Download Chapter 5: Transmission Lines

UNIVERSITI MALAYSIA PERLIS
EKT 241/4:
ELECTROMAGNETIC
THEORY
CHAPTER 5 – TRANSMISSION LINES
PREPARED BY: NORDIANA MOHAMAD SAAID
[email protected]
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.
The role of wavelength
• The impact of a transmission line on the current
and voltage in the circuit depends on the length
of line, l and 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
• Transmission lines may be classified into two
types:
a) Transverse electromagnetic (TEM)
transmission lines – waves propagating along
these lines having electric and magnetic field
that are entirely transverse to the direction of
propagation
b) Higher order transmission lines – waves
propagating along these lines have at least one
significant field component in the direction of
propagation
Propagation modes
Lumped- element model
Lumped- element model
• A transmission line is represented by a parallelwire configuration regardless of the specific
shape of the line, 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’ .
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
Lumped- element model
Note: µ, σ, ε pertain to the insulating material between conductors
Lumped- element model
• All TEM transmission lines share the following
relation:
L'C '  
G' 

C' 
Note: µ, σ, ε pertain to the insulating material between conductors
Transmission line equations
• 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)
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,
  Im  

 jL' jC '   
L' C ' and Z 0 
jL'
L'

jC '
C'
• The ratio is given by

20
C' 

 90.9 pF/m 
8
Z 0 2  7 10  50
• We get L’ from Z0
Z0  L' C'  L'  50  90.9 1012  227 nH/m 
2
Lossless transmission line
• 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
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
• Hence, load impedance, ZL:
 V0   V0 
Z L   

V

V
0
 0

Z 0


• Solving in terms of V0- :
V0

 Z L  Z0  
V0
 
 Z L  Z0 
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,
V0 Z L  Z 0 Z L Z 0  1
  

(dimension less)
Z L  Z0 Z L Z0 1
V0
Voltage reflection coefficient
• Z0 for lossless line is a real number while ZL in
general is a complex number. Hence,
   e jr
Where |Γ| = magnitude of Γ
θr = phase angle of Γ
• A load is matched to the line if ZL = Z0 because
there will be no reflection by the load (Γ = 0 and
V0−= 0.
Voltage reflection coefficient
• 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+.
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
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
l max 

where n  0
angle of Γ
4
2
• First voltage minimum occurs at:
lmin
lmax   / 4 if lmax   / 4

lmax   / 4 if lmax   / 4
 r  n
l

• For next voltage maximum: max 4  2 where n  0
n  1, 2, 3, ...... if  r  0
n  0, 1, 2, ...... if  r  0
Voltage standing wave ratio
• VSWR is the ratio of the maximum voltage
amplitude to the minimum voltage amplitude:
~
V
1 
max
VSWR  ~

(dimension less)
1 
V
min
• 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 = ∞.
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 L / Z 0  1 100  j50   50

 0.45e j 26.6
Z L / Z 0  1 100  j50   50
• VSWR is given by:
1 
1  0.45
VSWR 

 2.6
1   1  0.45
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
Z in  ~
 jZ 0 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:
Z o   Z insc Z inoc
• 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.
Solution to Example 4
• To eliminate reflections at terminal AA’, the input
impedance Zin looking into the quarter-wave line
should be equal to Z01 (the characteristic
impedance of the feedline). Thus, Zin = 50Ω .
2
Z 02
Z in 
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.
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
• Average power for incident wave;
Pavi 
V0
2
(W)
2Z 0
• Average power for reflected wave:
Pavr   
2
V0
2
2Z 0
2
   Pavi
Power flow on a lossless
transmission line
• The net average power delivered to the load:
Pav 
Pavi

Pavr

V0
2
2Z 0
1   
2
(W)
Smith Chart
• Smith chart is used to analyze & design
transmission line circuits.
• Impedances on Smith chart are represented by
normalized value, zL for example:
ZL
zL 
Z0
• the normalized load impedance, zL is
dimensionless.
Smith Chart
• Reflection coefficient, Γ :
Z L / Z0 1

Z L / Z0 1
ZL
• Since z L 
Z0
, Γ becomes:
zL 1

zL  1
1 
• Re-arrange in terms of zL: z L 
 rL  jxL
1 
• Normalized load admittance: y L  1  1  
zL
1 
Smith Chart
• The complex Γ plane.
Smith Chart
• The families of circle for rL and xL.
Smith Chart
• Plotting normalized impedance, zL = 2-j1
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 to Example 5
at B, zin = 0.6 –j0.66
VSWR, Voltage Maxima and
Voltage Minima
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.
• The distance between the load and the first
voltage minimum is lmin=(0.037+0.25)λ.
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 to Example 6
• The distance between successive minima is
equal to λ/2. Hence, λ = 40 cm.
• 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  j0.8
• De-normalize (multiplying by Z0) to get ZL:
Z L  500.6  j 0.8  30  j 40 
Solution to Example 6
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 paralllel with the first
section, hence it is called stub.
• The second section is terminated with either
short-circuit or open circuit.
Single- stub matching
Single- stub matching
• The distance d is chosen so as to transform the
load admittance, YL=1/ZL into an admittance of
the form Yd = Y0+jB when looking towards the
load at MM’.
• 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.
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 to Example 7
• The normalized load impedance is
Z
25  j50
(located at A).
zL  L 
 0.5  j
Z0
50
• 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).
Solution to Example 7
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
Solution to Example 7
Second matching point, D.
• At point D, yd  1  j1.58
• Distance B and C 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
Solution to Example 7