is:, (a), , 16.7, , 1, s2, , (b), , 1, s2, , (c), , s, s2, , (d), , s, s2, , Which of the following equations is called the state, equation?, #, (a) x A x B z, (b) y C x D z, , (e) None of the above, , (c) H(s) Y(s)Z(s), , If the transfer function of a system is, , (d) H(s) C(sI A)1B, , H(s) , , s2 s 2, s 4s2 5s 1, 3, , it follows that the input is X(s) s3 4s2 5s 1,, while the output is Y(s) s2 s 2., (a) True, 16.8, , 16.9, , 747, , 16.10 A single-input, single-output system is described by, the state model as:, #, x1 2x1 x2 3z, #, x2 4x2 z, y 3x1 2x2 z, , (b) False, , Which of the following matrices is incorrect?, , A network has its transfer function as, , (a) A c, , s1, H(s) , (s 2) (s 3), , (c) C [3, , The network is stable., (a) True, , 2, 0, , 1, d, 4, , (b) B c, , 2], , (d) D 0, , 3, d, 1, , Answers: 16.1b, 16.2d, 16.3c, 16.4b, 16.5b, 16.6a, 16.7b,, 16.8b, 16.9a, 16.10d., , (b) False, , Problems, Sections 16.2 and 16.3 Circuit Element Models, and Circuit Analysis, 16.1, , 16.3, , Find i(t) for t 7 0 for the circuit in Fig. 16.37., Assume is 24u (t) 12d(t) mA. (Hint: Can we, use superposition to help solve this problem?), , Determine i(t) in the circuit of Fig. 16.35 by means, of the Laplace transform., 1Ω, 1Ω, , u(t), , i (t), , +, −, , i, 2Ω, , is, , 1F, , 0.2 H, , Figure 16.37, , 1H, , For Prob. 16.3., , Figure 16.35, For Prob. 16.1., 16.2, , Using Fig. 16.36, design a problem to help other, students better understand circuit analysis using, Laplace transforms., L, , vs +, −, , 16.4, , The capacitor in the circuit of Fig. 16.38 is initially, uncharged. Find vo(t) for t 7 0., , C, +, vx, −, , R1, , i, , R2, , 15 (t) V, , Figure 16.36, , Figure 16.38, , For Prob. 16.2., , For Prob. 16.4., , +, −, , 4i, , 2Ω, +, vo, −, , 1F, , 1Ω
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ale29559_ch16.qxd, , 07/10/2008, , 04:12 PM, , Page 748, , Chapter 16, , 748, , 16.5, , Applications of the Laplace Transform, , If is(t) e2tu (t) A in the circuit shown in Fig. 16.39,, find the value of io(t)., , 16.10 Using Fig. 16.44, design a problem to help other, students understand how to use Thevenin’s theorem, (in the s-domain) to aid in circuit analysis., , i o(t), i s(t), , 2Ω, , 1H, , 0.5 F, L, , R1, , Figure 16.39, , v s(t) +, −, , For Prob. 16.5., 16.6, , Find v(t), t 7 0 in the circuit of Fig. 16.40. Let, vs 20 V., , +, −, , +, v (t), −, , 100 mF, , +, vo, −, , R2, , Figure 16.44, For Prob. 16.10., , t=0, , vs, , C, , 10 Ω, , 16.11 Solve for the mesh currents in the circuit of Fig. 16.45., You may leave your results in the s-domain., , Figure 16.40, For Prob. 16.6., 16.7, , 1Ω, , 2u(t) V, , 1Ω, , Find vo(t), for all t 7 0, in the circuit of Fig. 16.41., 1Ω, , +, −, , 1H, , 10u(t) V, +, vo, −, , 0.5 F, , u(t) A, , +, −, , 4Ω, 1, 4, , I1, , H, , I2, , 1H, , Figure 16.45, For Prob. 16.11., , Figure 16.41, For Prob. 16.7., 16.8, , 16.12 Find vo(t) in the circuit of Fig. 16.46., , If vo(0) 1 V, obtain vo(t) in the circuit of, Fig. 16.42., 1Ω, , 1Ω, +, vo, −, , +, 3u(t) −, , 1H, 10e −t u(t), , + 4u(t), −, , 0.5 F, , V +, −, , 2F, , +, v o (t), −, , 4Ω, , 3u(t) A, , Figure 16.46, For Prob. 16.12., , Figure 16.42, For Prob. 16.8., 16.9, , Find the input impedance Zin(s) of each of the, circuits in Fig. 16.43., , 16.13 Determine io(t) in the circuit of Fig. 16.47., , 1Ω, , 1H, 2Ω, , 1H, , 2Ω, , 1F, , 0.5 F, , io, , 1F, 1Ω, (a), , 2H, , 2Ω, , (b), , Figure 16.43, , Figure 16.47, , For Prob. 16.9., , For Prob. 16.13., , 5e −2tu(t) A, , 1Ω
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ale29559_ch16.qxd, , 07/10/2008, , 04:12 PM, , Page 749, , Problems, v s (t), 3V, , *16.14 Determine io(t) in the network shown in Fig. 16.48., 1Ω, , 749, , 4Ω, , io, +, −, , (15 + 30u(t)) V, , 1s, , 0, 2H, , t, (a), , 1, F, 4, 1Ω, , Figure 16.48, For Prob. 16.14., , v s (t), , +, −, , +, v o(t), −, , 1F, , 2Ω, , 16.15 Find Vx(s) in the circuit shown in Fig. 16.49., (b), 10 Ω, , 0.25 H, +, 3Vx, , +, −, , Vx, , Figure 16.52, For Prob. 16.18., , −, + 5e –2t u(t) V, −, , 0.2 F, , 16.19 Using Fig. 16.53, design a problem to help other, students better understand circuit analysis in the, s-domain with circuits that have dependent sources., , Figure 16.49, For Prob. 16.15., , ki, , R, , −+, i, , *16.16 Find io(t) for t 7 0 in the circuit of Fig. 16.50., vs +, −, , + vo −, , 2Ω, , 1Ω, , Figure 16.53, , 1F, 15e −2tu(t) V, , +, −, , 0.5v o, , +, −, , +, −, 1H, , +, vo, −, , C, , L, , For Prob. 16.19., 9u(−t) V, , 16.20 Find vo(t) in the circuit of Fig. 16.54 if vx(0) 2 V, and i(0) 1 A., , io, , Figure 16.50, , + vx −, , For Prob. 16.16., , i, , 1F, , 16.17 Calculate io(t) for t 7 0 in the network of Fig. 16.51., e −tu(t) A, , 1Ω, , 1Ω, , 1H, , 2e −tu(t) V, +−, 1F, , 1Ω, , Figure 16.54, , io 1 H, , 4u(t) A, , For Prob. 16.20., 1Ω, , 16.21 Find the voltage vo(t) in the circuit of Fig. 16.55 by, means of the Laplace transform., , Figure 16.51, , 1Ω, , For Prob. 16.17., 16.18 (a) Find the Laplace transform of the voltage shown, in Fig. 16.52(a). (b) Using that value of vs(t) in the, circuit shown in Fig. 16.52(b), find the value of vo(t)., , 10u(t) A, , Figure 16.55, * An asterisk indicates a challenging problem., , +, vo, −, , For Prob. 16.21., , 0.5 F, , 1H, , 2Ω, , 1F, , +, v, −o
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ale29559_ch16.qxd, , 07/14/2008, , 01:04 PM, , Page 750, , Chapter 16, , 750, , Applications of the Laplace Transform, , 16.22 Using Fig. 16.56, design a problem to help other, students better understand solving for node voltages, by working in the s-domain., L, , v1, , is, , 10 kΩ, 50 F, , v2, , 20 kΩ, , R2, , R1, , vs, , C, , −, +, , +, −, , vo, , Figure 16.60, For Prob. 16.26., , Figure 16.56, For Prob. 16.22., , 16.27 Find I1(s) and I2(s) in the circuit of Fig. 16.61., 1H, , 16.23 Consider the parallel RLC circuit of Fig. 16.57. Find, v(t) and i(t) given that v(0) 5 and i(0) 2 A., , i, 10 Ω, , 4u(t) A, , 1, 80, , 4H, , +, v, −, , F, , i1, , 2H, , 10e −3tu(t) V +, −, , i2, , 2H, , 1Ω, , 1Ω, , Figure 16.61, For Prob. 16.27., , Figure 16.57, For Prob. 16.23., , 16.24 The switch in Fig. 16.58 moves from position 1 to, position 2 at t 0. Find v(t), for all t 7 0., , 16.28 Using Fig. 16.62, design a problem to help other, students better understand how to do circuit analysis, with circuits that have mutually coupled elements by, working in the s-domain., M, R1, , 1, , t=0, v s(t) +, −, , 12 V, , +, v, −, , 2, , +, −, , 0.25 H, , 10 mF, , L1, , L2, , R2, , +, vo, −, , Figure 16.62, For Prob. 16.28., , Figure 16.58, , 16.29 For the ideal transformer circuit in Fig. 16.63,, determine io(t)., , For Prob. 16.24., , 1 Ω io, , 16.25 For the RLC circuit shown in Fig. 16.59, find the, complete response if v(0) 2 V when the switch is, closed., , 1:2, 10e, , t=0, , 2 cos 4t V, , +, −, , 6Ω, , 1H, 1, 9, , −tu(t), , V +, −, , 0.25 F, , 8Ω, , Figure 16.63, F, , +, v, −, , Figure 16.59, For Prob. 16.25., 16.26 For the op amp circuit in Fig. 16.60, find vo(t) for, t 7 0. Take vs 10e5tu (t) V., , For Prob. 16.29., , Section 16.4 Transfer Functions, 16.30 The transfer function of a system is, H(s) , , s2, 3s 1, , Find the output when the system has an input of, 4et3u (t).
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ale29559_ch16.qxd, , 07/10/2008, , 04:12 PM, , Page 751, , Problems, , 751, , 16.31 When the input to a system is a unit step function,, the response is 10 cos 2t u (t). Obtain the transfer, function of the system., 16.32 Design a problem to help other students better, understand how to find outputs when given a transfer, function and an input., 16.33 When a unit step is applied to a system at t 0, its, response is, 1, y(t) c 4 e3t e2t(2 cos 4t 3 sin 4t) d u (t), 2, What is the transfer function of the system?, 16.34 For the circuit in Fig. 16.64, find H(s) Vo(s)Vs(s)., Assume zero initial conditions., , i1, , kv s, , i2, , +, vx, −, , vs +, −, , 2H, +, −, , 0.5 F, , 4v x, , Figure 16.66, For Prob. 16.37., , 16.38 Refer to the network in Fig. 16.67. Find the following, transfer functions:, (a) H1(s) Vo(s)Vs(s), (b) H2(s) Vo(s)Is(s), (c) H3(s) Io(s)Is(s), (d) H4(s) Io(s)Vs(s), , is, 2Ω, , 3Ω, , 1Ω, , io, , 1Η, , 1H, , +, −, , 4Ω, , +, vo, −, , 0.1 F, , vs, , +, −, , 1F, , 1Ω, , 1F, , +, vo, −, , Figure 16.67, For Prob. 16.38., , Figure 16.64, For Prob. 16.34., , 16.39 Calculate the gain H(s) VoVs in the op amp circuit, of Fig. 16.68., 16.35 Obtain the transfer function H(s) VoVs for the, circuit of Fig. 16.65., , +, −, +, vs, , i, , vs, , 0.5 F, , +, −, , vo, C, , 1H, , 2i, , R, , +, −, , 3Ω, , +, vo, −, , Figure 16.65, , −, , Figure 16.68, For Prob. 16.39., , 16.40 Refer to the RL circuit in Fig. 16.69. Find:, , For Prob. 16.35., , (a) the impulse response h (t) of the circuit., (b) the unit step response of the circuit., 16.36 The transfer function of a certain circuit is, H(s) , , L, , 5, 3, 6, , , s1, s2, s4, vs +, −, , Find the impulse response of the circuit., 16.37 For the circuit in Fig. 16.66, find:, (a) I1Vs, , (b) I2Vx, , Figure 16.69, For Prob. 16.40., , R, , +, vo, −
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ale29559_ch16.qxd, , 07/10/2008, , 04:12 PM, , Page 752, , Chapter 16, , 752, , Applications of the Laplace Transform, , 16.41 A parallel RL circuit has R 4 and L 1 H. The, input to the circuit is is(t) 2etu (t) A. Find the, inductor current iL(t) for all t 7 0 and assume that, iL(0) 2 A., , 16.48 Develop the state equations for the following, differential equation., d 2 y(t), dt, , 16.42 A circuit has a transfer function, H(s) , , d 2 y(t), dt 2, , 16.43 Develop the state equations for Prob. 16.1., 16.44 Develop the state equations for the problem you, designed in Prob. 16.2., , d 3y(t), dt, , 3, , , , 7 d y(t), dz(t), 9y(t) , z(t), dt, dt, , dt, , 2, , , , 11 d y(t), 6y(t) z(t), dt, , 4, #, x c, 2, , 4, 0, d x c d u(t), 0, 2, , *16.52 Given the following state equation, solve for y1(t), and y2(t)., + v (t), − 2, , 2Ω, , 6 d 2 y(t), , y(t) [1 0] x, , 1H, , +, v o(t), −, , +, −, , , , *16.51 Given the following state equation, solve for y(t):, , 16.45 Develop the state equations for the circuit shown in, Fig. 16.70., , v 1(t), , 6 d y(t), 7y(t) z(t), dt, , *16.50 Develop the state equations for the following, differential equation., , Section 16.5 State Variables, , F, , , , *16.49 Develop the state equations for the following, differential equation., , s4, (s 1)(s 2)2, , Find the impulse response., , 1, 4, , 2, , 2, #, x c, 2, , 1, 1, d x c, 4, 4, , 1, u (t), d c, d, 0 2u (t), , 2, 1, , 2, 2, d x c, 0, 0, , u (t), 0, d, d c, 1 2u (t), , y c, , Figure 16.70, For Prob. 16.45., 16.46 Develop the state equations for the circuit shown in, Fig. 16.71., , Section 16.6 Applications, 16.53 Show that the parallel RLC circuit shown in Fig. 16.73, is stable., , 1H, , v s (t), , +, −, , +, −v o(t), , 2F, , 4Ω, , i s (t), Io, R, , Is, , Figure 16.71, , C, , L, , For Prob. 16.46., , Figure 16.73, 16.47 Develop the state equations for the circuit shown in, Fig. 16.72., , i1(t), , v1(t), , +, −, , Figure 16.72, For Prob. 16.47., , 1, 4, , F, , i 2(t), , 2Ω, , 1H, , For Prob. 16.53., , 16.54 A system is formed by cascading two systems as, shown in Fig. 16.74. Given that the impulse response, of the systems are, +, −, , v 2(t), , h1(t) 3etu (t),, , h2(t) e4tu (t), , (a) Obtain the impulse response of the overall, system., (b) Check if the overall system is stable.
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ale29559_ch16.qxd, , 07/10/2008, , 04:12 PM, , Page 754, , Chapter 16, , 754, , Applications of the Laplace Transform, , Comprehensive Problems, 16.60 Obtain the transfer function of the op amp circuit in, Fig. 16.80 in the form of, Vo(s), as, 2, Vi(s), s bs c, where a, b, and c are constants. Determine the, constants., 10 kΩ, , (a) Find Y(s)., (b) An 8-V battery is connected to the network via a, switch. If the switch is closed at t 0, find the, current i(t) through Y(s) using the Laplace, transform., 16.62 A gyrator is a device for simulating an inductor in, a network. A basic gyrator circuit is shown in, Fig. 16.81. By finding Vi (s)Io(s), show that the, inductance produced by the gyrator is L CR2., , 1 F, 0.5 ΩF, vi, , +, −, , 10 kΩ, , −, +, , R, , C, , vo, , Figure 16.80, For Prob. 16.60., 16.61 A certain network has an input admittance Y(s). The, admittance has a pole at s 3, a zero at s 1,, and Y() 0.25 S., , R, , vi +, −, , Figure 16.81, For Prob. 16.62., , −, +, , R, −, +, , io, R
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ale29559_ch17.qxd, , 07/10/2008, , 05:28 PM, , Page 755, , c h a p t e r, , The Fourier Series, , 17, , Research is to see what everybody else has seen, and think what, nobody has thought., —Albert Szent Györgyi, , Enhancing Your Skills and Your Career, ABET EC 2000 criteria (3.j), “a knowledge of contemporary, issues.”, Engineers must have knowledge of contemporary issues. To have a truly, meaningful career in the twenty-first century, you must have knowledge of contemporary issues, especially those that may directly affect, your job and/or work. One of the easiest ways to achieve this is to read, a lot—newspapers, magazines, and contemporary books. As a student, enrolled in an ABET-accredited program, some of the courses you take, will be directed toward meeting this criteria., , ABET EC 2000 criteria (3.k), “an ability to use the techniques,, skills, and modern engineering tools necessary for, engineering practice.”, The successful engineer must have the “ability to use the techniques,, skills, and modern engineering tools necessary for engineering practice.”, Clearly, a major focus of this textbook is to do just that. Learning to, use skillfully the tools that facilitate your working in a modern “knowledge capturing integrated design environment” (KCIDE) is fundamental to your performance as an engineer. The ability to work in a modern, KCIDE environment requires a thorough understanding of the tools, associated with that environment., The successful engineer, therefore, must keep abreast of the new, design, analysis, and simulation tools. That engineer must also use, those tools until he or she is comfortable with using them. The engineer, also must make sure software results are consistent with real-world, actualities. It is probably in this area that most engineers have the greatest difficulty. Thus, successful use of these tools requires constant, learning and relearning the fundamentals of the area in which the engineer is working., , Photo by Charles Alexander, , 755
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ale29559_ch17.qxd, , 756, , 07/10/2008, , 05:28 PM, , Page 756, , Chapter 17, , The Fourier Series, , Historical, Jean Baptiste Joseph Fourier (1768–1830), a French mathematician, first presented the series and transform that bear his name. Fourier’s, results were not enthusiastically received by the scientific world. He, could not even get his work published as a paper., Born in Auxerre, France, Fourier was orphaned at age 8. He attended, a local military college run by Benedictine monks, where he demonstrated great proficiency in mathematics. Like most of his contemporaries, Fourier was swept into the politics of the French Revolution., He played an important role in Napoleon’s expeditions to Egypt in the, later 1790s. Due to his political involvement, he narrowly escaped, death twice., , 17.1, , Introduction, , We have spent a considerable amount of time on the analysis of circuits, with sinusoidal sources. This chapter is concerned with a means of analyzing circuits with periodic, nonsinusoidal excitations. The notion of, periodic functions was introduced in Chapter 9; it was mentioned there, that the sinusoid is the most simple and useful periodic function. This, chapter introduces the Fourier series, a technique for expressing a periodic function in terms of sinusoids. Once the source function is expressed, in terms of sinusoids, we can apply the phasor method to analyze circuits., The Fourier series is named after Jean Baptiste Joseph Fourier, (1768–1830). In 1822, Fourier’s genius came up with the insight that, any practical periodic function can be represented as a sum of sinusoids. Such a representation, along with the superposition theorem,, allows us to find the response of circuits to arbitrary periodic inputs, using phasor techniques., We begin with the trigonometric Fourier series. Later we consider, the exponential Fourier series. We then apply Fourier series in circuit, analysis. Finally, practical applications of Fourier series in spectrum, analyzers and filters are demonstrated., , 17.2, , Trigonometric Fourier Series, , While studying heat flow, Fourier discovered that a nonsinusoidal periodic function can be expressed as an infinite sum of sinusoidal functions. Recall that a periodic function is one that repeats every T seconds., In other words, a periodic function f (t) satisfies, f (t) f (t nT), where n is an integer and T is the period of the function., , (17.1)
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ale29559_ch17.qxd, , 07/10/2008, , 05:28 PM, , Page 757, , 17.2, , Trigonometric Fourier Series, , 757, , According to the Fourier theorem, any practical periodic function of frequency 0 can be expressed as an infinite sum of sine or, cosine functions that are integral multiples of 0. Thus, f (t) can be, expressed as, f (t) a0 a1 cos 0 t b1 sin 0 t a2 cos 2 0 t, b2 sin 20 t a3 cos 30 t b3 sin 30 t p, , (17.2), , or, , , , (17.3), , i, , b, , f (t) a0 a (an cos n0 t bn sin n0 t), n1, dc, ac, , where 0 2 pT is called the fundamental frequency in radians per, second. The sinusoid sin n0 t or cos n0 t is called the nth harmonic, of f (t); it is an odd harmonic if n is odd and an even harmonic if n, is even. Equation 17.3 is called the trigonometric Fourier series of, f (t). The constants an and bn are the Fourier coefficients. The coefficient a0 is the dc component or the average value of f (t). (Recall, that sinusoids have zero average values.) The coefficients an and bn, (for n 0) are the amplitudes of the sinusoids in the ac component., Thus,, , The harmonic frequency n is an, integral multiple of the fundamental, frequency 0, i.e., n n 0., , The Fourier series of a periodic function f (t) is a representation that, resolves f (t) into a dc component and an ac component comprising, an infinite series of harmonic sinusoids., , A function that can be represented by a Fourier series as in Eq. (17.3), must meet certain requirements, because the infinite series in Eq. (17.3), may or may not converge. These conditions on f (t) to yield a convergent, Fourier series are as follows:, 1. f (t) is single-valued everywhere., 2. f (t) has a finite number of finite discontinuities in any one, period., 3. f (t) has a finite number of maxima and minima in any one, period., 4. The integral, , , , t0T, , t0, , 0 f (t) 0 dt 6 for any t0., , These conditions are called Dirichlet conditions. Although they are not, necessary conditions, they are sufficient conditions for a Fourier series, to exist., A major task in Fourier series is the determination of the, Fourier coefficients a0, an, and bn. The process of determining the, , Historical note: Although Fourier, published his theorem in 1822, it was, P. G. L. Dirichlet (1805–1859) who later, supplied an acceptable proof of the, theorem., A software package like Mathcad or, Maple can be used to evaluate the, Fourier coefficients.
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ale29559_ch17.qxd, , 758, , 07/10/2008, , 05:28 PM, , Page 758, , Chapter 17, , The Fourier Series, , coefficients is called Fourier analysis. The following trigonometric, integrals are very helpful in Fourier analysis. For any integers m, and n,, , , , T, , , , T, , sin n0 t dt 0, , (17.4a), , cos n0 t dt 0, , (17.4b), , sin n 0 t cos m0 t dt 0, , (17.4c), , 0, , 0, , , , T, , 0, , , , T, , sin n0 t sin m0 t dt 0,, , (m n), , (17.4d), , cos n0 t cos m0 t dt 0,, , (m n), , (17.4e), , 0, , , , T, , 0, , , , T, , , , T, , sin2 n0 t dt , , T, 2, , (17.4f), , cos2 n0 t dt , , T, 2, , (17.4g), , 0, , 0, , Let us use these identities to evaluate the Fourier coefficients., We begin by finding a0. We integrate both sides of Eq. (17.3) over, one period and obtain, , , , T, , f (t) dt , , 0, , , , T, , , , T, , 0, , , , 0, , , , c a0 a (an cos n0 t bn sin n0 t) d dt, n1, , , , , , a0 dt a c, n1, , T, , an cos n0 t dt, , (17.5), , 0, , , , , , T, , 0, , bn sin n0 t dt d dt, , Invoking the identities of Eqs. (17.4a) and (17.4b), the two integrals, involving the ac terms vanish. Hence,, , , , T, , f (t) dt , , 0, , , , T, , a0 dt a0 T, , 0, , or, , a0 , , 1, T, , , , T, , f (t) dt, , 0, , showing that a0 is the average value of f (t)., , (17.6)
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ale29559_ch17.qxd, , 07/10/2008, , 05:28 PM, , Page 759, , 17.2, , Trigonometric Fourier Series, , To evaluate an, we multiply both sides of Eq. (17.3) by cos m 0 t, and integrate over one period:, , , , T, , f (t) cos m0 t dt, , 0, , , , , , T, , , , T, , 0, , , , 0, , , , , , c a0 a (an cos n0 t bn sin n0 t) d cos m0 t dt, n1, , , , a0 cos m0 t dt a c, n1, , , , T, , , , T, , an cos n0 t cos m0 t dt, , 0, , bn sin n0 t cos m0 t dt d dt, , 0, , (17.7), , The integral containing a0 is zero in view of Eq. (17.4b), while the, integral containing bn vanishes according to Eq. (17.4c). The integral, containing an will be zero except when m n, in which case it is T2,, according to Eqs. (17.4e) and (17.4g). Thus,, , , , T, , 0, , T, f (t) cos m0 t dt an ,, 2, , for m n, , or, 2, an , T, , , , T, , f (t) cos n0 t dt, , (17.8), , 0, , In a similar vein, we obtain bn by multiplying both sides of Eq. (17.3), by sin m0t and integrating over the period. The result is, bn , , 2, T, , , , T, , f (t) sin n0t dt, , (17.9), , 0, , Be aware that since f (t) is periodic, it may be more convenient to carry, the integrations above from T2 to T2 or generally from t0 to t0 T, instead of 0 to T. The result will be the same., An alternative form of Eq. (17.3) is the amplitude-phase form, , , f (t) a0 a A n cos(n0 t fn), , (17.10), , n1, , We can use Eqs. (9.11) and (9.12) to relate Eq. (17.3) to Eq. (17.10),, or we can apply the trigonometric identity, cos(a b) cos a cos b sin a sin b, , (17.11), , to the ac terms in Eq. (17.10) so that, , , , , n1, , n1, , a0 a An cos(n0t fn) a0 a (An cos fn) cos n0t, (An sin fn) sin n0t, , (17.12), , 759
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ale29559_ch17.qxd, , 07/17/2008, , 01:31 PM, , Page 760, , Chapter 17, , 760, , The Fourier Series, , Equating the coefficients of the series expansions in Eqs. (17.3) and, (17.12) shows that, an An cos fn,, , bn An sin fn, , (17.13a), , An 2a2n b2n,, , fn tan1, , bn, an, , (17.13b), , or, , To avoid any confusion in determining fn, it may be better to relate, the terms in complex form as, Anlfn an jbn, , The frequency spectrum is also known, as the line spectrum in view of the, discrete frequency components., , Values of cosine, sine, and, exponential functions for, integral multiples of p., Function, , Value, , cos 2n p, sin 2n p, cos n p, sin n p, , 1, 0, (1)n, 0, , sin, , n even, n odd, , b, , (1)n2,, 0,, , np, 2, , b, , (1)(n1)2,, 0,, , 1, (1)n, , e jnp2, , b, , Thus, the Fourier analysis is also a mathematical tool for finding the, spectrum of a periodic signal. Section 17.6 will elaborate more on the, spectrum of a signal., To evaluate the Fourier coefficients a0, an, and bn, we often need, to apply the following integrals:, , n odd, n even, , n even, n odd, , (1)n2,, j(1)(n1)2,, , Example 17.1, , (17.15a), (17.15b), , 2, , (17.15c), , 2, , (17.15d), , It is also useful to know the values of the cosine, sine, and exponential functions for integral multiples of p. These are given in Table 17.1,, where n is an integer., , Determine the Fourier series of the waveform shown in Fig. 17.1., Obtain the amplitude and phase spectra., , f (t), 1, , –1, , cos at dt a sin at, 1, sin at dt a cos at, 1, 1, t cos at dt a cos at a t sin at, 1, 1, t sin at dt a sin at a t cos at, 1, , np, 2, , e j2np, e jnp, , –2, , The convenience of this relationship will become evident in Section 17.6., The plot of the amplitude An of the harmonics versus n0 is, called the amplitude spectrum of f (t); the plot of the phase fn versus, n0 is the phase spectrum of f (t). Both the amplitude and phase spectra form the frequency spectrum of f (t)., The frequency spectrum of a signal consists of the plots of the amplitudes and phases of the harmonics versus frequency., , TABLE 17.1, , cos, , (17.14), , Solution:, The Fourier series is given by Eq. (17.3), namely,, 0, , 1, , 2, , Figure 17.1, For Example 17.1; a square wave., , 3 t, , , , f (t) a0 a (an cos n0 t bn sin n0 t), n1, , (17.1.1)
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ale29559_ch17.qxd, , 07/10/2008, , 05:28 PM, , Page 761, , 17.2, , Trigonometric Fourier Series, , Our goal is to obtain the Fourier coefficients a0, an, and bn using, Eqs. (17.6), (17.8), and (17.9). First, we describe the waveform as, f (t) b, , 1,, 0,, , 0 6 t 6 1, 1 6 t 6 2, , (17.1.2), , 761, , 1, 2, , dc component, , t, , and f (t) f (t T). Since T 2, 0 2 pT p. Thus,, a0 , , 1, T, , , , T, , 1, c, 2, , f (t) dt , , 0, , , , 1, , 1 dt , , 0, , , , 2, , 1, , 1 1 1, 0 dt d t ` , 2 0 2, , (17.1.3), Fundamental ac component, , Using Eq. (17.8) along with Eq. (17.15a),, an , , , 2, T, , , , t, , (a), , T, , f (t) cos n0 t dt, , 0, , 2, c, 2, , 1, , , , 1 cos n p t dt , , 0, , , , 2, , 0 cos n p t dt, , (17.1.4), t, , 1, , 1, , , , 1, 1, sin n p t ` , 3sin n p sin (0)4 0, np, np, 0, , Sum of first two ac components, , From Eq. (17.9) with the aid of Eq. (17.15b),, bn , , 2, T, , , , T, , f (t) sin n0 t dt, , 0, , 2, c, 2, , , , 1, , 1 sin n p t dt , , 0, , , , 2, , 1, , , , 1, 1, cos n p t `, np, 0, , , , 1, (cos n p 1),, np, , 2, ,, 1, [1 (1)n] c n p, , np, 0,, , t, , 0 sin n p t dt d, , Sum of first three ac components, , (17.1.5), cos n p (1)n, t, , n odd, n even, , Sum of first four ac components, , Substituting the Fourier coefficients in Eqs. (17.1.3) to (17.1.5) into, Eq. (17.1.1) gives the Fourier series as, f (t) , , 1, 2, 2, 2, sin p t , sin 3 p t , sin 5 p t p, p, 2, 3p, 5p, , (17.1.6), , t, , Since f (t) contains only the dc component and the sine terms with the, fundamental component and odd harmonics, it may be written as, , Sum of first five ac components, , f (t) , , , , 2, 1, 1, , sin n p t,, a, p, n, 2, k1, , (b), , n 2k 1, , (17.1.7), , By summing the terms one by one as demonstrated in Fig. 17.2,, we notice how superposition of the terms can evolve into the original, square. As more and more Fourier components are added, the sum gets, closer and closer to the square wave. However, it is not possible in, practice to sum the series in Eq. (17.1.6) or (17.1.7) to infinity. Only, a partial sum (n 1, 2, 3, p , N , where N is finite) is possible. If we, plot the partial sum (or truncated series) over one period for a large N, , Figure17.2, Evolution of a square wave from its, Fourier components., , Summing the Fourier terms by hand, calculation may be tedious. A computer is helpful to compute the terms, and plot the sum like those shown in, Fig. 17.2.
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ale29559_ch17.qxd, , 07/10/2008, , 05:28 PM, , Page 762, , Chapter 17, , 762, , The Fourier Series, f (t), 1, , Historical note: Named after the mathematical physicist Josiah Willard Gibbs,, who first observed it in 1899., , 0, , 1, , 2, , t, , Figure 17.3, , Truncating the Fourier series at N 11;, Gibbs phenomenon., , 2, , , An, 0.5, , 2, 3, , 0, , , , 2, 5, , , 2 3 4 5 6, (a), , as in Fig. 17.3, we notice that the partial sum oscillates above and, below the actual value of f (t). At the neighborhood of the points of, discontinuity ( x 0, 1, 2, p ), there is overshoot and damped, oscillation. In fact, an overshoot of about 9 percent of the peak value, is always present, regardless of the number of terms used to, approximate f (t). This is called the Gibbs phenomenon., Finally, let us obtain the amplitude and phase spectra for the signal, in Fig. 17.1. Since an 0,, , , 0°, , , , An , , 2 3 4 5 6, , 2a 2n, , , , b 2n, , , , 2, ,, 0 bn 0 c n p, 0,, , n odd, , (17.1.8), , n even, , and, –90°, , fn tan1, , (b), , Figure 17.4, , Practice Problem 17.1, f (t), 1, , –1, , 1, –1, , 2, , 3, , (17.1.9), , Find the Fourier series of the square wave in Fig. 17.5. Plot the amplitude and phase spectra., Answer: f (t) , , 0, , n odd, n even, , The plots of An and fn for different values of n0 n p provide the, amplitude and phase spectra in Fig. 17.4. Notice that the amplitudes of, the harmonics decay very fast with frequency., , For Example 17.1: (a) amplitude and, (b) phase spectrum of the function, shown in Fig. 17.1., , –2, , bn, 90,, b, an, 0,, , , , spectra., , 4 1, a n sin n p t, n 2k 1. See Fig. 17.6 for the, p k1, , 4, , , An, , Figure 17.5, For Practice Prob. 17.1., , , 4, 3, , 0, , , , 0°, , 4, 5, , 2 3 4 5 6, (a), , , , , , 2 3 4 5 6, , , –90°, (b), , Figure 17.6, For Practice Prob. 17.1: amplitude and phase spectra for the function shown in, Fig. 17.5.
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ale29559_ch17.qxd, , 07/10/2008, , 05:28 PM, , Page 763, , 17.2, , Trigonometric Fourier Series, , 763, , Example 17.2, , Obtain the Fourier series for the periodic function in Fig. 17.7 and plot, the amplitude and phase spectra., f (t), , Solution:, The function is described as, , 1, , f (t) b, , t,, 0,, , 0 6 t 6 1, 1 6 t 6 2, , –2, , 1, T, , , , T, , 1, c, 2, , f (t) dt , , 0, , , , 1, , 0, , t dt , , , , 2, , 1, , 0 dt d , , 1 t2 1 1, ` , (17.2.1), 22 0 4, , To evaluate an and bn, we need the integrals in Eq. (17.15):, an , , , 2, T, , , , T, , f (t) cos n0 t dt, , 0, , , , 2, c, 2, , 1, , t cos n p t dt , , 0, , , , 2, , 1, , 0 cos n p t dt d, , (17.2.2), , 1, 1, t, c 2 2 cos n pt , sin npt d `, np, np, 0, (1)n 1, 1, 2 2 (cos n p 1) 0 , np, n2p2, , since cos n p (1)n; and, bn , , , 2, T, , , , c, , T, , f (t) sin n0 t dt, , 0, , 2, c, 2, , , , 1, , t sin n pt dt , , 0, , , , 2, , 1, , 0 sin n pt dt d, , (17.2.3), , 1, , 1, t, sin n pt , cos n pt d `, 2, n, p, np, 0, 2, , 0, , (1)n1, cos n p, , np, np, , Substituting the Fourier coefficients just found into Eq. (17.3) yields, f (t) , , , [(1)n 1], (1)n1, 1, a c, cos, n, pt, , sin n pt d, np, 4 n1, (n p)2, , To obtain the amplitude and phase spectra, we notice that, for even, harmonics, an 0, bn 1np, so that, Anlfn an jbn 0 j, , 1, np, , (17.2.4), , Hence,, An 0bn 0 , , 0, , For Example 17.2., , Since T 2, 0 2pT p. Then, a0 , , –1, , Figure 17.7, , 1, ,, n 2, 4, . . ., np, n 2, 4, . . ., fn 90,, , (17.2.5), , 1, , 2, , 3 t
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ale29559_ch17.qxd, , 07/10/2008, , 05:28 PM, , Page 764, , Chapter 17, , 764, An, , The Fourier Series, , For odd harmonics, an 2(n2p 2), bn 1(n p) so that, , 0.38, , 0.25, , Anlfn an jbn , 0.16, 0.11, 0.06 0.05, , , 2 3 4 5 6, , An 2a 2n b 2n , , , , (a), , 270°, , 4, 1, 2 2, 4, Bn p, np, 4, , 1, 2 2 24 n2p 2,, np, , (17.2.7), , n 1, 3, . . ., , 262.7°, , 258°, , 237.8°, , (17.2.6), , That is,, , 0.08, 0, , 2, 1, j, 2, np, np, 2, , From Eq. (17.2.6), we observe that f lies in the third quadrant,, so that, , 180°, 90°, , 90°, , , , 0, , fn 180 tan1, , 90°, , 90°, , 2 3 4 5 6, , , , np, ,, 2, , n 1, 3, . . ., , (17.2.8), , From Eqs. (17.2.5), (17.2.7), and (17.2.8), we plot An and fn for, different values of n0 n p to obtain the amplitude spectrum and, phase spectrum as shown in Fig. 17.8., , (b), , Figure 17.8, For Example 17.2: (a) amplitude, spectrum, (b) phase spectrum., , Practice Problem 17.2, f (t), , Answer: f (t) , , 3, , –2, , –1, , Determine the Fourier series of the sawtooth waveform in Fig. 17.9., , 0, , 1, , 2, , 3, 3 1, , a n sin 2 p nt., p n1, 2, , 3 t, , Figure 17.9, For Practice Prob. 17.2., , 17.3, , Symmetry Considerations, , We noticed that the Fourier series of Example 17.1 consisted only of, the sine terms. One may wonder if a method exists whereby one can, know in advance that some Fourier coefficients would be zero and, avoid the unnecessary work involved in the tedious process of calculating them. Such a method does exist; it is based on recognizing the, existence of symmetry. Here we discuss three types of symmetry:, (1) even symmetry, (2) odd symmetry, (3) half-wave symmetry., , 17.3.1 Even Symmetry, A function f (t) is even if its plot is symmetrical about the vertical axis;, that is,, f (t) f (t), , (17.16)
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ale29559_ch17.qxd, , 07/10/2008, , 05:28 PM, , Page 765, , 17.3, , Symmetry Considerations, , Examples of even functions are t 2, t 4, and cos t. Figure 17.10 shows, more examples of periodic even functions. Note that each of these, examples satisfies Eq. (17.16). A main property of an even function, fe(t) is that:, , , , T2, , , , fe(t) dt 2, , T2, , 765, f (t), –T, 2, , T, 2, , A, , –T, , 0, , –A, , T, , t, , T, , t, , 2, , t, , T2, , fe(t) dt, , (a), , (17.17), , 0, , g(t), , because integrating from T2 to 0 is the same as integrating from 0, to T2. Utilizing this property, the Fourier coefficients for an even function become, , A, , –T, , 0, (b), , 2, T, , a0 , , , , T2, , , , T2, , f (t) dt, , h(t), , 0, , 4, an , T, , A, , (17.18), f (t) cos n0 t dt, , –2, , , , T2, , f (t) dt , , T2, , 1, c, T, , , , 0, , f (t) dt , , T2, , , , T2, , 0, , f (t) dt d, , (17.19), , We change variables for the integral over the interval T2 6 t 6 0, by letting t x, so that dt dx, f (t) f (t) f (x), since f (t) is, an even function, and when t T2, x T2. Then,, a0 , , 1, c, T, , 1, c, T, , , , 0, , , , f (x)(dx) , , T2, , , , T2, , 0, , T2, , , , f (x) dx , , 0, , T2, , 0, , f (t) dt d, (17.20), , f (t) dt d, , showing that the two integrals are identical. Hence,, a0 , , 2, T, , , , T2, , f (t) dt, , (17.21), , 0, , as expected. Similarly, from Eq. (17.8),, an , , 2, c, T, , , , 0, , T2, , , , Figure 17.10, , Since bn 0, Eq. (17.3) becomes a Fourier cosine series. This makes, sense because the cosine function is itself even. It also makes intuitive, sense that an even function contains no sine terms since the sine function is odd., To confirm Eq. (17.18) quantitatively, we apply the property of an, even function in Eq. (17.17) in evaluating the Fourier coefficients in, Eqs. (17.6), (17.8), and (17.9). It is convenient in each case to integrate over the interval T2 6 t 6 T2, which is symmetrical about, the origin. Thus,, 1, T, , 0, (c), , bn 0, , a0 , , –, , 0, , f (t) cos n0 t dt , , , , T2, , 0, , f (t) cos n0 t dt d, , (17.22), , Typical examples of even periodic, functions.
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ale29559_ch17.qxd, , 07/10/2008, , 05:28 PM, , Page 766, , Chapter 17, , 766, , The Fourier Series, , We make the same change of variables that led to Eq. (17.20) and note, that both f (t) and cos n0 t are even functions, implying that f (t) , f (t) and cos(n0 t) cos n0 t. Equation (17.22) becomes, an , , , , 2, c, T, 2, c, T, 2, c, T, , , , 0, , , , 0, , , , T2, , f (x) cos(n0 x)(dx) , , T2, , , , T2, , 0, , , , f (x) cos(n0 x)(dx) , , T2, , T2, , f (t) cos n0 t dt d, , 0, , f (x) cos(n0 x) dx , , 0, , , , T2, , f (t) cos n0 t dt d, , f (t) cos n0 t dt d, , 0, , (17.23a), , or, an , , 4, T, , , , T2, , f (t) cos n0 t dt, , (17.23b), , 0, , as expected. For bn, we apply Eq. (17.9),, bn , , 2, c, T, , , , 0, , f (t) sin n0 t dt , , T2, , T2, , , , f (t) sin n0 t dt d, , 0, , (17.24), , We make the same change of variables but keep in mind that f (t) , f (t) but sin(n 0 t) sin n0 t. Equation (17.24) yields, bn , , , f (t), A, –T, , 0, –A, , –T, 2, , , , T t, , T, 2, , 2, c, T, , , , 0, , 2, c, T, , , , 0, , f (x) sin(n0 x)(dx) , , T2, , 0, , f (x) sin n0 x dx , , T2, , 2, c , T, , , , 0, , , , f (t) sin n0 t dt d, , T2, , f (x) sin(n0 x) dx , , 0, , f (t) sin n0 t dt d, , T2, , , , T2, , 0, , 0, , (a), , , , T2, , f (t) sin n0 t dt d, (17.25), , confirming Eq. (17.18)., , g(t), A, –T, , T, , 0, , 17.3.2 Odd Symmetry, , t, , –A, , A function f (t) is said to be odd if its plot is antisymmetrical about the, vertical axis:, , (b), h(t), , f (t) f (t), , (17.26), , A, –T, , 0, , T, –A, , (c), , Figure 17.11, Typical examples of odd periodic, functions., , t, , Examples of odd functions are t, t 3, and sin t. Figure 17.11 shows more, examples of periodic odd functions. All these examples satisfy, Eq. (17.26). An odd function fo(t) has this major characteristic:, , , , T2, , T2, , fo(t) dt 0, , (17.27)
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ale29559_ch17.qxd, , 07/10/2008, , 05:28 PM, , Page 767, , 17.3, , Symmetry Considerations, , because integration from T2 to 0 is the negative of that from 0 to, T2. With this property, the Fourier coefficients for an odd function, become, a0 0,, bn , , 4, T, , an 0, , T2, , , , (17.28), , f (t) sin n0 t dt, , 0, , which give us a Fourier sine series. Again, this makes sense because, the sine function is itself an odd function. Also, note that there is no, dc term for the Fourier series expansion of an odd function., The quantitative proof of Eq. (17.28) follows the same procedure, taken to prove Eq. (17.18) except that f (t) is now odd, so that, f (t) f (t). With this fundamental but simple difference, it is easy, to see that a0 0 in Eq. (17.20), an 0 in Eq. (17.23a), and bn in, Eq. (17.24) becomes, bn , , , , 2, c, T, , , , 0, , T2, , 2, c , T, 2, c, T, , , , f (x) sin(n0 x)(dx) , , , , T2, , 0, , , , 0, , , , f (x) sin n0 x dx , , T2, , T2, , 0, , T2, , f (x) sin(n0 x) dx , , 0, , , , T2, , 0, , bn , , 4, T, , , , f (t) sin n0 t dt d, , f (t) sin n0 t dt d, , f (t) sin n0 t dt d, , T2, , f (t) sin n0 t dt, , (17.29), , 0, , as expected., It is interesting to note that any periodic function f (t) with neither, even nor odd symmetry may be decomposed into even and odd parts., Using the properties of even and odd functions from Eqs. (17.16) and, (17.26), we can write, , e, , e, , 1, 1, f (t) [ f (t) f (t)] [ f (t) f (t)] fe(t) fo(t), 2, 2, even, , odd, , (17.30), , Notice that fe(t) 12[ f (t) f (t)] satisfies the property of an, even function in Eq. (17.16), while fo(t) 12[ f (t) f (t)] satisfies, the property of an odd function in Eq. (17.26). The fact that fe(t), contains only the dc term and the cosine terms, while fo(t) has only, the sine terms, can be exploited in grouping the Fourier series expansion of f (t) as, , , , , n1, , n1, , g, , e, , f (t) a0 a an cos n0 t a bn sin n0 t fe(t) fo(t), even, , odd, , (17.31), , It follows readily from Eq. (17.31) that when f (t) is even, bn 0, and, when f (t) is odd, a0 0 an., , 767
Page 800 :
ale29559_ch17.qxd, , 07/10/2008, , 05:28 PM, , Page 768, , Chapter 17, , 768, , The Fourier Series, , Also, note the following properties of odd and even functions:, 1. The product of two even functions is also an even function., 2. The product of two odd functions is an even function., 3. The product of an even function and an odd function is an odd, function., 4. The sum (or difference) of two even functions is also an even, function., 5. The sum (or difference) of two odd functions is an odd function., 6. The sum (or difference) of an even function and an odd function, is neither even nor odd., Each of these properties can be proved using Eqs. (17.16) and (17.26)., , 17.3.3 Half-Wave Symmetry, A function is half-wave (odd) symmetric if, f at , , T, b f (t), 2, , (17.32), , which means that each half-cycle is the mirror image of the next halfcycle. Notice that functions cos n0 t and sin n0 t satisfy Eq. (17.32), for odd values of n and therefore possess half-wave symmetry when n, is odd. Figure 17.12 shows other examples of half-wave symmetric, functions. The functions in Figs. 17.11(a) and 17.11(b) are also halfwave symmetric. Notice that for each function, one half-cycle is the, inverted version of the adjacent half-cycle. The Fourier coefficients, become, a0 0, 4, an c T, 0,, 4, bn c T, 0,, , , , T2, , , , T2, , f (t) cos n0 t dt,, , for n odd, , 0, , (17.33), , for n even, f (t) sin n0 t dt,, , for n odd, , 0, , for n even, , f (t), , g(t), , A, , A, T, , –T, , t, , 0, , –T, , –A, (a), , Figure 17.12, Typical examples of half-wave odd symmetric functions., , 0, , T, , –A, , (b), , t
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ale29559_ch17.qxd, , 07/10/2008, , 05:28 PM, , Page 769, , 17.3, , Symmetry Considerations, , showing that the Fourier series of a half-wave symmetric function contains only odd harmonics., To derive Eq. (17.33), we apply the property of half-wave symmetric functions in Eq. (17.32) in evaluating the Fourier coefficients in, Eqs. (17.6), (17.8), and (17.9). Thus,, 1, a0 , T, , , , T2, , 1, f (t) dt c, T, , T2, , , , 0, , f (t) dt , , T2, , , , T2, , 0, , f (t) dt d, , (17.34), , We change variables for the integral over the interval T2 6 t 6 0, by letting x t T2, so that dx dt; when t T2, x 0; and, when t 0, x T2. Also, we keep Eq. (17.32) in mind; that is,, f (x T2) f (x). Then,, a0 , , 1, c, T, , , , T2, , f ax , , 0, , , , 1, c , T, , T, b dx , 2, , T2, , , , f (x) dx , , 0, , , , 0, , T2, , 0, , T2, , f (t) dt d, , f (t) dt d 0, , (17.35), , confirming the expression for a0 in Eq. (17.33). Similarly,, an , , 2, c, T, , , , 0, , f (t) cos n0 t dt , , T2, , , , T2, , 0, , f (t) cos n0 t dt d, , (17.36), , We make the same change of variables that led to Eq. (17.35) so that, Eq. (17.36) becomes, an , , 2, c, T, , , , T2, , 0, , f ax , , , , , T, T, b cos n0 ax b dx, 2, 2, , T2, , f (t) cos n0 t dt d, , 0, , (17.37), , Since f (x T2) f (x) and, cos n0 ax , , T, b cos(n0 t n p), 2, cos n0 t cos n p sin n0 t sin n p, (1)n cos n0 t, , (17.38), , substituting these in Eq. (17.37) leads to, an , , 2, [1 (1)n], T, , 4, T, c, 0,, , , , , , T2, , f (t) cos n0 t dt, , 0, , (17.39), , T2, , f (t) cos n0 t dt,, , for n odd, , 0, , for n even, , confirming Eq. (17.33). By following a similar procedure, we can, derive bn as in Eq. (17.33)., Table 17.2 summarizes the effects of these symmetries on the, Fourier coefficients. Table 17.3 provides the Fourier series of some, common periodic functions., , 769
Page 802 :
ale29559_ch17.qxd, , 07/17/2008, , 01:31 PM, , Page 770, , Chapter 17, , 770, , The Fourier Series, , TABLE 17.2, , Effects of symmetry on Fourier coefficients., Symmetry, Even, Odd, Half-wave, , a0, , an, , bn, , Remarks, , a0 0, a0 0, a0 0, , an 0, an 0, a2n 0, a2n1 0, , bn 0, bn 0, b2n 0, b2n1 0, , Integrate over T2 and multiply by 2 to get the coefficients., Integrate over T2 and multiply by 2 to get the coefficients., Integrate over T2 and multiply by 2 to get, the coefficients., , TABLE 17.3, , The Fourier series of common functions., Function, , Fourier series, , 1. Square wave, f(t), , , A, , 0, , f (t) , , 4A, 1, a 2n 1 sin(2n 1) 0 t, p n1, , f (t) , , At, 2A, 1, npt, , a n sin T cos n0 t, T, T n1, , f (t) , , , sin n0 t, A, A, , a, p n1, n, 2, , f (t) , , A, 4A, 1, 2 a, cos(2n 1)0 t, 2, p n1 (2n 1)2, , f (t) , , A, A, 2A, 1, sin 0 t , cos 2n0 t, a, 2, p, p n1, 2, 4n 1, , f (t) , , 4A, 2A, 1, , cos n0 t, a, p, p n1, 4n2 1, , t, , T, , 2. Rectangular pulse train, f (t), A, , , −, 2, , T, , 0 , 2, , t, , 3. Sawtooth wave, f(t), A, , 0, , t, , T, , 4. Triangular wave, f(t), A, , , , 0, , t, , T, , 5. Half-wave rectified sine, f (t), , , A, , T, , 0, , t, , 6. Full-wave rectified sine, f (t), , , A, , 0, , T, , t
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ale29559_ch17.qxd, , 07/10/2008, , 05:28 PM, , Page 771, , 17.3, , Symmetry Considerations, , 771, , Example 17.3, , Find the Fourier series expansion of f (t) given in Fig. 17.13., , f (t), 1, , –5, , –4, , –3, , –2, , 0, , –1, , 1, , 2, , 3, , 4, , 5, , t, , –1, , Figure 17.13, For Example 17.3., , Solution:, The function f (t) is an odd function. Hence a0 0 an. The period, is T 4, and 0 2 pT p2, so that, bn , , , 4, T, , , , T2, , f (t) sin n0 t dt, , 0, , 4, c, 4, , , , 1, , 1 sin, , 0, , np, t dt , 2, , , , 2, , 0 sin, , 1, , np, t dt d, 2, , 1, , , , 2, np t, 2, np, cos, ` , a1 cos b, np, 2 0 np, 2, , Hence,, f (t) , , 2 1, np, np, a n a1 cos 2 b sin 2 t, p n1, , which is a Fourier sine series., , Find the Fourier series of the function f (t) in Fig. 17.14., , f (t), 4, , –2π, , –π, , 0, , π, , 2π, , –4, , Figure 17.14, For Practice Prob. 17.3., , Answer: f (t) , , 16 1, a n sin n t, n 2k 1., p k1, , 3π, , t, , Practice Problem 17.3
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ale29559_ch17.qxd, , 07/10/2008, , 772, , Example 17.4, , 05:28 PM, , Page 772, , Chapter 17, , The Fourier Series, , Determine the Fourier series for the half-wave rectified cosine function, shown in Fig. 17.15., f (t), 1, , –5, , –3, , –1, , 0, , 1, , 3, , 5, , t, , Figure 17.15, A half-wave rectified cosine function; for Example 17.4., , Solution:, This is an even function so that bn 0. Also, T 4, 0 2pT p2., Over a period,, 2 6 t 6 1, p, f (t) d cos t, 1 6 t 6 1, 2, 0,, 1 6 t 6 2, 0,, , a0 , , an , , , , 4, T, , 2, T, , , , T2, , f (t) dt , , 0, , 2, c, 4, , , , 1, , p, cos t dt , 2, 0, , , , 2, , 1, , 0 dt d, , 1 2, p 1, 1, sin t ` , p, p, 2, 2 0, , T2, , f (t) cos n0 t dt , , 0, , 4, c, 4, , , , 1, , p, npt, cos t cos, dt 0 d, 2, 2, 0, , 1, But cos A cos B [cos(A B) cos(A B)]. Then, 2, an , , 1, 2, , , , , , 1, , 1, , 0, , p, p, c cos (n 1)t cos (n 1)t d dt, 2, 2, , For n 1,, a1 , , 1, 2, , 0, , [cos p t 1] dt , , 1, 1 sin pt, 1, td ` , c, p, 2, 2, 0, , For n 7 1,, an , , 1, p, 1, p, sin (n 1) , sin (n 1), p(n 1), 2, p(n 1), 2, , For n odd (n 1, 3, 5, p), (n 1) and (n 1) are both even, so, p, p, sin (n 1) 0 sin (n 1),, 2, 2, , n odd, , For n even (n 2, 4, 6, p), (n 1) and (n 1) are both odd. Also,, p, p, np, sin (n 1) sin (n 1) cos, (1)n2,, 2, 2, 2, , n even, , Hence,, an , , (1)n2, (1)n2, 2(1)n2, ,, , , p(n 1), p(n 1), p(n2 1), , n even
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ale29559_ch17.qxd, , 07/10/2008, , 05:28 PM, , Page 773, , 17.3, , Symmetry Considerations, , 773, , Thus,, f (t) , , 1, 1, p, 2 (1)n2, np, cos t , cos t, a, 2, p, p neven (n 1), 2, 2, 2, , To avoid using n 2, 4, 6, p and also to ease computation, we can, replace n by 2k, where k 1, 2, 3, . . . and obtain, f (t) , , (1)k, 1, 1, p, 2 , cos t , a (4k 2 1) cos k p t, p, p k1, 2, 2, , which is a Fourier cosine series., , Find the Fourier series expansion of the function in Fig. 17.16., Answer: f (t) 2 , , Practice Problem 17.4, f (t), , 16 1, a n2 cos n t, n 2k 1., p 2 k1, , 4, , –2, , 0, , 2, , 4 t, , Figure 17.16, For Practice Prob. 17.4., , Example 17.5, , Calculate the Fourier series for the function in Fig. 17.17., f (t), , Solution:, The function in Fig. 17.17 is half-wave odd symmetric, so that, a0 0 an. It is described over half the period as, f (t) t,, , 1 6 t 6 1, , T 4, 0 2pT p2. Hence,, bn , , 4, T, , , , T2, , f (t) sin n0 t dt, , 0, , Instead of integrating f (t) from 0 to 2, it is more convenient to integrate, from 1 to 1. Applying Eq. (17.15d),, bn , , 4, 4, , , , 1, , 1, , t sin, , t cos n p t2 1, sin n pt2, npt, , dt c 2 2, d `, 2, n p2, n p 4, 1, , , , 4, np, np, 2, np, np, c cos, c sin, sin a b d , cos a b d, 2, n, p, 2, 2, 2, 2, np, , , , np, 8, sin, 2, 2, np, , 2, , 2, , since sin(x) sin x is an odd function, while cos(x) cos x is, an even function. Using the identities for sin n p2 in Table 17.1,, bn , , 8, (1)(n1)2, n odd 1, 3, 5, . . ., n p2, 2, , 1, , –2, , –1, , 0, –1, , Figure 17.17, For Example 17.5., , 1, , 2, , 3, , 4 t
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ale29559_ch17.qxd, , 07/10/2008, , 05:28 PM, , 774, , Page 774, , Chapter 17, , The Fourier Series, , Thus,, , , f (t) a bn sin np t, 2, n1,3,5, where bn is given above., , Practice Problem 17.5, , Determine the Fourier series of the function in Fig. 17.12(a). Take, A 2 and T 2p., Answer: f (t) , , 17.4, , 4 2, 1, a a n2p cos n t n sin n tb, n 2k 1., p k1, , Circuit Applications, , We find that in practice, many circuits are driven by nonsinusoidal periodic functions. To find the steady-state response of a circuit to a nonsinusoidal periodic excitation requires the application of a Fourier, series, ac phasor analysis, and the superposition principle. The procedure usually involves four steps., , Steps for Applying Fourier Series:, 1. Express the excitation as a Fourier series., 2. Transform the circuit from the time domain to the frequency, domain., 3. Find the response of the dc and ac components in the Fourier, series., 4. Add the individual dc and ac responses using the superposition, principle., , The first step is to determine the Fourier series expansion of the, excitation. For the periodic voltage source shown in Fig. 17.18(a), for, example, the Fourier series is expressed as, , , v(t) V0 a Vn cos(n0 t un), , (17.40), , n1, , (The same could be done for a periodic current source.) Equation (17.40), shows that v(t) consists of two parts: the dc component V0 and the ac, component Vn Vnlun with several harmonics. This Fourier series representation may be regarded as a set of series-connected sinusoidal, sources, with each source having its own amplitude and frequency, as, shown in Fig. 17.18(b)., The third step is finding the response to each term in the Fourier, series. The response to the dc component can be determined in the
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ale29559_ch17.qxd, , 07/10/2008, , 05:28 PM, , Page 775, , 17.4, , Circuit Applications, , 775, Io, , i(t), , i(t), , Linear, network, , v (t) +, −, Periodic, Source, , V0, , +, −, , V1 cos(0 t + 1), , +, −, , V2 cos(20 t + 2), , +, −, , Vn cos(n0 t + n ), , +, −, , (a), V0 +, −, , Z( = 0), , Linear, network, , +, I1, (b), , (b), , (a), , V1 1 +, −, , Figure 17.18, , Z(0), , (a) Linear network excited by a periodic voltage source, (b) Fourier series, representation (time-domain)., , +, I2, , frequency domain by setting n 0 or 0 as in Fig. 17.19(a), or in, the time domain by replacing all inductors with short circuits and all, capacitors with open circuits. The response to the ac component is, obtained by applying the phasor techniques covered in Chapter 9, as, shown in Fig. 17.19(b). The network is represented by its impedance, Z(n0) or admittance Y(n0). Z(n0) is the input impedance at the, source when is everywhere replaced by n0, and Y(n0) is the reciprocal of Z(n0)., Finally, following the principle of superposition, we add all the, individual responses. For the case shown in Fig. 17.19,, i(t) i0(t) i1(t) i2(t) p, , , I0 a 0In 0 cos(n0 t cn), , (17.41), , V2 2 +, −, , Z(20), , +, In, , Vn n +, −, , Z(n0), , n1, , where each component In with frequency n0 has been transformed to, the time domain to get in(t), and cn is the argument of In., , Figure 17.19, Steady-state responses: (a) dc component,, (b) ac component (frequency domain)., , Example 17.6, , Let the function f (t) in Example 17.1 be the voltage source vs(t) in the, circuit of Fig. 17.20. Find the response vo(t) of the circuit., Solution:, From Example 17.1,, vs(t) , , 1, 2 1, , a n sin n p t,, p k1, 2, , vs (t) +, −, , n 2k 1, , where n n0 n p rad/s. Using phasors, we obtain the response Vo, in the circuit of Fig. 17.20 by voltage division:, Vo , , jn L, j 2n p, Vs , Vs, R jn L, 5 j 2n p, , For the dc component (n 0 or n 0), Vs , , 1, 2, , 1, , 5Ω, , Vo 0, , Figure 17.20, For Example 17.6., , 2H, , +, vo (t), −
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ale29559_ch17.qxd, , 07/10/2008, , 05:28 PM, , Page 776, , Chapter 17, , 776, , The Fourier Series, , This is expected, since the inductor is a short circuit to dc. For the nth, harmonic,, Vs , , 2, l90, np, , (17.6.1), , and the corresponding response is, 2n pl90, , Vo , , , , 225 4n p, 2, , 2, , l tan, , 1, , a, , 2, l90b, n, 2n p5 p, (17.6.2), , 4l tan1 2n p5, 225 4n2p 2, , In the time domain,, , , vo(t) a, k1, , 4, 225 4n p, 2, , 2, , cos an p t tan1, , 2n p, b,, 5, , n 2k 1, , The first three terms (k 1, 2, 3 or n 1, 3, 5) of the odd harmonics, in the summation give us, | Vo |, , vo(t) 0.4981 cos(pt 51.49) 0.2051 cos(3 p t 75.14), 0.1257 cos(5 p t 80.96) p V, , 0.5, , 0.2, 0.13, 0.1, 0, , , , 2 3 4 5 6 7, , , , Figure 17.21, For Example 17.6: Amplitude spectrum of, the output voltage., , Practice Problem 17.6, 2Ω, , vs(t), , +, −, , 1F, , +, vo(t), −, , Figure 17.21 shows the amplitude spectrum for output voltage vo(t),, while that of the input voltage vs(t) is in Fig. 17.4(a). Notice that the, two spectra are close. Why? We observe that the circuit in Fig. 17.20, is a highpass filter with the corner frequency c RL 2.5 rad/s,, which is less than the fundamental frequency 0 p rad/s. The dc, component is not passed and the first harmonic is slightly attenuated,, but higher harmonics are passed. In fact, from Eqs. (17.6.1) and, (17.6.2), Vo is identical to Vs for large n, which is characteristic of a, highpass filter., , If the sawtooth waveform in Fig. 17.9 (see Practice Prob. 17.2) is the, voltage source vs(t) in the circuit of Fig. 17.22, find the response vo(t)., Answer: vo(t) , , 1 sin(2pnt tan1 4np), 1, , V., a, p n1, 2, n21 16n2p2, , Figure 17.22, For Practice Prob. 17.6., , Example 17.7, , Find the response io(t) of the circuit of Fig. 17.23 if the input voltage, v(t) has the Fourier series expansion, , 2(1)n, (cos nt n sin nt), v(t) 1 a, 2, n1 1 n
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ale29559_ch17.qxd, , 07/10/2008, , 05:28 PM, , Page 777, , 17.4, , Circuit Applications, , 777, i(t), , Solution:, Using Eq. (17.13), we can express the input voltage as, , , n1, , 2Ω, io(t), , 2(1)n, , v(t) 1 a, , 4Ω, , 1, , 21 n2, , cos(nt tan, , v (t) +, −, , n), , 1 1.414 cos(t 45) 0.8944 cos(2t 63.45), 0.6345 cos(3t 71.56) 0.4851 cos(4t 78.7) p, , 2H, , 2Ω, , Figure 17.23, For Example 17.7., , We notice that 0 1, n n rad/s. The impedance at the source is, Z 4 jn2 4 4 , , jn8, 8 jn8, , 4 jn2, 2 jn, , The input current is, I, , 2 jn, V, , V, Z, 8 jn8, , where V is the phasor form of the source voltage v(t). By current, division,, Io , , 4, V, I, 4 jn2, 4 jn4, , Since n n, Io can be expressed as, Io , , V, 421 n2l tan1 n, , For the dc component (n 0 or n 0), V1, , 1, , Io , , V, 1, , 4, 4, , For the nth harmonic,, V, , 2(1)n, 21 n2, , ltan1 n, , so that, Io , , 2(1)n, , 1, , ltan1 n , 2, , 421 n2ltan1 n 21 n, , (1)n, 2(1 n2), , In the time domain,, io(t) , , , (1)n, 1, a, cos nt A, 2, 4, n1 2(1 n ), , If the input voltage in the circuit of Fig. 17.24 is, v(t) , , , , 1, 1, 1, p, 2 a a 2 cos nt sin ntb V, n, 3, p n1 n, , determine the response io(t)., , Practice Problem 17.7
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 778, , Chapter 17, , 778, 2Ω, , Answer:, io(t), v (t) +, −, , The Fourier Series, , 1, 21 n2p 2, 2n, a, cos ant tan1, tan1 npb A., 9 n1 n2p 2 29 4n2, 3, , 1Ω, , 1F, , Figure 17.24, For Practice Prob. 17.7., , 17.5, , Average Power and RMS Values, , Recall the concepts of average power and rms value of a periodic signal that we discussed in Chapter 11. To find the average power, absorbed by a circuit due to a periodic excitation, we write the voltage, and current in amplitude-phase form [see Eq. (17.10)] as, , , v(t) Vdc a Vn cos(n0 t un), , (17.42), , n1, , , i(t) Idc a Im cos(m0 t fm), , (17.43), , m1, , Following the passive sign convention (Fig. 17.25), the average, power is, , i(t), +, v (t), , Linear, circuit, , −, , Figure 17.25, The voltage polarity reference and current, reference direction., , P, , 1, T, , T, , , , vi dt, , (17.44), , 0, , Substituting Eqs. (17.42) and (17.43) into Eq. (17.44) gives, P, , 1, T, , , , , , T, , Vdc Idc dt a, m1, , 0, , , , Vn Idc, a, T, n1, , , , , , , m1 n1, , , , T, , cos(m0 t fm) dt, , 0, , T, , cos(n0 t un) dt, , (17.45), , 0, , Vn Im, , a a, , ImVdc, T, , T, , , , T, , cos(n0 t un) cos(m0 t fm) dt, , 0, , The second and third integrals vanish, since we are integrating the, cosine over its period. According to Eq. (17.4e), all terms in the, fourth integral are zero when m n. By evaluating the first integral, and applying Eq. (17.4g) to the fourth integral for the case m n,, we obtain, , P Vdc Idc , , 1 , a Vn In cos(un fn), 2 n1, , (17.46), , This shows that in average-power calculation involving periodic voltage and current, the total average power is the sum of the average powers in each harmonically related voltage and current., Given a periodic function f (t), its rms value (or the effective value), is given by, Frms , , 1, BT, , , , T, , 0, , f 2(t) dt, , (17.47)
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 779, , 17.5, , Average Power and RMS Values, , 779, , Substituting f (t) in Eq. (17.10) into Eq. (17.47) and noting that, (a b)2 a2 2ab b2, we obtain, F 2rms , , 1, T, , , , T, , 0, , , , c a 20 2 a a0 An cos(n0 t fn), n1, , , , , , a a An Am cos(n0 t fn) cos(m0 t fm) d dt, n1 m1, , , , 1, T, , , , T, , 0, , , 1, a 20 dt 2 a a0 An, T, n1, , , , 1, a a An Am, T, n1 m1, , , , , , T, , cos(n0 t fn) dt, , 0, , T, , cos(n0 t fn) cos(m0 t fm) dt, , 0, , (17.48), Distinct integers n and m have been introduced to handle the product of the two series summations. Using the same reasoning as, above, we get, F 2rms a 20 , , 1 2, a An, 2 n1, , or, Frms , , B, , a 20 , , 1 2, a An, 2 n1, , (17.49), , In terms of Fourier coefficients an and bn, Eq. (17.49) may be written as, Frms , , B, , a 20 , , 1 , 2, 2, a (a n b n), 2 n1, , (17.50), , If f (t) is the current through a resistor R, then the power dissipated in, the resistor is, P RF 2rms, , (17.51), , Or if f (t) is the voltage across a resistor R, the power dissipated in the, resistor is, P, , F 2rms, R, , (17.52), , One can avoid specifying the nature of the signal by choosing a 1-, resistance. The power dissipated by the 1- resistance is, , P1 F 2rms a 20 , , 1 2, 2, a (a n b n), 2 n1, , (17.53), , This result is known as Parseval’s theorem. Notice that a 20 is the power, in the dc component, while 12 (a 2n b 2n) is the ac power in the nth harmonic. Thus, Parseval’s theorem states that the average power in a periodic signal is the sum of the average power in its dc component and, the average powers in its harmonics., , Historical note: Named after the French, mathematician Marc-Antoine Parseval, Deschemes (1755–1836).
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Chapter 17, , 780, , Example 17.8, , i(t), , Page 780, , +, v (t), −, , 10 Ω, , Determine the average power supplied to the circuit in Fig. 17.26 if, i(t) 2 10 cos(t 10) 6 cos(3t 35) A., 2F, , Solution:, The input impedance of the network is, Z 10 g, , Figure 17.26, For Example 17.8., , The Fourier Series, , 10(1j2), 1, 10, , , j2, 10 1j 2, 1 j 20, , Hence,, V IZ , , 10I, 21 4002l tan1 20, , For the dc component, 0,, I2A, , V 10(2) 20 V, , 1, , This is expected, because the capacitor is an open circuit to dc and the, entire 2-A current flows through the resistor. For 1 rad/s,, I 10l10, , V, , 1, , 10(10l10), 21 400l tan1 20, , 5l77.14, For 3 rad/s,, I 6l35, , 1, , V, , 10(6l35), 21 3600l tan1 60, , 1l54.04, Thus, in the time domain,, v(t) 20 5 cos(t 77.14) 1 cos(3t 54.04) V, We obtain the average power supplied to the circuit by applying, Eq. (17.46), as, P Vdc Idc , , 1 , a Vn In cos(un fn), 2 n1, , To get the proper signs of un and fn, we have to compare v and i in this, example with Eqs. (17.42) and (17.43). Thus,, 1, P 20(2) (5)(10) cos[77.14 (10)], 2, 1, (1)(6) cos[54.04 (35)], 2, 40 1.247 0.05 41.5 W, Alternatively, we can find the average power absorbed by the resistor as, V 2dc, 1 0Vn 0, 202, 1 52, 1 12, a, , , , R, 2 n1 R, 10, 2 10, 2 10, 40 1.25 0.05 41.5 W, 2, , P, , which is the same as the power supplied, since the capacitor absorbs, no average power.
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 781, , 17.6, , Exponential Fourier Series, , 781, , Practice Problem 17.8, , The voltage and current at the terminals of a circuit are, v(t) 128 192 cos 120 p t 96 cos(360 p t 30), i(t) 8 cos(120 p t 10) 3.2 cos(360 p t 60), Find the average power absorbed by the circuit., Answer: 22.23 kW., , Find an estimate for the rms value of the voltage in Example 17.7., , Example 17.9, , Solution:, From Example 17.7, v(t) is expressed as, v(t) 1 1.414 cos(t 45) 0.8944 cos(2t 63.45), 0.6345 cos(3t 71.56), 0.4851 cos(4t 78.7) p V, Using Eq. (17.49), we find, Vrms , , B, , , , a 20 , , 1 2, a An, 2 n1, , 1, 12 c(1.414)2 (0.8944)2 (0.6345)2 (0.4851)2 pd, B, 2, , 22.7186 1.649 V, This is only an estimate, as we have not taken enough terms of the, series. The actual function represented by the Fourier series is, v(t) , , pet, ,, sinh p, , p 6 t 6 p, , with v(t) v(t T). The exact rms value of this is 1.776 V., , Practice Problem 17.9, , Find the rms value of the periodic current, i(t) 8 30 cos 2t 20 sin 2t 15 cos 4t 10 sin 4t A, Answer: 29.61 A., , 17.6, , Exponential Fourier Series, , A compact way of expressing the Fourier series in Eq. (17.3) is to put, it in exponential form. This requires that we represent the sine and, cosine functions in the exponential form using Euler’s identity:, 1, cos n 0t [e jn0t ejn0t], 2, 1 jn0t, sin n 0 t [e, ejn0t], 2j, , (17.54a), (17.54b)
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ale29559_ch17.qxd, , 782, , 07/10/2008, , 05:29 PM, , Page 782, , Chapter 17, , The Fourier Series, , Substituting Eq. (17.54) into Eq. (17.3) and collecting terms, we obtain, f (t) a0 , , 1 , jn0t, (an jbn)ejn0t] (17.55), a [(an jbn)e, 2 n1, , If we define a new coefficient cn so that, c0 a0,, , cn , , (an jbn), ,, 2, , cn c*n , , (an jbn), 2, , (17.56), , then f (t) becomes, , , f (t) c0 a (cne jn0t cn ejn0t ), , (17.57), , n1, , or, , , f (t) a cne jn0t, , (17.58), , n, , This is the complex or exponential Fourier series representation of f (t)., Note that this exponential form is more compact than the sine-cosine, form in Eq. (17.3). Although the exponential Fourier series coefficients, cn can also be obtained from an and bn using Eq. (17.56), they can also, be obtained directly from f (t) as, , cn , , 1, T, , , , T, , f (t)ejn0t dt, , (17.59), , 0, , where 0 2 pT, as usual. The plots of the magnitude and phase of, cn versus n0 are called the complex amplitude spectrum and complex, phase spectrum of f (t), respectively. The two spectra form the complex frequency spectrum of f (t)., The exponential Fourier series of a periodic function f (t ) describes, the spectrum of f (t ) in terms of the amplitude and phase angle of ac, components at positive and negative harmonic frequencies., , The coefficients of the three forms of Fourier series (sine-cosine, form, amplitude-phase form, and exponential form) are related by, Anlfn an jbn 2cn, , (17.60), , or, cn 0 cn 0 lun , , 2a 2n b 2n, ltan1 bnan, 2, , if only an 7 0. Note that the phase un of cn is equal to fn., , (17.61)
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 783, , 17.6, , Exponential Fourier Series, , 783, , In terms of the Fourier complex coefficients cn, the rms value of, a periodic signal f (t) can be found as, F 2rms , , , , 1, T, , T, , f 2(t) dt , , 0, , , 1, a cn c, T, n, , , , T, , 0, , 1, T, , , , T, , 0, , , , f (t) c a cne jn0t d dt, n, , f (t)e jn0t dt d, , , , , , n, , n, , (17.62), , a cnc*n a 0cn 0 2, or, Frms , , , , 2, a 0cn 0, B n, , (17.63), , Equation (17.62) can be written as, , , F 2rms 0c0 0 2 2 a 0cn 0 2, , (17.64), , n1, , Again, the power dissipated by a 1- resistance is, , , P1 F 2rms a 0cn 0 2, , (17.65), , n, , which is a restatement of Parseval’s theorem. The power spectrum of, the signal f (t) is the plot of 0cn 0 2 versus n0. If f (t) is the voltage across, a resistor R, the average power absorbed by the resistor is F 2rmsR; if, f (t) is the current through R, the power is F 2rms R., As an illustration, consider the periodic pulse train of Fig. 17.27., Our goal is to obtain its amplitude and phase spectra. The period of, the pulse train is T 10, so that 0 2 pT p5. Using Eq. (17.59),, cn , , 1, T, , , , T2, , T2, , f (t)ejn0t dt , , 1, 10, , , , 1, , 1, , , , –11 –9, , 1, 1, ejn0t ` , (ejn0 e jn0), jn0, jn0, 1, sin n0, 2 e jn0 ejn0, ,, 2, n0, n0, 2j, , 2, , 10, , 10ejn0t dt, , 1, , , , f (t), , 0 , , –1 0 1, , 9 11 t, , Figure 17.27, The periodic pulse train., , (17.66), , p, 5, , sin np5, np5, , and, , , f (t) 2 a, n, , sin n p5 jnpt5, e, n p5, , (17.67), , Notice from Eq. (17.66) that cn is the product of 2 and a function of the, form sin xx. This function is known as the sinc function; we write it as, sinc(x) , , sin x, x, , (17.68), , Some properties of the sinc function are important here. For zero argument, the value of the sinc function is unity,, sinc(0) 1, , (17.69), , The sinc function is called the sampling, function in communication theory,, where it is very useful.
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 784, , Chapter 17, , 784, , The Fourier Series, , This is obtained by applying L’Hopital’s rule to Eq. (17.68). For an, integral multiple of p, the value of the sinc function is zero,, sinc(n p) 0,, , n 1, 2, 3, . . ., , (17.70), , Also, the sinc function shows even symmetry. With all this in mind,, we can obtain the amplitude and phase spectra of f (t). From Eq. (17.66),, the magnitude is, 0cn 0 2 `, , sin n p5, `, n p5, , (17.71), , while the phase is, np, 7 0, 5, un d, np, 180, sin, 6 0, 5, 0, sin, , Examining the input and output spectra allows visualization of the effect of, a circuit on a periodic signal., |cn |, , 2, 1.87, 1.51, , (17.72), , Figure 17.28 shows the plot of 0cn 0 versus n for n varying from 10, to 10, where n 0 is the normalized frequency. Figure 17.29, shows the plot of un versus n. Both the amplitude spectrum and phase, spectrum are called line spectra, because the values of 0cn 0 and un occur, only at discrete values of frequencies. The spacing between the lines, is 0. The power spectrum, which is the plot of 0 cn 0 2 versus n0, can, also be plotted. Notice that the sinc function forms the envelope of the, amplitude spectrum., , 1.0, , n, , 0.47, , 180°, , 0.43, 0.31, 0.38, 0.27, –10 –8 –6 – 4 –2 0 2 4 6 8 10 n, , Figure 17.28, The amplitude of a periodic pulse train., , Example 17.10, , –10 –8, , –6, , –4, , –2, , 0, , 2, , 4, , 6, , 8, , 10 n, , Figure 17.29, The phase spectrum of a periodic pulse train., , Find the exponential Fourier series expansion of the periodic function, f (t) et, 0 6 t 6 2 p with f (t 2 p) f (t)., Solution:, Since T 2 p, 0 2 pT 1. Hence,, cn , , , 1, T, , , , T, , f (t)ejn0t dt , , 0, , 1, 2p, , , , 2p, , etejnt dt, , 0, , 2p, 1, 1, 1, e(1jn)t ` , [e2 p ej2pn 1], 2 p 1 jn, 2 p(1 jn), 0, , But by Euler’s identity,, ej2pn cos 2 p n j sin 2 p n 1 j0 1
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 785, , 17.6, , Exponential Fourier Series, , 785, , Thus,, cn , , 1, 85, [e2 p 1] , 2 p(1 jn), 1 jn, , The complex Fourier series is, , 85, f (t) a, e jnt, n 1 jn, , We may want to plot the complex frequency spectrum of f (t). If we, let cn 0cn 0 lun , then, 0cn 0 , , 85, 21 n, , 2, , un tan1 n, , ,, , By inserting in negative and positive values of n, we obtain the amplitude, and the phase plots of cn versus n0 n, as in Fig. 17.30., |cn |, n, , 85, , 90°, 60.1, 38, 26.9, , 20.6, 16.7, –5, , –5, , –4, , –3, , –2, , –1, , 0, , 1, , 2, , 3, , 4, , –4, , –3, , –2, , –1, , 5 n0, , 0, , 1, , 2, , 3, , 4, , 5 n0, , (a), , –90°, (b), , Figure 17.30, The complex frequency spectrum of the function in Example 17.10: (a) amplitude spectrum, (b) phase spectrum., , Obtain the complex Fourier series of the function in Fig. 17.1., Answer: f (t) , , Practice Problem 17.10, , , j jnpt, 1, a, e ., 2 n n p, n0, nodd, , Find the complex Fourier series of the sawtooth wave in Fig. 17.9. Plot, the amplitude and the phase spectra., Solution:, From Fig. 17.9, f (t) t, 0 6 t 6 1, T 1 so that 0 2 pT 2 p., Hence,, T, 1, 1, 1, cn , (17.11.1), f (t)ejn0t dt , tej2npt dt, T 0, 1 0, , , , , , Example 17.11
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 786, , Chapter 17, , 786, , The Fourier Series, , But, , , , teat dt , , eat, (ax 1) C, a2, , Applying this to Eq. (17.11.1) gives, cn , , , 1, ej2npt, (j2n, p, t, , 1), `, (j2n p)2, 0, , (17.11.2), , ej2np (j2n p 1) 1, 4n2 p2, , Again,, ej2pn cos 2 p n j sin 2 p n 1 j0 1, so that Eq. (17.11.2) becomes, cn , , j2np, 4n p, 2, , 2, , , , j, 2np, , (17.11.3), , This does not include the case when n 0. When n 0,, c0 , , 1, T, , , , T, , f (t) dt , , 0, , 1, 1, , , , 1, , t dt , , 0, , t2 0, ` 0.5, 2 1, , (17.11.4), , Hence,, , , f (t) 0.5 a, n, n0, , j j2npt, e, 2n p, , (17.11.5), , and, 1, , 0cn 0 c 2 0n 0 p, 0.5,, , , n0, , ,, , un 90,, , n 0 (17.11.6), , n0, , By plotting 0cn 0 and un for different n, we obtain the amplitude spectrum, and the phase spectrum shown in Fig. 17.31., |cn |, 0.5, n, 90°, 0.16, , 0.16, , 0.08, 0.03 0.04 0.05, –50 –40 –30 –20 –0, , 0.08 0.05, 0.04 0.03, 0, , 0, , 20 30 40 50 , , (a), , Figure 17.31, For Example 17.11: (a) amplitude spectrum, (b) phase spectrum., , –50 – 40 –30 –20 –0, , 0, (b), , 0, , 20 30 40 50
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 787, , Fourier Analysis with PSpice, , 17.7, , 787, , Practice Problem 17.11, , Obtain the complex Fourier series expansion of f (t) in Fig. 17.17., Show the amplitude and phase spectra., , , Answer: f (t) a, n, n0, , j(1)n jnpt, e . See Fig. 17.32 for the spectra., np, |cn |, , n, , 0.32, , 0.32, , 0.16, , 90°, 0.16, , 0.11, , –3, 0.11, , 0.8, – 4 –3 –2, , –4, , 0.8, –1, , 0, , 1, , 2, , 3, , 4, , –1, , 1, , –2, , 3, , 0, , 2, , 4, , n, , −90°, , n, , (b), , (a), , Figure 17.32, For Practice Prob. 17.11: (a) amplitude spectrum, (b) phase spectrum., , 17.7, , Fourier Analysis with PSpice, , Fourier analysis is usually performed with PSpice in conjunction with, transient analysis. Therefore, we must do a transient analysis in order, to perform a Fourier analysis., To perform the Fourier analysis of a waveform, we need a circuit, whose input is the waveform and whose output is the Fourier decomposition. A suitable circuit is a current (or voltage) source in series with, a 1- resistor as shown in Fig. 17.33. The waveform is inputted as, vs(t) using VPULSE for a pulse or VSIN for a sinusoid, and the attributes of the waveform are set over its period T. The output V(1) from, node 1 is the dc level (a0) and the first nine harmonics (An) with their, corresponding phases cn; that is,, vo(t) a0 a An sin(n0 t cn), , (17.73), , n1, , where, cn fn , , is, , 1Ω, , 1, +, vo, −, , vs +, −, , 1Ω, , 0, , 0, , (a), , (b), , +, vo, −, , Figure 17.33, , 9, , An 2a 2n b 2n,, , 1, , p, ,, 2, , bn, fn tan1 a, n, , (17.74), , Notice in Eq. (17.74) that the PSpice output is in the sine and angle form, rather than the cosine and angle form in Eq. (17.10). The PSpice output, also includes the normalized Fourier coefficients. Each coefficient an is, normalized by dividing it by the magnitude of the fundamental a1, so, that the normalized component is ana1. The corresponding phase cn is, normalized by subtracting from it the phase c1 of the fundamental, so, that the normalized phase is cn c1., There are two types of Fourier analyses offered by PSpice for, Windows: Discrete Fourier Transform (DFT) performed by the PSpice, , Fourier analysis with PSpice using:, (a) a current source, (b) a voltage source.
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ale29559_ch17.qxd, , 07/10/2008, , 788, , 05:29 PM, , Page 788, , Chapter 17, , The Fourier Series, , program and Fast Fourier Transform (FFT) performed by the PSpice, A/D program. While DFT is an approximation of the exponential Fourier, series, FTT is an algorithm for rapid efficient numerical computation of, DFT. A full discussion of DFT and FTT is beyond the scope of this book., , 17.7.1 Discrete Fourier Transform, A discrete Fourier transform (DFT) is performed by the PSpice program, which tabulates the harmonics in an output file. To enable a, Fourier analysis, we select Analysis/Setup/Transient and bring up the, Transient dialog box, shown in Fig. 17.34. The Print Step should be a, small fraction of the period T, while the Final Time could be 6T. The, Center Frequency is the fundamental frequency f0 1T. The particular variable whose DFT is desired, V(1) in Fig. 17.34, is entered in, the Output Vars command box. In addition to filling in the Transient, dialog box, DCLICK Enable Fourier. With the Fourier analysis enabled, and the schematic saved, run PSpice by selecting Analysis/Simulate, as usual. The program executes a harmonic decomposition into Fourier, components of the result of the transient analysis. The results are sent, to an output file which can be retrieved by selecting Analysis/Examine, Output. The output file includes the dc value and the first nine harmonics by default, although you can specify more in the Number of, harmonics box (see Fig. 17.34)., Figure 17.34, Transient dialog box., , 17.7.2 Fast Fourier Transform, A fast Fourier transform (FFT) is performed by the PSpice A/D program and displays as a PSpice A/D plot the complete spectrum of a, transient expression. As explained above, we first construct the schematic, in Fig. 17.33(b) and enter the attributes of the waveform. We also need, to enter the Print Step and the Final Time in the Transient dialog box., Once this is done, we can obtain the FFT of the waveform in two ways., One way is to insert a voltage marker at node 1 in the schematic, of the circuit in Fig. 17.33(b). After saving the schematic and selecting Analysis/Simulate, the waveform V(1) will be displayed in the, PSpice A/D window. Double clicking the FFT icon in the PSpice A/D, menu will automatically replace the waveform with its FFT. From the, FFT-generated graph, we can obtain the harmonics. In case the FFTgenerated graph is crowded, we can use the User Defined data range, (see Fig. 17.35) to specify a smaller range., , Figure 17.35, X axis settings dialog box.
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ale29559_ch17.qxd, , 07/17/2008, , 01:31 PM, , Page 789, , 17.7, , Fourier Analysis with PSpice, , 789, , Another way of obtaining the FFT of V(1) is to not insert a voltage marker at node 1 in the schematic. After selecting Analysis/, Simulate, the PSpice A/D window will come up with no graph on it., We select Trace/Add and type V(1) in the Trace Command box and, DCLICKL OK. We now select Plot/X-Axis Settings to bring up the, X-Axis Setting dialog box shown in Fig. 17.35 and then select, Fourier/OK. This will cause the FFT of the selected trace (or traces), to be displayed. This second approach is useful for obtaining the FFT, of any trace associated with the circuit., A major advantage of the FFT method is that it provides graphical output. But its major disadvantage is that some of the harmonics, may be too small to see., In both DFT and FFT, we should let the simulation run for a large, number of cycles and use a small value of Step Ceiling (in the Transient, dialog box) to ensure accurate results. The Final Time in the Transient dialog box should be at least five times the period of the signal, to allow the simulation to reach steady state., , Example 17.12, , Use PSpice to determine the Fourier coefficients of the signal in Fig. 17.1., Solution:, Figure 17.36 shows the schematic for obtaining the Fourier coefficients., With the signal in Fig. 17.1 in mind, we enter the attributes of the, voltage source VPULSE as shown in Fig. 17.36. We will solve this, example using both the DFT and FFT approaches., , ■ METHOD 1 DFT Approach: (The voltage marker in Fig. 17.36, is not needed for this method.) From Fig. 17.1, it is evident that T 2 s,, f0 , , V, 1, V1=0, V2=1, TD=0, TF=1u, TR=1u, PW=1, PER=2, , + V3, −, , 1, , R1, , 0, , 1, 1, 0.5 Hz, T, 2, , Figure 17.36, Schematic for Example 17.12., , So, in the transient dialog box, we select the Final Time as 6T 12 s,, the Print Step as 0.01 s, the Step Ceiling as 10 ms, the Center, Frequency as 0.5 Hz, and the output variable as V(1). (In fact, Fig. 17.34, is for this particular example.) When PSpice is run, the output file, contains the following result:, FOURIER COEFFICIENTS OF TRANSIENT RESPONSE V(1), DC COMPONENT = 4.989950E-01, HARMONIC, NO, , FREQUENCY, (HZ), , FOURIER, COMPONENT, , NORMALIZED, COMPONENT, , PHASE, (DEG), , NORMALIZED, PHASE (DEG), , 1, 2, 3, 4, 5, 6, 7, 8, 9, , 5.000E-01, 1.000E+00, 1.500E+00, 2.000E+00, 2.500E+00, 3.000E+00, 3.500E+00, 4.000E+00, 4.500E+00, , 6.366E-01, 2.012E-03, 2.122E-01, 2.016E-03, 1.273E-01, 2.024E-03, 9.088E-02, 2.035E-03, 7.065E-02, , 1.000E+00, 3.160E-03, 3.333E-01, 3.167E-03, 1.999E-01, 3.180E-03, 1.427E-01, 3.197E-03, 1.110E-01, , -1.809E-01, -9.226E+01, -5.427E-01, -9.451E+01, -9.048E-01, -9.676E+01, -1.267E+00, -9.898E+01, -1.630E+00, , 0.000E+00, -9.208E+01, -3.619E-01, -9.433E+01, -7.239E-01, -9.658E+01, -1.086E+00, -9.880E+01, -1.449E+00
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ale29559_ch17.qxd, , 07/17/2008, , 01:31 PM, , Page 790, , Chapter 17, , 790, , The Fourier Series, , Comparing the result with that in Eq. (17.1.7) (see Example 17.1) or, with the spectra in Fig. 17.4 shows a close agreement. From Eq. (17.1.7),, the dc component is 0.5 while PSpice gives 0.498995. Also, the signal, has only odd harmonics with phase cn 90, whereas PSpice seems, to indicate that the signal has even harmonics although the magnitudes, of the even harmonics are small., , ■ METHOD 2 FFT Approach: With voltage marker in Fig. 17.36, in place, we run PSpice and obtain the waveform V(1) shown in, Fig. 17.37(a) on the PSpice A/D window. By double clicking the FFT, icon in the PSpice A/D menu and changing the X-axis setting to 0 to, 10 Hz, we obtain the FFT of V(1) as shown in Fig. 17.37(b). The FFTgenerated graph contains the dc and harmonic components within the, selected frequency range. Notice that the magnitudes and frequencies, of the harmonics agree with the DFT-generated tabulated values., 1.0 V, , 0 V, 0 s, , 2 s, , 4 s, , 6 s, Time, , V(1), , 8 s, , 10 s, , 12 s, , (a), 1.0 V, , 0 V, 0 Hz, V(1), , 2 Hz, , 4 Hz, 6 Hz, Frequency, , 8 Hz, , 10 Hz, , (b), , Figure 17.37, (a) Original waveform of Fig. 17.1, (b) FFT of the waveform., , Practice Problem 17.12, , Obtain the Fourier coefficients of the function in Fig. 17.7 using PSpice., Answer:, , FOURIER COEFFICIENTS OF TRANSIENT RESPONSE V(1), DC COMPONENT = 4.950000E-01, HARMONIC, NO, , FREQUENCY, (HZ), , FOURIER, COMPONENT, , NORMALIZED, COMPONENT, , PHASE, (DEG), , NORMALIZED, PHASE (DEG), , 1, 2, 3, , 1.000E+00, 2.000E+00, 3.000E+00, , 3.184E-01, 1.593E-01, 1.063E-01, , 1.000E+00, 5.002E-01, 3.338E-01, , -1.782E+02, -1.764E+02, -1.746E+02, , 0.000E+00, 1.800E+00, 3.600E+00, (continued)
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ale29559_ch17.qxd, , 07/17/2008, , 01:31 PM, , Page 791, , 17.7, , (continued), 4, 5, 6, 7, 8, 9, , 4.000E+00, 5.000E+00, 6.000E+00, 7.000E+00, 8.000E+00, 9.000E+00, , Fourier Analysis with PSpice, , 7.979E-02, 6.392E-01, 5.337E-02, 4.584E-02, 4.021E-02, 3.584E-02, , 2.506E-03, 2.008E-01, 1.676E-03, 1.440E-01, 1.263E-01, 1.126E-01, , 791, , -1.728E+02, -1.710E+02, -1.692E+02, -1.674E+02, -1.656E+02, -1.638E+02, , 5.400E+00, 7.200E+00, 9.000E+00, 1.080E+01, 1.260E+01, 1.440E+01, , Example 17.13, , If vs 12 sin(200 p t) u (t) V in the circuit of Fig. 17.38, find i(t)., , 1Ω, , Solution:, 1. Define. Although the problem appears to be clearly stated, it, might be advisable to check with the individual who assigned, the problem to make sure he or she wants the transient response, rather than the steady-state response; in the latter case the, problem becomes trivial., 2. Present. We are to determine the response i(t) given the input, vs(t), using PSpice and Fourier analysis., 3. Alternative. We will use DFT to perform the initial analysis. We, will then check using the FFT approach., 4. Attempt. The schematic is shown in Fig. 17.39. We may use the, DFT approach to obtain the Fourier coefficents of i(t). Since the, period of the input waveform is T 1100 10 ms, in the, Transient dialog box we select Print Step: 0.1 ms, Final Time:, 100 ms, Center Frequency: 100 Hz, Number of harmonics: 4,, and Output Vars: I(L1). When the circuit is simulated, the output, file includes the following:, , i(t), vs +, −, , 1Ω, , 1H, , Figure 17.38, For Example 17.13., R1, , I, , 1, VAMPL=12, FREQ=100 + V1, R2, −, VOFF=0, , 1, , 1H, , L1, , 0, , Figure 17.39, Schematic of the circuit in Fig. 17.38., , FOURIER COEFFICIENTS OF TRANSIENT RESPONSE I(VD), DC COMPONENT = 8.583269E-03, HARMONIC, NO, , FREQUENCY, (HZ), , FOURIER, COMPONENT, , NORMALIZED, COMPONENT, , PHASE, (DEG), , NORMALIZED, PHASE (DEG), , 1, 2, 3, 4, , 1.000E+02, 2.000E+02, 3.000E+02, 4.000E+02, , 8.730E-03, 1.017E-04, 6.811E-05, 4.403E-05, , 1.000E+00, 1.165E-02, 7.802E-03, 5.044E-03, , -8.984E+01, -8.306E+01, -8.235E+01, -8.943E+01, , 0.000E+00, 6.783E+00, 7.490E+00, 4.054E+00, , With the Fourier coefficients, the Fourier series describing, the current i(t) can be obtained using Eq. (17.73); that is,, i(t) 8.5833 8.73 sin(2 p 100t 89.84), 0.1017 sin(2 p 200t 83.06), 0.068 sin(2 p 300t 82.35) p mA, 5. Evaluate. We can also use the FFT approach to cross-check our, result. The current marker is inserted at pin 1 of the inductor as, shown in Fig. 17.39. Running PSpice will automatically produce, the plot of I(L1) in the PSpice A/D window, as shown in
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ale29559_ch17.qxd, , 07/17/2008, , 01:31 PM, , Page 792, , Chapter 17, , 792, , The Fourier Series, 20 mA, , –20 mA, 0 s, 20 ms, I (L1), , 40 ms 60 ms, Time, , 80 ms, , 100 ms, , (a), 10 mA, , 0 A, 0 Hz, 40 Hz, I (L1), , 80 Hz 120 Hz 160 Hz 200 Hz, Frequency, (b), , Figure 17.40, For Example 17.13: (a) plot of i(t), (b) the FFT of i(t)., , Fig. 17.40(a). By double clicking the FFT icon and setting the, range of the X-axis from 0 to 200 Hz, we generate the FFT of, I(L1) shown in Fig. 17.40(b). It is clear from the FFT-generated, plot that only the dc component and the first harmonic are, visible. Higher harmonics are negligibly small., One final observation, does the answer make sense? Let us, look at the actual transient response, i(t) (9.549e0.5t , 9.549) cos(200 pt) u (t) mA. The period of the cosine wave is, 10 ms while the time constant of the exponential is 2000 ms, (2 seconds). So, the answer we obtained by Fourier techniques, does agree., 6. Satisfactory? Clearly, we have solved the problem satisfactorily, using the specified approach. We can now present our results as, a solution to the problem., , Practice Problem 17.13, +, v (t), −, , is(t), , 10 Ω, , 2F, , A sinusoidal current source of amplitude 4 A and frequency 2 kHz is, applied to the circuit in Fig. 17.41. Use PSpice to find v(t)., Answer: v(t) 150.72 145.5 sin(4 p 103t 90) p mV., The Fourier components are shown below:, , Figure 17.41, For Practice Prob. 17.13., FOURIER COEFFICIENTS OF TRANSIENT RESPONSE V(R1:1), DC COMPONENT = -1.507169E-04, HARMONIC, NO, , FREQUENCY, (HZ), , FOURIER, COMPONENT, , NORMALIZED, COMPONENT, , PHASE, (DEG), , NORMALIZED, PHASE (DEG), , 1, 2, 3, 4, , 2.000E+03, 4.000E+03, 6.000E+03, 8.000E+03, , 1.455E-04, 1.851E-06, 1.406E-06, 1.010E-06, , 1.000E+00, 1.273E-02, 9.662E-03, 6.946E-02, , 9.006E+01, 9.597E+01, 9.323E+01, 8.077E+01, , 0.000E+00, 5.910E+00, 3.167E+00, -9.292E+00
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ale29559_ch17.qxd, , 07/17/2008, , 01:31 PM, , Page 793, , 17.8, , 17.8, , Applications, , 793, , Applications, , We demonstrated in Section 17.4 that the Fourier series expansion permits the application of the phasor techniques used in ac analysis to circuits containing nonsinusoidal periodic excitations. The Fourier series, has many other practical applications, particularly in communications, and signal processing. Typical applications include spectrum analysis,, filtering, rectification, and harmonic distortion. We will consider two, of these: spectrum analyzers and filters., , 17.8.1 Spectrum Analyzers, The Fourier series provides the spectrum of a signal. As we have seen,, the spectrum consists of the amplitudes and phases of the harmonics, versus frequency. By providing the spectrum of a signal f (t), the, Fourier series helps us identify the pertinent features of the signal. It, demonstrates which frequencies are playing an important role in the, shape of the output and which ones are not. For example, audible, sounds have significant components in the frequency range of 20 Hz, to 15 kHz, while visible light signals range from 105 GHz to 106 GHz., Table 17.4 presents some other signals and the frequency ranges of, their components. A periodic function is said to be band-limited if its, amplitude spectrum contains only a finite number of coefficients An or, cn. In this case, the Fourier series becomes, N, , N, , f (t) a cne jn0t a0 a An cos(n0 t fn), nN, , (17.75), , n1, , This shows that we need only 2N 1 terms (namely, a0, A1, A2, p , AN,, f1, f2, p , fN) to completely specify f (t) if 0 is known. This leads, to the sampling theorem: a band-limited periodic function whose, Fourier series contains N harmonics is uniquely specified by its values, at 2N 1 instants in one period., A spectrum analyzer is an instrument that displays the amplitude, of the components of a signal versus frequency. It shows the various, frequency components (spectral lines) that indicate the amount of, energy at each frequency., It is unlike an oscilloscope, which displays the entire signal (all, components) versus time. An oscilloscope shows the signal in the time, domain, while the spectrum analyzer shows the signal in the frequency, domain. There is perhaps no instrument more useful to a circuit analyst than the spectrum analyzer. An analyzer can conduct noise and spurious signal analysis, phase checks, electromagnetic interference and, filter examinations, vibration measurements, radar measurements, and, more. Spectrum analyzers are commercially available in various sizes, and shapes. Figure 17.42 displays a typical one., , 17.8.2 Filters, Filters are an important component of electronics and communications, systems. Chapter 14 presented a full discussion on passive and active, filters. Here, we investigate how to design filters to select the fundamental component (or any desired harmonic) of the input signal and, reject other harmonics. This filtering process cannot be accomplished, , TABLE 17.4, , Frequency ranges of typical, signals., Signal, , Frequency Range, , Audible sounds, AM radio, Short-wave radio, Video signals, (U.S. standards), VHF television,, FM radio, UHF television, Cellular telephone, Microwaves, Visible light, X-rays, , 20 Hz to 15 kHz, 540–1600 kHz, 3–36 MHz, dc to 4.2 MHz, 54–216 MHz, 470–806 MHz, 824–891.5 MHz, 2.4–300 GHz, 105–106 GHz, 108–109 GHz
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ale29559_ch17.qxd, , 794, , 07/10/2008, , 05:29 PM, , Page 794, , Chapter 17, , The Fourier Series, , Figure 17.42, A typical spectrum analyzer. © SETI Institute/SPL/Photo, Researchers, Inc., , without the Fourier series expansion of the input signal. For the purpose, of illustration, we will consider two cases, a lowpass filter and a bandpass filter. In Example 17.6, we already looked at a highpass RL filter., The output of a lowpass filter depends on the input signal, the, transfer function H () of the filter, and the corner or half-power frequency c. We recall that c 1RC for an RC passive filter. As shown, in Fig. 17.43(a), the lowpass filter passes the dc and low-frequency components, while blocking the high-frequency components. By making c, sufficiently large (c W 0, e.g., making C small), a large number of, the harmonics can be passed. On the other hand, by making c sufficiently small (c V 0), we can block out all the ac components and, pass only dc, as shown typically in Fig. 17.43(b). (See Fig. 17.2(a) for, the Fourier series expansion of the square wave.), , |H |, 1, 1, 2, 0, , 0 20 30 , , 0, , 0, , c, , 0 20 30 , , , , (a), A, , Lowpass, filter, , A, 2, , dc, , c << 0, (b), , Figure 17.43, (a) Input and output spectra of a lowpass filter, (b) the lowpass filter passes, only the dc component when c V 0.
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 795, , 17.8, , Applications, , 795, , Similarly, the output of a bandpass filter depends on the input signal,, the transfer function of the filter H (), its bandwidth B, and its center, frequency c. As illustrated in Fig. 17.44(a), the filter passes all the harmonics of the input signal within a band of frequencies (1 6 6 2), centered around c. We have assumed that 0, 20, and 30 are within, that band. If the filter is made highly selective (B V 0) and c 0,, where 0 is the fundamental frequency of the input signal, the filter, passes only the fundamental component (n 1) of the input and blocks, out all higher harmonics. As shown in Fig. 17.44(b), with a square wave, as input, we obtain a sine wave of the same frequency as the output., (Again, refer to Fig. 17.2(a).), , In this section, we have used c for the, center frequency of the bandpass filter, instead of 0 as in Chapter 14, to avoid, confusing 0 with the fundamental frequency of the input signal., , |H|, , 0, , 1, 1, 2, , 0 20 30 , , 0 0 20 30 , , 0 1, , c 2, , , , (a), Bandpass, filter, T, , c = 0, B << 0, , T, , (b), , Figure 17.44, (a) Input and output spectra of a bandpass filter, (b) the bandpass filter passes, only the fundamental component when B V 0., , If the sawtooth waveform in Fig. 17.45(a) is applied to an ideal lowpass, filter with the transfer function shown in Fig. 17.45(b), determine the, output., , –1, , x(t), , |H |, , 1, , 1, , 0, , 1, (a), , 2, , 3 t, , 0, , 10, , , , (b), , Figure 17.45, For example 17.14., , Solution:, The input signal in Fig. 17.45(a) is the same as the signal in Fig. 17.9., From Practice Prob. 17.2, we know that the Fourier series expansion is, x (t) , , 1, 1, 1, 1, sin 0t , sin 20 t , sin 30 t p, p, 2, 2p, 3p, , Example 17.14
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 796, , Chapter 17, , 796, , The Fourier Series, , where the period is T 1 s and the fundamental frequency is 0 , 2 p rad/s. Since the corner frequency of the filter is c 10 rad/s, only, the dc component and harmonics with n0 6 10 will be passed. For, n 2, n0 4 p 12.566 rad/s, which is higher than 10 rad/s, meaning, that second and higher harmonics will be rejected. Thus, only the dc, and fundamental components will be passed. Hence, the output of the, filter is, 1, 1, sin 2 p t, p, 2, , y (t) , , Practice Problem 17.14, |H |, 1, , Rework Example 17.14 if the lowpass filter is replaced by the ideal, bandpass filter shown in Fig. 17.46., Answer: y (t) , , 0, , 15, , 1, 1, 1, sin 30t , sin 40t , sin 50t., 3p, 4p, 5p, , 35 , , Figure 17.46, For Practice Prob. 17.14., , 17.9, , Summary, , 1. A periodic function is one that repeats itself every T seconds; that, is, f (t nT) f (t), n 1, 2, 3, p ., 2. Any nonsinusoidal periodic function f (t) that we encounter in electrical engineering can be expressed in terms of sinusoids using, Fourier series:, , , i, , b, , f (t) a0 a (an cos n0t bn sin n0t), n1, dc, ac, where 0 2 pT is the fundamental frequency. The Fourier series, resolves the function into the dc component a0 and an ac component containing infinitely many harmonically related sinusoids., The Fourier coefficients are determined as, a0 , , 1, T, , , , T, , an , , f (t) d t,, , 0, , bn , , 2, T, , , , 2, T, , , , T, , f (t) cos n0t d t, , 0, , T, , f (t) sin n0t dt, , 0, , If f (t) is an even function, bn 0, and when f (t) is odd, a0 0, and an 0. If f(t) is half-wave symmetric, a0 an bn 0 for, even values of n., 3. An alternative to the trigonometric (or sine-cosine) Fourier series, is the amplitude-phase form, , , f (t) a0 a An cos(n0t fn), n1
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 797, , 17.9, , Summary, , where, bn, fn tan1 a, , An 2a 2n b 2n,, , n, , 4. Fourier series representation allows us to apply the phasor method, in analyzing circuits when the source function is a nonsinusoidal, periodic function. We use phasor technique to determine the, response of each harmonic in the series, transform the responses, to the time domain, and add them up., 5. The average-power of periodic voltage and current is, P Vdc Idc , , 1 , a Vn In cos(un fn), 2 n1, , In other words, the total average power is the sum of the average, powers in each harmonically related voltage and current., 6. A periodic function can also be represented in terms of an exponential (or complex) Fourier series as, , , f (t) a cne jn0t, n, , where, cn , , 1, T, , , , T, , f (t)ejn0t d t, , 0, , and 0 2 pT. The exponential form describes the spectrum of, f(t) in terms of the amplitude and phase of ac components at positive and negative harmonic frequencies. Thus, there are three basic, forms of Fourier series representation: the trigonometric form, the, amplitude-phase form, and the exponential form., 7. The frequency (or line) spectrum is the plot of An and fn or 0cn 0 and, unversus frequency., 8. The rms value of a periodic function is given by, Frms , , B, , a20 , , 1 2, a An, 2 n1, , The power dissipated by a 1- resistance is, P1 F 2rms a 20 , , , 1 , (a 2n b 2n ) a 0cn 0 2, a, 2 n1, n, , This relationship is known as Parseval’s theorem., 9. Using PSpice, a Fourier analysis of a circuit can be performed in, conjunction with the transient analysis., 10. Fourier series find application in spectrum analyzers and filters., The spectrum analyzer is an instrument that displays the discrete, Fourier spectra of an input signal, so that an analyst can determine, the frequencies and relative energies of the signal’s components., Because the Fourier spectra are discrete spectra, filters can be, designed for great effectiveness in blocking frequency components, of a signal that are outside a desired range., , 797
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 798, , Chapter 17, , 798, , The Fourier Series, , Review Questions, 17.1, , Which of the following cannot be a Fourier series?, 2, , (a) t , , 3, , 4, , 17.6, , 5, , t, t, t, t, , 2, 3, 4, 5, , (b) 5 sin t 3 sin 2t 2 sin 3t sin 4t, (c) sin t 2 cos 3t 4 sin 4t cos 4t, , 17.7, , (d) sin t 3 sin 2.7t cos p t 2 tan p t, j p t, , (e) 1 e, 17.2, , 17.4, , 17.5, , (c) p, , (b) 2, , 17.8, , (a) t t 2, , (b) t 2 cos t, , (d) t 2 t 4, , (e) sinh t, , (c) e t, , (b) t sin t, (d) t 3 cos t, , (e) 1, , The function in Fig. 17.14 is half-wave symmetric., (b) False, , The plot of 0cn 0 versus n0 is called:, , (c) complex phase spectrum, 17.9, , Which of the following are odd functions?, (c) t ln t, , (d) 6, , (c) 9, , (b) complex amplitude spectrum, , 2, , (a) sin t cos t, , (b) 11, , (a) complex frequency spectrum, , (d) 2 p, , Which of the following are even functions?, , (a) 12, , (a) True, , If f (t) t, 0 6 t 6 p, f (t n p) f (t), the value, of 0 is, (a) 1, , 17.3, , ej2 p t, ej3 p t, , , 2, 3, , If f (t) 10 8 cos t 4 cos 3t 2 cos 5t p ,, the angular frequency of the 6th harmonic is, , When the periodic voltage 2 6 sin 0 t is applied, to a 1- resistor, the integer closest to the power, (in watts) dissipated in the resistor is:, (a) 5, , (b) 8, , (c) 20, , (d) 22, , (e) 40, , 17.10 The instrument for displaying the spectrum of a, signal is known as:, , (e) sinh t, , (a) oscilloscope, , (b) spectrogram, , If f (t) 10 8 cos t 4 cos 3t 2 cos 5t p ,, the magnitude of the dc component is:, , (c) spectrum analyzer, , (d) Fourier spectrometer, , (a) 10, , (b) 8, , (d) 2, , (e) 0, , (c) 4, , Answers: 17.1a,d, 17.2b, 17.3b,c,d, 17.4d,e, 17.5a, 17.6d,, 17.7a, 17.8b, 17.9d, 17.10c., , Problems, Section 17.2 Trigonometric Fourier Series, 17.1, , 17.3, , Evaluate each of the following functions and see if it, is periodic. If periodic, find its period., , Give the Fourier coefficients a0, an, and bn of the, waveform in Fig. 17.47. Plot the amplitude and, phase spectra., , (a) f (t) cos p t 2 cos 3 p t 3 cos 5 p t, , g(t), , (b) y(t) sin t 4 cos 2 p t, , 20, , (c) g(t) sin 3t cos 4t, (d) h(t) cos2 t, , 10, , (e) z(t) 4.2 sin(0.4 p t 10), 0.8 sin(0.6 p t 50), (f) p(t) 10, pt, , (g) q(t) e, 17.2, , Using MATLAB, synthesize the periodic waveform, for which the Fourier trigonometric Fourier series is, 1, 4, 1, 1, cos 5t p b, f (t) 2 acos t cos 3t , 2, 9, 25, p, , – 4 –3 –2 –1, , 0, , 1, , 2, , 3, , 4, , 5, , 6 t, , Figure 17.47, For Prob. 17.3., 17.4, , Find the Fourier series expansion of the backward, sawtooth waveform of Fig. 17.48. Obtain the, amplitude and phase spectra.
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 799, , Problems, , 17.9, , f (t), 30, , –4, , –2, , 0, , 2, , 4, , 799, , Determine the Fourier coefficients an and bn of the, first three harmonic terms of the rectified cosine, wave in Fig. 17.52., , 6 t, , f (t), , Figure 17.48, , 10, , For Probs. 17.4 and 17.66., 17.5, , Obtain the Fourier series expansion for the, waveform shown in Fig. 17.49., , −2, , 0, , 2, , 4, , 6, , 8, , 10, , t, , Figure 17.52, For Prob. 17.9., , z(t), , 17.10 Find the exponential Fourier series for the waveform, in Fig. 17.53., , 1, −, , , , 0, , 2, , 3, , t, v (t), , −2, , Vo, (a), , Figure 17.49, , , , 0, , For Prob. 17.5., , 2, , t, , 3, , Figure 17.53, 17.6, , Find the trigonometric Fourier series for, , f (t) e, , 20,, 40,, , 0 6 t 6 p, and, p 6 t 6 2p, , f (t 2p) f (t)., , *17.7 Determine the Fourier series of the periodic function, in Fig. 17.50., , For Prob. 17.10., 17.11 Obtain the exponential Fourier series for the signal, in Fig. 17.54., y (t), 1, , f (t), 6, , −1, , 0, , 1, , 2, , 3, , 4, , 5, , t, , Figure 17.54, –1, , 0, , 1, , 3 t(s), , 2, , –3, , For Prob. 17.11., *17.12 A voltage source has a periodic waveform defined, over its period as, , Figure 17.50, , v(t) t(2p t) V,, , For Prob. 17.7., , 0 6 t 6 2p, , Find the Fourier series for this voltage., 17.8, , Using Fig. 17.51, design a problem to help other, students better understand how to determine the, exponention Fourier series from a periodic wave shape., , 17.13 Design a problem to help other students better, understand obtaining the Fourier series from a, periodic function., 17.14 Find the quadrature (cosine and sine) form of the, Fourier series, , f (t), , , 10, np, cos a2nt , b, f (t) 2 a 3, 4, n, , 1, n1, , f (0), , 0, , t1, , t2, , t3, , t4, , Figure 17.51, For Prob. 17.8., , t5 t, , 17.15 Express the Fourier series, , 4, 1, cos 10nt 3 sin 10nt, f (t) 10 a 2, n, , 1, n, n1, , (a) in a cosine and angle form,, * An asterisk indicates a challenging problem., , (b) in a sine and angle form.
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 800, , Chapter 17, , 800, , The Fourier Series, , 17.16 The waveform in Fig. 17.55(a) has the following, Fourier series:, , f3(t), 2, , 1, 1, 4, 2 acos p t cos 3 p t, 2, 9, p, 1, cos 5 p t p b V, 25, , v1(t) , , 1, –4, , 0, –1, , –2, , 2, , t, , 4, , –2, , Obtain the Fourier series of v2(t) in Fig. 17.55(b)., , (c), , Figure 17.56, , v 1(t), , For Probs. 17.18 and 17.63., , 1, , –2, , –1, , 0, , 1, , 2, , 3, , 4 t, , 2, , 3, , 4 t, , (a), , 17.19 Obtain the Fourier series for the periodic waveform, in Fig. 17.57., , v 2 (t), 1, , –2, , –1, , 0, , 1, , –1, , –4, , –3, , –2, , –1, , 0, , 1, , 2, , 3, , 4, , 5, , 6, , Figure 17.57, , (b), , For Prob. 17.19., , Figure 17.55, For Probs. 17.16 and 17.69., , Section 17.3 Symmetry Considerations, 17.17 Determine if these functions are even, odd, or neither., (a) 1 t, , (b) t 2 1, , 17.20 Find the Fourier series for the signal in Fig. 17.58., Evaluate f (t) at t 2 using the first three nonzero, harmonics., , (c) cos n p t sin n p t, , t, , 2, , (d) sin p t, , (e) e, , f (t), , 17.18 Determine the fundamental frequency and specify, the type of symmetry present in the functions in, Fig. 17.56., –4, , f1(t), , –1, , –2, , 0, , 2, , 4, , 6, , 8 t, , Figure 17.58, For Probs. 17.20 and 17.67., , 2, , –2, , 4, , 0, , 1, , 2, , t, , 3, , 17.21 Determine the trigonometric Fourier series of the, signal in Fig. 17.59., , –2, (a), f2(t), , f (t), , 2, 1, , 2, , –2 –1, , 0, , 1, , 2, (b), , 3, , 4, , 5, , t, , –5 – 4 –3 –2 –1, , Figure 17.59, For Prob. 17.21., , 0, , 1, , 2, , 3, , 4, , 5 t
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 801, , Problems, , 801, f (t), , 17.22 Calculate the Fourier coefficients for the function in, Fig. 17.60., , 2, 1, f (t), , –2π, , –π, , π, , 0, , –1, , 2π, , 3π, , 4π, , t, , 4, –2, –5 – 4 –3 –2 –1, , 0, , 1, , 2, , 3, , 4, , Figure 17.62, , 5 t, , For Probs. 17.24 and 17.60., , Figure 17.60, For Prob. 17.22., , 17.25 Determine the Fourier series representation of the, function in Fig. 17.63., 17.23 Using Fig. 17.61, design a problem to help other, students better understand finding the Fourier series, of a periodic wave shape., , f (t), 1, –4, , f (t), , –2, , 0, , 2, , 4, , t, , –1, f (0), , Figure 17.63, For Prob. 17.25., , –t2, , –t1, , 0, , t1, , t2, , t3 t, , 17.26 Find the Fourier series representation of the signal, shown in Fig. 17.64., , –f (0), , Figure 17.61, f (t), , For Prob. 17.23., , 10, 5, , 17.24 In the periodic function of Fig. 17.62,, (a) find the trigonometric Fourier series coefficients, a2 and b2,, , – 4 –3 –2 –1, , 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , Figure 17.64, For Prob. 17.26., , (b) calculate the magnitude and phase of the, component of f (t) that has n 10 rad/s,, , 17.27 For the waveform shown in Fig. 17.65 below,, , (c) use the first four nonzero terms to estimate, f (p2),, , (a) specify the type of symmetry it has,, , (d) show that, , (b) calculate a3 and b3,, , 1, 1, 1, 1, 1, 1, p, , p, 4, 1, 3, 5, 7, 9, 11, , (c) find the rms value using the first five nonzero, harmonics., , f (t), 1, , –5, , –4, , –3, , –2, , –1, , 0, –1, , Figure 17.65, For Prob. 17.27., , 1, , 2, , 3, , 4, , 5, , t, , t(s)
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 802, , Chapter 17, , 802, , 17.28 Obtain the trigonometric Fourier series for the, voltage waveform shown in Fig. 17.66., , The Fourier Series, , 17.33 In the circuit shown in Fig. 17.69, the Fourier series, expansion of vs(t) is, vs(t) 3 , , v (t), 2, , –2, , –1, , 4 1, a n sin(n p t), p n1, , Find vo(t)., 0, , 1, , 2, , 3, , 4 t, , –2, 10 Ω, , Figure 17.66, For Prob. 17.28., 17.29 Determine the Fourier series expansion of the, sawtooth function in Fig. 17.67., , vs (t), , +, −, , 2H, , 1, 4, , +, vo(t), −, , F, , Figure 17.69, , f (t), , For Prob. 17.33., , , , –2, , –, , , , 0, , 2, , t, , –, , 17.34 Using Fig. 17.70, design a problem to help other, students better understand circuit responses to a, Fourier series., , Figure 17.67, For Prob. 17.29., L, , R, , 17.30 (a) If f (t) is an even function, show that, 2, cn , T, , , , T2, , f (t) cos no t dt, , j2, T, , +, vo(t), −, , C, , 0, , (b) If f (t) is an odd function, show that, cn , , v (t) +, −, , , , T2, , f (t) sin no t dt, , Figure 17.70, For Prob. 17.34., , 0, , 17.31 Let an and bn be the Fourier series coefficients of, f (t) and let o be its fundamental frequency., Suppose f (t) is time-scaled to give h(t) f (at)., Express the a¿n and b¿n, and ¿o, of h(t) in terms of, an, bn, and o of f (t)., , 17.35 If vs in the circuit of Fig. 17.71 is the same as, function f2(t) in Fig. 17.56(b), determine the dc, component and the first three nonzero harmonics, of vo(t)., , Section 17.4 Circuit Applications, 17.32 Find i(t) in the circuit of Fig. 17.68 given that, , 1, is(t) 1 a 2 cos 3n t A, n, n1, , 1Ω, , vs +, −, , 2Ω, i(t), is, , Figure 17.68, For Prob. 17.32., , 1Ω, , 1H, , 1F, , 1Ω, , +, vo, −, , Figure 17.71, For Prob. 17.35., , 2H, , *17.36 Find the response io for the circuit in Fig. 17.72(a),, where vs(t) is shown in Fig. 17.72(b).
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 803, , Problems, , 803, , 5Ω, , 1Ω, io, , +, −, , vs, , vs, , 1H, , +, −, , 1H, , +, vo, −, , (b), , 100 mF, , Figure 17.74, , (a), , For Prob. 17.38., , vs, , 17.39 If the periodic voltage in Fig. 17.75(a) is applied to, the circuit in Fig. 17.75(b), find io(t)., , 10, , vs(t), , 3 t, , 2, , 1, , 0, , (b), , 7.5, , Figure 17.72, For Prob. 17.36., 2.5, , 17.37 If the periodic current waveform in Fig. 17.73(a) is, applied to the circuit in Fig. 17.73(b), find vo., , 0, , 1, , 2, , 3, , t, , (a), , is(t), 20 Ω, , 6, , 40 Ω, io(t), , 4, vs +, −, , 2, , 100 mH, , 50 mF, , 0, –2, (b), 0, , 1, , 2, , Figure 17.75, , t, , 3, , For Prob. 17.39., , (a), 2Ω, +, 1Ω, , is, , 3H, , vo, −, , *17.40 The signal in Fig. 17.76(a) is applied to the circuit in, Fig. 17.76(b). Find vo(t)., vs(t), 6, , (b), , Figure 17.73, 0, , For Prob. 17.37., , 1, , 2, , 3, , 4, , 5, , t, , (a), , 17.38 If the square wave shown in Fig. 17.74(a) is applied, to the circuit in Fig. 17.74(b), find the Fourier series, for vo(t)., vs +, −, , vs (t) V, , 2vx, , 1Ω, , −+, +, vx, −, , 0.25 F, , 5, (b), 0, , 1, , 2, (a), , 3, , t, , Figure 17.76, For Prob. 17.40., , 3Ω, , +, vo, −
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 804, , Chapter 17, , 804, , 17.41 The full-wave rectified sinusoidal voltage in, Fig. 17.77(a) is applied to the lowpass filter in, Fig. 17.77(b). Obtain the output voltage vo(t) of, the filter., , The Fourier Series, , If the current entering the terminal at higher potential, is, i(t) 6 4 cos(60 p t 10), 2 cos(120 p t 60) A, find:, , v in(t), , (a) the rms value of the voltage,, , 5, , (b) the rms value of the current,, (c) the average power absorbed by the circuit., –, , , , 0, , 2 t, , (a), 2H, , v in(t) +, −, , +, vo, −, , 10 Ω, , 0.1 F, , *17.44 Design a problem to help other students better, understand how to find the rms voltage across, and the rms current through an electrical element, given a Fourier series for both the current and the, voltage. In addition, have them calculate the, average power delivered to the element and the, power spectrum., 17.45 A series RLC circuit has R 10 , L 2 mH, and, C 40 mF. Determine the effective current and, average power absorbed when the applied voltage is, , (b), , Figure 17.77, , v(t) 100 cos 1000t 50 cos 2000t, , For Prob. 17.41., , 25 cos 3000t V, 17.42 The square wave in Fig. 17.78(a) is applied to the, circuit in Fig. 17.78(b). Find the Fourier series, of vo(t)., , (a) 5 cos 3t 2 cos(3t p3), (b) 8 sin(p t p4) 10 cos(p t p8), , vs (t) V, 10, , 0, , 17.46 Use MATLAB to plot the following sinusoids for, 0 6 t 6 5:, , 1, , 2, , 3, , 17.47 The periodic current waveform in Fig. 17.79 is, applied across a 2-k resistor. Find the percentage, of the total average power dissipation caused by the, dc component., , t, , –10, (a), , i(t), 40 nF, 4, , 10 kΩ, , vs, , +, −, , vo, , –1, , +, −, , 0, , 1, , 2, , 3 t, , –2, , Figure 17.79, (b), , For Prob. 17.47., , Figure 17.78, For Prob. 17.42., , 17.48 For the circuit in Fig. 17.80,, , Section 17.5 Average Power and RMS Values, 17.43 The voltage across the terminals of a circuit is, v(t) 30 20 cos(60 p t 45), 10 cos(60 p t 45) V, , i(t) 20 16 cos(10t 45), 12 cos(20t 60) mA, (a) find v(t), and, (b) calculate the average power dissipated in the, resistor.
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 805, , Problems, , 805, i(t), , 100 F, , i(t), , 2 kΩ, , +, v (t), −, , sin t, , 1, , –2, , Figure 17.80, , –, , , , 0, , 3 t, , 2, , Figure 17.82, , For Prob. 17.48., , For Prob. 17.55., , 17.49 (a) For the periodic waveform in Prob. 17.5, find the, rms value., (b) Use the first five harmonic terms of the Fourier, series in Prob. 17.5 to determine the effective, value of the signal., (c) Calculate the percentage error in the estimated, rms value of z(t) if, % error a, , estimated value, 1b, exact value, , 17.56 The Fourier series trigonometric representation of a, periodic function is, , 1, n, f (t) 10 a a 2, cos n p t 2, sin n p tb, n, , 1, n, 1, n1, Find the exponential Fourier series representation, of f (t)., 17.57 The coefficients of the trigonometric Fourier series, representation of a function are:, bn 0,, , 100, , 6, ,, n3 2, , an , , n 0, 1, 2, . . ., , If n 50n, find the exponential Fourier series for, the function., , Section 17.6 Exponential Fourier Series, 17.50 Obtain the exponential Fourier series for f (t) 2t,, 1 6 t 6 1, with f (t 2n) f (t) for all integer, values of n., 17.51 Design a problem to help other students better, understand how to find the exponential Fourier series, of a given periodic function., 17.52 Calculate the complex Fourier series for f (t) et,, p 6 t 6 p, with f (t 2p n) f (t) for all, integer values of n., 17.53 Find the complex Fourier series for f (t) et,, 0 6 t 6 1, with f (t n) f (t) for all integer, values of n., , 17.58 Find the exponential Fourier series of a function that, has the following trigonometric Fourier series, coefficients:, a0 , , p, ,, 4, , bn , , (1)n, ,, n, , an , , (1)n 1, pn2, , Take T 2 p., 17.59 The complex Fourier series of the function in, Fig. 17.83(a) is, , jej(2n1) t, 1, f (t) a, 2 n (2n 1)p, Find the complex Fourier series of the function h(t), in Fig. 17.83(b)., , 17.54 Find the exponential Fourier series for the function, in Fig. 17.81., , f (t), 1, , f (t), 2, –2, , –, , –4, , –3, , –1, , , , 0, , 1, , 2, , 3 t, , 2, , 3, , (a), , 0, , 1, , 2, , 3, , 4, , 5, , 6, , t, , h(t), , –1, 2, , Figure 17.81, For Prob. 17.54., –2, , –1, , 0, , 1, –2, , 17.55 Obtain the exponential Fourier series expansion, of the half-wave rectified sinusoidal current of, Fig. 17.82., , (b), , Figure 17.83, For Prob.17.59., , t
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 806, , Chapter 17, , 806, , The Fourier Series, n, , 17.60 Obtain the complex Fourier coefficients of the signal, in Fig. 17.62., 17.61 The spectra of the Fourier series of a function are, shown in Fig. 17.84. (a) Obtain the trigonometric, Fourier series. (b) Calculate the rms value of the, function., , 90, , 40, 0, , An, , 20, , 80, 60, , n0, , 6, –90, 4, (b), , Figure 17.85, For Prob. 17.62., , 2, , 0, , 1, , 2, , 1, , 1, 2, , 3, , 4, , n (rad/s), , n, , 1, , 2, , 3, , 17.64 Design a problem to help other students better, understand the amplitude and phase spectra of a, given Fourier series., , 4, n (rad/s), , 0, , –25°, , 17.63 Plot the amplitude spectrum for the signal f2(t) in, Fig. 17.56(b). Consider the first five terms., , 17.65 Given that, , 20, 3, f (t) a a 2 2 cos 2n t , sin 2n tb, n, p, n, p, n1, , –20°, , nodd, , plot the first five terms of the amplitude and phase, spectra for the function., , –35°, –50°, , Section 17.7 Fourier Analysis with PSpice, , Figure 17.84, For Prob. 17.61., 17.62 The amplitude and phase spectra of a truncated, Fourier series are shown in Fig. 17.85., (a) Find an expression for the periodic voltage using, the amplitude-phase form. See Eq. (17.10)., (b) Is the voltage an odd or even function of t?, , 17.66 Determine the Fourier coefficients for the waveform, in Fig. 17.48 using PSpice., 17.67 Calculate the Fourier coefficients of the signal in, Fig. 17.58 using PSpice., 17.68 Use PSpice to find the Fourier components of the, signal in Prob. 17.7., 17.69 Use PSpice to obtain the Fourier coefficients of the, waveform in Fig. 17.55(a)., , An, 12, , 17.70 Design a problem to help other students better, understand how to use PSpice to solve circuit, problems with periodic inputs., , 10, , 17.71 Use PSpice to solve Prob. 17.39., 8, , Section 17.8 Applications, , 5, 3, 0 20 40 60 80, (a), , 17.72, n0, , The signal displayed by a medical device can, be approximated by the waveform shown in, Fig. 17.86. Find the Fourier series representation, of the signal.
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ale29559_ch17.qxd, , 07/10/2008, , 05:29 PM, , Page 807, , Comprehensive Problems, , (a) Express i(t) in amplitude-phase form., , f (t), , (b) If i(t) flows through a 2- resistor, how many, watts of average power will be dissipated?, , 10, , –6, , –4, , –2, , 807, , 2, , 0, , 4, , 6 t, , –10, , Figure 17.86, For Prob. 17.72., 17.73 A spectrum analyzer indicates that a signal is made, up of three components only: 640 kHz at 2 V,, 644 kHz at 1 V, 636 kHz at 1 V. If the signal is, applied across a 10- resistor, what is the average, power absorbed by the resistor?, 17.74 A certain band-limited periodic current has only, three frequencies in its Fourier series representation:, dc, 50 Hz, and 100 Hz. The current may be, represented as, i(t) 4 6 sin 100 p t 8 cos 100 p t, 3 sin 200 p t 4 cos 200 p t A, , 17.75 Design a lowpass RC filter with a resistance, R 2 k. The input to the filter is a periodic, rectangular pulse train (see Table 17.3) with, A 1 V, T 10 ms, and t 1 ms. Select C such, that the dc component of the output is 50 times greater, than the fundamental component of the output., 17.76 A periodic signal given by vs(t) 10 V for, 0 6 t 6 1 and 0 V for 1 6 t 6 2 is applied to the, highpass filter in Fig. 17.87. Determine the value of, R such that the output signal vo(t) has an average, power of at least 70 percent of the average power of, the input signal., 1H, , Vs, , +, −, , 10 Ω, , R, , +, Vo, −, , Figure 17.87, For Prob. 17.76., , Comprehensive Problems, 17.77 The voltage across a device is given by, v(t) 2 10 cos 4t 8 cos 6t 6 cos 8t, 5 sin 4t 3 sin 6t sin 8t V, Find:, (a) the period of v(t),, (b) the average value of v(t),, (c) the effective value of v(t)., 17.78 A certain band-limited periodic voltage has only, three harmonics in its Fourier series representation., The harmonics have the following rms values:, fundamental 40 V, third harmonic 20 V, fifth, harmonic 10 V., (a) If the voltage is applied across a 5- resistor, find, the average power dissipated by the resistor., (b) If a dc component is added to the periodic, voltage and the measured power dissipated, increases by 5 percent, determine the value of, the dc component added., 17.79 Write a program to compute the Fourier coefficients, (up to the 10th harmonic) of the square wave in, Table 17.3 with A 10 and T 2., 17.80 Write a computer program to calculate the exponential, Fourier series of the half-wave rectified sinusoidal, , current of Fig. 17.82. Consider terms up to the 10th, harmonic., 17.81 Consider the full-wave rectified sinusoidal current in, Table 17.3. Assume that the current is passed, through a 1- resistor., (a) Find the average power absorbed by the resistor., (b) Obtain cn for n 1, 2, 3, and 4., (c) What fraction of the total power is carried by the, dc component?, (d) What fraction of the total power is carried by the, second harmonic (n 2)?, 17.82 A band-limited voltage signal is found to have the, complex Fourier coefficients presented in the table, below. Calculate the average power that the signal, would supply a 4- resistor., , n 0, , 0 cn 0, , 0, , 2, 3, 4, 5, , 10.0, 8.5, 4.2, 2.1, 0.5, 0.2, , Un, 0, 15, 30, 45, 60, 75
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 809, , c h a p t e r, , Fourier Transform, , 18, , Planning is doing today to make us better tomorrow because the future, belongs to those who make the hard decisions today., —Business Week, , Enhancing Your Skills and Your Career, Career in Communications Systems, Communications systems apply the principles of circuit analysis. A, communication system is designed to convey information from a, source (the transmitter) to a destination (the receiver) via a channel (the, propagation medium). Communications engineers design systems for, transmitting and receiving information. The information can be in the, form of voice, data, or video., We live in the information age—news, weather, sports, shopping,, financial, business inventory, and other sources make information available to us almost instantly via communications systems. Some obvious, examples of communications systems are the telephone network,, mobile cellular telephones, radio, cable TV, satellite TV, fax, and radar., Mobile radio, used by police and fire departments, aircraft, and various businesses is another example., The field of communications is perhaps the fastest growing area in, electrical engineering. The merging of the communications field with, computer technology in recent years has led to digital data communications networks such as local area networks, metropolitan area networks,, and broadband integrated services digital networks. For example, the, Internet (the “information superhighway”) allows educators, business, people, and others to send electronic mail from their computers worldwide, log onto remote databases, and transfer files. The Internet has hit, the world like a tidal wave and is drastically changing the way people do, business, communicate, and get information. This trend will continue., A communications engineer designs systems that provide highquality information services. The systems include hardware for generating,, transmitting, and receiving information signals. Communications engineers, are employed in numerous communications industries and places where, communications systems are routinely used. More and more government, agencies, academic departments, and businesses are demanding faster and, more accurate transmission of information. To meet these needs, communications engineers are in high demand. Therefore, the future is in communications and every electrical engineer must prepare accordingly., , Photo by Charles Alexander, , 809
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 810, , Chapter 18, , 810, , 18.1, , Fourier Transform, , Introduction, , Fourier series enable us to represent a periodic function as a sum of, sinusoids and to obtain the frequency spectrum from the series. The, Fourier transform allows us to extend the concept of a frequency spectrum to nonperiodic functions. The transform assumes that a nonperiodic function is a periodic function with an infinite period. Thus, the, Fourier transform is an integral representation of a nonperiodic function that is analogous to a Fourier series representation of a periodic, function., The Fourier transform is an integral transform like the Laplace, transform. It transforms a function in the time domain into the frequency domain. The Fourier transform is very useful in communications systems and digital signal processing, in situations where the, Laplace transform does not apply. While the Laplace transform can, only handle circuits with inputs for t 7 0 with initial conditions, the, Fourier transform can handle circuits with inputs for t 6 0 as well as, those for t 7 0., We begin by using a Fourier series as a stepping stone in defining, the Fourier transform. Then we develop some of the properties of the, Fourier transform. Next, we apply the Fourier transform in analyzing, circuits. We discuss Parseval’s theorem, compare the Laplace and, Fourier transforms, and see how the Fourier transform is applied in, amplitude modulation and sampling., , 18.2, , p(t), A, , , , 0, , t, , (a), f (t), A, , −T, , , , 0, , T, , t, , (b), , Figure 18.1, (a) A nonperiodic function, (b) increasing, T to infinity makes f(t) become the nonperiodic function in (a)., , Definition of the Fourier Transform, , We saw in the previous chapter that a nonsinusoidal periodic function, can be represented by a Fourier series, provided that it satisfies the, Dirichlet conditions. What happens if a function is not periodic? Unfortunately, there are many important nonperiodic functions—such as a, unit step or an exponential function—that we cannot represent by a, Fourier series. As we shall see, the Fourier transform allows a transformation from the time to the frequency domain, even if the function, is not periodic., Suppose we want to find the Fourier transform of a nonperiodic, function p(t), shown in Fig. 18.1(a). We consider a periodic function, f (t) whose shape over one period is the same as p(t), as shown in, Fig. 18.1(b). If we let the period T S , only a single pulse of width, t [the desired nonperiodic function in Fig. 18.1(a)] remains, because, the adjacent pulses have been moved to infinity. Thus, the function, f (t) is no longer periodic. In other words, f (t) p(t) as T S . It is, interesting to consider the spectrum of f (t) for A 10 and t 0.2, (see Section 17.6). Figure 18.2 shows the effect of increasing T, on the spectrum. First, we notice that the general shape of the, spectrum remains the same, and the frequency at which the envelope, first becomes zero remains the same. However, the amplitude of the, spectrum and the spacing between adjacent components both, decrease, while the number of harmonics increases. Thus, over a, range of frequencies, the sum of the amplitudes of the harmonics, remains almost constant. As the total “strength” or energy of the
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 811, , 18.2, , Definition of the Fourier Transform, 2, , = 0.2, , −5, , 0, , 5 Hz, , 0, , 5 Hz, , T=1, 1, , = 0.2, , −5, T=2, = 0.2, , 0.4, , −5, , 0, , 5 Hz, , T=5, , Figure 18.2, Effect of increasing T on the spectrum of the periodic pulse, trains in Fig. 18.1(b)., L. Balmer, Signals and Systems: An Introduction [London: Prentice-Hall,, 1991], p. 229., , components within a band must remain unchanged, the amplitudes, of the harmonics must decrease as T increases. Since f 1T, as T, increases, f or decreases, so that the discrete spectrum ultimately, becomes continuous., To further understand this connection between a nonperiodic function and its periodic counterpart, consider the exponential form of a, Fourier series in Eq. (17.58), namely,, , , f (t) a cne jn0t, , (18.1), , n, , where, cn , , 1, T, , , , T2, , f (t)ejn0t dt, , (18.2), , 2p, T, , (18.3), , T2, , The fundamental frequency is, 0 , , and the spacing between adjacent harmonics is, ¢ (n 1)0 n0 0 , , 2p, T, , (18.4), , 811
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 812, , 812, , Chapter 18, , Fourier Transform, , Substituting Eq. (18.2) into Eq. (18.1) gives, , 1, f (t) a c, T, n, , , , f (t)ejn0t dt d e jn0t, , T2, , , , ¢, a c, n 2 p, , , T2, , , 1, a c, 2 p n, , , , T2, , f (t)ejn0t dt d e jn0t, , T2, , , , T2, , T2, , (18.5), , f (t)ejn0t dt d ¢e jn0t, , If we let T S , the summation becomes integration, the incremental, spacing ¢ becomes the differential separation d, and the discrete, harmonic frequency n0 becomes a continuous frequency . Thus, as, T S ,, , , , , 1, , , , ¢, , 1, , d, , n0, , 1, , , , a, n, , , , (18.6), , so that Eq. (18.5) becomes, f (t) , Some authors use F ( j) instead of F (), to represent the Fourier transform., , 1, 2p, , , , , , , c, , , , f (t)ejt dt d e jt d, , , , (18.7), , The term in the brackets is known as the Fourier transform of f(t) and, is represented by F(). Thus,, , F() F[ f (t)] , , , , , , f (t)ejt dt, , (18.8), , , , where F is the Fourier transform operator. It is evident from Eq. (18.8), that:, The Fourier transform is an integral transformation of f (t) from the time, domain to the frequency domain., , In general, F() is a complex function; its magnitude is called the, amplitude spectrum, while its phase is called the phase spectrum. Thus, F() is the spectrum., Equation (18.7) can be written in terms of F(), and we obtain the, inverse Fourier transform as, , f (t) F1[F()] , , 1, 2p, , , , , , F()e jt d, , (18.9), , , , The function f(t) and its transform F() form the Fourier transform pairs:, f (t), , 3, , F(), , since one can be derived from the other., , (18.10)
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 813, , 18.2, , Definition of the Fourier Transform, , 813, , The Fourier transform F() exists when the Fourier integral in, Eq. (18.8) converges. A sufficient but not necessary condition that f(t) has, a Fourier transform is that it be completely integrable in the sense that, , , , , , , , 0 f (t) 0 dt 6 , , (18.11), , For example, the Fourier transform of the unit ramp function tu(t) does, not exist, because the function does not satisfy the condition above., To avoid the complex algebra that explicitly appears in the Fourier, transform, it is sometimes expedient to temporarily replace j with s, and then replace s with j at the end., , Find the Fourier transform of the following functions: (a) d(t t0),, (b) e j0t, (c) cos 0 t., Solution:, (a) For the impulse function,, F() F[d(t t0)] , , , , , , d(t t0)ejt dt ejt0, , (18.1.1), , , , where the sifting property of the impulse function in Eq. (7.32) has, been applied. For the special case t0 0, we obtain, F[d(t)] 1, , (18.1.2), , This shows that the magnitude of the spectrum of the impulse function, is constant; that is, all frequencies are equally represented in the, impulse function., (b) We can find the Fourier transform of e j0t in two ways. If we let, F() d( 0), then we can find f(t) using Eq. (18.9), writing, f (t) , , 1, 2p, , , , , , d( 0) e jt d, , , , Using the sifting property of the impulse function gives, f (t) , , 1 j0t, e, 2p, , Since F() and f(t) constitute a Fourier transform pair, so too must, 2 p d( 0) and e j0t,, F[e j0t] 2 p d( 0), Alternatively, from Eq. (18.1.2),, d(t) F1[1], Using the inverse Fourier transform formula in Eq. (18.9),, d(t) F1[1] , , 1, 2p, , , , , , , , 1e jt d, , (18.1.3), , Example 18.1
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , 814, , Page 814, , Chapter 18, , Fourier Transform, , or, , , , , , e jt d 2 p d(t), , (18.1.4), , , , Interchanging variables t and results in, , , , , , e jt dt 2 p d(), , (18.1.5), , , , Using this result, the Fourier transform of the given function is, F[e j0t] , , , , , , e j0t ejt dt , , , , , , , , e j(0) dt 2 p d(0 ), , , , Since the impulse function is an even function, with d(0 ) , d( 0),, F[e j0t] 2 p d( 0), , (18.1.6), , By simply changing the sign of 0, we readily obtain, F[ej0t] 2 p d( 0), , (18.1.7), , Also, by setting 0 0,, F[1] 2 p d(), , (18.1.8), , (c) By using the result in Eqs. (18.1.6) and (18.1.7), we get, e j0t ej0t, d, 2, 1, 1, F[e j0 t] F[ej0t], 2, 2, p d( 0) p d( 0), , F[cos 0 t] F c, , (18.1.9), , The Fourier transform of the cosine signal is shown in Fig. 18.3., , f (t), , F(), , , 1, 0, , t, , −0, , , 0, , 0 , , Figure 18.3, , Fourier transform of f (t) cos 0 t., , Practice Problem 18.1, , Determine the Fourier transforms of the following functions: (a) gate, function g(t) 4u(t 1) 8u(t 2), (b) 4d(t 2), (c) 10 sin 0 t., Answer: (a) (4ej 8ej2 )j, (b) 4e j2,, (c) j10 p [d( 0) d( 0)].
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 815, , 18.2, , Definition of the Fourier Transform, , 815, , Example 18.2, , Derive the Fourier transform of a single rectangular pulse of width t, and height A, shown in Fig. 18.4., , f (t), , Solution:, , A, , F() , , , , t2, , Aejt dt , , t2, , t2, , A jt, 2, e, j, t2, −, 2, , 2 A e jt2 ejt2, a, , b, , 2j, sin t2, t, At, At sinc, t2, 2, , , 2, , 0, , t, , Figure 18.4, A rectangular pulse; for Example 18.2., |F()|, , If we make A 10 and t 2 as in Fig. 17.27 (like in Section 17.6),, then, , 20, , F() 20 sinc , whose amplitude spectrum is shown in Fig. 18.5. Comparing Fig. 18.4, with the frequency spectrum of the rectangular pulses in Fig. 17.28,, we notice that the spectrum in Fig. 17.28 is discrete and its envelope, has the same shape as the Fourier transform of a single rectangular, pulse., , Obtain the Fourier transform of the function in Fig. 18.6., Answer:, , −3 −2 − 0, , 2 3 , , Figure 18.5, Amplitude spectrum of the rectangular, pulse in Fig. 18.4: for Example 18.2., , Practice Problem 18.2, , 10(cos 1), ., j, , f (t), 5, 1, −1, , 0, , t, , −5, , Figure 18.6, For Practice Prob. 18.2., , Example 18.3, , Obtain the Fourier transform of the “switched-on” exponential function, shown in Fig. 18.7., f (t), 1, , Solution:, From Fig. 18.7,, f (t) eatu(t) b, , eat,, 0,, , e−at, , t 7 0, t 6 0, , Hence,, F() , , , , , , , , f (t)ejt dt , , , , , , eat ejt dt , , 0, , 1, 1 (aj)t , 2 , e, , a j, a j, 0, , , , , , 0, , e(aj)t dt, , 0, , Figure 18.7, For Example 18.3., , t
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 816, , Chapter 18, , 816, , Practice Problem 18.3, , Fourier Transform, , Determine the Fourier transform of the “switched-off ” exponential, function in Fig. 18.8., , f (t), 6, , Answer:, , 6, ., a j, , e at, , 18.3, 0, , Figure 18.8, For Practice Prob. 18.3., , t, , Properties of the Fourier Transform, , We now develop some properties of the Fourier transform that are useful in finding the transforms of complicated functions from the transforms of simple functions. For each property, we will first state and, derive it, and then illustrate it with some examples., , Linearity, If F1() and F2() are the Fourier transforms of f1(t) and f2(t), respectively, then, F[a1 f1(t) a2 f2(t)] a1F1() a2 F2(), , (18.12), , where a1 and a2 are constants. This property simply states that the, Fourier transform of a linear combination of functions is the same as, the linear combination of the transforms of the individual functions., The proof of the linearity property in Eq. (18.12) is straightforward., By definition,, F[a1 f1(t) a2 f2(t)] , , , , , , , , , , , , , [a1 f1(t) a2 f2(t)]ejt dt, a1 f1(t)ejt dt , , , , a1F1() a2 F2(), , , , , , a2 f2(t)ejt dt, , , , (18.13), For example, sin 0 t 2j1 (e j0 t ej0 t ). Using the linearity, property,, 1, [F(e j0t ) F(ej0t )], 2j, p, [d( 0) d( 0)], j, j p [d( 0) d( 0), , F[sin 0 t] , , (18.14), , Time Scaling, If F() F[ f (t)], then, F[ f (at)] , , 1, , Fa b, a, 0a 0, , (18.15), , where a is a constant. Equation (18.15) shows that time expansion, ( 0 a 0 7 1) corresponds to frequency compression, or conversely, time
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 817, , 18.3, , Properties of the Fourier Transform, , compression ( 0a 0 6 1) implies frequency expansion. The proof of the, time-scaling property proceeds as follows., F[ f (at)] , , , , , , f (at)ejt dt, , (18.16), , , , If we let x at, so that dx a dt, then, F[ f (at)] , , , , , , f (x)ejxa, , , , 1, , dx, Fa b, a, a, a, , (18.17), , For example, for the rectangular pulse p(t) in Example 18.2,, F[p(t)] At sinc, , t, 2, , (18.18a), , Using Eq. (18.15),, F[p(2t)] , , At, t, sinc, 2, 4, , (18.18b), , It may be helpful to plot p(t) and p(2t) and their Fourier transforms., Since, p(t) c, , t, t, 6 t 6, 2, 2, otherwise, , A, , 0,, , (18.19a), , then replacing every t with 2t gives, p(2t) c, , t, t, t, t, 6 2t 6, A, 6 t 6, 2, 2 c, 4, 4, otherwise, 0, otherwise, , A, , 0,, , (18.19b), , showing that p(2t) is time compressed, as shown in Fig. 18.9(b). To, plot both Fourier transforms in Eq. (18.18), we recall that the sinc, function has zeros when its argument is n p, where n is an integer., Hence, for the transform of p(t) in Eq. (18.18a), t2 2 p f t2 , n p S f nt, and for the transform of p(2t) in Eq. (18.18b), t4 , 2 p f t4 n p S f 2nt. The plots of the Fourier transforms are, shown in Fig. 18.9, which shows that time compression corresponds, with frequency expansion. We should expect this intuitively, because, when the signal is squashed in time, we expect it to change more, rapidly, thereby causing higher-frequency components to exist., , Time Shifting, If F() F[ f (t)], then, F[ f (t t0)] ejt0 F(), , (18.20), , that is, a delay in the time domain corresponds to a phase shift in the, frequency domain. To derive the time shifting property, we note that, F[ f (t t0)] , , , , , , , , f (t t0)ejt dt, , (18.21), , 817
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ale29559_ch18.qxd, , 818, , 07/11/2008, , 09:55 AM, , Page 818, , Chapter 18, , Fourier Transform, F[ p(t)], A, , p(t), A, , −, 2, , , 2, , 0, , t, , −2, , , −3, , , 0, , −1, , , 2, , , 1, , , 3, , , f, , (a), p(2t), , F[ p(2t)], A, 2, , A, , −, 4, , 0, , , 4, , −4, , , t, , −2, , , 2, , , 0, , 4 f, , , (b), , Figure 18.9, The effect of time scaling: (a) transform of the pulse, (b) time compression, of the pulse causes frequency expansion., , If we let x t t0 so that dx dt and t x t0, then, F[ f (t t0)] , , , , , , f (x)ej(xt0) dx, , , jt0, , e, , , , (18.22), , , , jx, , f (x)e, , jt0, , dx e, , F(), , , , Similarly, F[ f (t t0)] e jt0 F()., For example, from Example 18.3,, F[eatu(t)] , , 1, a j, , (18.23), , The transform of f (t) e(t2)u(t 2) is, F() Fe(t2)u(t 2) , , ej2, 1 j, , (18.24), , Frequency Shifting (or Amplitude Modulation), This property states that if F() F[ f (t)], then, F f (t)e j0t F( 0), , (18.25), , meaning, a frequency shift in the frequency domain adds a phase shift, to the time function. By definition,, F[ f (t)e j0t] , , , , , , , , , , , , , , , f (t)e j0t ejt dt, (18.26), j(0)t, , f (t)e, , dt F( 0)
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 819, , 18.3, , Properties of the Fourier Transform, , 819, , For example, cos 0 t 12 (e j0 t ej0 t ). Using the property in, Eq. (18.25),, 1, 1, F[ f (t) cos 0 t] F[ f (t)e j0 t ] F[ f (t)ej0 t ], 2, 2, 1, 1, F( 0) F( 0), 2, 2, , (18.27), , This is an important result in modulation where frequency components, of a signal are shifted. If, for example, the amplitude spectrum of f (t), is as shown in Fig. 18.10(a), then the amplitude spectrum of f (t) cos 0 t, will be as shown in Fig. 18.10(b). We will elaborate on amplitude modulation in Section 18.7.1., , |F[ f (t)]|, , |F[ f (t) cos 0 t]|, , A, , 1, F ( + 0), 2, , −B, , 0, , B, , , , −0 − B 0, , −0 + B, , 1, F ( − 0), 2, , A, 2, 0 – B, , 0, (b), , (a), , Figure 18.10, , Amplitude spectra of: (a) signal f (t), (b) modulated signal f (t) cos 0 t., , Time Differentiation, Given that F() F[ f (t)], then, F[ f ¿(t)] jF(), , (18.28), , In other words, the transform of the derivative of f(t) is obtained by, multiplying the transform of f(t) by j. By definition,, f (t) F1 [F()] , , 1, 2p, , , , , , F()e jt d, , (18.29), , , , Taking the derivative of both sides with respect to t gives, f ¿(t) , , j, 2p, , , , , , F()e jt d jF1[F()], , , , or, F[ f ¿(t)] jF(), , (18.30), , Repeated applications of Eq. (18.30) give, F f (n)(t) ( j)nF(), , (18.31), , 0, , 0 + B, ,
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ale29559_ch18.qxd, , 820, , 07/11/2008, , 09:55 AM, , Page 820, , Chapter 18, , Fourier Transform, , For example, if f (t) eatu(t), then, f ¿(t) aeatu(t) eatd(t) af (t) eatd(t), , (18.32), , Taking the Fourier transforms of the first and last terms, we obtain, jF() aF() 1, , F() , , 1, , 1, a j, , (18.33), , which agrees with the result in Example 18.3., , Time Integration, Given that F() F[ f (t)], then, Fc, , , , t, , , , f (t) dt d , , F(), p F(0) d (), j, , (18.34), , that is, the transform of the integral of f(t) is obtained by dividing the, transform of f (t) by j and adding the result to the impulse term that, reflects the dc component F(0). Someone might ask, “How do we know, that when we take the Fourier transform for time integration, we should, integrate over the interval [, t] and not [, ]?” When we integrate over [, ], the result does not depend on time anymore, and, the Fourier transform of a constant is what we will eventually get. But, when we integrate over [, t], we get the integral of the function, from the past to time t, so that the result depends on t and we can take, the Fourier transform of that., If is replaced by 0 in Eq. (18.8),, F(0) , , , , , , f (t) dt, , (18.35), , , , indicating that the dc component is zero when the integral of f(t) over, all time vanishes. The proof of the time integration in Eq. (18.34) will, be given later when we consider the convolution property., For example, we know that F[d(t)] 1 and that integrating the, impulse function gives the unit step function [see Eq. (7.39a)]. By, applying the property in Eq. (18.34), we obtain the Fourier transform, of the unit step function as, F[u(t)] F c, , , , t, , , , d(t) dt d , , 1, p d(), j, , (18.36), , Reversal, If F() F[ f (t)], then, F[ f (t)] F() F*(), , (18.37), , where the asterisk denotes the complex conjugate. This property states, that reversing f(t) about the time axis reverses F() about the frequency, axis. This may be regarded as a special case of time scaling for which, a 1 in Eq. (18.15).
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 821, , 18.3, , Properties of the Fourier Transform, , 821, , For example, 1 u(t) u(t). Hence,, F[1] F[u(t)] F[u(t)], 1, , p d(), j, 1, p d(), j, 2 p d(), , Duality, This property states that if F() is the Fourier transform of f(t), then, the Fourier transform of F(t) is 2 p f (); we write, F[ f (t)] F(), , F[F(t)] 2 p f (), , 1, , (18.38), , This expresses the symmetry property of the Fourier transform. To, derive this property, we recall that, 1, 2p, , f (t) F1 [F()] , , , , , , F()e jt d, , , , or, 2 p f (t) , , , , , , F()e jt d, , (18.39), , , , Replacing t by t gives, 2 p f (t) , , , , , , F()ejt d, , , , If we interchange t and , we obtain, 2 p f () , , , , , , F(t)ejt dt F[F(t)], , (18.40), , , , as expected., For example, if f (t) e0 t 0, then, F() , , 2, 1, 2, , (18.41), , By the duality property, the Fourier transform of F(t) 2(t 2 1) is, 2 p f () 2 p e0 0, , (18.42), , Figure 18.11 shows another example of the duality property. It illustrates the fact that if f (t) d(t) so that F() 1, as in Fig. 18.11(a),, then the Fourier transform of F(t) 1 is 2 p f () 2 p d() as shown, in Fig. 18.11(b)., , Convolution, Recall from Chapter 15 that if x(t) is the input excitation to a circuit, with an impulse function of h(t), then the output response y(t) is given, by the convolution integral, y(t) h(t) * x(t) , , , , , , , , h(l) x (t l) dl, , (18.43), , Since f (t) is the sum of the signals in, Figs. 18.7 and 18.8, F () is the sum, of the results in Example 18.3 and, Practice Prob. 18.3.
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 822, , Chapter 18, , 822, , Fourier Transform, , f(t), , F(), , F(t), , F [F(t)], , 1, , 1, , 1, , 2f(), , 0, , t, , 0, , , , 0, , t, , 0, , , , (b), , (a), , Figure 18.11, A typical illustration of the duality property of the Fourier transform: (a) transform of impulse,, (b) transform of unit dc level., , If X(), H(), and Y() are the Fourier transforms of x(t), h(t), and y(t),, respectively, then, Y() F[h(t) * x(t)] H()X(), , (18.44), , which indicates that convolution in the time domain corresponds with, multiplication in the frequency domain., To derive the convolution property, we take the Fourier transform, of both sides of Eq. (18.43) to get, , , Y() , , , , , c, , , , , , h(l) x (t l) dl d ejt dt, , (18.45), , Exchanging the order of integration and factoring h(l), which does not, depend on t, we have, Y() , , , , , , , , h(l) c, , , , , , , , x(t l)ejt dt d dl, , For the integral within the brackets, let t t l so that t t l, and dt dt. Then,, Y() , , , , , , , , , , , , , , , The important relationship in Eq. (18.46), is the key reason for using the Fourier, transform in the analysis of linear, systems., , h(l) c, , , , , , , , jl, , h(l)e, , x(t)ej(tl) dt d dl, dl, , , , , , (18.46), jt, , x(t)e, , dt H() X (), , , , as expected. This result expands the phasor method beyond what was, done with the Fourier series in the previous chapter., To illustrate the convolution property, suppose both h(t) and, x(t) are identical rectangular pulses, as shown in Fig. 18.12(a) and, 18.12(b). We recall from Example 18.2 and Fig. 18.5 that the Fourier, transforms of the rectangular pulses are sinc functions, as shown in, Fig. 18.12(c) and 18.12(d). According to the convolution property,, the product of the sinc functions should give us the convolution of, the rectangular pulses in the time domain. Thus, the convolution, of the pulses in Fig. 18.12(e) and the product of the sinc functions, in Fig. 18.12(f ) form a Fourier pair., In view of the duality property, we expect that if convolution in, the time domain corresponds with multiplication in the frequency
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 823, , 18.3, , Properties of the Fourier Transform, , 823, , h(t), , x(t), A, , A, Convolution, , −T0, , T0, , −T0, , t, , (a), , T0, , t, , (b), , h(t) ∗ x(t), 2 A2T0, , −2T0, , 2T0, , t, , (e), , H() X(), (2 AT0)2, , − 1, 2T0, , 1, 2T0, , , , (f ), H(), , X(), , 2 AT0, , 2 AT0, Multiplication, , , 1, 2T0, , 1, 2T0, (d), , (c), , Figure 18.12, Graphical illustration of the convolution property., E. O. Brigham, The Fast Fourier Transform [Englewood Cliffs, NJ: Prentice Hall, 1974], p. 60., , domain, then multiplication in the time domain should have a correspondence in the frequency domain. This happens to be the case. If, f (t) f1(t) f2(t), then, F() F[ f1(t) f2(t)] , , 1, F1() * F2(), 2p, , (18.47), , or, F() , , 1, 2p, , , , , , , , F1(l)F2( l) dl, , (18.48), ,
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ale29559_ch18.qxd, , 824, , 07/17/2008, , 01:35 PM, , Page 824, , Chapter 18, , Fourier Transform, , which is convolution in the frequency domain. The proof of Eq. (18.48), readily follows from the duality property in Eq. (18.38)., Let us now derive the time integration property in Eq. (18.34). If, we replace x(t) with the unit step function u(t) and h(t) with f(t) in, Eq. (18.43), then, , , , , , f (l) u (t l) dl f (t) * u(t), , (18.49), , , , But by the definition of the unit step function,, u(t l) b, , 1, t l 7 0, 0, t l 7 0, , We can write this as, u(t l) b, , 1,, 0,, , l 6 t, l 7 t, , Substituting this into Eq. (18.49) makes the interval of integration, change from [, ] to [, t], and thus Eq. (18.49) becomes, , , , t, , f (l) dl u(t) * f (t), , , , Taking the Fourier transform of both sides yields, Fc, , , , t, , , , f (l) dl d U()F(), , (18.50), , But from Eq. (18.36), the Fourier transform of the unit step function is, U() , , 1, p d(), j, , Substituting this into Eq. (18.50) gives, Fc, , , , t, , , , f (l) dl d a, , 1, p d()b F(), j, F(), , p F(0) d (), j, , (18.51), , which is the time integration property of Eq. (18.34). Note that in, Eq. (18.51), F() d () F(0) d (), since d() is only nonzero at 0., Table 18.1 lists these properties of the Fourier transform. Table 18.2, presents the transform pairs of some common functions. Note the similarities between these tables and Tables 15.1 and 15.2., , TABLE 18.1, , Properties of the Fourier transform., Property, , f(t), , F(), , Linearity, , a1 f1(t) a2 f2(t), , a1F1() a2F2(), , Scaling, , f (at), , Time shift, Frequency shift, , f (t a), e j0t f (t), , , 1, Fa b, a, 0a 0, ja, F(), e, F( 0)
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ale29559_ch18.qxd, , 07/17/2008, , 01:35 PM, , Page 825, , 18.3, , TABLE 18.1, , Properties of the Fourier Transform, , (continued), , Property, , f(t), , F(), , Modulation, , cos(0 t) f (t), , 1, [F( 0) F( 0)], 2, , Time differentiation, , Time integration, , df, dt, d nf, dt n, , , , jF(), ( j)nF(), , t, , f (t) dt, , , , n, , Frequency differentiation, , t f (t), , Reversal, Duality, Convolution in t, , f (t), F(t), f1(t) * f2(t), , Convolution in , , f1(t) f2(t), , F(), p F(0) d (), j, dn, F(), dn, F() or F*(), 2 p f (), F1()F2(), 1, F1() * F2(), 2p, , ( j)n, , TABLE 18.2, , Fourier transform pairs., f(t), , F(), , d(t), 1, , 1, 2 p d(), , u(t), , p d() , , u(t t) u(t t), , 2, , 0t 0, sgn(t), eatu(t), eatu(t), t neatu(t), ea0 t 0, e j0t, sin 0 t, cos 0 t, eat sin 0 tu(t), eat cos 0 tu(t), , 1, j, , sin t, , 2, 2, 2, j, 1, a j, 1, a j, n!, (a j)n1, 2a, a2 2, 2 p d( 0), j p [d( 0) d( 0)], p [d( 0) d( 0)], 0, (a j)2 20, a j, (a j)2 20, , 825
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 826, , Chapter 18, , 826, , Example 18.4, , Fourier Transform, , Find the Fourier transforms of the following functions: (a) signum, function sgn(t), shown in Fig. 18.13, (b) the double-sided exponential, ea 0 t 0, and (c) the sinc function (sin t)t., , sgn(t), 1, , Solution:, , 0, , t, −1, , Figure 18.13, The signum function of Example 18.4., , (a) We can obtain the Fourier transform of the signum function in three, ways., , ■ METHOD 1 We can write the signum function in terms of the unit, step function as, sgn(t) f (t) u(t) u(t), But from Eq. (18.36),, U() F[u(t)] p d() , , 1, j, , Applying this and the reversal property, we obtain, F[sgn(t)] U() U(), 1, 1, 2, ap d() b ap d() , b, j, j, j, , ■ METHOD 2 since d() d(), another way of writing the, signum function in terms of the unit step function is, f (t) sgn(t) 1 2u(t), Taking the Fourier transform of each term gives, F() 2 p d() 2 ap d() , , 1, 2, b, j, j, , ■ METHOD 3 We can take the derivative of the signum function, in Fig. 18.13 and obtain, f ¿(t) 2d(t), Taking the transform of this,, jF() 2, , 1, , F() , , 2, j, , as obtained previously., (b) The double-sided exponential can be expressed as, f (t) ea 0 t 0 eatu(t) eatu(t) y(t) y(t), where y(t) eatu(t) so that Y() 1(a j). Applying the reversal, property,, Fea 0 t 0 Y() Y() a, , 1, 1, 2a, , b 2, a j, a j, a 2, , (c) From Example 18.2,, sin(t2), t, t, t, t sinc, F c u at b u at b d t, 2, 2, t2, 2
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 827, , 18.3, , Properties of the Fourier Transform, , 827, , Setting t2 1 gives, F[u(t 1) u(t 1)] 2, , sin , , , Applying the duality property yields, Fc2, , sin t, d 2 p [U( 1) U( 1)], t, , Fc, , sin t, d p [U( 1) U( 1)], t, , or, , Determine the Fourier transforms of these functions: (a) gate function, g(t) u(t) u(t 1), (b) f (t) te2tu(t), and (c) sawtooth pulse, p(t) 10t[u(t) u(t 2)]., Answer: (a) (1 ej ) c p d() , (c), , 20j j2, 10(ej2 1), e, , ., , 2, , Practice Problem 18.4, , 1, 1, d , (b), ,, j, (2 j)2, , Find the Fourier transform of the function in Fig. 18.14., , Example 18.5, , Solution:, The Fourier transform can be found directly using Eq. (18.8), but it is, much easier to find it using the derivative property. We can express the, function as, , f (t), , f (t) b, , −1, , 1 t, 1 6 t 6 0, 1 t,, 0 6 t 6 1, , Figure 18.14, For Example 18.5., , Its first derivative is shown in Fig. 18.15(a) and is given by, f ¿(t) b, , 1, 1 6 t 6 0, 1,, 0 6 t 6 1, , f ¿(t), , f ¿¿(t), 1, , 1, , 1, , 1, −1, , 0, , t, , −1, , −1, , 0, , 1, , −2, (a), , 1, , (b), , Figure 18.15, First and second derivatives of f (t) in Fig. 18.14; for, Example 18.5., , t, , 0, , 1, , t
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 828, , Chapter 18, , 828, , Fourier Transform, , Its second derivative is in Fig. 18.15(b) and is given by, f –(t) d(t 1) 2d(t) d(t 1), Taking the Fourier transform of both sides,, ( j)2 F() e j 2 ej 2 2 cos , or, F() , , Practice Problem 18.5, , 2(1 cos ), 2, , Determine the Fourier transform of the function in Fig. 18.16., Answer: (8 cos 3 4 cos 4 4 cos 2)2., , f (t), 2, , −4, , −2, , 0, , 2, , 4, , t, , Figure 18.16, For Practice Prob. 18.5., , Example 18.6, , Obtain the inverse Fourier transform of:, (a) F() , , 10 j 4, , (b) G() , , ( j)2 6 j 8, , 2 21, 2 9, , Solution:, (a) To avoid complex algebra, we can replace j with s for the moment., Using partial fraction expansion,, F(s) , , 10s 4, A, B, 10s 4, , , , (s 4) (s 2), s4, s2, s 2 6s 8, , where, A (s 4)F(s) 0 s4 , , 10s 4, 36, `, , 18, (s 2) s4, 2, , B (s 2)F(s) 0 s2 , , 10s 4, 16, `, , 8, (s 4) s2, 2, , Substituting A 18 and B 8 in F(s) and s with j gives, F( j) , , 18, 8, , j 4, j 2, , With the aid of Table 18.2, we obtain the inverse transform as, f (t) (18e4t 8e2t ) u (t), (b) We simplify G() as, G() , , 12, 2 21, 1 2, 2, 9, 9
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 829, , 18.4, , Circuit Applications, , 829, , With the aid of Table 18.2, the inverse transform is obtained as, g(t) d(t) 2e3 0 t 0, , Practice Problem 18.6, , Find the inverse Fourier transform of:, 6(3 j2), (1 j)(4 j)(2 j), 2(1 j), 1, , (b) Y() p d() , j, (1 j)2 16, (a) H() , , Answer: (a) h(t) (2et 3e2t 5e4t ) u (t),, (b) y(t) (1 2et cos 4t) u (t)., , 18.4, , Circuit Applications, , The Fourier transform generalizes the phasor technique to nonperiodic, functions. Therefore, we apply Fourier transforms to circuits with nonsinusoidal excitations in exactly the same way we apply phasor techniques, to circuits with sinusoidal excitations. Thus, Ohm’s law is still valid:, V() Z() I (), , (18.52), , where V() and I() are the Fourier transforms of the voltage and current and Z() is the impedance. We get the same expressions for the, impedances of resistors, inductors, and capacitors as in phasor analysis, namely,, R, L, , 1, 1, , C, , 1, , R, jL, 1, jC, , (18.53), , Once we transform the functions for the circuit elements into the frequency domain and take the Fourier transforms of the excitations, we, can use circuit techniques such as voltage division, source transformation, mesh analysis, node analysis, or Thevenin’s theorem, to find the, unknown response (current or voltage). Finally, we take the inverse, Fourier transform to obtain the response in the time domain., Although the Fourier transform method produces a response that, exists for 6 t 6 , Fourier analysis cannot handle circuits with, initial conditions., The transfer function is again defined as the ratio of the output, response Y() to the input excitation X(); that is,, H() , , Y(), X(), , (18.54)
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 830, , Chapter 18, , 830, , Fourier Transform, , or, Y() H() X (), X(), , Y(), , H(), , Figure 18.17, Input-output relationship of a circuit in, the frequency domain., , (18.55), , The frequency domain input-output relationship is portrayed in, Fig. 18.17. Equation (18.55) shows that if we know the transfer function and the input, we can readily find the output. The relationship in, Eq. (18.54) is the principal reason for using the Fourier transform in, circuit analysis. Notice that H() is identical to H(s) with s j. Also,, if the input is an impulse function [i.e., x(t) d(t)], then X() 1, so, that the response is, Y() H() F[h(t)], , (18.56), , indicating that H() is the Fourier transform of the impulse response h(t)., , Example 18.7, , Find vo(t) in the circuit of Fig. 18.18 for vi (t) 2e3tu(t)., , 2Ω, , vi (t), , +, −, , 1F, , +, vo(t), −, , Figure 18.18, , Solution:, The Fourier transform of the input voltage is, Vi () , , 2, 3 j, , and the transfer function obtained by voltage division is, , For Example 18.7., , H() , , 1j, Vo(), 1, , , Vi (), 2 1j, 1 j2, , Hence,, Vo() Vi ()H() , , 2, (3 j) (1 j2), , or, Vo() , , 1, (3 j) (0.5 j), , Vo() , , 0.4, 0.4, , 3 j, 0.5 j, , By partial fractions,, , Taking the inverse Fourier transform yields, vo(t) 0.4(e0.5t e3t ) u (t), , Practice Problem 18.7, , Answer: 5 10(1 e4t ) u (t) V., , 1H, , vi (t), , +, −, , Figure 18.19, For Practice Prob. 18.7., , Determine vo(t) in Fig. 18.19 if vi(t) 5sgn(t) (5 10u(t)) V., , 4Ω, , +, vo(t), −
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 831, , 18.4, , Circuit Applications, , 831, , Example 18.8, , Using the Fourier transform method, find io(t) in Fig. 18.20 when, is(t) 10 sin 2t A., , io(t), , Solution:, By current division,, , is (t), , j, Io(), 2, H() , , , Is(), 2 4 2j, 1 j3, , 2Ω, , 4Ω, 0.5 F, , Figure 18.20, For Example 18.8., , If is(t) 10 sin 2t, then, Is() j p 10[d( 2) d( 2)], Hence,, Io() H() Is () , , 10 p [d( 2) d( 2)], 1 j3, , The inverse Fourier transform of Io() cannot be found using Table 18.2., We resort to the inverse Fourier transform formula in Eq. (18.9) and, write, 1, 2p, , io(t) F1[Io ()] , , , , , , , , 10 p [d( 2) d( 2)] jt, e d, 1 j3, , We apply the sifting property of the impulse function, namely,, d( 0) f () f (0), or, , , , , , d( 0)f () d f (0), , , , and obtain, io(t) , , 10 p, 2, 2 j2t, c, e j2t , e, d, 2 p 1 j6, 1 j6, , 10 c, , e j2t, 6.082e j80.54, , , , ej2t, 6.082 ej80.54, , d, , 1.644e j(2t80.54) ej(2t80.54), 3.288 cos(2t 80.54) A, Find the current io(t) in the circuit in Fig. 18.21, given that is (t) , 5 cos 4t A., Answer: 2.95 cos(4t 26.57) A., , Practice Problem 18.8, 2H, , io(t), 10 Ω, , is (t), 6Ω, , Figure 18.21, For Practice Prob. 18.8.
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 832, , 832, , Chapter 18, , Fourier Transform, , 18.5, , Parseval’s Theorem, , Parseval’s theorem demonstrates one practical use of the Fourier transform. It relates the energy carried by a signal to the Fourier transform of, the signal. If p(t) is the power associated with the signal, the energy carried by the signal is, W, , , , , , p(t) dt, , (18.57), , , , In order to be able to compare the energy content of current and voltage signals, it is convenient to use a 1- resistor as the base for energy, calculation. For a 1- resistor, p(t) v2(t) i 2(t) f 2(t), where f(t), stands for either voltage or current. The energy delivered to the 1-, resistor is, W1 , , , , , , f 2(t) dt, , (18.58), , , , Parseval’s theorem states that this same energy can be calculated in the, frequency domain as, W1 , , , , , , f 2(t) dt , , , , , , 1, 2p, , , , 0F() 0 2 d, , , , (18.59), , Parseval’s theorem states that the total energy delivered to a 1-, resistor equals the total area under the square of f (t ) or 12 p times, the total area under the square of the magnitude of the Fourier transform of f (t )., In fact, 0 F () 0 2 is sometimes known as, the energy spectral density of signal f (t )., , Parseval’s theorem relates energy associated with a signal to its Fourier, transform. It provides the physical significance of F(), namely, that, 0F() 0 2 is a measure of the energy density (in joules per hertz) corresponding to f (t)., To derive Eq. (18.59), we begin with Eq. (18.58) and substitute, Eq. (18.9) for one of the f (t)’s. We obtain, W1 , , , , , , , , f 2(t) dt , , , , , , f (t) c, , , , 1, 2p, , , , , , , , F()e jt d d dt, , (18.60), , The function f (t) can be moved inside the integral within the brackets,, since the integral does not involve time:, W1 , , , , , , 1, 2p, , , , , , f (t)F()e jt d dt, , (18.61), , , , Reversing the order of integration,, 1, 2p, , , , 1, , 2p, , , , W1 , , , , , , , F() c, , , , , , , , f (t)ej()t dt d d, , 1, F()F() d , 2, p, , , , , (18.62), , , , , , F()F*() d
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 833, , 18.5, , Parseval’s Theorem, , 833, , But if z x jy, zz* (x jy)(x jy) x 2 y 2 0 z 0 2. Hence,, W1 , , , , , , f 2(t) dt , , , , 1, 2p, , , , , , , , 0F() 0 2 d, , (18.63), , as expected. Equation (18.63) indicates that the energy carried by a, signal can be found by integrating either the square of f (t) in the, time domain or 12 p times the square of F() in the frequency, domain., Since 0F() 0 2 is an even function, we may integrate from 0 to , and double the result; that is,, W1 , , , , , , f 2(t) dt , , , , 1, p, , , , , , 0, , 0F() 0 2 d, , (18.64), , We may also calculate the energy in any frequency band 1 6 6 2 as, W1 , , 1, p, , , , 2, , 1, , 0F() 0 2 d, , (18.65), , Notice that Parseval’s theorem as stated here applies to nonperiodic functions. Parseval’s theorem for periodic functions was presented, in Sections 17.5 and 17.6. As evident in Eq. (18.63), Parseval’s theorem shows that the energy associated with a nonperiodic signal, is spread over the entire frequency spectrum, whereas the energy of, a periodic signal is concentrated at the frequencies of its harmonic, components., , The voltage across a 10- resistor is v(t) 5e3tu(t) V. Find the total, energy dissipated in the resistor., Solution:, 1. Define. The problem is well defined and clearly stated., 2. Present. We are given the voltage across the resistor for all time, and are asked to find the energy dissipated by the resistor. We note, that the voltage is zero for all time less than zero. Thus, we only, need to consider the time from zero to infinity., 3. Alternative. There are basically two ways to find this answer., The first would be to find the answer in the time domain. We, will use the second approach to find the answer using Fourier, analysis., 4. Attempt. In the time domain,, W10 0.1, , , , , , f 2(t) dt 0.1, , , , 2.5, , 6t, , , , , , 25e6t dt, , 0, , , , e, 2.5, 2 , 416.7 mJ, 6 0, 6, , Example 18.9
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ale29559_ch18.qxd, , 07/11/2008, , 11:02 AM, , 834, , Page 834, , Chapter 18, , Fourier Transform, , 5. Evaluate. In the frequency domain,, 5, 3 j, , F() V() , so that, , 0 F() 0 2 F()F()* , , 25, 9 2, , Hence, the energy dissipated is, 0.1, 2p, , W10 , , , , , , , , 0F() 0 2 d , , 0.1, p, , , , , , 25, d, 9 2, , 0, , 2.5 1 p, 2.5, 2.5 1 1 , a ba b , a tan, b 2 , 416.7 mJ, , p 3, p 3 2, 3 0, 6, 6. Satisfactory? We have satisfactorily solved the problem and can, present the results as a solution to the problem., , Practice Problem 18.9, , (a) Calculate the total energy absorbed by a 1- resistor with, i(t) 5e2 0 t 0 A in the time domain. (b) Repeat (a) in the frequency, domain., Answer: (a) 12.5 J, (b) 12.5 J., , Example 18.10, , Calculate the fraction of the total energy dissipated by a 1- resistor, in the frequency band 10 6 6 10 rad/s when the voltage across, it is v(t) e2tu(t)., Solution:, Given that f (t) v(t) e2tu(t), then, F() , , 1, 2 j, , 0F() 0 2 , , 1, , 1, 4 2, , The total energy dissipated by the resistor is, W1 , , 1, p, , , , , , 0, , 0F() 0 2 d , , 1, p, , , , , , 0, , d, 4 2, , 1 1, , 1 1 p, a tan1, 2 b a b 0.25 J, p 2, p, 2 0, 2 2, The energy in the frequencies 10 6 6 10 rad/s is, W, , 1, p, , , , 10, , 0, , 0F() 0 2 d , , 1, p, , , , 10, , 0, , 10, d, 1 1, 1 , , a, tan, 2, b, p 2, 2 0, 4 2, , 1, 1 78.69, , tan1 5 , a, pb 0.218 J, 2p, 2 p 180
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , 18.6, , Page 835, , Comparing the Fourier and Laplace Transforms, , 835, , Its percentage of the total energy is, 0.218, W, , 87.4 %, W1, 0.25, , A 2- resistor has i(t) 2etu(t) A. What percentage of the total, energy is in the frequency band 4 6 6 4 rad/s?, , Practice Problem 18.10, , Answer: 84.4 percent., , 18.6, , Comparing the Fourier and, Laplace Transforms, , It is worthwhile to take some moments to compare the Laplace and, Fourier transforms. The following similarities and differences should, be noted:, 1. The Laplace transform defined in Chapter 15 is one-sided in that, the integral is over 0 6 t 6 , making it only useful for positivetime functions, f(t), t 7 0. The Fourier transform is applicable to, functions defined for all time., 2. For a function f (t) that is nonzero for positive time only (i.e.,, f (t) 0, t 6 0) and, related by, , , , , , 0, , 0 f (t) 0 dt 6 , the two transforms are, , F() F(s) 0 sj, , (18.66), , This equation also shows that the Fourier transform can be, regarded as a special case of the Laplace transform with s j., Recall that s s j. Therefore, Eq. (18.66) shows that the, Laplace transform is related to the entire s plane, whereas the, Fourier transform is restricted to the j axis. See Fig. 15.1., 3. The Laplace transform is applicable to a wider range of functions, than the Fourier transform. For example, the function tu(t) has a, Laplace transform but no Fourier transform. But Fourier transforms exist for signals that are not physically realizable and have, no Laplace transforms., 4. The Laplace transform is better suited for the analysis of transient, problems involving initial conditions, since it permits the inclusion, of the initial conditions, whereas the Fourier transform does not., The Fourier transform is especially useful for problems in the, steady state., 5. The Fourier transform provides greater insight into the frequency, characteristics of signals than does the Laplace transform., Some of the similarities and differences can be observed by comparing Tables 15.1 and 15.2 with Tables 18.1 and 18.2., , In other words, if all the poles of F (s), lie in the left-hand side of the s plane,, then one can obtain the Fourier transform F () from the corresponding, Laplace transform F (s) by merely replacing s by j . Note that this is not, the case, for example, for u (t ) or, cos atu (t ).
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ale29559_ch18.qxd, , 836, , 07/11/2008, , 09:55 AM, , Page 836, , Chapter 18, , 18.7, , Fourier Transform, , Applications, , Besides its usefulness for circuit analysis, the Fourier transform is used, extensively in a variety of fields such as optics, spectroscopy, acoustics,, computer science, and electrical engineering. In electrical engineering,, it is applied in communications systems and signal processing, where, frequency response and frequency spectra are vital. Here we consider, two simple applications: amplitude modulation (AM) and sampling., , 18.7.1 Amplitude Modulation, Electromagnetic radiation or transmission of information through, space has become an indispensable part of a modern technological, society. However, transmission through space is only efficient and, economical at high frequencies (above 20 kHz). To transmit intelligent signals—such as for speech and music—contained in the lowfrequency range of 50 Hz to 20 kHz is expensive; it requires a huge, amount of power and large antennas. A common method of transmitting low-frequency audio information is to transmit a high-frequency, signal, called a carrier, which is controlled in some way to correspond to the audio information. Three characteristics (amplitude, frequency, or phase) of a carrier can be controlled so as to allow it to, carry the intelligent signal, called the modulating signal. Here we will, only consider the control of the carrier’s amplitude. This is known as, amplitude modulation., Amplitude modulation (AM) is a process whereby the amplitude of, the carrier is controlled by the modulating signal., , AM is used in ordinary commercial radio bands and the video portion, of commercial television., Suppose the audio information, such as voice or music (or the, modulating signal in general) to be transmitted is m(t) Vm cos mt,, while the high-frequency carrier is c(t) Vc cos ct, where c W m., Then an AM signal f (t) is given by, f (t) Vc[1 m(t)] cos ct, , (18.67), , Figure 18.22 illustrates the modulating signal m(t), the carrier c(t), and, the AM signal f(t). We can use the result in Eq. (18.27) together with the, Fourier transform of the cosine function (see Example 18.1 or Table 18.1), to determine the spectrum of the AM signal:, F() F[Vc cos ct] F[Vc m(t) cos c t], Vc p[d( c) d( c)], Vc, [M( c) M( c)], 2, , (18.68), , where M() is the Fourier transform of the modulating signal m(t)., Shown in Fig. 18.23 is the frequency spectrum of the AM signal. Figure 18.23 indicates that the AM signal consists of the carrier and two, other sinusoids. The sinusoid with frequency c m is known as the, lower sideband, while the one with frequency c m is known as, the upper sideband.
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 837, , 18.7, , Applications, , 837, , |M()|, Lower, sideband, , m(t), , Carrier, , Upper, sideband, , Vm, −m 0 m, , t, , , , 0 c − m, , c, , c + m, , Figure 18.23, , (a), , Frequency spectrum of AM signal., |C()|, c(t), Vc, −c, , t, , c, , 0, , , , (b), f (t), , |F()|, , −c, , t, |, , c, , 0, |, , |, , , |, , 2m, , 2m, , (c), , Figure 18.22, Time domain and frequency display of: (a) modulating signal, (b) carrier, signal, (c) AM signal., , Notice that we have assumed that the modulating signal is sinusoidal to make the analysis easy. In real life, m(t) is a nonsinusoidal,, band-limited signal—its frequency spectrum is within the range, between 0 and u 2 p fu (i.e., the signal has an upper frequency, limit). Typically, fu 5 kHz for AM radio. If the frequency spectrum, of the modulating signal is as shown in Fig. 18.24(a), then the frequency spectrum of the AM signal is shown in Fig. 18.24(b). Thus,, to avoid any interference, carriers for AM radio stations are spaced, 10 kHz apart., At the receiving end of the transmission, the audio information is recovered from the modulated carrier by a process known as, demodulation., , |F()|, | M()|, , Carrier, , m, , 0, (a), , , , 0, , c − m, , c, (b), , Figure 18.24, Frequency spectrum of: (a) modulating signal, (b) AM signal., , c + m, , , ,
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 838, , Chapter 18, , 838, , Example 18.11, , Fourier Transform, , A music signal has frequency components from 15 Hz to 30 kHz. If, this signal could be used to amplitude modulate a 1.2-MHz carrier, find, the range of frequencies for the lower and upper sidebands., Solution:, The lower sideband is the difference of the carrier and modulating, frequencies. It will include the frequencies from, 1,200,000 30,000 Hz 1,170,000 Hz, to, 1,200,000 15 Hz 1,199,985 Hz, The upper sideband is the sum of the carrier and modulating frequencies., It will include the frequencies from, 1,200,000 15 Hz 1,200,015 Hz, to, 1,200,000 30,000 Hz 1,230,000 Hz, , Practice Problem 18.11, , If a 2-MHz carrier is modulated by a 4-kHz intelligent signal, determine the frequencies of the three components of the AM signal that, results., Answer: 2,004,000 Hz, 2,000,000 Hz, 1,996,000 Hz., , g(t), , 18.7.2 Sampling, 0, , t, , (a), (t − nTs), , −Ts, , 0, , Ts 2Ts 3Ts, , ..., , t, , (b), , In analog systems, signals are processed in their entirety. However, in, modern digital systems, only samples of signals are required for processing. This is possible as a result of the sampling theorem given in, Section 17.8.1. The sampling can be done by using a train of pulses or, impulses. We will use impulse sampling here., Consider the continuous signal g(t) shown in Fig. 18.25(a). This, can be multiplied by a train of impulses d(t nTs) shown in, Fig. 18.25(b), where Ts is the sampling interval and fs 1Ts is the, sampling frequency or the sampling rate. The sampled signal gs(t) is, therefore, , , , , n, , n, , gs(t) g(t) a d(t nTs) a g(nTs) d (t nTs) (18.69), , g(t), , The Fourier transform of this is, −Ts, , 0, , Ts 2Ts 3Ts, (c), , Figure 18.25, (a) Continuous (analog) signal to be sampled, (b) train of impulses, (c) sampled, (digital) signal., , t, , , , , , n, , n, , Gs() a g(nTs)F[d(t nTs)] a g(nTs)ejnTs (18.70), It can be shown that, , , 1, jnTs, , a g(nTs)e, Ts, n, , , , a G( ns), n, , (18.71)
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 839, , 18.8, , Summary, , 839, , where s 2 pTs. Thus, Eq. (18.70) becomes, Gs() , , 1, Ts, , , , a G( ns), , (18.72), , n, , This shows that the Fourier transform Gs() of the sampled signal is, a sum of translates of the Fourier transform of the original signal at a, rate of 1Ts., In order to ensure optimum recovery of the original signal, what, must be the sampling interval? This fundamental question in sampling, is answered by an equivalent part of the sampling theorem:, A band-limited signal, with no frequency component higher than, W hertz, may be completely recovered from its samples taken at a frequency at least twice as high as 2W samples per second., , In other words, for a signal with bandwidth W hertz, there is no loss, of information or overlapping if the sampling frequency is at least twice, the highest frequency in the modulating signal. Thus,, 1, fs 2W, Ts, , (18.73), , The sampling frequency fs 2W is known as the Nyquist frequency, or rate, and 1fs is the Nyquist interval., , A telephone signal with a cutoff frequency of 5 kHz is sampled at a, rate 60 percent higher than the minimum allowed rate. Find the sampling rate., , Example 18.12, , Solution:, The minimum sample rate is the Nyquist rate 2W 2 5 10 kHz., Hence,, fs 1.60 2W 16 kHz, , An audio signal that is band-limited to 12.5 kHz is digitized into 8-bit, samples. What is the maximum sampling interval that must be used to, ensure complete recovery?, Answer: 40 ms., , 18.8, , Summary, , 1. The Fourier transform converts a nonperiodic function f(t) into a, transform F(), where, F() F[ f (t)] , , , , , , , , f (t)ejt dt, , Practice Problem 18.12
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 840, , Chapter 18, , 840, , Fourier Transform, , 2. The inverse Fourier transform of F() is, f (t) F1[F()] , , 1, 2p, , , , , , F()e jt d, , , , 3. Important Fourier transform properties and pairs are summarized, in Tables 18.1 and 18.2, respectively., 4. Using the Fourier transform method to analyze a circuit involves, finding the Fourier transform of the excitation, transforming the, circuit element into the frequency domain, solving for the unknown, response, and transforming the response to the time domain using, the inverse Fourier transform., 5. If H() is the transfer function of a network, then H() is the, Fourier transform of the network’s impulse response; that is,, H() F[h(t)], The output Vo() of the network can be obtained from the input, Vi () using, Vo() H()Vi (), 6. Parseval’s theorem gives the energy relationship between a function f (t) and its Fourier transform F(). The 1- energy is, W1 , , , , , , f 2(t) dt , , , , 1, 2p, , , , , , 0F() 0 2 d, , , , The theorem is useful in calculating energy carried by a signal, either in the time domain or in the frequency domain., 7. Typical applications of the Fourier transform are found in amplitude modulation (AM) and sampling. For AM application, a way, of determining the sidebands in an amplitude-modulated wave is, derived from the modulation property of the Fourier transform. For, sampling application, we found that no information is lost in sampling (required for digital transmission) if the sampling frequency, is equal to at least twice the Nyquist rate., , Review Questions, 18.1, , Which of these functions does not have a Fourier, transform?, (a) etu(t), (c) 1t, , 18.2, , 18.3, , (b) te3tu(t), (d) 0t 0 u(t), , 18.4, , (a) d(t), 18.5, , The Fourier transform of e j2t is:, 1, (a), 2 j, , 1, (b), 2 j, , (c) 2 p d( 2), , (d) 2 p d( 2), , The inverse Fourier transform of, 2t, , ej, is, 2 j, , 2t, , u(t 1), , (a) e, , (b) e, , (c) e2(t1), , (d) e2(t1)u(t 1), , The inverse Fourier transform of d() is:, , 18.6, , (b) u(t), , The inverse Fourier transform of j is:, (a) d¿(t), , (b) u¿(t), , (c) 1t, , (d) undefined, , Evaluating the integral, , , , , , , , (a) 0, 18.7, , (d) 12 p, , (c) 1, , (b) 2, , The integral, , , , , , , , (a) 0, , (b) 2, , 10 d(), 4 2, , (c) 2.5, 10 d( 1), 4 2, (c) 2.5, , d results in:, , (d) , d gives:, (d)
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 841, , Problems, , 18.8, , 18.9, , The current through an initially uncharged 1-F, capacitor is d(t) A. The voltage across the capacitor is:, (a) u(t) V, , (b) 12 u(t) V, , (c) etu(t) V, , (d) d(t) V, , A unit step current is applied through a 1-H inductor., The voltage across the inductor is:, (a) u(t) V, , (b) sgn(t) V, , (c) etu(t) V, , (d) d(t) V, , 841, , 18.10 Parseval’s theorem is only for nonperiodic functions., (a) True, , (b) False, , Answers: 18.1c, 18.2c, 18.3d, 18.4d, 18.5a, 18.6c,, 18.7b, 18.8a, 18.9d, 18.10b, , Problems, †, , Sections 18.2 and 18.3 Fourier Transform, and its Properties, , 18.1, , 18.4, , Find the Fourier transform of the waveform shown, in Fig. 18.29., , Obtain the Fourier transform of the function in, Fig. 18.26., , g(t), , f (t), 10, 1, 1, −2, , −1, , 2, , 0, , t, –1, , −1, , 0, , 1, , t, , Figure 18.29, For Prob. 18.4., , Figure 18.26, For Prob. 18.1., 18.2, , Using Fig. 18.27, design a problem to help other, students better understand the Fourier transform, given a wave shape., , 18.5, , Obtain the Fourier transform of the signal shown in, Fig. 18.30., , f (t), h(t), , f (0), , 1, 0, , t1, , t, , Figure 18.27, For Prob. 18.2., 18.3, , Calculate the Fourier transform of the signal in, Fig. 18.28., , –1, , 0, , 1, , t, , f (t), 1, , −1, , −2, , Figure 18.30, 0, , 2 t, , For Prob. 18.5., , −1, , 18.6, , Figure 18.28, , Find the Fourier transforms of both functions in, Fig. 18.31 on the following page., , For Prob. 18.3., † We have marked (with the MATLAB icon) the problems where we are asking the student to find the Fourier transform of a wave shape., We do this because you can use MATLAB to plot the results as a check.
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 842, , Chapter 18, , 842, f (t), 20, , g(t), 20, , 10, , 10, , Fourier Transform, , 18.9, , Determine the Fourier transforms of the signals in, Fig. 18.34., y (t), 2, , 0, , 1, , 2, , t, , 0, , 1, , (a), , 2, , t, , (b), 1, , Figure 18.31, For Prob. 18.6., –2, , 18.7, , Find the Fourier transforms of the signals in, Fig. 18.32., , f1(t), , f2(t), , 2, , 10, , –1, , 0, (a), , 1, , 2, , t, , z (t), , 0, , 1, , t, , 1, −2, 0, , 1, , 2, , t, , 2, , 0, , (a), , (b), , t, , Figure 18.34, , (b), , For Prob. 18.9., , Figure 18.32, For Prob. 18.7., , 18.10 Obtain the Fourier transforms of the signals shown, in Fig. 18.35., 18.8, , Obtain the Fourier transforms of the signals shown, in Fig. 18.33., , x(t), , y(t), , 1, , e −|t|, , 1, , e−t, f (t), 2, , 0, , −1, , t, , 0, , 1, , t, , (b), , (a), , Figure 18.35, For Prob. 18.10., 0, , 1, , 2, , t, , 18.11 Find the Fourier transform of the “sine-wave pulse”, shown in Fig. 18.36., , (a), , g (t), , f (t), , 2, , 5, , sin t, , 1, 0, –2, , –1, , 0, (b), , 1, , 2, , t, , Figure 18.33, , Figure 18.36, , For Prob. 18.8., , For Prob. 18.11., , 1, , 2, , t
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 843, , Problems, , 18.12 Find the Fourier transform of the following signals., 3t, , (a) f1(t) e, , sin(10t) u (t), , (b) Find the Fourier transform of the signal in, Fig. 18.37., , (b) f2(t) e4t cos(10t) u (t), , f (t), , 18.13 Find the Fourier transform of the following signals:, (a) f (t) cos(at p3),, , 6 t 6 , , (b) g(t) u(t 1) sin p t,, , 6 t 6 , , 6 t 6 ,, (c) h(t) (1 A sin at) cos bt,, where A, a, and b are constants, (d) i(t) 1 t,, , 0 6 t 6 4, , 18.14 Design a problem to help other students better, understand finding the Fourier transform of a variety, of time varying functions (do at least three)., , 1, , 0, , , , 2d(t 1) dt, , , , (c) f (t) d(3t) d¿(2t), *18.16 Determine the Fourier transforms of these functions:, , , , F() , , n, , has the Fourier transform, , , F() a cnd( n0), n, , where 0 2 pT., , * An asterisk indicates a challenging problem., , (b) f (2t 1), , d, f (t), dt, , (e), , , , (c) f (t) cos 2t, , t, , f (t) dt, , , , 18.24 Given that F[ f (t)] ( j)(ej 1), find the, Fourier transforms of:, (a) x(t) f (t) 3, , (b) y(t) f (t 2), , (c) h(t) f ¿(t), 2, 5, (d) g(t) 4 f a tb 10 f a tb, 3, 3, 18.25 Obtain the inverse Fourier transform of the, following signals., (a) G() , , 10, j 2, , (b) H() , , 6, 4, , (c) X() , , 5, ( j 1)( j 2), , , , f (t) a cne jn0t, , 10, (2 j)(5 j), , determine the transforms of the following:, , 18.19 Find the Fourier transform of, , 18.20 (a) Show that a periodic signal with exponential, Fourier series, , sin a, b., a, , 18.23 If the Fourier transform of f(t) is, , (d), , f (t) cos 2 p t[u(t) u(t 1)], , sin a 2, p, b d , a, a, , j, F[ f (t) sin 0 t] [F( 0) F( 0)], 2, , (b) sin 10tu(t), , d, F(), d, , a, , 18.22 Prove that if F() is the Fourier transform of f (t),, , (a) f (3t), , (d) F[t f (t)] j, , t, , 5, , F[u(t a) u(t a)] 2a a, , (a) cos 2tu(t), , df (t), d jF(), (b) F c, dt, (c) F[ f (t)] F(), , , , , , (b) g(t) 20(4 t ), , (a) F[ f (t t0)] ejt0 F(), , 4, , Hint: Use the fact that, , 2, , 18.18 Given that F() F[ f (t)], prove the following, results, using the definition of Fourier transform:, , 3, , 18.21 Show that, , (a) f (t) 10t 2, , 18.17 Find the Fourier transforms of:, , 2, , For Prob. 18.20(b)., , (a) f (t) d(t 3) d(t 3), , , , , , Figure 18.37, , 18.15 Find the Fourier transforms of the following functions:, , (b) f (t) , , 843, , 2, , 18.26 Determine the inverse Fourier transforms of the, following:, (a) F() , , ej2, 1 j, , (b) H() , , 1, ( j 4)2, , (c) G() 2u( 1) 2u( 1)
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 844, , Chapter 18, , 844, , Fourier Transform, , 18.27 Find the inverse Fourier transforms of the following, functions:, (a) F() , , F(), 20, , 100, j( j 10), , 10, , 10 j, (b) G() , (j 2)( j 3), , −2, , 60, 2, j40 1300, , (c) H() , , −1, , F() , , p d(), (a), (5 j)(2 j), , (a) x(t) f (3t 1), (b) y(t) f (t) cos 5t, (c) z(t) , , 5 p d(), 5, (d), , 5 j, j(5 j), *18.29 Determine the inverse Fourier transforms of:, (a) F() 4d( 3) d() 4d( 3), (b) G() 4u( 2) 4u( 2), 18.30 For a linear system with input x(t) and output y(t),, find the impulse response for the following cases:, y(t) u(t) u(t), y(t) e2tu(t), y(t) eat sin btu(t), , 18.31 Given a linear system with output y(t) and impulse, response h(t), find the corresponding input x(t) for, the following cases:, (a) y(t) teatu(t),, , e, j 1, 1, (c) F3() , (1 2)2, (a) F1() , , (e) i(t) t f (t), , 18.36 The transfer function of a circuit is, H() , , 2, j 2, , If the input signal to the circuit is vs(t) e4tu(t) V,, find the output signal. Assume all initial conditions, are zero., 18.37 Find the transfer function Io()Is() for the circuit, in Fig. 18.39., io(t), , h(t) d(t), , h(t) sgn(t), , *18.32 Determine the functions corresponding to the, following Fourier transforms:, j, , (d) h(t) f (t) * f (t), , h(t) eatu(t), , (b) y(t) u(t 1) u(t 1),, (c) y(t) eatu(t),, , d, f (t), dt, , Section 18.4 Circuit Applications, , (c) H() 6 cos 2, , (c) x(t) d(t),, , 1, 2 j, , Determine the Fourier transform of the following, signals:, , 10d( 2), (b), j( j 1), 20d( 1), (c), (2 j)(3 j), , (b) x(t) e u(t),, , t, , 18.35 A signal f(t) has Fourier transform, , 18.28 Find the inverse Fourier transforms of:, , t, , 2, , 1, , For Prob. 18.34., , d(), (d) Y() , ( j 1)( j 2), , (a) x(t) eatu(t),, , 0, , Figure 18.38, , (b) F2() 2e 0 0, (d) F4() , , d(), 1 j2, , 2Ω, , is (t), , Figure 18.39, For Prob. 18.37., 18.38 Using Fig. 18.40, design a problem to help other, students better understand using Fourier transforms, to do circuit analysis., , *18.33 Find f (t) if:, , R, , (a) F() 2 sin p[u( 1) u( 1)], j, 1, (b) F() (sin 2 sin ) (cos 2 cos ), , , 18.34 Determine the signal f(t) whose Fourier transform is, shown in Fig. 18.38. (Hint: Use the duality property.), , 4Ω, , 1H, , i, vs, , Figure 18.40, For Prob. 18.38., , +, −, , L
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 845, , Problems, , 18.39 Given the circuit in Fig. 18.41, with its excitation,, determine the Fourier transform of i(t)., , 845, , io(t), 2Ω, , i(t), , 1H, , vs(t), , Figure 18.44, For Prob. 18.42., , 2, , 18.43 Find vo(t) in the circuit of Fig. 18.45, where, is 5etu(t) A., 0, , 1, , t, , (a), , +, 40 Ω, , 1 kΩ, , is, , 20 mF, , −, , vo, , i(t), vs, , Figure 18.45, , +, −, , 1 mH, , For Prob. 18.43., 18.44 If the rectangular pulse in Fig. 18.46(a) is applied to, the circuit in Fig. 18.46(b), find vo at t 1 s., , (b), , Figure 18.41, For Prob. 18.39., , vs (t), 2Ω, , 10, , 18.40 Determine the current i(t) in the circuit of, Fig. 18.42(b), given the voltage source shown in, Fig. 18.42(a)., 0, , 2Ω, , v(t), 1, v (t), 0, , 2, , 1, , +, −, , +, −, , 2Ω, , 1H, , +, vo, −, , t, , 2, (a), , (b), , i(t), , Figure 18.46, , 1F, , For Prob. 18.44., , t, , (a), , vs, , 18.45 Use the Fourier transform to find i(t) in the circuit of, Fig. 18.47 if vs(t) 10e2tu(t)., , (b), , Figure 18.42, For Prob. 18.40., 18.41 Determine the Fourier transform of v(t) in the circuit, shown in Fig. 18.43., , V, , +, −, , +, v (t) −, , vs, , +, −, , 2Ω, , 1H, , Figure 18.47, , 2Ω, , e−2tu(t), , i, , 2Ω, , For Prob. 18.45., , 1F, , 0.5H, , 2(t) A, , 18.46 Determine the Fourier transform of io(t) in the circuit, of Fig. 18.48., 2Ω, , Figure 18.43, For Prob. 18.41., 5e −tu(t) V +, −, , 18.42 Obtain the current io(t) in the circuit of Fig. 18.44., (a) Let i(t) sgn(t) A., (b) Let i(t) 4[u(t) u(t 1)] A., , Figure 18.48, For Prob. 18.46., , 1, 4, , +, −, , F, , io, 2H, , 15δ(t) V
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 846, , Chapter 18, , 846, , Fourier Transform, , 18.47 Find the voltage vo(t) in the circuit of Fig. 18.49. Let, is(t) 8etu(t) A ., , 18.51 Find the energy dissipated by the resistor in the, circuit of Fig. 18.53., , 2Ω, , 2Ω, +, 1, F, 2, , 1Ω, , is, , 10etu(t)V +, −, , vo, −, , 1H, , Figure 18.53, , Figure 18.49, For Prob. 18.47., , For Prob. 18.51., , 18.48 Find io(t) in the op amp circuit of Fig. 18.50., , Section 18.5 Parseval’s Theorem, 18.52 For F() , , 20 F, 20 kΩ, , 2e −tu(t), , V, , 1, , find J , 3 j, , 20 t 0, , 18.53 If f (t) e, , −, +, , , find J , , , , io(t), , +, −, , , , , , 20 kΩ, , , , , , f 2(t) dt., , , , 0F() 0 2 d., , 18.54 Design a problem to help other students better, understand finding the total energy in a given signal., 18.55 Let f (t) 5e(t2)u(t). Find F() and use it to find, the total energy in f(t)., , Figure 18.50, For Prob. 18.48., 18.49 Use the Fourier transform method to obtain vo(t) in, the circuit of Fig. 18.51., , 18.56 The voltage across a 1- resistor is v(t) , te2tu(t) V. (a) What is the total energy absorbed by, the resistor? (b) What fraction of this energy, absorbed is in the frequency band 2 2?, 18.57 Let i(t) 5etu(t) A. Find the total energy carried, by i(t) and the percentage of the 1- energy in the, frequency range of 5 6 6 5 rad/s., , 1H, , Section 18.6 Applications, 2H, cos t V, , +, −, , 1H, , 2Ω, , 1Ω, , 18.58 An AM signal is specified by, , +, vo, −, , f (t) 10(1 4 cos 200 p t) cos p 104t, Determine the following:, , Figure 18.51, , (a) the carrier frequency,, , For Prob. 18.49., , (b) the lower sideband frequency,, (c) the upper sideband frequency., , 18.50 Determine vo(t) in the transformer circuit of, Fig. 18.52., , 0.5 H, , 18.59 For the linear system in Fig. 18.54, when the input, voltage is vi (t) 2d(t) V, the output is vo(t) , 10e2t 6e4t V. Find the output when the input is, vi (t) 4etu(t) V., , 1Ω, , 2(t), , +, −, , 1H, , 1H, , 1Ω, , +, vo, −, , h (t), vi (t), , Figure 18.52, , Figure 18.54, , For Prob. 18.50., , For Prob. 18.9., , vo (t)
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ale29559_ch18.qxd, , 07/11/2008, , 09:55 AM, , Page 847, , Comprehensive Problems, , 18.60 A band-limited signal has the following Fourier, series representation:, is(t) 10 8 cos(2 p t 30) 5 cos(4 p t 150)mA, If the signal is applied to the circuit in Fig. 18.55,, find v(t)., 2Ω, +, is (t), , 1H, , 1F, , v(t), −, , Figure 18.55, For Prob. 18.60., 18.61 In a system, the input signal x(t) is amplitudemodulated by m (t) 2 cos 0 t. The response, y(t) m(t)x(t). Find Y() in terms of X()., , 847, , 18.62 A voice signal occupying the frequency band of 0.4, to 3.5 kHz is used to amplitude-modulate a 10-MHz, carrier. Determine the range of frequencies for the, lower and upper sidebands., 18.63 For a given locality, calculate the number of stations, allowable in the AM broadcasting band (540 to, 1600 kHz) without interference with one another., 18.64 Repeat the previous problem for the FM broadcasting, band (88 to 108 MHz), assuming that the carrier, frequencies are spaced 200 kHz apart., 18.65 The highest-frequency component of a voice signal, is 3.4 kHz. What is the Nyquist rate of the sampler, of the voice signal?, 18.66 A TV signal is band-limited to 4.5 MHz. If samples, are to be reconstructed at a distant point, what is the, maximum sampling interval allowable?, *18.67 Given a signal g(t) sinc(200 p t), find the Nyquist, rate and the Nyquist interval for the signal., , Comprehensive Problems, 18.68 The voltage signal at the input of a filter is v(t) , 50e20 t| V. What percentage of the total 1-, energy content lies in the frequency range of, 1 6 6 5 rad/s?, 18.69 A signal with Fourier transform, F() , , 20, 4 j, , is passed through a filter whose cutoff frequency is, 2 rad/s (i.e., 0 6 6 2). What fraction of the, energy in the input signal is contained in the, output signal?
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ale29559_ch19.qxd, , 07/16/2008, , 05:48 PM, , Page 849, , c h a p t e r, , Two-Port Networks, , 19, , Never put off till tomorrow what you can do today., Never trouble another for what you can do yourself., Never spend your money before you have it., Never buy what you do not want because it is cheap., Pride costs us more than hunger, thirst, and cold., We seldom repent having eaten too little., Nothing is troublesome that we do willingly., How much pain the evils have cost us that have never happened!, Take things always by the smooth handle., When angry, count ten before you speak; if very angry, a hundred., —Thomas Jefferson, , Enhancing Your Career, Career in Education, While two thirds of all engineers work in private industry, some work, in academia and prepare students for engineering careers. The course, on circuit analysis you are studying is an important part of the preparation process. If you enjoy teaching others, you may want to consider, becoming an engineering educator., Engineering professors work on state-of-the-art research projects,, teach courses at graduate and undergraduate levels, and provide services, to their professional societies and the community at large. They are, expected to make original contributions in their areas of specialty. This, requires a broad-based education in the fundamentals of electrical engineering and a mastery of the skills necessary for communicating their, efforts to others., If you like to do research, to work at the frontiers of engineering, to, make contributions to technological advancement, to invent, consult, and/, or teach, consider a career in engineering education. The best way to start, is by talking with your professors and benefiting from their experience., A solid understanding of mathematics and physics at the undergraduate level is vital to your success as an engineering professor. If, you are having difficulty in solving your engineering textbook problems, start correcting any weaknesses you have in your mathematics, and physics fundamentals., Most universities these days require that engineering professors have, a doctor’s degree. In addition, some universities require that they be, actively involved in research leading to publications in reputable journals., , Photo by James Watson, , 849
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ale29559_ch19.qxd, , 07/16/2008, , 05:48 PM, , Page 850, , Chapter 19, , 850, , Two-Port Networks, , To prepare yourself for a career in engineering education, get as broad, an education as possible, because electrical engineering is changing rapidly and becoming interdisciplinary. Without doubt, engineering education is a rewarding career. Professors get a sense of satisfaction and, fulfillment as they see their students graduate, become leaders in their, profession, and contribute significantly to the betterment of humanity., , 19.1, I, +, Linear, network, , V, −, I, (a), I1, , I2, +, , +, Linear, network, , V1, , V2, , −, , −, I1, , I2, (b), , Introduction, , A pair of terminals through which a current may enter or leave a network is known as a port. Two-terminal devices or elements (such as, resistors, capacitors, and inductors) result in one-port networks. Most, of the circuits we have dealt with so far are two-terminal or one-port, circuits, represented in Fig. 19.1(a). We have considered the voltage, across or current through a single pair of terminals—such as the two, terminals of a resistor, a capacitor, or an inductor. We have also studied four-terminal or two-port circuits involving op amps, transistors,, and transformers, as shown in Fig. 19.1(b). In general, a network may, have n ports. A port is an access to the network and consists of a pair, of terminals; the current entering one terminal leaves through the other, terminal so that the net current entering the port equals zero., In this chapter, we are mainly concerned with two-port networks, (or, simply, two-ports)., A two-port network is an electrical network with two separate ports, for input and output., , Figure 19.1, (a) One-port network, (b) two-port, network., , Thus, a two-port network has two terminal pairs acting as access points., As shown in Fig. 19.1(b), the current entering one terminal of a pair, leaves the other terminal in the pair. Three-terminal devices such as, transistors can be configured into two-port networks., Our study of two-port networks is for at least two reasons. First,, such networks are useful in communications, control systems, power, systems, and electronics. For example, they are used in electronics to, model transistors and to facilitate cascaded design. Second, knowing, the parameters of a two-port network enables us to treat it as a “black, box” when embedded within a larger network., To characterize a two-port network requires that we relate the terminal quantities V1, V2, I1, and I2 in Fig. 19.1(b), out of which two are, independent. The various terms that relate these voltages and currents, are called parameters. Our goal in this chapter is to derive six sets of these, parameters. We will show the relationship between these parameters and, how two-port networks can be connected in series, parallel, or cascade., As with op amps, we are only interested in the terminal behavior of the, circuits. And we will assume that the two-port circuits contain no independent sources, although they can contain dependent sources. Finally,, we will apply some of the concepts developed in this chapter to the, analysis of transistor circuits and synthesis of ladder networks., , 19.2, , Impedance Parameters, , Impedance and admittance parameters are commonly used in the synthesis of filters. They are also useful in the design and analysis of, impedance-matching networks and power distribution networks. We
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ale29559_ch19.qxd, , 07/16/2008, , 05:48 PM, , Page 852, , Chapter 19, , 852, , Two-Port Networks, , Similarly, we obtain z12 and z22 by connecting a voltage V2 (or a current source I2) to port 2 with port 1 open-circuited as in Fig. 19.3(b), and finding I2 and V1; we then get, z12 , , 1, , 2, I, Reciprocal, two-port, , V +, −, , A, , (a), 1, , 2, , I, Reciprocal, two-port, , A, , + V, −, , (b), , Figure 19.4, Interchanging a voltage source at one port, with an ideal ammeter at the other port, produces the same reading in a reciprocal, two-port., , +, , I2, z22 – z12, z12, , V1, −, , z22 , , V2, I2, , (19.6), , The above procedure provides us with a means of calculating or measuring the z parameters., Sometimes z11 and z22 are called driving-point impedances, while, z21 and z12 are called transfer impedances. A driving-point impedance, is the input impedance of a two-terminal (one-port) device. Thus, z11, is the input driving-point impedance with the output port opencircuited, while z22 is the output driving-point impedance with the input, port open-circuited., When z11 z22, the two-port network is said to be symmetrical., This implies that the network has mirrorlike symmetry about some center line; that is, a line can be found that divides the network into two, similar halves., When the two-port network is linear and has no dependent sources,, the transfer impedances are equal (z12 z21), and the two-port is said, to be reciprocal. This means that if the points of excitation and response, are interchanged, the transfer impedances remain the same. As illustrated in Fig. 19.4, a two-port is reciprocal if interchanging an ideal voltage source at one port with an ideal ammeter at the other port gives the, same ammeter reading. The reciprocal network yields V z12I according to Eq. (19.1) when connected as in Fig. 19.4(a), but yields V z21I, when connected as in Fig. 19.4(b). This is possible only if z12 z21., Any two-port that is made entirely of resistors, capacitors, and inductors must be reciprocal. A reciprocal network can be replaced by the, T-equivalent circuit in Fig. 19.5(a). If the network is not reciprocal, a, more general equivalent network is shown in Fig. 19.5(b); notice that, this figure follows directly from Eq. (19.1)., , I1, z11 – z12, , V1, ,, I2, , I1, +, , +, , V2, , V1, , −, , −, , I2, z11, z12 I2, , z22, +, −, , +, −, , z21 I1, , +, V2, −, , (a), , (b), , Figure 19.5, (a) T-equivalent circuit (for reciprocal case only), (b) general equivalent circuit., I1, +, , I2, 1:n, , +, , V1, , V2, , −, , −, , Figure 19.6, An ideal transformer has no z parameters., , It should be mentioned that for some two-port networks, the z, parameters do not exist because they cannot be described by Eq. (19.1)., As an example, consider the ideal transformer of Fig. 19.6. The defining equations for the two-port network are:, 1, V1 V2,, n, , I1 n I2, , (19.7)
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ale29559_ch19.qxd, , 07/16/2008, , 05:48 PM, , Page 859, , 19.4, , Hybrid Parameters, , 859, , or in matrix form,, c, , h11 h12, I1, I1, V1, d c, d c d [h] c d, I2, h21 h22 V2, V2, , (19.15), , The h terms are known as the hybrid parameters (or, simply, h, parameters) because they are a hybrid combination of ratios. They, are very useful for describing electronic devices such as transistors, (see Section 19.9); it is much easier to measure experimentally the h, parameters of such devices than to measure their z or y parameters., In fact, we have seen that the ideal transformer in Fig. 19.6, described, by Eq. (19.7), does not have z parameters. The ideal transformer can, be described by the hybrid parameters, because Eq. (19.7) conforms, with Eq. (19.14)., The values of the parameters are determined as, , h11 , h21, , V1, `, ,, I1 V20, , I2, `, ,, I1 V2 0, , h12 , h22, , V1, `, V2 I10, , I2, , `, V2 I10, , (19.16), , It is evident from Eq. (19.16) that the parameters h11, h12, h21, and h22, represent an impedance, a voltage gain, a current gain, and an admittance, respectively. This is why they are called the hybrid parameters., To be specific,, h11, h12, h21, h22, , , , , , , Short-circuit input impedance, Open-circuit reverse voltage gain, Short-circuit forward current gain, Open-circuit output admittance, , (19.17), , The procedure for calculating the h parameters is similar to that used, for the z or y parameters. We apply a voltage or current source to the, appropriate port, short-circuit or open-circuit the other port, depending, on the parameter of interest, and perform regular circuit analysis. For, reciprocal networks, h12 h21. This can be proved in the same way, as we proved that z12 z21. Figure 19.20 shows the hybrid model of, a two-port network., A set of parameters closely related to the h parameters are the g, parameters or inverse hybrid parameters. These are used to describe, the terminal currents and voltages as, , I2, , h11, , +, V1, , +, h12V2, , +, −, , −, , h21I1, , h22, , V2, −, , Figure 19.20, , I1 g11V1 g12I2, V2 g21V1 g22I2, , (19.18), , I1, g11 g12 V1, V1, d c, d c d [g] c d, V2, g21 g22, I2, I2, , (19.19), , or, c, , I1, , The h-parameter equivalent network of a, two-port network.
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ale29559_ch19.qxd, , 07/16/2008, , 05:48 PM, , Page 863, , 19.5, , Transmission Parameters, , 863, , or, , I1, , g11 , , I1, 1, , V1, s1, , By voltage division,, V2 , , V1, , 1Ω, , (a), I1, , To obtain g12 and g22, we short-circuit the input port and connect a current, source I2 to the output port as in Fig. 19.29(b). By current division,, I1 , , 1, I2, s1, , or, g12 , , V2, −, , or, g21, , I2 = 0, +, , +, −, , 1, V1, s1, , V2, 1, , , V1, s1, , 1/s, , s, , 1/s, , s, , +, , +, 1Ω, , V1 = 0, , V2, , I2, , −, , −, (b), , Figure 19.29, Determining the g parameters in the s, domain for the circuit in Fig. 19.28., , I1, 1, , I2, s1, , Also,, 1, V2 I2 a s 7 1b, s, or, g22 , , V2, s, 1, s2 s 1, , , s, I2, s1, s(s 1), , Thus,, 1, s1, [g] ≥, 1, s1, , 1, s1, ¥, s2 s 1, s(s 1), , , For the ladder network in Fig. 19.30, determine the g parameters in the, s domain., , Practice Problem 19.7, 1H, , 1H, , s2, s2 3s 1, Answer: [g] ≥, 1, s2 3s 1, , 1, s2 3s 1, ¥., s(s 2), s2 3s 1, , , , 1Ω, , Figure 19.30, For Practice Prob. 19.7., , 19.5, , Transmission Parameters, , Since there are no restrictions on which terminal voltages and currents, should be considered independent and which should be dependent variables, we expect to be able to generate many sets of parameters., , 1Ω
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ale29559_ch19.qxd, , 07/16/2008, , 05:48 PM, , Page 864, , Chapter 19, , 864, , Two-Port Networks, , Another set of parameters relates the variables at the input port to those, at the output port. Thus,, V1 AV2 BI2, I1 CV2 DI2, , (19.22), , V1, A B, V2, V2, d c, d c, d [T] c, d, I1, C D I2, I2, , (19.23), , or, c, , I1, , –I2, , +, V1, , +, Linear, two-port, , −, , Figure 19.31, Terminal variables used to define the, ADCB parameters., , V2, −, , Equations (19.22) and (19.23) relate the input variables (V1 and I1) to, the output variables (V2 and I2). Notice that in computing the transmission parameters, I2 is used rather than I2, because the current is, considered to be leaving the network, as shown in Fig. 19.31, as, opposed to entering the network as in Fig. 19.1(b). This is done merely, for conventional reasons; when you cascade two-ports (output to input),, it is most logical to think of I2 as leaving the two-port. It is also customary in the power industry to consider I2 as leaving the two-port., The two-port parameters in Eqs. (19.22) and (19.23) provide a, measure of how a circuit transmits voltage and current from a source, to a load. They are useful in the analysis of transmission lines (such, as cable and fiber) because they express sending-end variables (V1 and, I1) in terms of the receiving-end variables (V2 and I2). For this reason, they are called transmission parameters. They are also known as, ABCD parameters. They are used in the design of telephone systems,, microwave networks, and radars., The transmission parameters are determined as, , A, , V1, `, ,, V2 I20, , I1, C, `, ,, V2 I20, , B, , V1, `, I2 V20, , I1, D `, I2 V20, , (19.24), , Thus, the transmission parameters are called, specifically,, A, B, C, D, , , , , , , Open-circuit voltage ratio, Negative short-circuit transfer impedance, Open-circuit transfer admittance, Negative short-circuit current ratio, , (19.25), , A and D are dimensionless, B is in ohms, and C is in siemens. Since, the transmission parameters provide a direct relationship between input, and output variables, they are very useful in cascaded networks., Our last set of parameters may be defined by expressing the, variables at the output port in terms of the variables at the input port., We obtain, V2 aV1 bI1, I2 cV1 dI1, , (19.26)
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ale29559_ch19.qxd, , 07/16/2008, , 05:48 PM, , Page 865, , 19.5, , Transmission Parameters, , 865, , or, c, , V2, a b, V1, V1, d c, d c, d [t] c, d, I2, c d I1, I1, , (19.27), , The parameters a, b, c, and d are called the inverse transmission, or t,, parameters. They are determined as follows:, , a, , V2, `, ,, V1 I10, , b, , I2, c, `, ,, V1 I10, , V2, `, I1 V10, , I2, d `, I1 V10, , (19.28), , From Eq. (19.28) and from our experience so far, it is evident that these, parameters are known individually as, a Open-circuit voltage gain, b Negative short-circuit transfer impedance, c Open-circuit transfer admittance, d Negative short-circuit current gain, , (19.29), , While a and d are dimensionless, b and c are in ohms and siemens,, respectively., In terms of the transmission or inverse transmission parameters, a, network is reciprocal if, AD BC 1,, , ad bc 1, , (19.30), , These relations can be proved in the same way as the transfer impedance relations for the z parameters. Alternatively, we will be able to, use Table 19.1 a little later to derive Eq. (19.30) from the fact that, z12 z21 for reciprocal networks., , Example 19.8, , Find the transmission parameters for the two-port network in Fig. 19.32., Solution:, To determine A and C, we leave the output port open as in Fig. 19.33(a), so that I2 0 and place a voltage source V1 at the input port. We have, V1 (10 20)I1 30I1, , and, , I1, , 10 Ω, , 20 Ω, , V2 20I1 3I1 17I1, , Thus,, A, , V1, 30I1, , 1.765,, V2, 17I1, , C, , I1, I1, , 0.0588 S, V2, 17I1, , To obtain B and D, we short-circuit the output port so that V2 0 as, shown in Fig. 19.33(b) and place a voltage source V1 at the input port., At node a in the circuit of Fig. 19.33(b), KCL gives, Va, V1 Va, , I2 0, 10, 20, , (19.8.1), , 3I1, +−, , Figure 19.32, For Example 19.8., , I2
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ale29559_ch19.qxd, , 07/16/2008, , 05:48 PM, , Page 870, , 870, , Chapter 19, , Two-Port Networks, , Also, given that z21 z12 for a reciprocal network, we can use the, table to express this condition in terms of other parameters. It can also, be shown that, [g] [h]1, , (19.42), , [t] [T]1, , (19.43), , but, , Example 19.10, , Find [z] and [g] of a two-port network if, [T] c, , 10 1.5 , d, 2S, 4, , Solution:, If A 10, B 1.5, C 2, D 4, the determinant of the matrix is, ¢ T AD BC 40 3 37, From Table 19.1,, ¢T, A, 10, 37, , 5,, z12 , , 18.5, C, 2, C, 2, 1, 1, D, 4, z21 0.5,, z22 2, C, 2, C, 2, ¢T, C, 2, 37, g11 , 0.2,, g12 , 3.7, A, 10, A, 10, 1, 1, B, 1.5, g21 , 0.1,, g22 , 0.15, A, 10, A, 10, z11 , , Thus,, [z] c, , Practice Problem 19.10, , 5 18.5, d ,, 0.5 2, , 0.2 S 3.7, d, 0.1 0.15 , , Determine [y] and [T] of a two-port network whose z parameters are, [z] c, Answer: [y] c, , Example 19.11, , [g] c, , 6 4, d , 4 6, , 0.3 0.2, 1.5, 5, d S, [T] c, d., 0.2 0.3, 0.25 S 1.5, , Obtain the y parameters of the op amp circuit in Fig. 19.37. Show that, the circuit has no z parameters., Solution:, Since no current can enter the input terminals of the op amp, I1 0,, which can be expressed in terms of V1 and V2 as, I1 0V1 0V2, , (19.11.1)
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ale29559_ch19.qxd, , 07/16/2008, , 05:48 PM, , Page 871, , 19.7, , Interconnection of Networks, , 871, , Comparing this with Eq. (19.8) gives, , I1, , y11 0 y12, , R3, , +, −, , +, , I2, +, , Io, , Also,, V1, , V2 R3I2 Io(R1 R2), where Io is the current through R1 and R2. But Io V1R1. Hence,, V1(R1 R2), V2 R3I2 , R1, , Io, , R2, , V2, , R1, , −, , −, , Figure 19.37, For Example 19.11., , which can be written as, I2 , , (R1 R2), V2, V1 , R1R3, R3, , Comparing this with Eq. (19.8) shows that, y21 , , (R1 R2), ,, R1R3, , y22 , , 1, R3, , The determinant of the [y] matrix is, ¢ y y11y22 y12y21 0, Since ¢ y 0, the [y] matrix has no inverse; therefore, the [z] matrix, does not exist according to Eq. (19.34). Note that the circuit is not, reciprocal because of the active element., , Find the z parameters of the op amp circuit in Fig. 19.38. Show that, the circuit has no y parameters., , Practice Problem 19.11, R2, , Answer: [z] c, , R1 0, d . Since [z]1 does not exist, [y] does not exist., R2 0, , I1, , R1, , +, , −, +, , V1, −, , 19.7, , Interconnection of Networks, , A large, complex network may be divided into subnetworks for the purposes of analysis and design. The subnetworks are modeled as twoport networks, interconnected to form the original network. The, two-port networks may therefore be regarded as building blocks that, can be interconnected to form a complex network. The interconnection, can be in series, in parallel, or in cascade. Although the interconnected, network can be described by any of the six parameter sets, a certain, set of parameters may have a definite advantage. For example, when, the networks are in series, their individual z parameters add up to give, the z parameters of the larger network. When they are in parallel, their, individual y parameters add up to give the y parameters of the larger, network. When they are cascaded, their individual transmission parameters can be multiplied together to get the transmission parameters of, the larger network., , Figure 19.38, For Practice Prob. 19.11., , I2, +, V2, −
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ale29559_ch19.qxd, , 07/16/2008, , 05:48 PM, , Page 876, , Chapter 19, , 876, , Practice Problem 19.13, , Obtain the y parameters for the network in Fig. 19.45., Answer: c, , j5 S, , – j5 S, , Two-Port Networks, , 27 j15 25 j10, d S., 25 j10, 27 j5, , 1S, , – j 10 S, 2S, , 2S, , Figure 19.45, For Practice Prob. 19.13., , Example 19.14, 4Ω, , Find the transmission parameters for the circuit in Fig. 19.46., 6Ω, , 8Ω, 1Ω, , 2Ω, , Solution:, We can regard the given circuit in Fig. 19.46 as a cascade connection, of two T networks as shown in Fig. 19.47(a). We can show that a, T network, shown in Fig. 19.47(b), has the following transmission, parameters [see Prob. 19.52(b)]:, , Figure 19.46, For Example 19.14., , A1, , R1, ,, R2, , C, , R1(R2 R3), R2, R3, D1, R2, , B R3 , , 1, ,, R2, , Applying this to the cascaded networks Na and Nb in Fig. 19.47(a), we get, Aa 1 4 5,, Ca 1 S,, , Ba 8 4 9 44 , Da 1 8 9, , or in matrix form,, 4Ω, , [Ta] c, , 6Ω, , 8Ω, 1Ω, , 2Ω, , and, Ab 1,, , Na, , Nb, (a), , R1, , R3, , 5 44 , d, 1S, 9, , Bb 6 ,, , Cb 0.5 S,, , Db 1 , , 6, 4, 2, , i.e.,, [Tb] c, , 1, 6, d, 0.5 S 4, , Thus, for the total network in Fig. 19.46,, R2, , [T] [Ta][Tb] c, , 5 44, 1 6, d c, d, 1 9, 0.5 4, , Figure 19.47, , c, , For Example 19.14: (a) Breaking the, circuit in Fig. 19.46 into two two-ports,, (b) a general T two-port., , 5 1 44 0.5 5 6 44 4, d, 1 1 9 0.5 1 6 9 4, , c, , 27 206 , d, 5.5 S, 42, , (b)
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ale29559_ch19.qxd, , 07/16/2008, , 05:48 PM, , Page 877, , 19.8, , Computing Two-Port Parameters Using PSpice, , 877, , Notice that, ¢ Ta ¢ Tb ¢ T 1, showing that the network is reciprocal., , Obtain the ABCD parameter representation of the circuit in Fig. 19.48., , Practice Problem 19.14, 30 Ω, , 29.25 2200 , d., Answer: [T] c, 0.425 S, 32, , 60 Ω, , 40 Ω, , 20 Ω, , 50 Ω, , 20 Ω, , Figure 19.48, For Practice Prob. 19.14., , 19.8, , Computing Two-Port Parameters, Using PSpice, , Hand calculation of the two-port parameters may become difficult, when the two-port is complicated. We resort to PSpice in such situations. If the circuit is purely resistive, PSpice dc analysis may be used;, otherwise, PSpice ac analysis is required at a specific frequency. The, key to using PSpice in computing a particular two-port parameter is to, remember how that parameter is defined and to constrain the appropriate port variable with a 1-A or 1-V source while using an open or, short circuit to impose the other necessary constraints. The following, two examples illustrate the idea., , Example 19.15, , Find the h parameters of the network in Fig. 19.49., Solution:, From Eq. (19.16),, , 5Ω, , h11 , , V1, `, ,, I1 V20, , h21 , , IDC, , R2, , I2, `, I1 V20, , H1, , 5, , For Example 19.15., , DC=1A, , H, , R13, , GAIN=4, , 10, , –5.000E–01, , R8 10, , 0, , 1.833E–01, , R5, 6, +, −, , +, −, I1, , 10 Ω, , Figure 19.49, , H1, , 6, , +−, , 10 Ω, , .8333, , R5, , 4ix, , ix, , showing that h11 and h21 can be found by setting V2 0. Also by setting, I1 1 A, h11 becomes V11 while h21 becomes I21. With this in mind,, we draw the schematic in Fig. 19.50(a). We insert a 1-A dc current source, , 10.0000, , 6Ω, , (a), , Figure 19.50, For Example 19.15: (a) computing h11 and h21, (b) computing h12 and h22., , H, R8 10, , 0, , R13, , GAIN=4, , (b), , 10, , +, 1V, −, , V8
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ale29559_ch19.qxd, , 07/16/2008, , 05:49 PM, , Page 878, , Chapter 19, , 878, , Two-Port Networks, , IDC to take care of I1 1 A, the pseudocomponent VIEWPOINT to, display V1 and pseudocomponent IPROBE to display I2. After saving, the schematic, we run PSpice by selecting Analysis/Simulate and note, the values displayed on the pseudocomponents. We obtain, h11 , , V1, 10 ,, 1, , h21 , , I2, 0.5, 1, , h22 , , I2, `, V2 I10, , Similarly, from Eq. (19.16),, h12 , , V1, `, ,, V2 I10, , indicating that we obtain h12 and h22 by open-circuiting the input port, (I1 0). By making V2 1 V, h12 becomes V11 while h22 becomes, I21. Thus, we use the schematic in Fig. 19.50(b) with a 1-V dc voltage, source VDC inserted at the output terminal to take care of V2 1 V., The pseudocomponents VIEWPOINT and IPROBE are inserted to, display the values of V1 and I2, respectively. (Notice that in Fig., 19.50(b), the 5- resistor is ignored because the input port is opencircuited and PSpice will not allow such. We may include the 5-, resistor if we replace the open circuit with a very large resistor, say,, 10 M.) After simulating the schematic, we obtain the values displayed, on the pseudocomponents as shown in Fig. 19.50(b). Thus,, h12 , , Practice Problem 19.15, , I2, 0.1833 S, 1, , Obtain the h parameters for the network in Fig. 19.51 using PSpice., , 6Ω, , 8Ω, , 4Ω, , h22 , , Answer: h11 4.238 , h21 0.6190, h12 0.7143, h22 , 0.1429 S., , 2vx, , 3Ω, , V1, 0.8333,, 1, , +, vx, −, , 4Ω, , Figure 19.51, For Practice Prob. 19.15., , Example 19.16, , Find the z parameters for the circuit in Fig. 19.52 at 106 rad/s., , 2 H, +, vx, −, , 8 kΩ, , 4 nF, , Figure 19.52, For Example 19.16., , vx, 20, , 2 kΩ, , Solution:, Notice that we used dc analysis in Example 19.15 because the circuit, in Fig. 19.49 is purely resistive. Here, we use ac analysis at, f 2 p 0.15915 MHz, because L and C are frequency dependent., In Eq. (19.3), we defined the z parameters as, z11 , , V1, `, ,, I1 I20, , z21 , , V2, `, I1 I20
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ale29559_ch19.qxd, , 07/16/2008, , 05:49 PM, , Page 880, , Chapter 19, , 880, , Two-Port Networks, , This leads to the schematic in Fig. 19.53(b). The only difference, between this schematic and the one in Fig. 19.53(a) is that the 1-A ac, current source IAC is now at the output terminal. We run the schematic, in Fig. 19.53(b) and obtain V1 and V2 from the output file. Thus,, z12 , , Practice Problem 19.16, 4Ω, , 8Ω, , V1, 19.70l175.7 ,, 1, , z22 , , V2, 19.56l175.7 , 1, , Obtain the z parameters of the circuit in Fig. 19.54 at f 60 Hz., Answer: z11 3.987l175.5 , z21 0.0175l2.65 ,, z12 0, z22 0.2651l91.9 ., , ix, +, −, , 0.2 H, , 10ix, , 10 mF, , Figure 19.54, , 19.9, , For Practice Prob. 19.16., , Applications, , We have seen how the six sets of network parameters can be used to, characterize a wide range of two-port networks. Depending on the way, two-ports are interconnected to form a larger network, a particular, set of parameters may have advantages over others, as we noticed in, Section 19.7. In this section, we will consider two important application, areas of two-port parameters: transistor circuits and synthesis of ladder, networks., , 19.9.1 Transistor Circuits, I1, , Zs, , I2, , +, Vs, , +, −, , V1, −, Zin, , +, Two-port, network, , V2, −, Zout, , ZL, , The two-port network is often used to isolate a load from the excitation of a circuit. For example, the two-port in Fig. 19.55 may represent an amplifier, a filter, or some other network. When the two-port, represents an amplifier, expressions for the voltage gain Av, the current, gain Ai, the input impedance Zin, and the output impedance Zout can be, derived with ease. They are defined as follows:, Av , , V2(s), V1(s), , (19.62), , Ai , , I2(s), I1(s), , (19.63), , Zin , , V1(s), I1(s), , (19.64), , V2(s), `, I2(s) Vs0, , (19.65), , Figure 19.55, Two-port network isolating source and, load., , Zout , , Any of the six sets of two-port parameters can be used to derive the, expressions in Eqs. (19.62) to (19.65). However, the hybrid (h) parameters are the most useful for transistors; they are easily measured and, are often provided in the manufacturer’s data or spec sheets for transistors. The h parameters provide a quick estimate of the performance, of transistor circuits. They are used for finding the exact voltage gain,, input impedance, and output impedance of a transistor.
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ale29559_ch19.qxd, , 07/16/2008, , 05:49 PM, , Page 881, , 19.9, , Applications, , 881, , The h parameters for transistors have specific meanings expressed, by their subscripts. They are listed by the first subscript and related to, the general h parameters as follows:, hi h11,, , hr h12,, , hf h21,, , ho h22, , (19.66), , The subscripts i, r, f, and o stand for input, reverse, forward, and output., The second subscript specifies the type of connection used: e for common emitter (CE), c for common collector (CC), and b for common base, (CB). Here we are mainly concerned with the common-emitter connection. Thus, the four h parameters for the common-emitter amplifier are:, hie , hre , hfe , hoe , , Base input impedance, Reverse voltage feedback ratio, Base-collector current gain, Output admittance, , (19.67), , These are calculated or measured in the same way as the general h, parameters. Typical values are hie 6 k, hre 1.5 104, hfe , 200, hoe 8 mS. We must keep in mind that these values represent ac, characteristics of the transistor, measured under specific circumstances., Figure 19.56 shows the circuit schematic for the common-emitter, amplifier and the equivalent hybrid model. From the figure, we see that, Vb hie Ib hreVc, Ic hfe Ib hoeVc, Ic, Ib, , B, +, Vb, , (19.68a), (19.68b), , C, , B, , +, , +, , Vc, , Vb, , Ib, , Ic, , hie, , +, hreVc, , +, −, , hfeI b, , −, , −, , −, , −, , E, , E, , E, (b), , Figure 19.56, Common emitter amplifier: (a) circuit schematic, (b) hybrid model., , Consider the transistor amplifier connected to an ac source and a, load as in Fig. 19.57. This is an example of a two-port network embedded within a larger network. We can analyze the hybrid equivalent circuit, as usual with Eq. (19.68) in mind. (See Example 19.6.) Recognizing, Two-port network, Rs Ib, , Ic, , hie, +, , +, −, , hoe Vc, , E, (a), , Vs, , C, , Vb, −, , +, hreVc, , +, −, , hfe I b, , hoe, , Z in, , Figure 19.57, Transistor amplifier with source and load resistance., , Vc, −, , Z out, , RL
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ale29559_ch19.qxd, , 07/16/2008, , 05:49 PM, , Page 883, , 19.9, , Rs, , Applications, , 883, , Ic, , hie, , +, , Ib, , +, −, , hreVc, , hfe Ib, , hoe Vc, , + 1V, −, , −, , Figure 19.58, Finding the output impedance of the amplifier circuit in, Fig. 19.57., , 1-V source at the output terminals, we obtain the circuit in Fig. 19.58, from, which Zout is determined as 1Ic. Since Vc 1 V, the input loop gives, hre(1) Ib(Rs hie), , Ib , , 1, , hre, Rs hie, , (19.76), , For the output loop,, Ic hoe(1) hfe Ib, , (19.77), , Substituting Eq. (19.76) into Eq. (19.77) gives, Ic , , (Rs hie)hoe hre hfe, , (19.78), , Rs hie, , From this, we obtain the output impedance Zout as 1Ic; that is,, Zout , , Rs hie, (Rs hie)hoe hre hfe, , (19.79), , Consider the common-emitter amplifier circuit of Fig. 19.59. Determine, the voltage gain, current gain, input impedance, and output impedance, using these h parameters:, hie 1 k,, , hre 2.5 104,, , hfe 50,, , hoe 20 mS, , Find the output voltage Vo., 0.8 kΩ, 3.2 0° mV, , +, −, , 1.2 kΩ, , +, Vo, −, , Figure 19.59, For Example 19.17., , Solution:, 1. Define. In an initial look at this problem, it appears to be clearly, stated. However, when we are asked to determine the input, impedance and the voltage gain, do they refer to the transistor or, the circuit? As far as the current gain and the output impedance, are concerned, they are the same for both cases., We ask for clarification and are told that we should calculate, the input impedance, the output impedance, and the voltage gain, , Example 19.17
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ale29559_ch19.qxd, , 884, , 07/16/2008, , 05:49 PM, , Page 884, , Chapter 19, , Two-Port Networks, , for the circuit and not the transistor. It is interesting to note that, the problem can be restated so that it becomes a simple design, problem: Given the h parameters, design a simple amplifier that, has a gain of 60., 2. Present. Given a simple transistor circuit, an input voltage of 3.2 mV,, and the h parameters of the transistor, calculate the output voltage., 3. Alternative. There are a couple of ways we can approach the, problem, the most straightforward being to use the equivalent, circuit shown in Fig. 19.57. Once you have the equivalent circuit, you can use circuit analysis to determine the answer. Once you, have a solution, you can check it by plugging in the answer into, the circuit equations to see if they are correct. Another approach, is to simplify the right-hand side of the equivalent circuit and, work backward to see if you obtain approximately the same, answer. We will use that approach here., 4. Attempt. We note that Rs 0.8 k and RL 1.2 k. We treat, the transistor of Fig. 19.59 as a two-port network and apply, Eqs. (19.70) to (19.79)., hie hoe hre hfe 103 20 106 2.5 104 50, 7.5 103, hfe RL, , Av , , hie (hie hoe hre hfe) RL, 59.46, , , , 50 1200, 1000 7.5 103 1200, , Av is the voltage gain of the amplifier VoVb. To calculate the, gain of the circuit we need to find VoVs. We can do this by using the, mesh equation for the circuit on the left and Eqs. (19.71) and (19.73)., Vs RsIb Vb 0, or, 20 106, 1, , Vo, 50, 59.46, 50, 1 20 106 1.2 103, 0.03047 Vo., , Vs 800, , Thus, the circuit gain is equal to 32.82. Now we can calculate, the output voltage., Vo gain Vs 105.09l0 mV., Ai , , hfe, 1 hoe RL, , , , Zin hie , , 50, 48.83, 1 20 106 1200, hre hfe RL, , 1 hoe RL, , 1000 , , 2.5 104 50 1200, 1 20 106 1200, , 985.4 , You can modify Zin to include the 800-ohm resistor so that, Circuit input impedance 800 985.4 1785.4 .
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ale29559_ch19.qxd, , 07/16/2008, , 05:49 PM, , Page 885, , 19.9, , Applications, , 885, , (Rs hie)hoe hre hfe, (800 1000) 20 106 2.5 104 50 23.5 103, Zout , , Rs hie, 800 1000, , 76.6 k, (Rs hie)hoe hre hfe, 23.5 103, , 5. Evaluate. In the equivalent circuit, hoe represents a resistor of, 50,000 . This is in parallel with a load resistor equal to 1.2 k., The size of the load resistor is so small relative to the hoe, resistor that hoe can be neglected. This then leads to, Ic hfeIb 50Ib,, , Vc 1200Ic,, , and the following loop equation from the left-hand side of the, circuit:, 0.0032 (800 1000)Ib (0.00025)(1200)(50)Ib 0, Ib 0.0032(1785) 1.7927 mA., Ic 50 1.7927 89.64 mA and Vc 1200 89.64 106, 107.57 mV, This is a good approximation to 105.09 mV., Voltage gain 107.573.2 33.62, Again, this is a good approximation to 32.82., Circuit input impedance 0.0321.7927 106 1785 , which clearly compares well with the 1785.4 we obtained, before., For these calculations, we assumed that Zout . Our, calculations produced 72.6 k. We can test our assumption by, calculating the equivalent resistance of this and the load, resistance., 72,600 1200(72,600 1200) 1,180.5 1.1805 k, Again, we have a good approximation., 6. Satisfactory? We have satisfactorily solved the problem and, checked the results. We can now present our results as a solution, to the problem., , For the transistor amplifier of Fig. 19.60, find the voltage gain, current, gain, input impedance, and output impedance. Assume that, hie 6 k,, , hre 1.5 104,, , hfe 200,, , Practice Problem 19.17, 150 kΩ, , hoe 8 mS, , Answer: 123.61 for the transistor and 4.753 for the circuit, 194.17,, 6 k for the transistor and 156 k for the circuit, 128.08 k., , 2 0° mV, , +, −, , Figure 19.60, For Practice Prob. 19.17., , 19.9.2 Ladder Network Synthesis, Another application of two-port parameters is the synthesis (or building) of ladder networks, which are found frequently in practice and, , 3.75 kΩ
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ale29559_ch19.qxd, , 07/16/2008, , 05:49 PM, , Page 886, , Chapter 19, , 886, L1, , L3, C2, , Ln, C4, (a), , L1, , L3, C2, , Ln – 1, C4, , Cn, , (b), , Figure 19.61, LC ladder networks for lowpass filters of:, (a) odd order, (b) even order., , Two-Port Networks, , have particular use in designing passive lowpass filters. Based on our, discussion of second-order circuits in Chapter 8, the order of the filter is the order of the characteristic equation describing the filter and, is determined by the number of reactive elements that cannot be combined into single elements (e.g., through series or parallel combination). Figure 19.61(a) shows an LC ladder network with an odd, number of elements (to realize an odd-order filter), while Fig. 19.61(b), shows one with an even number of elements (for realizing an evenorder filter). When either network is terminated by the load impedance, ZL and the source impedance Zs, we obtain the structure in Fig. 19.62., To make the design less complicated, we will assume that Zs 0. Our, goal is to synthesize the transfer function of the LC ladder network., We begin by characterizing the ladder network by its admittance, parameters, namely,, I1 y11V1 y12V2, I2 y21V1 y22V2, , I1, , Zs, , +, −, , I2, LC ladder, network, , +, Vs, , (19.80a), (19.80b), , V1, y11 y12, y21 y22, , −, , +, , +, , V2, , ZL Vo, , −, , −, , Figure 19.62, LC ladder network with terminating impedances., , (Of course, the impedance parameters could be used instead of the, admittance parameters.) At the input port, V1 Vs since Zs 0. At the, output port, V2 Vo and I2 V2ZL VoYL. Thus, Eq. (19.80b), becomes, VoYL y21Vs y22Vo, or, H(s) , , y21, Vo, , Vs, YL y22, , (19.81), , We can write this as, H(s) , , y21YL, 1 y22YL, , (19.82), , We may ignore the negative sign in Eq. (19.82) because filter requirements are often stated in terms of the magnitude of the transfer function., The main objective in filter design is to select capacitors and inductors, so that the parameters y21 and y22 are synthesized, thereby realizing the, desired transfer function. To achieve this, we take advantage of an important property of the LC ladder network: all z and y parameters are ratios, of polynomials that contain only even powers of s or odd powers
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ale29559_ch19.qxd, , 07/16/2008, , 05:49 PM, , Page 887, , 19.9, , Applications, , 887, , of s—that is, they are ratios of either Od(s)Ev(s) or Ev(s)Od(s), where, Od and Ev are odd and even functions, respectively. Let, H(s) , , No Ne, N(s), , D(s), Do De, , (19.83), , where N(s) and D(s) are the numerator and denominator of the transfer function H(s); No and Ne are the odd and even parts of N; Do and, De are the odd and even parts of D. Since N(s) must be either odd or, even, we can write Eq. (19.83) as, No, , (Ne 0), Do De, H(s) µ, Ne, , (No 0), Do De, , (19.84), , and can rewrite this as, NoDe, , (Ne 0), 1 DoDe, H(s) µ, NeDo, , (No 0), 1 DeDo, , (19.85), , Comparing this with Eq. (19.82), we obtain the y parameters of the, network as, No, , (Ne 0), y21, De, µ, Ne, YL, , (No 0), Do, , (19.86), , Do, , (Ne 0), y22, De, µ, De, YL, , (No 0), Do, , (19.87), , and, , The following example illustrates the procedure., , Design the LC ladder network terminated with a 1- resistor that has, the normalized transfer function, H(s) , , 1, s3 2s2 2s 1, , (This transfer function is for a Butterworth lowpass filter.), Solution:, The denominator shows that this is a third-order network, so that the, LC ladder network is shown in Fig. 19.63(a), with two inductors and, one capacitor. Our goal is to determine the values of the inductors and, , Example 19.18
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ale29559_ch19.qxd, , 07/16/2008, , 05:49 PM, , Page 889, , 19.10, , Summary, , Realize the following transfer function using an LC ladder network terminated in a 1- resistor:, H(s) , , 2, s3 s2 4s 2, , Answer: Ladder network in Fig. 19.63(a) with L1 L3 1.0 H and, C2 0.5 F., , 19.10, , Summary, , 1. A two-port network is one with two ports (or two pairs of access, terminals), known as input and output ports., 2. The six parameters used to model a two-port network are the, impedance [z], admittance [y], hybrid [h], inverse hybrid [g],, transmission [T], and inverse transmission [t] parameters., 3. The parameters relate the input and output port variables as, c, c, , V1, I1, d [z] c d ,, V2, I2, , I1, V1, d [g] c d ,, V2, I2, , c, c, , I1, V1, d [y] c d ,, I2, V2, , c, , V1, V2, d [T] c, d,, I1, I2, , V1, I1, d [h] c d, I2, V2, c, , V2, V1, d [t] c, d, I2, I1, , 4. The parameters can be calculated or measured by short-circuiting, or open-circuiting the appropriate input or output port., 5. A two-port network is reciprocal if z12 z21, y12 y21, h12 , h21, g12 g21, ¢ T 1 or ¢ t 1. Networks that have dependent, sources are not reciprocal., 6. Table 19.1 provides the relationships between the six sets of, parameters. Three important relationships are, [y] [z]1,, , [g] [h]1,, , [t] [T]1, , 7. Two-port networks may be connected in series, in parallel, or in, cascade. In the series connection the z parameters are added, in the, parallel connection the y parameters are added, and in the cascade, connection the transmission parameters are multiplied in the correct order., 8. One can use PSpice to compute the two-port parameters by constraining the appropriate port variables with a 1-A or 1-V source, while using an open or short circuit to impose the other necessary, constraints., 9. The network parameters are specifically applied in the analysis of, transistor circuits and the synthesis of ladder LC networks. Network parameters are especially useful in the analysis of transistor, circuits because these circuits are easily modeled as two-port networks. LC ladder networks, important in the design of passive lowpass filters, resemble cascaded T networks and are therefore best, analyzed as two-ports., , 889, , Practice Problem 19.18
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ale29559_ch19.qxd, , 07/16/2008, , 05:49 PM, , Page 890, , Chapter 19, , 890, , Two-Port Networks, , Review Questions, 19.1, , For the single-element two-port network in, Fig. 19.64(a), z11 is:, (a) 0, , (b) 5, , (d) 20, , (e) undefined, , 19.6, , (c) 10, , 10 Ω, , 19.7, , 10 Ω, , (b), , (a), , 19.8, , For the single-element two-port network in, Fig. 19.64(b), B is:, (a) 0, , (b) 5, , (d) 20, , (e) undefined, , (c) 10, , When port 1 of a two-port circuit is short-circuited,, I1 4I2 and V2 0.25I2. Which of the following, is true?, (a) y11 4, , (b) y12 16, , (c) y21 16, , (d) y22 0.25, , A two-port is described by the following equations:, , Figure 19.64, , V1 50I1 10I2, , For Review Questions., , V2 30I1 20I2, , 19.2, , 19.3, , 19.4, , (a) 0, , (b) 5, , (c) 10, , (d) 20, , (e) undefined, , 19.9, , For the single-element two-port network in, Fig. 19.64(a), y11 is:, (a) 0, , (b) 5, , (c) 10, , (d) 20, , (e) undefined, , For the single-element two-port network in, Fig. 19.64(b), h21 is:, (a) 0.1 (b) 1, (d) 10, , 19.5, , Which of the following is not true?, , For the single-element two-port network in, Fig. 19.64(b), z11 is:, , (c) 0, , (e) undefined, , For the single-element two-port network in, Fig. 19.64(a), B is:, (a) 0, , (b) 5, , (d) 20, , (e) undefined, , (a) z12 10, , (b) y12 0.0143, , (c) h12 0.5, , (d) A 50, , If a two-port is reciprocal, which of the following is, not true?, (a) z21 z12, , (b) y21 y12, , (c) h21 h12, , (d) AD BC 1, , 19.10 If the two single-element two-port networks in, Fig. 19.64 are cascaded, then D is:, (a) 0, , (b) 0.1, , (c) 2, , (d) 10, , (e) undefined, , Answers: 19.1c, 19.2e, 19.3e, 19.4b, 19.5a, 19.6c, 19.7b,, 19.8d, 19.9c, 19.10c., , (c) 10, , Problems, Section 19.2 Impedance Parameters, 19.1, , Obtain the z parameters for the network in, Fig. 19.65., 1Ω, , *19.2 Find the impedance parameter equivalent of the, network in Fig. 19.66., 1Ω, , 1Ω, , 4Ω, , 6Ω, , 1Ω, , 2Ω, , 1Ω, , 1Ω, , 1Ω, 1Ω, , 1Ω, 1Ω, , 1Ω, , Figure 19.66, For Prob. 19.2., , Figure 19.65, For Probs. 19.1 and 19.28, , * An asterisk indicates a challenging problem., , 1Ω
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ale29559_ch19.qxd, , 07/16/2008, , 05:49 PM, , Page 891, , Problems, , 19.3, , Find the z parameters of the circuit in Fig. 19.67., , 19.8, , 891, , Find the z parameters of the two-port in Fig. 19.72., , − j 10 Ω, , 4Ω, , j4 Ω, , − j2 Ω, , 5Ω, , j6 Ω, j8 Ω, , j6 Ω, , Figure 19.67, For Prob. 19.3., 19.4, , Using Fig. 19.68, design a problem to help other, students better understand how to determine z, parameters from an electrical circuit., , 10 Ω, , Figure 19.72, , jXL, , For Prob. 19.8., – jXC, , R, , 19.9, , The y parameters of a network are:, [ y] c, , Figure 19.68, For Prob. 19.4., 19.5, , Obtain the z parameters for the network in Fig. 19.69, as functions of s., 2Ω, 2Ω, , 2H, , 0.5 F, , Determine the z parameters for the network., 19.10 Construct a two-port that realizes each of the, following z parameters., (a) [z] c, , 0.5 F, , For Prob. 19.5., Compute the z parameters of the circuit in Fig. 19.70., I1, , 5Ω, , 10 Ω, , +, , 4I1, , 1, , − +, , −, , 1, s, , Calculate the impedance-parameter equivalent of the, circuit in Fig. 19.71., , 2Ω, , 10 6, d , 4 12, , I1, , I2, , +, , 100 Ω, 3A, +, 50 Ω, , 4Ω, , V1, −, , 60 Ω, , −, , +, [z], , V2, , 10 Ω, , −, , Figure 19.73, −, +, , For Prob. 19.7 and 19.80., , ¥ , , Find I1, I2, V1, and V2., , For Prob. 19.6 and 19.73., , Figure 19.71, , 1, s, , 6 j3 5 j2, d, 5 j2 8 j, , [z] c, , Figure 19.70, , vx, , 2s , , 19.12 For the circuit shown in Fig. 19.73, let, , −, , 20 Ω, , 1, s, , [z] c, , +, V2, , 3, s, , 19.11 Determine a two-port network that is represented by, the following z parameters:, , I2, , 20 Ω, , V1, , 19.7, , 25 20, d , 5 10, , (b) [z] ≥, , Figure 19.69, 19.6, , 0.25 0.1, dS, 0.1 0.2, , For Prob. 19.12., 12vx, , 19.13 Determine the average power delivered to ZL , 5 j4 in the network of Fig. 19.74. Note: The, voltage is rms.
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ale29559_ch19.qxd, , 07/16/2008, , 05:49 PM, , Page 892, , Chapter 19, , 892, , Two-Port Networks, , 10 Ω, , Section 19.3 Admittance Parameters, z11 = 40 Ω, z12 = 60 Ω, z21 = 80 Ω, z22 = 100 Ω, , +, −, , 50 0° V, , ZL, , *19.17 Determine the z and y parameters for the circuit in, Fig. 19.78., , 4Ω, , Figure 19.74, For Prob. 19.13., 2Ω, , 19.14 For the two-port network shown in Fig. 19.75, show, that at the output terminals,, ZTh z22 , , z12z21, z11 Zs, , 6Ω, , Figure 19.78, , and, , For Prob. 19.17., VTh, , z21, , Vs, z11 Zs, , I1, , Zs, , +, −, , 19.18 Calculate the y parameters for the two-port in, Fig. 19.79., , I2, , +, Vs, , 8Ω, , 6Ω, , +, Two-port, network, , V1, , V2, , −, , 3Ω, , ZL, , 6Ω, , −, , 3Ω, , Figure 19.75, , Figure 19.79, , For Probs. 19.14 and 19.41., , For Probs. 19.18 and 19.37., , 19.15 For the two-port circuit in Fig. 19.76,, , 19.19 Using Fig. 19.80, design a problem to help other, students better understand how to find y parameters, in the s-domain., , [z] c, , 40 60, d , 80 120, , (a) Find ZL for maximum power transfer to the load., , R1, , (b) Calculate the maximum power delivered to the, load., C, , L, , 10 Ω, , 120 V rms, , R2, , +, −, , ZL, , [z], , Figure 19.80, For Prob. 19.19., , Figure 19.76, , 19.20 Find the y parameters for the circuit in Fig. 19.81., , For Prob. 19.15., 19.16 For the circuit in Fig. 19.77, at 2 rad/s,, z11 10 , z12 z21 j6 , z22 4 . Obtain, the Thevenin equivalent circuit at terminals a-b and, calculate vo., 5Ω, 15 cos 2t V +, −, , 3ix, , 2Ω, ix, , a, [z], , 2H, , +, vo, −, , 4Ω, , b, , Figure 19.77, , Figure 19.81, , For Prob. 19.16., , For Prob. 19.20., , 6Ω
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ale29559_ch19.qxd, , 07/16/2008, , 05:49 PM, , Page 893, , Problems, , 893, 4Ω, , 19.21 Obtain the admittance parameter equivalent circuit, of the two-port in Fig. 19.82., 0.2V1, +, , +, Vx, −, , +, 5Ω, , V1, , 2Vx, , 2Ω, , 10 Ω, , V2, , −, , −, , 1Ω, , Figure 19.85, For Prob. 19.26., , Figure 19.82, For Prob. 19.21., , 19.27 Find the y parameters for the circuit in Fig. 19.86., I1, , 19.22 Obtain the y parameters of the two-port network in, Fig. 19.83., , I2, , 4Ω, , +, I1, +, , 0.1V2, , V1, , I2, , 5Ω, , +, −, +, , 10 Ω, , 20I1, , V2, , −, , −, , +, 5Ω, , V1, , Figure 19.86, , 2 Ω V2, , 0.5V2, , −, , For Prob. 19.27., , −, , 19.28 In the circuit of Fig. 19.65, the input port is, connected to a 1-A dc current source. Calculate the, power dissipated by the 2- resistor by using the y, parameters. Confirm your result by direct circuit, analysis., , Figure 19.83, For Prob. 19.22., 19.23 (a) Find the y parameters of the two-port in Fig. 19.84., (b) Determine V2(s) for vs 2u(t) V., , 19.29 In the bridge circuit of Fig. 19.87, I1 10 A and, I2 4 A., (a) Find V1 and V2 using y parameters., , 1, , (b) Confirm the results in part (a) by direct circuit, analysis., 1, s, , 1, , 3Ω, , +, vs, , +, −, , V1, −, , 1, , s, , +, V2, −, , 3Ω, , 2, , 3Ω, , +, V1, , I1, , +, 1Ω, , V2, , −, , Figure 19.84, , I2, , −, , Figure 19.87, , For Prob. 19.23., , For Prob. 19.29., 19.24 Find the resistive circuit that represents these y, parameters:, , Section 19.4 Hybrid Parameters, , 1, 1, , 4, 8, ¥S, [y] ≥, 1, 3, , 8 16, 19.25 Draw the two-port network that has the following y, parameters:, , 19.30 Find the h parameters for the networks in Fig. 19.88., , 1, 0.5, dS, 0.5 1.5, 19.26 Calculate [y] for the two-port in Fig. 19.85., [y] c, , 40 Ω, , 10 Ω, , 60 Ω, , (a), , Figure 19.88, For Prob. 19.30., , 20 Ω, , (b)
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ale29559_ch19.qxd, , 07/16/2008, , 05:49 PM, , Page 894, , Chapter 19, , 894, , Two-Port Networks, 1Ω, , 19.31 Determine the hybrid parameters for the network in, Fig. 19.89., I1, , 1Ω, , 2Ω, , 1Ω, , +, , 4Ω, 1:2, , I2, +, , 2Ω, , V1, , 4I1, , −, , V2, −, , Figure 19.89, , Figure 19.93, For Prob. 19.35., 19.36 For the two-port in Fig. 19.94,, , For Prob. 19.31., 19.32 Using Fig. 19.90, design a problem to help other, students better understand how to find the h and g, parameters for a circuit in the s-domain., L1, , R, , L2, , [h] c, , 16 , 3, d, 2 0.01 S, , Find:, (a) V2V1, , (b) I2I1, , (c) I1V1, , (d) V2I1, , C, , I1, , 4Ω, , I2, , +, , Figure 19.90, , 10 V +, −, , For Prob. 19.32., , +, , V1, , V2, , [h], , −, , 19.33 Obtain the h parameters for the two-port of Fig. 19.91., , 25 Ω, , −, , Figure 19.94, For Prob. 19.36., , 4Ω, , j6 Ω, , − j3 Ω, , 5Ω, , Figure 19.91, , 19.37 The input port of the circuit in Fig. 19.79 is, connected to a 10-V dc voltage source while the, output port is terminated by a 5- resistor. Find the, voltage across the 5- resistor by using h parameters, of the circuit. Confirm your result by using direct, circuit analysis., 19.38 The h parameters of the two-port of Fig. 19.95 are:, , For Prob. 19.33., 19.34 Obtain the h and g parameters of the two-port in, Fig. 19.92., , 600 0.04, d, 30, 2 mS, Given the Zs 2 k and ZL 400 , find Zin and, Zout., [h] c, , 300 Ω, 10 Ω, , Zs, +, , +, , V1, , V2, , 50 Ω, +, Vx, −, , 100 Ω, , −, +, , Vs +, −, , −, , 10Vx, Zin, , [h], , −, , ZL, , Zout, , Figure 19.92, , Figure 19.95, , For Prob. 19.34., , For Prob. 19.38., , 19.35 Determine the h parameters for the network in, Fig. 19.93., , 19.39 Obtain the g parameters for the wye circuit of, Fig. 19.96.
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ale29559_ch19.qxd, , 07/16/2008, , 05:49 PM, , Page 895, , Problems, I1, , R1, , R3, , I2, , jXL, , +, , +, R2, , V1, , 895, , – jXC1, , – jXC2, , V2, , −, , −, , R, , Figure 19.96, For Prob. 19.39., , Figure 19.99, , 19.40 Using Fig. 19.97, design a problem to help other, students better understand how to find g parameters, in an ac circuit., – j XC, , For Prob. 19.44., 19.45 Find the ABCD parameters for the circuit in, Fig. 19.100., – j4 Ω, , j XL, , 8Ω, , R, , Figure 19.100, , Figure 19.97, For Prob. 19.40., , For Prob. 19.45., , 19.41 For the two-port in Fig. 19.75, show that, , 19.46 Find the transmission parameters for the circuit in, Fig. 19.101., , g21, I2, , I1, g11ZL ¢ g, , 1Ω, , 1Ω, Ix, , g21ZL, V2, , Vs, (1 g11Zs)(g22 ZL) g21g12Zs, , 2Ω, , 4Ix, , where ¢ g is the determinant of [g] matrix., 19.42 The h parameters of a two-port device are given by, h11 600 ,, , h12 103,, , h21 120,, , h22 2 106 S, Draw a circuit model of the device including the, value of each element., , Figure 19.101, For Prob. 19.46., 19.47 Obtain the ABCD parameters for the network in, Fig. 19.102., 6Ω, , Section 19.5 Transmission Parameters, 19.43 Find the transmission parameters for the singleelement two-port networks in Fig. 19.98., , 1Ω, , 4Ω, , +, Vx, −, , Z, , 2Ω, , +, −, , 5Vx, , Y, , Figure 19.102, For Prob. 19.47, (a), , (b), , Figure 19.98, For Prob. 19.43., 19.44 Using Fig. 19.99, design a problem to help other, students better understand how to find the transmission, parameters of an ac circuit., , 19.48 For a two-port, let A 4, B 30 , C 0.1 S,, and D 1.5. Calculate the input impedance, Zin V1I1, when:, (a) the output terminals are short-circuited,, (b) the output port is open-circuited,, (c) the output port is terminated by a 10- load.
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ale29559_ch19.qxd, , 07/16/2008, , 05:49 PM, , Page 897, , Problems, , 19.60 Design a T network necessary to realize the, following z parameters at 106 rad/s., 4 j3, 2, [z] c, d k, 2, 5j, , 897, 40 kΩ, , I1, , 20 kΩ, , 10 kΩ, , +, , +, , 1 F, , V1, , 19.61 For the bridge circuit in Fig. 19.108, obtain:, , I2, , −, +, , V2, , −, , −, , (a) the z parameters, (b) the h parameters, , Figure 19.111, , (c) the transmission parameters, , For Prob. 19.64., , 2Ω, 2Ω, , Section 19.7 Interconnection of Networks, 19.65 What is the y parameter presentation of the circuit in, Fig. 19.112?, , 2Ω, 2Ω, , 2Ω, , I1, , I2, , Figure 19.108, , 2Ω, , +, , For Prob. 19.61., , +, V2, , V1, −, , 19.62 Find the z parameters of the op amp circuit in, Fig. 19.109. Obtain the transmission parameters., , 1Ω, , −, , 1Ω, , Figure 19.112, For Prob. 19.65., I1, , 10 kΩ, +, −, , +, , 40 kΩ, , I2, +, , 50 kΩ, V1, , 30 kΩ, , V2, 20 kΩ, , −, , 19.66 In the two-port of Fig. 19.113, let y12 y21 0,, y11 2 mS, and y22 10 mS. Find VoVs., , −, , 60 Ω, +, [y], , Figure 19.109, For Prob. 19.62., , Vs, , +, −, , Vo, 100 Ω, , 19.63 Determine the z parameters of the two-port in, Fig. 19.110., , 300 Ω, , −, , Figure 19.113, For Prob. 19.66., 1:3, , 4Ω, , 9Ω, , 19.67 If three copies of the circuit in Fig. 19.114 are, connected in parallel, find the overall transmission, parameters., 30 Ω, , Figure 19.110, , 40 Ω, , For Prob. 19.63., 10 Ω, , 19.64 Determine the y parameters at 1,000 rad/s for the, op amp circuit in Fig. 19.111. Find the corresponding h, parameters., , Figure 19.114, For Prob. 19.67.
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ale29559_ch19.qxd, , 07/16/2008, , 05:49 PM, , Page 898, , Chapter 19, , 898, , Two-Port Networks, 8Ω, , 19.68 Obtain the h parameters for the network in Fig. 19.115., , 6Ω, 10 Ω, , 2Ω, 2Ω, , 4Ω, , 1:2, , 2Ω, , 5Ω, 2Ω, , 1Ω, 1Ω, , Figure 19.118, , 1Ω, , For Prob. 19.71., , Figure 19.115, For Prob. 19.68., , *19.72 A series-parallel connection of two two-ports is, shown in Fig. 19.119. Determine the z parameter, representation of the network., , *19.69 The circuit in Fig. 19.116 may be regarded as two, two-ports connected in parallel. Obtain the y, parameters as functions of s., , I1, h11 = 25 Ω, h12 = 4, h21 = – 4, h22 = 1 S, , +, 1F, , 2Ω, 2:1, , V1, , I2, +, V2, −, , h11 = 16 Ω, h12 = 1, h21 = –1, h22 = 0.5 S, , −, , Figure 19.119, , 2Ω, , For Prob. 19.72., 1H, , 19.73 Three copies of the circuit shown in Fig. 19.70 are, connected in cascade. Determine the z parameters., , Figure 19.116, For Prob. 19.69., , *19.70 For the parallel-series connection of the two, two-ports in Fig. 19.117, find the g parameters., , *19.74 Determine the ABCD parameters of the circuit in, Fig. 19.120 as functions of s. (Hint: Partition the, circuit into subcircuits and cascade them using the, results of Prob. 19.43.), , 1H, , 1H, , I2, I1, +, V1, −, , z11 = 25 Ω, z12 = 20 Ω, z21 = 5 Ω, z22 = 10 Ω, z11 = 50 Ω, z12 = 25 Ω, z21 = 25 Ω, z22 = 30 Ω, , +, , V2, , 1Ω, , 1F, , 1Ω, , 1F, , Figure 19.120, For Prob. 19.74., , −, , Figure 19.117, For Prob. 19.70., , *19.71 Determine the z parameters for the network in, Fig. 19.118., , *19.75 For the individual two-ports shown in Fig. 19.121, where,, [za] c, , 8 6, 8 4, d [yb] c, d S, 4 5, 2 10, , (a) Determine the y parameters of the overall two-port., (b) Find the voltage ratio VoVi when ZL 2 .
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ale29559_ch19.qxd, , 07/16/2008, , 05:49 PM, , Page 899, , Problems, , Vi, , +, −, , Nb, , Na, , ZL, , +, Vo, −, , 899, , 19.79 Use PSpice to determine the z parameters of the, circuit in Fig. 19.125. Take 2 rad/s., 1Ω, , 0.25 F, , 2Ω, , Figure 19.121, For Prob. 19.75., 4Ω, , 2H, , Section 19.8 Computing Two-Port Parameters, Using PSpice, 19.76 Use PSpice to obtain the z parameters of the network, in Fig. 19.122., , 4Ω, , 19.80 Use PSpice to find the z parameters of the circuit in, Fig. 19.71., , 10 Ω, , 19.82 Use PSpice to rework Prob. 19.31., , 1Ω, , 4Ω, , 10 Ω, , For Prob. 19.79., , 19.81 Repeat Prob. 19.26 using PSpice., , 8Ω, 1Ω, , Figure 19.125, , 19.83 Rework Prob. 19.47 using PSpice., , 10 Ω, , 6Ω, , 6Ω, , 19.84 Using PSpice, find the transmission parameters for, the network in Fig. 19.126., 1Ω, , 2Ω, , + V −, o, 1Ω, , Figure 19.122, , 1Ω, , For Prob. 19.76., Vo, 2, , 2Ω, , 19.77 Using PSpice, find the h parameters of the network, in Fig. 19.123. Take 1 rad/s., , 2Ω, , Figure 19.126, For Prob. 19.84., , +, , 1Ω, , 1F, −, , +, , −, , 19.85 At 1 rad/s, find the transmission parameters of, the network in Fig. 19.127 using PSpice., , 2Ω, , 1Ω, 1Ω, , 1H, , Figure 19.123, , 1F, , 1H, , For Prob. 19.77., 19.78 Obtain the h parameters at 4 rad/s for the circuit, in Fig. 19.124 using PSpice., , 1H, , 1F, , Figure 19.127, For Prob. 19.85., 19.86 Obtain the g parameters for the network in Fig. 19.128, using PSpice., , 1H, , ix, 4Ω, , 4Ω, 1, 8, , 2Ω, 1Ω, , F, , Figure 19.124, , Figure 19.128, , For Prob. 19.78., , For Prob. 19.86., , 3Ω, , 2A, , 5ix
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ale29559_ch19.qxd, , 07/16/2008, , 05:49 PM, , Page 900, , Chapter 19, , 900, , Two-Port Networks, , 19.87 For the circuit shown in Fig. 19.129, use PSpice to, obtain the t parameters. Assume 1 rad/s., , *19.92 Determine Av, Ai, Zin, and Zout for the amplifier, shown in Fig. 19.131. Assume that, , j2 Ω, 1Ω, , 1Ω, , hie 4 k,, , hre 104, , hfe 100,, , hoe 30 mS, , 1Ω, , – j2 Ω, , – j2 Ω, 1.2 kΩ, , Figure 19.129, , 4 kΩ, 240 Ω, , Vs +, −, , For Prob. 19.87., , Section 19.9 Applications, 19.88 Using the y parameters, derive formulas for Zin, Zout,, Ai, and Av for the common-emitter transistor circuit., 19.89 A transistor has the following parameters in a, common-emitter circuit:, , For Prob. 19.92., *19.93 Calculate Av, Ai, Zin, and Zout for the transistor, network in Fig. 19.132. Assume that, , hre 2.6 104, , hie 2,640 ,, hfe 72,, , Figure 19.131, , hoe 16 mS,, , hre 2.5 104, , hie 2 k,, , RL 100 k, , hfe 150,, , hoe 10 mS, , What is the voltage amplification of the transistor?, How many decibels gain is this?, 19.90 A transistor with, hfe 120,, , hie 2 k, , hre 104,, , hoe 20 mS, , 1 kΩ, 3.8 kΩ, 0.2 kΩ, , Vs +, −, , is used for a CE amplifier to provide an input, resistance of 1.5 k., (a) Determine the necessary load resistance RL., (b) Calculate Av, Ai, and Zout if the amplifier is driven, by a 4-mV source having an internal resistance of, 600 ., (c) Find the voltage across the load., , Figure 19.132, For Prob. 19.93., 19.94 A transistor in its common-emitter mode is specified, by, [h] c, , 19.91 For the transistor network of Fig. 19.130,, hfe 80,, , hie 1.2 k, 4, , hre 1.5 10 ,, , hoe 20 mS, , Determine the following:, (a) voltage gain Av VoVs,, , Two such identical transistors are connected in, cascade to form a two-stage amplifier used at audio, frequencies. If the amplifier is terminated by a 4-k, resistor, calculate the overall Av and Zin., 19.95 Realize an LC ladder network such that, , (b) current gain Ai IoIi,, (c) input impedance Zin,, , y22 , , (d) output impedance Zout., Ii, 2 kΩ, Vs +, −, , Io, 2.4 kΩ, , +, Vo, −, , For Prob. 19.91., , s3 5s, s 10s2 8, 4, , 19.96 Design an LC ladder network to realize a lowpass, filter with transfer function, H(s) , , 1, s 2.613s 3.414s2 2.613s 1, 4, , 2, , 19.97 Synthesize the transfer function, H(s) , , Figure 19.130, , 200 , 0, d, 100 106 S, , Vo, s3, 3, Vs, s 6s 12s 24, , using the LC ladder network in Fig. 19.133.
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ale29559_ch19.qxd, , 07/16/2008, , 05:49 PM, , Page 901, , Comprehensive Problem, C1, , If ZL 20 k, find the required value of Vs to, produce Vo 16 V., , C3, , +, , +, L2, , Vs, , 1Ω, , Vo, , −, , 901, , 1Ω, , −, Vs, , Figure 19.133, , +, −, , [hb ], , [ha], , ZL, , For Prob. 19.97., 19.98 A two-stage amplifier in Fig. 19.134 contains two, identical stages with, [h] c, , Figure 19.134, For Prob. 19.98., , 2 k 0.004, d, 200 500 mS, , Comprehensive Problem, 19.99 Assume that the two circuits in Fig. 19.135 are, equivalent. The parameters of the two circuits must, be equal. Using this factor and the z parameters,, derive Eqs. (9.67) and (9.68)., Z1, a, , Zb, a, , c, Zc, , Za, , Z2, , n, , c, , b, , d, (b), , Z3, , Figure 19.135, For Prob. 19.99., , b, , d, (a), , +, Vo, −
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ale29559_appA.qxd, , 07/16/2008, , 06:22 PM, , A-2, , Page A-2, , Appendix A, , Simultaneous Equations and Matrix Inversion, , An alternative method of obtaining the determinant of a 3 3 matrix, is by repeating the first two rows and multiplying the terms diagonally, as follows., a11 a12 a13, a21 a22 a23, ¢, 5 a31 a32 a33 5, a11 a12 a13 , a21 a22 a23 , , , a11a22 a33 a21a32 a13 a31a12 a23 a13 a22 a31 a23 a32 a11, a33 a12 a21, (A.11), In summary:, , The solution of linear simultaneous equations by Cramer’s rule boils, down to finding, xk , , ¢k, ,, ¢, , k 1, 2, . . . , n, , (A.12), , where ¢ is the determinant of matrix A and ¢ k is the determinant of, the matrix formed by replacing the k th column of A by B., , One may use other methods, such, as matrix inversion and elimination., Only Cramer’s method is covered, here, because of its simplicity and, also because of the availability of, powerful calculators., , Example A.1, , You may not find much need to use Cramer’s method described in, this appendix, in view of the availability of calculators, computers, and, software packages such as MATLAB, which can be used easily to solve, a set of linear equations. But in case you need to solve the equations, by hand, the material covered in this appendix becomes useful. At any, rate, it is important to know the mathematical basis of those calculators and software packages., , Solve the simultaneous equations, 4x1 3x2 17,, , 3x1 5x2 21, , Solution:, The given set of equations is cast in matrix form as, c, , 4 3 x1, 17, d c d c, d, 3, 5 x2, 21, , The determinants are evaluated as, ¢2, , 4 3, 2 4 5 (3)(3) 11, 3, 5, , ¢1 2, , 17 3, 2 17 5 (3)(21) 22, 21, 5, , ¢2 2, , 4, 17, 2 4 (21) 17 (3) 33, 3 21
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ale29559_appA.qxd, , 07/16/2008, , 06:22 PM, , Page A-3, , Appendix A, , Simultaneous Equations and Matrix Inversion, , A-3, , Hence,, x1 , , ¢1, 22, , 2,, ¢, 11, , x2 , , ¢2, 33, , 3, ¢, 11, , Find the solution to the following simultaneous equations:, 3x1 x2 4,, , Practice Problem A.1, , 6x1 18x2 16, , Answer: x1 1.833, x2 1.5., , Determine x1, x2, and x3 for this set of simultaneous equations:, 25x1 5x2 20x3 50, 5x1 10x2 4x3 0, 5x1 4x2 9x3 0, Solution:, In matrix form, the given set of equations becomes, 50, 25 5 20 x1, £ 5 10 4 § £ x2 § £ 0 §, 0, 5 4, 9 x3, We apply Eq. (A.11) to find the determinants. This requires that we, repeat the first two rows of the matrix. Thus,, 25 5 20, 5 10 4, 5 5 4, 95, 25 5 20, 5 10 4, , 25 5 20, ¢ 3 5 10 4 3 , 5 4, 9, , , , , , , 25(10)9 (5)(4)(20) (5)(5)(4), (20)(10)(5) (4)(4)25 9(5)(5), 2250 400 100 1000 400 225 125, Similarly,, 50 5 20, 0 10 4, 5 0 4, 95, 50 5 20, 0 10 4, , 50 5 20, ¢ 1 3 0 10 4 3 , 0 4, 9, , , , , , , 4500 0 0 0 800 0 3700, , Example A.2
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ale29559_appA.qxd, , 07/16/2008, , 06:23 PM, , A-4, , Page A-4, , Appendix A, , Simultaneous Equations and Matrix Inversion, , 25 50 20, ¢ 2 3 5 0 4 3 , 5 0, 9, , , , 0 0 1000 0 , , 25 50 20, 5 0 4, 5 5 0, 95, 25 50 20, 5 0 4, 0 2250 3250, , 25 5, 25 5 50, 5 10, ¢ 3 3 5 10 0 3 , 5 5 4, 5 4 0, 25 5, 5 10, , 0 1000 0 2500 0 0, , , , , , 50, 0, 05, 50, 0, , , , , 3500, , Hence, we now find, ¢1, 3700, , 29.6, ¢, 125, ¢2, 3250, x2 , , 26, ¢, 125, ¢2, 3500, , 28, x3 , ¢, 125, x1 , , Practice Problem A.2, , Obtain the solution of this set of simultaneous equations:, 3x1 x2 2x3 1, x1 6x2 3x3 0, 2x1 3x2 6x3 6, Answer: x1 3 x3, x2 2., , A.2, , Matrix Inversion, , The linear system of equations in Eq. (A.3) can be solved by matrix, inversion. In the matrix equation AX B, we may invert A to get X,, i.e.,, X A1B, , (A.13), , where A1 is the inverse of A. Matrix inversion is needed in other, applications apart from using it to solve a set of equations., By definition, the inverse of matrix A satisfies, A1A AA1 I, , (A.14)
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ale29559_appA.qxd, , 07/16/2008, , 06:23 PM, , Page A-7, , Appendix A, , Simultaneous Equations and Matrix Inversion, , A-7, , Hence,, X A1B , , 1 3 10 2, 1 64, 4, c, d c d , c, d c, d, 16 1, 2 7, 16, 16, 1, , i.e., x1 4 and x2 1., , Practice Problem A.3, , Solve the following two equations by matrix inversion., 2y1 y2 4, y1 3y2 9, Answer: y1 3, y2 2., , Determine x1, x2, and x3 for the following simultaneous equations using, matrix inversion., x1 x2 x3 5, x1 2x2 9, 4x1 x2 x3 2, Solution:, In matrix form, the equations become, 1 1, 1 x1, 5, £ 1 2, 0 § £ x2 § £ 9 §, 4 1 1 x3, 2, or, AX B ¡ X A1B, where, 1 1, 1, A £ 1 2, 0§,, 4 1 1, , x1, X £ x2 § ,, x3, , 5, B £ 9§, 2, , We now find the cofactors, c11 2, , 2, 0, 1, 0, 1 2, 2 2, c12 2, 2 1, c13 2, 2 9, 1 1, 4 1, 4 1, , c21 2, c31 2, , 1, 1, 1, 1, 2 2, c22 2, 2 5,, 1 1, 4 1, , 1 1, 2 2,, 2 0, , c32 2, , c23 2, , 1 1, 23, 4 1, , 1 1, 1 1, 2 1, c33 2, 23, 1 0, 1 2, , Example A.4
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ale29559_appA.qxd, , 07/16/2008, , 06:23 PM, , A-8, , Page A-8, , Appendix A, , Simultaneous Equations and Matrix Inversion, , The adjoint of matrix A is, 2 1 9 T, 2, 2 2, adj A £ 2 5, 3 § £ 1 5 1 §, 2 1, 3, 9, 3, 3, We can find the determinant of A using any row or column of A. Since, one element of the second row is 0, we can take advantage of this to, find the determinant as, 0A 0 1c21 2c22 (0)c23 1(2) 2(5) 12, Hence, the inverse of A is, A1 , , 2, 2 2, 1, £ 1 5 1 §, 12, 9, 3, 3, , 2, 2 2, 5, 1, 1, XA B, £ 1 5 1 § £ 9 § £ 4 §, 12, 9, 3, 3 2, 2, 1, , i.e., x1 1, x2 4, x3 2., , Practice Problem A.4, , Solve the following equations using matrix inversion., y1 y3 1, 2y1 3y2 y3 1, y1 y2 y3 3, Answer: y1 6, y2 2, y3 5.
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ale29559_appB.qxd, , 07/16/2008, , 06:53 PM, , Page A-9, , Appendix B, Complex Numbers, The ability to manipulate complex numbers is very handy in circuit, analysis and in electrical engineering in general. Complex numbers are, particularly useful in the analysis of ac circuits. Again, although calculators and computer software packages are now available to manipulate complex numbers, it is still advisable for a student to be familiar, with how to handle them by hand., , B.1, , Representations of Complex, Numbers, , A complex number z may be written in rectangular form as, z x jy, , (B.1), , where j 11; x is the real part of z while y is the imaginary part, of z; that is,, x Re(z),, , y Im(z), , (B.2), , The complex number z is shown plotted in the complex plane in, Fig. B.1. Since j 11,, 1, j, j, j 2 1, , Im, z, , jy, r, , j j j j, j4 j2 j2 1, j5 j j4 j, o, n4, j, jn, 3, , The complex plane looks like the, two-dimensional curvilinear coordinate, space, but it is not., , 2, , (B.3), , y, , , 0, , x, , Re, , Figure B.1, Graphical representation of a, complex number., , A second way of representing the complex number z is by specifying its magnitude r and the angle u it makes with the real axis, as, Fig. B.1 shows. This is known as the polar form. It is given by, z 0z 0 lu rlu, , (B.4), , where, r 2x2 y2,, , u tan 1, , y, x, , (B.5a), , or, x r cos u,, , y r sin u, , (B.5b), , that is,, z x jy rlu r cos u jr sin u, , (B.6), A-9
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ale29559_appB.qxd, , 07/16/2008, , 06:53 PM, , A-10, , Page A-10, , Appendix B, , Complex Numbers, , In converting from rectangular to polar form using Eq. (B.5), we must, exercise care in determining the correct value of u. These are the four, possibilities:, , In the exponential form, z re ju so, that dzd u jre j u jz., , y, x, , z x jy,, , u tan1, , z x jy,, , u 180 tan 1, , y, x, , (2nd Quadrant), , z x jy,, , u 180 tan 1, , y, x, , (3rd Quadrant), , z x jy,, , u 360 tan 1, , y, x, , (4th Quadrant), , (1st Quadrant), , (B.7), , assuming that x and y are positive., The third way of representing the complex z is the exponential, form:, z re ju, , (B.8), , This is almost the same as the polar form, because we use the same, magnitude r and the angle u., The three forms of representing a complex number are summarized as follows., z x jy,, , (x r cos u, y r sin u), , Rectangular form, , z rlu,, , y, ar 2x2 y2, u tan 1 b, x, , Polar form, , z re ju,, , y, ar 2x2 y2, u tan 1 b Exponential form, x, (B.9), , The first two forms are related by Eqs. (B.5) and (B.6). In Section B.3, we will derive Euler’s formula, which proves that the third form is also, equivalent to the first two., , Example B.1, , Express the following complex numbers in polar and exponential form:, (a) z1 6 j8, (b) z2 6 j8, (c) z3 6 j8, (d) z4 6 j8., Solution:, Notice that we have deliberately chosen these complex numbers to fall, in the four quadrants, as shown in Fig. B.2., (a) For z1 6 j8 (1st quadrant),, r1 262 82 10,, , 8, u1 tan 1 53.13, 6, , Hence, the polar form is 10l53.13 and the exponential form is 10e j53.13., (b) For z2 6 j8 (4th quadrant),, r2 262 (8)2 10,, , 8, u2 360 tan 1 306.87, 6
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ale29559_appB.qxd, , 07/16/2008, , 06:53 PM, , A-14, , Practice Problem B.4, , Page A-14, , Appendix B, , Complex Numbers, , Evaluate these complex fractions:, (a), , 6l30 j5 3, 1 j 2e, , (b) B, , j45, , (15 j7)(3 j2)* *, R, (4 j6)*(3l70), , Answer: (a) 3.387l5.615, (b) 2.759l287.6., , B.3, , Euler’s Formula, , Euler’s formula is an important result in complex variables. We derive, it from the series expansion of ex, cos u, and sin u. We know that, ex 1 x , , x3, x4, x2, , , p, 2!, 3!, 4!, , (B.19), , u2, u3, u4, j , p, 2!, 3!, 4!, , (B.20), , Replacing x by ju gives, e ju 1 ju , Also,, u2, u4, u6, , , p, 2!, 4!, 6!, u3, u5, u7, sin u u , , , p, 3!, 5!, 7!, , cos u 1 , , (B.21), , so that, cos u j sin u 1 ju , , u2, u3, u4, u5, j , j p, 2!, 3!, 4!, 5!, , (B.22), , Comparing Eqs. (B.20) and (B.22), we conclude that, e ju cos u j sin u, , (B.23), , This is known as Euler’s formula. The exponential form of representing a complex number as in Eq. (B.8) is based on Euler’s formula., From Eq. (B.23), notice that, cos u Re(e ju),, , sin u Im(e ju), , (B.24), , and that, 0e ju 0 2 cos2 u sin2 u 1, , Replacing u by u in Eq. (B.23) gives, , eju cos u j sin u, , (B.25), , Adding Eqs. (B.23) and (B.25) yields, 1, cos u (e ju eju), 2, , (B.26)
Page 950 :
ale29559_appC.qxd, , 07/16/2008, , 06:44 PM, , Page A-16, , Appendix C, Mathematical Formulas, This appendix—by no means exhaustive—serves as a handy reference. It, does contain all the formulas needed to solve circuit problems in this, book., , C.1, , Quadratic Formula, , The roots of the quadratic equation ax2 bx c 0 are, x1, x2 , , C.2, , b 2b2 4ac, 2a, , Trigonometric Identities, sin (x) sin x, cos(x) cos x, 1, ,, sec x , cos x, tan x , , sin x, ,, cos x, , 1, sin x, 1, cot x , tan x, csc x , , sin(x 90) cos x, cos(x 90) sin x, sin(x 180) sin x, cos(x 180) cos x, cos2 x sin2 x 1, a, b, c, , , sin A, sin B, sin C, a2 b2 c2 2bc cos A, tan 12 (A, tan 12 (A, , B), B), , , , ab, ab, , (law of sines), (law of cosines), (law of tangents), , sin(x y) sin x cos y cos x sin y, cos(x y) cos x cos y sin x sin y, tan x tan y, tan(x y) , 1 tan x tan y, 2 sin x sin y cos(x y) cos(x y), 2 sin x cos y sin(x y) sin(x y), 2 cos x cos y cos(x y) cos(x y), sin 2x 2 sin x cos x, A-16
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ale29559_appC.qxd, , 07/16/2008, , 06:44 PM, , Page A-19, , Appendix C, , Mathematical Formulas, , x sin ax dx a (sin ax ax cos ax) C, 1, x cos ax dx a (cos ax ax sin ax) C, 1, , 2, , 2, , x sin ax dx a (2ax sin ax 2 cos ax a x, 1, , 2, , 2 2, , 3, , cos ax) C, , x, , 2, , cos ax dx , , 1, (2ax cos ax 2 sin ax a2x2 sin ax) C, a3, , e, , ax, , sin bx dx , , eax, (a sin bx b cos bx) C, a b2, , , , eax cos bx dx , , sin ax sin bx dx , , 2, , eax, (a cos bx b sin bx) C, a2 b2, sin(a b)x, sin(a b)x, , C,, 2(a b), 2(a b), , sin ax cos bx dx , cos ax cos bx dx , a, , 2, , a, , cos(a b)x, cos(a b)x, , C,, 2(a b), 2(a b), , sin(a b)x, sin(a b)x, , C,, 2(a b), 2(a b), , a2 b2, a2 b2, a2 b2, , dx, 1, x, tan1 C, 2, a, a, x, , x2 dx, x, x a tan1 C, 2, a, x, , 2, , (a, , 2, , dx, 1, x, 1, x, 2a 2, tan1 b C, 2 2, 2, a, a, x), 2a x a, , C.6, , Definite Integrals, , If m and n are integers,, , , , 2p, , sin ax dx 0, , 0, , , , 2p, , cos ax dx 0, , 0, , , , p, , sin2 ax dx , , 0, , , , p, , sin mx sin nx dx , , 0, , , , , , p, , , , p, , 0, , p, , cos mx cos nx dx 0,, , 0,, , m n even, , 2m, ,, 2, m n2, , m n odd, , sin mx cos nx dx c, , 2p, , 0, , p, 2, mn, , 0, , 0, , , , cos2 ax dx , , sin mx sin nx dx , , , , p, , p, , 0,, p,, , sin mx sin nx dx b, , mn, mn, , A-19
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ale29559_appC.qxd, , A-20, , 07/16/2008, , 06:44 PM, , Page A-20, , Appendix C, , Mathematical Formulas, , , , , , 0, , C.7, , p, ,, 2, sin ax, dx e 0,, x, p, ,, 2, , a 7 0, a0, a 6 0, , L’Hopital’s Rule, , If f (0) 0 h(0), then, lim, , xS0, , f (x), f ¿(x), lim, h(x) xS0 h¿(x), , where the prime indicates differentiation.
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , Page A-21, , Appendix D, PSpice for Windows, There are several computer software packages, such as Spice, Mathcad,, Quattro, MATLAB, and Maple, which can be used for circuit analysis. The, most popular is Spice, which stands for Simulation Program with, Integrated-Circuit Emphasis. Spice was developed at the Department of, Electrical and Computer Engineering at the University of California at, Berkeley in the 1970s for mainframe computers. Since then about 20 versions have been developed. PSpice, a version of Spice for personal computers, was developed by MicroSim Corporation in California and made, available in 1984 and later by OrCAD and cadence. PSpice has been made, available in different operating systems (DOS, Windows, Unix, etc.)., If you do not have access to PSpice, you can find out how to obtain, a free student copy by going to the text website (www.mhhe.com/, alexander). The instructions and examples in this appendix were developed for version 9.1, but also work for later versions., Assuming that you are using Windows and have the PSpice software, installed in your computer, you can access PSpice by clicking the Start, icon on the left-hand corner of your PC; drag the cursor to Programs, to, PSpice students and to Schematics, and then click as shown in Fig. D.1., The objective of this appendix is to provide a short tutorial on, using the Windows-based PSpice on an IBM PC or equivalent., PSpice can analyze up to roughly 130 elements and 100 nodes. It, is capable of performing three major types of circuit analysis: dc analysis, transient analysis, and ac analysis. In addition, it can also perform, transfer function analysis, Fourier analysis, and operating point analysis.The circuit can contain resistors, inductors, capacitors, independent, and dependent voltage and current sources, op amps, transformers,, transmission lines, and semiconductor devices., We will assume that you are familiar with using the Microsoft, Windows operating system and that PSpice for Windows is already, installed in your computer. As with any standard Windows application,, PSpice provides an on-line help system., , The student version of PSpice can be, obtained free of charge., , If you need help on any topic at any level, click Help, click Help Topics,, click Search, and type in the topic., , D.1, , Design Center for Windows, , In earlier versions of PSpice prior to Windows 95, PSpice for Windows, is formally known as the MicroSim Design Center, which is a computer environment for simulating electric circuits. The Design Center, for Windows includes the following programs:, Schematics: This program is a graphical editor used to draw the, circuit to be simulated on the screen. It allows the user to enter, A-21
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , A-22, , Page A-22, , Appendix D, , PSpice for Windows, , Figure D.1, Accessing PSpice on Windows., , the components, wire the components together to form the circuit, and specify the type of analysis to be performed., Pspice: This program simulates the circuit created using Schematics. By simulation, we mean a method of analysis described in, a program by which a circuit is represented by mathematical, models of the components comprising the circuit., Orcad PSpice: This program provides a graphic display of the output generated by the PSpice program. It can be used to observe, any voltage or current in the circuit., One may think of Schematics as the computer breadboard for setting, up the circuit topology, PSpice as the simulator (performing the computation), and Orcad PSpice as the oscilloscope. Using the Schematics program is perhaps the hardest part of circuit simulation using PSpice. The, next section covers the essential skills needed to operate the Schematics., , D.2, , Creating a Circuit, , For a circuit to be analyzed by PSpice, we must take three steps: (1) create the circuit, (2) simulate it, and (3) print or plot the results. In this, section, we learn how to create the circuit using the Schematics program., Before we discuss how to use the Schematics capture, we need to, know how to use the mouse to select an object and perform an action., One uses the mouse in Schematics in conjunction with the keyboard to
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , Page A-23, , Appendix D, , PSpice for Windows, , carry out various instructions. Throughout this text, we will use the following terms to represent actions to be performed by the mouse:, • CLICKL : click the left button once to select an item., • CLICKR : click the right button once to abort a mode., • DCLICKL : double-click the left button to edit a selection or end, a mode., • DCLICKR : double-click the right button to repeat an action., • CLICKLH : click the left button, hold down, and move the mouse, to drag a selected item. Release the left button after the item has, been placed., • DRAG : drag the mouse (without clicking) to move an item., When the term “click” is used, it means that you quickly press and release, the left mouse button. To select an item requires CLICKL, while to perform an action requires DCLICKL. Also, to avoid writing “click” several, times, the menu to be clicked will be highlighted in bold. For example,, “click Draw, click Get New Part” will be written as Draw/Get New, Part. Of course, we can always press the <Esc> key to abort any action., Assuming that you are using Windows, you can access Pspice by, clicking the Start icon on the left-hand corner of your PC, drag the cursor, to Programs, PSpice student; and to Schematics, as shown in Fig. D.1., Alternatively, you have the PSpice icon on your screen. DCLICK on it., Either way, a blank screen will appear as shown in Fig. D.2. The file, , Figure D.2, Schematics window., , A-23
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , Page A-24, , Appendix D, , A-24, , PSpice for Windows, , name [Schematic1 p.1] next to PSpice Schematics is assigned to a circuit which is yet to be saved. You can change it by pulling down the, File menu., To create a circuit using Schematics requires three steps: (1) placing, the parts or components of the circuit, (2) wiring the parts together to, form the circuit, and (3) changing attributes of the parts., , Step 1: Placing the Parts, Each circuit part is retrieved by following this procedure:, C2, , R2, , • Select Draw/Get New Part to pull down the Draw menu (or type, <Ctrl-G>)., • Use scroll bar to select the part (or type the part name, e.g., R for, resistor, in the PartName box). Figures D.3 to D.5 show some part, names and symbols for circuit elements and independent voltage, and current sources., • Click Place & Close (or press <Enter>)., • DRAG part to the desired location on the screen., • CLICKR to terminate the placement mode., , L2, , 1k, , 1n, , 10uH, , (a), , (b), , (c), , Figure D.3, Part symbols and attributes for circuit, elements: (a) a resistor, (b) a capacitor,, (c) an inductor., , +, OV, , OV, , VDC, −, , (a) A dc only source, , +, −, , VAC, , (b) An ac only source, , +, VSIN, −, (c) An ac or dc source, , + VSRC, −, (d) An ac, dc, or transient source, , Figure D.4, Part symbols and attributes for independent voltage sources., , OA, , IDC, , (a) A dc only source, , OA, , +, IAC, −, , (b) An ac only source, , +, ISIN, −, (c) An ac or dc source, , ISRC, , (d) An ac, dc, or transient source, , Figure D.5, Part symbols and attributes for independent, current sources., , Sometimes, we want to rotate a part 90. To rotate a resistor, for example, select the part R and click Edit/Rotate (or type <Ctrl R>). To, delete a part, CLICKL to select (highlight red) the part, then click, Edit/Cut (or press <Delete>)., , Step 2: Wiring Parts Together, We complete the circuit by wiring the parts together. We first select, Draw/Wire (or type <Ctrl-W>) to be in wiring mode. A pencil cursor, will appear in place of an arrow cursor. DRAG the pencil cursor to the, first point you want to connect and CLICKL. Next, DRAG the pencil cursor to the second point and CLICKL to change the dashed line, to a solid line. (Only solid lines are wires.) CLICKR to end the wiring, mode. To resume the wiring mode, press the <Space bar>. Repeat the, above procedure for each connection in the circuit until all the parts, are wired. The wiring is not complete without adding a ground connection (part AGND) to a schematic; PSpice will not operate without, it. To verify that the parts are actually connected together, the Junctions, option available in the Options/Set Display Level menu should be in, the on position when wiring the parts. By default, the Junctions option, is marked with a checksign (✓) in the dialog box, indicating that it, is on., Some of the connections have a black dot indicating a connection., Although it is not necessary to have a dot where a wire joins a pin,
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , Page A-25, , Appendix D, , PSpice for Windows, , A-25, , having the dot shows the presence of a connection. To be sure a dot, appears, make sure the wire overlaps the pin., If you make a mistake, you can delete the part or wire by, highlighting it (select CLICKL) and pressing the <Delete> key., Typing <Ctrl-L> will erase the fragments that are not really on the, shematic., , Step 3: Changing Attributes of Parts, As shown in Figs. D.3 to D.5, each component has an attribute in addition to its symbol. Attributes are the labels for parts. Each attribute consists of a name and its designated value. For example, R and VSRC, are the names of resistor and voltage source (dc, ac, or transient, source), while 2k and DC 10V are the designated values of the, resistor and voltage source, respectively., As parts are placed on the screen, they are automatically, assigned names by successive numbers (R1, R2, R3, etc.). Also,, some parts are assigned some predetermined values. For example,, all resistors are placed horizontally and assigned a value of 1 k., We may need to change the attributes (names and values) of a part., Although there are several ways of changing the attributes, the following is one simple way., To change the name R3 to RX, for example, DCLICKL on the, text R3 to bring up the Edit Reference Designator dialog box of, Fig. D.6(a). Type the new name RX and click the OK button to accept, the change. The same procedure can be used to change VDC to V1 or, whatever., To change the value 1k to 10Meg, for example, DCLICKL on, the 1k attribute (not the symbol) to open up the Set Attribute Value, dialog box of Fig. D.6(b). Type the new value 10Meg (no space, between 10 and Meg) and click the OK button to accept the change., Similarly, to change the default value 0V to 15kV for voltage source, VDC, DCLICKL the symbol for VDC to bring up the PartName dialog box. DCLICKL on the DC attribute and type 15kV in the, value box. For convenience, one can express numbers with the scale, factors in Table D.1. For example, 6.6 108 can be written as 66N, or 0.066U., Except for the ground, which is automatically assigned node 0,, every node is either given a name (or number) or is assigned one in, the netlist. A node is labeled by giving a name to a wire connected to, that node. DCLICKL the wire to open up the Set Attribute Value dialog box, and type the label., To obtain a hard copy of the screen/schematic, click File/Print/OK., To save the schematic created, select File/Save As and type Filename., Click OK or press <Enter>. This creates a file named “filename” and, saves it with extension .sch., , Draw the circuit in Fig. D.7 using Schematics., Solution:, We will follow the three steps mentioned above. We begin by doubleclicking the Schematics icon. This provides us with a blank screen as, , (a), , (b), , Figure D.6, (a) Changing name R3 to RX,, (b) changing 1k to 10Meg., A component may have several attributes; some are displayed by default. If, need be, we may add more attributes, for display, but we should hide unimportant attributes to avoid clutter., TABLE D.1, , Scale factors., Symbol, , Value, , Name of suffix, , T, G, MEG, K, M, U, N, P, F, , 1012, 109, 106, 103, 103, 106, 109, 1012, 1015, , tera, giga, mega, kilo, milli, micro, nano, pico, femto, , It is always expedient to number the, nodes by numbering the wires. Otherwise, Schematics will label the nodes, its own way, and one may not understand which node is which in the, output results., , Example D.1
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , Page A-26, , Appendix D, , A-26, 5 kΩ, , 12 V +, −, , 2 kΩ, , For Example D.1., , 1k, + V1, −, , R2, , 1k, , R2, , 1k, , (a), R1, k1, + V1, −, , 0, (b), R1, 5k, + V1, −, , Click Draw/Get New Part (or type <Ctrl-G>)., Type VSRC in the Part Browser Basic box., Click OK (or type <Enter>)., DRAG the part to the desired location on the screen., CLICKL to place VSRC and CLICKR to terminate, placement mode., , At this point, only the voltage source V1 in Fig. D.8(a) is shown on, the screen, highlighted red. To place the resistors, we need to:, , R1, , 12 V, , a worksheet to draw the circuit on. We now take the following steps, to create the circuit in Fig. D.7., To place the voltage source, we need to:, 1., 2., 3., 4., 5., , Figure D.7, , PSpice for Windows, , R2, , 2k, , 0, (c), , Figure D.8, Creating the circuit in Fig. D.7:, (a) placing the parts, (b) wiring the parts, together, (c) changing the attributes., , 1., 2., 3., 4., 5., 6., 7., 8., , Click Draw/Get New Part., Type R in the Part Browser Basic box., Click OK., DRAG resistor to R1’s location on the screen., CLICKL to place R1., CLICKL to place R2 and CLICKR to terminate placement mode., DRAG R2 to its location., Edit/Rotate (or type <Ctrl-R>) to rotate R2., , At this point, the three parts have been created as shown in Fig. D.8(a)., The next step is to connect the parts by wiring. To do this:, 1. Click Draw/Wire to be in wiring mode, indicated by the, pencil cursor., 2. DRAG the pencil cursor to the top of V1., 3. CLICKL to join the wire to the top of V1., 4. DRAG the dotted wire to the top corner., 5. CLICKL to turn wire segment solid, and anchor at corner., 6. DRAG dotted wire to left of R1., 7. CLICKL to turn wire segment solid and anchor to left of R1., 8. CLICKR to end placement mode., Follow the same steps to connect R1 with R2 and V1 with R2., (You can resume the wiring mode by pressing <Space bar>.) At this, point, we have the circuit in Fig. D.8(b), except that the ground symbol, is missing. We insert the ground by taking the following steps:, 1., 2., 3., 4., 5., , Click Draw/Get New Part., Type AGND in the Part Browser Basic box., Click OK., DRAG the part to the desired location on the screen., CLICKL to place AGND and CLICKR to terminate, placement mode., , The last thing to be done is to change or assign values to the attributes., To assign the attribute 12V to V1, we take these steps:, 1. DCLICKL on the V1 symbol to open up the PartName, dialog box., 2. DCLICKL on the DC attribute., 3. Type 12V (or simply 12) in the Value box., 4. Click Save Attr., 5. Click OK.
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , Page A-27, , Appendix D, , PSpice for Windows, , A-27, , To assign 5k to R1, we follow these steps:, 1. DCLICKL on 1k attribute of R1 to bring up the Set Attribute, Value dialog box., 2. Type 5k in the Value box., 3. Click OK., Use the same procedure in assigning value 2k to R2. Figure D.8(c) shows, the final circuit., , Practice Problem D.1, , Construct the circuit in Fig. D.9 with Schematics., Answer: See the schematic Fig. D.10., R1, , 3 kΩ, , 3k, 5V, , +, −, , 10 kΩ, , 1 MΩ, , 5 V +, −, , V1, , R2, , R3, , 2k, , 1Meg, , 0, , D.3, , Figure D.9, , Figure D.10, , For Practice Prob. D.1., , For Practice Prob. D.1., , DC Analysis, , DC analysis is one of the standard analyses that we can perform using, PSpice. Other standard analyses include transient, AC, and Fourier., Under DC analysis, there are two kinds of simulation that PSpice can, execute: DC nodal analysis and DC sweep., , 1. DC Nodal Analysis, PSpice allows dc nodal analysis to be performed on sources with an, attribute of the form DC value and provides the dc voltage at each, node of the circuit and dc branch currents if required. To view dc node, voltages and branch currents requires adding two kinds of additional, parts, shown in Fig. D.11. The symbol VIEWPOINT is connected to, each node at which the voltage is to be viewed, while the symbol, IPROBE is connected in the branch where the current is to be displayed. This necessitates modifying the schematic. For example, let us, consider placing voltage VIEWPOINTS and current IPROBES to the, schematic in Fig. D.8(c). To add VIEWPOINTS, we take the following steps:, 1., 2., 3., 4., 5., 6., , Click Draw/Get New Part (or type <Ctrl-G>)., Type VIEWPOINT in the Part Browser Basic box., Click OK (or type <Enter>)., DRAG to locate VIEWPOINT above V1 and CLICKL., DRAG to locate VIEWPOINT above R2 and CLICKL., CLICKR to end placement mode., , (a), , (b), , Figure D.11, Symbols for: (a) voltage VIEWPOINT,, (b) current IPROBE.
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , Page A-28, , Appendix D, , A-28, , Figure D.12 shows the two voltage VIEWPOINTS. Since the IPROBE, symbol must be connected in series with a branch element, we need to, move R2 down by clicking and dragging R2 and the wires. Once this, is done, we add IPROBE as follows:, , R1, 5k, , + V1, −, , 12 V, , R2, , 2k, , 0, , Figure D.12, Placing VIEWPOINTS and IPROBES., , E1, +, −, , F1, , +, −, e, , f, , (a), , (b), , G1, , H1, , +, , +, −, , −, , PSpice for Windows, , g, , h, , (c), , (d), , Figure D.13, Dependent sources:, (a) voltage-controlled voltage source, (VCVS),, (b) current-controlled current source, (CCCS),, (c) voltage-controlled current source, (VCCS),, (d) current-controlled voltage source, (CCVS)., , A netlist can be generated manually or, automatically by Schematics., There are two kinds of common errors, in PSpice: (1) errors involving wiring of, the circuit, and (2) errors that occur, during simulation., , 1., 2., 3., 4., 5., 6., , Click Draw/Get New Part (or type <Ctrl-G>)., Type IPROBE in the Part Browser Basic box., Click OK (or type <Enter>)., DRAG to locate IPROBE above R2 and CLICKL., CLICKR to end placement mode., Use wiring to join all gaps., , The schematic becomes that shown in Fig. D.12. We are ready to simulate the circuit. At this point, we must save the schematic—PSpice, will not run without first saving the schematic to be simulated. Before, learning how to run PSpice, note the following points:, 1. There must be a reference node or ground connection (part, AGND) in the schematic. Any node can be used as ground, and, the voltages at other nodes will be measured with respect to the, selected ground., 2. Dependent sources are found in the Parts library. Obtain them by, selecting Draw/Get New Part and typing the part name. Figure D.13, shows the part name for each type, with the gain. E is a voltagecontrolled voltage source with gain e; F is a current-controlled, current source with gain f; G is a voltage-controlled current source, with a transconductance gain g; and H is a current-controlled voltage source with transresistance gain h., 3. By convention, we assume in dc analysis that all capacitors are, open circuits and all inductors are short circuits., We run PSpice by clicking Analysis/Simulate. This invokes the, electric rule check (ERC), which generates the netlist. The ERC performs a connectivity check on the schematic before creating the netlist., The netlist is a list describing the operational behavior of each component in the circuit and its connections. Each line in the netlist represents a single component of the circuit. The netlist can be examined, by clicking Analysis/Examine Netlist from the Schematics window., If there are errors in the schematic, an error window will appear. Click, OK (or type <Enter>) to display the error list. After noting the errors,, exit from the error list and go back to Schematics to correct the errors., If no errors are found, the system automatically enters PSpice and performs the simulation (nodal analysis). When the analysis is complete,, the program displays Bias point calculated, and creates the result/, output file with extension .out., To examine the output file, click Analysis/Examine Output from, the Schematics window (or click File/Examine Output from the, PSpice window). To print the output file, click File/Print, and to exit, the output file, click File/Exit., We can also examine the results of the simulation by looking at, the values displayed on the VIEWPOINTS and IPROBES parts of the, schematics after the simulation is complete. The values displayed with, VIEWPOINTS and IPROBES should be the same as those in the output file.
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , Page A-29, , Appendix D, , PSpice for Windows, , 2. DC Sweep, DC nodal analysis allows simulation for DC sources with fixed voltages or currents. DC sweep provides more flexibility in that it allows, the calculation of node voltages and branch currents of a circuit when, a source is swept over a range of values. As in nodal analysis, we, assume capacitors to be open circuits and inductors to be short circuits., Suppose we desire to perform a DC sweep of voltage source V1, in Fig. D.12 from 0 to 20 volts in 1-volt increments. We proceed as, follows:, 1., 2., 3., 4., 5., 6., 7., 8., , Click Analysis/Setup., CLICKL DC Sweep button., Click Name box and type V1., Click Start Value box and type 0., Click End Value box and type 20., Click Increment box and type 1., Click OK to end the DC Sweep dialog box and save parameters., Click Close to end the Analysis Setup menu., , Figure D.14 shows the DC Sweep dialog box. Notice that the default, setting is Voltage Source for the Swept Var. Type, while it is Linear for, Sweep Type. If needed, other options can be selected by clicking the, appropriate buttons., , Figure D.14, DC sweep analysis dialog box., , To run DC sweep analysis, click Analysis/Simulate. Schematics, will create a netlist and then run PSpice if no errors are found. If errors, are found in the schematic, check for them in the Error List and correct them as usual. If no errors are found, the data generated by PSpice, is passed to Orcad PSpice. The Orcad PSpice window will appear, displaying a graph in which the X axis is by default set to the DC sweep, variable and range, and the Y axis is blank for now. To display some, specific plots, click Trace/Add in the Orcad PSpice menu to open the, Add Traces dialog box. The box contains traces, which are the output, variables (node voltages and branch currents) in the data file available, for display. Select the traces to be displayed by clicking or typing them,, and click OK. The selected traces will be plotted and displayed on the, , A-29
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , Page A-30, , Appendix D, , A-30, , PSpice for Windows, , screen. As many traces as you want may be added to the same plot or, on different windows. Select a new window by clicking Window/New., To delete a trace, click the trace name in the legend of the plot to highlight it and click Edit/Delete (or press <Delete>)., It is important to understand how to interpret the traces. We must, interpret the voltage and current variables according to the passive sign, convention. As parts are initially placed horizontally in a schematic as, shown typically in Fig. D.3, the left-hand terminal is named pin 1 while, the right-hand terminal is pin 2. When a component (say R1) is rotated, counterclockwise once, pin 2 would be on the top, since rotation is about, pin 1. Therefore, if current enters through pin 2, the current I(R1), through R1 would be negative. In other words, positive current implies, that the current enters through pin 1, and negative current means that the, current enters through pin 2. As for voltage variables, they are always, with respect to the ground. For example, V(R1:2) is the voltage (with, respect to the ground) at pin 2 of resistor R1; V(V1:) is the voltage, (with respect to the ground) at the positive terminal of voltage source, V1; and V(E2:1) is the voltage at pin 1 of component E2 with respect, to ground, regardless of the polarity., , Example D.2, , For the circuit in Fig. D.15, find the dc node voltages and the current io., 1, , 12 kΩ, , 1 kΩ, , 2, , 3, , io, 28 V +, −, , 4 kΩ, , 7 mA, , 3 kΩ, , Figure D.15, For Example D.2., , Solution:, We use Schematics to create the circuit. After saving the circuit, click, Analysis/Simulate to simulate the circuit. We obtain the results of, the dc analysis from the output file or from the VIEWPOINT AND, IPROBE parts, as shown in Fig. D.16. The netlist file is shown in, Fig. D.17. Notice that the netlist contains the name, value, and, , R2, , 1, , 13.000, 2, , R3, , 15.000, , 3, , * Schematics Netlist *, , Figure D.16, , V_V1, 1, R_R1, 0, R_R2, 1, R_R3, 2, R_R4, 0, I_I1, 0, v_V2, 2, Figure D.17, , For Example D.2; schematic for the circuit in Fig. D.15., , The Netlist file for Example D.2., , 12k, , −, , IDC, , 4, , +, 28, , 1k, 3.250E − 03, I1, R4, , V1, , R1, , 3k, , 4k, , 0, , 7mA, , 0, 4, 2, 3, 3, 3, 4, , 28, 4k, 12k, 1k, 3k, DC 7mA, 0
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , Page A-31, , Appendix D, , PSpice for Windows, , A-31, , connection for each element in the circuit. First example, the first, line shows that the voltage source V1 has a value of 28 V and is, connected between nodes 0 and 1. Figure D.18 shows the edited, version of the output file. The output file also contains the Netlist, file, but this was removed from Fig. D.18. From IPROBE or the, output file, we obtain io as 3.25 mA., , **** 11/26/99 20:56:05 ********* NT Evaluation PSpice (Nov. 1999) *********, * C:\ MSIMEV63\ examd2.sch, **** CIRCUIT DESCRIPTION, ****************************************************************************, * Schematics Version 6.3 - April 1996, * Sat Jul 26 20:56:04 1997, **** INCLUDING examd2.als ****, * Schematics Aliases *, .ALIASES, V_V1, V1(+=1, R_R1, R1(1=0, R_R2, R2(1=1, R_R3, R3(1=2, R_R4, R4(1=0, I_I1, I1(+=0, v_V2, V2(+=2, _ _(1=1), _ _(2=2), _ _(3=3), .ENDALIASES, , -=0, 2=4, 2=2, 2=3, 2=3, -=3, -=4, , ), ), ), ), ), ), ), , .probe, .END, NODE VOLTAGE, , NODE VOLTAGE, , NODE VOLTAGE, , NODE VOLTAGE, , ( 1) 28.0000, , ( 2) 13.0000, , ( 3) 15.0000, , ( 4) 13.0000, , VOLTAGE SOURCE CURRENTS, NAME, CURRENT, V_V1, v_V2, , -1.250E-03, 3.250E-03, , TOTAL POWER DISSIPATION 3.50E-02 WATTS, Figure D.18, Output file (edited version) for Example D.2.
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , A-32, , Practice Problem D.2, , Page A-32, , Appendix D, , PSpice for Windows, , Use PSpice to determine the node voltages and the current ix in the, circuit of Fig. D.19., 2 kΩ, , 1, , 3 kΩ, , 2, , 3, ix, , 50 V +, −, , 6 mA, , 4 kΩ, , 9 kΩ, , 0, , Figure D.19, For Practice Prob. D.2., , Answer: V1 50, V2 37.2, V3 27.9, ix 3.1 mA., , Example D.3, , Plot I1 and I2 if the dc voltage source in Fig. D.20 is swept from 2 V, to 10 V., 4Vx, , 2 kΩ, , −+, , +V −, x, 2–10 V, , +, −, , 4 kΩ, , I1, , 6 kΩ, , I2, , Figure D.20, For Example D.3., , Solution:, We draw the schematic of the circuit and set the attributes as shown, in Fig. D.21. Notice how the voltage-controlled voltage source E1 is, connected. After completing the schematic, we select Analysis/Setup, and input the start, end, and increment values as 2, 10, and 0.5,, respectively. By selecting Analysis/Simulate, we bring up the Orcad, PSpice window. We select Trace/Add and click I(R1) and I(R3) to, , E1, , −, , 1, , 3, , 2k, , +, OV, , R1, , +, − −, , +, , E, , 2, , V1, , R2, , 4k, , R3, , 6k, , 0, , Figure D.21, The schematic for the circuit in Fig. D.20.
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , Page A-33, , Appendix D, , PSpice for Windows, , A-33, , be displayed. (The negative sign is needed to make the current through, R3 positive.) Figure D.22 shows the result., 10 mA, , I2, 5 mA, I1, , 0 A, 2 V, , 4 V, I(R1), , 6 V, –I(R3), V_V1, , 8 V, , 10 V, , Figure D.22, Plots of I1 and I2 against V1., , Use PSpice to obtain the plots of ix and io if the dc voltage source in, Fig. D.23 is swept from 2 V to 10 V., , Practice Problem D.3, , Answer: The plots of ix and io are displayed in Fig. D.24., 2 kΩ, , 4.0 m A, , 4 kΩ, ix, , 2–10 V +, −, , 5 kΩ, , io, 2ix, , 1 kΩ, 2.0 m A, iO, ix, , Figure D.23, For Practice Prob. D.3., 0 A, 2 V, , 4 V, 6 V, –I(R3), –I(R4), V_V1, , 8 V, , 10 V, , Figure D.24, Plots of ix and io versus V1., , D.4, , Transient Analysis, , In PSpice, transient analysis is generally used to examine the behavior, of a waveform (voltage or current) as time varies. Transient analysis, solves some differential equations describing a circuit and obtains voltages and currents versus time. Transient analysis is also used to obtain, Fourier analysis. To perform transient analysis on a circuit using PSpice, , Transient analysis is used to view the, transient response of inductors and, capacitors.
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ale29559_appD.qxd, , A-34, , 07/17/2008, , 09:18 AM, , Page A-34, , Appendix D, , PSpice for Windows, , usually involves these steps: (1) drawing the circuit, (2) providing specifications, and (3) simulating the circuit., , 1. Drawing the Circuit, In order to run a transient analysis on a circuit, the circuit must first, be created using Schematics and the source must be specified. PSpice, has several time-varying functions or sources that enhance the performance of transient analysis. Sources used in the transient analysis, include:, • VSIN, ISIN: damped sinusoidal voltage or current source, e.g.,, v(t) 10e0.2t sin(120 p t 60)., • VPULSE, IPULSE: voltage or current pulse., • VEXP, IEXP: voltage or current exponential source, e.g., i(t) , 6[1 exp(0.5t)]., • VPWL, IPWL: piecewise linear voltage or current function, which, can be used to create an arbitrary waveform., It is expedient to take a close look at these functions., VSIN is the exponentially damped sinusoidal voltage source, for, example,, v(t) Vo Vmea(ttd) sin[2 p f (t td) f], , (D.1), , The VSIN source has the following attributes, which are illustrated in, Fig. D.25 and compared with Eq. (D.1):, VOFF Offset voltage, Vo, VAMPL Amplitude, Vm, TD Time delay in seconds, td, FREQ Frequency in Hz, f, DF Damping factor (dimensionless), a, PHASE Phase in degrees, f, , 6.0 V, , VAMPL, , 4.0 V, TD, , 1/FREQ, , VOFF, 2.0 V, , 0 V, 0 s, , 5.0s, Time, , Figure D.25, Sinusoidal voltage source VSIN., , (D.2)
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , Page A-35, , PSpice for Windows, , Appendix D, , Attributes TD, DF, and PHASE are set to 0 by default but can be, assigned other values if necessary. What has been said about VSIN is, also true for ISIN., The VPULSE source has the following attributes, which are portrayed in Fig. D.26:, V1 Low voltage, V2 High voltage, TD Initial time delay in seconds, TR Rise time in seconds, TF Fall time in seconds, PW Pulse width in seconds, PER Period in seconds, , (D.3), , PW, TR, , TF, , 4.0 V, , 3.0 V, , 2.0 V, , V2, TD, , PER, , 1.0 V, V1, 0 V, 0 s, , 5 s, Time, , 10 s, , Figure D.26, Pulse voltage source VPULSE., , Attributes V1 and V2 must be assigned values. By default, attribute TD, is assigned 0; TR and TF are assigned the print step value; and PW, and PER are assigned the final time value. The values of the print time, and final time are obtained as default values from the specifications, provided by the user in the Transient Analysis/Setup, to be discussed, a little later., The exponential voltage source VEXP has the following attributes,, typically illustrated in Fig. D.27:, V1 Initial voltage, V2 Final voltage, TD1 Rise delay in seconds, TC1 Rise time constant in seconds, TD2 Fall delay in seconds, TC2 Fall time in seconds, , (D.4), , A-35
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , Page A-36, , Appendix D, , A-36, , PSpice for Windows, 6.0 V, , TC1, 4.0 V, , TC2, V2, , 2.0 V, V1, TD2, TD1, 0 V, 0 s, , 4.0 s, Time, , 8.0 s, , Figure D.27, Exponential voltage source VEXP., , f(t), V2, , V1, , T1, , T2, , T3, , T4 t, , Figure D.28, Piecewise linear voltage source VPWL., , v (t), 4, , 0, , 2, , 4, , 6, , 8, , t, , −2, , Figure D.29, An example of a piecewise linear voltage, source VPWL., , The piecewise linear voltage source VPWL, such as shown in, Fig. D.28, requires specifying pairs of TN, VN, where VN is the voltage at time TN for N 1, 2, p , 10. For example, for the function in, Fig. D.29, we will need to specify the attributes T1 0, V1 0,, T2 2, V2 4, T3 6, V3 4, and T4 8, V4 2., To obtain information about other sources, click Help/Search for, Help on . . . and type in the name of the source. To add a source to the, schematic, take the following steps:, 1., 2., 3., 4., , Select Draw/Get New Part., Type the name of the source., Click OK and DRAG the symbol to the desired location., DCLICKL the symbol of the source to open up the PartName, dialog box., 5. For each attribute, DCLICKL on the attribute, enter the value, and, click Save Attr to accept changes., 6. Click OK to accept new attributes., In step 5, the attributes may not be shown on the schematic after entering their values. To display an attribute, select Change Display/Both, Name and Value in the PartName dialog box., In addition to specifying the source to be used in transient analysis,, there may be need to set initial conditions on capacitors and inductors, in the circuit. To do so, DCLICKL the part symbol to bring up the, PartName dialog box, click IC and type in the initial condition. The, IC attribute allows for setting the initial conditions on a capacitor or, inductor. The default value of IC is 0. The attributes of open/close, switches (with part names Sw_tClose and Sw_tOpen) can be changed, in a similar manner., , 2. Providing Specifications, After the circuit is drawn and the source is specified with its attributes, we need to add some specifications for the transient analysis.
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , Page A-37, , Appendix D, , PSpice for Windows, , A-37, , For example, suppose we want the analysis to run the simulation from, 0 to 10 ms with a print interval of 2 ns; we enter these specifications, as follows:, 1. Select Analysis/Setup/Transient to open up the Transient Analysis, dialog box., 2. CLICKL Print Step and type 2 ns., 3. CLICKL Final Time and type 10 ms., 4. CLICKL Step Ceiling and type 5 s., 5. CLICKL OK/Close to accept specifications., These specifications control the simulation and the display of output, variables. Final Time specifies how long the simulation should run. In, other words, the simulation runs from t 0 to t Final Time. Print, Step refers to the time interval the print part will print out; it controls, how often simulation results are written to the output file. The value, of Print Step can be any value less than the Final Time, but it cannot, be zero. Step Ceiling is the maximum time between simulation points;, specifying its value is optional. By selecting 10 ms as Final Time and, 5 ms as Step Ceiling, the simulation will have a minimum of, 10 ms/5 ms 2000 points. When Step Ceiling is unspecified, PSpice, selects its own internal time step—the time between simulation points., The time step is selected as large as possible to reduce simulation, time. If the user has no idea of what the plot may look like, it is recommended that the value of Step Ceiling be unspecified. If the plot is, jagged as a result of a large time step assumed by PSpice, the user, may now specify a Step Ceiling that will smooth the plot. Keep in, mind that a smaller value gives more points in the simulation but takes, more time., , 3. Simulating the Circuit, After the circuit is drawn, the specifications for the transient analysis are given, and the circuit is saved, we are ready to simulate, it. To perform transient analysis, we select Analysis/Simulate. If, there are no errors, the Orcad PSpice window will automatically, appear. As usual, the time axis (or X axis) is drawn but no curves, are drawn yet. Select Trace/Add and click on the variables to be, displayed., An alternative way of displaying the results is to use markers., Although there are many types of markers, we will discuss only, voltage and current markers. A voltage marker is used to display, voltage at a node relative to ground; a current marker is for, displaying current through a component pin. To place a voltage, marker at a node, take the following steps while in the Schematics, window:, 1. Select Markers/Mark Voltage/Level., 2. DRAG the voltage marker to the desired node., 3. CLICKL to place the marker and CLICKR to end the placement, mode., This will cause two things to happen immediately. The voltage marker, becomes part of the circuit and the appropriate node voltage is automatically displayed by the Orcad PSpice window when the Schematics is, , To obtain the Fourier component of a, signal, we enable the Fourier option in, the Transient Analysis dialog box., Chapter 17 has more on this.
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , Page A-38, , Appendix D, , A-38, , PSpice for Windows, , run. To place a current marker at a component pin, take the following, steps in the Schematics window:, 1. Select Markers/Mark current into pin., 2. DRAG the current marker to the desired pin., 3. CLICKL to place the marker and CLICKR to end the placement, mode., This will automatically add the current through the pin to your graph., It is important that the current marker be placed at the pin of the component; otherwise the system would reject the marker. You can place, as many voltage and current markers as you want on a circuit. To, remove the markers from the circuit as well as the plots from the Orcad, PSpice window, select Markers/Clear All from Schematics window., , Example D.4, , Assuming that i(0) 10A, plot the zero-input response i(t) in the circuit, of Fig. D.30 for 0 6 t 6 4 s using PSpice., , 4Ω, i(t), +, −, , 2Ω, , 0.5 H, , 3i, , Solution:, The circuit is the same as the one for Example 7.3, where we obtained, the solution as, i(t) 10e(23)t, , Figure D.30, , For PSpice analysis, the schematic is in Fig. D.31, where the currentcontrolled source H1 has been wired to agree with the circuit in, Fig. D.30. The voltage of H1 is 3 times the current through inductor, L1. Therefore, for H1, we set GAIN 3 and for the inductor L1, we, set the initial condition IC 10. Using the Analysis/Setup/Transient, dialog box, we set Print Step 0.25 s and Final Time 4 s. After, simulating the circuit, the output is taken as the inductor current i(t),, which is plotted in Fig. D.32., , For Example D.4., , R1, , 4, , H1, R2, 2, , +, −, , 10 A, , h, L1, , 0.5H, , 0, , 5 A, , Figure D.31, The schematic of the circuit in Fig. D.30., , 0 A, 0 s, , 1.0 s, I(L1), , 2.0 s, Time, , Figure D.32, Output plot for Example D.4., , 3.0 s, , 4.0 s
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , Page A-39, , Appendix D, , PSpice for Windows, , Using PSpice, plot the source-free response v(t) in the circuit of, Fig. D.33, assuming that v(0) 10 V., Answer: Figure D.34 shows the, 10e0.25t cos 0.5t 5e0.25t sin 0.5t V., , plot., , Note, , that, , A-39, , Practice Problem D.4, 2H, , 1Ω, , v(t) , , +, 1.6 F, , −, , v (t), , Figure D.33, , 4.0 KV, , For Practice Prob. D.4., 2.0 KV, , 0 V, , −2.0 KV, , −4.0 KV, 0 s, , 200 ms 400 ms 600 ms, -V(R1:2), Time, , Figure D.34, Output plot for Practice Prob. D.4., , Plot the forced response vo(t) in the circuit of Fig. D.35(a) for, 0 6 t 6 5 s if the source voltage is as shown in Fig. D.35(b)., v s (t)(V), 1, , vs, , +, −, , 10 kΩ, , 2, , 50 F, , 30 mH, , 20 kΩ, , 12, , 3, +, vo, −, , 0, (a), , 1, , 2, , 3 t(s), , (b), , Figure D.35, For Example D.5., , Solution:, We draw the circuit and set the attributes as shown in Fig. D.36. We, enter in the data in Fig. D.35(b) by double-clicking the symbol of the, voltage source V1 and typing in T1 0, V1 0, T2 1 ns, V2 12,, T3 1 s, V3 12, T4 1.001 s, V4 0, T5 2 s, V5 0,, T6 2.001 s, V6 12, T7 3 s, V7 12, T8 3.001 s, V8 0. In, the Analysis/Setup/Transient dialog box, we set Print Step 0.2 s and, Final Time 5 s. When the circuit is simulated and we are in the Orcad, PSpice window, we close or minimize the window to go back to the, , Example D.5
Page 974 : ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , Page A-40, , Appendix D, , A-40, , PSpice for Windows, , Schematics window. We place two voltage markers as shown in Fig. D.36, to get the plots of input vs and output vo. We press <Alt-Esc> to get into, the Orcad PSpice window and obtain the plots shown in Fig. D.37., R1, , L1, , 10k, , 30mH, , V, , 15 V, , 20k, , 10 V, , V, +, −, , V1, , C1, , 50u, , R2, , 0, 5 V, , Figure D.36, The schematic of the circuit in Fig. D.35., , 0 V, 0 s, , 2.0 s, 4.0 s, V(L1:2), V(V1:+), Time, , 6.0 s, , Figure D.37, Output plot for Example D.5., , Practice Problem D.5, , Obtain the plot of v(t) in the circuit in Fig. D.38 for 0 6 t 6 0.5 s if, is 2et sin 2 p (5)t A., Answer: See Fig. D.39., , 20 mH, , is, , 10 F, , 10 F, , 5.0 KV, , 5 kΩ, , +, v (t), −, 0 V, , Figure D.38, For Practice Prob. D.5., , -5.0 KV, 0 s, , 200 ms 400 ms 600 ms, -V(R1:2), Time, , Figure D.39, Output plot for Practice Prob. D.5., , D.5, , AC Analysis/Frequency Response, , Using AC sweep, PSpice can perform AC analysis of a circuit for a single frequency or over a range of frequencies in increments that can vary, linearly, by decade, or by octave. In AC sweep, one or more sources
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , Page A-41, , Appendix D, , PSpice for Windows, , are swept over a range of frequencies while the voltages and currents, of the circuit are calculated. Thus, we use AC sweep both for phasor, analysis and for frequency response analysis: it will output Bode gain, and phase plots. (Keep in mind that a phasor is a complex quantity, with real and imaginary parts or with magnitude and phase.), While transient analysis is done in the time domain, AC analysis is, performed in the frequency domain. For example, if vs 10cos(377t , 40), transient analysis can be used to display ys as a function of time,, whereas AC sweep will give the magnitude as 10 and phase as 40., To perform AC sweep requires taking three steps similar to those for, transient analysis: (1) drawing the circuit, (2) providing specifications,, and (3) simulating the circuit., , 1. Drawing the Circuit, We first draw the circuit using Schematics and specify the source(s)., Sources used in AC sweep are AC sources VAC and IAC. The sources, and attributes are entered into the Schematics as stated in the previous, section. For each independent source, we must specify its magnitude, and phase., , 2. Providing Specifications, Before simulating the circuit, we need to add some specifications for, AC sweep. For example, suppose we want a linear sweep at frequencies 50, 100, and 150 Hz. We enter these parameters as follows:, 1. Select Analysis/Setup/AC Sweep to open up the dialog box for, AC Sweep., 2. CLICKL Linear for the X axis to have a linear scale., 3. Type 3 in the Total Pts box., 4. Type 50 in the Start Freq box., 5. Type 150 in the End Freq box., 6. CLICKL OK/Close to accept specifications., A linear sweep implies that simulation points are spread uniformly, between the starting and ending frequencies. Note that the Start Freq, cannot be zero because 0 Hz corresponds to DC analysis. If we want, the simulation to be done at a single frequency, we enter 1 in step 3, and the same frequency in steps 4 and 5. If we want the AC sweep to, simulate the circuit from 1 Hz to 10 MHz at 10 points per decade, we, CLICKL on Decade in step 2 to make the X axis logarithmic, enter, 10 in the Total Pts box in step 3, enter 1 in the Start Freq box, and, enter 10Meg in the End Freq box. Keep in mind that a decade is a factor of 10. In this case, a decade is from 1 Hz to 10 Hz, from 10 Hz to, 100, from 100 to 1 kHz, and so forth., , 3. Simulating the Circuit, After providing the necessary specifications and saving the circuit, we, perform the AC sweep by selecting Analysis/Simulate. If no errors, are encountered, the circuit is simulated. At the end of the simulation,, the system displays AC analysis finished and creates an output file, with extension .out. Also, the Orcad PSpice program will automatically, run if there are no errors. The frequency axis (or X axis) is drawn but, no curves are shown yet. Select Trace/Add from the Orcad PSpice, menu bar and click on the variables to be displayed. We may also use, , A-41
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , A-42, , To generate Bode plots involves using, the AC sweep and the dB command in, Orcad PSpice., , Page A-42, , Appendix D, , PSpice for Windows, , current or voltage markers to display the traces as explained in the previous section. To use advanced markers such as vdb, idb, vphase,, iphase, vreal, and ireal, select Markers/Mark Advanced., In case the resolution of the trace is not good enough, we may, need to check the data points to see if they are enough. To do so, select, Tools/Options/Mark Data Points/OK in the Orcad PSpice menu and, the data points will be displayed. If necessary, we can improve the resolution by increasing the value of the entry in the Total Pts box in the, Analysis/Setup/AC Sweep and Noise Analysis dialog box for AC, sweep., Bode plots are separate plots of magnitude and phase versus frequency. To obtain Bode plots, it is common to use an AC source, say, V1, with 1 volt magnitude and zero phase. After we have selected, Analysis/Simulate and have the Orcad PSpice program running, we, can display the magnitude and phase plots as mentioned above. Suppose we want to display a Bode magnitude plot of V(4). We select, Trace/Add and type dB(V(4)) in Trace Command box. dB(V(4)) is, equivalent to 20log(V(4)), and because the magnitude of V1 or V(R1:1), is unity, dB(V(4)) actually corresponds to dB(V(4)/V(R1:1)), which is, the gain. Adding the trace dB(V(4)) will give a Bode magnitude/gain, plot with the Y axis in dB., Once a plot is obtained in the Orcad PSpice window, we can add, labels to it for documentation purposes. To add a title to the plot, select, Edit/Modify Title in the Orcad PSpice menu and type the title in the, dialog box. To add a label to the Y axis, select Plot/Y Axis Settings,, type the label, and CLICKL OK. Add a label to the X axis in the same, manner., As an alternative approach, we can avoid running the Orcad, PSpice program by using pseudocomponents to send results to the, output file. Pseudocomponents are like parts that can be inserted into, a schematic as if they were circuit elements, but they do not correspond to circuit elements. We can add them to the circuit for specifying initial conditions or for output control. In fact, we have already, used two pseudocomponents, VIEWPOINT and IPROBE, for DC, analysis. Other important pseudocomponents and their usage are, shown in Fig. D.40 and listed in Table D.2. The pseudocomponents, are added to the schematic. To add a pseudocomponent, select, Draw/Get New Parts in the Schematics window, select the pseudocomponent, place it at the desired location, and add the appropriate, attributes as usual. Once the pseudocomponents are added to the, schematic, we select Analysis/Setup/AC Sweep and enter the specifications for the AC sweep, and finally, select Analysis/Simulate to, perform AC sweep. If no errors are encountered, the voltages and currents specified in the pseudocomponents will be saved in the output, file. We obtain the output file by selecting Analysis/Examine Output, in the Schematics window., |}, , VPRINT1, PRINTDGTLCHG, , VPRINT2, IPRINT, , Figure D.40, Print and plot pseudocomponents., , VPLOT1, , |}, , VPLOT2, IPLOT
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ale29559_appD.qxd, , 07/17/2008, , 09:18 AM, , Page A-43, , Appendix D, , PSpice for Windows, , A-43, , TABLE D.2, , Print and plot pseudocomponents., Symbol, , Description, , IPLOT, IPRINT, VPLOT1, , Plot showing branch current; symbol must be placed in series, Table showing branch current; symbol must be placed in series, Plot showing voltages at the node to which the symbol is, connected, Plot showing voltage differentials between two points to which, the symbol is connected, Table showing voltages at the node to which the symbol is, connected, Table showing voltage differentials between two points to, which the symbol is connected, , VPLOT2, VPRINT1, VPRINT2, , Example D.6, , Find current i in the circuit in Fig. D.41., , 5 cos 2t A, , 1, , 2Ω, , 0.5 H, , 2, , 3, , i, 20 sin 2t V, , +, −, , 0.25 F, , 2i, , 4Ω, , 0, , Figure D.41, For Example D.6., , Solution:, Recall that 20 sin 2t 20 cos(2t 90) and that f 2 p 22 p , 0.31831. The schematic is shown in Fig. D.42. The attributes of V1 are, set as ACMAG 20, ACPHASE 90; while the attributes of IAC, are set as AC 5. The current-controlled current source is connected, in such a way as to conform with the original circuit in Fig. D.41; its, gain is set equal to 2. The attributes of the pseudocomponent IPRINT are, set as AC yes, MAG yes, PHASE ok, REAL , and IMAG ., Since this is a single-frequency ac analysis, we select Analysis/, Setup/AC Sweep and enter Total Pts 1, Start Freq 0.31831, and, Final Freq 0.31831. We save the circuit and select Analysis/Simulate, for simulation. The output file includes, FREQ, , IM(V_PRINT3), , IP(V_PRINT3), , 3.183E-01, , 7.906E+00, , 4.349E+01, , From the output file, we obtain I 7.906l43.49 A or i(t) 7.906 cos, (2t 43.49) A. This example is for a single-frequency ac analysis;, Example D.7 is for AC sweep over a range of frequencies.
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ale29559_appD.qxd, , 07/17/2008, , 09:19 AM, , A-44, , Page A-44, , PSpice for Windows, , Appendix D, , IAC, , 5A, , +, −, I1, L1, , R1, , 0.5 H, , 2, AC = yes, MAG = yes, PHASE = ok, 20 +, −, , IPRINT, F1, R2, , V1, , 4, , F, C1, , 0.25, GAIN = 2, 0, , Figure D.42, The schematic of the circuit in Fig. D.41., , Practice Problem D.6, , Find ix 1t2 in the circuit in Fig. D.43., 10 Ω, , 1H, ix, , 20 cos 4t V, , +, −, , 0.1 F, , 2ix, , 0.5 H, , Figure D.43, For Practice Prob. D.6., , Answer: From the output file, Ix 7.59l108.43 or, ix 7.59 cos(4t 108.43) A., , Example D.7, , For the RC circuit shown in Fig. D.44, obtain the magnitude plot of, the output voltage yo for frequencies from 1 Hz to 10 kHz. Let, R 1 k and C 4 mF., Solution:, The schematic is shown in Fig. D.45. We assume that the magnitude of, V1 is 1 and its phase is zero; we enter these as the attributes of V1. We, R1, , 1, , R, , 2, , 1k, v i (t), , +, −, , +, C, , −, , v o(t), , ACMAG = 1V, +, ACPHASE = 0 −, , Figure D.44, For Example D.7., , C1, , V1, , 0, , Figure D.45, The schematic of the circuit in, Fig. D.44., , 4u
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ale29559_appD.qxd, , 07/17/2008, , 09:19 AM, , Page A-45, , Appendix D, , PSpice for Windows, , A-45, , also assume 10 points per decade. For the AC sweep specifications, we, select Analysis/Setup/AC Sweep and enter 10 in the Total Pts box, 1, in the Start Freq box, and 10k in the Final Freq box. After saving the, circuit, we select Analysis/Simulate. From the Orcad PSpice menu,, we obtain the plot in Fig. D.46(a) by selecting Traces/Add and clicking, V(2). Also, by selecting Trace/Add and typing dB(V(2)) in the Trace, Command box, we obtain the Bode plot in Fig. D.46(b). The two plots, in Fig. D.46 indicate that the circuit is a lowpass filter: low frequencies, are passed while high frequencies are blocked by the circuit., 0, , 1.0 V, , 0.5 V, , 0 V, 1.0 Hz, 10 Hz, V(2), , 100 Hz, , 1.0 kHz, , 10 kHz, , -50, 10 Hz, 1.0 Hz, dB(V(2)), , 100 Hz, , Frequency, , Frequency, , (a), , (b), , 1.0 kHz, , 10 kHz, , Figure D.46, Result of Example D.7: (a) linear, (b) Bode plot., , For the circuit in Fig. D.44, replace the capacitor C with an inductor, L 4 mH and obtain the magnitude plot (both linear and Bode) for vo, for 10 6 f 6 100 MHz., , Practice Problem D.7, , Answer: See the plots in Fig. D.47., 1.0 V, , 0, , 0.5 V, , -40, , 0 V, 10 Hz, V(2), , 1.0 Hz, , 100 kHz, , 10 MHz, , Frequency, (a), , Figure D.47, Result of Practice Prob. D.7: (a) linear plot, (b) Bode plot., , -80, 10 Hz, 1.0 Hz, 100 kHz, dB(V(2)), Frequency, (b), , 10 MHz
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ale29559_appE.qxd, , 07/17/2008, , 09:35 AM, , Page A-46, , Appendix E, MATLAB, MATLAB has become a powerful tool of technical professionals worldwide. The term MATLAB is an abbreviation for MATrix LABoratory,, implying that MATLAB is a computational tool that uses matrices and, vectors (or arrays) to carry out numerical analysis, signal processing,, and scientific visualization tasks. Because MATLAB uses matrices as, its fundamental building blocks, one can write mathematical expressions involving matrices just as easily as one would on paper. MATLAB, is available for Macintosh, Unix, and Windows operating systems. A, student version of MATLAB is available for personal computers (PCs)., A copy of MATLAB can be obtained from, The Mathworks, Inc., 3 Apple Hill Drive, Natick, MA 01760-2098, Phone:(508) 647-7000, Website: http://www.mathworks.com, A brief introduction to MATLAB is presented in this appendix., What is presented is sufficient for solving problems in this book. More, about MATLAB can be found in MATLAB books and from on-line help., The best way to learn MATLAB is to work with it after having learned, the basics., , E.1, , MATLAB Fundamentals, , The Command window is the primary area where you interact with, MATLAB. A little later, we will learn how to use the text editor to create M-files, which allow for execution of sequences of commands. For, now, we focus on how to work in the Command window. We will first, learn how to use MATLAB as a calculator., , Using MATLAB as a Calculator, The following are algebraic operators used in MATLAB:, +, *, ^, /, \, , Addition, Subtraction, Multiplication, Exponentiation, Right division (a/b means a b), Left division (a\b means b a), , To begin to use MATLAB, we use these operators. Type commands to the MATLAB prompt “>>” in the Command window, A-46
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ale29559_appE.qxd, , 07/17/2008, , 09:35 AM, , Page A-47, , Appendix E, , MATLAB, , (correct any mistakes by backspacing) and press the Enter key. For, example,, >> a = 2; b = 4;c = -6;, >> dat = b^2 - 4*a*c, dat =, 64, >> e = sqrt(dat)/10, e =, 0.8000, The first command assigns the values 2, 4, and 6 to the variables a, b,, and c, respectively. MATLAB does not respond because this line ends with, a colon. The second command sets dat to b2 4ac and MATLAB returns, the answer as 64. Finally, the third line sets e equal to the square root of, dat and divides by 10. MATLAB prints the answer as 0.8. Other mathematical functions, listed in Table E.1, can be used similarly to how the function sqrt is used here. Table E.1 provides just a tiny sample of MATLAB, functions. Others can be obtained from the on-line help. To get help, type, >> help, A long list of topics will come up. For a specific topic, type the command name. For example, to get help on “log to base 2,” type, >> help log2, A help message on the log function will be displayed. Note that, MATLAB is case sensitive, so sin(a) is not the same as sin(A)., TABLE E.1, , Typical elementary math functions., Function, , Remark, , abs(x), acos, acosh(x), , Absolute value or complex magnitude of x, Inverse cosine and inverse hyperbolic cosine of x in, radians, Inverse cotangent and inverse hyperbolic cotangent, of x in radians, Phase angle (in radian) of a complex number x, Inverse sine and inverse hyperbolic sine of x in radians, Inverse tangent and inverse hyperbolic tangent of x, in radians, Complex conjugate of x, Cosine and hyperbolic cosine of x in radians, Cotangent and hyperbolic cotangent of x in radians, Exponential of x, Round toward zero, Imaginary part of a complex number x, Natural logarithm of x, Logarithm of x to base 2, Common logarithms (base 10) of x, Real part of a complex number x, Sine and hyperbolic sine of x in radians, Square root of x, Tangent and hyperbolic tangent of x in radians, , acot, acoth(x), angle(x), asin, asinh(x), atan, atanh(x), conj(x), cos, cosh(x), cot, coth(x), exp(x), fix, imag(x), log(x), log2(x), log10(x), real(x), sin, sinh(x), sqrt(x), tan, tanh, , A-47
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ale29559_appE.qxd, , 07/17/2008, , 09:35 AM, , Page A-48, , Appendix E, , A-48, , MATLAB, , Try the following examples:, >> 3^(log10(25.6)), >> y = 2* sin(pi/3), >> exp(y+4-1), In addition to operating on mathematical functions, MATLAB, allows one to work easily with vectors and matrices. A vector (or array), is a special matrix with one row or one column. For example,, >> a = [1 -3 6 10 -8 11 14];, is a row vector. Defining a matrix is similar to defining a vector. For, example, a 3 3 matrix can be entered as, >> A = [1 2 3; 4 5 6; 7 8 9], or as, >> A = [ 1 2 3, 4 5 6, 7 8 9], , TABLE E.2, , Matrix operations., Operation, , Remark, , A’, , Finds the transpose of, matrix A, Evaluates the determinant, of matrix A, Calculates the inverse of, matrix A, Determines the eigenval, ues of matrix A, Finds the diagonal, elements of matrix A, , det(A), inv(A), eig(A), diag(A), , In addition to the arithmetic operations that can be performed on a, matrix, the operations in Table E.2 can be implemented., Using the operations in Table E.2, we can manipulate matrices as, follows:, >> B = A’, B =, 1, 4 7, 2, 5 8, 3, 6 9, >> C = A + B, C =, 2, 6 10, 6 10 14, 10 14 18, >> D = A^3 - B*C, D =, 372, 432 492, 948 1131 1314, 1524 1830 2136, >> e = [1 2; 3 4], e =, 1 2, 3 4, >> f = det(e), f =, -2, >> g = inv(e), g =, -2.0000 1.0000, 1.5000 -0.5000
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ale29559_appE.qxd, , 07/17/2008, , 09:35 AM, , Page A-49, , Appendix E, , MATLAB, , A-49, , TABLE E.3, , Special matrices, variables, and constants., Matrix, Variable, Constant, , Remark, , eye, ones, zeros, i or j, pi, NaN, inf, eps, rand, , Identity matrix, An array of 1s, An array of 0s, Imaginary unit or sqrt(-1), 3.142, Not a number, Infinity, A very small number, 2.2e - 16, Random element, , >> H = eig(g), H =, -2.6861, 0.1861, Note that not all matrices can be inverted. A matrix can be inverted if, and only if its determinant is nonzero. Special matrices, variables, and, constants are listed in Table E.3. For example, type, >> eye(3), ans =, 1 0 0, 0 1 0, 0 0 1, to get a 3 3 identity matrix., , Plotting, To plot using MATLAB is easy. For a two-dimensional plot, use the, plot command with two arguments as follows:, >> plot(xdata,ydata), where xdata and ydata are vectors of the same length containing, the data to be plotted., For example, suppose we want to plot y = 10*sin(2*pi*x), from 0 to 5*pi. We will proceed with the following commands:, % x is a vector, 0 <= x <= 5*pi, increments of pi/100, % creates a vector y, % creates the plot, With this, MATLAB responds with the plot in Fig. E.1., MATLAB will let you graph multiple plots together and distinguish them with different colors. This is obtained with the format, plot(xdata, ydata, ‘color’), where the color is indicated, by using a character string from the options listed in Table E.4., For example,, >> plot (x1,y1, ‘r’, x2,y2, ‘b’, x3,y3, ‘--’);, will graph data (x1, y1) in red, data (x2, y2) in blue, and data, (x3, y3) in dashed line all on the same plot., , >> x = 0:pi/100:5*pi;, >> y = 10*sin(2*pi*x);, >> plot(x,y);
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ale29559_appE.qxd, , 07/17/2008, , 09:35 AM, , Page A-50, , Appendix E, , A-50, , MATLAB, 10, , TABLE E.4, , 8, , Various color and line types., y, m, c, r, g, b, w, k, , Yellow, Magenta, Cyan, Red, Green, Blue, White, Black, , ., o, x, +, *, :, –., ––, , 6, 4, , Point, Circle, x mark, Plus, Solid, Star, Dotted, Dashdot, Dashed, , 2, 0, –2, –4, –6, –8, –10, , 0, , 2, , 4, , 6, , 8, , 10, , 12, , 14, , 16, , Figure E.1, MATLAB plot of y = 10*sin(2*pi*x)., , MATLAB also allows for logarithm scaling. Rather than using the, plot command, we use, loglog log(y) versus log(x), semilogx y versus log(x), semilogy log(y) versus x, Three-dimensional plots are drawn using the functions mesh, and meshdom (mesh domain). For example, to draw the graph of, z = x*exp( - x^2 - y^2)over the domain -1 < x, y < 1, we, type the following commands:, >>, >>, >>, >>, >>, , xx = -1:.1:1;, yy = xx;, [x,y] = meshgrid(xx,yy);, z = x.*exp(-x.^2 -y.^2);, mesh(z);, , (The dot symbol used in x. and y. allows element-by-element multiplication.) The result is shown in Fig. E.2., , 0.5, , 0, , – 0.5, 30, 20, 10, 0, , Figure E.2, A three-dimensional plot., , 0, , 5, , 10, , 15, , 20, , 25
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ale29559_appE.qxd, , 07/17/2008, , 09:35 AM, , Page A-51, , Appendix E, , MATLAB, , A-51, , Programming MATLAB, So far we have used MATLAB as a calculator. You can also use, MATLAB to create your own program. The command line editing in, MATLAB can be inconvenient if one has several lines to execute. To, avoid this problem, you can create a program that is a sequence of, statements to be executed. If you are in the Command window, click, File/New/M-files to open a new file in the MATLAB Editor/Debugger, or simple text editor. Type the program and save it in a file with an, extension .m, say filename.m; it is for this reason that it is called, an M-file. Once the program is saved as an M-file, exit the Debugger, window. You are now back in the Command window. Type the file, without the extension .m to get results. For example, the plot that was, made in Fig. E.2 can be improved by adding title and labels and being, typed as an M-file called example1.m., % x is a vector, 0 <= x <= 5*pi, increments of pi/100, % creates a vector y, % create the plot, % label the x axis, % label the y axis, % title the plot, % add grid, , x = 0:pi/100:5*pi;, y = 10*sin(2*pi*x);, plot(x,y);, xlabel(‘x (in radians)’);, ylabel(‘10*sin(2*pi*x)’);, title(‘A sine functions’);, grid, , Once the file is saved as example1.m and you exit the text editor,, type, >> example1, in the Command window and hit Enter to obtain the result shown in, Fig. E.3., To allow flow control in a program, certain relational and logical, operators are necessary. They are shown in Table E.5. Perhaps the most, commonly used flow control statements are for and if. The for, , A sine function, , 10, 8, , 10*sin(2*pi*x), , 6, 4, 2, 0, –2, –4, –6, –8, –10, 0, , 2, , 4, , 6, 8, 10, x (in radians), , 12, , 14, , Figure E.3, MATLAB plot of y = 10*sin(2*pi*x) with title, and labels., , 16, , TABLE E.5, , Relational and logical, operators., Operator, , Remark, , <, <=, >, >=, ==, ~=, &, |, ~, , less than, less than or equal, greater than, greater than or equal, equal, not equal, and, or, not
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ale29559_appE.qxd, , A-52, , 07/17/2008, , 09:35 AM, , Page A-52, , Appendix E, , MATLAB, , statement is used to create a loop or a repetitive procedure and has the, general form, for x = array, [commands], end, The if statement is used when certain conditions need to be met before, an expression is executed. It has the general form, if expression, [commands if expression is True], else, [commands if expression is False], end, For example, suppose we have an array y(x) and we want to, determine the minimum value of y and its corresponding index x. This, can be done by creating an M-file as shown here., % example2.m, % This program finds the minimum y value and, its corresponding x index, x = [1 2 3 4 5 6 7 8 9 10]; %the nth term in y, y = [3 9 15 8 1 0 -2 4 12 5];, min1 = y(1); for k = 1:10, min2 = y(k);, if(min2 < min1), min1 = min2;, xo = x(k);, else, min1 = min1;, end, end, diary, min1, xo, diary off, Note the use of the for and if statements. When this program is, saved as example2.m, we execute it in the Command window and, obtain the minimum value of y as -2 and the corresponding value of, x as 7, as expected., >> example2, min1 =, -2, xo =, 7
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ale29559_appE.qxd, , 07/17/2008, , 09:35 AM, , Page A-53, , MATLAB, , Appendix E, , If we are not interested in the corresponding index, we could do the, same thing using the command, >> min(y), The following tips are helpful in working effectively with, MATLAB:, • Comment your M-file by adding lines beginning with a %, character., • To suppress output, end each command with a semicolon (;); you, may remove the semicolon when debugging the file., • Press the up and down arrow keys to retrieve previously executed, commands., • If your expression does not fit on one line, use an ellipse (. . .) at, the end of the line and continue on the next line. For example,, MATLAB considers, y = sin(x + log10(2x + 3)) + cos(x + ..., log10(2x + 3));, as one line of expression., • Keep in mind that variable and function names are case sensitive., , Solving Equations, Consider the general system of n simultaneous equations:, a11x1 a12 x2 p a1n xn b1, a21x1 a22 x2 p a2n xn b2, o, an1x1 an2 x2 p ann xn bn, or in matrix form, AX B, where, a11 a12, a21 a22, A ≥ p p, , p, p, p, , a1n, a2n, p ¥, , x1, x2, X ≥p¥, , an1 an2 an3 an4, , xn, , b1, b2, B ≥p¥, bn, , A is a square matrix and is known as the coefficient matrix, while X, and B are vectors. X is the solution vector we are seeking to get. There, are two ways to solve for X in MATLAB. First, we can use the backslash operator(\) so that, X = A\B, Second, we can solve for X as, X A1B, which in MATLAB is the same as, X = inv(A)*B, , A-53
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ale29559_appE.qxd, , 07/17/2008, , 09:35 AM, , A-54, , Example E.1, , Page A-54, , Appendix E, , MATLAB, , Use MATLAB to solve Example A.2., Solution:, From Example A.2, we obtain matrix A and vector B and enter them, in MATLAB as follows., >> A = [25 -5 -20; -5 10 -4; -5 -4 9], A =, 25 -5 -20, -5 10, -4, -5 -4, 9, >> B = [50 0 0]’, B =, 50, 0, 0, >> X = inv(A)*B, X =, 29.6000, 26.0000, 28.0000, >> X = A\B, X =, 29.6000, 26.0000, 28.0000, Thus, x1 = 29.6, x2 = 26, and x3 = 28., , Practice Problem E.1, , Solve the problem in Practice Prob. A.2 using MATLAB., Answer: x1 = 3 = x3, x2 = 2., , E.2, , DC Circuit Analysis, , There is nothing special in applying MATLAB to resistive dc circuits., We apply mesh and nodal analysis as usual and solve the resulting, simultaneous equations using MATLAB as is described in Section E.1., Examples E.2 to E.5 illustrate., , Example E.2, , Use nodal analysis to solve for the nodal voltages in the circuit of Fig. E.4., Solution:, At node 1,, 2, , V1 0, V1 V2, , S 16 3V1 2V2, 4, 8, , (E.2.1)
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ale29559_appE.qxd, , 07/17/2008, , 09:35 AM, , Page A-57, , Appendix E, , MATLAB, , A-57, , Putting Eqs. (E.3.1) to (E.3.4) together in matrix form, we have, 9 4, 0 2, I1, 6, 4 15 4 6, I2, 12, ≥, ¥ ≥ ¥ ≥, ¥, 0 4 10 2, I3, 12, 2 6 2 20, I4, 0, or AI B, where the vector I contains the unknown mesh currents., We now use MATLAB to determine I as follows:, >> A = [9 -4 0 -2; -4 15 -4 -6;, 0 -4 10 -2; -2 -6 -2 20], A =, 9 -4, 0 -2, -4 15 -4 -6, 0 -4 10 -2, -2 -6 -2 20, >> B = [6 -12 12 0]’, B=, 6, -12, 12, 0, >> I = inv(A)*B, I=, 0.5203, -0.3555, 1.0682, 0.0522, Thus, I1 0.5203, I2 0.3555, I3 1.0682, and I4 0.0522 A., , Find the mesh currents in the circuit in Fig. E.7 using MATLAB., 2Ω, , I1, , 2Ω, , 4Ω, , 4Ω, , 10 V, , +, −, , I3, , I2, , +, −, , 8V, , 4Ω, , 4Ω, , I4, , 2Ω, , 2Ω, , Figure E.7, For Practice Prob. E.3., , Answer: I1 0.2222, I2 0.6222, I3 1.1778, and I4 0.2222 A., , Practice Problem E.3
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ale29559_appE.qxd, , 07/17/2008, , 09:35 AM, , Page A-58, , Appendix E, , A-58, , E.3, , MATLAB, , AC Circuit Analysis, , Using MATLAB in ac circuit analysis is similar to how MATLAB is, used for dc circuit analysis. We must first apply nodal or mesh analysis to the circuit and then use MATLAB to solve the resulting system, of equations. However, the circuit is in the frequency domain, and we, are dealing with phasors or complex numbers. So in addition to what, we learned in Section E.2, we need to understand how MATLAB handles complex numbers., MATLAB expresses complex numbers in the usual manner, except, that the imaginary part can be either j or i representing 11. Thus,, 3 j4 can be written in MATLAB as 3 - j4, 3 - j*4, 3 - i4, or, 3 - I*4. Here are the other complex functions:, abs(A), angle(A), conj(A), imag(A), real(A), , Absolute value of magnitude of A, Angle of A in radians, Complex conjugate of A, Imaginary part of A, Real part of A, , Keep in mind that an angle in radians must be multiplied by 180p to, convert it to degrees, and vice versa. Also, the transpose operator (‘), gives the complex conjugate transpose, whereas the dot-transpose (.‘), transposes an array without conjugating it., , Example E. 4, 20 mF, , v, , +, −, , 2H, , v1, , 10 Ω, , Figure E.8, For Example E.4., , In the circuit of Fig. E.8, let v 4 cos(5t 30) V and i 0.8 cos 5t A., Find v1 and v2., v2, , 20 Ω, , i, , Solution:, As usual, we convert the circuit in the time-domain to its frequencydomain equivalent., v 4 cos(5t 30) ¡ V 4l30, 5, i 0.8 cos 5t ¡ I 8l0, 2 H ¡ jL j5 2 j10, 1, 1, 20 mF ¡, , j10, jC, j10 103, Thus, the frequency-domain equivalent circuit is shown in Fig. E.9. We, now apply nodal analysis to this., – j10 Ω, , 4 –30°, , +, −, , V1, , j10 Ω V, 2, , 10 Ω, , 20 Ω, , Figure E.9, The frequency-domain equivalent circuit of the, circuit in Fig. E.8., , 0.8 A
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ale29559_appE.qxd, , 07/17/2008, , 09:35 AM, , Page A-59, , Appendix E, , MATLAB, , At node 1,, 4l30 V1, j10, , , , V1, V1 V2, , ¡ 4l30 3.468 j2, 10, j10, jV1 V2, (E.4.1), , At node 2,, 0.8 , , V2, V2 V1, , ¡ j16 2V1 (2 j)V2 (E.4.2), 20, j10, , Equations (E.4.1) and (E.4.2) can be cast in matrix form as, c, , j, 1, V1, 3.468 j2, d c d c, d, 2 (2 j) V2, j16, , or AV B. We use MATLAB to invert A and multiply the inverse by, B to get V., >> A = [-j 1; -2 (2 + j)], A =, 0 - 1.0000i 1.000, -2.0000 2.0000 + 1.000 i, >> B = [(3.468 - 2j) 16j].’ %note the dot-transpose, B=, 3.4680 - 2.0000i, 0 + 16.0000i, >> V = inv(A)*B, V =, 4.6055 - 2.4403i, 5.9083 + 2.6055i, >> abs(V(1)), ans =, 5.2121, >> angle(V(1))*180/pi %converts angle from, radians to degrees, ans =, -27.9175, >> abs(V(2)), ans =, 6.4573, >> angle(V(2))*180/pi, ans =, 23.7973, Thus,, V1 4.6055 j2.4403 5.212l27.92, V2 5.908 j2.605 6.457l23.8, , A-59
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ale29559_appE.qxd, , 07/17/2008, , 09:35 AM, , A-62, , Page A-62, , Appendix E, , MATLAB, , >> abs (I(2)), ans =, 21.0426, >> angle(I(2))*180/pi, ans =, -53.8864, >> abs(I(3)), ans =, 8.2841, >> angle(I(3))*180/pi, ans =, -46.8833, >> abs(IbB), ans =, 11.3709, >> angle(IbB)*180/pi, ans =, -113.8001, Thus, I1 18.23l21.21,, , I2 21.04l58.89,, , I3 8.284l46.88, and IbB 11.37l113.8A., , Practice Problem E.5, , In the unbalanced wye-wye three-phase system in Fig. E.12, find the, line currents I1, I2, and I3 and the phase voltage VCN ., 220 0° V, , 2 + j1 Ω, , I1, , 7 + j10 Ω, , 220 –120° V 2 – j 0.5 Ω, B, –+, , I2, , 8 + j6 Ω, , 220 120° V, , I3, , –+, , –+, , 2 + j1 Ω, , A, , C, , N, , 10 – j12 Ω, , Figure E.12, For Practice Prob. E.5., , Answer: 22.66l26.54 A, 6.036l150.48 A, 19.93l138.9 A,, 94.29l159.3 V., , E.4, , Frequency Response, , Frequency response involves plotting the magnitude and phase of the, transfer function H(s) D(s)N(s) or obtaining the Bode magnitude, and phase plots of H(s). One hard way to obtain the plots is to generate
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ale29559_appE.qxd, , 07/17/2008, , 09:35 AM, , Page A-63, , Appendix E, , MATLAB, , A-63, , data using the for loop for each value of s j for a given range of, and then plot the data as we did in Section E.1. However, there is, an easy way that allows us to use one of two MATLAB commands:, freqs and bode. For each command, we must first specify H(s), as num and den, where num and den are the vectors of coefficients, of the numerator N(s) and denominator D(s) in descending powers, of s, i.e., from the highest power to the constant term. The general, form of the bode function is, bode(num, den, range);, where range is a specified frequency interval for the plot. If range is, omitted, MATLAB automatically selects the frequency range. The range, could be linear or logarithmic. For example, for 1 6 6 1000 rad/s, with 50 plot points, we can specify a linear range as, range = linspace(1,1000,50);, For a logarithmic range with 102 6 6 104 rad/s and 100 plot, points in between, we specify range as, range = logspace(-2,4,100);, For the freqs function, the general form is, hs = freqs(num, den, range);, where hs is the frequency response (generally complex). We still need, to calculate the magnitude in decibels as, mag = 20*log 10(abs(hs)), and phase in degrees as, phase = angle(hs)*180/pi, and plot them, whereas the bode function does it all at once. We illustrate with an example., , Use MATLAB to obtain the Bode plots of, G (s) , , s3, s 14.8s 38.1s 2554, 3, , 2, , Solution:, With the explanation previously given, we develop the MATLAB code, as shown here., % for example e.6, num=[1 0 0 0];, den = [1 14.8 38.1 2554];, w = logspace(-1,3);, bode(num, den, w);, title(‘Bode plot for a highpass filter’), Running the program produces the Bode plots in Fig. E.13. It is, evident from the magnitude plot that G(s) represents a highpass filter., , Example E.6
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ale29559_appE.qxd, , 07/17/2008, , 09:35 AM, , A-64, , Page A-64, , Appendix E, , MATLAB, Bode plot for a highpass filter, , Phase (deg); magnitude (dB), , 0, –50, –100, , 20, 0, –20, –40, –60, –80, 10–1, , 100, , 101, Frequency (rad/s), , 102, , 103, , Figure E.13, For Example E.6., , Use MATLAB to determine the frequency response of, H(s) , , 10(s 1), s 6s 100, 2, , Answer: See Fig. E.14., Bode diagrams, 0, –10, Phase (deg); magnitude (dB), , Practice Problem E.6, , –20, –30, –40, 50, 0, –50, 10–1, , 100, , Figure E.14, For Practice Prob. E.6., , 101, Frequency (rad/s), , 102, , 103
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ale29559_appF.qxd, , 07/17/2008, , 09:40 AM, , Page A-65, , Appendix F, KCIDE for Circuits, Engineers of the twenty-first century will need to be able to work in, a “knowledge capturing integrated design environment” (known as, KCIDE). Essentially, engineers will go to their computer where they, will do their work on a platform, much like Windows, where their various software packages, laboratory work, and other support software, packages (such as Word) will all come together and interact with, each other to help them with their work. These platforms will capture, the work being done by the engineer and make the data available to, be used in any manner the engineer chooses (such as preliminary, design reports, user manuals, papers, books, proposals, or requests for, proposals)., A detailed presentation of all the elements associated with learning how to work in such an environment is beyond the scope of this, book. However, a platform to begin the process of training engineers, to work in this environment is included in this textbook. KCIDE for, Circuits was designed to assist the circuits student to learn how to work, in a simplified KCIDE environment designed especially for electrical, circuits students. The software that is used in the platform includes, PSpice, MATLAB, Excel, Word, and PowerPoint., In this appendix, we will help you to understand the KCIDE for, Circuits platform and how to use it. The software can be obtained,, free of any charges, from the website http://KCIDE.FennResearch.org., More details and examples are also included at the website. In addition, we will also have support services available at the website., , F.1, , How to Work with KCIDE for Circuits, , The structure of the platform and how it is effectively used follows, the problem-solving process used throughout the text. This is essentially a systems approach to problem solving that uses a structured, process to capture your work and present it in two different formats., It will be helpful to work through an example to see how to use the, platform., , Use the KCIDE for Circuits platform to solve Example 3.2., , Example F.1, , Solution:, Opening the software, we see the screen shown in Fig. F.1, where we, define a new project. Although we can name the project anyway we, A-65
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ale29559_appF.qxd, , A-66, , 07/17/2008, , 09:40 AM, , Page A-66, , Appendix F, , KCIDE for Circuits, , Figure F.1, Creating a new project in KCIDE for Circuits., , wish to, we name it KCIDE Example F-1 050626 (see Fig. F.2). Note, that the last six digits are: year/month/day. The reason for this is that, if we create different files corresponding to different dates, the files will, always appear in chronological order., , Figure F.2, Naming the project., , We now enter the problem statement into the screen shown in, Fig. F.3. After we have entered the problem statement, we can click on, the button to Open PSpice. The next screen, Fig. F.4, shows what is, seen when the Open PSpice button is clicked. To open the PSpice, schematic capture, we need to click on the page 1 icon. In Schematic,, we create the circuit representing our problem. This is shown in, Fig. F.5., We now need to enter all we know about the problem by entering, our problem analysis into the text box and then identifying the number
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ale29559_appF.qxd, , 07/17/2008, , 09:40 AM, , Page A-67, , Appendix F, , Figure F.3, Entering the problem statement., , Figure F.4, How to open the schematic capture of PSpice., , Figure F.5, Circuit for Example F.1., , KCIDE for Circuits, , A-67
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ale29559_appF.qxd, , A-68, , 07/17/2008, , 09:40 AM, , Page A-68, , Appendix F, , KCIDE for Circuits, , of unknown nodes and loops for the circuit (see Fig. F.6). We continue, this process by going to the next screen and entering the requested, information (see Fig. F.7)., , Figure F.6, Presenting what we know about the problem, part 1., , Figure F.7, Identifying the unknown node voltages and unknown loop currents,, part 2., , We now proceed to selecting the method of solution. We do this, by entering the requested information into the screen shown in Fig. F.8., Now we can develop the equations that will generate a solution for, the problem. Since nodal analysis is required for the solution for the, node voltages, all we need to do is to write the nodal equations., Once we have the appropriate equations, as shown in Fig. F.9, we can, select a solution technique. In this case we chose Excel to solve our, simultaneous equations.
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ale29559_appF.qxd, , 07/17/2008, , 09:40 AM, , Page A-69, , Appendix F, , KCIDE for Circuits, , Figure F.8, Selecting the method of solving the problem., , Figure F.9, Solving for the unknown node voltages., , Now, when we go to the next screen, the Evaluate portion of the, solution, we actually open PSpice again. We need to open page 1 to, retrieve our original circuit (see Fig. F.10). Once we have our original, PSpice circuit, we need to prepare it for solving for our unknowns. The, first step in this task is to go to the PSpice button and select New, Simulation Profile (see Fig. F.11)., We next need to assign a name to the new simulation profile, (Fig. F.12). Clicking on the Create button produces the screen shown, in Fig. F.13. For this problem, we select Bias Point for the Analysis, type., Clicking on OK returns the screen to the original condition. Now, we go to the PSpice button and select Run from the dropdown menu,, , A-69
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ale29559_appF.qxd, , A-70, , 07/17/2008, , 09:40 AM, , Page A-70, , Appendix F, , KCIDE for Circuits, , Figure F.10, Opening PSpice again., , Figure F.11, Setting up our circuit for solution by PSpice., , Figure F.12, Setting up our circuit for solution by PSpice.
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ale29559_appF.qxd, , 07/17/2008, , 09:40 AM, , Page A-71, , Appendix F, , KCIDE for Circuits, , Figure F.13, Setting up our circuit for solution by PSpice., , Fig. F.14. Running PSpice produces the screen shown in Fig. F.15. We, can immediately see that the voltages agree with the solution we, obtained by using nodal analysis. Clicking on Next leads to our being, asked, as in Fig. F.16, if we have any graphics to export. For this, problem, we have no graphs., , Figure F.14, Setting up our circuit for solution by PSpice., , We are now approaching the end of the process. We are asked to, comment about the solution, Fig. F.17. And, we are asked if the, answers agree with the PSpice solution. The answers do agree and, we can proceed to determining what we want to export, Fig. F.18., We can generate Word and/or PowerPoint files, Fig. F.19. In this case,, we select both but will only show the output of the Word file, Fig., F.20. Note: This output was modified so that it could be presented on, two pages., , A-71
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ale29559_appF.qxd, , A-72, , 07/17/2008, , 09:40 AM, , Page A-72, , Appendix F, , KCIDE for Circuits, , Figure F.15, Problem solution using PSpice., , Figure F.16, Screen for exporting graphs., , Figure F.17, Determining if the problem has been solved correctly.
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ale29559_appF.qxd, , 07/17/2008, , 09:40 AM, , Page A-73, , Appendix F, , KCIDE for Circuits, , Figure F.18, Determining if you want to generate Word and/or PowerPoint documents., , Figure F.19, Generating Word and PowerPoint files for Example F.1., , Figure F.20, Word file output., , A-73
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ale29559_appF.qxd, , 07/17/2008, , 09:40 AM, , A-74, , Page A-74, , Appendix F, , KCIDE for Circuits, , We have now completed a detailed example. We suggest that you, first try to do this with the platform and look at your output in both, Word and PowerPoint. To help you to continue to develop your facility, with the platform, try working the following practice problem, using, mesh analysis. For more examples, please go to the website., , Practice Problem F.1, , Use the KCIDE for Circuits platform to solve Practice Prob. 3.2., Answer: v1 80 V, v2 64 V, and v3 156 V.
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ale29559_appG.qxd, , 07/17/2008, , 10:04 AM, , Page A-75, , Appendix G, Answers to Odd-Numbered Problems, Chapter 1, , 1.27 (a) 43.2 kC, (b) 475.2 kJ, (c) 1.188 cents, , 1.1, , (a) 0.1038 C, (b) 0.19865 C, (c) 3.941 C,, (d) 26.08 C, , 1.3, , (a) 3t 1 C, (b) t 2 5t mC,, (c) 2 sin(10t p6) 1 mC,, (d) e30t [0.16 cos 40t 0.12 sin 40t] C, , 1.29 39.6 cents, 1.31 $42.05, 1.33 6 C, 1.35 2.333 MWh, , 1.5, , 25 C, , 1.7, , 25 A, 0 6 t 6 2, i • 25 A, 2 6 t 6 6, 25 A, 6 6 t 6 8, , 1.37 29.84 kWh, 1.39 24 cents, , See the sketch in Fig. G.1., , Chapter 2, i(t) A, , 2.1, , This is a design problem with several answers., , 2.3, , 184.3 mm, , 2.5, , n 9, b 15, l 7, , 2.7, , (a) 6 branches and 5 nodes, and (b) 7 branches, and 5 nodes., , 2.9, , 14 A, 2 A, 10 A, , 25, , 0, , 2, , 4, , 6, , 8, , t (s), , −25, , Figure G.1, For Prob. 1.7., , 2.11 6 V, 3 V, 1.9, , (a) 10 C, (b) 22.5 C, (c) 30 C, , 2.13 12 A, 10 A, 5 A, 2 A, , 1.11 3.672 kC, 4.406 kJ, , 2.15 10 V, 2 A, , 1.13 164.5 mW, 78.34 mJ, , 2.17 2 V, 22 V, 10 V, , 1.15 (a) 1.297 C, (b) 90e4t W, (c) 22.5 J, , 2.19 2 A, 12 W, 8 W, 40 W, 20 W, , 1.17 70 W, , 2.21 4.167 W, , 1.19 3 A, , 2.23 2 V, 1.92 W, , 1.21 2.696 1023 electrons, 43,200 C, , 2.25 0.1 A, 2 kV, 0.2 kW, , 1.23 $1.35, , 2.27 6.4 V, , 1.25 21.6 cents, , 2.29 1.625 , A-75
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ale29559_appG.qxd, , 07/17/2008, , 10:04 AM, , Page A-76, , A-76, , Appendix G, , Answers to Odd-Numbered Problems, , 2.31 11.2 A, 1.6 A, 9.6 A, 6.4 A, 3.2 A, , 2.79 75 , , 2.33 3 V, 6 A, , 2.81 38 k, 3.333 k, , 2.35 8 V, 0.2 A, , 2.83 3 k, (best answer), , 2.37 2.5 , 2.39 (a) 727.3 , (b) 3 k, , Chapter 3, , 2.41 16 , , 3.1, , This is a design problem with several answers., , 2.43 (a) 12 , (b) 16 , , 3.3, , 4 A, 2 A, 1.3333 A, 0.667 A, 40 V, , 2.45 (a) 59.8 , (b) 32.5 , , 3.5, , 20 V, , 2.47 24 , , 3.7, , 5.714 V, , 2.49 (a) 4 , (b) R1 18 , R2 6 , R3 3 , , 3.9, , 39.67 mA, , 2.51 (a) 9.231 , (b) 36.25 , , 3.11 293.9 W, 177.79 W, 238 W, , 2.53 (a) 142.32 , (b) 33.33 , , 3.13 8 V, 8 V, , 2.55 997.4 mA, 2.57 12.21 , 1.64 A, 2.59 1.2 A, 2.61 Use R1 and R3 bulbs, 2.63 0.4 , 1 W, 2.65 4 k, 2.67 (a) 4 V, (b) 2.857 V, (c) 28.57%, (d) 6.25%, 2.69 (a) 1.278 V (with), 1.29 V (without), (b) 9.30 V (with), 10 V (without), (c) 25 V (with), 30.77 V (without), 2.71 10 , , 3.15 29.45 A, 144.6 W, 129.6 W, 12 W, 3.17 1.73 A, 3.19 10 V, 4.933 V, 12.267 V, 3.21 1 V, 3 V, 3.23 22.34 V, 3.25 25.52 V, 22.05 V, 14.842 V, 15.055 V, 3.27 625 mV, 375 mV, 1.625 V, 3.29 0.7708 V, 1.209 V, 2.309 V, 0.7076 V, 3.31 4.97 V, 4.85 V, 0.12 V, 3.33 (a) and (b) are both planar and can be redrawn as, shown in Fig. G.2., , 2.73 45 , 2.75 (a) 19.9 k, (b) 20 k, 2.77 (a) Four 20- resistors in parallel, (b) One 300- resistor in series with a 1.8- resistor, and a parallel combination of two 20- resistors, (c) Two 24-k resistors in parallel connected in, series with two 56-k resistors in parallel, (d) A series combination of a 20- resistor, 300- , resistor, 24-k resistor, and a parallel, combination of two 56-k resistors, , 3Ω, , 6Ω, , 5Ω, , 1Ω, 4Ω, , 2Ω, 2A, , (a)
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ale29559_appG.qxd, , 07/17/2008, , 10:04 AM, , Page A-77, , Appendix G, , Answers to Odd-Numbered Problems, , 6, 9 3 4 0, i1, 4, i2, 3 8, 0, 0, 3.73 ≥, ¥, ¥ ≥ ¥ ≥, i3, 2, 4 0, 6 1, i4, 3, 0, 0 1 2, , 4Ω, 3Ω, 12 V +, −, , 5Ω, , A-77, , 2Ω, , 1Ω, (b), , Figure G.2, For Prob. 3.33., 3.35 20 V, , 3.75 3 A, 0 A, 3 A, 3.77 3.111 V, 1.4444 V, 3.79 5.278 V, 10.28 V, 694.4 mV, 26.88 V, 3.81 26.67 V, 6.667 V, 173.33 V, 46.67 V, 3.83 See Fig. G.3; 12.5 V, , 3.37 1.1111 V, 1, , 20 Ω, , 70 Ω, , 2, , 3, , 3.39 0.8 A, 0.9 A, 20 V, , 3.41 1.188 A, , +, −, , 50 Ω, , 2A, , 30 Ω, , 0, , 3.43 1.7778 A, 53.33 V, 3.45 8.561 A, , Figure G.3, For Prob. 3.83., , 3.47 10 V, 4.933 V, 12.267 V, , 3.85 9 , , 3.49 33.78 V, 10.67 A, , 3.87 8, , 3.51 20 V, , 3.89 30 mA, 12 V, , 3.53 1.6196 mA, 1.0202 mA, 2.461 mA, 3 mA,, 2.423 mA, , 3.91 0.61 mA, 8.641 V, 49 mV, , 3.55 1 A, 0 A, 2 A, 3.57 3.23 k, 28 V, 72 V, 3.59 1.344 kV, 5.6 A, 3.61 0.3, 3.63 4 V, 2.105 A, , Chapter 4, 4.1, , 0.1 A, 1 A, , 4.3, , (a) 0.5 V, 0.5 A, (b) 5 V, 5 A, (c) 5 V, 500 mA, , 4.5, , 4.5 V, , 4.7, , 888.9 mV, , 4.9, , 7V, , 3.65 2.17 A, 1.9912 A, 1.8119 A, 2.094 A, 2.249 A, 4.11 17.99 V, 1.799 A, 3.67 12 V, 4.13 8.696 V, 1.75 0.25 1, V1, 20, 3.69 £ 0.25, 1, 0.25 § £ V2 § £ 5 §, 1 0.25 1.25, 5, V3, , 4.15 1.875 A, 10.55 W, , 3.71 2.085 A, 653.3 mA, 1.2312 A, , 4.19 26.67 V, , 4.17 8.571 V
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ale29559_appG.qxd, , 07/17/2008, , 10:04 AM, , Page A-78, , A-78, , Appendix G, , Answers to Odd-Numbered Problems, , 4.21 This is a design problem with multiple answers., , 4.81 3.3 , 10 V (Note, values obtained graphically), , 4.23 2 A, 32 W, , 4.83 8 , 12 V, , 4.25 6.6 V, , 4.85 (a) 24 V, 30 k, (b) 9.6 V, , 4.27 48 V, , 4.87 (a) 10 mA, 8 k, (b) 9.926 mA, , 4.29 3 V, , 4.89 (a) 99.99 mA, (b) 99.99 mA, , 4.31 3.652 V, , 4.91 (a) 100 , 20 , (b) 100 , 200 , , 4.33 (a) 8 , 16 V, (b) 20 , 50 V, , 4.93, , 4.35 125 mV, 4.37 10 , 1 A, 4.39 20 , 16.4 V, 4.41 4 , 8 V, 2 A, , Vs, Rs (1 b)Ro, , 4.95 5.333 V, 66.67 k, 4.97 2.4 k, 4.8 V, , Chapter 5, , 4.43 10 , 0 V, , 5.1, , (a) 1.5 M, (b) 60 , (c) 98.06 dB, , 4.45 3 , 3 A, , 5.3, , 10 V, , 5.5, , 0.9999990, , 4.49 28 , 3.286 A, , 5.7, , 100 nV, 10 mV, , 4.51 (a) 2 , 7 A, (b) 1.5 , 12.667 A, , 5.9, , (a) 2 V, (b) 3 V, , 4.53 3 , 1 A, , 5.11 This is a design problem with multiple answers., , 4.55 100 k, 20 mA, , 5.13 2.7 V, 288 mA, , 4.47 476.2 m, 1.9841 V, 4.176 A, , 4.57 10 , 166.67 V, 16.667 A, 4.59 22.5 , 40 V, 1.7778 A, , 5.15 (a) aR1 R3 , , R1R3, b, (b) 92 k, R2, , 4.61 1.2 , 9.6 V, 8 A, , 5.17 (a) 1.2, (b) 8, (c) 200, , 4.63 3.333 , 0 A, , 5.19 0.375 mA, , 4.65 V0 (48 5I0) V, , 5.21 4 V, , 4.67 25 , 7.84 W, , 5.23 , , 4.69 (theoretically), , Rf, R1, , 5.25 1.25 V, , 4.71 8 k, 1.152 W, 5.27 1.8 V, 4.73 20.77 W, R2, R1, , 4.75 RL 10 , PL tends toward infinity., , 5.29, , 4.77 (a) 3.8 , 4 V, (b) 3.2 , 15 V, , 5.31 727.2 mA, , 4.79 10 , 167 V, , 5.33 6 mA, 108 mW
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ale29559_appG.qxd, , 07/17/2008, , 10:04 AM, , Page A-81, , Appendix G, , Answers to Odd-Numbered Problems, , 6.63 See Fig. G.7., , A-81, , vo , , v o (t) (V), , v, , 1, R1C, , 1, , dt , , 1, R2C, , v, , 2, , dt , , 1, R2C, , v, , 2, , dt, , For the given problem, C 2mF : R1 500 k,, R2 125 k, R3 50 k., , 6, 4, , 6.73 Consider the op amp as shown in Fig. G.10., , 2, 0, 2, , 3, , 4, , 5, , 6, , R, , t (s), , –2, R, , v, , –4, , a, , –6, , R, , Figure G.7, For Prob. 6.63., 6.65 (a) 40 J, 40 J, (b) 80 J, (c) 5 105(e200t 1) , 4 A, 1.25 105(e200t 1) 2 A, (d) 6.25 105(e200t 1) 2 A, , +, , R, , v, , vo, , b, , +, −, , vi, , −, +, , −, , Figure G.10, For Prob. 6.73., , 6.67 200 cos(50t) mV, Let va vb v. At node a,, , 6.69 See Fig. G.8., , v v0, 0v, , S 2v v0 0, R, R, , v (t) (V), 5, , At node b,, , 2.5, , v v0, vi v, dv, , C, R, R, dt, , vi 2v vo RC, , 0, 1, , 2, , 3, , 4, , 5, , 6, , 7, , t (s), , –5, , dv, dt, , (2), , Combining Eqs. (1) and (2),, vi vo vo , , –2.5, , (1), , RC dvo, 2 dt, , vo , , or, , 2, RC, , v dt, i, , showing that the circuit is a noninverting integrator., 6.75 30 mV, , –7.5, , Figure G.8, 6.77 See Fig. G.11., , For Prob. 6.69., 6.71 By combining a summer with an integrator, we have, the circuit shown in Fig. G.9., R1, , v i (t) (V), 8, 4, , C, , 4, , 0, R2, , 1, , −, +, , –4, , R3, , –8, , Figure G.9, , Figure G.11, , For Prob. 6.71., , For Prob. 6.77., , 2, , 3, , t (s)
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ale29559_appG.qxd, , 07/17/2008, , 10:04 AM, , Page A-82, , Appendix G, , A-82, , Answers to Odd-Numbered Problems, , 7.11 1.4118e3t A, , 6.79 See Fig. G.12., , 7.13 (a) 5 k, 5 H, 1 ms, (b) 25.28 mJ, 7.15 (a) 0.25 s, (b) 0.5 ms, 1V t=0, −, +, C, , 7.17 2e16tu (t) V, R, , R, dy/dt, , R/4, , −, +, , –y, , 7.19 2e5tu (t) A, , R, , −, +, , R, , 7.21 13.333 , , −, +, dy/dt, , R, f (t), , 7.23 2e4t V, t 7 0, 0.5e4t V, t 7 0, 7.25 This is a design problem with multiple answers., , Figure G.12, , 7.27 3 10 u (t 1) 20u (t) 50u (t 1) , 30u(t 2) 4 V, , For Prob. 6.79., , 7.29 (a) See Fig. G.14(a). (b) See Fig. G.14(b)., (c) z(t) 5 cos 4t d (t 1) 5 cos 4d (t 1) , 3.268d(t 1), which is sketched in, Fig. G.14(c)., , 6.81 See Fig. G.13., , x (t), C, R, , −, +, , d 2v/dt2, , C, R, –dv/dt, , −, +, , R, R/5, v, , 1.8395, , −, +, , R/2, , d 2v/dt2, , f (t), 0, , Figure G.13, , 1, , For Prob. 6.81., , t, (a), , y (t), 54.36, , 6.83 Eight groups in parallel with each group made up of, two capacitors in series, 6.85 1.25 mH inductor, , 0, , Chapter 7, 7.1, , (a) 0.7143 mF, (b) 5 ms, (c) 3.466 ms, , 7.3, , 3.222 ms, , 7.5, , This is a design problem with multiple answers., , 7.7, , vo(t) 36 2et20 4 V for all t 7 0., vo(t) 4e5t V for all t 7 0., , t, (b), , z (t), , 0, , 7.9, , 1, , 1, , t, , –3.268 (0), (c), , Figure G.14, For Prob. 7.29.
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ale29559_appG.qxd, , 07/17/2008, , 10:04 AM, , Page A-83, , Appendix G, , Answers to Odd-Numbered Problems, , 7.31 (a) 112 109, (b) 7, , or, , 7.33 2u (t 2) A, , A-83, , i VSR, ett, I0 VSR, i(t) , , 7.35 (a) e2tu (t) V, (b) 2e1.5tu (t) A, , VS, VS, aI0 b ett, R, R, , which is the same as Eq. (7.60)., , 7.37 (a) 4 s, (b) 10 V, (c) (10 8et4) u (t) V, , 7.53 (a) 5 A, 5et2u (t) A, (b) 6 A, 6e2t3u (t) A, , 7.39 (a) 4 V, t 6 0, 20 16et8, t 7 0,, (b) 4 V, t 6 0, 12 8et6 V, t 7 0., , 7.55 96 V, 96e4tu (t) V, , 7.41 This is a design problem with multiple answers., 7.57 4.8e2tu (t) A, 1.2e5tu (t) A, , 7.43 0.8 A, 0.8et480u (t) A, , 7.59 3e4tu (t) V, , 7.45 (4 3e14.286t ) u (t) V, 7.47 e, , 0 6 t 6 1, 48 (1 et) V,, (60 29.66e(t1) ) V,, t 7 1, , 7.49 e, , 8 (1 et5) V, 0 6 t 6 1, 1.45e(t1)5 V,, t 7 1, , 7.51 VS Ri L, or L, , 7.61 20e80tu (t) V, (5 (5 5e80t) u (t)) A, 7.63 16e8tu (t) V, (4 (4 4e8t )u (t)) A, 7.65 e, , di, dt, , 7.67 10e100t3u (t) A, , VS, di, R ai b, dt, R, , 7.69 48 (et3000 1) u (t) V, , di, R, , dt, i VSR, L, , 7.71 6 (1 e5t ) u (t) V, , Integrating both sides,, ln ai , ln a, , 0 6 t 6 1, 4 (1 e2t) A, 3.458e2(t1) A t 7 1, , 7.73 50u(t) mA, , VS i(t) R, b ` , t, R I0, L, , 7.75 (6 3e50t) u (t) V, 0.2 mA, , i VSR, t, b, t, I0 VSR, , 7.77 See Fig. G.15., , –12 V, , –16 V, , –20 V, , –24 V, 0s, , 1.0 s, V(R2:2, R4:2), , Figure G.15, For Prob. 7.77., , 2.0 s, , 3.0 s, Time, , 4.0 s, , 5.0 s
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ale29559_appG.qxd, , 07/17/2008, , 10:04 AM, , Page A-84, , Appendix G, , A-84, , Answers to Odd-Numbered Problems, , 7.79 (0.5 4.5e80t3) u (t) A, 7.81 See Fig. G.16., 4.0 A, , 3.0 A, , 2.0 A, , 1.0 A, , 0 A, 0 s, , 0.5 s, , 1.0 s, , 1.5 s, , 2.0 s, , 2.5 s, , 3.0 s, , I(L1), Time, , Figure G.16, For Prob. 7.81., , 7.83 6.278 m/s, 7.85 (a) 659.7 ms, (b) 16.636 s, , 8.7, , overdamped, , 8.9, , (2 10t)e5tu(t) A, , 8.11 20 (1 t)et V for t 7 0., 7.87 441 mA, , 8.13 120 , , 7.89 L 6 200 mH, , 8.15 750 , 200 mF , 25 H, , 7.91 1.271 , , 8.17 (64.65e2.679t 4.641e37.32t ) V, 8.19 18 sin(0.5t) V for t 7 0., , Chapter 8, , 8.21 18et 2e9t V, , 8.1, , (a) 2 A, 12 V, (b) 4 As, 5 Vs, (c) 0 A, 0 V, , 8.3, , (a) 0 A, 10 V, 0 V, (b) 0 A/s, 8 Vs, 8 Vs,, (c) 400 mA, 6 V, 16 V, , 8.25 This is a design problem with multiple answers., , (a) 0 A, 0 V, (b) 4 As, 0 Vs, (c) 2.4 A, 9.6 V, , 8.27 (6 6(cos(2t) sin(2t)e2t)u(t)) V, , 8.23 40 mF, , 8.5
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ale29559_appG.qxd, , 07/17/2008, , 10:04 AM, , Page A-86, , Appendix G, , A-86, , Answers to Odd-Numbered Problems, , 8.73 This is a design problem with multiple answers., , 9.13 (a) 1.2749 j0.1520 , (b) 2.083 , (c) 35 j14, , 8.75 See Fig. G.19., , 9.15 (a) 6 j11 , (b) 120.99 j4.415 , (c) 1, 9.17 15.62 cos(50t 9.8) V, , 0.1 Ω, , 9.19 (a) 3.32 cos(20t 114.49) ,, , 2F, 0.5 H, 24 A, 0.25 Ω, , 12 A, , (b) 64.78 cos(50t 70.89) ,, (c) 9.44 cos(400t 44.7), 9.21 (a) f (t) 8.324 cos(30t 34.86) ,, (b) g (t) 5.565 cos(t 62.49) ,, , Figure G.19, For Prob. 8.75., , (c) h (t) 1.2748 cos(40t 168.69), 9.23 (a) 43.49 cos(t 6.59) V,, , 8.77 See Fig. G.20., , (b) 18.028 cos(t 78.69) A, 1Ω, 1F, 4, 1H, , 9.25 (a) 0.8 cos(2t 98.13) A ,, , 1Ω, 2, −, +, , (b) 0.745 cos(5t 4.56) A, 1, Ω, 3, , 12 A, , 5V, , 9.27 0.289 cos(377t 92.45) V, 9.29 2 sin(106t 65), , Figure G.20, For Prob. 8.77., , 9.31 78.3 cos(2t 51.21) mA, , 8.79 434 mF, , 9.33 69.82 V, , 8.81 2.533 mH, 625 mF, , 9.35 4.789 cos(200t 16.7) A, , 8.83, , vs, R dv, R, 1 diD, d 2v, , , iD , , 2, L, dt, LC, C, dt, LC, dt, , 9.37 (500 j50) mS, 9.39 9.135 j 27.47 ,, 414.5 cos(10t 71.6) mA, , Chapter 9, 9.1, , (a) 50 V , (b) 209.4 ms , (c) 4.775 Hz ,, (d) 44.48 V, 0.3 rad, , 9.3, , (a) 4 cos(t 120), (b) 2 cos(6t 90),, (c) 10 cos(t 110), , 9.5, , 20, v1 lags v2, , 9.7, , Proof, , 9.9, , (a) 50.88l15.52, (b) 60.02l110.96, , 9.11 (a) 21l15 V, (b) 8l160 mA,, (c) 120l140 V , (d) 60l190 mA, , 9.41 15.812 cos(t 18.43) V, 9.43 499.7l28.85 mA, 9.45 5 A, 9.47 1.8428 cos(2,000t 52.63) A, 9.49 1.4142 sin(200t 45) V, 9.51 12.5 cos(2t 53.13) A, 9.53 8.873l21.67 A, 9.55 2.798 j16.403
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ale29559_appG.qxd, , 07/17/2008, , 10:04 AM, , Page A-87, , Appendix G, , Answers to Odd-Numbered Problems, , A-87, , 9.57 0.3171 j 0.1463 S, , 10.9, , 9.59 2.707 j 2.509, , 10.11 498.7l86.87 mA, , 9.61 1 j 0.5 , , 10.13 29.36l62.88 V, , 9.63 34.69 j6.93 , , 10.15 15.812l43.49 A, , 9.65 17.35l0.9 A, 6.83 j1.094 , , 10.17 13.875l162.12 A, , 9.67 (a) 14.8l20.22 mS , (b) 19.7l74.57 mS, , 10.19 7.682l50.19 V, , 9.69 1.661 j 0.6647 S, , 10.21 (a) 1, 0, , , 9.71 1.058 j 2.235 , 10.23, , 9.73 0.3796 j1.46 , 9.75 Can be achieved by the RL circuit shown in, Fig. G.21., 10 Ω, , 10 Ω, , +, Vi, , 6.154 cos(103t 70.26) V, , j10 Ω, , j10 Ω, , j L, j L, , (b) 0, 1,, RA C, RA C, , (1 2LC)Vs, 1 2LC jRC(2 2LC), , 10.25 2.828 cos(2t 45) A, 10.27 4.698l95.24 A, 0.9928l37.71 A, 10.29 This is a design problem with multiple answers., , +, Vo, −, , −, , Figure G.21, For Prob. 9.75., , 10.31 2.179l61.44 A, 10.33 15.92l43.49 A, 10.35 985.5l2.1 mA, 10.37 2.38l96.37 A, 2.38l143.63 A, 2.38l23.63 A, , 9.77 (a) 51.49 lagging , (b) 1.5915 MHz, 9.79 (a) 140.2 , (b) leading, (c) 18.43 V, , 10.39 0.6357l109.6 A, 0.5738l124.4 A,, 0.2425l60.42 A, 0.1675l48.5 A, , 9.81 1.8 k, 0.1 mF, , 10.41 2.122 cos (2t 45) 7.156 sin (4t 25.56) V, , 9.83 104.17 mH, , 10.43 9.902 cos(2t 129.17) A, , 9.85 Proof, , 10.45 3989.1 cos(10t 21.47), 499 sin(4t 176.57)4 mA, , 9.87 38.21l8.97 , 9.89 8.05 mH, , 10.47 [4 0.504 sin(t 19.1), 0.3352 cos(3t 76.43)] A, , 9.91 235 pF, , 10.49 8.944 sin (200t 56.56) A, , 9.93 1.7958l38.66 A, , 10.51 109.3l30 mA, , Chapter 10, 10.1, , 1.9704 cos(10t 5.65) A, , 10.3, , 7.67 cos(4t 35.02) V, , 10.5, , 12.398 cos(4 103t 4.06) mA, , 10.7, , 124.08l154 V, , 10.53 (3.529 j5.883) V, 10.55 (a) ZN ZTh 22.63l63.43 ,, VTh 50l30 V, IN 2.236l273.4 A ,, (b) ZN ZTh 10l26 ,, VTh 33.92l58 V, IN 3.392l32 A, 10.57 This is a design problem with multiple answers.
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ale29559_appG.qxd, , 07/17/2008, , 10:04 AM, , Page A-88, , Appendix G, , A-88, , Answers to Odd-Numbered Problems, , 10.59 6 j38 , , 11.15 0.5 j 0.5 , 9 kW, , 10.61 24 j12 V, 8 j6 , , 11.17 20 , 5 W, , 10.63 1 k, 5.657 cos(200t 75) A, , 11.19 2.576 , 3.798 W, , 10.65 This is a design problem with multiple answers., , 11.21 19.58 , , 10.67 4.945l69.76 V, 0.4378l75.24 A,, 11.243 j1.079 , , 11.23 This is a design problem with multiple answers., , 10.69 jRC, Vm cos t, , 11.25 8.165, , 10.71 96 cos (2t 44.52) V, , 11.27 2.887 A, , 10.73 21.21l45 k, , 11.29 5.773 A, 400 W, , 10.75 0.12499l180, , 11.31 2.944 V, , 10.77, , R2 R3 jC2R2R3, (1 jR1C1)(R3 jC2R2R3), , 10.79 3.578 cos(1,000t 26.56) V, 10.81 11.27l128.1 V, 10.83 6.611 cos (1,000t 159.2) V, , 11.33 6.665, 11.35 21.6 V, 11.37 This is a design problem with multiple answers., 11.39 (a) 0.7592, 6.643 kW, 5.695 kVAR ,, (b) 312 mF, , 10.85 This is a design problem with multiple answers., 11.41 (a) 0.5547 (leading), (b) 0.9304 (lagging), 10.87 15.91l169.6 V, 5.172l138.6 V, 2.27l152.4 V, 10.89 Proof, 10.91 (a) 180 kHz ,, (b) 40 k, 10.93 Proof, 10.95 Proof, , Chapter 11, (Assume all values of currents and voltages are rms unless, otherwise specified.), 11.1, , 800 1,600 cos(100t 60) W, 800 W, , 11.3, , 13.333 W, , 11.5, , P1 11.33 W, P2 40.79 W,, P3H P0.25F 0, , 11.7, , 160 W, , 11.9, , 1.794 mW, , 11.43 This is a design problem with multiple answers., 11.45 (a) 46.9 V, 1.061 A, (b) 20 W, 11.47 (a) S 112 j194 VA,, average power 112 W,, reactive power 194 VAR, (b) S 226.3 j 226.3 VA,, average power 226.3 W,, reactive power 226.3 VAR, (c) S 110.85 j 64 VA, average power , 110.85 W, reactive power 64 VAR, (d) S 7.071 j 7.071 kVA, average power , 7.071 kW, reactive power 7.071 kVAR, 11.49 (a) 4 j 2.373 kVA,, (b) 1.6 j1.2 kVA ,, (c) 0.4624 j1.2705 kVA ,, (d) 110.77 j166.16 VA, , 11.11 12.751 mW, , 11.51 (a) 0.9956 (lagging),, (b) 1.751 kW,, (c) 164.9 VAR,, (d) 1.7587 kVA,, (e) (1,751 j164.9) VA, , 11.13 (a) 120 j60 , (b) 12.605 W, , 11.53 (a) 93.97l29.8 A , (b) 1.0 (lagging)
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ale29559_appG.qxd, , 07/17/2008, , 10:04 AM, , Page A-89, , Appendix G, , Answers to Odd-Numbered Problems, , 11.55 This is a design problem with multiple, answers., 11.57 (50.45 j33.64) VA, , Chapter 12, (Assume all values of currents and voltages are rms unless, otherwise specified.), 12.1, , 11.59 j339.3 VAR, j1.4146 kVAR, , (a) 231l30, 231l150, 231l90 V ,, (b) 231l30, 231l150, 231l90 V, , 11.61 33.1l92.4 A, 6.62l2.4 kVA, , 12.3, , abc sequence, 208l250 V, , 11.63 443.3l28.13 A, , 12.5, , 260 cos (t 62) V, 260 cos (t 58) V,, 260 cos (t 182) V, , 12.7, , 44l53.13 A, 44l66.87 A, 44l173.13 A, , 12.9, , 4.8l36.87 A, 4.8l156.87 A, 4.8l83.13 A, , 11.65 80 mW, 11.67 36l36.86 mVA, 12.042 mW, 11.69 (a) 0.6402 (lagging),, (b) 295.1 W,, (c) 130.4 mF, , 12.11 207.8 V, 99.85 A, , 11.71 (a) 50.14 j1.7509 m ,, (b) 0.9994 lagging,, (c) 2.392l2 kA, , 12.15 13.66 A, , 11.73 (a) 12.21 kVA, (b) 50.86l35 A ,, (c) 4.083 kVAR, 188.03 mF , (d) 43.4l16.26 A, 11.75 (a) 1,835.9 j114.68 VA , (b) 0.998 (leading),, (c) no correction is necessary, , 12.13 40.85 A,15.02 kW, , 12.17 5.773l5 A, 5.773 l115 A,, 5.773 l125 A, 12.19 5.47l18.43 A, 5.47l138.43 A, 5.47l101.57 A,, 9.474l48.43 A, 9.474l168.43 A,, 9.474l71.57 A, 12.21 34.36l98.66 A, 59.51l171.34 A, , 11.77 157.69 W, 11.79 50 mW, 11.81 This is a design problem with multiple, answers., 11.83 (a) 688.1 W, (b) 840 VA,, (c) 481.8 VAR, (d) 0.8191 (lagging), 11.85 (a) 20 A, 17.85l163.26 A, 5.907l119.5 A ,, (b) 4,451 j 617 VA, (c) 0.9904 (lagging), , 12.23 (a) 13.995 A,, (b) 2.448 kW, 12.25 8.87l4.78, 8.87l115.22, 8.87l124.78 A, 12.27 91.79 V, 12.29 1.3 j1.1465 kVA, 12.31 (a) 6.144 j 4.608 ,, (b) 18.04 A, (c) 207.2 mF, 12.33 15.385 A, 360.3 V, , 11.87 0.5333, 11.89 (a) 12 kVA, 9.36 j7.51 kVA ,, (b) 2.866 j2.3 , 11.91 0.9775, 104 mF, 11.93 (a) 7.328 kW, 1.196 kVAR, (b) 0.987, , A-89, , 12.35 (a) 14.61 j5.953 A ,, (b) 3.361 j1.368 kVA ,, (c) 0.9261, 12.37 55.51 A, 1.298 j 1.731 , 12.39 431.1 W, 12.41 9.021 A, , 11.95 (a) 2.814 kHz,, (b) 431.8 mW, , 12.43 4.373 j1.145 kVA, , 11.97 547.3 W, , 12.45 2.109l24.83 kV
Page 1024 :
ale29559_appG.qxd, , 07/17/2008, , 10:04 AM, , Page A-90, , A-90, , Appendix G, , Answers to Odd-Numbered Problems, , 12.47 39.19 A (rms), 0.9982 (lagging), , 12.85 ZY 2.133 , , 12.49 (a) 5.808 kW, (b) 1.9356 kW, , 12.87 1.448l176.6 A, 1,252 j 711.6 VA,, 1,085 j 721.2 VA, , 12.51 (a) 19.2 j14.4 A, 42.76 j 27.09 A,, 12 j 20.78 A ,, (b) 31.2 j 6.38 A, 61.96 j 41.48 A,, 30.76 j 47.86 A, 12.53 This is a design problem with multiple, answers., , Chapter 13, (Assume all values of currents and voltages are rms unless, otherwise specified.), 13.1, , 10 H, , 13.3, , 150 mH, 50 mH, 25 mH, 0.2887, , 13.5, , (a) 123.7 mH, (b) 24.31 mH, , 12.57 Ia 1.9585l18.1 A, Ib 1.4656l130.55 A,, Ic 1.947l117.8 A, , 13.7, , 540.5l144.16 mV, , 13.9, , 4.148l21.12 V, , 12.59 220.6l34.56, 214.1l81.49, 49.91l50.59 V,, assuming that N is grounded ., , 13.11 412.3 cos (600t 140.43) mA, , 12.61 11.15l37 A, 230.8l133.4 V, assuming N is, grounded ., , 13.15 1 j19.5 , 1.404l9.44 A, , 12.63 18.67l158.9 A, 12.38l144.1 A, , 13.19 See Fig. G.22., , 12.55 9.6l90 A, 6l120 A, 8l150 A,, 3.103 j 3.264 kVA, , 13.13 4.308 j6.538 , , 13.17 13.073 j 25.86 , , 12.65 11.02l12 A, 11.02l108 A, 11.02l132 A, j65 Ω, , 12.67 (a) 97.67 kW, 88.67 kW, 82.67 kW,, (b) 108.97 A, 12.69 Ia 94.32l62.05 A, Ib 94.32l177.95 A,, Ic 94.32l57.95 A, 28.8 j 18.03 kVA, , j55 Ω, , –j25 Ω, , Figure G.22, For Prob. 13.19., , 12.71 (a) 2,590 W, 4,808 W,, (b) 8,335 VA, 12.73 2,360 W, 632.8 W, , 13.21 This is a design problem with multiple, answers., , 12.75 (a) 20 mA,, (b) 200 mA, , 13.23 50.68 cos (10t 52.54) A,, 27.19 cos(10t 100.89) A, 1.5 kJ, , 12.77 320 W, , 13.25 2.2 sin (2t 4.88) A, 1.5085l17.9 , 13.27 1.567 W, , 12.79 17.15l19.65, 17.15l139.65, 17.15l100.35 A,, 223l2.97, 223l117.03, 223l122.97 V, , 13.29 0.984, 130.5 mJ, , 12.81 516 V, , 13.31 This is a design problem with multiple, answers., , 12.83 183.42 A, , 13.33 12.769 j 7.154
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ale29559_appG.qxd, , 07/17/2008, , 10:04 AM, , Page A-91, , Appendix G, , Answers to Odd-Numbered Problems, , 13.35 10.143l21.4 A, 532.8l134.85 mA,, 529.4l110.41 mA, , A-91, , 13.83 21.6l33.91 A, 302.8l34.21 V, 13.85 100 turns, , 13.37 (a) 5, (b) 104.17 A, (c) 20.83 A, 13.87 0.5, 13.39 15.7l20.31 A, 78.5l20.31 A, 13.41 7.857 A, 23.57 A, , 13.89 0.5, 41.67 A, 83.33 A, 13.91 (a) 1,875 kVA, (b) 7,812 A, , 13.43 4.186 V, 16.744 V, 13.93 (a) See Fig. G.23(a). (b) See Fig. G.23(b)., , 13.45 58.72 W, 13.47 118.03 cos (3t 59.93) V, 13.49 0.937 cos (2t 51.34) A, , 110 V, , 14V, , 13.51 8 j1.5 , 29.49l10.62 A, 13.53 (a) 5, (b) 8 W, 13.55 1.6669 , 13.57, , (a), , (a) 25.9l69.96, 12.95l69.96 A (rms),, (b) 21.06l147.4, 42.12l147.4,, 42.12l147.4 V(rms), (c) 1554l20.04 VA, , 220 V, 50 V, , 13.59 P10 395 W, P12 266.6 W,, P20 49.39 W, 13.61 6 A, 0.36 A, 60 V, , (b), , Figure G.23, 13.63 3.795l18.43 A, 1.8975l18.43 A, 0.6325l161.6 A, , For Prob. 13.93., , 13.65 11.05 W, 13.95 (a) 160, (b) 139 mA, 13.67 (a) 160 V, (b) 31.25 A, (c) 12.5 A, 13.69 (1.2 j 2) k, 5.333 W, , Chapter 14, , 13.71 [1 (N1N2)] ZL, 2, , 13.73 (a) three-phase ¢ -Y transformer,, (b) 8.66l156.87 A, 5l83.13 A,, (c) 1.8 kW, , 14.1, , jo, 1, , o , 1 jo, RC, , 14.3, , 5, s2 8s 5, , 14.5, , (a), , sRL, (R Rs)Ls RRs, , (b), , R, LRCs2 Ls R, , 13.75 (a) 0.11547, (b) 76.98 A, 15.395 A, 13.77 (a) a single-phase transformer, 1 : n, n 1110,, (b) 7.576 mA, 13.79 7.836l68.01 A, 2.441l77.86 A,, 8.016l54.92 A, 13.81 104.5l13.96 mA, 29.54l143.8 mA,, 208.8l24.4 mA, , 14.7, , (a) 1.005773, (b) 0.4898, (c) 1.718 105, , 14.9, , See Fig. G.24.
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ale29559_appG.qxd, , 07/17/2008, , 10:04 AM, , Page A-92, , Appendix G, , A-92, , Answers to Odd-Numbered Problems, , 14.13 See Fig. G.26., , |H|, , |G|, 1, , 0.1, , 100 (rad/s), , 10, , 20, −20, 0.1, , 1, , 10, , 100 (rad/s), , −40, −20, arg H, , −40, , 1, , 0.1, , 100 (rad/s), , 10, , (a), arg G, , −90°, 0.1, , 1, , 10, , 100 (rad/s), , 10, , 100 (rad ⁄s), , 10 20, , 100 (rad ⁄s), , −180°, −90°, , Figure G.24, For Prob. 14.9., , −180°, (b), , 14.11 See Fig. G.25., , Figure G.26, For Prob. 14.13., , HdB, 40, 34, 20, 14, 0.1, , 14.15 See Fig. G.27., , 1, , 10, , , , 100, , HdB, , –20, 6.021, – 40, 0.1, , (a), , 12, , , , (a), , 90°, , , , 45°, 0.1, , 1, , 10, , –45°, , 100, , , 0.2, 0.1, , –90°, , 1 2, , −90°, (b), , (b), , Figure G.25, , Figure G.27, , For Prob. 14.11., , For Prob. 14.15: (a) magnitude plot, (b) phase plot.
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ale29559_appG.qxd, , 07/17/2008, , 10:05 AM, , Page A-93, , Appendix G, , Answers to Odd-Numbered Problems, , 14.17 See Fig. G.28., , 14.21 See Fig. G.30., 104 j, , 14.23, GdB, , 20, 0.1, , A-93, , 1, , 10, , (10 j)(100 j)2, , 14.25 2 k, 2 j0.75 k, 2 j 0.3 k, 2 j 0.3 k,, 2 j 0.75 k, , 100, , , –12, –20, , 14.27 R 1 , L 0.1 H, C 25 mF, 14.29 4.082 krad/s, 38.67, 105.55 rad/s, , –40, , 14.31 8.796 106 rad/s,, , (a), , , 14.33 14.21 mH, 56.84 pF, , 90°, 0.1, , 10, , 14.35 40 , 2.5 mH, 10 mF, 2.5 krad/s, 198.75 krad/s,, 202.25 krad/s, , , , 100, , –90°, –180°, , 1, , 14.37, , 2LC R2C2, , (b), , Figure G.28, , 14.39 (a) 19.89 nF (b) 164.45 mH, (c) 552.9 krad/s,, (d) 25.13 krad/s, (e) 22, , For Prob. 14.17., , 14.41 This is a design problem with multiple answers., 14.43 (a) 2.357 krad/s, (b) 1 , , 14.19 See Fig. G.29., , H, 20 log j, , 0.1, , 1, , 10, , 20, , 40, , 100, , 20 log 1 , , 20 log 1, 80, , 20, 40, , 60, (a), 90, , 0.1, , 1, , 2, , 4, , 10, , (b), , Figure G.29, For Prob. 14.19., , 20, , 40, , 100, , 200, , 400, , j, 10
Page 1028 :
ale29559_appG.qxd, , 07/17/2008, , 10:05 AM, , Page A-94, , Appendix G, , A-94, , Answers to Odd-Numbered Problems, , H&, 40, 20 log i, 20 log 1 j/20, , 20, , 0.1, , 1, , 10, , 20, , , , 100, , 20 log 0.05, , – 20, , –20 log 1j, , – 40, , –60, –20 log 1, , ( ), , j, j, , 40, 20, , 2, , – 80, , Figure G.30, For Prob. 14.21., , 14.45 (a), , j, 2(1 j), , 2, , 14.77 (a) 1,200 H, 0.5208 mF, (b) 2 mH, 312.5 nF,, (c) 8 mH, 7.81 pF, , , (b) 0.25, , 14.47 796 kHz, , 14.79 (a) 8s 5 , , 14.49 This is a design problem with multiple answers., , 10, ,, s, , (b) 0.8s 50 , , 104, , 111.8 rad/s, s, , 14.51 1.256 k, 14.53 18.045 k. 2.872 H, 10.5, , 14.81 (a) 0.4 , 0.4 H, 1 mF, 1 mS,, (b) 0.4 , 0.4 mH, 1 mF, 1 mS, , 14.55 1.56 kHz 6 f 6 1.62 kHz, 25, , 14.83 0.1 pF, 0.5 pF, 1 M, 2 M, , 14.57 (a) 1 rad/s, 3 rad/s, (b) 1 rad/s, 3 rad/s, , 14.85 See Fig. G.31., , 14.59 2.408 krad/s, 15.811 krad/s, , 14.87 See Fig. G.32; highpass filter, f0 1.2 Hz., , 1, ,, 1 jRC, jRC, (b), 1 jRC, , 14.89 See Fig. G.33., , 14.61 (a), , 14.91 See Fig. G.34; fo 800 Hz., 14.93, , 14.63 10 M, 100 k, , RCs 1, RCs 1, , 14.95 (a) 0.541 MHz 6 fo 6 1.624 MHz,, (b) 67.98, 204.1, , 14.65 Proof, 14.67 If Rf 20 k, then Ri 80 k and, C 15.915 nF., 14.69 Let R 10 k, then Rf 25 k, C 7.96 nF., 4, , 3, , 14.71 Kf 2 10 , Km 5 10, , 14.97, , s3LRLC1C2, (sRiC1 1)(s2LC2 sRLC2 1) s2LC1(sRLC2 1), , 14.99 8.165 MHz, 4.188 106 rad/s, 14.101 1.061 k, , 14.73 9.6 M, 32 mH, 0.375 pF, 14.75 200 , 400 mH, 1 mF, , 14.103, , R2(1 sCR1), R1 R2 sCR1R2
Page 1029 :
ale29559_appG.qxd, , 07/17/2008, , 10:05 AM, , Page A-95, , Appendix G, , Answers to Odd-Numbered Problems, , A-95, , 15 V, , 10 V, , 5V, , 0V, 100 Hz, , 300 Hz, , 1.0 K Hz, , 3.0 K Hz, , 10 K Hz, , 3.0 K Hz, , 10 K Hz, , Frequency, , VP (R2:2), , (a), 0d, , –50 d, , –100 d, 100 Hz, , 300 Hz, , VP (R2:2), , 1.0 K Hz, Frequency, (b), , Figure G.31, For Prob. 14.85., , 1.0 V, , 0.5 V, , 0 V, 100 mHz 300 mHz 1.0 Hz 3.0 Hz 10 Hz, VP(R3:1), Frequency, , Figure G.32, For Prob. 14.87., , 30 Hz, , 100 Hz
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ale29559_appG.qxd, , 07/17/2008, , 10:05 AM, , Page A-96, , Appendix G, , A-96, , Answers to Odd-Numbered Problems, , 10 V, , 0 V, 100 Hz, , 200 Hz, , 300 Hz, , 400 Hz 500 Hz 600 Hz, , 800 Hz, , V(L1:1), Frequency, , Figure G.33, For Prob. 14.89., , 1.0 KV, , 0.5 KV, , 0 V, 10 Hz, V(C1:1), , 100 Hz, , 1.0 KHz, , 10 KHz, , Frequency, , Figure G.34, For Prob. 14.91., , Chapter 15, 15.1, , 15.3, , 15.5, , 4, 3, 4, 2, , (b) , ,, 2, s, s, s, , 2, s, s2, 8s 18, (c) 2, , (d) 2, s 9, s 4s 12, , 15.7, , (a), , s2, 4, , (b), ,, (s 2)2 9, (s 2)2 16, 1, s3, (c), (d), ,, (s 3)2 4, (s 4)2 1, 4(s 1), (e), [(s 1)2 4]2, , 15.9, , (a), , 8 12 23s 6s2 23s3, (a), ,, (s2 4)3, 72, 2, (b), , (c) 2 4s,, 5, s, (s 2), 2e, 5, 18, (d), , (e) , (f), , (g) sn, s, s1, 3s 1, , 15.11 (a), , s, ,, s2 a2, a, (b) 2, s a2, (a), , (a), , 2es, 2e2s, e2s, 2 , (b) 4, ,, 2, s, s, e (s 4), 2.702s, 8.415, 2, ,, (c) 2, s 4, s 4, (d), , (b), (c), , 6 2s 6 4s, e, e, s, s, 6(s 1), , ,, s 2s 3, 24(s 2), 2, , (s2 4s 12)2, , ,, , e(2s6)[(4e2 4e2)s (16e2 8e2)], s2 6s 8
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ale29559_appG.qxd, , 07/17/2008, , 10:05 AM, , Page A-97, , Appendix G, , 15.13 (a), (b), , s2 1, ,, (s2 1)2, 2(s 1), , (s2 2s 2)2, b, (c) tan1a b, s, , 15.15 5, , Answers to Odd-Numbered Problems, , (c) e2(t4)u (t 4) ,, 10, 10, (d) a cos t cos 2tb u (t), 3, 3, , ,, , 15.39 (a) (1.6et cos 4t 4.05et sin 4t, 3.6e2t cos 4t (3.45e2t sin 4t) u (t),, (b) [0.08333 cos 3t 0.02778 sin 3t, 0.0944e0.551t 0.1778e5.449t] u (t), , 1 es ses, s2(1 e3s), , 15.17 This is a design problem with multiple, answers., , 15.19, , 15.21, , 1, 1 es, (2ps 1 e2ps), 2ps, , 2ps (1 e, 2, , 15.23 (a), (b), , ), , (1 es)2, , ,, s(1 e2s), 2(1 e2s) 4se2s(s s2), s3(1 e2s), , 15.25 (a) 5 and 0, (b) 5 and 0, 15.27 (a) u (t) 2etu (t), (b) 3d(t) 11e4tu(t),, (c) (2et 2e3t ) u (t),, (d) (3e4t 3e2t 6te2t) u (t), 2t, , 15.29 (1 e, , A-97, , 2t, , cos(3t) (13) e, , 8t,, 16 8t,, 16,, 15.41 z(t) f, 8t 80,, 112 8t,, 0,, , 0 6 t 6 2, 2 6 t 6 6, 6 6 t 6 8, 8 6 t 6 12, 12 6 t 6 14, otherwise, , 1 2, t ,, 0 6 t 6 1, 2, 1 2, 15.43 (a) y(t) f t 2t 1, 1 6 t 6 2, 2, 1,, t 7 2, 0,, otherwise, (b) y (t) 2(1 et), t 7 0,, 1, 1 2, t t ,, 2, 2, 1, 1, t2 t ,, 2, (c) y(t) g 2, 1 2, 9, t 3t ,, 2, 2, 0,, , 1 6 t 6 0, 0 6 t 6 2, 2 6 t 6 3, otherwise, , sin(3t)) u (t), , 15.31 (a) (5et 20e2t 15e3t) u (t) ,, t2, (b) aet a1 3t b e2t b u (t) ,, 2, (c) (0.2e2t 0.2et cos (2t), 0.4et sin (2t)) u (t), 15.33 (a) (3et 3 cos (t) 3 sin (t)) u (t) ,, (b) sin (t p) u (t p),, (c) 3u (t)[1 et tet 0.5t 2et], 15.35 (a) [2e(t6) e2(t6)] u (t 6),, 1, 4, (b) u (t)[et e4t] u (t 2)[e(t2) e4(t2)],, 3, 3, 1, (c), u (t 1)[3e3(t1) 3 cos 2(t 1), 13, 2 sin 2(t 1)], , 15.45 (4e2t 8te2t) u (t), 15.47 (a) (et 2e2t) u (t), (b) (et e2t) u (t), 1, eat, t, 15.49 (a) a (eat 1) 2 2 (at 1)b u (t) ,, a, a, a, (b) [0.5 cos (t)(t 0.5 sin (2t)), 0.5 sin(t) (cos(t) 1)] u (t), 15.51 (3et 4e2t 5e3t) u(t)V, 15.53 cos (t) sin (t) or 1.4142 cos (t 45), 15.55 a, , 1, 3 4t, 3, 1, e, et cos (2t), e2t , 40, 20, 104, 65, 2, et sin (2t)b u (t), 65, , 15.57 This is a design problem with multiple answers., 15.37 (a) (2 e2t) u (t) ,, (b) [0.4e3t 0.6et cos t 0.8et sin t] u (t) ,, , 15.59 [2.5et 12e2t 10.5e3t] u (t)
Page 1033 :
ale29559_appG.qxd, , 07/17/2008, , 10:05 AM, , Page A-99, , Appendix G, , Answers to Odd-Numbered Problems, , A-99, , 16.61 See Fig. G.35., , Bode Diagrams, , Phase (deg); magnitude (dB), , –20, –40, –60, –80, , –50, –100, –150, 10–1, , 100, 101, Frequency (rad/s), , 102, , Figure G.35, For Prob. 16.61., , 16.63 See Fig. G.36., , Step Response, 07, 06, , Amplitude, , 05, 04, 03, 02, 01, 0, 0, , 1, , 2, , 3, Time (sec), , 4, , 5, , 6, , Figure G.36, For Prob. 16.63., , 16.65 See Fig. G.37., 16.67 a 100, b 400, c 20,000, 16.69 Proof, , Chapter 17, 17.1, , (a) periodic, 2, (b) not periodic,, (c) periodic, 2 p, (d) periodic, p,, (e) periodic, 10, (f) not periodic,, (g) not periodic
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ale29559_appG.qxd, , 07/17/2008, , 10:05 AM, , Page A-100, , Appendix G, , A-100, , Answers to Odd-Numbered Problems, , 0.18, 0.16, 0.14, 0.12, 0.1, 0.08, 0.06, 0.04, 0.02, 0, –0.02, 0, , 0.5, , 1, , 1.5, , 2, , 2.5, , 3, , 3.5, , 4, , 4.5, , 5, , Figure G.37, For Prob. 16.65., , 17.3, , a0 7.5, 10, n odd,, (1)(n1)2,, np, n even, 10, np, bn , c 3 2 cos n p cos, d, np, 2, an e, , 17.5, , , 16, 1, 17.15 (a) 10 a, 6, 2, 2, B, (n 1), n, n1, 2, 1 n 1, cos a10 nt tan, b,, 4 p3, , 16, 1, (b) 10 a, 6, 2, B, (n, , 1), n, n1, 4n3, b, sin a10 n t tan1 2, n 1, , , 6, sin n t, 0.5 a, n1 n p, nodd, , 17.7, , , 9, 4n p, 2n p t, 3 a c, sin, cos, 3, 3, n0 n p, , , , 9, 4np, 2n p t, a1 cos, b sin, d, np, 3, 3, , 17.17 (a) neither odd nor even, (b) even, (c) odd, (d) even,, (e) neither odd nor even, , 10, 5, sin n p2 , (cos p n cos n p2), 2, n, n o, o, cos p n2, 5, 2, 2 2 (sin p n sin n p2) , cos n p , n, , n o, n o, o, , 17.19, 17.9, , 17.11, , a0 3.183, a1 10, a2 6.362, a3 0,, b1 0 b2 b3, , 1, a n2p2 [1 j( jn p2 1) sin n p2, n, n p sin n p2]e jn p t2, , 17.13 This is a design problem with multiple, answers., , 17.21, , 2, , , 1, 8, np, npt, a 2 2 c 1 cos a, b d cos a, b, 2, 2, 2, n1 n p, , 17.23 This is a design problem with multiple, answers.
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ale29559_appG.qxd, , 07/17/2008, , 10:05 AM, , Page A-101, , Appendix G, , Answers to Odd-Numbered Problems, , 17.25, 2pn, 2, 2pn, 2pn, sin a, c 2 2 acos a, b 1b , bd cos a, b, , pn, 3, 3, 3, pn, d, t, a, 3, 2pn, 2, 2pn, 2pn, n1, nodd c, cos a, sin a, b, bd sin a, b, np, 3, 3, 3, p2n2, , In , , 3, , 17.41, , 17.27 (a) odd, (b) 0.045, (c) 0.3829, 2, 1, 17.29 2 a c 2 cos (n t) sin (n t) d , n 2k 1, n, k1 n p, , , , , 10, a An cos (2nt un),, p, n1, 100, ,, An , 2, p(4n 1)216n2 40n 29, , 17.43 (a) 33.91 V,, (b) 6.782 A,, (c) 203.1 W, , 2p, 2p, , ao, T¿, Ta, 2, T¿, , a¿n , , 1, n2(804np)2 (2n2p2 1,200), , un 90 tan1 (2n 2.5), , , , 17.31 ¿o , , A-101, , T¿, , f (at) cos n¿o t dt, , 17.45 4.263 A, 181.7 W, , 0, , Let at l, dt dla, and aT¿ T. Then, 2a, T, , a¿n , , T, , f (l) cos n l dla a, o, , 17.47 10%, , n, , 0, , Similarly, b¿n bn, , 17.49 (a) 1.5326,, (b) 1.7086,, (c) 3.061%, , , , 17.33 vo(t) a An sin (n pt un) V,, n1, , An , , 8(4 2n2p2), 2(20 10n2p2)2 64n2p2, , un 90 tan1 a, , 17.35, , ,, , 17.51 This is a design problem with multiple, answers., , 8n p, b, 20 10n2p2, , , , 17.53, , 0.6321e j2npt, n 1 j2np, , 17.55, , , 1 ejnp jnt, a 2p(1 n2) e, n, , , 3, 2p n, a An cos a, un b, where, 8 nodd, 3, , An , un , , , , 17.37 a, n1, , 2n p, 6, sin, np, 3, 29p2n2 (2p2n23 3)2, , ,, , p, 2n p, 1, b, tan1 a, , np, 2, 9, , 8(1 cos p n), 21 n2p2, , cos (n p t tan1 n p) V, , a, , 17.57 3 , , , , 17.59 a, , n, n 0, , 17.39, , 1, 200 , , a In sin (npt un), n 2k 1,, p k1, 20, p2 1,200, ,, 802np, , 2, , un 90 tan, , 1 2n, , , , a, , n , n, , 0, , 3, e j50nt, n3 2, , j4ej(2n1)pt, (2n1)p, , 17.61 (a) 6 2.571 cos t 3.83 sin t 1.638 cos 2t, 1.147 sin 2t 0.906 cos 3t 0.423 sin 3t, 0.47 cos 4t 0.171 sin 4t, (b) 6.828
Page 1036 :
ale29559_appG.qxd, , 07/17/2008, , 10:05 AM, , Page A-102, , Appendix G, , A-102, , Answers to Odd-Numbered Problems, , 17.63 See Fig. G.38., , 1.333, An, , 0.551, , 0.275, 0.1378 0.1103, 0, 0, , 1, , 2, , 3, , 4, , n, , 5, , Figure G.38, For Prob. 17.63., , 17.65 See Fig. G.39., , 2.24, , An, 0.39, , 0.208, , 0.143, , 0.109, 0, , 0, , 2, , 6, , 10, , 14, , 18, , 2, , 6, , 10, , −25.23°, , −54.73°, , −60°, , −67°, , −90°, , For Prob. 17.65., , 18, n, , −30°, , Figure G.39, , 14, , n, , n, , −73.14°, , −76.74°
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ale29559_appG.qxd, , 07/17/2008, , 10:05 AM, , Page A-104, , Appendix G, , A-104, , 17.73 300 mW, , Answers to Odd-Numbered Problems, , 4, 2, sin 2 sin ,, , , 2ej, 2, (b) 2 , (1 j), , 2, , 18.9, , (a), , 17.77 (a) p, (b) 2 V, (c) 11.02 V, , 18.11, , 5p, (ej2 1), 2 p2, , 17.79 See below for the program in MATLAB and the, results., % for problem 17.79, a 10;, c 4.*api, for n 1:10, b(n)=c/(2*n-1);, end, diary, n, b, diary off, , 18.13 (a) pejp3d( a) pe jp3d( a),, e j, (b) 2, , (c) p[d( b) d( b)], 1, jpA, [d( a b) d( a b), , 2, d( a b) d( a b)],, 1, ej4, ej4, (d) 2 j 2 ( j4 1), , , 17.75 24.59 mF, , n, , 18.15 (a) 2j sin 3, (b), , bn, , j, 1, 2ej, , (c) , j, 3, 2, , j, p, [d( 2) d( 2)] 2, ,, 2, 4, jp, 10, [d( 10) d( 10)] 2, (b), 2, 100, , 18.17 (a), 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, , 12.7307, 4.2430, 2.5461, 1.8187, 1.414, 1.1573, 0.9793, 0.8487, 0.7488, 0.6700, , 18.19, , j, 4p2, 2, , (ej 1), , 18.21 Proof, 30, ,, (6 j)(15 j), j2, 20e, ,, (b), (4 j)(10 j), 5, , (c), [2 j( 2)][5 j( 2)], 5, ,, [2 j( 2)][5 j( 2)], j10, ,, (d), (2 j)(5 j), 10, pd(), (e), j(2 j)(5 j), , 18.23 (a), A2, , (b) 0 c1 0 2A/(3p), 0c2 0 2A(15p),, 2, 0 c3 0 2A/(35p), 0c4 0 2A/(63p) (c) 81.1%, (d) 0.72%, , 17.81 (a), , Chapter 18, 18.1, , 18.3, , 2(cos 2 cos ), j, j, , , 2, , (2 cos 2 sin 2), , 18.25 (a) 10e2tu(t),, (b) 1.5e2|t|,, (c) 5etu(t) 5e2tu(t), , 18.5, , 18.7, , 2j, 2j, 2 sin , , , , (a), , 5ej2, 5, 2 ej ej2, , (b), (1 j2) 2, 2, j, , , , 18.27 (a) 5 sgn (t) 10e10tu(t),, (b) 4e2tu(t) 6e3tu(t),, (c) 2e20t sin (30t) u (t), (d), , 1, p, 4
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ale29559_appG.qxd, , 07/17/2008, , 10:05 AM, , Page A-105, , Appendix G, , Answers to Odd-Numbered Problems, , 1, 4 sin 2t, (1 8 cos 3t), (b), ,, 2p, pt, (c) 3d(t 2) 3d(t 2), , 18.29 (a), , 18.31 (a) x(t) eatu(t),, (b) x(t) u(t 1) u(t 1),, 1, a, (c) x(t) d(t) eat u(t), 2, 2, 18.33 (a), , 2 j sin t, t2 p2, , , (b) u(t 1) u(t 2), , ej3, 1, 1, 1, , (b) c, , d,, 6 j, 2 2 j( 5), 2 j( 5), j, 1, 1, (c), , (e), , (d), 2 j, (2 j)2, (2 j)2, , 18.35 (a), , 18.37, , 18.39, , 18.41, , 18.65 6.8 kHz, 18.67 200 Hz, 5 ms, 18.69 35.24%, , Chapter 19, 19.1, , c, , 4, 1, , 19.3, , c, , 4 j6, j6, , 1, 1, 1, 2x, 2 2 ej b, a, , , 106 j j, , 18.43 1000(e1t e1.25t ) u (t) V, , j6, d, j4, , s3 2s2 3s 1, ≥, 2, s3 2s2 3s 1, , 19.7, , c, , 29.88, 70.37, , 19.9, , c, , 5, 2.5, , 2j(4.5 j 2), (2 j)(4 22 j), , 1, d, 1.667, , 2(s2 s 1), 19.5, , j, 4 j3, , A-105, , 2, s3 2s2 3s 1, ¥ , 2(s2 2s 2), s3 2s2 3s 1, , 3.704, d, 11.11, , 2.5, d, 6.25, , 19.11 See Fig. G.41., , 18.45 5(et e2t ) u (t) A, j5 Ω, , 1Ω, t, , 18.47 16(e, , 2t, , e, , 3Ω, , j1 Ω, , ) u (t) V, 5Ω, , 18.49 0.542 cos (t 13.64) V, , – j2 Ω, , 18.51 16.667 J, , Figure G.41, 18.53 p, , For Prob. 19.11., , 18.55 682.5 J, , 19.13 329.9 W, , 18.57 12.5 J, 87.41%, , 19.15 24 , 384 W, , 18.59 (16et 20e2t 4e4t ) u (t) V, , 19.17 c, , 4.8, 0.4, , 0.4, 0.21, d , c, 4.2, 0.02, , 0.02, dS, 0.24, , 18.61 2X() 0.5X( 0) 0.5X( 0), 18.63 106 stations, , 19.19 This is a design problem with multiple, answers.
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ale29559_appG.qxd, , 07/17/2008, , 10:05 AM, , Page A-106, , Appendix G, , A-106, , 19.21 See Fig. G.42., , 19.43 (a) c, , I1, , I1, , +, , +, 0.4 S, , V1, , Answers to Odd-Numbered Problems, , 0.1 S, , 0.2V1, , −, , Z, 1, d , (b) c, 1, Y, , 1, 0, , 19.45 c, , 1 j 0.5, 0.125 S, , 19.47 c, , 0.3235, 0.02941, , 0, d, 1, , j4 , d, 1, , V2, −, , Figure G.42, , 1.176, d, 0.4706, , For Prob. 19.21., , 19.23, , £, , s2, , (s 1), 0.8(s 1), s2 s 1 § , 2, s 1.8s 1.2, s, , (s 1), , 19.51 c, , 19.25 See Fig. G.43., , 0.5 S, , 1S, , 1, , s, ¥, 1, 2, s, , 2 j5, d, 2 j, A, AD BC, 1, D, , z12 , , z21 , z22 , C, C, C, C, , 19.55 Proof, 7, 1, 20, , 1, 20, 20, 7, d , ≥, ¥ S, ≥, 7, 3, 1, 1, 20, 20, 7, 1, 1, S, 3, 3, 7, 20 , ¥, c, ≥, d, 20, 1, 1S, 3, , 3, 3, , 3, 19.57 c, 1, , Figure G.43, For Prob. 19.25., , 0.25, 5, , 2, j, , 19.53 z11 , , 0.5 S, , 19.27 c, , 2s 1, s, 19.49 ≥, (s 1)(3s 1), S, s, , 0.025, dS, 0.6, , 1, 7, ¥,, 1, S, 7, , 19.29 (a) 22 V, 8 V, (b) same, 19.31 c, , 3.8 , 3.6, , 19.33 c, , 3.077 j1.2821, 0.3846 j 0.2564, , 2, 19.35 c, 0.5, , 0.1, 0.2, 16.667 6.667, d , c, d S,, 3.333 3.333, 0.1, 0.5, 10 , 2, 5, 10 , c, d, c, d, 1, 0.3 S, 0.3 S, 1, , 19.59 c, , 0.4, d, 0.2 S, 0.3846 j 0.2564, d, 0.0769 j 0.2821, , 0.5, d, 0, , 10, 3, 19.61 ≥, 8, 3, , 8, 6, , 3, 5, ¥ , ≥, 4, 10, 3, 5, , 19.63 c, , 2.4, d, 7.2, , 19.37 1.1905 V, R2, 1, , g12 , R1 R2, R1 R2, R2, R1R2, , , g22 R3 , R1 R2, R1 R2, , 19.39 g11 , g21, , 19.41 Proof, , 0.8, 2.4, , 1, 3, 19.65 ≥, 1, , 3, , , , 1, 3, ¥ S, 2, 3, , 4, 5, 5, 4, ¥, ≥, 3, 3, S, S, 10, 8, , 3, , 2, ¥, 5, 4
Page 1041 :
ale29559_appG.qxd, , 07/17/2008, , 10:05 AM, , Page A-107, , Appendix G, , 19.67 c, , 4, 0.1576, , 63.29, d, 4.994, (3s 2), 2(s 2), ¥, 5s2 4s 4, 2s(s 2), , s1, s2, 19.69 ≥, (3s 2), 2(s 2), 19.71 c, , 2, 3.334, , 3.334, d, 20 .22, , 19.73 c, , 14.628, 5.432, , 3.141, d, 19.625, , 19.75 (a) c, 19.77 c, , Answers to Odd-Numbered Problems, , 0.3015, 0.0588, , 19.85 c, , 1.581l71.59, j, , 19.87 c, , j1,765, j888.2, , j, d, 5.661 104, , j1,765, d, j888.2, , 19.89 1,613, 64.15 dB, 19.91 (a) 25.64 for the transistor and 9.615 for the, circuit., 19.93 17.74, 144.5, 31.17 , 6.148 M, , 0.1765, d , (b) 0.0051, 10.94, , 0.9488l161.6, 0.3163l161.6, , 4.669l136.7, 19.79 c, 2.53l108.4, , A-107, , 0.3163l18.42, d, 0.9488l161.6, , 19.95 See Fig. G.44., , 0.425 F, , 1.471 H, , 0.2 F, , 2.53l108.4, d, 1.789l153.4, , Figure G.44, 19.81 c, , 1.5, 3.5, , 0.5, dS, 1.5, , For Prob. 19.95., , 19.97 0.25 F, 0.3333 H, 0.5 F, 0.3235, 19.83 c, 0.02941, , 1.1765, d, 0.4706, , 1H, , 19.99 Proof
Page 1042 :
ale29559_appG.qxd, , 07/17/2008, , 10:05 AM, , Page A-108
Page 1043 :
ale29559_bib.qxd, , 07/17/2008, , 01:12 PM, , Page B-1, , Bibliography, Aidala, J. B., and L. Katz. Transients in Electric Circuits., Englewood Cliffs, NJ: Prentice Hall, 1980., Angerbaur, G. J. Principles of DC and AC Circuits. 3rd ed., Albany, NY: Delman Publishers, 1989., Attia, J. O. Electronics and Circuit Analysis Using MATLAB., Boca Raton, FL: CRC Press, 1999., Balabanian, N. Electric Circuits. New York: McGraw-Hill,, 1994., Bartkowiak, R. A. Electric Circuit Analysis. New York:, Harper & Row, 1985., Blackwell, W. A., and L. L. Grigsby. Introductory Network, Theory. Boston, MA: PWS Engineering, 1985., Bobrow, L. S. Elementary Linear Circuit Analysis. 2nd ed., New York: Holt, Rinehart & Winston, 1987., Boctor, S. A. Electric Circuit Analysis. 2nd ed. Englewood, Cliffs, NJ: Prentice Hall, 1992., Boylestad, R. L. Introduction to Circuit Analysis. 10th ed., Columbus, OH: Merrill, 2000., Budak, A. Circuit Theory Fundamentals and Applications., 2nd ed. Englewood Cliffs, NJ: Prentice Hall, 1987., Carlson, B. A. Circuit: Engineering Concepts and Analysis, of Linear Electric Circuits. Boston, MA: PWS Publishing,, 1999., Chattergy, R. Spicey Circuits: Elements of Computer-Aided, Circuit Analysis. Boca Raton, FL: CRC Press, 1992., Chen, W. K. The Circuit and Filters Handbook. Boca Raton,, FL: CRC Press, 1995., Choudhury, D. R. Networks and Systems. New York: John, Wiley & Sons, 1988., Ciletti, M. D. Introduction to Circuit Analysis and Design., New York: Oxford Univ. Press, 1995., Cogdeil, J. R. Foundations of Electric Circuits. Upper Saddle, River, NJ: Prentice Hall, 1998., Cunningham, D. R., and J. A. Stuller. Circuit Analysis. 2nd ed., New York: John Wiley & Sons, 1999., Davis, A., (ed.). Circuit Analysis Exam File. San Jose, CA:, Engineering Press, 1986., Davis, A. M. Linear Electric Circuit Analysis. Washington,, DC: Thomson Publishing, 1998., DeCarlo, R. A., and P. M. Lin. Linear Circuit Analysis. 2nd ed., New York: Oxford Univ. Press, 2001., Del Toro, V. Engineering Circuits. Englewood Cliffs, NJ:, Prentice Hall, 1987., Dorf, R. C., and J. A. Svoboda. Introduction to Electric, Circuits. 4th ed. New York: John Wiley & Sons, 1999., Edminister, J. Schaum’s Outline of Electric Circuits. 3rd ed., New York: McGraw-Hill, 1996., , Floyd, T. L. Principles of Electric Circuits. 7th ed. Upper, Saddle River, NJ: Prentice Hall, 2002., Franco, S. Electric Circuits Fundamentals. Fort Worth, FL:, Saunders College Publishing, 1995., Goody, R. W. Microsim PSpice for Windows. Vol. 1. 2nd ed., Upper Saddle River, NJ: Prentice Hall, 1998., Harrison, C. A. Transform Methods in Circuit Analysis., Philadelphia, PA: Saunders, 1990., Harter, J. J., and P. Y. Lin. Essentials of Electric Circuits., 2nd ed. Englewood Cliffs, NJ: Prentice Hall, 1986., Hayt, W. H., and J. E. Kemmerly. Engineering Circuit, Analysis. 6th ed. New York: McGraw-Hill, 2001., Hazen, M. E. Fundamentals of DC and AC Circuits. Philadelphia, PA: Saunders, 1990., Hostetter, G. H. Engineering Network Analysis. New York:, Harper & Row, 1984., Huelsman, L. P. Basic Circuit Theory. 3rd ed. Englewood, Cliffs, NJ: Prentice Hall, 1991., Irwin, J. D. Basic Engineering Circuit Analysis. 7th ed., New York: John Wiley & Sons, 2001., Jackson, H. W., and P. A. White. Introduction to Electric, Circuits. 7th ed. Englewood Cliffs, NJ: Prentice Hall, 1997., Johnson, D. E. et al. Electric Circuit Analysis. 3rd ed. Upper, Saddle River, NJ: Prentice Hall, 1997., Karni, S. Applied Circuit Analysis. New York: John Wiley &, Sons, 1988., Kraus, A. D. Circuit Analysis. St. Paul, MN: West Publishing,, 1991., Madhu, S. Linear Circuit Analysis. 2nd ed. Englewood Cliffs,, NJ: Prentice Hall, 1988., Mayergoyz, I. D., and W. Lawson. Basic Electric Circuits, Theory. San Diego, CA: Academic Press, 1997., Mottershead, A. Introduction to Electricity and Electronics:, Conventional and Current Version. 3rd ed. Englewood, Cliffs, NJ: Prentice Hall, 1990., Nasar, S. A. 3000 Solved Problems in Electric Circuits., (Schaum’s Outline) New York: McGraw-Hill, 1988., Neudorfer, P. O., and M. Hassul. Introduction to Circuit, Analysis. Englewood Cliffs, NJ: Prentice Hall, 1990., Nilsson, J. W., and S. A. Riedel. Electric Circuits. 5th ed., Reading, MA: Addison-Wesley, 1996., O’Malley, J. R. Basic Circuit Analysis. (Schaum’s Outline), New York: McGraw-Hill, 2nd ed., 1992., Parrett, R. DC-AC Circuits: Concepts and Applications., Englewood Cliffs, NJ: Prentice Hall, 1991., Paul, C. R. Analysis of Linear Circuits. New York:, McGraw-Hill, 1989., B-1
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ale29559_bib.qxd, , B-2, , 07/17/2008, , 01:12 PM, , Page B-2, , Bibliography, , Poularikas, A. D., (ed.). The Transforms and Applications, Handbook. Boca Raton, FL: CRC Press, 2nd ed., 1999., Ridsdale, R. E. Electric Circuits. 2nd ed. New York: McGrawHill, 1984., Sander, K. F. Electric Circuit Analysis: Principles and Applications. Reading, MA: Addison-Wesley, 1992., Scott, D. Introduction to Circuit Analysis: A Systems Approach., New York: McGraw-Hill, 1987., Smith, K. C., and R. E. Alley. Electrical Circuits: An Introduction. New York: Cambridge Univ. Press, 1992., Stanley, W. D. Transform Circuit Analysis for Engineering and, Technology. 3rd ed. Upper Saddle River, NJ: Prentice Hall,, 1997., Strum, R. D., and J. R. Ward. Electric Circuits and Networks., 2nd ed. Englewood Cliffs, NJ: Prentice Hall, 1985., , Su, K. L. Fundamentals of Circuit Analysis. Prospect Heights,, IL: Waveland Press, 1993., Thomas, R. E., and A. J. Rosa. The Analysis and Design of, Linear Circuits. 3rd ed. New York: John Wiley & Sons,, 2000., Tocci, R. J. Introduction to Electric Circuit Analysis. 2nd ed., Englewood Cliffs, NJ: Prentice Hall, 1990., Tuinenga, P. W. SPICE: A Guide to Circuit Simulation., Englewood Cliffs, NJ: Prentice Hall, 1992., Whitehouse, J. E. Principles of Network Analysis. Chichester,, U.K.: Ellis Horwood, 1991., Yorke, R. Electric Circuit Theory. 2nd ed. Oxford, U.K.:, Pergamon Press, 1986.
Page 1045 :
ale29559_index.qxd, , 07/25/2008, , 12:41 PM, , Page I-1, , Index, A, A, 5, a, 5, abc sequence, 507, ABCD parameters, 864, ac, 7, ac bridge, 398–402, ac circuits, 369, ac power analysis, 457–501, apparent power, 471, average power, 459, complex power, 473–477, conservation of ac power, 477–480, effective value, 467, electricity consumption cost, 486–488, instantaneous power, 458, maximum average power transfer, 464–467, power factor, 471, power factor correction, 481–483, power measurement, 483–486, rms value, 468, ac voltage, 10, acb sequence, 507, Active bandpass filter, 643–645, Active bandreject filter, 645–648, Active element, 15, Active filter, 637, 642–648, Additivity property, 128, Adjoint of A, A–5, Admittance, 388, Admittance parameters, 855–858, Air-core transformers, 568, Alexander, Charles K., 127, 313, Alternating current (ac), 7, 369, AM, 818–819, 836–838, Ammeter, 62, Ampere, Andre-Marie, 7, Amplitude modulation (AM), 818–819, 836–838, Amplitude-phase form of Fourier series, 759, Amplitude spectrum, 760, 812, Analog computer, 237–240, , Analog meter, 63, Analysis. See Methods of analysis, Analyzing ac circuits, 414, Apparent power, 471, 475, Asking questions, 715–716, Attenuator, 173, Atto-, 5, Audio transformer, 568, Automobile ignition circuit, 298–299, Automobile ignition system, 353–355, Autotransformer, 581, Average power, 459, 778, Averaging amplifier, 207, , B, Bacon, Francis, 3, Bailey, F. J., 253, Balanced, 158, Balanced delta-delta connection, 514–516, Balanced delta/wye circuit, 393, Balanced delta-wye connection, 516–519, Balanced network, 55, Balanced three-phase voltages, 505–508, Balanced wye-delta connection, 512–514, Balanced wye-wye connection, 509–512, Bandpass filter, 639, 643–645, Bandreject filter, 640, 645–648, Bandwidth, 631, Bandwidth of rejection, 640, Bardeen, John, 108, Barkhausen criteria, 439, Bell, Alexander Graham, 618, Bilateral Laplace transform, 677, Binary weighted ladder, 196, Bipolar junction transistor (BJT), 107–109, Bode, Hendrik W., 619, Bode plot, 619–629, Bode straight-line magnitude and phase plots, 623, Branch, 35, Brattain, Walter, 108, Braun, Karl Ferdinand, 18, , I-1
Page 1046 :
ale29559_index.qxd, , 07/25/2008, , 12:41 PM, , I-2, , Page I-2, , Index, , Break frequency, 621, Butterworth filter, 670, Buxton, W. J. Wilmont, 81, Byron, Lord, 175, , C, C, 5, c, 5, Capacitance, 216, Capacitance multiplier, 437–439, Capacitor, 216–222, Capacitors and inductors, 215–252, analog computer, 237–240, capacitor, 216–222, characteristics, 232, differentiator, 235–236, energy stored in, 241, inductors, 226–230, integrator, 234–235, series and parallel capacitors, 222–225, series and parallel inductors, 230–233, special properties, 234, Careers, communications systems, 809, computer engineering, 253, control systems, 613, education, 849, electromagnetics, 555, electronic instrumentation, 175, electronics, 81, power systems, 457, software engineering, 413, Cascaded op amp circuits, 191–194, cd, 5, Centi-, 5, Ceramic capacitor, 218, Characteristic equation, 320, Chassis ground, 83, Chip inductor, 226, Circuit analysis, 722–725, Circuit applications, 774–778, 829–831, Circuit element models, 716–722, Circuit stability and synthesis, 737–745, Circuit theorems, 127–173, linearity, 128–129, maximum power transfer, 150–152, Norton’s theorem, 145–148, 149–150, PSpice, 152–155, resistance measurement, 158–160, source modeling, 155–157, , source transformation, 135–138, superposition, 130–135, Thevenin’s theorem, 139–145, 149–150, Closed-loop gain, 178, Coefficient of coupling, 565, 566, Cofactors of A, A–5, Colpitts oscillator, 455, Common-base current gain, 109, Common emitter amplifier, 881, Common-emitter amplifier circuit, 173, Common-emitter current gain, 109, Communication skills, 127, Communications systems, 809, Complete response, 275, Completing the square, 692, Complex amplitude spectrum, 782, Complex conjugate, A–12, Complex Fourier series, 782, Complex numbers, 378, complex conjugate, A–12, Euler’s formula, A–14, exponential form, A–10, identities, A–15, mathematical operations, A–12 to A–14, polar form, A–9, real part/imaginary part, A–9, rectangular form, A–9, Complex phase spectrum, 782, Complex power, 473–477, Computer engineering, 253, Computer programs, KCIDE for Circuits;, MATLAB; PSpice, Conductance, 33, 388, Conductance matrix, 101, Conservation of ac power, 477–480, Control systems, 613, Convolution, 821–824, Convolution integral, 697–705, Convolution of two signals, 698, Copper wound dry power transformer, 568, Corner frequency, 621, Coulomb, 6, Coupling coefficient, 566, Cramer’s rule, 85, 87, 92, 97, A, Critically damped case, source-free parallel RLC circuit, 327, source-free series RLC circuit, 321–322, step response of parallel RLC circuit, 337, step response of series RLC circuit, 332, Crossover network, 661–663, Current, 6
Page 1047 :
ale29559_index.qxd, , 07/25/2008, , 12:41 PM, , Page I-3, , Index, , Current divider, 46, Current division, 391–392, Current division principle, 46, Current flow, 8, Cutoff frequency, 638, Cyclic frequency, 372, , D, d, 5, da, 5, DAC, 196–197, Damped natural frequency, 323, Damping, 323, Damping factor, 321, Damping frequency, 323, d’Arsonval meter movement, 61, Darwin, Francis, 215, Datum node, 82, dB, 618, dc, 7, dc meters, 60–63, dc transistor circuit, 107–109, dc voltage, 10, Deci-, 5, Decibel (dB), 618, Decibel scale, 617–619, Definite integrals, A–19 to A–20, Deka-, 5, Delay circuit, 293–295, Delta-delta three-phase transformer connection, 584, Delta function, 267, Delta-to-wye conversion, 53–54, 392, Delta-wye three-phase transformer, connection, 585, Demodulation, 837, Dependent current source, 15, Dependent source, 15, Dependent voltage source, 15, Derivatives, A–17 to A–18, Deschemes, Marc-Antoine Parseval, 779, Determinant of A, A–5, Difference amplifier, 187–190, Differentiator, 235–236, Digital meter, 63, Digital-to-analog computer (DAC), 196–197, Dinger, J. E., 413, Direct current (dc), 7, Dirichlet conditions, 757, Distribution transformers, 595, Dot convention, 559, 560, , I-3, , Driving-point impedance, 852, Duality, 350–352, 821, , E, E, 5, Earth ground, 83, Edison, Thomas, 14, 59, 369, 504, 505, Education, 849, Effective value, 467, 1884 International Electrical Exhibition, 14, Electric charge, 6, Electric circuit, 4, Electric current, 6, Electrical isolation, 592, Electrical lighting systems, 58–60, Electrical system safety guidelines, 542, Electricity bills, 19, Electricity consumption cost, 486–488, Electrolytic capacitor, 218, Electromagnetics, 555, Electronic instrumentation, 175, Electronics, 81, Element, 4, Elimination technique, 85, 86, Energy, 12, Equivalent circuit, 135, Equivalent conductance, parallel-connected capacitors, 223, resistors in parallel, 46, resistors in series, 65, series-connected capacitors, 224, Equivalent inductance, parallel inductors, 231, series-connected inductors, 231, Equivalent resistance, combination of resistors in series and parallel, 47, resistors in parallel, 45, resistors in series, 44, Ethical responsibility, 503, Euler’s formula, A–14, Euler’s identities, 323, 781, Even symmetry, 764–766, Exa-, 5, Exponential Fourier series, 781–787, , F, f, 5, Faraday, Michael, 217, 457, Faraday’s law, 557, 574
Page 1048 :
ale29559_index.qxd, , I-4, , 07/25/2008, , 12:41 PM, , Page I-4, , Index, , Femto-, 5, Filmtrim capacitor, 218, Filter, active, 642–648, defined, 637, design, 793–796, limitations/advantages, 642, passive, 637–642, Final-value theorem, 686, First-order circuits, 253–312, automobile ignition circuit, 298–299, delay circuit, 293–295, first-order op amp circuit, 284–289, natural response, 255, photoflash unit, 295–296, PSpice, 289–293, relay circuit, 296–298, singularity functions, 265–273, source-free RC circuit, 254–259, source-free RL circuit, 259–265, step response of RC circuit, 273–279, step response of RL circuit, 280–284, time constant, 256, unit impulse function, 267–268, unit ramp function, 268, unit step function, 266–267, First-order highpass filter, 643, First-order lowpass filter, 643, First-order op amp circuit, 284–289, Fixed capacitor, 218, Fixed resistor, 32, Forced response, 275, Four-bit DAC, 196, Fourier, Jean Baptiste Joseph, 756, Fourier analysis, 758, Fourier coefficients, 757, Fourier cosine series, 765, Fourier series, 755–807. See also Fourier, transform, amplitude-phase form, 759, amplitude spectrum, 760, applying, steps in, 774, average power, 778, circuit applications, 774–778, defined, 757, Dirichlet conditions, 757, exponential Fourier series, 781–787, filters, 793–795, Fourier analysis, 758, Fourier coefficients, 757, , Fourier cosine series, 765, Fourier sine series, 767, frequency spectrum, 760, full-wave rectified wine, 770, Gibbs phenomenon, 762, half-wave rectified wine, 770, Parseval’s theorem, 779, phase spectrum, 760, PSpice, 787–792, rectangular pulse train, 770, rms value, 779, sawtooth wave, 770, sinc function, 783, spectrum analyzer, 793, square wave, 770, symmetry considerations, 764–770, triangular wave, 770, trigonometric Fourier series, 757, Fourier sine series, 767, Fourier theorem, 757, Fourier transform. See also Fourier series, amplitude modulation (AM), 818–819, 836–838, circuit applications, 829–831, convolution, 821–824, defined, 812, duality, 821, frequency shifting, 818–819, inverse, 812, Laplace transform, compared, 835, linearity, 816, Parseval’s theorem, 832–835, properties, listed, 824–825, reversal, 820–821, sampling, 838–839, time differentiation, 819–820, time integration, 820, time scaling, 816–817, Fourier transform pairs, 825, Franklin, Benjamin, 6, 613, Frequency differentiation, 684, Frequency domain, 380, Frequency mixer, 658, Frequency of rejection, 640, Frequency response, 613–673, active filters, 642–648, Bode plot, 619–629, crossover network, 661–663, decibel scale, 617–619, defined, 614, MATLAB, 655–657
Page 1049 :
ale29559_index.qxd, , 07/25/2008, , 12:41 PM, , Page I-5, , Index, , parallel resonance, 634–637, passive filters, 637–642, PSpice, 652–655, radio receiver, 657–659, scaling, 648–651, series resonance, 629–634, touch-tone telephone, 660–661, transfer function, 614–617, Frequency scaling, 650, Frequency shift, 681–682, Frequency shifting, 818–819, Frequency spectrum, 760, Frequency translation, 681, Fundamental frequency, 757, , G, G, 5, g parameters, 859, Ganged tuning, 658, Gate function, 269, General second-order circuit, 339–343, Generalized node, 89, GFCI, 542, Gibbs phenomenon, 762, Giga-, 5, Ground, 82, Ground-fault circuit interrupter (GFCI), 542, Györgyi, Albert Szent, 755, Gyrator, 754, , H, h, 5, h parameters, 859, Half-power frequencies, 631, Half-wave symmetry, 768–769, Hartley oscillator, 455, Heaviside, Oliver, 691, Heaviside’s theorem, 691, Hecto-, 5, Henry, Joseph, 227, Herbert, G., 503, Hertz, Heinrich Rudorf, 372, Heterodyne circuit, 658, High-Q circuit, 632, Highpass filter, 639, 643, 670, Homogeneity property, 128, Hybrid parameters, 858–863, Hyperbolic functions, A–17, , I, Ibn, Al Halif Omar, 457, Ideal autotransformer, 581–584, Ideal current source, 23, Ideal dependent (controlled) source, 15, Ideal independent source, 15, Ideal op amp, 179–181, Ideal transformer, 573–580, Ideal voltage source, 23, Immittance parameters, 855, Impedance, 387, Impedance combinations, 390–396, Impedance matching, 576, 593, Impedance parameters, 850–854, Impedance triangle, 475, Impulse function, 267–268, Indefinite integrals, A–18 to A–19, Independent current source, 15, Independent source, 15, Independent voltage source, 15, Inductance, 226, Inductance simulator, 454, Inductive, 387, Inductors, 226–230. See also Capacitors and, inductors, Initial-value theorem, 685, Inspection, 100–104, Instantaneous power, 11, 458, Instrumentation amplifier, 187–188, 198–199, Integrator, 234–235, Integrodifferential equations, 705–707, International System of Units (SI), 5, Inverse Fourier transform, 812, Inverse hybrid parameters, 859, Inverse Laplace transform, 690–697, Inverse transmission parameters, 865, Inverting amplifier, 181–183, Isolation transformer, 575, , J, Jefferson, Thomas, 849, , K, K, 5, k, 5, KCIDE for Circuits, A–65 to A–74, KCL, 37–39, , I-5
Page 1050 :
ale29559_index.qxd, , I-6, , 07/25/2008, , 12:41 PM, , Page I-6, , Index, , kg, 5, Kilo-, 5, Kirchhoff, Gustav Robert, 38, Kirchhoff’s current law (KCL), 37–39, Kirchhoff’s laws in frequency domain, 389–390, Kirchhoff’s voltage law (KVL), 39–40, KVL, 39–40, , L, Ladder method, 726, Ladder network synthesis, 885–889, Lagging power factor, 471, Lamme, B. G., 369, Laplace, Pierre Simon, 676, Laplace transform, 675–754, applying, steps in, 716, circuit analysis, 722–725, circuit element models, 716–722, convolution integral, 697–705, defined, 677, final value, 686, Fourier transform, compared, 835, frequency differentiation, 684, frequency shift, 681–682, initial value, 685, integrodifferential equations, 705–707, inverse transform, 690–697, linearity, 680, network stability, 737–740, network synthesis, 740–745, one-sided/two-sided, 677, properties, listed, 687, scaling, 680, significance, 676, state variables, 730–737, time differentiation, 682, time integration, 683–684, time periodicity, 684–685, time shift, 680–681, transfer function, 726–730, Laplace transform pairs, 687, Law of conservation of charge, 6, Law of conservation of energy, 12, Law of cosines, A–16, Law of sines, A–16, Law of tangents, A–16, Leading power factor, 471, Least significant bit (LSB), 196, L’Hopital’s rule, 784, A–20, , Lighting systems, 58–60, Line spectra, 784, Linear capacitor, 218, Linear circuit, 128, 129, Linear inductor, 227, Linear resistor, 33, Linear transformer, 567–573, Linearity, 128–129, 680, 816, Loading effect, 156, Local oscillator, 658, Logarithm, 617, Loop, 36, Loop analysis, 94, Loosely coupled, 566, Lowpass filter, 638–639, 643, LSB, 196, , M, M, 5, m, 5, Magnetically coupled circuits, 555–612, dot convention, 559, 560, energy in coupled circuit, 564–567, ideal autotransformer, 581–584, ideal transformer, 573–580, linear transformer, 567–573, mutual inductance, 557–563, power distribution, 595–596, PSpice, 586–591, three-phase transformer, 584–586, transformer as isolation device, 592–593, transformer as matching device, 593–594, Magnitude plot, 622–628, Magnitude scaling, 649, Mathematical formulas, definite integrals, A–19 to A–20, derivatives, A–17 to A–18, hyperbolic functions, A–17, indefinite integrals, A–18 to A–19, l’Hopital’s rule, A–20, quadratic formula, A–16, trigonometric identities, A–16 to A–17, MATLAB, A–46 to A–49, AC circuit analysis, A–58 to A–62, calculator, as, A–46 to A–49, color and line types, A–50, DC circuit analysis, A–54 to A–57, elementary math functions, A–47, frequency response, 655–657, A–62 to A–64
Page 1051 :
ale29559_index.qxd, , 07/25/2008, , 12:41 PM, , Page I-7, , Index, , matrix operations, A–48, plotting, A–49 to A–50, programming, A–51 to A–53, relational/logical operators, A–51, solving equations, A–53 to A–54, special matrices, variables, constants, A–49, Matrix inversion, A–4 to A–8, Maximum average power transfer, 464–467, Maximum average power transfer theorem, 465, Maximum power theorem, 150, Maximum power transfer, 150–152, Maxwell, James Clerk, 556, Maxwell bridge, 411, Mega-, 5, Megger tester, 158, Mesh, 93, Mesh analysis, 93–95, 417–421, Mesh analysis with current sources, 98–100, Mesh current, 94, Method of algebra, 693, Methods of analysis, 81–126, dc transistor circuit, 107–109, inspection, 100–104, mesh analysis, 93–95, mesh analysis with current sources,, 98–100, nodal analysis, 82–84, nodal analysis with voltage sources, 88–90, nodal vs. mesh analysis, 104–105, PSpice, 105–107, Mho, 33, Micro-, 5, Milli-, 5, Milliohmmeter, 158, Morse, Samuel F. B., 63, Most significant bit (MSB), 196, MSB, 196, Multidisciplinary teams, 369, , 5, Mutual inductance, 557–563, Mutual voltage, 558, , N, n, 5, nth harmonic, 757, Nano-, 5, Natural frequencies, 321, Natural response, 255, Negative current flow, 8, , Negative sequence, 507, Neper frequency, 321, Network function, 614, Network stability, 737–740, Network synthesis, 740–745, Nodal analysis, 82–84, 414–417, Nodal analysis with voltage sources, 88–90, Nodal vs. mesh analysis, 104–105, Node, 35, Noninverting amplifier, 183–185, Nonlinear capacitor, 218, Nonlinear inductor, 227, Nonlinear resistor, 33, Nonplanar, 93, Normalized Butterworth lowpass filter, 650, Norton, E. L., 145, Norton equivalent circuit, 426–430, Norton’s theorem, 145–148, 149–150, Notch filter, 640, 645–648, npn transistor, 108, nth harmonic, 757, Nyquist frequency, 839, Nyquist interval, 839, , O, Odd symmetry, 766–768, Ohm, Georg Simon, 31, Ohm’s law, 31, 120/240 household power system, 541, One-sided Laplace transform, 677, Op amp ac circuits, 431–432, Open circuit, 32, Open-circuit impedance parameters, 851, Open delta, 585, Open-loop voltage gain, 177, Operational amplifier (op amp), 175–213, ac circuit, 431–432, cascaded op amp circuits, 191–194, defined, 176, difference amplifier, 187–190, digital-to-analog computer (DAC), 196–197, feedback, 178, first order circuits, 284–289, ideal op amp, 179–181, instrumentation amplifier, 198–199, inverting amplifier, 181–183, noninverting amplifier, 183–185, PSpice, 194–195, second order circuits, 344–346, , I-7
Page 1052 :
ale29559_index.qxd, , 07/25/2008, , 12:41 PM, , Page I-8, , I-8, , Index, , Operational amplifier (continued), summary of basic circuits, 200, summing amplifier, 185–187, terminals, 176, voltage follower, 184, Oscillator, 439–441, Overdamped case, source-free parallel RLC circuit, 327, source-free series RLC circuit, 321, step response of parallel RLC circuit, 337, step response of series RLC circuit, 332, , P, P, 5, p, 5, Parallel, 36, Parallel capacitors, 222–225, Parallel inductors, 230–233, Parallel resistors, 45–47, Parallel resonance, 634–637, Parallel RLC circuit, source-free, 326–331, step response, 336–339, Parameters, ABCD, 864, admittance, 855–858, defined, 850, g, 859, h, 859, hybrid, 858–863, immittance, 855, impedance, 850–854, inverse hybrid, 859, inverse transmission, 865, relationship between, 868–871, 889, T, 864, 889, t, 865, transmission, 863–867, y, 855, z, 851, Parseval’s theorem, 779, 832–835, Partial fraction expansion, 690, Passive element, 15, Passive filters, 637–642, Passive sign convention, 11, Perfectly coupled, 566, Period, 372, Periodic function, 372, 756, Peta-, 5, , Phase lot, 622–628, Phase sequence, 507, Phase-shifters, 396–398, Phase spectrum, 760, 812, Phase voltages, 506, Phasor, 376–384, Phasor diagram, 379, Phasor relationships for circuit elements,, 385–386, Photoflash unit, 295–296, Pico-, 5, Planar, 93, Pole, 615, 620, 621, Polyester capacitor, 218, Polyphase, 504, Port, 850, Positive current flow, 8, Positive sequence, 507, Potentiometer (pot), 32, Power, 11, Power analysis. See ac power analysis, Power distribution, 595–596, Power factor, 471, Power factor angle, 471, Power factor correction, 481–483, Power grid, 595, Power in balanced three-phase system, 519–525, Power measurement, 483–486, Power spectrum, 783, Power systems, 457, Primary winding, 568, Principle of current division, 46, Principle of voltage division, 44, Problem-solving technique, 20–21, Professional responsibility, 503, PSpice, A–21 to A–45, ac analysis, 433–437, A–40 to A–45, analysis of magnetically coupled transformers,, 586–591, analysis of RLC circuits, 346–349, circuit analysis, 105–107, creating a circuit, A–22 to A–27, DC nodal analysis, A–27 to A–28, DC sweep, A–29 to A–33, design center for Windows, A–21 to A–22, Fourier analysis, 787–792, frequency response, 652–655, A–40 to, A–45, op amp circuit analysis, 194–195, print and plot pseudocomponents, A–43
Page 1053 :
ale29559_index.qxd, , 07/25/2008, , 12:41 PM, , Page I-9, , Index, , pseudocomponents, A–42, A–43, three-phase circuits, 529–534, transient analysis, 289–293, A–33 to A–40, two-port networks, 877–880, verifying circuit theorems, 152–155, , Q, Quadratic formula, A–16, Quadratic pole/zero, 621, Quadrature power, 474, Quality factor, 632, , R, Radio receiver, 657–659, RC delay circuit, 293, RC phase-shifting circuits, 396–398, Reactance, 387, Reactive power, 474, 475, Real power, 474, 475, Reference node, 82, Reflected impedance, 569, 576, Relay, 296, Relay circuit, 296–298, Relay delay time, 297, Residential wiring, 540–542, Residue method, 691, Residues, 691, Resistance, 30, 387, Resistance bridge, 158, Resistance matrix, 101, Resistance measurement, 158–160, Resistivity, 30, Resistor, 30, 232, Resonance, 630, Resonant frequency, 321, 630, Resonant RLC circuits, 629–635, Reversal, 820–821, rms value, 468, 779, Rolloff frequency, 639, Root-mean-square (rms) value, 468, , S, s, 5, Safety guidelines, electrical systems, 542, Sampling, 268, 838–839, Sampling frequency, 838, , Sampling function, 783, Sampling interval, 838, Sampling rate, 838, Sampling theorem, 793, Sawtooth function, 270, Scaling, 648–651, 680, Schockley, William, 108, Scott, C. F., 369, Second-order circuits, 313–367, automobile ignition system, 353–355, characteristic equation, 320, duality, 350–352, general second-order circuit, 339–343, initial/final values, 314–319, PSpice, 346–349, second-order differential equation, 320, second order op amp circuit, 344–346, smoothing circuits, 355–356, source-free parallel RLC circuit, 326–331, source-free series RLC circuit, 319–326, step response of parallel RLC circuit,, 336–339, step response of series RLC circuit,, 331–336, Second-order differential equation, 320, Second order op amp circuit, 344–346, Secondary winding, 568, Selectivity, 632, Self-inductance, 557, Sensitivity, 64, Series, 36, Series and parallel capacitors, 222–225, Series and parallel inductors, 230–233, Series resistors, 43–44, Series resonance, 629–634, Series RLC circuit, source-free, 319–326, step response, 331–336, Short circuit, 32, SI prefixes, 5, SI units, 5, Siemens, 33, Sifting, 268, Signal, 10, Simultaneous equations, A to A–4, Sinc function, 783, Sine wave oscillator, 439, Single-phase equivalent circuit, 511, Single-phase three-wire residential, wiring, 541, , I-9
Page 1054 :
ale29559_index.qxd, , I-10, , 07/25/2008, , 12:41 PM, , Page I-10, , Index, , Singularity functions, 265–273, Sinusoid, 371–376, cyclic frequency, 372, defined, 370, phase, 373, phasor representation, 379, sine/cosine form, 373, Sinusoid-phasor transformation, 380, Sinusoidal steady-state analysis, 413–455, capacitance multiplier, 437–439, mesh analysis, 417–421, nodal analysis, 414–417, Norton equivalent circuit, 426–430, op amp ac circuits, 431–432, oscillator, 439–441, PSpice, 433–437, source transformation, 424–426, superposition theorem, 421–424, Thevenin equivalent circuit, 426–430, Sinusoidal steady-state response, 371, Smoothing circuits, 355–356, Software engineering, 413, Software packages, KCIDE for Circuits;, MATLAB; PSpice, Solenoidal wound inductor, 226, Source-free circuits, 254, Source-free parallel RLC circuit, 326–331, Source-free RC circuit, 254–259, Source-free RL circuit, 259–265, Source-free series RLC circuit, 319–326, Source modeling, 155–157, Source transformation, 135–138, 424–426, Spectrum, 812, Spectrum analyzer, 793, Sprague, Frank, 14, Stability, 737–740, State variables, 730–737, Steady-state response, 276, Steinmetz, Charles Proteus, 377, Step-down autotransformer, 581, Step-down transformer, 575, Step response, 273, Step response of parallel RLC circuit,, 336–339, Step response of RC circuit, 273–279, Step response of RL circuit, 280–284, Step response of series RLC circuit,, 331–336, Step-up autotransformer, 581, Step-up transformer, 575, Storage elements, 216, , Strength of impulse function, 267, Summer, 186, Summing amplifier, 185–187, Superheterodyne receiver, 658, Supernode, 89, Superposition, 130–135, Superposition principle, 130, Superposition theorem, 421–424, Susceptance, 388, Switching functions, 265, Symmetry, even, 764–766, half-wave, 768–769, odd, 766–768, summary, 770, Synthesis of ladder networks, 880, System, 716, System design, 215, , T, T, 5, T parameters, 864, 889, t parameters, 865, Television picture tube, 17, Tera-, 5, Tesla, Nikola, 369, 505, Thevenin, M. Leon, 139, Thevenin equivalent circuit, 139, 288, 426–430, Thevenin resistance, 279, Thevenin’s theorem, 139–145, 149–150, Thompson, Elihu, 14, Three-phase circuits, 503–504, balanced delta-delta connection, 514–516, balanced delta-wye connection, 516–519, balanced three-phase voltages, 505–508, balanced wye-delta connection, 512–514, balanced wye-wye connection, 509–512, importance, 504, power in balanced system, 519–525, PSpice, 529–534, residential wiring, 540–542, three-phase power measurement, 535–540, unbalanced three-phase system, 525–528, Three-phase four-wire system, 504, Three-phase power measurement, 535–540, Three-phase transformer, 584–586, Three-stage cascaded connection, 191, Three-wattmeter method, 535, Three-wire type single-phase system, 504, Three-wire Y-Y system, 511
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ale29559_index.qxd, , 07/25/2008, , 12:41 PM, , Page I-11, , Index, , Tightly coupled, 566, Time constant, 256, Time-delay property, 681, Time differentiation, 682, 819–820, Time integration, 683–684, 820, Time periodicity, 684–685, Time scaling, 816–817, Time shift, 680–681, Toroidal inductor, 226, Total response, 275, Touch-tone telephone, 660–661, Transfer function, 614–617, 726–730, Transfer impedance, 852, Transformation ratio, 574, Transformer, air-core, 568, defined, 568, distribution, 595, ideal, 573–580, isolation, 575, isolation device, as, 592, linear, 567–573, matching device, as, 593, step-down, 575, step-up, 575, three-phase, 584–586, uses, 591, Transformer bank, 584, Transient response, 276, Transistor, 107, Transistor amplifier, 881, Transistor circuits, 880–885, Transmission parameters, 863–867, Transresistance amplifiers, 183, Trigonometric Fourier series, 757, Trigonometric identities, A–16 to A–17, Trimmer capacitor, 218, Turns ratio, 574, TV picture tube, 17, Two-phase three-wire system, 504, Two-port networks, 849–901, admittance parameters, 855–858, cascade connection, 873, defined, 850, hybrid parameters, 858–863, impedance parameters, 850–854, interconnection of networks, 871–877, ladder network synthesis, 885–889, parallel connection, 872, parameters. See Parameters, PSpice, 871–877, , reciprocal network, 852, series connection, 872, symmetrical network, 852, transistor circuits, 880–885, transmission parameters, 863–867, Two-sided Laplace transform, 677, Two-wattmeter method, 535, Two-wire type single-phase system, 504, , U, Unbalanced, 158, Unbalanced three-phase system, 525–528, Unbalanced three-phase Y-connected load, 525, Undamped natural frequency, 321, 323, Underdamped case, source-free parallel RLC circuit, 327, source-free series RLC circuit, 323–324, step response of parallel RLC circuit, 337, step response of series RLC circuit, 332, Unilateral Laplace transform, 677, Unit impulse function, 267–268, Unit impulse response, 727, Unit ramp function, 268, Unit step function, 266–267, Unity gain amplifier, 184, Unloaded source, 156, , V, VAR, 474, Variable capacitor, 218, Variable resistor, 32, Volt-ampere reactive (VAR), 474, Volta, Alessandro Antonio, 10, Voltage, 9–10, Voltage divider, 44, Voltage division, 391, Voltage division principle, 44, Voltage follower, 184, Voltmeter, 62, , W, Watson, James A., 715, Watson, Thomas A., 618, Wattmeter, 483, Westinghouse, George, 369, 505, Weston, Edward, 14, Wheatstone, Charles, 158, Wheatstone bridge, 158, , I-11
Page 1056 :
ale29559_index.qxd, , I-12, , 07/25/2008, , 12:41 PM, , Page I-12, , Index, , Wheatstone bridge circuit, 172, Wien bridge, 411, Wien-bridge oscillator, 439–441, Winding capacitance, 228, Winding resistance, 228, Wiring diagram of a room, 541, Wye-delta three-phase transformer connection, 585, Wye-to-delta conversion, 54–55, 392, Wye-wye three phase transformer connection, 584, , Y, y parameters, 855, , Z, z parameters, 851, Zero, 615, 620, 621, Zworykin, Vladimir K., 18
