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This gets us to "B", and we find that $\frac{{Z}_{\mathrm{L}}()}{{Z}_{0}()}=1+1.2i$ .Now this is a very interesting
result. Suppose we take the load off the line, and add, in series, an additional capacitor, whose reactance is $\frac{1}{j\times \omega \times C}=-(i\times 1.2{Z}_{0})$ .The capacitor and the inductor just cancel each other out (series resonance) and so the apparent load for the line is just ${Z}_{0}()$ , the magnitude of the reflection coefficient (Γ)= 0 and the $\mathrm{VSWR}=1.0$ ! All of the energy flowing down the line is coupled to the load resistor, and nothing is reflected back towards the load.
We were lucky that the real part of $\frac{{Z}_{\mathrm{L}}()}{{Z}_{0}()}=1$ . If there were not that case, we would not be able to "match" the load to the line, right? Not completely. Let'sconsider another example. The next figure shows a line with a ${Z}_{0}()=50$ , terminated with a $25(\Omega )$ resistor. ${\Gamma}_{\mathrm{L}}()=\frac{-1}{3}$ , and we end up with the VSWR circle shown in the subsequent figure .
How could we match this load? We could add another 25Ω in series with the first resistor, but if we want to maximize thepower we deliver to the first one, this would not be a very satisfactory approach. Let's move down the line a ways. If we goto point "B", we find that
at this spot, $\frac{{Z}_{\mathrm{s}}()}{{Z}_{0}()}=1+0.8i$ . Once again we have an impedance with a normalized real part equals 1! How far do we go? It looks like it's a littlemore than $0.15(\lambda )$ . If we add a negative reactance in series with the line at this point, with a normalized value of $-(0.8i)$ , then from that point on back to the generator, the line would "look" like it was terminated with a matched load.
There's one awkward feature to this solution, and that is we have to cut the line to insert the capacitor. It would be a loteasier if we could simply add something across the line, instead of having to cut it. This is easily done, if we go over into theadmittance world.
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