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ECE 344
Microwave Fundamentals
Spring 2017
Lectre 03:
Smith Charts
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2
Smith Chart
History.
Construction & Transformation.
Scales.
Examples of how to use Smith Chart to calculate:
Reflection Coefficient.
Standing wave ratio.
Input impedance.
Location of the first maximum and minimum.
Admittance.
ILOS
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The Smith chart developed in 1939 by P. Smith, in the Bell Telephone
laboratories.
It is a graphical procedure for solving impedance transformation problems to
reduce the computational effort required.
In practice, high-frequency circuits often contain two or more transmission
lines interspersed with series and shunt elements.
The Smith chart technique can significantly reduce the numerical and
algebraic manipulations required to solve such problems.
Introduction
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The Smith chart is a specially constructed impedance/admittance diagram
used in solving TL problems. Usefull characteristics of the Smith chart are:
a) all possible values of impedance and admittance can be plotted on the
chart,
b) an easy method for converting impedance to admitance (and vice
versa) is available,
c) the Smith chart provides a simple graphical method for determining
the impedance transformation due to a length of TL
Introduction
The Smith chart, shown in the figure below, is a graphical aid that can be very
useful for solving transmission line problems.
Introduction
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7
Recall:
2
0 0
20 0
0 0
2
0 02
1 1
1 1
11
1 1
z z z
L
z z z
L
z
L
z
L
V z V e e V e z
V VI z e e e z
Z Z
V z zeZ z Z Z
I z e z
Generalized reflection Coefficient: 2 z
Lz e
Generalized Reflection Coefficient
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I (z)
+ Z L
- 0 ,Z V (z)
z=0 z 0
0
LL
L
Z Z
Z Z
2
2L
z
L
j z
L
R I
z e
e e
z j z
Lossless transmission line ( = 0)
2Lj z
Lz e
0
0
LL
L
Z Z
Z Z
Generalized Reflection Coefficient (cont.)
Re 0
1
L
L
Z
For
0
0
0
0
L L
L
L L
L L
L L
R jX Z
R jX Z
R Z jX
R Z jX
2 22 0
2 2
0
L L
L
L L
R Z X
R Z X
Proof:
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Complex Plane
2
2
2
L
L
R I
j z
L
j z
L
j d
L
z
j
e
e
e
Increasing d (toward
generator)
d
Re
Im Decreasing d (toward load)
L
L
L
2L d
Lossless line
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z d
d = distance from load
Smith Chart Construction
It is a transformation from the impedance graph, to a reflection coefficient graph.
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Resistance circles
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Reactance circles
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3) There are two wavelength scales on the periphery of the chart. One is labeled as Wavelengths
toward Generator and the other Wavelengths toward the Load. It is used to find determine the
impedance at a point nearer the input than the known impedance.
The clockwise rotation is referred to as mowing toward the generator and counterclockwise rotation is
referred to as mowing toward the load.
Important notes about smith chart
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Important notes about smith chart
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5.025.00
jZ
Zz L
L
Γ circle |Γ1|
|Γ2|
|Γ1| = |Γ2|
L1L1L1 xrzfor j
0rzfor L2L2 j
2
2
2
||1
||1LrS
As proofed from slide 24
||1
||1
S
88.4SrL2 Circle
xL2 =0
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Solutions steps
1) find the normalized load impedance
2) Locate rL and xL circles on the smith chart.
3) intersection of rL and xL circles will give you a point zL
4) draw the ΓL circle to pass through the point zL, where the center of ΓL circle
is the center of the smith chart.
5) To calculate ΓL , use you ruler to measure the distance 0 zL then project
this distance on the linear scale of reflection coefficient to calculate ΓL
6) the intersection of the ΓL circle with the horizontal line of the smith chart (Γr
axis) gives you the value of SWR as explained in slide 24.
LLL
L jxrZ
Zz
0
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EXAMPLE 2.2 BASIC SMITH CHART OPERATIONS
A load impedance of 40 + j70 Ω terminates a 100 Ω transmission line that is
0.3λ long. Find
1- the reflection coefficient at the load,
2- the reflection coefficient at the input to the line,
3- the input impedance,
4- the standing wave ratio on the line, and
5- the return loss.
Solution
The normalized load impedance is
7.04.00
jZ
Zz L
L
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LL
7.04.0
jxr
jzL
a
b
c
104
59.086.2
69.1||
,L
Lac
ab
9.3SWR
106.0
406.0
3.0
1.615.36
611.0365.0
0 jzZZ
jz
inin
in
the reflection coefficient at
the load
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59.0||
,in
in
The reflection
coefficient at the input
RL = 4.6 dB.
(Calculating Reflection Coefficient) Example 1
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(Calculating input impedance) Example 1 (cont.)
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(Calculating location of Vmin and Vmax) Example 1 (cont.)
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Admittance (Y) Calculations
As an alternative, we can continue to use the original plane,
and add admittance curves to the chart.
1
1nn n
zY z
zG z jB z
Admittance (Y) Chart
1
1n n nR z jX z
zZ z
z
Compare with previous Smith chart derivation, which started with this equation:
If (Rn Xn) = (a, b) is some point on the Smith chart corresponding to = 0,
Then (Gn Bn) = (a, b) corresponds to a point located at = - 0 (180o rotation).
Side note: A 180o rotation on a Smith chart makes a normalized impedance become its reciprocal. 33
Rn circles, rotated 180o, becomes Gn circles.
Xn circles, rotated 180o, becomes Bn circles.
Admittance (Y) Chart (cont.)
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Open ckt.
Match pt.
Gn = 0
Short ckt.
Inductive (Bn < 0)
Capacitive (Bn > 0)
Gn = 1
Bn = +1
Bn = -1
Bn = 0
plane plane
Short-hand version
plane
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Admittance (Y) Chart (cont.)
Gn = 1
Bn = -1
Bn = 1
Impedance and Admittance (ZY) Chart
Short-hand version
plane
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Gn = 1 Rn = 1
Xn = 1
Xn = -1
Bn = -1
Bn = 1
0 50
100 50L
Z
Z j
Find Z(-d)
/ 1/ 4, 3/ 8, 1/ 2gd at
,
0
2 1LL n
ZZ j
Z
/ 4
0.4 0.2
/ 4 20 10
g
n
g
d
Z j
Z j
Example 2
a
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nLZ
3 / 8 gd
/ 4gd
a
b
0
1/ 2 gd
or
Impedance chart
plane
V (-d) +
- z
ZL
z = 0
I (-d)
z = -d
3 / 8
0.5 0.5
3 / 8 25 25
g
n
g
d
Z j
Z j
b
/ 2
2 1
/ 2 100 50
g
n
g
d
Z j
Z j
c
Example 2 (cont.)
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3 / 8 0.5 0.212 0.087
nLZ
3 / 8 gd
/ 4gd
a
b
0/ 2
gd
0.087g
c0.5g
0.462g
0.212gImpedance chart
plane
V (-d) +
- z
ZL
z = 0
I (-d)
z = -d
0 050 20mS
8mS 4mSL
Z Y
Y j
Find Y(-d)
/ 1/ 4, 3/ 8, 1/ 2gd at
,
0
0.4 0.2LL n
YY j
Y
/ 4
2 1
/ 4 40mS 20mS
g
n
g
d
Y j
Y j
Example 3
a
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nLY
3 / 8 gd
/ 4gd
a
b
0
1/ 2 gd
or
c
Admittance chart
plane
V (-d) +
- z
ZL
z = 0
I (-d)
z = -d
3 / 8
1 1
3 / 8 20mS 20mS
g
n
g
d
Y j
Y j
/ 2
0.4 0.2
/ 2 8mS 4mS
g
n
g
d
Y j
Y j
Example 3 (cont.)
b
c
40
nLY
3 / 8 gd
/ 4gd
a
b
0
1/ 2 gd
c
Admittance chart
plane
V (-d) +
- z
ZL
z = 0
I (-d)
z = -d
Simple answer:
* When adding elements in series use Z-chart
* When adding elements in parallel use Y-chart
A B CZ Z Z
A B CY Y Y
Which Chart to Use?
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ZA ZB ZC
YA YB YC
Use a short-circuited section of air-filled TEM, 50 transmission line
( = k0, g =0) to create an impedance of Zin = -j25 at f = 10 GHz.
,
251/ 2
50in nZ j j
0.426 0 0.426g g gL
Example 4
L = 1.28 cm
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0
0 0 0
2 2c
f k
0 3.0 cm
Impedance chart
plane
SC
-1/2
0 .426 g
50
0 g L
SC 50, k0
25inZ j
At this link:
http://www.sss-mag.com/topten5.html
Download the following zip file:
smith_v191.zip
Extract the following files:
smith.exe mith.hlp smith.pdf
This is the application file
Electronic Smith Chart
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