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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.