>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.
>
Then on to Part II: what is the parallel-equivalent resistance of 200nH
in parallel with 200 ohms at 100MHz? (presumably using the method
described below)
>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.
>
It is evident to me that the evidence will become clear when the
calculations are presented to the jury.
cheers
Rich...
R. L. Measures, 805-386-3734, AG6K
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