# 6.11 Finding zl

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How to find the load impedence using the Smith Chart and VSWR circle.

Let's move on to some other Smith Chart applications. Suppose, somehow, we can obtain a plot of $V(s)$ on a line with some unknown load on it. The data might look like . What can we tell from this plot? Well, $V(\mathrm{max})=1.7$ and $V(\mathrm{min})=0.3$ which means

$\mathrm{VSWR}=\frac{1.7}{0.3}=5.667$
and hence
$\left|\Gamma \right|=\frac{\mathrm{VSWR}-1}{\mathrm{VSWR}+1}=\frac{4.667}{6.667}=0.7$
Since $\left|r(s)\right|=\left|\Gamma \right|$ , we can plot $r(s)$ on the Smith Chart, as shown here . We do this by setting the compass at a radius of 0.7 and drawing a circle! Now, $\frac{{Z}_{L}}{{Z}_{0}}$ is somewhere on this circle. We just do not know where yet! There is more information to begleaned from the VSWR plot however. Firstly, we note that the plot has a periodicity of about 10 cm. This means thatλthe wavelength of the signal on the line is 20 cm. Why? According to this equation, $\left|V(s)\right|$ goes as $\cos \phi (s)$ and $\phi (s)={\theta }_{\Gamma }-2\beta s$ and $\beta =\frac{2\pi }{\lambda }$ , thus $\left|V(s)\right|$ goes as $\cos \left(\frac{4\pi s}{\lambda }\right)$ . Thus each $\frac{\lambda }{2}$ , we are back to where we started.

Secondly, we note that there is a voltage minima at about 2.5 cm away from the load. Where on would we expect to find a voltage minima? It would be where $r(s)$ has a phase angle of ${180}^{°}$ or point "A" shown in here . The voltage minima is always where the VSWR circle passes through the real axis on the left hand side. (Conversely a voltagemaxima is where the circle goes through the real axis on the right hand side.) We don't really care about $\frac{Z(s)}{{Z}_{0}}$ at a voltage minima, what we want is $\frac{Z(s=0)}{{Z}_{0}}$ , the normalized load impedance. This should be easy! If we start at "A" and go $\frac{2.5}{20}=0.125\lambda$ towards the load we should end up at the point corresponding to $\frac{{Z}_{L}}{{Z}_{0}}$ . The arrow on the mini-Smith Chart says "Wavelengths towards generator" If we start at A, and want to go towards the load , we had better go around the opposite direction from the arrow. (Actually, as you can see on a real Smith Chart, there are arrows pointing in both directions, and they are appropriately marked for yourconvenience.)

So we start at "A" go $0.125\lambda$ in a counter-clockwise direction, and mark a new point "B" which represents our $\frac{{Z}_{L}}{{Z}_{0}}$ which appears to be about $0.35+-0.95i$ or so . Thus, the load in this case (assuming a $50\Omega$ line impedance) is a resistor, again by co-incidence of about $50\Omega$ , in series with a capacitor with a negative reactance of about $47.5\Omega$ . Note that we could have started at the minima at 12.5 cm or even 22.5 cm, and then have rotated $\frac{12.5}{20}=0.625\lambda$ or $\frac{22.5}{20}=1.125\lambda$ towards the load. Since $\frac{\lambda }{2}=0.5\lambda$ means one complete rotation around the Smith Chart, we would have ended up at the same spot, with the same $\frac{{Z}_{L}}{{Z}_{0}}$ that we already have! We could also have started at a maxima, at say 7.5 cm, marked our starting point on the right handside of the Smith chart, and then we would go $0.375\lambda$ counterclockwise and again, we'd end up at "B". Now, here is another example. In this case the $\mathrm{VSWR}=\frac{1.5}{0.5}=3$ , which means $\left|\Gamma \right|=0.5$ and we get a circle as shown in . The wavelength $\lambda =2(25-10)=30\mathrm{cm}$ . The first minima is thus a distance of $\frac{10}{30}=0.333\lambda$ from the load. So we again start at the minima, "A" and now rotate as distance $0.333\lambda$ towards the load .

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