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EELE 3332 – Electromagnetic II Chapter 11 Transmission Lines Islamic University of Gaza Electrical Engineering Department Dr. Talal Skaik 201 1 11.1 Introduction Wave propagation in unbounded media is used in radio or TV broadcasting, where the information being transmitted is meant for everyone who may be interested. Another means of transmitting power or information is by guided structures. Guided structures serve to guide (or direct) the propagation of energy from the source to the load. Typical examples of transmission lines such and structures are waveguides. Waveguides are discussed in the next chapter; transmission lines are considered in this chapter. 2 Introduction Transmission lines are commonly used in power distribution (at low frequencies) and in communications (at high frequencies). A transmission line basically consists of two or more parallel conductors used to connect a source to a load. Typical transmission lines include coaxial cable, a two-wire line, a parallel-plate line, and a microstrip line. Coaxial cables are used in connecting TV sets to TV antennas. Microstrip lines are particularly important in integrated circuits. Transmission line problems are usually solved using EM field 3 theory and electric circuit theory, the two major theories on Introduction Figure 11.1 Typical transmission lines in cross-sectional view: (a) coaxial line, (b) two-wire line, (c) planar line, (d) wire above conducting plane, (e) microstrip line. 4 11.2 Transmission Line Parameters Transmission line parameters are: R: Resistance per unit lengt(Ω/m) L: Inductance per unit length. (H/m) G: Conductance per unit length. (S/m) C: Capacitance per unit length. (F/m) Distributed parameters of a twoconductor transmission line 5 Transmission Line Parameters 6 Transmission Line Parameters The line parameters R, L, G, and C are uniformly distributed along the entire length of the line. For each line, the conductors are characterized by σc,µc,εc,=ε0, and the homogeneous dielectric separating the conductors is characterized by σ,µ,ε. G ≠1/R; R is the ac resistance per unit length of the conductors comprising the line and G is the conductance per unit length due to the dielectric medium separating the conductors. For each line: LC and G C 7 Fields inside transmission line •Transmission lines transmit TEM waves. •V proportional to E, •I proportional to H 8 11.3 Transmission Line Equations Two-conductor transmission lines support a TEM wave; E and H are perpendicular to each other and transverse to the direction of propagation. E and H are related to V and I: V E.dl, I= H.dl Using V and I in solving the transmission line problem is simpler than solving E and H (requires Maxwell’s equations). Examine an incremental portion of length Δz of a two- conductor transmission line. 9 Transmission Line Representation 10 Transmission Line I (z, t) Equations Using KVL:- V (z, t) RzI (z, t) Lz V (z z,t) t V (z z,t) V (z, t) I (z, t) RI (z, t) L or z t Taking the limit as z 0 leads to: V (z, t) I (z, t) RI (z, t) L z t equivalent circuit model of a two-conductor T.L. of differential length z. 11 NNNNNNNN Using KCL:- I (z, t) I (z z, t) I V (z z,t) t I (z z, t) I (z, t) V (z z,t) GV (z z, t) C or z t Taking the limit as z 0 leads to: I (z, t)=I (z z, t) GzV (z z, t) Cz I (z, t) V (z, t) GV (z, t) C z t 12 Transmission Line Equations The time domain form of the transmission line equations: V (z, t) I (z, t) RI (z, t) L z t I (z, t) V (z, t) GV (z, t) C z t If we assume harmonic time dependence so that: V(z, t)=Re[Vs (z)ejt ] I(z, t)=Re[Is (z)e jt ] where Vs and Is are the phasor forms of V (z, t) and I (z, t), dVs (R jL)I s dz dIs (G jC)Vs dz 13 Transmission Line Equations dVs dIs (R j L)I s , (G jC)Vs dz dz To solve the previous equations, take second derivative of Vs gives d 2V s (R jL)(G jC)Vs 2 dz Now take second derivative of Is gives d2 Is (G jC)(R jL)Is 2 dz Hence, the wave equations for voltage and current become d 2Vs 2 Vs 0 2 dz , where j (R jL)(G jC) d 2I s 2 Is 0 2 14 dz j (R jL)(G jC) : is the propagation constant : attenuation constant (Np/m or dB/m) : phase constant ( rad/m) 2 = wavelength is: , wave velocity is: u f The solutions to the wave equations are: Vs V0e z +z V0e z -z , I s I 0e z +z I 0e z -z where V0 ,V0 , I 0 , I 0 are wave amplitudes. sign wave traveling along +z direction. - sign wave traveling along -z direction. 15 Transmission Line Equations Vs V0e z V0e z I s I 0e z I 0e z In time domain: V (z, t) Re[Vs (z)e j t ] V (z,t) V0 e z cos(t z) V0 e z cos(t z) Similarly for current: I (z, t) I 0 e z cos(t z) I 0 e z cos(t z) 16 Characteristic Impedance, Z0 The Characteristic Impedance Z0 of the line is the ratio of the positively travelling voltage wave to the current wave at any point on the line. V (z) V e z , I (z) I e z 0 0 dV (z) since (R jL)I (z), dz ( V0 e z ) (R jL)I 0 e z V V R jL R jL 0 Zo Ro jX o 0 I0 I0 G jC R jL Ro jX o , Zo G jC Characterestic Admittance 1 Y0 Z0 17 Lossless Line A transmission line is said to be lossless if the conductors of (R=G=0) the line are perfect (σc ≈ ∞) and the dielectric medium separating them is lossless (σ ≈ 0) For lossless line, R=G=0 Since j = (R jL)(G jC) 0, j , LC 2 1 u f , LC Xo 0 Zo Ro L C 18 Distortionless Line Any signal that carries significant information must has some non-zero bandwidth. In other words, the signal energy (as well as the information it carries) is spread across many frequencies. If the different frequencies that comprise a signal travel at different velocities, that signal will arrive at the end of a transmission line distorted. We call this phenomenon signal dispersion. Recall for lossless lines, however, the phase velocity is independent of frequency—no dispersion will occur! u 1/ LC Of course, a perfectly lossless line is impossible, but we find phase velocity is approximately constant if the line is low-loss. 19 Distortionless Line (R/L=G/C) A distortionless line is one in which the attenuation constant α is frequency independent while the phase constant β is linearly dependent on frequency. A distorionless line results if the line parameters are such that R G L C 1 jL 1 jC RG = (R jL)(G jC)= R G α does not depend on jC = RG 1 j frequency, whereas β is a G or RG , LC linear function of frequency. Thus, u / 1/ LC (frequency independent) 20 R Z0 G Distortionless Line R 1 jL/ R R j(R/L=G/C) L jC G 1 jC / G G L C (Real) 1 u= LC Notes: Shape distortion of signals happen if α and u are frequency dependent. u and Z0 for distortionless line are the same as lossless line. A lossless line is also a distortionless line, but a distortionless line is not necessarily lossless. Lossless lines are desirable in power transmission, and telephone 21 lines are required to be distortionless. Distortionless Line – Practical use To achieve the required condition of R/L=G/C for a transmission line, L may be increased by loading the cable with a metal with high magnetic permeability (μ). A common practice is to replace repeaters in long lines to maintain the desired shape and duration of pulses for long distance transmission. 22 Summary General Lossless Distortionless Zo (R jL)(G jC) R jL Zo G jC 0 j LC L Z o Ro C RG j LC Zo Ro L C Example An air line has characteristic 11.1 impedance of 70 Ω and a phase constant of 3 rad/m at 100 MHz. Calculate the inductance per meter and the capacitance per meter of the line. An air line can be regarded as lossless line because and c . Hence 0 and =0 RG0 L Z0 =R0 = C LC Deviding the two equations yields: R0 1 C 3 or C 68.2 pF 6 R0 2 10010 (70) L=R02C (70)2 (68.2 1012 ) 334.2 nH/m 2 4 Example A distortionless line has11.2 Z0=60 Ω, α=20 mNp/m, u=0.6c, where c is the speed of light. Find R,L,G , C and λ at 100 MHz. RC A distortionless line has RC GL or G ,u L 1 LC L C R , = RG R Z0 C L Z0 R Z0 (2010 3 )(60) 1.2/m 2 400106 Since = RG G 333 S/m R 1.2 L 1 gives by u Dividing Z 0 C LC 60 Z0 333 nH/m L= 8 u 0.6(310 ) 25 Example 11.2 – solution continued L by u Multiplying Z 0 C uZ 0 = 1 gives LC 1 1 1 C 92.59 pF/m 8 C uZ 0 0.6(310 )60 u 0.6(3108 ) 1.8 m = 8 f 10 26 11.4 Input impedance, standing wave ratio, powera transmission line of length l, characterised by and Z0, Consider connected to a load ZL. Generator sees the line with the load as an input impedance Zin. 27 Input impedance Vs (z) V0e z V0e z V0 I s (z) e Z0 z V0 Z0 e z , V0 V0 Z0 = I0 I0 At generator terminals (sending end): Let V0 V (z 0), I0 I (z 0), Substitute in prev. equs.: 1 V V0 Z 0 I 0 2 ... (1) V0 V0 I0 1 V0 V0 Z 0 I 0 Z0 Z0 2 If the input impedance at the terminals is Zin , V0 V0 V0 Zin then V0 Vg , Z in Z g 0 I0 Vg Z in Z g 28 Input impedance Vs (z) V0 e z V0 e z , V V I s (z) 0 e z 0 e z Z0 Z0 At the load: Let VL V (z l), I L I (z l), Substitute in prev. equs. VL V0e l V0e l V I L 0 e l Z0 V0 Z0 l e 1 V VL Z 0 I L e l 2 ... (2) 1 V0 VL Z 0 I L e l 2 0 Now determine the input impedance Zin =Vs (z) / I s (z) atany point on the line. 29 Input impedance V0 V At the generator, recall V0 V0 V0 , I0 0 , then Z0 Z0 Z 0 V0 V0 Vs (z) 0V Zin = I s (z) I 0 V0 V 0 Substituting eq. 2 and utilizing the fact that: e l e l 2 cosh l, e l e l 2 sinh l , l l e e sinh l or tanh l cosh l e l e l we get Z L Z0 tanh l Z in Z 0 (General - Lossy Line) Z 0 Z L tanh l 30 Input impedance (Lossless Line) Z L Z0 tanh l (General - Lossy Line) Z in Z 0 Z 0 Z L tanh l For a lossless line, =j , tanh j l j tan l, then Z L jZ0 tan l Z Z in (Lossless Line) 0 Z 0 jZ L tan l βl is known as electrical length, in degrees or radians Note: To find Zin at a distance l ' from load, replace l by l ' :- Z L jZ0 tan l ' Z in Z 0 Z jZ tan l ' L 0 31 Reflection Coefficient, (at load) Define L as the voltage reflection coefficient (at the load), as the ratio of the voltage reflection wave to the incident wave at the load, l V e 0 L l V0 e 1 VL Z 0 I L el 2 Since 1 V0 VL Z 0 I L e l 2 V0 L , and VL Z L I L Z L Z0 (Voltage Reflection coefficient at load) ZL Z0 32 Reflection Coefficient, (at generator) Define 0 (at z 0) as the voltage reflection coefficient at the source, as the ratio of the voltage reflection wave to the incident wave at source, 0 V e V 0 0 0 0 V0 e V0 1 V V0 Z 0 I 0 2 Since , and V0 Zin I0 1 V0 V0 Z 0 I 0 2 0 0 Zin Z0 (Voltage Reflection coefficient at source) Z in Z 0 33 Reflection Coefficient The voltage reflection coefficient at any point on the line is the ratio of the reflected voltage wave to that of the incident wave. z V e V That is: (z) 0 z 0 e 2 z V0 e V0 The current reflection coefficient at any point on the line is the negative of the voltage reflection coefficient at that point. Thus the current reflection coefficient at the load is I e l / I e l 0 0 L 34 Standing Wave Ratio Whenever there is a reflected wave, a standing wave will form out of the combination of incident and reflected waves. The standing wave ratio s is defined as: (as we did for plane waves) Vmax Imax 1 L s= Vmin Imin 1 L Z L Z0 L ZL Z0 When load is perfectly matched (Z L Z0 ) Total Transmission L 0 s 1 When load is a short circuit : Total Reflection L 1 L 1 s When load is an open circuit : Total Reflection L 1 L 1 s 35 Power The time-average power flow along the line at the point z is: Pave 1 Re[Vs(z)I s* (z)]. 2 Pave 2 0 V 2Z 0 1 , 2 For a lossless line, this can be reduced to: P V ave 0 2 / 2Z V 2 0 Incident Power (Pi) 2 0 / 2Z 0 Reflected Power (Pr) The average power flow is constant at any point on the lossless line. The total power delivered to the load (Pav ) is equal to the incident power ( V0 If 2 2 0 / 2Z 0 ) minus the reflected power ( V / 2Z 0 ) 2 0, maximum power is delivered to the load, while no power is delivered for 1. The above discussion assumes that the generator is matched. 3 6 Special Cases , ZL=0, ZL=∞, ZL=Z0 Short Circuited Line (ZL=0) Open Circuited Line (ZL=∞) Matched Line (ZL=Z0) 37 Shorted Line (ZL=0) Z L jZ 0 tan l Zin Z0 jZ 0 tan l Z 0 jZ L tan l Z L Z0 L 1, s ZL Z0 (Total Reflection) 38 Open-Circuited Line (ZL=∞) Z in Z 0 Z L jZ0 tan l , Z0 jZL tan l (Z L ) Z0 1 j Z tan l L Z 0 jZ cot l Z in Z 0 0 j tan l Z0 j tan l ZL L Z L Z0 1, s ZL Z0 (Total Reflection) 39 Matched Line (ZL=Z0) Most desired case from practical point of view. Since Zin Z0 Zin Z 0 Z L Z0 0, s 1 ZL Z0 The whole wave is transmitted, and there is no reflection. L The incident power is fully absorbed by the load. (Maximum power transfer) 40 Example 11.3 A certain transmission line 2 m long operating at ω=106 rad/s has α=8 dB/m, β=1 rad/m, and Z0= 60+j40 Ω. If the line is connected to a source of 10∟00 V, Zg=40 Ω and transmitted by a load of 20+j50 Ω, determine (a) The input impedance (b) The sending end current (c) The current at the middle of the line. Solution (a) Since 1 Np=8.686 dB 8 = 0.921 Np / m 8.686 =+j 0.921 j1 l=2(0.921 j1) 1.84 j2 41 Example 11.3 – Solution continued tanh l tanh 1.84 j2 1.033 j0.03929 Z L Z0 tanh l Z in Z 0 Z Z tanh l L 0 20 j50 (60 j40)(1.033 j0.03929) (60 j40) 60 j40 (20 j50)(1.033 j0.03929) Z in Zin 60.25 j38.79 (b) The sending end current is I (z 0) I 0 Vg 10 I0 93.03 21.15 mA Z in Z g 60.25 j38.79 40 42 Example 11.3 – Solution continued (c) To find the current at any point, we need V and V . But 0 0 I 0 93.03 21.15 mA V Z I (71.6632.770 )(0.09303 21.150 ) 6.66711.620 0 in 0 1 1 V V0 Z 0 I 0 6.66711.62 (60 j40)(0.09303 21.15) 2 2 6.68712.080 0 1 1 V V0 Z 0 I 0 6.66711.62 (60 j40)(0.09303 21.15) 2 2 0.05182600 0 43 Example 11.3 – Solution continued At the middle of the line, z l / 2, z= l / 2 0.921+j1, Hence the current at this point is: V I s (z l /2) 0 e Z0 V l/2 0 l/2 e 0 0 Z (6.687e j12.08 )e00.921 j1 (0.0518e j 260 )e0.921 j1 = 60 j40 60 j40 Note that j1 is in radians and is equivalent to j57.30 ,(j1180/ ):0 I s (z l / 2)= (6.687e j12.08 )e0.921e j57.3 0 72.1e j33.69 0.0369e 0 j78.91 0 0 0.001805e 0 (0.0518e j260 )e0.921e j57.3 0 72.1e j33.69 0 j 283.61 6.673 j34.456 mA =35.102810 mA 44