That's a worthwhile educational exercise.

You might find a single equation, but it gets a bit messy. However, it's
easily done in a few steps, if you're comfortable with complex
arithmetic. I use an HP48GX calculator for this sort of thing, since it
readily and directly handles complex numbers.

Call the capacitor on the input side of the network C1 and the on the
output side C2, and the inductor L. First calculate the reactance of
each of them at the frequency of interest, Xc1, Xl, and Xc2. You'll find
formulas for those in the Handbook. Note that Xc1 and Xc2 will be
negative and Xl will be positive.

First calculate the parallel combination of the load impedance (Zl) and
Xc2. That will be the impedance seen looking toward the load from the
output side of the inductor: Za = Zl || Xc2 = (Rl + jXl) || (jXc2),
where Xc2 is negative. The combined impedance of two impedances Z1 and
Z2 in parallel are 1 / ((1/Z1) + (1/Z2)) = (Z1 * Z2) / (Z1 + Z2).

Now add Xl to your result to find the Z looking into the input side of
the inductor: Zb = Za + jXl. (If the inductor has appreciable loss, use
Zl = Rl + jXl instead of just jXl.) Finally, calculate the parallel
combination of that impedance and the impedance of the input capacitor
to find the impedance looking into the network: Zin = Zb || jXc1,
remembering that Xc1 will be negative.

You can of course combine all this into one equation, but it gets pretty
big. I prefer to do it in steps, one of the reasons being that I
understand exactly what I'm doing at each step rather than just dumping
numbers into an equation and hoping that what comes out is right. A
Smith Chart gives you an even better feel for what's going on, and this
would be a good opportunity to get acquainted with that valuable tool.
You might solve the problem both arithmetically and by using the Smith
Chart and compare results.

If you aren't comfortable with complex arithmetic, the equations get
considerably more complicated since you'll have to deal with the
resistance and reactance separately. If that's the case, I second the
advice already given that you either solve it using a Smith Chart or
consider just letting one of the other folks on this group do the
calculation for you.

Roy Lewallen, W7EL

wrote:
I have an MFJ-269 which I can measure the input to the coax and get a
direct impedance measurement.

Then I would put the tuner in line with the MFJ-269 as the source and
adjust the tuner until I see 50 + j0 on the MFJ.

I would follow by measuring the pi-network components also using the
MFJ as it will measure values at the desired RF frequency.

I would keep the signal frequency low so that stray inductance and
capacitance do not affect the measurement.

What I would like to do is see by using the values I obtain, calculate
the cable load and compare with MFJ direct measurement.

I will review the handbook to see if there is a standard equation
knowing the values and assume that the Q of the inductor is greater
than 10.




All in the fun of the hobby.


de KJ4UO