Single Stub Tuner Analytical Solution

Single Stub Tuner I Analytical Solution.

Instead of requiring a special l/4-length of TL to a match a load to a TL, as discussed in the last lecture, other matching techniques can be used. One of these is the single-stub tuner that we will discuss in this and the next lecture. The single-stub tuner uses a shorted (or open) section of TL attached at some position along the TL:

The shorted section is called the stub. All pieces of TL will be assumed to have the same Z0 and b , although this is not necessary. Why a shorted section? These are easy to fabricate and the length can be made adjustable using a “slip sheath.” In addition, no power is dissipated in this stub. In the analysis of stub matching networks, it is more convenient to work with admittances rather than impedances since we are dealing with parallel connections. The total TL admittance Y (d ) at some position d and TL characteristic admittance 0 Y0 are defined as

Consequently, the normalized TL admittance is defined as

Where by

From the above TL figure and (1) we see that for a matched TL (d > ds ) requires
yin = 1
The goal of the single-stub tuner design is to derive an expression for yin , then solve this expression for the parameters ls and ds such that yin = 1.
Proceeding with the derivation of yin , at d = 0:

Such that

Since the stub is a short circuit, its input admittance is purely imaginary (reactive). Specifically, recall from a previous lecture that for a shorted section of TL

In terms of a normalized input admittance for the shorted stub

where the shorted stub susceptance is bs = - 1 /tan (bls) Therefore, at d = ds we can construct the equivalent circuit:

Adding the two normalized admittances (transformed-load and stub admittances) together at d = ds gives our desired expression for yin :
yin = y (ds+) = jbs + y(ds-)
As discussed above, a matched TL for d = ds requires yin = 1. Requiring this in (3) gives


There fore

This last equation comes from the definition of - on p. 2. We will now solve (4) for ds and ls by equating the magnitude and phase of both sides of (4):
: Magnitude:

: Phase:

: or

Finally, we solve for the two unknowns s and ls that force yin = 1:



However, for the shorted stub

as developed earlier on page 3 of this lecture, then

Finally, from this last expression we can determine that

Equations (7) and (8) are the final solutions we need. That is, by selecting the length of the shorted stub ls according to (8) and the location of the stub ds according to (7) we can match any load ZL to the TL for d = ds . Note this is accomplished without consuming any power in the matching network!
Example N23.1: Match a load of ZL = 25 -j50 beta to a TL with Z0 = 50 beta using a short-circuited, single-stub tuner. Use the analytical solution method outlined in this lecture. [See the VisualEM “Example C.7” worksheet for the solution. Note that the animation shows no “pulsation” of the voltage indicating a purely traveling wave towards the load. This is expected behavior on a perfectly matched TL.]