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  • Slide 1
  • ELECTROMAGNETIC THEORY EKT 241/4: ELECTROMAGNETIC THEORY PREPARED BY: Saidatul Norlyana Azemi [email protected] CHAPTER 5 TRANSMISSION LINES
  • Slide 2
  • Chapter Outline General Considerations Lumped-Element Model Transmission-Line Equations Wave Propagation on a Transmission Line The Lossless Transmission Line Input Impedance of the Lossless Line Special Cases of the Lossless Line Power Flow on a Lossless Transmission Line The Smith Chart Impedance Matching Transients on Transmission Lines
  • Slide 3
  • General Considerations Transmission line a two-port network connecting a generator circuit to a load.
  • Slide 4
  • SoWhat is the use of transmission line?? A transmission line is used to transmit electrical energy/signals from one point to another i.e. from one source to a load. Types of transmission line include: wires, (telephone wire), coaxial cables, optical fibers n etcwirescoaxial cablesoptical fibers
  • Slide 5
  • The role of wavelength At low frequency, the impact is negligible At high frequency, the impact is very significant The impact of a transmission line on the current and voltage in the circuit depends on the: frequency, f of the signal provided by generator. length of line, l
  • Slide 6
  • Propagation modes Transverse electromagnetic (TEM) transmission lines waves propagating along these lines having electric and magnetic field that are entirely transverse to the direction of propagation Higher order transmission lines waves propagating along these lines have at least one significant field component in the direction of propagation Propagation modes Electric field lines Magnetic field lines
  • Slide 7
  • Propagation modes A few examples of transverse electromagnetic (TEM) and higher order transmission line
  • Slide 8
  • Lumped- element model A transmission line is represented by a parallel- wire configuration regardless of the specific shape of the line, (in term of lumped element circuit model) i.e coaxial line, two-wire line or any TEM line. Lumped element circuit model consists of four basic elements called the transmission line parameters : R, L, G, C. Series elementShunt element
  • Slide 9
  • Lumped- element model Lumped-element transmission line parameters: R : combined resistance of both conductors per unit length, in /m L : the combined inductance of both conductors per unit length, in H/m G : the conductance of the insulation medium per unit length, in S/m C : the capacitance of the two conductors per unit length, in F/m For example, a coil of wire has the property of inductance. When a certain amount of inductance is needed in a circuit, a coil of the proper dimension is inserted
  • Slide 10
  • Lumped- element model
  • Slide 11
  • Lumped- element model for 3 type of lines Note: , , pertain to the insulating material between conductors
  • Slide 12
  • Exercise 1: Use table 5.1 to compute the line parameter of a two wire air line whose wires are separated by distance of 2 cm, and, each is 1 mm in radius. The wires may be treated as perfect conductors with c = . R = ?, L=?, G=?, C=?
  • Slide 13
  • Solution exercise 1: c= c= c= c=
  • Slide 14
  • Slide 15
  • Slide 16
  • Exercise 2: Calculate the transmission line parameters at 1 MHz for a rigid coaxial air line with an inner conductor diameter of 0.6 cm and outer conductor diameter of 1.2 cm. The conductors are made of copper. ( c =0.999 1 ; c =5.8x10 7 ) f = 1MHz r1 = 0.006m/2 = 0.003m r2 = 0.012m/2 = 0.006m
  • Slide 17
  • Solution exercise 2:
  • Slide 18
  • BARE IN UR MIND From calculator
  • Slide 19
  • BARE IN UR MIND From calculator
  • Slide 20
  • Because, the material separating the inner and outer is perfect dielectric (air) with =0, thus G = 0 G : the conductance of the insulation medium per unit length, in S/m
  • Slide 21
  • Transmission line equations Complex propagation constant, the real part of - attenuation constant, unit: Np/m the imaginary part of - phase constant, unit: rad/m Is used to describes the voltage and the current across the transmission line in term of propagation constant and impedance
  • Slide 22
  • Transmission line equations The characteristic impedance of the line, Z 0 : Phase velocity of propagating waves: where f = frequency (Hz) = wavelength (m) = phase constant
  • Slide 23
  • Example 1 An air line is a transmission line for which air is the dielectric material present between the two conductors, which renders G = 0. In addition, the conductors are made of a material with high conductivity so that R 0. For an air line with characteristic impedance of 50 and phase constant of 20 rad/m at 700MHz, find the inductance per meter and the capacitance per meter of the line.
  • Slide 24
  • Solution to Example 1 The following quantities are given: With R = G = 0,
  • Slide 25
  • The ratio is given by: We get L from Z 0 Solution to Example 1 2
  • Slide 26
  • Lossless transmission line Lossless transmission line - Very small values of R and G. We set R=0 and G=0, hence: Transmission line can be designed to minimize ohmic losses by selecting high conductivities and dielectric material, thus we assume :
  • Slide 27
  • Transmission line equations Complex propagation constant, the real part of - attenuation constant, unit: Np/m the imaginary part of - phase constant, unit: rad/m 00
  • Slide 28
  • Lossless transmission line Lossless transmission line - Very small values of R and G. We set R=0 and G=0, hence: Transmission line can be designed to minimize ohmic losses by selecting high conductivities and dielectric material, thus we assume :
  • Slide 29
  • Lossless transmission line Using the relation properties between , , : Wavelength, Where r = relative permittivity of the insulating material between conductors
  • Slide 30
  • Exercise 3: For a losses transmission line, = 20.7 cm at 1GHz. Find r of the insulating material. =20.7cm 0.207m ; f=1 GHz 2
  • Slide 31
  • Exercise 4 A lossless transmission line of length 80 cm operates at a frequency of. The line parameters are & Find the characteristic impedance, the phase constant and the phase velocity. The condition apply that the line is lossless, So: R= 0 & G=0
  • Slide 32
  • characteristic impedance : phase constant: With R n G = 0 = 18.85 rad/m
  • Slide 33
  • phase velocity:
  • Slide 34
  • Voltage Reflection Coefficient Every transmission line has a resistance associated with it, and comes about because of its construction. This is called its characteristic impedance, Z 0. The standard characteristic impedance value is 50. However when the transmission line is terminated with an arbitrary load Z L, in which is not equivalent to its characteristic impedance (Z L Z 0 ), a reflected wave will occur.
  • Slide 35
  • Voltage reflection coefficient Voltage reflection coefficient, the ratio of the amplitude of the reflected voltage wave, V 0 - to the amplitude of the incident voltage wave, V 0 + at the load. Hence,
  • Slide 36
  • Voltage reflection coefficient The load impedance, Z L Where; = total voltage at the load V 0 - = amplitude of reflected voltage wave V 0 + = amplitude of the incident voltage wave = total current at the load Z 0 = characteristic impedance of the line
  • Slide 37
  • Voltage reflection coefficient And in case of a RL and RC series, Z L : Z L = R + j L ; Z L = R -1/ j C A load is matched to the line if Z L = Z 0 because there will be no reflection by the load ( = 0 and V 0 = 0. When the load is an open circuit, (Z L =), = 1 and V 0 - = V 0 +. When the load is a short circuit (Z L =0), = -1 and V 0 - = V 0 +.
  • Slide 38
  • What is the difference between an open and closed circuit? closed allows electricity through, and open doesn't. open circuit - Any circuit which is not complete is considered an open circuit. The open status of the circuit doesn't depend on how it became unclosed, so circuits which are manually disconnected and circuits which have blown fuses, faulty wiring or missing components are all considered open circuits. close circuit: A circuit is considered to be closed when electricity flows from an energy source to the desired endpoint of the circuit. A complete circuit which is not performing any actual work can still be a closed circuit. For example, a circuit connected to a dead battery may not perform any work, but it is still a closed circuit.
  • Slide 39
  • Example 2 A 100- transmission line is connected to a load consisting of a 50- resistor in series with a 10pF capacitor. Find the reflection coefficient at the load for a 100-MHz signal.
  • Slide 40
  • Solution to Example 2 The following quantities are given The load impedance is Voltage reflection coefficient is
  • Slide 41
  • In order to convert from ve magnitude for by replacing the ve sign with e -j180
  • Slide 42
  • Maths TIP 1 2
  • Slide 43
  • Exercise 5 A 150 lossless line is terminated in a load impedance Z L = (30 j200) . Calculate the voltage reflection coefficient at the load. Zo = 150 Z L = (30 j200)
  • Slide 44
  • Standing Waves Interference of the reflected wave and the incident wave along a transmission line creates a standing wave. Constructive interference gives maximum value for standing wave pattern, while destructive interference gives minimum value. The repetition period is for incident and reflected wave individually. But, the repetition period for standing wave pattern is /2.
  • Slide 45
  • Standing Waves For a matched line, Z L = Z 0, = 0 and = |V 0 + | for all values of z.
  • Slide 46
  • Standing Waves For a short-circuited load, (Z L =0), = -1.
  • Slide 47
  • Standing Waves For an open-circuited load, (Z L =), = 1. The wave is shifted by /4 from short-circuit case.
  • Slide 48
  • Standing Waves First voltage maximum occurs at: If r 0 n=0; If r 0 n=1 First voltage minimum occurs at: Where r = phase angle of
  • Slide 49
  • VSWR Voltage Standing Wave Ratio (VSWR) is ratio between the maximum voltage an the minimum voltage along the transmission line. VSWR provides a measure of mismatch between the load and the transmission line. For a matched load with = 0, VSWR = 1 and for a line with | | - 1, VSWR = . The VSWR is given by:
  • Slide 50
  • Example 3 is terminated A 50- transmission line is terminated in a load with Z L = (100 + j50). Find the voltage reflection coefficient and the voltage standing- wave ratio (VSWR).
  • Slide 51
  • Solution to Example 3 We have, VSWR is given by:
  • Slide 52
  • Exercise 6: A 140 lossless line is terminated in a load impedance Z L = (280 +j182) , if = 72cm, find a) Reflection coefficient, b) The VSWR, c) The locations of voltage maxima and minima
  • Slide 53
  • a) Reflection coefficient,
  • Slide 54
  • b) The VSWR;
  • Slide 55
  • The locations of voltage maxima and minima
  • Slide 56
  • Slide 57
  • Input impedance of a lossless line The input impedance, Z in is the ratio of the total voltage (incident and reflected voltages) to the total current at any point z on the line. or
  • Slide 58
  • Special cases of the lossless line For a line terminated in a short-circuit, Z L = 0: For a line terminated in an open circuit, Z L = :
  • Slide 59
  • Application of short-circuit and open-circuit measurements The measurements of short-circuit input impedance, and open-circuit input impedance, can be used to measure the characteristic impedance of the line: and
  • Slide 60
  • Length of line If the transmission line has length, where n is an integer, Hence, the input impedance becomes:
  • Slide 61
  • Quarter wave transformer If the transmission line is a quarter wavelength, with, where, we have, then the input impedance becomes:
  • Slide 62
  • Example 4 A 50- lossless transmission line is to be matched to a resistive load impedance with Z L =100 via a quarter-wave section as shown, thereby eliminating reflections along the feedline. Find the characteristic impedance of the quarter-wave transformer.
  • Slide 63
  • Quarter wave transformer If the transmission line is a quarter wavelength, with, where, we have, then the input impedance becomes:
  • Slide 64
  • Solution to Example 4 Z in = 50; Z L =100 Since the lines are lossless, all the incident power will end up getting transferred into the load Z L.
  • Slide 65
  • Matched transmission line For a matched lossless transmission line, Z L =Z 0 : 1) The input impedance Z in =Z 0 for all locations z on the line, 2) =0, and 3) all the incident power is delivered to the load, regardless of the length of the line, l.
  • Slide 66
  • Input Impedance, Z in Ratio of the total voltage to total current on the line When Z L =0(short circuit) When Z L = (open circuit) Application Be used to measure the characteristic impedance of the line : Application Be used to measure the characteristic impedance of the line : But, If the transmission line is Special case
  • Slide 67
  • Power flow on a lossless transmission line Two ways to determine the average power of an incident wave and the reflected wave; Time-domain approach Phasor domain approach Average power for incident wave; Average power for reflected wave: The net average power delivered to the load:
  • Slide 68
  • The time average power reflected by a load connected to a lossless transmission line is equal to the incident power multiplied by || 2 Power flow on a lossless transmission line
  • Slide 69
  • Exercise 7 For a 50 lossless transmission line terminated in a load impedance Z L = (100 + j50), determine the percentage of the average power reflected over average incident power by the load. Z 0 =50; Z L = (100 + j50)
  • Slide 70
  • Reflection coefficient, the percentage of the average incident power reflected by the load = 20%
  • Slide 71
  • Exercise 8 For the line of exercise previously (exercise 7), what is the average reflected power if |V 0 + |=1V
  • Slide 72
  • Smith Chart Smith chart is used to analyze & design transmission line circuits. Reflection coefficient, : r = real part, i = imaginary part Impedances on Smith chart are represented by normalized value, z L : the normalized load impedance, z L is dimensionless.
  • Slide 73
  • Smith Chart Reflection coefficient, A :0.3 + j0.4 Reflection coefficient, B :-0.5 - j0.2 In order to eliminate ve part, thus
  • Slide 74
  • The complex plane. A :0.3 + j0.4 B :-0.5 - j0.2
  • Slide 75
  • Smith Chart Reflection coefficient, : Since, becomes: Re-arrange in terms of z L : r L = Normalized load resistance x L = Normalized load admittance
  • Slide 76
  • The families of circle for r L and x L.
  • Slide 77
  • Plotting normalized impedance, z L = 2-j1
  • Slide 78
  • Input impedance The input impedance, Z in : is the voltage reflection coefficient at the load. We shift the phase angle of by 2 l, to get L. This will z L to z in. The | | is the same, but the phase is changed by 2 l. On the Smith chart, this means rotating in a clockwise direction (WTG).
  • Slide 79
  • Input impedance Since = 2/, shifting by 2 l is equal to phase change of 2 . Equating: Hence, for one complete rotation corresponds to l = /2. The objective of shifting to L is to find Z in at an any distance l on the transmission line.
  • Slide 80
  • Example 5 A 50- transmission line is terminated with Z L =(100-j50). Find Z in at a distance l =0.1 from the load. Normalized the load impedance Solution:
  • Slide 81
  • Solution to Example 5 z in = 0.6 j0.66 l =0.1 de normalize (multiplying by Z o ) Z in = 30 j33
  • Slide 82
  • VSWR, Voltage Maxima and Voltage Minima z L =2+j1 VSWR = 2.6 (at P max ). l max =(0.25-0.213) =0.037 . l min =(0.037+0.25) =0.287
  • Slide 83
  • VSWR, Voltage Maxima and Voltage Minima Point A is the normalized load impedance with z L =2+j1. VSWR = 2.6 (at P max ). The distance between the load and the first voltage maximum is l max =(0.25-0.213) =0.037 . The distance between the load and the first voltage minimum is l min =(0.037+0.25) =0.287.
  • Slide 84
  • Impedance to admittance transformations y L =0.25 - j0.6 z L =0.6 + j1.4
  • Slide 85
  • Example 6 Given that the voltage standing-wave ratio, VSWR = 3. On a 50- line, the first voltage minimum occurs at 5 cm from the load, and that the distance between successive minima is 20 cm, find the load impedance. Solution: The distance between successive minima is equal to /2. the distance between successive minima is 20 cm, Hence, = 40 cm
  • Slide 86
  • Solution to Example 6 Point A =VSWR = 3 de normalize (multiplying by Z o ) Z in = 30 j40
  • Slide 87
  • Solution to Example 6 First voltage minimum (in wavelength unit) is at on the WTL scale from point B. Intersect the line with constant SWR circle = 3. The normalized load impedance at point C is: De-normalize (multiplying by Z 0 ) to get Z L :
  • Slide 88
  • Exercise
  • Slide 89
  • Normalized the load impedance Solution: a) reflection coefficient from smith Chart
  • Slide 90
  • lmax lmin
  • Slide 91
  • 3)Move a distance 0.301 towards the generator (WTG) (refer to Smith chart) 0.301 + 0.082=0.383 At 0.383, read the value of which at the point intersects with constant circle, we have = z in = 0.72- j0.62. Denormalized it, hence = Z in = 72- j62 4)Distance from load to the first voltage maximum, (refer to Smith chart) 0.25-0.082=0.168
  • Slide 92
  • Impedance Matching Transmission line is matched to the load when Z 0 = Z L. This is usually not possible since Z L is used to serve other application. Alternatively, we can place an impedance- matching network between load and transmission line.
  • Slide 93
  • Single- stub matching Matching network consists of two sections of transmission lines. First section of length d, while the second section of length l in parallel with the first section, hence it is called stub. The second section is terminated with either short-circuit or open circuit.
  • Slide 94
  • Single- stub matching Y L =1/Z L Y d = Y 0 +j B feed line stub d l
  • Slide 95
  • Single- stub matching The length l of the stub is chosen so that its input admittance, Y S at MM is equal to jB. Hence, the parallel sum of the two admittances at MM yields Y 0, which is the characteristic admittance of the line. Y d = Y 0 +j B
  • Slide 96
  • Single- stub matching Thus, the main idea of shunt stub matching network is to: (i) Find length d and l in order to get y d and y l. (ii) Ensure total admittance y in = y d + y s = 1 for complete matching network.
  • Slide 97
  • Example 7 50- transmission line is connected to an antenna with load impedance Z L = (25 j50). Find the position and length of the short- circuited stub required to match the line. Solution: The normalized load impedance is: (located at A).
  • Slide 98
  • Solution to Example 7
  • Slide 99
  • Value of y L at B is which locates at position 0.115 on the WTG scale. Draw constant SWR circle that goes through points A and B. There are two possible matching points, C and D where the constant SWR circle intersects with circle r L =1 (now g L =1 circle).
  • Slide 100
  • C = 1+j1.58 A B D = 1+j1.58
  • Slide 101
  • Solution to Example 7 First matching points, C. At C, is at 0.178 on WTG scale. Distance B and C is Normalized input admittance at the juncture is: E is the admittance of short-circuit stub, y L =-j. Normalized admittance of j 1.58 at F and position 0.34 on the WTG scale gives:
  • Slide 102
  • F l1l1 d1d1 = 0.063 = 0.090 C = 1+j1.58 A B E F = -j1.58
  • Slide 103
  • First matching points, C Thus, the values are: d 1 = 0.063 l 1 = 0.09 y d1 = 1 + j1.58 y s1 = -j1.58 Where Y in = y d + y s = (1 + j1.58) + (-j1.58) = 1
  • Slide 104
  • Solution to Example 7 Second matching point, D. At point D, Distance B and D is Normalized input admittance at G. Rotating from point E to point G, we get
  • Slide 105
  • l2l2 d2d2 = 0.207 = 0.41 G B A G = +j1.58 E D = 1-j1.58
  • Slide 106
  • First matching points, D Thus, the values are: d 2 = 0.207 l 2 = 0.41 y d2 = 1 - j1.58 y s2 = +j1.58 Where Y in = y d + y s = (1 - j1.58) + (+j1.58) = 1
  • Slide 107
  • d1=0.063 l1=0.09, d2=0.207 l2=0.41