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EELE 3332 – Electromagnetic IIChapter 11
Transmission Lines
Prof. Hala J. EL‐KhozondarIslamic University of Gaza
Electrical Engineering Department
2016 1
2
11.1 Introduction
Wave propagation in unbounded media is used in radio or TVbroadcasting, where the information being transmitted is meant foreveryone who may be interested.
Another means of transmitting power or information is by guidedstructures. Guided structures serve to guide (or direct) thepropagation of energy from the source to the load.
Typical examples of such structures are transmission lines andwaveguides.
Waveguides are discussed in the next chapter; transmission linesare considered in this chapter.
3
Transmission lines are commonly used in power distribution (atlow frequencies) and in communications (at high frequencies).
A transmission line basically consists of two or more parallelconductors used to connect a source to a load.
Typical transmission lines include coaxial cable, a two‐wire line, aparallel‐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 fieldtheory and electric circuit theory, the two major theories on whichelectrical engineering is based.
Introduction
4
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.
Introduction
Transmission line parameters are:
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11.2 Transmission Line Parameters
R: Resistance per unit length. (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 two‐conductor transmission line
6
Transmission Line Parameters
7
The line parameters R, L, G, and C are uniformly distributedalong 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 ischaracterized by σ,µ,ε.
G ≠1/R; R is the ac resistance per unit length of the conductorscomprising the line and G is the conductance per unit length due tothe dielectric medium separating the conductors.
For each line:
Transmission Line Parameters
and GLCC
8
Fields inside transmission line•Transmission lines transmit TEM waves.•V proportional to E, •I proportional to H
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11.3 Transmission Line EquationsTwo‐conductor transmission lines support a TEM wave; E and Hare perpendicular to each other and transverse to the direction ofpropagation.
E and H are related to V and I:
Using V and I in solving the transmission line problem is simplerthan solving E and H (requires Maxwell’s equations).
Examine an incremental portion of length Δz of a two‐conductortransmission line.
E. , I= H.V dl dl
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Transmission Line Representation
11
Transmission Line Equations( , )Using KVL:- ( , ) ( , ) ( , )
( , ) ( , ) ( , ) ( , )
Taking the limit as z 0 leads to: ( , ) ( , ) ( , )
I z tV z t R zI z t L z V z z tt
V z z t V z t I z tor RI z t Lz t
V z t I z tRI z t Lz t
equivalent circuit model of a two‐conductor T.L. of differential length z.
12
Using KCL:- ( , ) ( , )( , ) ( , )= ( , ) ( , )
( , ) ( , ) ( , ) ( , )
Taking the limit as z 0 leads to: (
I z t I z z t IV z z tI z t I z z t G zV z z t C z
tI z z t I z t V z z tor GV z z t C
z t
I z
, ) ( , )( , )t V z tGV z t Cz t
The time domain form of the transmission line equations:
( , ) ( , )( , )
( , ) ( , )( , )
If we assume harmonic time dependence so that:V( , )
V z t I z tRI z t Lz t
I z t V z tGV z t Cz t
z t
j ts
j ts
s s
=Re[ (z) ]
I( , )=Re[ (z) ]where and are the phasor forms of ( , ) and ( , ),
( )
( )
ss
ss
V e
z t I eV I V z t I z t
dV R j L IdzdI G j C Vdz
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Transmission Line Equations
s2
2
s2
2
( ) , ( )
To solve the previous equations, take second derivative of gives
( )( )
Now take second derivative of gives
( )(
s ss s
ss
s
dV dIR j L I G j C Vdz dz
V
d V R j L G j C Vdz
I
d I G j C R jdz
22
2
22
2
)
, the for voltage and current become
0 , where ( )( )
0
s
ss
ss
L I
Hence
d V Vdz j R j L G j Cd I Idz
wave equations
14
Transmission Line Equations
( )( )
: is the propagation constant : attenuation constant (Np/m or dB/m)
: phase constant ( rad/m)
2wavelength is: = , wave velocity is:
The soluti
j R j L G j C
u f
0 0 0 0
0 0 0 0
ons to the wave equations are:
,
where , , , are wave amplitudes. wave traveling along +z direction.
- sign wave traveling along -z direction
z z z zs sV V e V e I I e I e
V V I Isign
. 15
+z -z +z -z
16
0 0
0 0
0 0
In time domain:( , ) Re[ ( ) ]
( , ) cos( ) cos(
z zs
z zs
j ts
z z
V V e V e
I I e I e
V z t V z e
V z t V e t z V e
0 0
)
Similarly for current: ( , ) cos( ) cos( )z z
t z
I z t I e t z I e t z
Transmission Line Equations
0 0
0 0
0 0
0 0
0
( ) , ( )( )since ( ) ( ),
( ) ( )
Characterestic 1 , Admittance
z z
z z
o o o
o o o
V z V e I z I edV z R j L I z
dzV e R j L I e
V R j L R j L VZ R jXI G j C I
R j LZ R jX YG j C Z
017
Characteristic Impedance, Z0The Characteristic Impedance Z0 of the line is the ratio of thepositively travelling voltage wave to the current wave at any point onthe line.
For lossless line, R=G=0
Since = ( )( )
0, ,
1 2 ,
0
o o o
j R j L G j C
j LC
u fLC
R j LZ R jXG j C
LX Z R
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Lossless Line (R=G=0)A transmission line is said to be lossless if the conductors of the lineare perfect (σc ≈ ∞) and the dielectric medium separating them islossless (σ ≈ 0)
and GLCC
19
Any signal that carries significant information must has somenon-zero bandwidth. In other words, the signal energy (as well asthe information it carries) is spread across many frequencies.
If the different frequencies that comprise a signal travel atdifferent velocities, that signal will arrive at the end of atransmission line distorted. We call this phenomenon signaldispersion.
Recall for lossless lines, however, the phase velocity isindependent of frequency—no dispersion will occur!
Of course, a perfectly lossless line is impossible, but we findphase velocity is approximately constant if the line is low-loss.
Distortionless Line
1/u LC
A distorionless line results if the line parameters are such that
, = ( )( )= 1 1
= 1
R GL C
j L j CThus R j L G j C RGR G
j CRG jG
or RG
,
/ 1/ (frequency independent)
LC
u LC
20
Distortionless Line (R/L=G/C)A distortionless line is one in which the attenuation constant α isfrequency independent while the phase constant β is linearlydependent on frequency.
α does not depend onfrequency, whereas β is a linearfunction of frequency.
Notes:Shape distortion of signals happen if α and u are frequencydependent.
u and Z0 for distortionless line are the same as lossless line.
A lossless line is also a distortionless line, but a distortionless lineis not necessarily lossless.
Lossless lines are desirable in power transmission, and telephonelines are required to be distortionless. 21
Distortionless Line (R/L=G/C) 0
1 / (Real)
1 /
1u=
R j L RR j L R LZG j C G j C G G C
LC
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 highmagnetic permeability (μ).
A common practice is to replace repeaters in long lines to maintain thedesired shape and duration of pulses for long distance transmission. 22
Distortionless Line – Practical use
Zo
General
Lossless
Distortionless
CjGLjRZo
o oLZ RC
))(( CjGLjR
CLRZ oo
Summary
0 j LC
RG j LC
24
Example 11.1An air line has characteristic impedance of 70 Ω and a phaseconstant of 3 rad/m at 100 MHz. Calculate the inductance per meterand the capacitance per meter of the line.
0 0
0
60
An air line can be regarded as lossless line because 0 . 0 and =0
= =
1Deviding the two equations yields:
3 62 100 10 (70)
cand Hence R G
LZ RC
LCR
C
or CR
2 2 120
8.2 pF
= (70) (68.2 10 ) 334.2 nH/mL R C
00
30
2 6
0
1A distortionless line has or ,
, =
(20 10 )(60) 1.2 /m
400 10 = 333 S/m 1.2
1 by gi
RCRC GL G uL LC
L C RZ RG RC L Z
R Z
Since RG GR
LDividing Z uC LC
08
ves
60= 333 nH/m0.6(3 10 )
ZLu
25
Example 11.2A distortionless line has Z0=60 Ω, α=20 mNp/m, u=0.6c, where c isthe speed of light. Find R,L,G , C and λ at 100 MHz.
26
0
0 80
8
8
1 by gives
1 1 1= 92.59 pF/m 0.6(3 10 )60
0.6(3 10 )= 1.8 m10
LMultiplying Z uC LC
uZ CC uZ
uf
Example 11.2 – solution continued
27
11.4 Input impedance, standing wave ratio, power
Consider a transmission line of length l, characterised by and Z0,connected to a load ZL. Generator sees the line with the load as aninput impedance Zin.
0 0
0 0 0 00
0 0 0 0
( )
At generator terminals (sen
( ) , =
ding e
n
Let
d):
z zs
z zs
V z V e V e
V V V VI z e e ZZ Z I I
0 0
0 0 0 0 0 0 0
0 00
0 0 0 00 0
in
0 0
( 0), ( 0), Substitute in prev. equs.:
1 2 ... (1) 1 2
If the input impedance at the terminals is Z ,
then , ging
in g
V V z I I z
V V V V V Z IV VI V V Z IZ Z
VZV V IZ Z Z
in gZ 28
Input impedance
0 00 0
0 0
0 0 0 0
0 0
00 0
( ) , ( )
Let ( ), ( ), Substitute in prev. At the load:
equs.
1 2 1 2
z z z zs s
L L
l l lL L L
l lL
V VV z V e V e I z e eZ Z
V V z l I I z l
V V e V e V V Z I eV VI e e V VZ Z
0
in
... (2)
Now determine the input impedance Z = ( ) / ( ) at anypoint on the line.
lL L
s s
Z I e
V z I z
29
Input impedance
0 00 0 0 0
0 0
0 0 00in
0 0 0
firs , recall , ,
( )Z =( )
Substituting eq. 2 and utilizing the fact that:
cosh , sinh 2 2
t:At the generator
s
s
l l l l
V VV V V I thenZ Z
Z V VV z VI z I V V
e e e el l
00
0
,
sinhor tanhcosh
tanhwe get (General - Lossy Line)
Although this has been derived for the input impedance at the generator end
tan
,
h
l l
l l
Lin
L
l e ell e e
Z Z lZ ZZ Z l
it is a general expression. 30
Input impedance
00
0
00
0
tanh (General - Lossy Line) tanh
For a lossless line, = , tanh tan , then
tan tan
Lin
L
Lin
L
Z Z lZ ZZ Z l
j j l j l
Z jZ lZ ZZ jZ l
( )
Note: To find at a distance
Loss
less L
' from load,
e
rep
in
inZl
00
0
lace by ' :-
tan ' tan '
Lin
L
l l
Z jZ lZ ZZ jZ l
31
Input impedance (Lossless Line)
βl is known as electrical length, in degrees or radians
0
0
0 0
0 0
Define as the voltage reflection coefficient (at the load), as the ratio of the voltage reflection wave to the incident wave at the load,
12 , an12
L
l
L l
lL L
lL L
V eV e
V V Z I eSince
V V Z I e
0
0
Volt
d
age Reflection coefficient at l
oa d ( )
L L L
LL
L
V Z I
Z ZZ Z
32
Reflection Coefficient, (at load)
0
00 0
0 00 0
0 0 0 0
0 0
Define (at 0) as the voltage reflection coefficient at the source, as the ratio of the voltage reflection wave to the incident wave at source,
12 12
z
V e VV e V
V V Z ISince
V V Z
0 0
0 0
00
0
V
, and
oltage Reflection coefficient at
( )source
in
in
in
V Z II
Z ZZ Z
33
Reflection Coefficient, (at generator)
20 0
0 0
0 0
That is: ( )
Thus the current reflection coefficient at the load is /
zz
z
l lL
V e Vz eV e V
I e I e
34
Reflection Coefficient
The current reflection coefficient at any point on the line is thenegative of the voltage reflection coefficient at that point.
The voltage reflection coefficient at any point on the line is the ratioof the reflected voltage wave to that of the incident wave.
max max 0
min min 0
0
The standing wave ratio s is defined as: (as we did for plane waves)
1=
1
When load is ( ) Total Transmission
L LL
L L
L
V I Z ZsV I Z Z
perfectly matched Z Z
0 1When load is a :
Total Reflection1 1
When load is an :Total Reflection
1 1
L
L L
L L
sshort circuit
sopen circuit
s
35
Whenever there is a reflected wave, a standing wave will formout of the combination of incident and reflected waves.
Standing Wave Ratio
36
*
**0
00
The time-average power flow along the line at the point is: 1 Re[ ( ) ( )]. For a line, this can be reduced to:21 Re[ ( ) ( )2
ave s s
j l j l j l j lave
z
P V z I z
VP V
lossless
e e e ez
2
20
0
2
0 0
]
1 , 2
The average power flow is constant at any point on the line. The total power delivered to the load ( ) is equal to the incident
power minu( / 2 ) thes
ave
av
V
lossl
P
eP
V Z
s
Z
s
2 20 0 reflected power ( / 2 )
If 0, maximum power is delivered to the load, while no poweris delivered for 1.
The above discussion assumes that the generator is matched.
V Z
Power
Incident Power (Pi)
Reflected Power (Pr)
2 220 0 0 0/ 2 / 2aveP V Z V Z
Short Circuited Line
(ZL=0)
Open Circuited Line
(ZL=∞)
Matched Line
(ZL=Z0)
37
Special Cases , ZL=0, ZL=∞, ZL=Z0
38
Shorted Line (ZL=0)
00 0
0
0
0
tan tantan
1, (Total Reflection)
Lin
L
LL
L
Z jZ lZ Z jZ lZ jZ l
Z Z sZ Z
39
Open-Circuited Line (ZL=∞)
00
0
0
00 0
0
0
0
tan , ( )tan
1 tancot
tantan
1, (Total Reflection)
Lin L
L
Lin
L
LL
L
Z jZ lZ Z ZZ jZ l
Zj lZ ZZ Z jZ l
j lZ j lZ
Z Z sZ Z
40
Matched Line (ZL=Z0)
00
0
0 0
0
0
Most desired case from practical point of view.tan , tan
0, 1
The whole wave is transmitted, and there is no reflection.The incident power is f
Lin
L
L in
LL
L
Z jZ lZ ZZ jZ l
Since Z Z Z ZZ Z sZ Z
ully absorbed by the load. (Maximum power transfer)
( ) Since 1 Np=20 log (e)=8.686 dB8 = 0.921 /
8.686 = + 0.921 1 =2
Solutiona
Np m
j jl
(0.921 1) 1.84 2j j
41
Example 11.3A certain transmission line 2 m long operating at ω=106 rad/s hasα=8 dB/m, β=1 rad/m, and Z0= 60+j40 Ω. If the line is connectedto a source of 10∟00 V, Zg=40 Ω and transmitted by a load of20+j50 Ω, determine(a) The input impedance(b) The sending end current(c) The current at the middle of the line.
00
0
tanh tanh 1.84 2 1.033 0.03929
tanhtanh
20 50 (60 40)(1.033 0.03929)(60 40) 60 40 (20 50)(1.033 0.03929)
60.25 38.79
( ) The sending end cu
Lin
L
in
in
l j j
Z Z lZ ZZ Z l
j j jZ jj j j
Z j
b
0
0
rrent is ( 0)10 93.03 21.15
60.25 38.79 40g
in g
I z IV
I mAZ Z j
42
Example 11.3 – Solution continued
22 2
22 2
0 0
00 0 0
0 0
0 0 0 0
( ) To find the current at any point, we need V and V . 93.03 21.15
(71.66 32.77 )(0.09303 21.15 ) 6.667 11.62
1 1V 6.667 11.62 (60 40)(0.09303 21.15)2 2
6.687 12.
in
c ButI mA
V Z I
V Z I j
0
0 0 0 0
0
081 1V 6.667 11.62 (60 40)(0.09303 21.15)2 2
0.0518 260
V Z I j
43
Example 11.3 – Solution continued
0 0
/2 /20 0
0 0
12.08 0.921 1 260 0.921 1
At the middle of the line, / 2, z= / 2 0.921+j1, Hence thecurrent at this point is:
V V( / 2)
(6.687 ) (0.0518 ) =60 40 60 40
No
l ls
j j j j
z l l
I z l e eZ Z
e e e ej j
0 0 0 0
0 0
0
0
12.08 0.921 57.3 260 0.921 57.3
33.69 33.69
78.91
te that 1 is in radians and is equivalent to j57.3 ,( 1 180/ ):-
(6.687 ) (0.0518 )( / 2)=72.1 72.1
0.0369 0.001805
j j j j
s j j
j
j j
e e e e e eI z le e
e
0283.61
0
6.673 34.456 mA =35.10 281 mA
jej
44
Example 11.3 – Solution continued
