Bear,
I tried to answer you on the GCC Loop Group. For some reason the
post didn't appear. So:
Bear,
My shoggoth upchucked allowing me to provide the following example:
An impedance analyzer gives an impedance of 3 + j215 ohms at 1.6 MHz
for a certain Pioneer AM loop. Let's make the proximal capacitor
(Cp) 7900 pF and the distal capacitor (Cd) 490 pF. That gives
capacitive reactances –j12.6 ohms and –j203 ohms respectively at 1.6
MHz.
The loop in series with Cd gives
3 + j215 – j203 = 3 + j12
Using the parallel impedance formula gives
(0 – j12.6)(3 + j12)/[ (0 – j12.6)+(3 + j12)] = 50.9 – j2.4 ohms
That's not too bad for the shoggoth and me making some wild guesses.
Cp could be a 5100 pF cap with a dip switch to add 200, 400, 800,
1600, and 3300 pF in parallel. Cd could be a 360 pF variable with
200 pF in parallel.
Actually the shoggoth used
and transformed the negative inductor in the upper right network
into a capacitor.
Regards,
Steven
