382 Lecture 24

8/13/2019 382 Lecture 24

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Whites, EE 382 Lecture 24 of 11

© 2012 Keith W. Whites

Lecture 24: Single-Stub Tuner II –

Smith Chart Solution.

We will next consider single-stub tuner analysis using the Smith

chart. Before looking at this, however, we must first understand

that the Smith chart can be used as an admittance chart as well

as an impedance chart.

To see this, in Lecture 21 we derived the mapping upon which

the Smith chart is based [ z d d ] from the normalizedTL impedance

1

1

d z d

d

From this, we can express the normalized TL admittance as

111

d y d z d d

(1)

We can repeat the construction of the Smith chart with

y g jb and p jq , as we did originally for the

impedance chart. Substituting these quantities into (1) we find2 2

2 1

1 1

g p q

g g

(2)

and 2 2

2 1 11 p q

b b

(3)

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A Smith admittance chart can be constructed based on these two

equations for circles in the p-q plane:

Re p d

d q Im

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This Smith admittance chart looks very similar to the Smith

impedance chart. In fact, if we rotate one chart by 180º we

obtain the other.

This is actually an easily proved result. Consider the definition

of the negative generalized reflection coefficient

2

2 2

24

j d j d

L L

j d

L

d e e

e

That is,

4

d d

(4)

If we now substitute (4) into (1) we find that

1

44

14

d

y d z d

d

(5)

But what is / 4d ? It’s a half rotation around the Smith chart.

Discussion

From (5) we can deduce that:

1. If z d is known, then y d is the point on the constant

VSWR circle that is diametrically opposite the z d point

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on the Smith chart. (In this context, remember that a QWT

is an impedance inverter device.)

2.

The Smith chart can be used either as an impedance chart or as an admittance chart. Rather than keeping these two

types of charts around, we can use one for either impedance

or admittance calculations. The following example should

help you understand this.

3. One subtlety with these mixed Smith charts is that

generalized reflection coefficients are only correctlyrepresented on impedance charts when plotting

normalized impedances and on admittance charts when

plotting normalized admittances. You’ll read negative

generalized reflection coefficients otherwise (for

admittances on impedance charts and impedances on

admittance charts).

Example N24.1: Use the Smith chart to compute the normalized

input admittance of the TL shown below.

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0

50 500 1 1

50

L Z j z j

Z

p.u.

10 0.5 0.50

y j z

p.u.S

Rotating 0.362 “towards generator ,” we can read from the

Smith chart:

1.7 1.1 z L j and 0.42 0.28 y L j

[Exact: 1.682 1.103 z L j and 0.4157 0.2727 y L j .]

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Single-Stub Matching with the Smith Chart

As we saw in the previous lecture, the single-stub tuner

geometry attached to a TL is

Recall that the operation of the single-stub tuner requires that

1. A length sd is chosen such that 1 y has a real part = 1.2. The imaginary part of 1 y is negated by the stub

susceptance after choosing the proper lengths

L .

We can perform these steps using only the Smith chart as our

calculator . This process will be illustrated by an example.

Example N24.2: Using the Smith chart, design a shorted single-

stub tuner to match the load 35 47.5 L Z j to a TL with

characteristic resistance 0 50 Z .

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The normalized load impedance and admittance are:

0.70 0.95 L z j p.u. and 0.50 0.68

L y j p.u.S

Steps:

1. Locate 0.50 0.68 L

y j p.u.S on the Smith

admittance chart.

2. Draw the constant VSWR circle using a compass.

3. Draw the line segment from the origin to L y [this is

the vector 0 / 4 ]. Rotate this vector towardsthe source until it intersects the unit conductance

circle. Along this circle Re 1 y d .

This is really the intersection of the constant VSWR

circle for this load with the unit conductance circle.

There will be two solutions. Both of these give

1 11 y jb .

For this example, we find from the Smith chart that

(I) 11 1.2 y j

(II) 11 1.2 y j

4. From these rotations we can determines

d as

(I) 0.168 0.109 0.059sd

(II) 0.332 0.109 0.223sd

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5. Next, find the stub lengthss L .

(I) want 1.2s

b

(II) want 1.2sb

When either of these two susceptances is added to1 y ,

then in 1 0 y j .

The stub lengths can also be determined directly from

the Smith chart. Consider the shorted stub

On the Smith admittance chart, L y is located at p

= 1, q = 0. From there, rotate “towards generator” to:

(I) 1.2 0.361 0.25 0.111s s

b L

(II) 1.2 0.25 0.139 0.389s s

b L

That’s it. The final two solutions are:(I) 0.059

sd and 0.111

s L

(II) 0.223sd and 0.389s

L

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We will check these two solutions using the results of the

analytical analysis from the last lecture:

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2

21.191

1

L

s

L

b

for 0.5116 1.367 L rad.

11 1tan 2

2 2 4

/ 4 1.904 2 1/ 4 0

/ 4 0.8299 2 1/ 4 0

L

ss

s

s

bd n

n b

n b

Then

for 1.191s

b with n = -1 (so that 0s

d ) gives 0.2235s

d ,

for 1.191s

b with n = 0 gives 0.05896s

d .

and

for 1.191sb gives

2

1 11 tan 0.38882 2

L

s

L

L

for 1.191sb gives

2

111

tan 0.11122

L

s

L

L

382 Lecture 24

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