Credit Hours Engineering Programme
ECE 261; Engineering Electromagnetics
Smith Chart
Prof. Hani Ghali
S1, Academic Year 2016/2017
December 2016
o Objectives;
Development of a graphical tool for solving transmission line problems
Practice Smith Chart in transmission line analysis
Solve matching problems using Smith Chart
ECE_261; Smith Chart
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o Reflection Coefficient; Parametric Equations
Graphical representation of the reflection coefficient; Smith Chart
o In general, the reflection coefficient is complex value; e j r ji
r cos & i sin
( r & i 1)
The Smith Chart lies in the complex plane;
Point A; =0.3 + j0.4 0.553º
Point B; =-0.5 - j0.2 0.54 202º
Short circuit; =-1 1180º
Open circuit; =1 10º
ECE_261; Smith Chart
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o Parametric Equations; (Cont.)
Rewrite the reflection coefficient as; e j r ji
Z
o Define; zL L
Zo
normalized load impedance; dimensionless
R jX L
Z
Z Zo ZL / Zo 1 zL 1
zL L L
rL jxL & L
Zo
Zo
ZL Zo ZL / Zo 1 zL 1
r jxL 1
(1 r ) ji
r ji L
Or rr jxL
rL jxL 1
(1 r ) ji
r jxL 1
Solving the above equation; r ji L
rL jxL 1
Expressions (and a plot) for r and i (x-y plane) in terms of rL and xL
ECE_261; Smith Chart
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o Parametric Equations; (Cont.)
rL 2
1 2
) i2 (
)
1 rL
1 rL
1 2
1
(r 1)2 (i
) ( )2
xL
xL
(r
o These equations are represented by family of circles of radius
& their centers located at (
ECE_261; Smith Chart
rL
1
, 0 ) or (1,
)
1 rL
xL
1
1
or
1 rL xL
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o Smith Chart;
Plot of; (r
rL 2
1 2
) i2 (
)
1 rL
1 rL
o Circles with centers located on the x-axis (i=0); (
& radius
1
1 rL
Plot of; (r 1)2 (i
rL
, 0)
1 rL
1 2
1
) ( )2
xL
xL
o Circles with centers located on the y-axis (r=0);
& radius
ECE_261; Smith Chart
1
xL
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o Smith Chart; (Cont.)
Positive xL circles
rL circles
Negative xL circles
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o Smith Chart; (Cont.)
For any load impedance normalized impedance, two circles are required to
define the real part and the imaginary part of the load
Two infinite number of circles for all possible values of rL and xL
The family of circles represent the so called; Smith Chart
http://em7e.eecs.umich.edu/ch2/mod2_6/Chart.html
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o Smith Chart; (Cont.)
Once the load, normalized load impedance, has been located on Smith Chart,
the vector from the origin to the load point represents the reflection
coefficient
1. The magnitude of this vector is the magnitude of the reflection
coefficient
2. The angle which the vector makes with the x-axis (r) is the phase of the
reflection coefficient
z 1 2 j1 1 1 j1
rr jxL 2 j1 L
0.45 26.6
Case (1);
z L 1 2 j1 1 3 j1
http://em7e.eecs.umich.edu/ch2/mod2_6/Chart.html
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o Smith Chart; (Cont.)
The angle of the reflection coefficient is measured on one of the outer scales
called angle of reflection coefficient
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o Smith Chart; Input Impedance
The input impedance of a transmission line is given by;
1 e j 2 l
• Zin (z l ) Zo
j
2
l
1 e
1 e j 2 l
& the normalized input impedance is; zin (z l )
1 e j 2 l
The quantity e j is the reflection coefficient @ the load
Define the input reflection coefficient, i.e.; the reflection coefficient measured
at the input of a transmission line of length (l)
l e 2 jl e je 2 jl e j( 2l )
zin (z l )
ECE_261; Smith Chart
1 l
1 l
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o Smith Chart; Input Impedance (Cont.)
Based on the two forms; zin (z l )
1
1 l
& zL
1
1 l
If (@ the load) is transformed into l , zL will be transformed to zin
o Using Smith chart, transforming to l means keeping the magnitude of
constant and decreasing the phase by 2l, which corresponds to rotation in a
clockwise direction on the Smith chart
One full rotation on Smith chart is gained for a length of l/2
The outermost scale around the perimeter is called wavelength toward
generator: WTG
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o Input Impedance; Class Activity
Case (2); Consider a 50- lossless transmission line terminated in a load
ZL=(100-j50), find Zin at a distance l=0.1l from the load toward the
generator
100 j50
Z
2 j1
Hints; zL L
Zo
50
Moving on a constant circle a distance 0.1 l (Point B)
The impedance @ B represents the input impedance at 0.1l from the load;
zin (at B ) 0.6 j0.66
Zin (0.1l ) zin Zo ( 30 j33)
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o Input Impedance; Class Activity (Cont.)
Case (2); Consider a 50- lossless transmission line terminated in a load
ZL=(100-j50), find Zin at a distance l=0.1l from the load toward the
generator
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o Smith Chart; Maxima and Minima
Case (3); Consider a load zL 2 j1 @ Smith chart at Point A
1. When this point moves towards the generator a distance l=0.037l till it
intersects with the +ve real axis (i=0) at point Pmax, the input reflection
coefficient l becomes purely +ve real value & the total voltage on the line
will be maximum (current is minimum).
2. Moving further (0.037l+0.25l) till it intersects with the -ve real axis (i=0) at
point Pmin, the input reflection coefficient l becomes purely -ve real value
& the total voltage on the line will be minimum (current is maximum).
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o Smith Chart; Maxima and Minima (Cont.)
Case (3); Consider a load zL 2 j1 @ Smith chart at Point A
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o Smith Chart; VSWR
Case (4); Consider the load zL 2 j1 @ Smith chart at Point A
When this point moves towards generator a distance l=0.037l till it intersects
with the +ve real axis (i=0) at point Pmax
r 1
S1
l r l
&
rl 1
S1
Similarity of these two equations the value of rl on the +ve real axis
(rl>1 at Pmax) is equal to VSWR
ECE_261; Smith Chart
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o Smith Chart; Class Activity
Case (5); A 50- lossless line is terminated in a load impedance ZL=(25+j50),
use Smith chart to find: a), b) VSWR, c) distance of the first maximum, d)
distance to first minimum, e) input impedance of the line, given that the line
is 3.3l, and input admittance of the line.
25 j50
Z
0.5 j1 Point A on Smith chart
1. zL L
Zo
50
2. To get ; the ratio OA/OO’= 0.62 which is the magnitude of the reflection
coefficient
3. To get the phase, the line passing from O to A crosses the outermost scale @
83º
4. The VSWR (constant ) circle with center at O crosses the real axis at point B
@ 4.26
SWR = 4.26
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o Smith Chart; Class Activity (Cont.)
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o Smith Chart; Class Activity (Cont.)
5. To get the first maximum at point B, and first minimum a point C, measure
on the WTG circle the distance from A to B, and the distance from A to C,
respectively
Lmax=(0.25-0.135)l= 0.115l & Lmin=(0.5-0.135)l= 0.365l
6. To get the input impedance at 3.3l, remove multiple of 0.5l as these lengths
are transparent to the input impedance, leaving only 0.3l
Moving 0.3l towards the generator (Point D); zin 0.28 j0.4
Zin zin .Zo (0.28 j0.4 ).50 (14 j20 )
& The input admittance is found by moving 0.25l (image point E);
yin 1.15 j1.7
Yin yin .Yo
ECE_261; Smith Chart
(1.15 j1.7 )
(0.023 j0.034 )S
50
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o Smith Chart; Matching
The line is said to be matched to the load when the load impedance is equal
to the characteristic impedance of the line; ZL=Zo=
In case of matching, the power delivered to the load is a maximum
In general, the load impedance characteristic impedance of the line
Use of matching network between the load and the transmission line
ECE_261; Smith Chart
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o Smith Chart; Matching (Cont.)
The load impedance ZL@ plane A-A’ is transformed to Zin @ M-M’ plane
The role of the matching network is to let Zin = Zo so that the load is now
seen as matched to the line
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o Matching Networks; Examples
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o Lumped Element Matching;
Choice of d and Ys to achieve matching at M-M’
Yin=Yd + Ys Yin=(Gd+jBd) + jBs
To achieve matching at M-M’, it is necessary that yin=1+0;
yin=gd + j(bd+bs)
bs=-bd
ECE_261; Smith Chart
& gd=1
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o Lumped Element Matching; Class Activity
Case (6); At 100 MHz, a load impedance ZL=(25-j50) is connected to a 50-
lossless transmission line. Insert a shunt element to eliminate reflections, find
the location (in wavelength) and the type and the value of the element
1
Z
z L L 0 .5 j 1 y L
0.4 j0.8 (Point B)
z
Zo
L
Using Smith Chart; d=(0.178-0.115)l=0.063l
yin=1 + j1.58 ys=-j1.58
1
Zo
j31.62
Zs
ys Yo j1.58
The required element is an inductor
& L=50 nH
ECE_261; Smith Chart
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o Single Stub Matching; Shunt Stub
Typical network having two adjustable parameters consists of a shunt-shortcircuited (or open-circuited) transmission line of length (l) called stub
connected in parallel shunt
The two adjustable parameters (degrees of freedom) are;
l (stub length)
d (position where the stub is connected)
The requirement on “d” is; Yd
& The requirement on “l” is; Ys
plane MM' Yo jB
plane MM' jB
The total input admittance @ plane MM’ = Yo
ECE_261; Smith Chart
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o Shunt Stub Matching; Class Activity
Case (7); At 100 MHz, a load impedance ZL=(25 - j50) is connected to a 50-
lossless transmission line. Use short-circuited shunt stub to eliminate
reflections, find the location (in wavelength) and the length of the stub
d1=0.063l & l10.09l
d2=0.207l
ECE_261; Smith Chart
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o Impedance Matching; Frequency Analysis
l/4 transformer;
ECE_261; Smith Chart
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o Frequency Analysis; (Cont.)
Single-stub, series or shunt
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o References;
Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, “Fundamentals of
Applied Electromagnetics”, Prentice Hall; Seventh Edition, October 11, 2014
David M. Pozar, "Microwave Engineering", John Wiley and Sons, 4th edition,
ISBN: 978-0-470-63155-3, 2012
Branislave M. Notaroš, “Electromagnetics”, Pearson; 1 edition (June 5, 2010),
ISBN-978-0132433846, 2010
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