Download EMII Chapter 11 P1-converted

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.
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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
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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 
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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 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.
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Transmission Line Representation
10
Transmission Line
I (z, t)
Equations
Using KVL:- V (z, t)  RzI (z, t)  Lz
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)  GzV (z  z, t)  Cz

I (z, t)
V (z, t)
 GV (z, t)  C
z
t
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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)ejt ]
I(z, t)=Re[Is (z)e
jt
]
where Vs and Is are the phasor forms of V (z, t) and I (z, t),
 dVs
 (R  jL)I s
dz
 dIs
 (G  jC)Vs
dz
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Transmission Line Equations
 dVs
 dIs
 (R  j L)I s ,
 (G  jC)Vs
dz
dz
To solve the previous equations, take second derivative of Vs gives
d 2V
s
 (R  jL)(G  jC)Vs
2
dz
Now take second derivative of Is gives
d2 Is
 (G  jC)(R  jL)Is
2
dz
Hence, the wave equations for voltage and current become
d 2Vs
2


Vs  0
2
dz
, where     j  (R  jL)(G  jC)
d 2I s
2


Is  0
2
14
dz
    j  (R  jL)(G  jC)
 : 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  V0e z
+z
 V0e z
-z
,
I s  I 0e z

+z
I 0e z
-z
where V0 ,V0 , I 0 , I 0 are wave amplitudes.
 sign  wave traveling along +z direction.
- sign  wave traveling along -z direction.
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Transmission Line Equations
Vs  V0e z  V0e z
I s  I 0e z 
I 0e 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)
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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  jL)I (z),
dz
 ( V0 e  z )  (R  jL)I 0 e  z


V
V
R  jL
R  jL

0
Zo   
 Ro  jX o  0

I0
I0
G  jC

R  jL
 Ro  jX o ,
Zo 
G  jC
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  jL)(G  jC)

  0,   j ,    LC
  2

1

u

 f ,

LC
Xo  0
Zo  Ro 
L
C
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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.
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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 jL  1 jC 
RG
 = (R  jL)(G  jC)=



R
G



α does not depend on
jC 

 = RG  1
    j
frequency, whereas β is a
G 

or
  RG ,    LC linear function of frequency.
Thus,
u   /   1/ LC (frequency independent)
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R
Z0 
G
Distortionless Line
R 1  jL/ R
R
j(R/L=G/C)
L


jC
G 1  jC / 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.
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Summary

General
Lossless
Distortionless
Zo
  (R  jL)(G  jC)
R  jL
Zo 
G  jC
  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
RG0
L
Z0 =R0 =
C
   LC
Deviding the two equations yields:
R0
1

 C
3

or C 

 68.2 pF
6
R0 2 10010 (70)
L=R02C  (70)2 (68.2 1012 )  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  (2010 3 )(60)  1.2/m
2
400106
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(310 )
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(310 )60
u 0.6(3108 )
 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)  V0e z  V0e 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  V0e l  V0e 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 el
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  (71.6632.770 )(0.09303  21.150 )  6.66711.620
0
in 0
1
1
V  V0  Z 0 I 0   6.66711.62  (60  j40)(0.09303  21.15)
2
2
 6.68712.080

0
1
1
V  V0  Z 0 I 0   6.66711.62  (60  j40)(0.09303  21.15)
2
2
 0.05182600

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 )e00.921 j1 (0.0518e j 260 )e0.921 j1
=

60  j40
60  j40
Note that j1 is in radians and is equivalent to j57.30 ,(j1180/ ):0
I s (z  l / 2)=
(6.687e j12.08 )e0.921e 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
=35.102810 mA
44