Single Stub Tuning
The second matching network we’ll discuss is the single-stub tuner (SST). The single-stub tuner uses a shorted or open section of TL attached at some position along another TL:
This is an example of a parallel SST, which is the only type we’ll study. (A series SST is shown in Fig. 5.4b of the text.) The shunt-connected section is called the stub. Although not necessary, all sections of TL will be assumed to have the same Z_{0} and b .
Why an open or shorted section^{'} Because these are easy to fabricate, the length can easily be made adjustable and little to no power is dissipated in the stub. (An open stub is sometimes easier to fabricate than a short.)
We will study the SST from two perspectives. First, we will develop an analytical solution, followed by a Smith chart graphical solution. Referring to the figure above, the transformed load impedance at the stub position z = -d is
where t º tan(b d ). With a shunt connection, it is much simpler to work with admittances than impedances. So, we’ll define the transformed load admittance as Y =1/ Z = G + jB. The distance d is chosen so that G = Y_{0} =(1/ Z_{0}) . As shown in the text, this condition leads to the solutions
Where
and
Z_{L} = R_{L} + jX_{L}
With this location of the stub, the transformed load admittance has a real part = Y_{0}, which is almost a matched state. In general, however, this transformed Y_{L} will also have an imaginary part B. The length of the stub, l_{s}, is chosen so that its input susceptance B_{s} = -B. Consequently, the parallel combination of the stub input susceptance and the transformed load admittance yield an input admittance Y_{in} = Y_{0} , as seen from the source end of the TL. As shown in the text, this second condition provides the solutions
Or
where B is the transformed load susceptance at z = -d. Lengths of TL that are integer multiples of l/2 can be added or subtracted from (2), (4), and (5) without altering the tuning.
Example N8.1: Match the load Z_{L} = 35-j 47.5 beta to a TL with Z_{0} = 50 beta using a shunt, short-circuited single-stub tuner.
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
Recall that the operation of the single stub tuner requires that
1. A distance d is chosen such that y_{0} ^{'} has a real part = 1.
2. The imaginary part of y_{1} ^{'} is negated by the stub
susceptance after choosing the proper length l_{s} . This produces y_{in} = 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 Z_{L} = 35 - j47.5 beta to a TL with characteristic impedance Z_{0} = 50 beta. The normalized load impedance and admittance are: z_{L} =0.70 - j0.95 p.u.beta and y_{L} =0.50 + j0.68 p.u.beta
Steps:
1. Locate y_{L} = 0.50 + j0.68 p.u.beta 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 y_{L} . Rotate this
vector towards the source until it intersects the unit conductance circle. Along this circle Re [y(z)] =1. 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 y_{1} ^{'} =1+ jb_{1} .
For this example, we find from the Smith chart that
(I) y_{1} ^{'} =1+ j1.2
(II) y_{1} ^{'} =1- j1.2
4. From these rotations we can compute d as
(I) d = 0.168l - 0.109l = 0.059l
(II) d = 0.332l - 0.109l = 0.223l
5. Next, find the stub lengths l_{s} :
(I) want b_{s} = -1.2
(II) want b_{s} =1.2
When either of these two susceptances is added to y_{1} ^{'}, then y_{in} =1.
The stub lengths can be determined directly from the Smith chart. Consider the shorted stub
On the Smith admittance chart, y_{L} = ¥ is located at
Re{T} =1, Jm{T} = 0. From there, rotate “wavelengths towards generator” to:
(I) b_{s} = -1.2 Þ l_{s} = 0.361l – 0.25l =0.111l
(II) b_{s} = +1.2 Þ l_{s} = 0.25l +0.139l = 0.389l
That’s it. The final two solutions are:
(I) d = 0.059l and 0.111 l_{s} = l
(II) d = 0.223l and 0.389 l_{s} = l