EELE 3332 Electromagnetic II Chapter 11. Transmission Lines. Islamic University of Gaza Electrical Engineering Department Dr.

Transcription

1 EEE 333 Electromagnetic II Chapter 11 Transmission ines Islamic University of Gaza Electrical Engineering Department Dr. Talal Skaik 1 1

2 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 such structures are transmission lines and waveguides. Waveguides are discussed in the next chapter; transmission lines are considered in this chapter.

3 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 theory and electric circuit theory, the two major theories on which electrical engineering is based. 3

4 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

5 11. Transmission ine Parameters Transmission line parameters are: R: Resistance per unit length. (Ω/m) : Inductance per unit length. (H/m) G: Conductance per unit length. (S/m) C: Capacitance per unit length. (F/m) Distributed parameters of a two-conductor transmission line 5

6 Transmission ine Parameters 6

7 Transmission ine Parameters The line parameters R,, G, and C are uniformly distributed along the entire length of the line. For each line, the conductors are characterized by σc,µc,εc,=ε, 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: C and G C 7

8 Fields inside transmission line Transmission lines transmit TEM waves. V proportional to E, I proportional to H 8

9 11.3 Transmission ine 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: 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. E., I= H. V dl dl 9

10 Transmission ine Representation 1

11 Transmission ine Equations I ( z, t) Using KV:- V ( z, t) RzI( z, t) z V ( z z, t) t V ( z z, t) V ( z, t) I ( z, t) or RI( z, t) z t Taking the limit as z leads to: V ( z, t) I ( z, t) RI( z, t) z t equivalent circuit model of a two-conductor T.. of differential length z. 11

12 Using KC:- I( z, t) I( z z, t) I V ( z z, t) I( z, t)= I( z z, t) GzV ( z z, t) Cz t I( z z, t) I( z, t) V ( z z, t) or GV ( z z, t) C z t Taking the limit as z leads to: I( z, t) V ( z, t) GV ( z, t) C z t 1

13 The time domain form of the transmission line equations: V ( z, t) I ( z, t) RI( z, t) z t I ( z, t) V ( z, t) GV ( z, t) C z t If we assume harmonic time dependence so that: jt V( zt, ) =Re[ V (z) e ] jt I( z, t)=re[ Is(z) e ] s s Transmission ine Equations where V and I are the phasor forms of V ( z, t) and I( z, t), s dvs ( R j) I dz dis ( G jc) V dz s s 13

14 dvs d Is Transmission ine Equations dvs dis ( R j) Is, ( G jc) V dz dz To solve the previous equations, take second derivative of ( R j)( G jc) Vs dz Now take second derivative of I s gives ( G jc)( R ji ) s dz Hence, the wave equations for voltage and current become s V s gives dvs dz d Is dz V I s s, where j ( R j)( G jc) 14

15 j ( R j)( G jc) : is the propagation constant : attenuation constant (Np/m or db/m) : phase constant ( rad/m) wavelength is: =, wave velocity is: u f The solutions to the wave equations are: V V e V e, I I e I e s z z z z s +z -z +z -z where V, V, I, I are wave amplitudes. sign wave traveling along +z direction. - sign wave traveling along -z direction. 15

16 Transmission ine Equations V V e V e s s z I I e I e z z z In time domain: j t V ( z, t) Re[ V ( z) e ] s V ( z, t) V e cos( t z) V e cos( t z z z) Similarly for current: z I( z, t) I e cos( t z) I e cos( t z) z 16

17 V ( z) V e, I( z) I e Characteristic Impedance, The Characteristic Impedance of the line is the ratio of the positively travelling voltage wave to the current wave at any point on the line. z z dv ( z) since ( R j) I( z), dz ( V e ) ( R j) I e z z V R j R j V R jx o o o I G jc I R j Characterestic 1 R jx, Y o o o G jc Admittance 17

18 ossless ine (R=G=) A transmission line is said to be lossless if the conductors of the line are perfect (σc ) and the dielectric medium separating them is lossless (σ ) For lossless line, R=G= Since j = ( R j)( G jc), j, C 1 u f, C X R o o o C 18

19 Distortionless ine 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/ C Of course, a perfectly lossless line is impossible, but we find phase velocity is approximately constant if the line is low-loss. 19

20 Distortionless ine (R/=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 C j jc Thus, = ( R j)( G jc)= RG 1 1 R G jc α does not depend on = RG 1 j G frequency, whereas β is a linear or RG, C function of frequency. u / 1/ C (frequency independent)

21 u= Notes: Shape distortion of signals happen if α and u are frequency dependent. u and 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. Distortionless ine (R/=G/C) R j R 1 j / R R G jc G 1 jc / G G C 1 C ossless lines are desirable in power transmission, and telephone lines are required to be distortionless. 1 (Real)

22 Distortionless ine Practical use To achieve the required condition of R/=G/C for a transmission line, 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.

23 o General ossless Distortionless C j G j R o o o R C ) )( ( C j G j R C R o o Summary j C RG j C

24 Example 11.1 An air line has characteristic impedance of 7 Ω and a phase constant of 3 rad/m at 1 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. Hence R G and = c = R = C C R 1 Deviding the two equations yields: C or C pf 6 R 1 1 (7) = R C (7) (68. 1 ) 334. nh/m 1 4

25 A distortionless line has =6 Ω, α= mnp/m, u=.6c, where c is the speed of light. Find R,,G, C and λ at 1 MHz. RC 1 A distortionless line has RC G or G, u C C R, = RG R C R Example ( 1 )(6) 1. /m 6 41 Since = RG G 333 S/m R 1. 1 Dividing by u C C 6 = 333 nh/m u 8.6(3 1 ) gives 5

26 Example 11. solution continued Multiplying u 1 by u gives C C = C 9.59 pf/m.6(3 1 )6 8 C u 8 u.6(31 ) = 1.8 m 8 f 1 6

27 11.4 Input impedance, standing wave ratio, power Consider a transmission line of length l, characterised by and, connected to a load. Generator sees the line with the load as an input impedance in. 7

28 V ( z) V e V e z V V V V Is( z) e e, = I I At generator terminals (sending end): et s z z z V V ( z ), I I( z ), Substitute in prev. equs.: V V V 1 V V I V V 1 V V I I If the input impedance at the terminals is, then V V, I Input impedance in g g in g in g V in... (1) 8

29 V V V ( z) V e V e, I ( z) e e s At the load: z z z z s et V V ( z l), I I( z l), Substitute in prev. V V e V e 1 V V I e equs. l l l V l V l 1 V V l I e I e e Input impedance... () Now determine the input impedance = V ( z) / I ( z) at any point on the line. in s s 9

30 At the generator, recall,, s in = s ( ) V ( z) V V V I z I V V Input impedance Substituting eq. and utilizing the fact that: l l l l e e e e cosh l, sinh l, l l sinh l e e or tanh l l l cosh l e e V V V V V I then we get tanh l in tanh l (General - ossy ine) 3

31 Input impedance (ossless ine) tanh l in (General - ossy ine) tanh l For a lossless line, = j, tanh jl j tan l, then j tan l in j tan l ( ossless ine) βl is known as electrical length, in degrees or radians Note: To find in at a distance l ' from load, replace l by l' :- j tan l ' ' in j tan l 31

32 Reflection Coefficient, (at load) Define as the voltage reflection coefficient (at the load), as the ratio of the voltage reflection wave to the incident wave at the load, Since V e V e l l 1 V V I e 1 V V I e l l, and V I ( Voltage Reflection coefficient at load) 3

33 Define (at z ) as the voltage reflection coefficient at the source, as the ratio of the voltage reflection wave to the incident wave at source, V e V V e V Since Reflection Coefficient, (at generator) 1 V V I 1 V V I, and V I in in in ( Voltage Reflection coefficient at source) 33

34 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. V e V That is: ( z) V e z z V e z 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 / I e l l 34

35 The standing wave ratio s is defined as: (as we did for plane waves) V I 1 s= V I max max min min 1 When load is perfectly matched ( ) Total Transmission s 1 When load is a short circuit : Total Reflection 1 1 s When load is an open circuit : Total Reflection s Standing Wave Ratio Whenever there is a reflected wave, a standing wave will form out of the combination of incident and reflected waves.

36 The time-average power flow along the line at the point z is: 1 Re[ ( ) * ave s s ( )]. For a line, this can be reduced to: P V z I z lossless P ave V 1, The average power flow is constant at any point on the lossless The total power delivered to the load ( P ) is equal to the incident minus power ( V / ) the reflected power ( V / ) If, maximum power is delivered to the load, while no power is delivered for 1. Power / / P V V ave Incident Power (Pi) The above discussion assumes that the generator is matched. av Reflected Power (Pr) line. 36

37 Special Cases, =, =, = Short Circuited ine (=) Open Circuited ine (= ) Matched ine (=) 37

38 Shorted ine (=) j tan l j tan l in j tan l s 1, (Total Reflection) 38

39 Open-Circuited ine (= ) j tan l, ( ) in j tan l 1 j tan l in jcot l jtan l jtan l 1, (Total Reflection) s 39

40 Matched ine (=) Most desired case from practical point of view. Since in in s, 1 The whole wave is transmitted, and there is no reflection. The incident power is fully absorbed by the load. ( Maximum power transfer) 4

41 Solution ( a) Since 1 Np=8.686 db Example 11.3 A certain transmission line m long operating at ω=1 6 rad/s has α=8 db/m, β=1 rad/m, and = 6+j4 Ω. If the line is connected to a source of 1 V, g =4 Ω and transmitted by a load of +j5 Ω, determine (a) The input impedance (b) The sending end current (c) The current at the middle of the line. 8 =.91 Np / m = + j.91 j1 l=(.91 j1) 1.84 j 41

42 tanh l tanh 1.84 j 1.33 j.399 tanh l in tanh l in in Example 11.3 Solution continued j5 (6 j4)(1.33 j.399) (6 j4) 6 j4 ( j5)(1.33 j.399) 6.5 j38.79 ( b) The sending end current is I( z ) I Vg j in g I ma 4

43 ( c) To find the current at any point, we need V and V. But I V ma in Example 11.3 Solution continued I ( )( ) V (6 4)( ) V I j 1 1 V (6 4)( ).5186 V I j 43

44 Example 11.3 Solution continued At the middle of the line, z l /, z= l /.91+j1, Hence the current at this point is: V V Is( z l / ) e e l/ l/ (6.687 e ) e (.518 e ) e = 6 j4 6 j4 j j1 j6.91 j1 Note that j1 is in radians and is equivalent to j57.3,( j118/ ):- (6.687 e ) e e (.518 e ) e e Is( z l / )= j33.69 j e 7.1e j j57.3 j6.91 j j ma j e.185 = ma e j