Transcript Fundamentals of Applied Electromagnetics

EKT241 – ELECTROMAGNETICS THEORY
Chapter 5 Transmission Lines
Chapter Objectives




Introduction to transmission lines
Lump-element model that represent TEM lines
Lossless line
Smith Chart to analyze transmission line problem
Chapter Outline
5-1)
5-2)
5-3)
5-4)
5-5)
5-6)
5-7)
5-8)
5-9)
5-10)
5-11)
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
5-1 General Considerations
•
Transmission lines connect a generator circuit
to a load circuit at the receiving end.
•
Transverse electromagnetic (TEM) lines have
waves that propagate transversely.
5-2 Lumped-Element Model
•
Transmission lines can be represented by a
lumped-element circuit model.
5-2 Lumped-Element Model
•
1.
2.
3.
4.
Lumped-element circuit model consists 4
transmission line parameters:
R’  (Ω/m)
L’  (H/m)
G’  (S/m)
C’  (F/m)
5-2 Lumped-Element Model
•
In summary,
•
All TEM transmission lines share the relations:
L'C '  
G' 

C' 
where µ, σ, ε = properties of conductor
5-3 Transmission-Line Equations
•
Transmission line equations in phasor form is
given as
~
dV z 
~

 R' jL'I z 
dz
~
dI z 
~

 G ' jC 'V z 
dz
5-4 Wave Propagation on a Transmission Line
•
The wave equation is derived as
~
d V z  2 ~
  V z   0 where  
2
dz
2
R'  jL'G'  jC '
Complex propagation constant
•
γ has real part α (attenuation constant) and
imaginary part β (phase constant).
 R' jL'G' jC ' (Np/m)
  Jm   Jm R' jL'G ' jC '  (rad/m)
  R e   R e
5-4 Wave Propagation on a Transmission Line
•
Characteristic impedance Z0 of the line is
Z0 
•
R' jL'

R' jL'

G ' jC '
 
Phase velocity for propagating wave is

u p  f 

where f = frequency (Hz)
λ = wavelength (m)
Example 5.1 Air Line
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 50 and
phase constant of 20 rad/m at 700 MHz, find the
inductance per meter and the capacitance per meter of
the line.
Solution 5.1 Air Line
The following quantities are given:
Z 0  50,   20 rad/m, f  700 MHz  7 108 Hz
With R’ = G’ = 0,
  Jm  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
5-5 The Lossless Transmission Line
•
Low R’ and G’ for transmission line is called
lossless transmission line.
 0
(lossless line)
   L' C ' (lossless line)
Z0 
•
L'
C'
(lossless line)
Using relation properties,
    (rad/m)
p 
1

(m/s)
5-5 The Lossless Transmission Line
•
Wavelength is given by
p
0
c 1



f
f r
r
•
where εr = relative permittivity
For the lossless line, there are 2 unknowns in the
equations for the total voltage and current on the line.
5-5.1 Voltage Reflection Coefficient
•
The relations for lossless are
V0 Z L  Z 0 Z L Z 0  1
  

V0
Z L  Z0 Z L Z0  1
I 0
V0
    

I0
V0
   e jr
•
A load that is matched to the line when ZL = Z0, Γ
= 0 and V0−= 0.
Example 5.2 Reflection Coefficient of a Series RC Load
A 100-Ω transmission line is connected to a load
consisting of a 50-Ω resistor in series with a 10-pF
capacitor. Find the reflection coefficient at the load for
a 100-MHz signal.
Solution 5.2 Reflection Coefficient of a Series RC Load
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
 50  j
1
 50  j159 
8
11
2  10  10
Voltage reflection coefficient is

Z L / Z 0  1 0.5  j1.59  1

 0.67e j119.3  0.76  60.7
Z L / Z 0  1 0.5  j1.59  1
5-5.2 Standing Waves
• 3 types of voltage standing-wave patterns:
(a) Matched load
(b) Short-circuited line
(c) Open-circuited line
5-5.2 Standing Waves
•
To find maximum and minimum values of voltage
magnitude, we have
 r  2n  r  n


2
4
2
 n  1,2... if  r  0

n  0,1,2... if  r  0
 z  lmax 
5-5.2 Standing Waves
•
First voltage maximum occurs at
lmax
•
First voltage minimum occurs at
lmin
•
r

where n  0
4
lmax   / 4 if lmax   / 4

lmax   / 4 if lmax   / 4
Voltage standing-wave ratio S is defined as
~
V
1 
max
S ~

(dimension less)
1 
V
min
Example 5.4 Standing-wave Ratio
A 50- transmission line is terminated in a load with
ZL = (100 + j50)Ω . Find the voltage reflection
coefficient and the voltage standing-wave ratio (SWR).
Solution
We have,
Z L / Z 0  1 100  j50   50


 0.45e j 26.6
Z L / Z 0  1 100  j50   50
S is given by
1 
1  0.45
S

 2.6
1   1  0.45
5-6 Input Impedance of the Lossless Line
•
•
Voltage to current ratio is called input
impedance Zin.
The input impedance at z = −l is given as
 Z L cos l  jZ0 sin l 
 Z L  jZ0 tan l 
  Z 0 

Z in  l   Z 0 
 Z 0 cos l  jZL sin l 
 Z 0  jZL tan l 
and
~

V
Z in 
1

g


V0  

 Z g  Z in  e jl  e  jl 


Example 5.6 Complete Solution for v(z, t) and i(z, t)
A 1.05-GHz generator circuit with series impedance
Zg = 10Ω and voltage source given by
vg t   10 sin t  30 V 
is connected to a load ZL = (100 + j50) through a
50-Ω, 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 v(z, t) and i(z, t) on the line.
Solution 5.6 Complete Solution for v(z, t) and i(z, t)
We find the wavelength from
0.7  3  10 8


 0.2m
f
1.05  10 9
up
and
 2l 
 2 0.67 
tan l   tan 
  tan 
  tan 126
  
 0.2 
The voltage reflection coefficient at the load is

Z L  Z 0 100  j50   50

 0.45e j 26.6
Z L  Z 0 100  j50   50
The input impedance of the line
 Z  jZ0 tan l 
Z in  Z 0  L
  21.9  j17.4
 Z 0  jZ0 tan l 
Solution 5.6 Complete Solution for v(z, t) and i(z, t)
Rewriting the expression for the generator voltage,
vg t   10 sin t  30

 10 cost  60  e 10e 60 e jwt
Thus the phasor voltage is
 V
~
Vg  10e  j 60  10  60 V 
The voltage on the line is
~


V
1

g Z in 

j159

V0 

10
.
2
e
 10.2159  V 
 jl

 Z g  Z in  e  e  jl 


and phasor voltage on the line is



~
V z   V0 e jz  e jz  10.2e j159 e jz  0.45e j 26.6e jz

Solution 5.6 Complete Solution for v(z, t) and i(z, t)
The instantaneous voltage and current is


~
vz, t   e V z e jt  10.2 cost  z  159
 4.55 cost  z  185.6 V 
i  z , t   0.20 cost   z  159
 0.091 cost  z  5.6
A 
5-7 Special Cases of the Lossless Line
•
Special cases
has useful properties.
5-7 .1 Short-Circuited Line
•
For short-circuited line at
z = −l,
Z insc
~
Vsc  l 
~
 jZ0 tan l
I sc  l 
Example 5.7 Equivalent Reactive Elements
Choose the length of a shorted 50- lossless transmission
line (Fig. 5-16) such that its input impedance at 2.25
GHz is equivalent to the reactance of a capacitor with
capacitance Ceq = 4 pF. The wave velocity on the line is
0.75c.
Solution 5.7 Equivalent Reactive Elements
We are given
u p  0.75c  2.25  10 8 m/s
Z 0  50
f  2.25  10 9 Hz
Ceq  4  10 12 F
The phase constant is
2nd
quadrant is

2


l1  2.8 rad or l1 
4th quadrant is l2  5.94 rad
2f
1
 62.8 rad/m, tan l  
 0.354
up
Z 0Ceq
2.8
or l2 


2.8
 4.46 cm
62.8
5.94
 9.46 cm
62.8
Any length l = 4.46 cm + nλ/2, where n is a positive
integer, is also a solution.
5-7.2 Open-Circuited Line
•
With ZL = ∞, it forms an
open-circuited line.
V  l 
Z inoc  ~oc
  jZ0 cot l
I oc  l 
5-7.3 Application of Short-Circuit and Open-Circuit Measurements
• Product and ratio of SC and OC equations give
the following results:
Z o   Z insc Z inoc
tan  l 
 Z insc
Z inoc
• Radio-frequency (RF) instruments measure the
impedance of any load.
Example 5.8 Measuring Z0 and β
Find Z0 and β of a 57-cm-long lossless transmission
line whose input impedance was measured as Zscin =
j40.42Ω when terminated in a short circuit and as
Zocin = −j121.24Ω when terminated in an open circuit.
From other measurements, we know that the line is
between 3 and 3.25 wavelengths long.
Solution 5.8 Measuring Z0 and β
We have,
Z 0   Z insc Z inoc 
 j 40.42  j121 .24   70
 Z insc
1
tan l 

oc
3
Z in
True value of βl is
l  6 

6
 19.4 rad 
and
19.4

 34 rad/m 
0.57
Example 5.9 Quarter-Wave Transformer
A 50-Ω lossless transmission line is to be matched to a
resistive load impedance with ZL = 100Ω via a quarterwave section as shown, thereby eliminating reflections
along the feedline. Find the characteristic impedance of
the quarter-wave transformer.
Solution 5.9 Quarter-Wave Transformer
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 = 50 .
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.
5-8 Power Flow on a Lossless Transmission Line
•
We shall examine the flow of power carried by
incident and reflected waves.
5-8.1 Instantaneous Power
•
Instantaneous power is the product of
instantaneous voltage and current.
5-8.2 Time-Average Power
•
More interested in time-averaged power flow.
5-8.2 Time-Average Power
• There are 2 types of approach:
1) Time-Domain Approach
• Incident power and reflected wave power are
Pavi 
•
V0
2
2Z 0
(W)
Pavr   
2
V0
2
2Z 0
2
   Pavi
For net average power delivered to the load,
Pav  Pavi  Pavr 
V0
2
2Z 0
1   
2
(W)
5-8.2 Time-Average Power
2) Phasor-Domain Approach
• Time-average power for any propagating wave is

1
~ ~*
Pav  R e V  I
2

5-9 Smith Chart
•
The Smith Chart is used for analyzing and
designing transmission-line circuits.
5-9 Smith Chart
•
•
Impedances represented by normalized values, Z0.
Reflection coefficient is
zL 
•
1 
1 
Normalized load
admittance is
1 1 
yL 

(dimension less)
zL 1  
Example 5.11 Determining ZL using the Smith Chart
Given that the voltage standing-wave ratio is S = 3 on
a 50-Ω line, that 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 first voltage minimum is at
5
l min 
 0.125
40
Solution 5.11 Determining ZL using the Smith Chart
From Smith Chart,
rL  S  3
The normalized load
impedance at point C is
z L  0.6  j 0.8
Multiplying by Z0 = 50Ω ,
we obtain
Z L  500.6  j 0.8  30  j 40 
5-10 Impedance Matching
•
•
Transmission line is matched to the load when
Z0 = ZL.
Alternatively, place an impedance-matching
network between load and transmission line.
Example 5.12 Single-Stub Matching
50-Ω transmission line is connected to an antenna
with load impedance ZL = (25 − j50). Find the position
and length of the short-circuited stub required
to match the line.
Solution
The normalized load impedance is
zL 
Z L 25  j50

 0.5  j
Z0
50
Located at point A.
Solution 5.12 Single-Stub Matching
Value of yL at B is yL  0.4 
0.115λ on the WTG scale.
At C,
yd  1  j1.58
j 0.8
which locates at position
located at 0.178λ on the WTG scale.
Distant B and C is d  0.178  0.155  0.063
Normalized input admittance at
the juncture is
yin  ys  yd
1  j 0  ys  1  j1.58
ys   j1.58
Solution 5.12 Single-Stub Matching
Normalized admittance of −j 1.58 at F and position
0.34λ on the WTG scale gives l1  0.34  0.25  0.09
At point D, yd  1  j1.58
Distant 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 5.12 Single-Stub Matching
5-11 Transients on Transmission Lines
•
•
Transient response is a time record of voltage
pulse.
An example of step function is shown below.
5-11.1 Transient Response
•
Steady-state voltage V∞ for d-c analysis of the
circuit is
V 
Vg Z L
Rg  Z L
where Vg = DC voltage source
•
Steady-state current is
Vg
V
I 

Z L Rg  Z L