Rich Measures wrote:
>? 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.
>>
Hard to guess remotely, but here are a few thoughts...
When there are no terminating resistors, it's easy to calculate the
resonant frequency. But when there is one or more terminating resistor,
you have to guess at the answer and do the calculation from the other
end.
1. C1 and R1 are in parallel. Convert to equivalent series-form Rs1-jXs1
at the assumed frequency (21.49 -j220.0 at 7.00MHz... but it would be
different at another frequency)
2. Convert C2 and R2 to equivalent series-form Rs2-jXs2 at the assumed
frequency (24.5 -j25.0 at 7.00MHz)
3. Calculate XL at the assumed frequency (245.0 at 7.00MHz).
4. Add them all together to give (Rs1 + Rs2) +j(XL -Xs1 -Xs2).
At the resonant frequency, the j() term goes to zero.
73 from Ian G3SEK Editor, 'The VHF/UHF DX Book'
'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.demon.co.uk/g3sek
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