[AMPS] Another arc question

[Ian White, G3SEK]

Rich Measures wrote:
>
>Sample problem:   Part I:   Ls = 200nH, Rs = 200 ohms, f = 100MHz.  What 
>is the Admittance, Y, Mr.  White?  A running description of your solution 
>would be helpful.  

No, I won't bite. There is absolutely no need to drag admittance into
this discussion. 

The whole calculation can be done in terms of the equations for
converting parallel Rp and Xp into the series equivalent, Rs and Xs at a
given frequency.

Start from the PHYSICAL circuit of the LR suppressor - a resistor in
parallel with an inductor. Call that Rp in parallel with reactance Xp. 

To get from paralleled Rp and Xp to the series equivalent at a given
frequency, namely Rs and Xs, the standard equations are:

Rs = Rp * Xp * Xp / (Rp * Rp + Xp * Xp)
Xs = Rp * Rp * Xp / (Rp * Rp + Xp * Xp)

To get back from series-connected Rs and Xs to parallel-connected Rp and
Xp, it goes:

Rp = (Rs * Rs + Xs * Xs) / Rs
Xp = (Rs * Rs + Xs * Xs) / Xs

The equations for the Q of this little network are:

Q = Xs/Rs
1/Q = Xp/Rp

Similar equations using admittance do exist, but they are NOT NECESSARY
for this problem. Insistence on mixing-in admittance leads to confusion,
evidently.

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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