# Connexions

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# Odds and Ends

Module by: Bill Wilson. E-mail the author

Summary: Some relevant notes on double stub matching.

Just a few odds and ends. Consider the following which is called a "cascaded line" problem. These are problems where we have two different transmission lines, with different characteristic impedances. Since we will give all of the distances in wavelengths, λ, we will assume that the λ we are talking about is the appropriate one for the line involved. If the phase velocities on the two lines is the same, then the physical lengths would correspond as well. The approach is relatively straight-forward. First let's plot Z L Z 0 Z L Z 0 on the Smith Chart. Then we have to rotate 0.2λ 0.2 λ so that we can find Z A Z 0 1 Z A Z 0 1 , the normalized impedance at point A, the junction between the two lines Figure 3.

Thus, we find Z A Z 0 1 =0.32+0.6i Z A Z 0 1 0.320.6 . Now we have to renormalize the impedance so we can move to the line with the new impedance Z 0 2 Z 0 2 . Since Z 0 1 =300Ω Z 0 1 300 Ω , Z A =96+-180i Z A 96-180 . This is the load for the second length of line, so let's find Z A Z 0 2 Z A Z 0 2 , which is easily found to be 1.9+-3.6i 1.9-3.6, so this can be plotted on the Smith Chart. Now we have to rotate around another 0.15λ 0.15 λ so that we can find Z in Z 0 2 Z in Z 0 2 . This appear to have a value of about 0.15+-0.45i 0.15-0.45, so Z in =7.5+-22.5iΩ Z in 7.5-22.5 Ω Figure 5. There is one application of the cascaded line problem that is used quite a bit in practice. Consider the following: We assume that we have a matched line with impedance Z 0 2 Z 0 2 and we connect it to another line whose impedance is Z 0 InvisibleComma; 1 Z 0 InvisibleComma; 1 Figure 6. If we connect the two of them together directly, we will have a reflection coefficient at the junction given by
Γ= Z 0 2 Z 0 1 Z 0 2 + Z 0 1 Γ Z 0 2 Z 0 1 Z 0 2 Z 0 1
(1)
Now let's imagine that we have inserted a section of line with length l=λ4 l λ 4 and impedance Z m Z m Figure 7. At point A, the junction between the first line and the matchng section, we can find the normalized impedance as
Z A Z M = Z 0 2 Z m Z A Z M Z 0 2 Z m
(2)
We take this impedence and rotate around on the Smith Chart λ4 λ 4 to find Z B Z M Z B Z M
Z B Z M = Z m Z 0 2 Z m Z B Z M Z m Z 0 2 Z m
(3)
where we have taken advantage of the fact that when we go half way around the Smith Chart, the impedance we get is just the inverse of what we had originally (half way around turns rs r s into rs r s ).

Thus

Z B = Z m 2 Z 0 2 Z B Z m 2 Z 0 2
(4)
If we want to have a match for line with impedence Z 0 1 Z 0 1 , then Z B Z B should equal Z 0 1 Z 0 1 and hence:
Z B = Z 0 1 = Z m 2 Z 0 2 Z B Z 0 1 Z m 2 Z 0 2
(5)
or
Z m = Z 0 1 Z 0 2 Z m Z 0 1 Z 0 2
(6)
This piece of line is called a quarter wave matching section and is a convenient way to connect two lines of different impedance.

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