Because the math can get a little hairy in complex impedance situations, we will take a quick look at an example matching two pure resistances. Let us assume our desired transformation is from a 50OMEGA; transmission line to a 100OMEGA; resistive load at 1000kHz. We know right off the bat that the shunt leg of the network will be on the load side, and since we will limit ourselves to networks with both a capacitor and inductor, there are two possible solutions. The first step is to determine the Q:

The first case has the capacitor in the shunt leg and the inductor in series. The values for the components are as follows:

Moving the inductor to the shunt leg and the capacitor to the series leg results in the following:

When using these equations, it is important to keep track of the decimal places. The frequency denoted by "f" is in terms of hertz, and the derived capacitance will be in farads, and inductance in henries. Because 1000kHz is 1MHz, and our desired component values are microhenries or microfarads, we can simplify the terms somewhat.

In this case the solution with the smaller inductor is probably the more cost effective solution; however, in a pinch you can use what is on the shelf. Note also that if you add an element in series on the load side of the network, you have created a T network. This additional element should be used to reduce the complex impedance to a pure resistance, then employ the above equations to make a match. Thus, with some components on the shelf, you could easily construct a matching network to return a non-directional AM station to air in a crisis.

- continued on page 3