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Whites, EE 481 Lecture 8 Page 3 of 8

With this location of the stub, the transformed load admittance

has a real part = Y0, which is almosta matched state. In general,

however, this transformed YLwill also have an imaginary part B.

The length of the stub, ls, is chosen so that its input susceptance

sB B . Consequently, the parallel combination of the stub

input susceptance and the transformed load admittance yield an

input admittance in 0Y Y , as seen from the source end of the TL.

As shown in the text, this second condition provides the

solutions

1 01 tan2

sl Y

B short-circuit stub (5.11b),(4)

or

1

0

1

tan2

ol B

Y open-circuit stub (5.11a),(5)

where Bis the transformed load susceptance at z= -d. Lengths

of TL that are integer multiples of /2 can be added or

subtractedfrom (2), (4), and (5) without altering the tuning.

Example N8.1: Match the load 35 47.5LZ j to a TL with

0 50Z using a shunt, short-circuited, single-stub tuner.

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Whites, EE 481 Lecture 8 Page 4 of 8

Single-Stub Tuning Using the Smith Chart

We will now solve the single-stub tuner problem using the

Smith chart. In terms of quantities normalized to the

characteristic impedance or admittance, the geometry is

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Whites, EE 481 Lecture 8 Page 5 of 8

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

1. A distance dis chosen such that 1y has a realpart = 1.

2. The imaginary part of 1y is negated by the stub

susceptance after choosing the proper length sl .

This producesyin= 1, which is the matched state.

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

calculator. This process will be illustrated by an example.

Example N8.2: Using the Smith chart, design a shorted shunt,

single-stub tuner to match the load 35 47.5LZ j to a TL

with characteristic impedance 0 50Z .

The normalized load impedance and admittance are:

0.70 0.95L

z j p.u. and 0.50 0.68L

y j p.u.S.

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Whites, EE 481 Lecture 8 Page 6 of 8

Steps:

1. Locate 0.50 0.68Ly j p.u.S. on the Smith admittance

chart. (See the chart on the next page.)

2. Draw the constant VSWR circle using a compass.

3. Draw the line segment from the origin to Ly . Rotate this

vector towards the source until it intersects the unit

conductance circle. Along this circle 1e y z .

This is really the intersection of the constant VSWR circlefor this load with the unit conductance circle.

There will be two solutions. Both of these give 1 11y jb .

For this example, we find from the Smith chart that

(I) 1 1 1.2y j

(II)1

1 1.2y j

4. From these rotations we can compute das

(I) 0.168 0.109 0.059d

(II) 0.332 0.109 0.223d

5. Next, find the stub lengths sl :

(I) want 1.2sb

(II) want 1.2sb

When either of these two susceptances is added to 1y , then

in 1y .

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Whites, EE 481 Lecture 8 Page 7 of 8

The stub lengths can be determined directly from the Smith

chart. Consider the shorted stub

ls

Y0, yL=

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Whites, EE 481 Lecture 8 Page 8 of 8

On the Smith admittance chart, Ly is located at

1e , 0m . From there, rotate wavelengths

towards generator to:

(I) 1.2 0.361 0.25 0.111s sb l

(II) 1.2 0.25 0.139 0.389s sb l

Thats it. The final two solutionsare:

(I) 0.059d and 0.111sl

(II) 0.223d and 0.389sl

Compare these Smith chart solutions with the analytical ones we

obtained earlier in Example N8.1.