Transcript Chapter 4 - Transmission Lines

CHAPTER 4
TRANSMISSION
LINES
TRANSMISSION LINES
4.1
INTRODUCTION
4.2
SMITH CHART
4.3
IMPEDANCE MATCHING
2
4.1 INTRODUCTION

One of electromagnetic theory application:
 Power
lines, telephone lines and cable
TV.

Develop equation for wave propagation on a
transmission lines.

Introduce Smith chart for the study of
transmission lines and use it to develop
impedance matching networks.
3
INTRODUCTION (Cont’d)
A sinusoidal voltage is
dropped across the resistor.
If the supply and resistor is
connected
by
an
ideal
(negligible length) conductor,
it will be in same phase.
4
INTRODUCTION (Cont’d)
When a quarter wavelength is added between
the supply and the resistor, the voltage at the
resistor is 900 out of phase with the supply
voltage.
5
a sinusoidal voltage is dropped across a resistor. The supply and resistor are
INTRODUCTION (Cont’d)
Transmission lines examples:
Twin lead
Coaxial Cable
6
INTRODUCTION (Cont’d)
Microstrip
with
schematic
cross sections.
A quarterAisquarter is
s along
with schematic
cross sections.
s of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Copyright © 2005 by John Wiley & Sons. All rights reserved.
7
INTRODUCTION (Cont’d)
The distributed
parameters for a
differential segment
of transmission line.
R’ (resistance/meter)
L’ (inductance/meter)
G’ (conductance/meter)
C’ (capacitance/meter)
distributed parameters for a differential segment of transmission line.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
* Primes indicate per unit length or distributed values.
8
INTRODUCTION (Cont’d)
For example, the distributed parameters for a
coaxial cable can be determined by using these
formulas:
2 d
,
ln b a 
2
C'
ln b a 
G' 
L' 

ln b a 
2
R' 
1  1 1  f
,
  
2  a b   c
Where,
d
c
Conductance of dielectric
Conductor conductivity
9
INTRODUCTION (Cont’d)
If the transmission line is modeled using
instantaneous voltage and current,
The distributed-parameter model including instantaneous voltage and current.
10
INTRODUCTION (Cont’d)
Telegraphist’s Equation is used to determine the
basic characteristics for transmission line, which
are:

R' jL'G ' jC '
Propagation constant
   j
R' jL'
Z0 
G ' jC
L'
Z0 
C'
Characteristic impedance
Characteristic impedance for
lossless line
11
INTRODUCTION (Cont’d)
Section of transmission line for attenuation
calculation:
Figure 6-5 (p. 273)
Section of T-line for attenuation calculations.
12
INTRODUCTION (Cont’d)
The power ratio can be expressed as a gain G(dB)
on logarithmic scale, called decibel scale.
 Pout 

G (dB)  10 log 
 Pin 
Or in G(dBm) scale, where to represent absolute
power levels with reference to 1mW
 P 
G (dBm )  10 log

 1mW 
13
INTRODUCTION (Cont’d)
Most of practical problems involving transmission
lines relate to what happens when the line is
terminated.
A T-line terminated with load impedance ZL.
14
INTRODUCTION (Cont’d)
The load is simply the ratio of the voltage and the
current at the load.


V0  V0
Z L  Z0 

V0  V0
It can be rearrange as:
V0

Z L  Z0 

V0
Z L  Z0
15
INTRODUCTION (Cont’d)
If the load is unequal to the characteristic
impedance of the line, the wave must be reflected
back to the load. The degree of impedance
mismatch
is
represented
by
the
reflection
coefficient at the load,

V0
Z L  Z0
L   
Z L  Z0
V0
Range from 0 to 1
16
INTRODUCTION (Cont’d)
Generally, the reflection coefficient at any point
along transmission line is ratio of the reflected
wave to incident wave. Superposition of this two
waves creates a standing wave pattern or voltage
standing wave ratio (VSWR) or ratio of max. to
min. voltage amplitude.
VSWR 
1  L
1  L
Range from 1 to ∞
17
INTRODUCTION (Cont’d)
18
INTRODUCTION (Cont’d)
The terminated T-line can be replaced by an equivalent lumped-element input
impedance.
At any point along the transmission line,
we can find the ratio of the total voltage
to the total current.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
19
INTRODUCTION (Cont’d)
It is known as input impedance,
Z L  Z 0 tanh l 
Z in  Z 0
Z 0  Z L tanh l 
For a special lossless case, it becomes:
Z L  jZ0 tan l 
Z in  Z 0
Z 0  jZ L tan l 
20
EXAMPLE 1
A source with 50  source impedance drives
a 50  transmission line that is 1/8 of
wavelength long, terminated in a load
ZL = 50 – j25 . Calculate:
(i) The reflection coefficient, ГL
(ii) VSWR
(iii) The input impedance seen by the source.
21
SOLUTION TO EXAMPLE 1
It can be shown as:
22
SOLUTION TO EXAMPLE 1 (Cont’d)
(i) The reflection coefficient,
Z L  Z0
L 
Z L  Z0

50  j 25   50

 0.242 e  j 76
50  j 25  50
0
(ii) VSWR
1  L
VSWR 
 1.64
1  L
23
SOLUTION TO EXAMPLE 1 (Cont’d)
(iii) The input impedance seen by the source, Zin
Need to calculate
Therefore,
2  
 

 8 4
 tan

4
1
Z L  jZ 0 tan 
Z in  Z 0
Z 0  jZ L tan 
50  j 25  j 50
 50
50  j 50  25
 30.8  j 3.8
24
4.2 SMITH CHART
25
SMITH CHART (Cont’d)
• Graphical tool for use with transmission line
circuits and microwave circuit elements.
• Only lossless transmission line will be
considered.
• Two graphs in one ;

Plots normalized impedance at any point.

Plots reflection coefficient at any point.
26
SMITH CHART (Cont’d)
The transmission
line calculator,
commonly
referred as the
Smith Chart
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
27
HOW TO USE SMITH CHART?
The Smith Chart is a plot of normalized
impedance. For example, if a Z0 = 50 Ω
transmission line is terminated in a load
ZL = 50 + j100 Ω as below:
28
SMITH CHART (Cont’d)
To locate this point on Smith Chart, normalize the
load impedance, ZNL = ZL/ZN to obtain ZNL = 1 + j2 Ω
29
SMITH CHART (Cont’d)
The normalized load
impedance is located
at the intersection of
the r =1 circle and the
x =+2 circle.
(c)
A T-line terminated in a load (a) shown with values normalized to Z0 in (b). (c) The
location of the normalized load impedance is found on the Smith Chart.
30
SMITH CHART (Cont’d)
The reflection coefficient has a magnitude L
and an angle   :
  L e
j 
Where the magnitude can be measured using a
scale for magnitude of reflection coefficient
provided below the Smith Chart, and the angle is
indicated on the angle of reflection coefficient
scale shown outside the L  1 circle on chart.
31
SMITH CHART (Cont’d)
Scale for magnitude of reflection coefficient
amentals of Electromagnetics
With Engineering
Applications by Stuart M. Wentworth
Scale for angle
of
Copyright
© 2005 by John Wiley & Sons. All rights reserved.
reflection
coefficient
32
SMITH CHART (Cont’d)
For this example,
  L e
 0.7 e
j 
j 45 0
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
33
SMITH CHART (Cont’d)
After locating the normalized impedance point,
draw the constant L e j circle. For example,
the line is 0.3λ length:

34
SMITH CHART (Cont’d)
• Move along the constant
L e j circle is akin

to moving along the transmission line.

Moving away from the load (towards
generator) corresponds to moving in the
clockwise direction on the Smith Chart.

Moving towards the load corresponds to
moving in the anti-clockwise direction on the
Smith Chart.
35
SMITH CHART (Cont’d)
• To find ZIN, move towards the generator by:
Drawing
a line from the center of chart to
outside Wavelengths Toward Generator (WTG)
scale, to get starting point a at 0.188λ
Adding
0.3λ moves along the constant L e j

circle to 0.488λ on the WTG scale.
Read
the corresponding normalized input
impedance point c, ZNIN = 0.175 - j0.08Ω
36
SMITH CHART (Cont’d)
Denormalizing, to find
an input impedance,
Z IN  Z NIN Z 0
Z IN  8.75  j 4
VSWR is
at point b,
VSWR  5.9
37
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
SMITH CHART (Cont’d)
For Z0 = 50Ω ,
a  ZL = 0 (short cct)
b  ZL = ∞ (open cct)
c  ZL = 100 + j100 Ω
d  ZL = 100 - j100 Ω
e  ZL = 50 Ω
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
38
Take out your Smith Chart,
pencil and compass!
LETS TRY!!
39
EXAMPLE 2
Repeat Example 1 using the Smith Chart.
40
SOLUTION TO EXAMPLE 2
(i) Locate the normalized load, and label it as
point a, where it corresponds to
Z NL  1  j 0.5
(ii) Draw constant L e j circle.

(iii) It can be seen that
L  0.245e
j  760
and
VSWR  1.66
41
SOLUTION TO EXAMPLE 2 (Cont’d)
(iv) Move from point a (at 0.356λ) on the WTG
scale, clockwise toward generator a distance
λ/8 or 0.125λ to point b, which is at 0.481λ.
We could find that at this point, it corresponds
to
Z NIN  0.62  j 0.07
Denormalizing it,
Z IN  31  j 3.5
42
EXAMPLE 3
The input impedance for a 100 Ω lossless
transmission line of length 1.162 λ is
measured as 12 + j42Ω. Determine the
load impedance.
44
SOLUTION TO EXAMPLE 3
(i) Normalize the input impedance:
Z in 12  j 42
zin 

 0.12  j 0.42
Z0
100
(ii) Locate the normalized input impedance and
label it as point a
45
SOLUTION TO EXAMPLE 3 (Cont’d)
(iii) Take note the value of wavelength for point a at
WTL scale.
At point a, WTL = 0.436λ
(iv) Move a distance 1.162λ towards the load to point b
WTL = 0.436λ + 1.162λ
= 1.598λ
But, to plot point b, 1.598λ – 1.500λ = 0.098λ
Note: One complete rotation of WTL/WTG = 0.5λ
46
SOLUTION TO EXAMPLE 3 (Cont’d)
(v) Read the point b:
Z NL  0.15  j 0.7
Denormalized it:
Z L  Z NL Z0
 15  j 70
47
EXAMPLE 4
On a 50  lossless transmission line,
the VSWR is measured as 3.4. A voltage
maximum is located 0.079λ away from
the load (towards generator). Determine
the load.
49
SOLUTION TO EXAMPLE 4
j
(i) Use the given VSWR to draw a constant L e

circle.
(ii) Then move from maximum voltage at
WTG = 0.250λ (towards the load) to point a
at WTG = 0.250λ - 0.079λ = 0.171λ.
(iii) At this point we have ZNL = 1 + j1.3 Ω,
or ZL = 50 + j65 Ω.
50
EXAMPLE 5 (TRY THIS!)
Use Smith Chart to determine the input impedance
Zin of the two line configuration shown as below:
52
ANSWER FOR EXAMPLE 5
ZIN = 65.7 – j 124.7Ω
53
4.3 IMPEDANCE MATCHING
• The transmission line is said to be matched
when Z0 = ZL which no reflection occurs.
• The purpose of matching network is to
transform the load impedance ZL such that the
input impedance Zin looking into the network is
equal to Z0 of the transmission line.
54
IMPEDANCE MATCHING (Cont’d)
Adding an impedance matching networks
ensures that all power make it or delivered to
the load.
Adding an impedance-matching network ensures that all power will make it to the
load.
55
IMPEDANCE MATCHING (Cont’d)

Techniques of impedance matching :

Quarter-wave transformer

Single / double stub tuner

Lumped element tuner

Multi-section transformer
56
QUARTER WAVE TRANSFORMER
The
quarter
wave
transformer
matching
network only can be constructed if the load
impedance is all real (no reactive component)
Quarter-wave transformer.
57
QUARTER WAVE TRANSF. (Cont’d)
To find the impedance looking into the quarter
wave long section of lossless ZS impedance line
terminated in a resistive load RL:
RL  jZ S tan l
Zin  Z S
Z S  jRL tan l
2  
 ,
But, for quarter wavelength, l 
 4 2
tan l  
58
QUARTER WAVE TRANSF. (Cont’d)
So,
2
ZS
Z in 
 Z0
RL
Rearrange to get impedance matched line,
Z S  Z 0 RL
59
IMPEDANCE MATCHING (Cont’d)
• It much more convenient to add shunt
elements rather than series elements  Easier
to work in terms of admittances.
• Admittance:
1
Y
Z
60
IMPEDANCE MATCHING (Cont’d)
Adding shunt elements using admittances:
Figure 6-25 (p. 299)
(a) Admittance
(b) Adding
elements
using
With
Smithrelationship
chart,to impedance.
it is easy
toshunt
find
normalized
admittances.
admittance – move to a point on the opposite
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
side of the constant L e j circle.

61
d
Smith
SHUNT STUB MATCHING NETWORK
The matching network has to transform the real part
of load impedance, RL to Z0 and reactive part, XL to
zero  Use two adjustable parameters – e.g. shuntstub.
(a) The generic layout of the shorted shunt-stub matching network.
63
SHUNT STUB MATCHING NET. (Cont’d)
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 ytot = yd + yl = 1 for
complete matching network.
64
SHUNT STUB USING SMITH CHART
• Locate the normalized load impedance ZNL.
• Draw constant SWR circle and locate YNL.
• Move clockwise (WTG) along L e j circle to intersect
with 1 ± jB  value of yd.

• The length moved from YNL towards yd is the through
line length, d.
• Locate yl at the point  jB .
• Depends on the shorted/open stub, move along the
periphery of the chart towards yl (WTG).
• The distance traveled is the length of stub, l .
65
SHORTED SHUNT STUB MATCHING
Generic layout of the shorted shunt stub
matching network:
Figure 6-28a (p. 302)
(a) The generic layout of the shorted shunt-stub matching network.
66
EXAMPLE 6 (TRY THIS!)
Construct the shorted shunt stub
matching network for a 50Ω line
terminated in a load ZL = 20 – j55Ω
67
SOLUTION TO EXAMPLE 6
1.
Locate the normalized load impedance,
ZNL = ZL/Z0 = 0.4 – j1.1Ω
2.
Draw constant L e j circle.
3.
Locate YNL. (0.112λ at WTG)
4.
Moving to the first intersection with the

1 ± jB circle, which is at 1 + j2.0  yd
5. Get the value of through line length, d
 from 0.112λ to 0.187λ on the WTG scale,
so d = 0.075λ
68
SOLUTION TO EXAMPLE 6 (Cont’d)
6. Locate the location of short on the Smith Chart
(note: when short circuit, ZL = 0, hence YL = ∞)
on the right side of the chart with WTG=0.25λ
7. Move clockwise (WTG) until point 
jB, which
is at 0 - j2.0, located at WTG= 0.324λ  yl
8. Determine the stub length, l
 0.324λ – 0.25λ = 0.074 λ
69
SOLUTION TO EXAMPLE 6 (Cont’d)
Thus, the values are:
d = 0.075 λ
l = 0.074 λ
yd = 1 + j2.0 Ω
yl = -j2.0 Ω
Where YTOT = yd + yl = (1 + j2.0) + (-j2.0) = 1
70
OPEN END SHUNT STUB MATCHING
Generic layout of the open ended shunt stub
matching network:
(a) The generic layout of the open-ended shunt-stub matching network.
72
EXAMPLE 7 (TRY THIS!)
Construct an open ended shunt stub
matching network for a 50Ω line
terminated in a load ZL = 150 + j100 Ω
73
SOLUTION TO EXAMPLE 7
1.
Locate the normalized load impedance,
ZNL = ZL/Z0 = 3.0 + j2.0Ω
2.
Draw constant L e j
3.
Locate YNL. (0.474λ at WTG)
4.
Moving to the first intersection with the

circle.
1 ± jB circle, which is at 1 + j1.6  yd
5. Get the value of through line length, d
 from 0.474λ to 0.178λ on the WTG scale,
so d = 0.204λ
74
SOLUTION TO EXAMPLE 7 (Cont’d)
6. Locate the location of open end on the Smith
Chart (note: when short circuit, ZL = ∞, hence
YL = 0) on the left side with WTG = 0.00λ
7. Move clockwise (WTG) until point 
jB, which
is at 0 – j1.6, located at WTG= 0.339λ  yl
8. Determine the stub length, l
 0.339λ – 0.00λ = 0.339 λ
75
SOLUTION TO EXAMPLE 7 (Cont’d)
Thus, the values are:
d = 0.204 λ
l = 0.339 λ
yd = 1 + j1.6 Ω
yl = -j1.6 Ω
Where YTOT = yd + yl = (1 + j1.6) + (-j1.6) = 1
76
p. 304)
on to
IMPORTANT!!
In both previous example, we chose the first
intersection with the1 ± jB circle in designing our
matching network. We could also have continued
on to the second intersection.
Thus, try both intersection to determine which
solution produces max/min length of through
line, d or length of stub, l.
78
EXERCISE (TRY THIS!)
Determine the through line length and stub
length for both example above by using second
intersection.
For shorted shunt stub (example 6):
d = 0.2 λ and l = 0.426 λ
For open ended shunt stub (example 7):
d = 0.348 λ and l = 0.161 λ
79
CHAPTER 4
END