6.9 Simple calculations with the smith chart

So, what do we do for $Z_L$ ? A quick glance at a transmission line problem shows that at the load we have a resistor and an inductor in parallel. Thiswas done on purpose, to show you one of the powerful aspects of the Smith Chart. Based on what you know from circuit theory youwould calculate the load impedance by using the formula for two impedances in parallel $Z_L=\frac{i\omega LR}{i\omega L+R}$ which will be somewhat messy to calculate.

Let's remember the formula for what the Smith Chart represents in terms of the phasor $r(s)$ .

We plot this on the Smith Chart by first finding the real part = 0.25 circle, and then we go down onto the lower half of the chart since that iswhere all the negative reactive parts are, and we find the curve which represents $-0.25i$ and where they intersect, we put a dot, and mark the location as $\frac{Y_L}{Y_0}$ . Now to find $\frac{Z_L}{Z_0}$ , we simply reflect half way around to the opposite side of the chart, which happens to be about $\frac{Y_L}{Y_0}=2+2i$ , and we mark that as well. Note that we can take the length of the line from the center of the Smith Chart to our $\frac{Z_L}{Z_0}$ and move it down to the $\left|\Gamma \right|$ scale and find that the reflection coefficient has a magnitude of about 0.6. On a real Smith Chart, there is also aphase angle scale on the outside of the circle (where our distance scale is) which you can use to read off the phase angleof the reflection coefficient as well. Putting that scale on the "mini Smith Chart" would clog things up too much, but thephase angle of $\Gamma$ is about ${3.0}^{°}$ .

Now the wavelength of the signal on the line is given as

For this example, $\beta L=\frac{2\pi }{\lambda }\times 0.215\lambda =1.35\text{radians}$ and ${\theta }_{\Gamma }=\Gamma =0.52\text{radians}$ . Thus we have:

Many times we don't care about $V^{+}$ itself, but are more interested in how much power is being delivered to the load. Note that power delivered to the inputof the line is also the amount of power which is delivered to the load! Finding $I_{\mathrm{in}}$ is easy, it's just $\frac{V_{\mathrm{in}}}{Z_{\mathrm{in}}}$ . All we have to do is change $Z_{\mathrm{in}}$ to polar form.