Post on 30-Nov-2021
HCMUT / 2019Dung Trinh, PhD
Dept. of Telecoms Engineering 1
Trinh Xuan Dung, PhD
dung.trinh@hcmut.edu.vn
Department of Telecommunications
Faculty of Electrical and Electronics Engineering
Ho Chi Minh city University of Technology
Chapter 2
Smith Chart and Impedance Matching
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Dept. of Telecoms Engineering
Contents
2
1. Introduction
2. Smith Chart
3. Smith Chart Applications
4. Y Smith Chart
5. Impedance Matching
Problems
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1. Introduction
3
Many of calculations required to solve T.L. problems involve the use of
complicated equations.
Smith Chart, developed by Phillip H. Smith in 1939, is a graphical aid that can
be very useful for solving T.L. problems.
The Smith chart, however, is more than just a graphical technique as it provides
a useful way of visualizing transmission line phenomenon without the need for
detailed numerical calculations.
A microwave engineer can develop a good intuition about transmission line
and impedance-matching problems by learning to think in terms of the Smith
chart.
From a mathematical point of view, the Smith chart is simply a representation
of all possible complex impedances with respect to coordinates defined by the
reflection coefficient.
The domain of definition of the reflection coefficient is a circle of radius 1 in
the complex plane. This is also the domain of the Smith chart.
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1. Introduction
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2. Smith Chart
5
We start from the general definition of reflection coefficient:
π(π)
π€ =π β π0π + π0
= π π0
β 1
π π0+ 1
=π§ β 1
π§ + 1
Now z can be written as:
π§ =1 + π€
1 β π€=1 + Re π€ + π Im π€
1 β Re π€ β π Im π€=1 β Re2 π€ β Im2 π€ + 2π Im π€
1 β Re π€ 2 + Im2 π€
π§ = π + ππ₯where: . Then: π =1 β Re2 π€ β Im2 π€
1 β Re π€ 2 + Im2 π€π₯ =
2 Im π€
1 β Re π€ 2 + Im2 π€
These equations can be re-arranged into:
Re π€ βπ
1 + π
2
+ Im2 π€ =1
1 + π
2
Re π€ β 1 2 + Im π€ β1
π₯
2
=1
π₯
2
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2. Smith Chart
Re π€ βπ
1 + π
2
+ Im2 π€ =1
1 + π
2
: Resistance circles
Center: π
1+π, 0
Radius: 1
1+π
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2. Smith Chart
: Reactance circles
Center: 1,1
π₯
Radius: 1
π₯
Re π€ β 1 2 + Im π€ β1
π₯
2
=1
π₯
2
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2. Smith Chart
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2. Smith ChartFor the constant r circles:1. The centers of all the constant r circles are on the horizontal axis β real part of the reflection coefficient.2. The radius of circles decreases when rincreases.3. All constant r circles pass through the point r =1, i = 0.4. The normalized resistance r = is at the point r =1, i = 0.
For the constant x (partial) circles:1. The centers of all the constant x circles are on the r =1 line. The circles with x > 0 (inductive reactance) are above the r
axis; the circles with x < 0 (capacitive) are below the r axis.
2. The radius of circles decreases when absolute value of x increases.3. The normalized reactances x = are at the point r =1, i = 0
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2. Smith Chart
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2. Smith Chart
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2. Smith Chart
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2. Smith Chart
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2. Smith Chart
Constant circle
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2. Smith Chart
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2. Smith Chart
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2. Smith Chart
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3. Smith Chart Applications
A. Given π π , find Ξ π or Given Ξ π , find π π .
B. Given ΞπΏ πππ ππΏ, find Ξ π and π π .
Given Ξ π and π π , find Ξπ πππ ππ .
C. Find dmax and dmin (maximum and minimum locations for the VSW pattern).
D. Find the VSWR.
E. Given π π , find π π or Given π π , find π π .
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3. Smith Chart ApplicationsA. Given π π , find π π
1. Normalize the impedance:
2. Find the circle of constant normalized resistance r.
3. Find the circle of constant normalized reactance x.
4. Find the interaction of the two curves indicates the reflection coefficient in the
complex plane. The chart provides directly magnitude and the phase angle of
Ξ π .
Example 1: Find Ξ π given π π = 25 + π100Ξ© and π0 = 50Ξ©
π π =π π
ππ=
πΉ
ππ+ π
πΏ
ππ= π + ππ
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3. Smith Chart Applications
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3. Smith Chart Applications
A. Given π π , find π π
1. Determine the complex point representing the given reflection coefficient Ξ πon the chart.
2. Read the value of normalized resistance r and the normalized reactance x that
correspond to the reflection coefficient point.
3. The normalized impedance is:
4. The actual impedance is:
π π = π + ππ
π π = π π ππ
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3. Smith Chart ApplicationsB. Given ππ³ πππ ππ³, find π π and π π
The magnitude of the reflection coefficient is constant along a lossless T.L.
terminated by a specific load, since:
1. Identify the load reflection coefficient ΞπΏ and the normalized load impedance ππΏon the Smith Chart.
2. Draw the circle of constant coefficient amplitude Ξ π = ΞπΏ
3. Starting from the point representing the load, travel on the circle in the
clockwise direction by an angle π = 2π½π.
4. The new location on the chart corresponds to location d on the T.L. Here the
value of Ξ π and π π can be read from the chart.
Example: Find Ξ π and π π given ππΏ = 25 + π100Ξ©, π0 = 50Ξ© and π = 0.18π
Ξ π = ΞπΏπβπ2π½π = ΞπΏ
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3. Smith Chart Applications
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3. Smith Chart Applications
Example 3: Find Ξ π and π πgiven ππ = 100 β π50Ξ© , π0 =50Ξ© and π = 0.1π
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3. Smith Chart ApplicationsC. Given ππ³ πππ ππ³, find π πππ and π πππ
1. Identify the load reflection coefficient ΞπΏ and the normalized load impedance ππΏon the Smith Chart.
2. Draw the circle of constant coefficient amplitude Ξ π = ΞπΏ
3. The circle intersects the real axis of the reflection coefficient at two points
which identify dmax (when Ξ π = real positive) and dmin (when Ξ π = real
negative).
4. The Smith chart provides an outer graduation where the distances normalized to
the wavelength can be read directly.
Example 4: Find dmax and dmin for ππΏ = 100 + π50Ξ©, π0 = 50Ξ© and π = 0.18π
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3. Smith Chart Applications
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3. Smith Chart ApplicationsD. Given ππ³ πππ ππ³, find π½πΊπΎπΉ
The VSWR is defined as:
The normalized impedance at the maximum location of the SW pattern is given by:
This quantity is always real and greater than 1. The VSWR is simply obtained on the
Smith Chart by reading the value of real normalized impedance at the location dmax
where Ξ is real and positive.
Example 5: Find VSWR for ππΏ = 25 Β± π100Ξ©, π0 = 50Ξ©.
ππππ =ππππ₯
ππππ=1 + ΞπΏ1 β ΞπΏ
π§ ππππ₯ =1 + Ξ ππππ₯
1 β Ξ ππππ₯=1 + ΞπΏ1 β ΞπΏ
= ππππ
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3. Smith Chart ApplicationsE. Given π π , find π π
The normalized impedance and admittance are defined as:
Since:
The actual values are given by:
Example 6: Find YL given ππΏ = 25 Β± π100Ξ©, π0 = 50Ξ©.
π§ π =1 + Ξ π
1 β Ξ ππ¦ π =
1 β Ξ π
1 + Ξ π
Ξ π +π
4= βΞ π β π§ π +
π
4= π¦ π
π π +π
4= π0π§ π +
π
4π π +
π
4= π0π¦ π +
π
4=π¦ π +
π4
π0
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3. Smith Chart Applications
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3. Smith Chart: Y Smith Chart The reflection coefficient is written as: π€ =
π§ β 1
π§ + 1=
1π¦ β 1
1π¦ + 1
= βπ¦ β 1
π¦ + 1
π€ =π§ β 1
π§ + 1
βπ€ =π¦ β 1
π¦ + 1
: Z Smith Chart
: Y Smith Chart
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Dept. of Telecoms Engineering 31
3. Smith Chart: Y Smith Chart
π€ =π§ β 1
π§ + 1
βπ€ =π¦ β 1
π¦ + 1
: Z Smith Chart
: Y Smith Chart
Since related impedance and
admittance are on opposite
sides of the same Smith
Chart, the imaginary parts
always have different sign.
Numerically we have:
π§ = π + ππ₯ π¦ = π + ππ =1
π§
π =π
π2 + π₯2
π = βπ₯
π2 + π₯2
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Dept. of Telecoms Engineering 32
3. Smith Chart Applications
C110p
R50
L
22.5nH
C212p
Z
Example 7: Find impedance of a
complex circuit using Smith Chart where
π 0 = 50Ξ© and π = 109 πππ/π .
1
1
0
1/1 2RC
R j Cz j
R
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3. Smith Chart Applications
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3. Smith Chart Applications
Example 8: A 50Ξ© lossless
T.L. of length 3.3π is
terminated by a load impedance
ππΏ = (25 + π50)Ξ©.
a. Find Ξ.
b. Find VSWR.
c. Find dmax and dmin.
d. Find Zin of T.L.
e. Find Yin.
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4. Impedance Matching
Impedance Matching
Maximum power transfer
ZL
Impedance
Matching
Network
ZS
What are Applications ?
T.L. Amplifier Design PA, LNA Component Design Equipment Interfaces
Using lump elements
Using transmission lines
ADS Smith Chart tool
Matching with Lumped Elements Single-Stub Matching Networks Quarter-wave Transformer
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4. Impedance Matching The purpose of the matching network is
to eliminate reflections at terminal MMβ
for wave incident from the source. Even
though multiple reflections may occur
between AAβ and MMβ, only a forward
travelling wave exists on the feedline.
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4. Impedance MatchingA. Quarter wavelength Transformer Matching:
In case of complex impedance:
ππΏ = 40 Ξ©π0 = 50
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4. Impedance MatchingExample:
π0 =60
πln
8β
π+π
4βπππ π β€ β
A simple but accurate equation for
Microstrip Characteristic Impedance:
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4. Impedance MatchingB. Lumped-Element Matching: choose d and YS to achieve a match at MMβ.
The input admittance at MMβ can be written as:
πππ = ππ + ππ = πΊπ + ππ΅π + ππ΅π
To achieve a matched condition at MMβ, it is necessary that π¦ππ = 1, which translates
into two specific conditions, namely:
ππ = 1
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4. Impedance MatchingB. Lumped-Element Matching: choose d and YS to achieve a match at MMβ.
Example 9: A load impedance
ππΏ = 25 β π50Ξ© is connected to
a 50Ξ© T.L. Insert a shunt
element to eliminate reflections
towards the sending end of the
line. Specify the insert location d
(in wavelengths), the type of
element and its value, given that
π = 100ππ»π§.
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4. Impedance MatchingC. Single Stub Matching: choose d and length of stub l to achieve a match at MMβ.
Example 10: Repeat Example 9
but use a shorted stub to match
the load impedance.
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More ExamplesExample 11: A 50Ξ© lossless line
0.6 long is terminated in a load
with ππΏ = (50 + π25)Ξ© . At 0.3
from load, a resistor with resistance
π = 30Ξ© is connected as shown in
following figure. Use the Smith
Chart to find πππ.
πππ = (ππ β πππ)π΄
ππ³ = π + ππ. π
ππ¨ = π. ππ + ππ. ππ
π. ππππ
π. ππππ
ππ© = π. ππ + ππ. ππ
πππ = π. π β ππ. π
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More ExamplesExample 12: Use the Smith Chart to find πππ of the 50Ξ© feedline shown in following
figure.
πππ = (ππ. π β πππ. π)π΄
ππ = π + π
ππ = π β π
π. ππππ
π. ππππ
ππππ = π. ππ + ππ. ππ
ππππ = π. ππ β ππ. ππ
πππππ = π. ππ
πππ = π. ππ β ππ. ππ
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More ExamplesExample 13: A 50Ξ© lossless line is to be matched to an antenna with ππΏ = (75 βπ20)Ξ© using a shorted stub. Use the Smith Chart to determine the stub length and
distance between the antenna and stub.
π π = π. ππππ, ππ = π. ππππ π π = π. ππππ, ππ = π. ππππ
ππ = π. π β ππ. π
π. ππππ
π = βππ. ππ
ππ = π. π β ππ. π
π. ππππ
π. ππππ
π = ππ. ππ
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Q&A
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