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Sby, , 13. Electromagnetic Waves and, Communication System
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❖ EM Waves : Waves that are caused by the acceleration of charged particles and consist of electric and magnetic fields vibrating sinusoidally, at right angles to each other and to the direction of propagation are called EM waves or EM radiations., , Sby
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Sby, , ❖ Characteristics of EM waves : ▪ The electric and magnetic fields , E and B are always perpendicular to each other and also to direction of propagation of the EM, waves., ▪ Electromagnetic waves transverse in nature ., ▪ The cross product E × B gives the direction in which the EM wave travels., ▪ The E and B fields vary sinusoidally and are in vibrate in phase., ▪ EM waves produced by accelerated electric charges. [ In nuclear transition or in the annihilation of an electron and a positron ], ▪ EM waves do not require any material medium for their propagation. They can travel through vacuum as well as through solids ,, liquids and gases., ▪ In free space , EM waves travel with velocity C, equal to that of light in free space. C = 3×108 m/s, 1, C = -----μ0Ɛ0, , = 3 × 108 m/s, , Where, μ0 = ( 4π × 10 -7 Tm / A ) is permeability, Ɛ0 = ( 8.85 × 10-12 C2/Nm2 ) is permittivity of free space ., , ▪ In a given material medium , the velocity ( vm) of EM waves is given by, 1, Vm = ---μƐ, , Where μ = Permeability ,, Ɛ = Permittivity of the given medium., , ▪ EM waves obey the principle of Superposition.
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Sby, ▪ The ratio of the amplitudes of electric and magnetic fields is always constant and it is equal to velocity of the electromagnetic, waves., 1, E0, =, ------ = 3 × 108 m/s, or c =, ----E 0 = c B0, μ0Ɛ0, B0, E0, , and B0 are the amplitudes of E and B respectively, , ▪ As the electric field vector E 0 is more prominent than the magnetic field vector B 0, it is responsible for optical effects due to, EM waves . For this reason electric vector is called light vector., ▪ The intensity of wave is proportional to the square of its amplitude and is given by the eq n, 1, Ɛ 0 E 02, ---IE =, 2, , 1 B02, IB = ---- ----2 μ0, , ▪ The energy of electromagnetic waves is equally distributed between the electric and magnetic field vector I E = IB, ▪ Electromagnetic waves can be polarized., , ▪ They exhibit phenomena of interference and diffraction., ▪ The electromagnetic waves obey the laws of reflection and refraction., , ▪ EM waves propagate in the form of time varying electric and magnetic field.
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❖ Electromagnetic Spectrum : The orderly distribution ( sequential arrangement ) of EM waves according to their wavelengths or frequencies in the form of, distinct groups having different properties is called the EM spectrum, , Sby
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❖ Radio Waves : Production –, •, , Radio waves are produced by accelerated motion of charge in a conducting wire., , •, , The frequency of waves produced by the circuit depends upon the magnitudes of the inductance and the, capacitance., Thus, by choosing suitable values of the inductance and the capacitance , radio waves of desired frequency can be produced., , •, , Properties : •, , They have very long wavelength ranging from a few centimeters to a few hundreds of kilometers., , •, , The frequency range of AM Band – 530KHz to 1710 KHz, , •, , Frequency of the wave used for TV transmission 54 MHz to 890 MHz, , •, , FM radio 88 MHz to 108 MHz, , Uses –, •, , Wireless communication purpose., , •, , Radio broadcasting and transmission of TV signals., , •, , In cellular phones uses radio waves to transmit voice communication in the ultra high frequency [ UHF ] band., , Sby
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❖ Microwaves : Production : •, , These waves are discovered of by H.Hertz in 1888., , •, , Microwaves are produced by oscillator electric circuits containing a capacitor and an inductor., , •, , They can be produced by special vacuum tubes., , Properties :•, , They heat certain substances on which they are incident., , •, , They can be detected by crystal detectors., , Uses :•, , Transmission of TV signals., , •, , Long distance telephone communication., , •, , Microwave ovens are used for cooking., , •, , Used in radar system for the location of distant object like ships , airplane's, , •, , They are used in the study of atomic and molecular structure., , Sby
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❖ Infrared waves :Production : •, , These waves were discovered by William Herschel [ 1737 – 1822 ] in 1800., , •, , All hot bodies are sources of infrared rays., , •, , About 60 % of the solar radiations are infrared in nature., , •, , Thermocouples Thermopile and bolometers are used to detect infrared rays., , Properties : •, , When infrared rays are incident an any object the object gets heated., , •, , These rays are strongly absorbed by glass., , •, , They can penetrate through thick columns of fog mist and cloud cover., , Uses :•, , Used in remote sensing., , •, , Used in diagnosis of superficial tumours and varicose veins., , •, , Used in solar water heater and cookers., , •, , Used to keep green house warm., , •, , Used in remote controls of TV , VCR., , •, , Infrared binoculars and thermal imaging cameras are used in military applications for night vision., , Sby
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Sby, , ❖ Visible Light : •, , It is the most familiar form of EM waves., , •, , These waves are detected by human eye. Therefore this wavelength range is called the visible light., , •, , The visible light is emitted due to atomic excitations., , •, , Visible light is emitted or reflected from objects around us provides us information's about those objects and hence about, the surroundings., , •, , Different wavelengths give rise to different colours. As shown in the table given below., , Colours, , Wavelength, , Violet, , 380-450 nm, , Blue, , 450-495 nm, , Green, , 495-570 nm, , Yellow, , 570-590 nm, , Orange, , 590-620 nm, , Red, , 620-750 nm
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❖ Ultraviolet rays :-, , Sby, , Productions :•, , Ultraviolet light can be produced by the mercury vapour lamp, electric spark and carbon arc lamp., , •, , They can also be obtained by striking electrical discharge in hydrogen and xenon gas tubes., , •, , The sun is the most important natural source of ultraviolet rays , most of which are absorbed by the ozone layer in the earths, atmosphere., , Properties : •, , They produce fluorescence in certain materials such as phosphors ., , •, , They cause photoelectric effect. When they are incident on metal surfaces,electronsare emitted by these surfaces., , •, , Ultraviolet rays cannot pass through glass but they can pass through quartz ,fluorite, rock salt etc., , •, , They possess the property of synthesizing vitamin D. When skin is exposed to them., , Uses :•, , UV rays destroy germs and bacteria and hence they are used for sterilizing surgical instruments and for purifications of water., , •, , Used in burglar alarms and security systems., , •, , Used to distinguish real and fake gems., , •, , Used to detect forgery., , •, , Used in analysis of chemical compounds.
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❖ X – rays : - [1895 ] W.C Rontgen., Production :•, , German physicist W.C Rontgen discovered x-rays while studding cathode rays. X- rays also called Rontgen rays., , •, , Cathode ray is a stream of electrons emitted by the cathode in a vacuum tube., , •, , X – ray are produced when cathode rays are suddenly stopped by an obstacle, , Properties :•, , They are high energy E.M. waves., , •, , They are not deflected by electric and magnetic fields., , •, , X-rays ionize the gases through which they pass., , •, , They have high penetrating power., , •, , Their over dose can kill living plant and animal tissues and hence are harmful., , Uses :•, , X- rays are used to study the structure of crystals., , •, , X-rays photographs are useful to detect bone fracture . X-rays have many other medical uses such as CT scan., , •, , X-rays are used to detect flaws or cracks in metals., , •, , These are used for detection of explosives opium etc., , •, , X-rays are used to skin diseases and to destroy tumours in the body of a patient., , •, , They are used to distinguish real diamonds , gems from artificial ones, , Sby
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Sby, , ❖ Propagation of EM waves :-, , Earth and atmospheric layer, , Different layers of the atmosphere, , Atmosphere : •, , The earth is surrounded by various layer of gases. The envelope of these gases around the earth is called atmosphere., , •, , Earth atmosphere is due to gravitational attraction of earth.
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❖ Troposphere : •, , It is the lower layer of the earth atmosphere close to earths surface., , •, , It extends up to about 12 km above the surface of the earth., , •, , This layer plays an important role in the weather phenomenon that affects our environment., , •, , Most of the weather vapour in the atmosphere is present in the, troposphere . It is the source of air., , •, , Dust, smoke, pollen grains , salt organic materials are present in it, , •, , The temperature of this layer decreases form about 280 k 220 k., , Sby
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Sby, ❖ Stratosphere : -, , •, , IT is the next layer to troposphere which extends to about 50 km above the surface of, the earth., , •, , It contains very little moistures and dust., , •, , The region between 15km to 50 km in stratosphere contains ozone gas. This layer is called, ozone layer.
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❖ Mesosphere : •, , The layer above the stratosphere is called the mesosphere., , •, , It extends from 50 km to 80 km above the earth surface, , •, , At a height of 80 km temperature is about 290 k to 180k Ionosphere, , •, , In this layer the temperature of the layer decrease's with increase in height., , Sby
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Sby, ❖ Ionosphere, •, , It is the part of atmosphere that extends from 80 km to thousand of KM, , •, , Beyond the mesosphere , the temperature begins to rise due to the partial, absorption of solar radiation by the molecules of air . This layer is called, thermosphere., , •, , This layer plays an important role in radio communication and, telecommunications.
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❖ Ground wave propagation : •, , When a radio wave from transmitting antenna propagates near surface of the earth , so as to reach the receiving antenna , the, wave propagation is called ground wave or surface wave propagation., , •, , In this mode , radio waves travel close to the surface of the earth and move along its curved surface from transmitter to receiver., , •, , The radio waves induce currents in the ground and loss their energy by absorption . Therefore the signal can not be transmitted, over large distances., , •, , Radio waves having frequency less than 2 MHz [ in the medium frequency band ] are transmitted by ground wave propagation., , •, , This is suitable for local broadcasting only for TV or FM signal ( very high frequency) ground wave propagation can not be used., , Applications Ground Wave Propagation, •, , These can be used for one-way communication from the military to submerged submarines as they penetrate to a significant, depth into seawater., , •, , AM, FM and television broadcasting can be done with the help of ground waves., , Sby
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Sby, ❖ Space wave propagation : •, , When the radio waves from the transmitting antenna reach the receiving antenna either directly along a straight line or after, reflection from the ground or satellite or after reflection from troposphere , the wave propagation is called space wave propagation., , •, , The radio waves reflected from troposphere are called tropospheric waves., , •, , Radio waves with frequency greater than 30 MHz can pass through the ionosphere [ 60Km to 1000 Km ] after suffering a small, deviation .Hence these waves can not be transmitted by space wave propagation except by using a satellite., , •, , Also for TV signals which have high frequency transmission over long distance is not possible by means of space wave propagation., , Range of the signal : •, , The maximum distance over which a signal can reach is called its range., , •, , For large TV coverage , the height of the transmitting antenna should be as large as possible . This is the reason why the transmitting, and receiving antennas are mounted on top of high rise buildings., , •, , Range is the straight line distance from the point of transmission [ the top of the antenna ] to the point on earth where the wave will, hit while travelling along a straight line.
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•, , Let the height of the transmitting antenna ( AA’ ) situated at A be h . B represents the point on the surface of the earth at which the, space wave hits the earth., The trianglr OA’B is a right angle triangle, From ΔOA’B, (OA’) = A’B2 + OB2, (R + h )2 = d2 + R2, R2 + h2 + 2Rh = d2 + R2, As h << R , neglecting h2, d = 2𝑅ℎ, •, , The range can be increased by mounting the reaceiver at a height h’ at point C on the surface of the earth., The range increases to d + d’, Where d’ is 2Rh’, Thus ,, Total range = d + d’, = 2𝑅ℎ + 2𝑅ℎ’, , Sby
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Sby, ❖ Sky wave propagation : •, , When radio waves from a transmitting antenna reach the receiving antenna after reflection in the ionosphere , the wave propagation is, called sky wave propagation., , •, , The sky wave include waves of frequency between 3MHz and 30 MHz., , •, , These waves can suffer multiple reflections between the ionosphere and the earth. Therefore they can be transmitted over large, distances., , Critical frequency :Critical frequency is the maximum value of the frequency of radio wave which can be reflected back to the earth from the, ionosphere when the waves are directed normally to ionosphere., Skip distance : Skip distance is the shortest distance from a transmitter measured along the surface of the earth at which a sky wave o f fixed, frequency will be returned to the earth so that no sky waves can be received within the skip distance.
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Sby, ❖ Introduction to communication system :-, , •, , There are three basic elements of every communication system, 1 . Transmitter, 2 . Communication channel, 3 . Receiver, , •, , In a communication system , the transmitter is located at one place and the receiver at another place., , •, , The communication channel is a passage through which signals transfer in between a transmitter and a receiver., , •, , This channel may be in the form of wires or cables or may also be wireless , depending on the types of communication system.
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Sby, , ❖ Different Modes of communications :•, , There are two basic modes of communications, 1 . Point to Point Communications, 2 . Broadcast Communications, , •, , In point to point communication mode communication takes place over a link between a single transmitter and a receiver, e.g. : - telephony., , •, , In the broadcast mode , there are large number of receivers corresponding to the single transmitter., e.g. : - Radio and Television transmission.
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❖ Commonly used terms in electronics communication System : • Signal : The information converted into electrical form that is suitable for transmission is called a signal., In a radio station , music and speech are converted into electrical form by a microphone for transmission into space . This, electrical form of sound is the space., A signal can be ,, , 1 ] Analog Signal, 2 ] Digital Signal., , 1 ] Analog Signal :A continuously varying signal [ voltage and current ] is called an analog signal., Ex . Sound and picture signals in TV are analog in nature., , 2] Digital Signal : A signal [ voltage and current ] that can have only two discrete values is called a digital signal., Ex . A square wave is a digital signal .It has two values viz, +5V and 0 V, , Sby
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Sby, • Transmitter : A transmitter converts the signal produced by a source of information into a form suitable for transmission through, a channel and subsequent reception., , • Transducer : A device that converts one form of energy into another form of energy is called a transducer., A loudspeaker is a transducer which converts electrical energy into sound energy., , • Receiver : The receiver receives the messages signal at the channel output , reconstructs it in recognizable form of the original message, for delivering into the user of information., , • Attenuation : The loss of strength of the signal while propagating through the channel is known as attenuation., It occurs because the channel distorts , reflects and refracts the signals as it passes through it.
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Sby, • Amplification : Amplification is the process of raising the strength of the signal, using an electronic circuits called amplifier., , • Range : The maximum distance between a source and a destination up to which the signal can be received with sufficient strength, , is termed as range., , • Repeater : It is a combination of a transmitter and a receiver receives the signal from the transmitter , amplifies it and transmits it to the, next repeater . Repeaters are used to increase the range of a communication system., , • Noise : A random unwanted signal is called noise., The source generating the noise may be located inside or outside the system . Efforts should be made to minimize the, noise level in a communication system., , • Bandwidth : The bandwidth of an electronic circuits is the range of frequencies over which it operates efficiently.
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Sby, • Modulation : The signals in communication system are low frequency signals and can not be transmitted over large distances, it is, superimposed on a high frequency wave [ called carrier wave ] . This process is called modulation., , • Demodulation : The process of regaining signal from a modulated wave is called modulation. This is the reverse process of modulation.
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1. Calculate the frequency in MHz of a radio wave of, wavelength 250 m . Remember that the speed of all EM waves in, vacuum is 3.0 × 108 m/s ., , Ans : - Given data ,, , λ = 250 m, , c = 3 × 108 m/s, , Frequency ν = ?, , Frequency ν = ?, , c=νλ, 𝑐, 3 ×108, ν = λ = 6.5 × 10 −7, , c=νλ, ν=, , 𝑐, λ, , =, , 3 ×108, 250, , = 1.2 × 106 MHz, The frequency of the radio wave is 1.2 MHz, , 2. Calculate the wavelength in nm of an X – ray wave of, frequency 2.0 × 1018 Hz ., , Formula , c = ν λ, , λ=, , =, , = 1.5 ×, , 4 . Calculate the wavelength of a microwave of frequency 8.0, GHz., , ν = 8 GHz = 8 × 10 9 Hz, C = 3 × 108 m/s, Wavelength λ = ?, Formula , c = ν λ, , C = 3 × 108 m/s, Wavelength λ = ?, 3 × 108, 2 × 1018, , = 4.6 × 10 14 Hz, The frequency of the red light is 4.6 × 10 14 Hz, , Ans : - Given data ,, , Ans : - Given data , ν = 2 × 1018 Hz, , 𝑐, ν, , Ans : Given data , λ = 6.5 × 10 -7 m, , c = 3 × 108 m/s, Formula ,, , Sby, , 3. The speed of light is 3 × 108 m/s . Calculate the frequency of, red light of wavelength of 6.5 × 10-7 m., , 10-10, , λ=, , 𝑐, ν, , =, , 3 × 108, 8 × 109, , = 3.75 × 10 - 2, = 3.75 cm, , = 0.15 nm, The wavelength of an X- ray is 0.15 nm, , The wavelength of microwave is 3.75 cm
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5. In a EM wave the electric field oscillates sinusoidally at a, frequency of 2 × 1010 Hz. What is the wave length of the wave ?, , 7 . A radio can be tune into any station in the 7.5 MHz band ., What is the corresponding wavelength band ., , Ans : -, , Ans : -, , Given data , ν1 = 7.5 MHz = 7.5 × 10 6 Hz, , ν = 2 × 1010 Hz, , Given data ,, , ν2 = 12 MHz = 12 × 10 6 Hz, , C = 3 × 10 8 m/s, Wavelength λ = ?, , Wavelength band λ = ?, , Formula , c = ν λ, , λ=, =, , 𝑐, ν, 3 × 108, 2 × 1010, , = 1.5 × 10-2, , = 1.5 × 10-2 m, The wavelength of the wave is 1.5 × 10 -2 m ., 6. A radio wave of frequency of 1.0 × 107 Hz propagates with, speed 3 × 108 m/s. Calculates its length., , Ans : -, , Hz, Given data , ν = 1 ×, C = 3 × 10 8 m/s, Wavelength λ = ?, 107, , 𝑐, ν, , =, , 3 × 108, 1.0 × 107, , λ=, , λ1 =, , 3 × 108, 7.5 × 106, , λ2 =, , 3 × 108, 12 × 106, , 𝑐, ν, , = 40 m, = 25 m, , The corresponding wavelength band is 40 m to 25 m ., 8. The amplitude of the magnetic field part of a harmonic Em, wave in vacuum is B0 = 5 × 10-7 T . What is the amplitude of the, electric field part of the wave ?, , Ans : B0 = 5 × 10 -7 T, , Given data ,, , C = 3 × 10 8 m/s, , Amplitude of electric field ( E0 ) = ?, Formula , c =, , Formula , c = ν λ, , λ=, , Sby, , = 30 m, , The wavelength of the radio wave is 30 m ., , 𝐸0, 𝐵0, , E0 = c × B0, = 3 × 108 × 5 × 10-7, , = 150 V / m, , The amplitude of electric field is 150 V / m
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Sby, 7 . A TV tower has a height of 200 m. How much population is, covered by TV transmission if the average population density, around the tower is 1000/km2 ? ( Radius of the earth = 6.4 × 106 m ), , 8. Height of a TV tower is 600 m at a given place. Calculate its, coverage range if the radius of the earth is 6400 km . What, should be the height to get the double coverage area ?, , Ans : -, , Ans : -, , Given data ,, h = 200 m, Population density ( n ) = 1000/km2 = 1000 10 -6/m2 = 10-3/m2, , Given data ,, h = 200 m, , R = 6.4 ×, m, Population covered = ?, , R = 6.4 × 106 m, Range d = ?, Height to get the double coverage (h’) = ?, , Formulae , 1 . A = π d2 = π ( 2𝑅ℎ )2 = 2πRh, , Formulae , d =, , 106, , 2 . Population covered = n A, , d=, , 1 . A = 2πRh, = 2 × 3.14 × 6.4 × 106 ×200, = 8 × 10, , 9, , m2, = 10-3 × 8 × 109, = 8 × 106, , The population of 8 ×, , = 87.6 km, , π, , (d’2), , =2(, , πd2), , (d’)2 = 2 d2, h =, , 106 is, , 2 × 600 × 6.4 × 106, , = 87.6 × 10 3 m, Now, for A’ = 2A, , 2 . Population covered = n A, , 2𝑅ℎ, , covered by TV transmission ., , =, , 𝑑′ 2, 2𝑅, 2𝑑2, 2𝑅, , = 2×h, = 2 × 600, h = 1200 m, The coverage range is 87.6 km, and the height of antenna to, double the area should be 1200 m.
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9 . A transmitting antenna at the top of a tower has a height 32 m, and that of the receiving antenna is 50 m. What is the maximum, distance between them for satisfactory communication in line of, sight mode ? Given radius of earth is 6.4 × 106 m ., , Ans : Given data ,, h t = 32 m, , h r = 50 m, , R = 6.4 × 106 m, Maximum distance or range d = ?, Formulae ,, d=, , 2𝑅ℎ, , d t = 2𝑅ℎ𝑡, =, , 2 × 32 × 6.4 × 106, , = 20.238 × 10 3 m, , = 20.238 km, dr =, =, , 2𝑅ℎ𝑟, 2 × 50 × 6.4 × 106, , = 25.298 × 10 3 m, = 25.298 km, , d = dr + dt, d = 20.238 + 25.298, d = 45.536 km, The maximum distance for, satisfactory communication, is 45.536 km., , Sby
