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From: Bret Cahill on 8 Aug 2010 18:45
If you already have the frequency response of the impedance of a
circuit over a bandwidth, it should be possible to largely correct for
differences in the reactance of a similar but less-than-perfectly-
known circuit by applying a wave form which is a sum of one or several
frequencies that fall within the known bandwidth.

Coefficients measured on the first circuit at the frequencies applied
to the 2nd circuit are multiplied onto the response of the 2nd
circuit.

If the phase angle is small, however, is may be possible to
approximate the reactance another if somewhat less accurate way:

The first term would be some coefficient -- nothing to do with the
coefficient above -- measured on the known circuit and multiplied by
phi^2 ( i"/i').

Does phi^2 or i"/i' look familiar to anyone?

The phi term comes from knowing the difference between two sinusoidal
curves separated by a small phase angle is close to the derivative of
either curve times the phase angle.


Bret Cahill








From: Bret Cahill on 9 Aug 2010 23:26
> If you already have the frequency response of the impedance of a
> circuit over a bandwidth, it should be possible to largely correct for
> differences in the reactance of a similar but less-than-perfectly-
> known circuit by applying a wave form which is a sum of one or several
> frequencies that fall within the known bandwidth.
>
> Coefficients measured on the first circuit at the frequencies applied
> to the 2nd circuit are multiplied onto the response of the 2nd
> circuit.
>
> If the phase angle is small, however, is may be possible to
> approximate the reactance another if somewhat less accurate way:
>
> The first term would be some coefficient -- nothing to do with the
> coefficient above -- measured on the known circuit and multiplied by
> phi^2 ( i"/i').
>
> Does phi^2 or i"/i' look familiar to anyone?

Maybe just a phi^4.

The chances that this approximation hasn't already been done are just
about nil.


Bret Cahill


From: Bret Cahill on 11 Aug 2010 09:50
> > If you already have the frequency response of the impedance of a
> > circuit over a bandwidth, it should be possible to largely correct for
> > differences in the reactance of a similar but less-than-perfectly-
> > known circuit by applying a wave form which is a sum of one or several
> > frequencies that fall within the known bandwidth.
>
> > Coefficients measured on the first circuit at the frequencies applied
> > to the 2nd circuit are multiplied onto the response of the 2nd
> > circuit.
>
> > If the phase angle is small, however, is may be possible to
> > approximate the reactance another if somewhat less accurate way:
>
> > The first term would be some coefficient -- nothing to do with the
> > coefficient above -- measured on the known circuit and multiplied by
> > phi^2 ( i"/i').
>
> > Does phi^2 or i"/i' look familiar to anyone?
>
> Maybe just a phi^4.

For a simple sinusoidal the first term is just proportional to phi.


Bret Cahill







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