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Transmission Lines Basic Theories
1 Introduction
At high frequencies, the wavelength is much smaller than
the circuit size, resulting in different phases at different
locations in the circuit.
Quasi-static circuit theory cannot be applied. We need to
use transmission line theory.
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A transmission line is a two-port network connecting
a generator circuit at the sending end to a load at the
receiving end.
Unlike in circuit theory, the length of a transmission line
is of utmost importance in transmission line analysis.
z0
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3 AC Steady-State Analysis3.1 Distributed parameter representation
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R = resistance per unit length, (/m)
L = inductance per unit length, (H/m)
G = conductance per unit length, (S/m)
C = capacitance per unit length, (F/m)
z= increment of length, (m)
We use the following distributed parameters to
characterize the circuit properties of a transmission line.
These parameters are related to the physical properties of
the material filling the space between the two wires.
where, , = permittivity, permeability, conductivity
of the surrounding medium.
=''CL
='
'
C
G(See Text Book No.3,
pp. 432-433)
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For the coaxial and two-wire transmission lines, thedistributed parameters are related to the physical
properties and geometrical dimensions as follows:
Surface
resistivity ofthe conductors
(See Text
Book No.3,
pp. 445-447)
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3.2 Equations and solutionsConsider a short section z of a transmission line
(dropping the primes onR,L, G, Chereafter) :
Using KVL and KCL circuit theorems, we canderive the following differential equations for this
section of transmission line.
Generator Load
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( ) ( )( , ), ( , ) , 0
( , )( , ) ( , ) ( , ) 0
i z tv z t R zi z t L z v z z t t
v z z t i z t G zv z z t C z i z z t
t
+ =
+ + + =
By letting z0, these lead to coupled equations:
( , ) ( , )( , )
( , ) ( , )( , )
v z t i z t Ri z t Lz t
i z t v z t Gv z t C
z t
= +
= +
General Transmission Line Equations Coupled!
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For sinusoidal varying voltages and currents, we can usephasor forms.
( ) ( ) tjezVtzv Re, =
( ) ( ) tjezItzi Re, =
V(z) and I(z) are called phasors of v(z,t) and i(z,t). In
terms of phasors, the coupled equations can be written as:
( )( ) ( )
( )( ) ( )
dV zR j L I z
dzdI z
G j C V z dz
= +
= +
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After decoupling,
( )
( )
22
2
22
2
( )
( )
d V zV z
dz
d I zI z
dz
=
=
( )( )j R j L G j C = + = + +
is the complex propagation constant whose real part is
the attenuation constant (Np/m) and whose imaginarypart
is the phase constant (rad/m). Generally, these
quantities are functions of .
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Solutions to transmission line equations:
0 0
0 0
( ) ( ) ( )
( ) ( ) ( )
z z
z
V z V z V z
V e V e
I z I z I z
I e I e
+
+
+
+
= +
= +
= +
= +
Forwardtravelling
wave.
Backwardtravelling
wave.
0 0 0 0, , ,V V I I
+ +
= wave amplitudes in the forward andbackward directions atz= 0. (They
are complex numbers in general.)
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Z0 and are the two most important parameters of
a transmission line. They depend on the
distributed parameters (RLGC) of the line itself
and but not the length of the line.
CjG
LjR
CjG
LjRZ
+
+=
+=
+=0
( )( )CjGLjRj ++=+=
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For lossless transmission lines,R = G = 0.Parameters for lossless transmission lines
==
=
LC
0
11
velocityphase ==== LCup
jkjfjjj =====
=
22
constantnpropagatiocomplex
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LCff
ff
up 121
lineiontransmissthealongwavelength
=====
=
C
L
CjG
LjR
Z
=
++=
=
impedancesticcharacteri0
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Voltage and current along the line:
( )
( ) jkzjkz
jkzjkz
eIeIzI
eVeVzV
+
+
+=
+=
00
00
LjLjk
jk
L
eVV
eVeV
z
z
===
=
==
+
+
0
00
0
0
0
0atvoltageincident
0atvoltagereflected
Define a reflection coefficient atz= 0 as L:
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In terms of the reflection coefficient L, the totalvoltage and current can be written as:
( )
( )jkzLjkz
jkzjkz
jkzjkz
eeV
eV
VeV
eVeVzV
2
0
2
0
00
00
1
1
+=
+=
+=
+
+
+
+
( )
( )jkzLjkz
jkzjkz
jkzjkz
eeI
eV
Ve
Z
V
eZ
Ve
Z
VzI
2
0
2
0
0
0
0
0
0
0
0
1
1
=
=
=
+
+
+
+
In subsequent analyses, we will consider only lossless
transmission lines.
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5 Infinitely Long Transmission LineFor an infinitely long transmission line, there can be no
reflected wave (backward travelling wave). So for an
infinite long transmission line, there is only a forwardtravelling wave.
( )
( )
( )
( ) 000)( Z
zI
zV
zI
zVzZ ===
+
+
0=L
( ) ( )
( ) ( ) jkz
jkz
eIzIzI
eVzVzV
++
++
==
==
0
0
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6 Terminated Transmission Line
Note the two coordinate systems and their relation:z= measuring from the left to the right
= measuring from the right to the left
loadsource
= -z
i
L
(
)Z()
z = -d
= dz = 0
= 0
z
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In thezcoordinate system,
( )
VeVeV jkjk =+ +00
( )zIeIeI jkzjkz =+ + 00
In the (= -z) coordinate system,
( ) IeIeI jkjk =+ + 00
( )zVeVeV jkzjkz =+ + 00
We will use the coordinate system in subsequentanalyses.
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( ) Ljkjk
eV
eV ===+
0
0
000
0
0
0 ZI
V=
+
+
The characteristic impedance in the coordinate system is:
The reflection coefficient at = 0 in the coordinate
system is:
As L is obtained at = 0 (the load position), it iscalled the reflection coefficient at the load.
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Putting the expressions for V0+
and V0-
into the equationsfor the voltage and current, we have:
( ) ( ) ( )[ ]( ) ( )[ ]
kjZkZI
eeZeeZIV
LL
jkjkjkjkLL
sincos
2
1
0
0
+=
++=
( ) ( ) ( )[ ]
( ) ( )[ ]
kjZkZZ
I
eeZeeZZ
II
LL
jkjkjkjk
LL
sincos
2
1
0
0
0
0
+=
++=
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Using V() andI(
), we can obtain the impedanceZ(
) at
an arbitrary point on the transmission line as:
( )( ) ( )( )
kjZZkjZZZ
IVZ
L
L
tantan)(
0
00
++==
The reflection coefficient at the loadL can be expressedas:
( )
( ) 00
0
0
0
0
2
12
1
ZZZZ
ZZI
ZZI
VV
L
L
LL
LLL
+=
+
== +
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In fact, we can further define a reflection coefficient ()at any point on the transmission line by:
( )
kj
L
kj
jk
jk
eeV
V
eV
eV 22
0
0
0
0
pointatvoltageincident
pointatvoltagereflected
+
+
===
=
( ) ( )( )
( ) ( )( )00
00
2
12
1
ZZIeV
ZZIeV
jk
jk
=
+=
+
As we know (by solving the two equations on page 22
with 0):
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( )( ) ( )[ ]
( ) ( )[ ]( )( )
0
0
0
0
2
12
1
ZZZZ
ZZI
ZZI
+=
+
=
Therefore, alternatively we can write,
( ) ( )( )
+=
11
0ZZ
Then,
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At the position of the generator (= d),
( )( )kdjZZkdjZZ
ZdZZL
Li
tan
tan)(
0
00
+
+===
( ) kdjLi
ii e
ZZ
ZZd 2
0
0 =+
===
i
Vg
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Example 1A 100- transmission line is connected to a load consisted
of a 50- resistor in series with a 10-pF capacitor.
(a) Find the reflection coefficient L at the load for a 100-MHz signal.
(b) Find the impedance Zin at the input end of the
transmission line if its length is 0.125.
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(b) d=0.125
( )
( )
( )
( )
0
00
00
0
( 0.125 )
tan 4
tan 4
14 3717 25 5544
29.32 60.65
in
L
L
L
L
Z Z
Z jZ
Z Z jZ
Z jZZ
Z jZ. - j .
= =
+
= +
+=
+
=
=
See animation Transmission Line Impedance Calculation
Normalizedzin = 0.1437-j 0.2555
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6.1 Voltage/current maxima and minima
( )
( )
kj
L
jk
kjjk
jkjk
eeV
eV
VeV
eVeVV
2
0
2
0
00
00
1
1
+
+
+
+
+=
+=
+=
( )
( )
2
0
2
0
0
1
1
1
L
j k
L
j k
L
V V e
V e
V
+
+
+
= +
= +
= +
|L|1
( )2
a complex number
Lj k
Le
=
=
Lj
LL e
=
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=L-2k
1 +
1
0
Complex plane of (1+)
Re
Im
( )2 'L kzLe
=1 L
1 L+
0=m
M
( )V
See animation Transmission Line Voltage Maxima and Minima
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( ) ( )( )
[ )
max
is maximum when 1 1 2 2
, 0,1,2,4 2
Note: has to be specified in the range , .
L
L
L
M
L
VV k n
n
n
+ = +
= =
= + =
( ) ( )( ) ( )
( )
[ )
min
is minimum when 1 1
2 2 1
2 1 , 0,1, 2,4 4
Note: has to be specified in the range , .
L
L
Lm
L
V
V k n
n n
= = = +
+ = + =
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( ) 20
0
0
1
1
j kLI I e
V
Z
+
+
=
=
As current is
Current is maximum when voltage is minimum and
minimum when voltage is maximum.
( ) ( )
=
++= L
LM n
nI with,,2,1,0,
4
12
4at
max
( )
=+= L
Lm n
nI with,,2,1,0,
24at
min
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( )
( )
( )( )
0max
0min
voltage standing wave ratio (VSWR)
1 1
(dimensionless)11
L L
LL
S
V V
V V
+
+
=
+ +
= = =
1
1
+
=
S
S
L
|V(z)|
|V|max
|V|min
|I|max
|I|min
|I(z)|
load load
lmax lmax
Define a voltage standing wave ratio (VSWR) as:
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6.2 Power flow in a transmission line
Power flow at any pointzon a transmission line
is given by:
( ) ( ) ( ){ }zIzVzPav *Re21=
Power delivered by the source:
{ }*Re2
1igs IVP =
Power dissipated in the source impedanceZg:
{ } { } { }giiigZZZ ZIIIZIVP ggg Re21
Re2
1Re
2
1 2** ===
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Power input to the transmission line:
( ) ( ) ( ){ }
{ } { } { }
=
=
===
==
*
2
*
*
2**
*
1Re
2
1Re
2
1
Re
2
1Re
2
1Re
2
1
Re21
i
i
i
ii
iiiiiii
avi
ZV
Z
VV
ZIIIZIV
dIdVdPP
( ) ( ) ( ){ }
{ } { } { }
=
=
===
==
*
2
*
*
2**
*
1Re
2
1Re
2
1
Re2
1
Re2
1
Re2
1
00Re2
10
L
L
L
LL
LLLLLLL
avL
ZV
Z
VV
ZIIIZIV
IVPPPower dissipated in the terminal impedance:
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Solutions
The following information is given:
m310
Hz10MHz100
,5050,50,V60
m,1.5,50
8
8
0
===
==+===
==
ccu
f
jZZV
dZ
p
Lgg
1.110
0
50 50 500.2 0.4 0.45
50 50 50
jLL
L
Z Z j j e
Z Z j
+ = = = + =+ + +
The reflection coefficient at the load is:
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rad11.1,45.0Therefore, ==
LL
load)(from them27.009.04
11.1
0when,24
Then,
===
=+=
n
nLM
( )( )kdjZZkdjZZZZ
L
Li
tantan
0
00
++=
The input impedance Zi looking at the input to thetransmission line is:
/
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A24.048.0505050
60
:islineiontransmisstheinput toat thecurrentThe
jjZZ
VI
ig
g
i =++
=+
=
{ } W2.750288.02
1Re
2
1 2==== iiiL ZIPP
As the transmission line is lossless, power delivered to the
loadPL is equal to the power input to the transmission line
Pi. Hence,
( )+=
++
++
= 5050
5.13
2tan505050
5.13
2
tan50505050 j
jj
jjZi
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giig ZIVV +=
Vi andIi are related to the source voltage Vgas:
From the expressions ofVi, Ii, and Vg, we can find V0+
.
( )( )kdjLgg
jkd
g
eZZ
eZVV
2
0
0
01
+
+=
kdj
L
jkd
i eeVV2
0 1 + += ( )kdjLjkdi ee
Z
VI 2
0
0 1 +
=
At = d, V(d) = Vi andI(d) =Ii.
tcoefficienreflectionsource0
0 =+
=
ZZ
ZZ
g
g
g
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( )
( )( ) ( )
( )( )( )
( )
kj
L
jk
kdj
Lgg
jkd
g
kj
L
jk
kdj
Lgg
jkd
g
eeeZZ
eVI
ee
eZZ
eZVV
2
2
0
2
2
0
0
11
1
1
+
=
+
+
=
Putting V0+ into the expressions of V() andI(), we have:
Now the voltage and current on the transmission line are
expressed in terms of the known parameters of thetransmission line.
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Example 3
A 1.05-GHz generator circuit with a series impedanceZg= 10and voltage source given by:
is connected to a load ZL = (100 + j50) through a 50-, 67-cm-long lossless transmission line. The phase velocity of the line is
0.7c, where c is the velocity of light in a vacuum. Find the
instantaneous voltage and current v(,t) and i(
,t) on the line and
the average power delivered to the load.
( ) ( ) ( )V30sin10g += ttv
Z0 = 50 ZL
dZg
vg Zi
0
IiA
A
Vi
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Solutions
35.32.0
67.0cm67
m2.01005.11037.0 9
8
p
===
===
d
fu
32
50105010
tcoefficienreflectionsource
0
0=+=+
=
ZZZZ
g
g
g
46.0
0
0 45.05050100
5050100
tcoefficienreflectionload
j
L
L
L
ej
j
ZZ
ZZ=
++
+=
+
=
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( ) ( )( ) ( )( )[ ]46.077.2
22
0
45.020.0
11
=
+=
kjjkj
kjL
jkkdj
Lgg
jkd
g
eee
eeeZZ
eVI
Therefore instantaneous forms are:
( ) ( ){ }( )[ ]{ }
( ) ( )23.3cos58.477.2cos18.1045.018.10Re
Re,
46.077.2
++++=+=
=
ktkt
eeee
eVtv
tjkjjkj
tj
( ) ( ){ } ( )[ ]{ }( ) ( )23.3cos09.077.2cos20.0
45.020.0Re
Re,
46.077.2
+++=
=
=
ktkt
eeee
eIti
tjkjjkj
tj
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( )( )
( )
( )
4.179.21
35.32
tan5010050
35.32
tan5050100
50
tan
tan
0
00
j
jj
jj
kdjZZ
kdjZZZZ
L
Li
+=
++
++
=
+
+=
55.13/
28.04.179.2110
10 jj
ig
g
i ej
e
ZZ
V
I
=++=+=
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Power delivered to the load
= power input to the transmission line at AA
{ }
{ }
{ }
{ }
Watt86.0
4.179.21Re28.0
2
1
Re21
Re2
1
Re
2
1
2
2
*
*
=
+=
=
=
=
j
ZI
IZI
IV
ii
iii
ii
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7 Special Cases of Terminations in a Transmission Line
For a matched line,ZL = Z0. Then,
7.1 Matched line
( ) ( )
( )
( ) ( )( )
linetheoflengthanyfor
0
tan
tan
0
0
0
00
000
=+=
=+
+=
ZZZZ
ZkjZZ
kjZZZZ
Thus, there is no reflection on a matched line. There isonly an incident voltage. It is same as the case of an
infinitely long line.
Note =-z
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Z0
Zin
1
-1
1
-1
Normalized current magnitude
1
-1
Normalized impedance (Zin/Z0)
Z0
z
z
z
z0
0
Note:
Normalized voltage = voltage/max. |voltage|
Normalized current = current/max. |current|
Normalized voltage magnitude
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For a short circuit,ZL = 0. Then
7.2 Short-circuited line
( ) ( )kzjZkjZZ tantan 00sc
in ==
Normalized impedance (=-tan(kz))
Normalized voltage magnitude
Normalized current magnitude
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7.3 Open-circuited line
For an open circuit,ZL = . Then
( ) ( )kzjZkjZZ cotcot 00oc
in ==
Normalized voltage magnitude
Normalized impedance (=cot(kz))
Note that:
( )[ ] ( )[ ]20
00
oc
in
sc
in
cottan
Z
kjZkjZZZ
=
=
( )[ ] ( )[ ]
( )
k
kjZkjZZZ
2
00
oc
in
sc
in
tan
cottan
=
=
k.andcompute,and,,Given 0oc
in
sc
in ZZZ
Normalized current magnitude
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Hon Tat Hui Transmission Lines Basic Theories
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7.4 /4 transmission line terminated inZL
7.5 /2 transmission line terminated inZL
( )( ) LL
Lin
Z
Z
jZZ
jZZZZZ
2
0
0
00
2tan
2tan)4( =
+
+===
( )( ) LL
Lin Z
jZZ
jZZZZZ =
+
+===
tan
tan)2(
0
00
Zin
Z0 ZL
Zin
Z0 ZL
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Example 4
The open-circuit and short-circuit impedances measured atthe input terminals of a lossless transmission line of length
1.5 m (which is less than a quarter wavelength) are j54.6
andj103 , respectively.
(a) FindZ0 and kof the line.
(b) Without changing the operating frequency, find theinput impedance of a short-circuited line that is twice
the given length.
(c) How long should the short-circuited line be in orderfor it to appear as an open circuit at the input
terminals?
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S l ti
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Solution
The given quantities are
(a)
(b) For a line twice as long, = 3 m and k =1.884 rad,
m5.1=
mrad628.0tan1 oc
in
sc
in
1
==
ZZk
232tan0sc
in jkjZZ ==
m102
==k
103scin jZ =
6.54ocin jZ =
75scinoc
in0 == ZZZ
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