The tools required for calculating the vales in an L Network come from the following equations that can be found in the ARRL Handbook.
XS = QRs and Xp=Rp/Q
Q=Xs/Rs and Q=Rp/Xp
Rp=Rs(Q^2+1)
Rs=Rp/〖(Q〗^2+1)
Q=√(RP/Rs-1)
The L network is based on a technique known as series to parallel transformations.
Suppose we have measured the impedance of an antenna we have constructed with an antenna analyzer and found it to be 21 + J5Ω. We are going to match this impedance to a 50Ω line.
We first give the larger impedance the designation of Rp then determine the Q. We will put the +J5Ω aside for the moment.
So Q= √(50/21-1) = 1.715
We now find the parallel reactance required to be placed across the 50Ω .
Xp=Rp/Q =50/1.715 =42.55 We now have our parallel form 50Ω in parallel with a 42.55Ω reactance. This reactance could be chosen as an inductor or a capacitor however in some circumstances the values may become quite ridiculous so try the other option. What is a 42.55Ω reactance? I shall choose a working frequency of 14.2MHz for my example. I am going to choose a capacitor for my reactance. The reactance of a capacitor can be calculated using the following formula C=1/2∏FXc where Xc is the reactance an f in this case 14.2MHz. In our case this gives a capacitor of 260 PF.
Now we convert to series equivalent. Xs = QRs Rs is the 21Ω we are trying to match. So Xs =1.175 x 21 = 24.675Ω. As we chose a capacitor for our parallel reactance we will now place a series inductor with a reactance of 24.675Ω in series to cancel the remaining reactance. The reactance of an inductor is calculated using the following equation Xs=2∏FL rearranging we have L= Xs/2∏F. In our case at 14.2MHz this is an inductance of 276nH. Remember we put the +J5Ω aside now when we take it into consideration we can subtract this reactance from our required reactance. The reactance of 5 ohms we need to subtract from the 24.675Ω we calculated for Xs. So in fact our required reactance is actually 19.675Ω or an inductor of 220 nH.
To summarize we place a 260PF capacitor across the 50 ohms and form an L network by placing a 220 nH inductor in series. The 21+J5Ω with the L network in place will have a combined impedance of 50+J0Ω.