Transmission Lines - Smith Chart and Impedance Matching

Hon Tat Hui Transmission Lines Smith Chart & Impedance Matching

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Transmission Lines Smith Chart & Impedance Matching

1 Smith Chart

Smith chart is a graphical plot of the normalized resistance and reactance functions in the complex reflection-coefficient plane. It is a graph showing both the normalized impedance and the reflection coefficient.

Smith chart is convenient for transmission line and circuit calculations. It is also a useful tool in impedance matching circuit design.


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Recall that:

( ) ( )( ) 00

ZZZZ

+= AAA

( ) ( )( )AAA

+=

11

0ZZ

Now normalize the impedance Z() by Z0.

( )0Z

Zz A=


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imre jzz +=+=

11

( )( ) imre

imre

jjjxrz

++=

+=+=11

11

In terms of the normalized impedance z (drop the dependence), we can write:

From the last equation, we have

2 2re im

2 2re im

1(1 )

r = +

2im2 2

re im

2(1 )

x= +


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The last two equations of r and x define two families of circles in the complex plane of reflection coefficient .

1=r

=r

0=r (short)

(open)

0 10.5

im

re 10.50

1

re

im

The Smith char is the superposition of these two families of circles together in the complex plane of reflection coefficient .


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The Smith Chart


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A point in the Smith chart gives the values of the normalized impedance z and the complex reflection coefficient at the same point on a transmission line.

or zzimzre

S


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S=+=

+

11

11

When the angle of is zero, is real and =||. Then,

Thus, the value of S is same as r when the angle of is zero and can be read out directly from the Smith chart by noting the r value (S = r).

rz ==+

11

But when the angle of is zero,

Since all points on the dotted black circle have the same ||, they must also have the same S. This circle is known as the constant VSWR circle.


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r = 3

All points on this circle have a S = r =3

For example,


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Example 1Plot the following impedances on to the Smith chart.

= 281 3


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SolutionsZ1

Z2

Z3

Z4

Z5

Z6

Z7

Z8


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i

L

()Z()

z = -d = d

z = 0 = 0

z

1.1 Smith chart and transmission lines

towards generator

( ) AA kjLe 2== L =

Recall that on a transmission line:


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Hence, can be obtained from L by moving clockwise along a constant circle on the Smith chart with a radius |L| through an angle -2k which is equivalent to / wavelengths measured towards the generator on the periphery of the Smith chart.

L

l=/=-2k

l

This circle is also known as the constant VSWR circle. All the points on this circle has the same S and same ||.


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Several scales around the outside of the Smith chart are used to determine the distance along the line. Some Smith charts have a number of scales at the bottom of the chart for measuring the reflection coefficient magnitude and others.

1.2 Reading on Smith chart


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Two scales on the periphery (in wavelengths):-Wavelengths towards generator (WTG scale),

clockwise sense-Wavelengths towards load (WTL scale),

anticlock sense

Note also that a complete turn around the Smithchart corresponds to a total length of /2. Because:( ) ( ) ( )

( ) ( )( ) ===

==

36025.0222

:is tofrom turnedphase the,5.0When

12

21

12

212

12

AAAA

AAAA AA

k

e kj


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Any point reflected through the centre point converts an impedance to an admittance and vice versa.

Top Half: inductive reactance, XL = Lor

capacitive susceptance , BC = C

Bottom Half: capacitive reactance , XC = 1/Cor

inductive susceptance , BL = 1/L

Impedance admittance

z y = 1/z

z=1.8+j2

y=0.25-j0.28= 1/(1.8+j2)


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Example 2Use Smith chart to find the input impedance Zin looking at the input of a transmission line.

(a) Actual circuit(b) Normalized circuit

0.3

Zin


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47.0

457.0

j

L

e==

Example 2 (contd):


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in

Example 2 (contd):

8118387658

c)point (at 076201775011

7.07.0

0

95.0

3.0224

'2

.j.ZzZ

.-j.

z

eee

e

inin

in

inin

j

jj

kzjLin

===

+===

=

See animation Transmission Line Impedance Calculation


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Voltage Maxima and Minima in Smith Chart

Voltage maxima occur when the angle of therelfection coefficient (()) = -2n (n = 0, 1, 2, ). This corresponds to the right-most point in the Smith chart. Voltage minima occur when the angle of the relfection coefficient (()) = -2(n+1) (n = 0, 1, 2, ). This corresponds to the left-most point in the Smith chart. See an example shown on next page.


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0=AmAMA

( )AV

Mm = 0

zL


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2 Impedance Matching

Reasons for impedance matching:1. Maximize power transfer to the load2. The input impedance remains constant at the value Z0.

Therefore, the input impedance is independent of the length of transmission line.

3. VSWR = 1. Therefore there are no voltage peaks on the transmission line.

Meaning of impedance matchingImpedance matching is to eliminate the reflected voltage or current on a transmission line.

Two matching techniques:1. Quarter-wave transformer2. Single-stub matching network


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2.1 Quarter-wave transformerFor a transmission line of length d=/4, characteristic impedance = Zs, and terminated in an impedance RL,

( )( ) L

s

Ls

sLsi R

ZjRZjZRZZZ

2

2tan2tan)4( =+

+=== A

When Zs is real, we can change Zs to achieve a desired Zi.

Zi


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Example 3A signal generator has an internal impedance of 50 . It needs to feed equal power through a lossless 50 transmission line to two separate resistive loads of 64 and 25 at a frequency of 10 MHz. Quarter-wave transformers are used to match the loads to the 50 line, as shown below. (a) Determine the required characteristic impedances and the physical lengths of the quarter-wavelength lines assuming the phase velocities of the waves traveling on the them is 0.5c. (b) Find the standing-wave ratios on the matching line sections.


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Solutions(a) As the two quarter-wave transformers are connected in parallel to the 50- line, if equal powers are required to the two loads, the input impedances of the two branches looking at the junction from the 50-line must be equal to 100 so that when they add together in parallel, the total impedance is 50 .

Rin1=100 100 100

50

Rin2=100

R0

R0


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m 152m 30ers transform thealong wavelength

42

11

0

00

====

====

r

rr

pcu

Physical length of the transformers = /4 = 3.75 m

Therefore, 100 100

22

11

====

inin

inin

RZRZ

The characteristic impedances R01 and R02 can be found by:

5025100'

8064100'

2202

1101

======

Lin

Lin

RRR

RRR

Solutions (contd):


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Solutions (contd):


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2.2 Single-stub matching network

parallel stub (short-circuited)

B-B = matching pointd = matching position = length of the matching studyB = normalized admittance of line at B-Bys = normalized admittance of the stubyL = normalized admittance of the load

load to be matched

What is a stub?A stub is a short section of transmission line (shorted or open at one end) whose input impedance can be changed by varying its length.


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After matched, there is no reflection on the line to the left of B-B. But there are reflections on the line to the right of B-B and on the stub.

B

B

R0

jx

-jx

R0Matched

When a transmission line is matched at the matching point,

0or 1 YYYYyyy sBisBi =+==+=


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Use of admittanceFor parallel stub matching, the stub is connected in parallel with the transmission line. Hence it will be more convenient to use admittance rather than impedance. In Smith chart, when a impedance z is known, the corresponding admittance, y = 1/z, can be obtained by a reflection through the centre of the Smith chart. The admittance y is represented on the same Smith chart but its position is different from that of z. When every impedance point on the Smith chart is reflected in this way, we transform the impedance Smith chart to an admittance Smith chart in which every point now represents a normalized admittance. For parallel-stub matching, we work in the admittance Smith chart.


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On plotting into the Smith chart, all values have to be normalized by the characteristic impedance Z0 (or the characteristic admittance Y0 = 1/ Z0) first. Normalized values are usually represented by small letters while un-normalized values by CAPITAL LETTERS. For example:

Yinyin = YL / Y0

Zinzin = Zin / Z0

YLyL = YL / Y0

ZLzL= ZL / Z0Un-normalized quantityNormalized quantity

Reminder


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( )( ) BL

LB jbkdjZZ

kdjZZdz

y +=++=== 1tan

tan)(

1

0

0

A

Choose d such that:

( ) ( ) Bs jbkjkjy === AA cottan1

Choose such that:

So that:

1=+= sBi yyy

Method to determine d and


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Steps in single-stub matching (using normalized z and y):1. Convert the load impedance zL to an equivalent

admittance yL = 1/zL.2. Use a line of length d and a characteristic impedance Z0

(characteristic admittance Y0 = 1/Z0) to transform yL to yB = 1 + jbB at B-B.

3. Connect a parallel stub of length and characteristic impedance Z0 at B-B with an input admittance ys=-jbB=-jcot(2/).

4. Then, the total admittance at B-B is:

matched 11 =+=+= BBsBi jbjbyyy


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Example 4A 50 lossless transmission line is connected to a load impedance ZL=35-j47.5. Find the position d and length l of a short-circuit stub required to match the load at a frequency of 200 MHz. Assume that the transmission line is a coaxial line filled with a dielectric material for which r = 9.Solutions

The detailed matching steps in the Smith chart will be explained by using the example shown below.


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Reasons for Choosing a Shorter-Line SolutionSometimes, out of the two basic sets of solutions, (d1, l1) and (d2 , l2), it is advantageous to choose a shorter combination of d and l. There are at least two reasons for this:1. Shorter lines can always reduce the inevitable loss along the lines, though this is

small.2. Shorter lines result in a smaller Q-factor for the resonant circuit consisting of the

stub and the transmission line to the right of the matching point. A smaller Q-factor produces a wider matching bandwidth, i.e., the matching condition being less sensitive to frequency change.

Drawings on Smith chart shown on next page


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scPO

1P 4P

'4P"3P

3P

'3P

2P

"4P'2P

1=g

Solutions on Smith chart for Example 4

d1

d2

1

2zL

yL1+j1.2

1-j1.2

-j1.2

j1.2

See animation Parallel Stub Matching Short

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Transmission Lines - Smith Chart and Impedance Matching

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