6.13: Matching

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This gets us to "B", and we find that $\frac{Z_{L}}{Z_{0}} = 1 + 1.2 i$ . Now this is a very interesting result.

A load resistor of resistance Z_0 and a load inductor of impedance 1.2 j times Z_0 are attached in series to the right end of a transmission line of impedance Z_0. — Figure $\PageIndex{1}$: The load impedance

Suppose we take the load off the line, and add, in series, an additional capacitor, whose reactance is $\frac{1}{j \times ω \times C} = - (i 1.2 Z_{0})$ .

The load resistor and load inductor in series from Figure 1 above have a capacitor attached to them in series. The capacitor has impedance of -1.2 j times 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 $VSWR = 1.0$ ! 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 $\frac{Z_{L}}{Z_{0}} = 1$ . If there were not that case, we would not be able to "match" the load to the line, right? Not completely. Let's consider another example. The next figure, Figure $\PageIndex{3}$, shows a line with a $Z_{0} = 50$ , terminated with a $25 (Ω)$ resistor. $Γ_{L} = \frac{- 1}{3}$ , and we end up with the VSWR circle shown in the Figure $\PageIndex{4}$.

A transmission line of impedance 50 Ohms contains a capacitor of impedance -40j Ohms located 0.152 lambda away from a terminating load resistor of 25 Ohms. — Figure $\PageIndex{3}$: Matching with a series capacitor

A VSWR circle on the mini Smith Chart has Z_L/Z_0 as its leftmost point which intersects with the horizontal axis at point A. — Figure $\PageIndex{4}$: Plotting $\frac{Z_{L}}{Z_{0}}$

How could we match this load? We could add another $25 \ \Omega$ in series with the first resistor, but if we want to maximize the power we deliver to the first one, this would not be a very satisfactory approach. Let's move down the line a ways. If we go to point "B", we find that at this spot, $\frac{Z_{s}}{Z_{0}} = 1 + 0.8i$.

Moving clockwise on the VSWR circle from point A to point B, a distance of slightly more than 0.15 lambda. — Figure $\PageIndex{5}$: Moving to the "right spot"

Once again, we have an impedance with a normalized real part equals 1! How far do we go? It looks like it's a little more than $0.15 (λ)$ . If we add a negative reactance in series with the line at this point, with a normalized value of $- (0.8 i)$ , 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 lot easier if we could simply add something across the line, instead of having to cut it. This is easily done, if we go over into the admittance world.