Re: DRSSTC stablity/ closed loop response

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: "Bob (R.A.) Jones" <a1accounting@xxxxxxxxxxxxx>

Hi Ed

> Original poster: Esondrmn@xxxxxxx
>
> In a message dated 2/1/05 3:59:27 PM Pacific Standard Time,
> tesla@xxxxxxxxxx writes:
>
>> As long as Terry started this, I also have questions about f1 & f2, or
> poles above and below the resonant frequency.  First, is this a function
of
> any resonator?  Does any specific combined L and C have this
> characteristic?  Or is it primarily a function of a two resonator coupled
> system?

Its a function of two resonators that are coupled in particular way. In our
case it is such that the equivalent circuit reduce to a series resonant
circuit in series with a parallel one. The  f0 peak of one resonant circuit
cancels the dip of the other so fo disappears (assuming high Qs). Below fo
the series resonant circuit looks like C while the parallel one looks like L
at some frequency their impedances are equal and that produces a new
resonance f1. Above resonance the series circuit looks like L while the
parallel one looks like C at some frequency above fo their equal and that
produces a second new resonance f2.
That's the graphic version the equation version involves finding the complex
roots of a fourth order equation yuck!!

>If the resonant frequency is say 100 khz, how far above and below
> this will the poles be?  If the degree of coupling has an impact, what is
> it?  If we looked at the amplitude of the three frequencies (F0, F1 and
> F2), does the amplitude of F1 and F2 go way down as coupling is reduced,
or
> does the distance in hertz away from F0 get larger?  The more I think
about
> this, the more questions that come to mind.
>
The tighter the coupling the further out either side of the original fo you
have to be for the two new resonances.
Note also that there is no particular requirement for the peak and dip to be
at the same frequency.

Robert