UNIVERSITI MALAYSIA PERLIS
EKT 241/4:
ELECTROMAGNETIC
THEORY
CHAPTER 5 – TRANSMISSION LINES
PREPARED BY: Saidatul Norlyana Azemi
snorlyana@unimap.edu.my
Chapter Outline
 General Considerations
 Lumped-Element Model
 Transmission-Line Equations
 Wave Propagation on a Transmission Line
 The Lossless Transmission Line
 Input Impedance of the Lossless Line
 Special Cases of the Lossless Line
 Power Flow on a Lossless Transmission Line
 The Smith Chart
 Impedance Matching
 Transients on Transmission Lines
General Considerations
• Transmission line – a two-port network
connecting a generator circuit to a load.
So…What is the use of
transmission line??
• A transmission line is used to transmit
electrical energy/signals from one point to
another
– i.e. from one source to a load.
• Types of transmission line include: wires,
(telephone wire), coaxial cables, optical
fibers n etc…
The role of wavelength
length of line, l
The impact of a transmission line
on the current and voltage in the
circuit depends on the:
frequency, f of the
signal provided by
generator.
• At low frequency, the impact is negligible
• At high frequency, the impact is very significant
Propagation modes
Electric field lines
Magnetic field lines
Transverse
electromagnetic
(TEM)
transmission lines
waves propagating along
these lines having electric
and magnetic field that are
entirely transverse to the
direction of propagation
Higher order
transmission
lines
waves propagating along these
lines have at least one significant
field component in the direction of
propagation
Propagation
modes
Propagation modes
A few examples of transverse electromagnetic (TEM) and higher order transmission line
Lumped- element model
• A transmission line is represented by a parallelwire configuration regardless of the specific
shape of the line, (in term of lumped element
circuit model)
– i.e coaxial line, two-wire line or any TEM line.
• Lumped element circuit model consists of four
basic elements called ‘the transmission line
parameters’ : R’ , L’ , G’ , C’ .
Series element
Shunt element
Lumped- element model
 Lumped-element transmission line parameters:
– R’ : combined resistance of both conductors per unit length, in
Ω/m
– L’ : the combined inductance of both conductors per unit length,
in H/m
– G’ : the conductance of the insulation medium per unit length, in
S/m
– C’ : the capacitance of the two conductors per unit length, in F/m
• For example, a coil of wire has the property of
inductance. When a certain amount of inductance is
needed in a circuit, a coil of the proper dimension is
inserted
Lumped- element model
Lumped- element model for 3 type of lines
Note: µ, σ, ε pertain to the insulating material between conductors
Exercise 1:
• Use table 5.1 to compute the line
parameter of a two wire air line whose
wires are separated by distance of 2 cm,
and, each is 1 mm in radius. The wires
may be treated as perfect conductors with
σc= .
R’ = ?, L’=?, G’=?, C’=?
Solution exercise 1:
Rs
R' 
a
Rs 
σc= 
fo
o
Rs 
fo

 Rs  0
 R' 0
 

L'  ln (d / 2a)  (d / 2a) 2  1
 


G' 
C'


2
ln (d / 2a )  (d / 2a )  1) 





2
ln (d / 2a )  (d / 2a )  1) 


σc= 
G ' 0
 

L'  ln (d / 2a)  (d / 2a) 2  1
 

d  2cm  0.02 m
a  1mm  0.001m
  0.02
0.02 2 
L'  ln (
) (
)  1
  2(0.001)
2(0.001)

 L'  1.20H / m
C'



ln (d / 2a )  (d / 2a ) 2  1) 


d  2cm  0.02 m
a  1mm  0.001m
C' 

 0.02
0.02 2 
ln (
) (
)  1
2(0.001)
 2(0.001)

C ' 9.29 pF / m
Exercise 2:
• Calculate the transmission line parameters
at 1 MHz for a rigid coaxial air line with
an inner conductor diameter of 0.6 cm
and outer conductor diameter of 1.2 cm.
The conductors are made of copper.
(μc=0.9991 ; σc=5.8x107)
f = 1MHz
r1 = 0.006m/2 = 0.003m
r2 = 0.012m/2 = 0.006m
Solution exercise 2:
Rs  1 1 
R' 
  
2  a b 
Rs 
Rs 
f
o
 (1Mhz) 
5.8 x10 7
 Rs  2.608 x10  4
2.608 x10 4  1
1 
R' 



2
 0.003 0.006 
 R '  0.0208  / m

L' 
ln( b / a)
2
a  0.003 m
b  0.006 m
  0.006 
L' 
ln 
2  0.003 
 L'  0.138 H / m
BARE IN
UR MIND
  o  r
o  (const33)
From calculator
 r  from appendixB
(pg238)
C'
2
ln b / a 
d  2cm  0.02 m
a  1mm  0.001m
2
C'
 0.006 
ln 

 0.003 
BARE IN
UR MIND
   o r
 o  (const32)
From calculator
 r  from appendixB
(pg 237)
C ' 80.3 pF / m
G' 
2
b
ln  
a
G ' 0
Because, the material
separating the inner and
outer is perfect dielectric
(air) with σ=0, thus G’ = 0
G’ : the conductance of the
insulation medium per unit length,
in S/m
Transmission line equations
Is used to describes the voltage and the current across the
transmission line in term of propagation constant and
impedance
• Complex propagation constant, γ

R'  jL'G'  jC '
   j
• α – the real part of γ
- attenuation constant, unit: Np/m
• β – the imaginary part of γ
- phase constant, unit: rad/m
Transmission line equations
• The characteristic impedance of the line, Z0 :
Z0 
R ' jL'
G ' jC '
 
• Phase velocity of propagating waves:

u p  f 

where f = frequency (Hz)
λ = wavelength (m)
β = phase constant
  2f
Example 1
An air line is a transmission line for which air is
the dielectric material present between the two
conductors, which renders G’ = 0.
In addition, the conductors are made of a
material with high conductivity so that R’ ≈0.
For an air line with characteristic impedance of
50Ω and phase constant of 20 rad/m at 700MHz,
find the inductance per meter and the
capacitance per meter of the line.
Solution to Example 1
• The following quantities are given:
Z 0  50,   20 rad/m, f  700 MHz  7 108 Hz
• With R’ = G’ = 0,
• propagation constant,  





R' jL'G ' jC ' 



2
 jL' jC '    L' C ' 


   L' C '
and
R' jL'
L'
• Z0 

G ' jC '
C'
Solution to Example 1

• The ratio is given by: Z 0
 L' 
2
     2 Z o 2 ( L' C ' )
 C' 

 L' C '
L'
C'
2
L'

 Z 0 L' C
C'
22ZZoo22CC22
22 
   Z oC

20
C' 

 90.9 pF/m 
8
Z 0 2  7 10  50
• We get L’ from Z0
Z0 
2
12


L' C'  L'  50  90.9 10  227 nH/m 
Lossless transmission line
Transmission line can be designed to minimize ohmic losses by
selecting high conductivities and dielectric material, thus we assume :
• Lossless transmission line - Very small values of R’
and G’.
• We set R’=0 and G’=0, hence:
 0
(lossless line)
   L' C ' (lossless line)
Transmission line equations
• Complex propagation constant, γ

R'
0  jL'G'
0  jC '
   j
• α – the real part of γ
- attenuation constant, unit: Np/m
• β – the imaginary part of γ
- phase constant, unit: rad/m
Lossless transmission line
Transmission line can be designed to minimize ohmic losses by
selecting high conductivities and dielectric material, thus we assume :
• Lossless transmission line - Very small values of R’
and G’.
• We set R’=0 and G’=0, hence:
 0
(lossless line)
   L' C ' (lossless line)
Z0 
R ' jL'
G ' jC '
since R'  0 and G'  0,
Z0 
L'
C'
(lossless line)
Lossless transmission line
• Using the relation properties between μ, σ, ε :
    (rad/m)
up 
1

(m/s)
• Wavelength, λ
0
c 1



f
f r
r
up
Where εr = relative permittivity of the insulating
material between conductors
Exercise 3:
• For a losses transmission line, λ = 20.7 cm
at 1GHz. Find εr of the insulating material.
λ=20.7cm 0.207m ;
0
c 1



f
f r
r
up
f=1 GHz
3x108 1
0.207 
1GHz  r
2
3x108 1
r 
1GHz 0.207
 r  1.449
 r  2 .1
Exercise 4
• A lossless transmission line of length 80
cm operates at a frequency of . The line
parameters are C  100 pF/m & L  0.25 μH/m
Find the characteristic impedance, the
phase constant and the phase velocity.
The condition apply that the line is
lossless, So: R= 0 & G=0
• characteristic
impedance :
L  0.25 μH/m
C  100 pF/m
L
Z0 
C
 Z0 

6
0.25 x10
100 x10 12
 50
• phase constant:   Im    R' jL'G' jC ' 
With R n G = 0
    L' C '
 2 (600 x10 6 ) (0.25 x10 6 )(100 x10 12 )
= 18.85 rad/m

• phase velocity: u p  f 


vp  

6
2 (600 x10 )
8
 2 x10 m / s
18.85
  2f
Voltage Reflection Coefficient
• Every transmission line has a resistance
associated with it, and comes about because of
its construction. This is called its characteristic
impedance, Z0.
• The standard characteristic impedance value is
50Ω. However when the transmission line is
terminated with an arbitrary load ZL, in which is
not equivalent to its characteristic impedance (ZL
≠ Z0), a reflected wave will occur.
Voltage reflection coefficient
• Voltage reflection coefficient, Γ – the ratio of the
amplitude of the reflected voltage wave, V0- to
the amplitude of the incident voltage wave, V0+ at
the load.
• Hence,
Z L / Z0  1
V Z Z
0
 0  L
V0 Z L  Z 0

Z L / Z0  1
(dimentionless)
Where   reflectioncoefficient
Z L  load impedance
Z 0  characteristic impedance
Voltage reflection coefficient
~
VL
ZL  ~
IL
• The load impedance, ZL
Where;
~


VL  V0  V0


V0
~ V0
IL 

Z0
Z0
~
V L = total voltage at the load
V0- = amplitude of reflected voltage wave
V0+ = amplitude of the incident voltage wave
~
I L = total current at the load
Z0 = characteristic impedance of the line
Voltage reflection coefficient
• And in case of a RL and RC series, ZL :
ZL = R + jL ;
ZL = R -1/ jC
• A load is matched to the line if ZL = Z0 because
there will be no reflection by the load (Γ = 0 and
V0−= 0.
• When the load is an open circuit, (ZL=∞), Γ = 1
and V0- = V0+.
• When the load is a short circuit (ZL=0), Γ = -1
and V0- = V0+.
What is the difference between
an open and closed circuit?
• closed allows electricity through, and open doesn't.
• open circuit - Any circuit which is not complete is considered an
open circuit. The open status of the circuit doesn't depend on how it
became unclosed, so circuits which are manually disconnected and
circuits which have blown fuses, faulty wiring or missing
components are all considered open circuits.
• close circuit: A circuit is considered to be closed when electricity
flows from an energy source to the desired endpoint of the circuit. A
complete circuit which is not performing any actual work can still be
a closed circuit. For example, a circuit connected to a dead battery
may not perform any work, but it is still a closed circuit.
Example 2
• A 100-Ω transmission line is connected to a
load consisting of a 50-Ω resistor in series with a
10pF capacitor. Find the reflection coefficient at
the load for a 100-MHz signal.
Solution to Example 2
• The following quantities are given
RL  50, CL  10 11 F, Z 0  100, f  100 MHz  108 Hz
• The load impedance is
Z L  RL  j / CL
1
 50  j
 50  j159 
8
11
2  10  10
• Voltage reflection coefficient is
Z L / Z 0  1 0.5  j1.59  1


 0.76  60.7
Z L / Z 0  1 0.5  j1.59  1
Z / Z  1 0.5  j1.59  1
 L 0

 0.76  60.7
Z L / Z 0  1 0.5  j1.59  1

  1 0.5  
2
2
  0.5  1.59    tan 
 
0.5  j1.59  1

 
 1.59  

0.5  j1.59  1 
  1 1.5  
2
2
 
 1.5  1.59    tan 

 
 1.59  
  1.5772.6 


 2.19  46.7 


In order to convert from –ve
magnitude for Г by replacing the
–ve sign with e-j180
 0.76119 .3
 0.76e j119 .3
 0.76e j119 .3 (e j180 )
 0.76e j  60 .7
   0.76
; r  60.7
Math’s TIP…
1
2
Exercise 5
• A 150 Ω lossless line is terminated in a
load impedance ZL= (30 –j200) Ω.
Calculate the voltage reflection coefficient
at the load.
Zo = 150 Ω
ZL= (30 –j200) Ω
(30  j 200)  150

(30  j 200)  150
Z L  Z0
 
Z L  Z0
o

j
72
.
95
   0.867 e
Standing Waves
• Interference of the reflected wave and the
incident wave along a transmission line creates
a standing wave.
• Constructive interference gives maximum value
for standing wave pattern, while destructive
interference gives minimum value.
• The repetition period is λ for incident and
reflected wave individually.
• But, the repetition period for standing wave
pattern is λ/2.
Standing Waves
• For a matched line, ZL = Z0, Γ = 0 and
~
V  z  = |V0+| for all values of z.
Standing Waves
• For a short-circuited load, (ZL=0), Γ = -1.
Standing Waves
• For an open-circuited load, (ZL=∞), Γ = 1.
The wave is shifted by λ/4 from short-circuit case.
Standing Waves
• First voltage maximum occurs at:
 r  n
Where θr = phase
lmax 

wheren  0
angle of Γ
4
2
• If θr ≥ 0  n=0;
• If θr ≤ 0  n=1
• First voltage minimum occurs at:
lmin
lmax   / 4 if lmax   / 4

lmax   / 4 if lmax   / 4
VSWR
• Voltage Standing Wave Ratio
(VSWR) is ratio between the
maximum voltage an the
minimum voltage along the
transmission line.
• VSWR provides a measure of
mismatch between the load
and the transmission line.
• For a matched load with Γ = 0,
VSWR = 1 and for a line with
|Γ| - 1, VSWR = ∞.
The VSWR is given by:
VSWR 
1 |  |
1 |  |
Z Z
Where,  L
ZL  Z 0
Where   reflectioncoefficient
Z L  load impedance
Z 0  characteristic impedance
Example 3
A 50- transmission line is terminated in a load
with ZL = (100 + j50)Ω . Find the voltage
reflection coefficient and the voltage standingwave ratio (VSWR).
Solution to Example 3
• We have,
Z  Z 0 100  j50   50
 L

 0.45e j 26.6
Z L  Z 0 100  j50   50
• VSWR is given by:
1 
1  0.45
VSWR 

 2.6
1   1  0.45
Exercise 6:
• A 140 Ω lossless line is terminated in a
load impedance ZL= (280 +j182) Ω, if λ =
72cm, find
a) Reflection coefficient, Г
b) The VSWR,
c) The locations of voltage maxima and
minima
• a) Reflection coefficient, Г
Z L  Z0
 
Z L  Z0
(280  j182)  (140)

(280  j182)  (140)
 1 182 
2
2
140  182   tan 

 140 



 182 
420 2  182 2   tan 1

 420 

140  j182

420  j182

230 52.4o
457 23.43o
 0.528.97 o
• b) The VSWR;
1 |  |
VSWR 
1 |  |
  0.528.97 o
VSWR 
1 | 0.528.97  |
1 | 0.528.97  |
1  0.5
VSWR 
3
1  0.5
• The locations of voltage maxima and
minima
 r  n
lmax 

wheren  0
4
2
(0.5)(72) n
lmax 

4
2
n
 2.9cm 
2
lmax   / 4 if lmax   / 4
lmin  
lmax   / 4 if lmax   / 4
  72cm
72cm / 4  18cm
 lmax   / 4
lmin  lmax   / 4
 72
 (2.9  n ) 
2
4
 20.9  n

2
Input impedance of a lossless
line
• The input impedance, Zin is the ratio of the total
voltage (incident and reflected voltages) to the
total current at any point z on the line.
~
V ( z)
Z in ( z )  ~
I ( z)
• or
1  e j 2 z 
 Z0 
j 2 z 
1   e

 Z L cos l  jZ0 sin l 
 Z L  jZ0 tan l 



Z in  l   Z 0 
 Z 0 

 Z 0 cos l  jZL sin l 
 Z 0  jZL tan l 
Special cases of the lossless
line
• For a line terminated in a short-circuit, ZL = 0:
~
Vsc  l 
sc
Zin  ~
 jZ0 tan l
I sc  l 
• For a line terminated in an open circuit, ZL = ∞:
Z inoc
Voc  l 
~
  jZ0 cot l
I oc  l 
Application of short-circuit and
open-circuit measurements
• The measurements of short-circuit input
impedance, Z insc and open-circuit input
impedance, Z inoc can be used to measure the
characteristic impedance of the line:
sc oc
Z o   Z in
Z in
• and
tan  l 
 Z insc
Z inoc
Length of line
• If the transmission line has length
where n is an integer,
l  n / 2 ,
tan l  tan 2 /  n / 2 
 tan n   0
• Hence, the input impedance becomes:
Zin  ZL
for l  n / 2
Quarter wave transformer
• If the transmission line is a quarter wavelength,
with l   / 4  n / 2 , where n  0 or any positive integer ,
we have l   2      , then the input
   4 
2
impedance becomes:
2
Z0
Z in 
ZL
for l   / 4  n / 2
Example 4
A 50-Ω lossless transmission line is to be matched
to a resistive load impedance with ZL=100Ω via a
quarter-wave section as shown, thereby eliminating
reflections along the feedline. Find the
characteristic impedance of the quarter-wave
transformer.
Quarter wave transformer
• If the transmission line is a quarter wavelength,
with l   / 4  n / 2 , where n  0 or any positive integer ,
we have l   2      , then the input
   4 
2
impedance becomes:
2
Z0
Z in 
ZL
for l   / 4  n / 2
Solution to Example 4
• Zin = 50Ω; ZL=100Ω
2
Z 02
2
Z in 
 Z 02  (50)(100 )
ZL
Z 02  50  100  70.7
• Since the lines are lossless, all the incident
power will end up getting transferred into the
load ZL.
Matched transmission line
• For a matched lossless transmission line, ZL=Z0:
1) The input impedance Zin=Z0 for all locations z
on the line,
2) Γ =0, and
3) all the incident power is delivered to the load,
regardless of the length of the line, l.
When ZL=0(short circuit)
Ratio of the total
voltage to total
current on the line
sc  jZ tan l
Zin
0
Special case
When ZL=(open circuit)
Input
Impedance, Zin
oc   jZ cot l
Zin
0
l
Application
Be used to measure the
characteristic impedance of
the line :
sc Z oc tan  l 
Z o   Zin
in
sc
 Z in
oc
Z in

2
l  0
Z in  Z L
But, If the
transmission line is
l

4
l 

2
Zin  Z0 2 Z L
Power flow on a lossless
transmission line
• Two ways to determine the average power of an incident
wave and the reflected wave;
– Time-domain approach
– Phasor domain approach
V0
2
i
• Average power for incident wave; Pav  2 Z
0
• Average power for reflected wave:
Pavr   
(W)

2 V0
2Z 0
• The net average power delivered to the load:
i  Pr 
Pav  Pav
av
V0
2
1   2 

2 Z 0 
(W)
2
2
   Pavi
Power flow on a lossless
transmission line
• The time average power reflected by a load connected to
a lossless transmission line is equal to the incident
power multiplied by |Г|2
Exercise 7
• For a 50Ω lossless transmission line terminated in
a load impedance ZL = (100 + j50)Ω, determine the
percentage of the average power reflected over
average incident power by the load.
Z0=50Ω; ZL = (100 + j50)Ω
2
r
i
Pav   Pav 
(W)
r
Pav
2

 
(W)
i
Pav
• Reflection coefficient, Г
Z L  Z0
 
Z L  Z0
(100  j50)  (50)

(100  j50)  (50)
 1 50 
2
2
50  50   tan  
 50 



 50 
150 2  50 2   tan 1

 150 

50  j50

150  j50

70.745 o
158 .118.4o
 0.4526.6o
2
  0.2
the percentage of the average incident
power reflected by the load = 20%
Exercise 8
• For the line of exercise previously
(exercise 7), what is the average reflected
power if |V0+|=1V
Pavr
 

V
0
2
2
2Z 0
2
   Pavi
2 1
r
 Pav   0.45
2
2(50)
 2mW
Smith Chart
• Smith chart is used to analyze & design
transmission line circuits.
• Reflection coefficient, Γ :    e j r  r  ji
Гr = real part, Гi = imaginary part
• Impedances on Smith chart are represented by
normalized value, zL : z  Z L
L
Z0
• the normalized load impedance, zL is
dimensionless.
Smith Chart
• Reflection coefficient, ΓA :0.3 + j0.4


1/ 2
2
2
  0.3  0.4 
 0. 5
 r  tan10.4 / 0.3  53
Reflection coefficient, ΓB :-0.5 - j0.2


1/ 2
2
2
  0.5  0.2 
 0.54
r
1


  tan
0.5 / 0.2  202
In order to eliminate –ve part, thus
 r  360   202   158
The complex Γ plane.
ΓA :0.3 + j0.4
ΓB :-0.5 - j0.2
Smith Chart
• Reflection coefficient, Γ :   Z L / Z 0  1
Z L / Z0 1
• Since
zL 1
ZL
zL 
, Γ becomes:   z  1
Z0
L
1 
 rL  jxL
• Re-arrange in terms of zL: z L 
1 
rL = Normalized load resistance
xL = Normalized load admittance
The families of circle for rL and xL.
Plotting normalized impedance, zL = 2-j1
(2  j1)  1

(2  j1)  1

12  12
32  12
 0.45
 r  tan11/ 2  26.6
Input impedance
• The input impedance, Zin:
1  e  j 2 l
Z in Z 0 
1  e  j 2 l
• Γ is the voltage reflection coefficient at the load.
• We shift the phase angle of Γ by 2βl, to get ΓL.
This will zL to zin. The |Γ| is the same, but the
phase is changed by 2βl.
• On the Smith chart, this means rotating in a
clockwise direction (WTG).
Input impedance
• Since β = 2π/λ, shifting by 2 βl is equal to phase
change of 2π.
• Equating: 2  l  2 2 l  2

• Hence, for one complete rotation corresponds to
l = λ/2.
• The objective of shifting Γ to ΓL is to find Zin at
an any distance l on the transmission line.
Example 5
• A 50-Ω transmission line is terminated with
ZL=(100-j50)Ω. Find Zin at a distance l =0.1λ
from the load.
Solution: Normalized the load impedance
Z L 100  j 50
zL 

Z0
50
 zL  2  j
Solution to Example 5
A2 j
l =0.1λ
zin = 0.6 –j0.66
de normalize
(multiplying by Zo)
Zin = 30 –j33
VSWR, Voltage Maxima and
Voltage Minima
zL=2+j1
VSWR = 2.6
(at Pmax).
lmax=(0.25-0.213)λ
=0.037λ.
lmin=(0.037+0.25)λ
=0.287λ
VSWR, Voltage Maxima and
Voltage Minima
• Point A is the normalized load impedance with
zL=2+j1.
• VSWR = 2.6 (at Pmax).
• The distance between the load and the first
voltage maximum is lmax=(0.25-0.213)λ=0.037λ.
• The distance between the load and the first
voltage minimum is lmin=(0.037+0.25)λ =0.287λ.
Impedance to admittance
transformations
zL=0.6 + j1.4
yL=0.25 - j0.6
Example 6
• Given that the voltage standing-wave ratio, VSWR = 3.
On a 50-Ω line, the first voltage minimum occurs at 5 cm
from the load, and that the distance between
successive minima is 20 cm, find the load
impedance.
Solution:
The distance between successive minima is equal to
λ/2.
the distance between successive minima is 20 cm,
Hence, λ = 40 cm
20   / 2
   2(20)
Solution to Example 6
Point A =VSWR = 3
5
l min 
 0.125
40
z L  0.6  j 0.8
de normalize
(multiplying by Zo)
Zin = 30 –j40
Solution to Example 6
• First voltage minimum (in wavelength unit) is at
5
l min 
 0.125 on the WTL scale from point B.
40
• Intersect the line with constant SWR circle = 3.
• The normalized load impedance at point C is:
z L  0.6  j 0.8
• De-normalize (multiplying by Z0) to get ZL:
Z L  500.6  j 0.8  30  j 40 
Exercise
Solution: Normalized the load impedance
ZL
zL 
 0.6  j 0.4
Z0
a) reflection coefficient from smith Chart

r  121
(0.6  j 0.4)  1

(0.6  j 0.4)  1

 0.4 2  0.4 2
1.6 2  0.4 2
 0.34
e
j
 0.34e
j121
 r  121
  0.082
• 0.25 - 0.082
 0.168
z L  0.6  j 0.4
lmin
lmax
length :
0.301 
Z in
• 0.72 - j0.62
• 0.301 0.082
 0.383
3) Move a distance 0.301λ towards the generator (WTG)
(refer to Smith chart)
• → 0.301λ + 0.082λ=0.383λ
• At 0.383λ, read the value of which at the point intersects
with constant circle, we have = zin = 0.72- j0.62.
• Denormalized it, hence = Zin = 72- j62
4) Distance from load to the first voltage maximum, (refer
to Smith chart)
→ 0.25λ-0.082λ=0.168λ
Impedance Matching
• Transmission line is matched to the load when
Z0 = ZL.
• This is usually not possible since ZL is used to
serve other application.
• Alternatively, we can place an impedancematching network between load and
transmission line.
Single- stub matching
• Matching network consists of two sections of
transmission lines.
• First section of length d, while the second
section of length l in parallel with the first
section, hence it is called stub.
• The second section is terminated with either
short-circuit or open circuit.
Single- stub matching
feed line
stub
YL=1/ZL
l
d
Yd = Y0+jB
Single- stub matching
• The length l of the stub is chosen so that its
input admittance, YS at MM’ is equal to –jB.
• Hence, the parallel sum of the two admittances
at MM’ yields Y0, which is the characteristic
admittance of the line.
Yd = Y0+jB
Single- stub matching
• Thus, the main idea of shunt stub matching network is
to:
• (i) Find length d and l in order to get yd and yl .
• (ii) Ensure total admittance yin = yd + ys = 1 for complete
matching network.
Example 7
50-Ω transmission line is connected to an
antenna with load impedance ZL = (25 − j50)Ω.
Find the position and length of the shortcircuited stub required to match the line.
Solution:
The normalized load impedance is:
Z L 25  j50
zL 

 0.5  j
Z0
50
(located at A).
Solution to Example 7
B  yL
 0.4  j0.8
y L  load admittance
A  0.5  j
Solution to Example 7
• Value of yL at B is yL  0.4  j 0.8 which locates
at position 0.115λ on the WTG scale.
• Draw constant SWR circle that goes through
points A and B.
• There are two possible matching points, C
and D where the constant SWR circle intersects
with circle rL=1 (now gL =1 circle).
B  0.115
B
C = 1+j1.58
D = 1+j1.58
A
Solution to Example 7
First matching points, C.
• At C, yd  1  j1.58 is at 0.178λ on WTG scale.
• Distance B and C is d  0.178  0.155  0.063
• Normalized input admittance yin  ys  yd
1  j 0  ys  1  j1.58
at the juncture is:
ys   j1.58
E is the admittance of short-circuit stub, yL=-j∞.
Normalized admittance of −j 1.58 at F and
position 0.34λ on the WTG scale gives:
l1  0.34  0.25  0.09
B  0.115
d1 = 0.063λ
d  0.178  0.115  0.063
B
C = 1+j1.58
E
l1 = 0.090λ
l  0.34  0.25  0.09
A
F
F = -j1.58
yin  ys  yd
1  j 0  ys  1  j1.58
ys   j1.58
First matching points, C
• Thus, the values are:
•
d1 = 0.063 λ
•
l1 = 0.09 λ
•
yd1 = 1 + j1.58 Ω
•
ys1 = -j1.58 Ω
• Where Yin = yd + ys = (1 + j1.58) + (-j1.58) = 1
Solution to Example 7
Second matching point, D.
• At point D, yd  1  j1.58
• Distance B and D is d2  0.322  0.115  0.207
• Normalized input admittance ys   j1.58 at G.
• Rotating from point E to point G, we get
l2  0.25  0.16  0.41
d  0.322  0.115
 0.207
l  0.25  0.16
 0.41
B
G G = +j1.58
d2 = 0.207λ
E
l2
= 0.41λ
D = 1-j1.58
A
First matching points, D
• Thus, the values are:
•
d2 = 0.207 λ
•
l2 = 0.41 λ
•
yd2 = 1 - j1.58 Ω
•
ys2 = +j1.58 Ω
• Where Yin = yd + ys = (1 - j1.58) + (+j1.58) = 1
d1=0.063 λ
d2=0.207 λ
l1=0.09λ,
l2=0.41 λ