CHAPTER 4
TRANSMISSION LINES
2
TRANSMISSION LINES
4.1 INTRODUCTION
4.2 SMITH CHART
4.3 IMPEDANCE MATCHING
3
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.
4.1 INTRODUCTION
4
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.
5
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.
INTRODUCTION (Cont’d)
6
Transmission lines examples:
INTRODUCTION (Cont’d)
Twin lead
Coaxial Cable
7
INTRODUCTION (Cont’d)
Microstrip
8
The distributed
parameters for a
differential segment
of transmission line.
R’ (resistance/meter) L’ (inductance/meter)
G’ (conductance/meter) C’ (capacitance/meter)
* Primes indicate per unit length or distributed values.
INTRODUCTION (Cont’d)
9
,11
2
1'
ln2
'
ln
2'
,ln
2'
c
d
f
baR
abL
abC
abG
INTRODUCTION (Cont’d)
For example, the distributed parameters for a
coaxial cable can be determined by using these
formulas:
Where,
Conductance of dielectric
Conductor conductivity
dc
10
If the transmission line is modeled using
instantaneous voltage and current,
INTRODUCTION (Cont’d)
11
Telegraphist’s Equation is used to determine
the basic characteristics for transmission line,
which are:
'
'
'
''
''''
0
0
C
LZ
CjG
LjRZ
j
CjGLjR
Propagation constant
Characteristic impedance
Characteristic impedance for
lossless line
INTRODUCTION (Cont’d)
12
Section of transmission line for attenuation calculation:
INTRODUCTION (Cont’d)
13
in
out
P
PdBG log10)(
The power ratio can be expressed as a gain
G(dB) on logarithmic scale, called decibel scale.
mW
PdBG m 1
log10)(
Or in G(dBm) scale, where to represent
absolute power levels with reference to 1mW
INTRODUCTION (Cont’d)
14
Most of practical problems involving
transmission lines relate to what happens when
the line is terminated.
INTRODUCTION (Cont’d)
15
The load is simply the ratio of the voltage and
the current at the load.
00
000
VV
VVZZL
It can be rearrange as:
00
00 V
ZZ
ZZV
L
L
INTRODUCTION (Cont’d)
16
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,
0
0
0
0
ZZ
ZZ
V
V
L
LL
INTRODUCTION (Cont’d)
Range from 0 to 1
17
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.
L
LVSWR
1
1
INTRODUCTION (Cont’d)
Range from 1 to ∞
18
INTRODUCTION (Cont’d)
19
At any point along the transmission line, we can find the ratio of the total voltage to the total current.
INTRODUCTION (Cont’d)
20
lZZ
lZZZZ
L
Lin
tanh
tanh
0
00
It is known as input impedance,
For a special lossless case, it becomes:
ljZZ
ljZZZZ
L
Lin
tan
tan
0
00
INTRODUCTION (Cont’d)
21
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.
22
SOLUTION TO EXAMPLE 1
It can be shown as:
23
(i) The reflection coefficient,
076
0
0
242.0502550
502550 j
L
LL
ej
j
ZZ
ZZ
(ii) VSWR
64.11
1
L
LVSWR
SOLUTION TO EXAMPLE 1 (Cont’d)
24
(iii) The input impedance seen by the source, Zin
8.38.30
255050
50255050
tan
tan
0
00
j
j
jj
jZZ
jZZZZ
L
Lin
48
2 1
4tan
Need to calculate
Therefore,
SOLUTION TO EXAMPLE 1 (Cont’d)
25
4.2 SMITH CHART
26
• 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.
SMITH CHART (Cont’d)
27
The transmission
line calculator,
commonly
referred as the
Smith Chart
SMITH CHART (Cont’d)
28
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:
29
To locate this point on Smith Chart, normalize the
load impedance, ZNL = ZL/ZN to obtain ZNL = 1 + j2
Ω
SMITH CHART (Cont’d)
30
The normalized load
impedance is located
at the intersection of
the r =1 circle and the
x =+2 circle.
SMITH CHART (Cont’d)
31
The reflection coefficient has a magnitude
and an angle : j
L e
L
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
circle on chart.
1L
SMITH CHART (Cont’d)
32
SMITH CHART (Cont’d)
Scale for magnitude of reflection coefficient
Scale for angle of reflection coefficient
33
For this example,
0457.0 j
jL
e
e
SMITH CHART (Cont’d)
34
After locating the normalized impedance
point, draw the constant circle. For
example, the line is 0.3λ length:
jL e
SMITH CHART (Cont’d)
35
• Move along the constant 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.
jL e
SMITH CHART (Cont’d)
36
• 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
circle to 0.488λ on the WTG scale.
Read the corresponding normalized input
impedance point c, ZNIN = 0.175 - j0.08Ω
jL e
SMITH CHART (Cont’d)
37
Denormalizing, to find an input impedance,
475.80
jZ
ZZZ
IN
NININ
SMITH CHART (Cont’d)
VSWR is at point b,
9.5VSWR
38
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
SMITH CHART (Cont’d)
39
Take out your Smith Chart, pencil and compass!
LETS TRY!!
40
EXAMPLE 2
Repeat Example 1 using the Smith Chart.
41
SOLUTION TO EXAMPLE 2
(i) Locate the normalized load, and label it as point a, where it corresponds to
(ii) Draw constant circle.
(iii)It can be seen that
5.01 jZNL
jL e
076245.0 j
L e and 66.1VSWR
42
(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
Denormalizing it,
07.062.0 jZNIN
5.331 jZ IN
SOLUTION TO EXAMPLE 2 (Cont’d)
44
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.
45
SOLUTION TO EXAMPLE 3
(i) Normalize the input impedance:
(ii) Locate the normalized input impedance and
label it as point a
42.012.0100
4212
0
jj
Z
Zz inin
46
(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
SOLUTION TO EXAMPLE 3 (Cont’d)
47
(v) Read the point b:
7.015.0 jZNL
Denormalized it:
70150
j
ZZZ NLL
SOLUTION TO EXAMPLE 3 (Cont’d)
49
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.
50
(i) Use the given VSWR to draw a constant
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.
SOLUTION TO EXAMPLE 4
jL e
52
EXAMPLE 5 (TRY THIS!)
Use Smith Chart to determine the input
impedance Zin of the two line configuration shown
as below:
53
ZIN = 65.7 – j 124.7Ω
ANSWER FOR EXAMPLE 5
54
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.
55
Adding an impedance matching networks ensures that all power make it or delivered to the load.
IMPEDANCE MATCHING (Cont’d)
56
Techniques of impedance matching :
Quarter-wave transformer
Single / double stub tuner
Lumped element tuner
Multi-section transformer
IMPEDANCE MATCHING (Cont’d)
57
QUARTER WAVE TRANSFORMER
The quarter wave transformer matching
network only can be constructed if the load
impedance is all real (no reactive component)
58
To find the impedance looking into the quarter
wave long section of lossless ZS impedance
line terminated in a resistive load RL:
ljRZ
ljZRZZ
LS
SLSin
tan
tan
, 24
2 lBut, for quarter wavelength,
l tan
QUARTER WAVE TRANSF. (Cont’d)
59
So,
0
2
ZR
ZZ
L
Sin
Rearrange to get impedance matched line,
LS RZZ 0
QUARTER WAVE TRANSF. (Cont’d)
60
• It much more convenient to add shunt elements rather than series elements Easier to work in terms of admittances.
• Admittance:Z
Y1
IMPEDANCE MATCHING (Cont’d)
61
Adding shunt elements using admittances:
With Smith chart, it is easy to find normalized
admittance – move to a point on the opposite
side of the constant circle. jL e
IMPEDANCE MATCHING (Cont’d)
63
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. shunt-stub.
64
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.
SHUNT STUB MATCHING NET. (Cont’d)
65
• Locate the normalized load impedance ZNL.
• Draw constant SWR circle and locate YNL.
• Move clockwise (WTG) along 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 .
SHUNT STUB USING SMITH CHART
jL e
66
SHORTED SHUNT STUB MATCHING
Generic layout of the shorted shunt stub
matching network:
67
EXAMPLE 6 (TRY THIS!)
Construct the shorted shunt stub
matching network for a 50Ω line
terminated in a load ZL = 20 –
j55Ω
68
1. Locate the normalized load impedance,
ZNL = ZL/Z0 = 0.4 – j1.1Ω
2. Draw constant 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
SOLUTION TO EXAMPLE 6
jL e
69
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
SOLUTION TO EXAMPLE 6 (Cont’d)
70
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
SOLUTION TO EXAMPLE 6 (Cont’d)
72
OPEN END SHUNT STUB MATCHING
Generic layout of the open ended shunt
stub matching network:
73
EXAMPLE 7 (TRY THIS!)
Construct an open ended shunt stub
matching network for a 50Ω line
terminated in a load ZL = 150 + j100
Ω
74
SOLUTION TO EXAMPLE 7
1. Locate the normalized load impedance,
ZNL = ZL/Z0 = 3.0 + j2.0Ω
2. Draw constant circle.
3. Locate YNL. (0.474λ at WTG)
4. Moving to the first intersection with the
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
jL e
75
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
SOLUTION TO EXAMPLE 7 (Cont’d)
76
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
SOLUTION TO EXAMPLE 7 (Cont’d)
78
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.
IMPORTANT!!
79
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
EXERCISE (TRY THIS!)
CHAPTER 4
END