Single stub matching smith chart steps

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 )«T(d )] from the normalized TL impedance

From this, we can express the normalized TL admittance as

We can repeat the construction of the Smith chart with y = g + jb and T = p + jq, as we did originally for the impedance chart. Substituting these quantities into (1) we find

And

A Smith admittance chart can be constructed based on these two equations for circles in the p-q plane:

This Smith admittance chart looks very similar to the Smith impedance chart. In fact, if we rotated one 180º we obtain the other. This is actually an easily proved result. Consider the definition of the negative generalized reflection coefficient

That is

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

But what is d +l / 4? It’s a half rotation around the Smith chart.
Discussion
From (3) 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 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 correctly represented 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.

Rotating 0.362 +l “towards generator,” we can read from the Smith chart: z (L) =1.7 + j1.1 and y (L) = 0.42 - j0.28 [Exact: z (L) =1.682 + j1.103 and y (L) = 0.4157 - j0.2727 .]

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 ds is chosen such that y 1 has a real part = 1.
2. The imaginary part of y 1 is negated by the stub susceptance after choosing the proper length ls . 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 singlestub tuner to match the load ZL = 35 - j47.5 beta to a TL with characteristic resistance Z0 = 50 beta The normalized load impedance and admittance are:
ZL = 0.70 - j0.95 p.u.beta and yL =0.50 + j0.68 p.u.S
Steps:
1. Locate yL = 0.50 +j0.68 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 yL [this is the vector T(0 +l / 4)].
Rotate this vector towards the source until it intersects the unit conductance circle. Along this circle Re 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
y1 = 1+ jb1.
For this example, we find from the Smith chart that
(I) y1 = 1 + j1.2
(II) y1 = 1 - j1.2
4. From these rotations we can determine s d as
(I) ds = 0.168 0 l = 0 .109 l0.059 l
(II) ds = 0.332 l = 0.109 l0.223 l
5. Next, find the stub lengths ls .
(I) want bs = -1.2
(II) want bs =1.2
When either of these two susceptances is added to then yin = 1.
The stub lengths can also be determined directly from the Smith chart. Consider the shorted stub

On the Smith admittance chart, yL = ¥ is located at p = 1, q = 0. From there, rotate “towards generator” to:
(I) bs = -1.2 Þ ls = 0.361 l - 025. l = 0.111l
(II) bs = +1.2 Þ ls = 0.25 l +139. l = 0.389l
That’s it. The final two solutions are:
(I) ds =0.059 l and ls = 0.111 l
(II) ds= 0.223 l and ls = 0.389 l

We will check these two solutions using the results of the analytical analysis from last lecture:

And