Chapter 2 Smith Chart and Impedance cuu duong than cong

HCMUT / 2019Dung Trinh, PhD

Dept. of Telecoms Engineering 1

Trinh Xuan Dung, PhD

[email protected]

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:

𝒁(𝒙)

𝛤 =𝑍 − 𝑍0𝑍 + 𝑍0

= 𝑍 𝑍0

− 1

𝑍 𝑍0+ 1

=𝑧 − 1

𝑧 + 1

Now z can be written as:

𝑧 =1 + 𝛤

1 − 𝛤=1 + Re 𝛤 + 𝑗 Im 𝛤

1 − Re 𝛤 − 𝑗 Im 𝛤=1 − Re2 𝛤 − Im2 𝛤 + 2𝑗 Im 𝛤

1 − Re 𝛤 2 + Im2 𝛤

𝑧 = 𝑟 + 𝑗𝑥where: . Then: 𝑟 =1 − Re2 𝛤 − Im2 𝛤

1 − Re 𝛤 2 + Im2 𝛤𝑥 =

2 Im 𝛤

1 − Re 𝛤 2 + Im2 𝛤

These equations can be re-arranged into:

Re 𝛤 −𝑟

1 + 𝑟

2

+ Im2 𝛤 =1

1 + 𝑟

2

Re 𝛤 − 1 2 + Im 𝛤 −1

𝑥

2

=1

𝑥

2


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2. Smith Chart

Re 𝛤 −𝑟

1 + 𝑟

2

+ Im2 𝛤 =1

1 + 𝑟

2

: Resistance circles

Center: 𝑟

1+𝑟, 0

Radius: 1

1+𝑟


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2. Smith Chart

: Reactance circles

Center: 1,1

𝑥

Radius: 1

𝑥

Re 𝛤 − 1 2 + Im 𝛤 −1

𝑥

2

=1

𝑥

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 – 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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A. Given 𝑍 𝑑 , find Γ 𝑑 or Given Γ 𝑑 , find 𝑍 𝑑 .

B. Given Γ𝐿 𝑎𝑛𝑑 𝑍𝐿, find Γ 𝑑 and 𝑍 𝑑 .

Given Γ 𝑑 and 𝑍 𝑑 , find Γ𝑅 𝑎𝑛𝑑 𝑍𝑅.

C. Find dmax and dmin (maximum and minimum locations for the VSW pattern).

D. Find the VSWR.

E. Given 𝑍 𝑑 , find 𝑌 𝑑 or Given 𝑌 𝑑 , find 𝑍 𝑑 .


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3. Smith Chart ApplicationsA. Given 𝒁 𝒅 , find 𝜞 𝒅

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

Γ 𝑑 .

Example 1: Find Γ 𝑑 given 𝑍 𝑑 = 25 + 𝑗100Ω and 𝑍0 = 50Ω

𝒛 𝒅 =𝒁 𝒅

𝒁𝟎=

𝑹

𝒁𝟎+ 𝒋

𝑿

𝒁𝟎= 𝒓 + 𝒋𝒙


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A. Given 𝜞 𝒅 , find 𝒁 𝒅

1. Determine the complex point representing the given reflection coefficient Γ 𝑑on 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:

𝒛 𝒅 = 𝒓 + 𝒋𝒙

𝒁 𝒅 = 𝒛 𝒅 𝒁𝟎


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3. Smith Chart ApplicationsB. Given 𝜞𝑳 𝒂𝒏𝒅 𝒁𝑳, find 𝜞 𝒅 and 𝒁 𝒅

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 Γ𝐿 and the normalized load impedance 𝑍𝐿on the Smith Chart.

2. Draw the circle of constant coefficient amplitude Γ 𝑑 = Γ𝐿

3. Starting from the point representing the load, travel on the circle in the

clockwise direction by an angle 𝜃 = 2𝛽𝑑.

4. The new location on the chart corresponds to location d on the T.L. Here the

value of Γ 𝑑 and 𝑍 𝑑 can be read from the chart.

Example: Find Γ 𝑑 and 𝑍 𝑑 given 𝑍𝐿 = 25 + 𝑗100Ω, 𝑍0 = 50Ω and 𝑑 = 0.18𝜆

Γ 𝑑 = Γ𝐿𝑒−𝑗2𝛽𝑑 = Γ𝐿


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Example 3: Find Γ 𝑑 and 𝑍 𝑑given 𝑍𝑅 = 100 − 𝑗50Ω , 𝑍0 =50Ω and 𝑑 = 0.1𝜆


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3. Smith Chart ApplicationsC. Given 𝜞𝑳 𝒂𝒏𝒅 𝒁𝑳, find 𝒅𝒎𝒂𝒙 and 𝒅𝒎𝒊𝒏

1. Identify the load reflection coefficient Γ𝐿 and the normalized load impedance 𝑍𝐿on the Smith Chart.

2. Draw the circle of constant coefficient amplitude Γ 𝑑 = Γ𝐿

3. The circle intersects the real axis of the reflection coefficient at two points

which identify dmax (when Γ 𝑑 = real positive) and dmin (when Γ 𝑑 = 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 𝑍𝐿 = 100 + 𝑗50Ω, 𝑍0 = 50Ω and 𝑑 = 0.18𝜆


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3. Smith Chart ApplicationsD. Given 𝜞𝑳 𝒂𝒏𝒅 𝒁𝑳, find 𝑽𝑺𝑾𝑹

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 Γ is real and positive.

Example 5: Find VSWR for 𝑍𝐿 = 25 ± 𝑗100Ω, 𝑍0 = 50Ω.

𝑉𝑆𝑊𝑅 =𝑉𝑚𝑎𝑥

𝑉𝑚𝑖𝑛=1 + Γ𝐿1 − Γ𝐿

𝑧 𝑑𝑚𝑎𝑥 =1 + Γ 𝑑𝑚𝑎𝑥

1 − Γ 𝑑𝑚𝑎𝑥=1 + Γ𝐿1 − Γ𝐿

= 𝑉𝑆𝑊𝑅


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3. Smith Chart ApplicationsE. Given 𝒁 𝒅 , find 𝒀 𝒅

The normalized impedance and admittance are defined as:

Since:

The actual values are given by:

Example 6: Find YL given 𝑍𝐿 = 25 ± 𝑗100Ω, 𝑍0 = 50Ω.

𝑧 𝑑 =1 + Γ 𝑑

1 − Γ 𝑑𝑦 𝑑 =

1 − Γ 𝑑

1 + Γ 𝑑

Γ 𝑑 +𝜆

4= −Γ 𝑑 → 𝑧 𝑑 +

𝜆

4= 𝑦 𝑑

𝑍 𝑑 +𝜆

4= 𝑍0𝑧 𝑑 +

𝜆

4𝑌 𝑑 +

𝜆

4= 𝑌0𝑦 𝑑 +

𝜆

4=𝑦 𝑑 +

𝜆4

𝑍0


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3. Smith Chart: Y Smith Chart The reflection coefficient is written as: 𝛤 =

𝑧 − 1

𝑧 + 1=

1𝑦 − 1

1𝑦 + 1

= −𝑦 − 1

𝑦 + 1

𝛤 =𝑧 − 1

𝑧 + 1

−𝛤 =𝑦 − 1

𝑦 + 1

: Z Smith Chart

: Y Smith Chart


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3. Smith Chart: Y Smith Chart

𝛤 =𝑧 − 1

𝑧 + 1

−𝛤 =𝑦 − 1

𝑦 + 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:

𝑧 = 𝑟 + 𝑗𝑥 𝑦 = 𝑔 + 𝑗𝑏 =1

𝑧

𝑔 =𝑟

𝑟2 + 𝑥2

𝑏 = −𝑥

𝑟2 + 𝑥2


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C110p

R50

L

22.5nH

C212p

Z

Example 7: Find impedance of a

complex circuit using Smith Chart where

𝑅0 = 50Ω and 𝜛 = 109 𝑟𝑎𝑑/𝑠.

1

1

0

1/1 2RC

R j Cz j

R


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Example 8: A 50Ω lossless

T.L. of length 3.3𝜆 is

terminated by a load impedance

𝑍𝐿 = (25 + 𝑗50)Ω.

a. Find Γ.

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 MM’

for wave incident from the source. Even

though multiple reflections may occur

between AA’ and MM’, 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:

𝑍𝐿 = 40 Ω𝑍0 = 50


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4. Impedance MatchingExample:

𝑍0 =60

𝜀ln

8ℎ

𝑊+𝑊

4ℎ𝑓𝑜𝑟 𝑊 ≤ ℎ

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 MM’.

The input admittance at MM’ can be written as:

𝑌𝑖𝑛 = 𝑌𝑑 + 𝑌𝑠 = 𝐺𝑑 + 𝑗𝐵𝑑 + 𝑗𝐵𝑠

To achieve a matched condition at MM’, it is necessary that 𝑦𝑖𝑛 = 1, which translates

into two specific conditions, namely:

𝑔𝑑 = 1

𝑏𝑠 = −𝑏𝑑CuuDuongThanCong.com https://fb.com/tailieudientucntt

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4. Impedance MatchingB. Lumped-Element Matching: choose d and YS to achieve a match at MM’.

Example 9: A load impedance

𝑍𝐿 = 25 − 𝑗50Ω is connected to

a 50Ω 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

𝑓 = 100𝑀𝐻𝑧.


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4. Impedance MatchingC. Single Stub Matching: choose d and length of stub l to achieve a match at MM’.

Example 10: Repeat Example 9

but use a shorted stub to match

the load impedance.


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More ExamplesExample 11: A 50Ω lossless line

0.6 long is terminated in a load

with 𝑍𝐿 = (50 + 𝑗25)Ω . At 0.3

from load, a resistor with resistance

𝑅 = 30Ω is connected as shown in

following figure. Use the Smith

Chart to find 𝑍𝑖𝑛.

𝒁𝒊𝒏 = (𝟗𝟓 − 𝒋𝟕𝟎)𝜴

𝒛𝑳 = 𝟏 + 𝒋𝟎. 𝟓

𝒚𝑨 = 𝟏. 𝟑𝟕 + 𝒋𝟎. 𝟒𝟓

𝟎. 𝟏𝟗𝟒𝝀

𝟎. 𝟑𝟗𝟒𝝀

𝒚𝑩 = 𝟑. 𝟎𝟒 + 𝒋𝟎. 𝟒𝟓

𝒛𝒊𝒏 = 𝟏. 𝟗 − 𝒋𝟏. 𝟒


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More ExamplesExample 12: Use the Smith Chart to find 𝑍𝑖𝑛 of the 50Ω feedline shown in following

figure.

𝒁𝒊𝒏 = (𝟖𝟐. 𝟓 − 𝒋𝟖𝟗. 𝟓)𝜴

𝒛𝟏 = 𝟏 + 𝒋

𝒛𝟐 = 𝟏 − 𝒋

𝟎. 𝟒𝟏𝟐𝝀

𝟎. 𝟎𝟖𝟖𝝀

𝒚𝒊𝒏𝟏 = 𝟏. 𝟗𝟕 + 𝒋𝟏. 𝟎𝟐

𝒚𝒊𝒏𝟐 = 𝟏. 𝟗𝟕 − 𝒋𝟏. 𝟎𝟐

𝒚𝒋𝒖𝒏𝒄 = 𝟑. 𝟗𝟒

𝒛𝒊𝒏 = 𝟏. 𝟔𝟓 − 𝒋𝟏. 𝟕𝟗


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More ExamplesExample 13: A 50Ω lossless line is to be matched to an antenna with 𝑍𝐿 = (75 −𝑗20)Ω using a shorted stub. Use the Smith Chart to determine the stub length and

distance between the antenna and stub.

𝒅𝟏 = 𝟎. 𝟏𝟎𝟒𝝀, 𝒍𝟏 = 𝟎. 𝟏𝟕𝟑𝝀 𝒅𝟐 = 𝟎. 𝟑𝟏𝟒𝝀, 𝒍𝟐 = 𝟎. 𝟑𝟐𝟕𝝀

𝒛𝟏 = 𝟏. 𝟓 − 𝒋𝟎. 𝟒

𝟎. 𝟎𝟒𝟏𝝀

𝒚 = −𝒋𝟎. 𝟓𝟐

𝒛𝟏 = 𝟏. 𝟓 − 𝒋𝟎. 𝟒

𝟎. 𝟎𝟕𝟕𝝀

𝟎. 𝟑𝟐𝟕𝝀

𝒚 = 𝒋𝟎. 𝟓𝟐


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Chapter 2 Smith Chart and Impedance cuu duong than cong