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[AMPS] Pi-Net math

To: <amps@contesting.com>
Subject: [AMPS] Pi-Net math
From: amps@txrx.demon.co.uk (Steve Thompson)
Date: Wed, 11 Aug 1999 07:41:16 +0100

-----Original Message-----
From: Rich Measures <measures@vcnet.com>
To: amps@contesting.com <amps@contesting.com>
To: <amps@contesting.com>
Date: 11 August 1999 01:25
Subject: Re: [AMPS] Pi-Net math




Dick

>Rich...
>
> ........
>........... Time to do your homework.
>
?  I did the homework, Dick.  I built the Pi-network test
model.  I used
Ian's calculated values  for:  2000-ohms RL / Ro=50-ohms /
7.00MHz / Q=10
or so.
-  Ian's results:
  "The accurate formula - in ARRL Handbooks since 1995 -
gives
C1=102.1pF,  L=5.57uH, C2=463.9pF. When analysed with the
load present,
this checks out with Q=9.98. "

-  My results .  Without RL and Ro, the resonant freq. was
7.37MHz, which
checks out mathematically.  When I connected the 2000 ohm
and 50 ohm
resistors, the resonant frequency did not appear to drop to
7.00MHz as
had been predicted.
Is there perhaps something that I am doing wrong, Dick?  .
Ian? .  Tom?
.  Peter?   Any suggestions would be greatly appreciated.



In normal use, you don't have resistors loading the circuit
at both ends - you put a load on one end and use the circuit
to transform the impedance so that looking into the network
you see the desired impedance. If you remove one of the
resistors, I think the resonant frequency will decrease.
Without doing any maths, I think I agree that the resonant
frequency doesn't change if you load the circuit with the
designed resistor values at both ends.

Steve


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