Re: Selling the Ampeg. What will I get now?

The BorgMan <me@xxxxxx> wrote:
"RichL" <rpleavitt@xxxxxxxxx> wrote in
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The BorgMan <me@xxxxxx> wrote:
"RichL" <rpleavitt@xxxxxxxxx> wrote in
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The BorgMan <me@xxxxxx> wrote:
"RichL" <rpleavitt@xxxxxxxxx> wrote in news:Pt-
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The BorgMan <me@xxxxxx> wrote:

Correct - RMS voltage/current across a resistance/impedance is
"average power", waveform is irrelevant... because as you note,
the RMS voltage/current calculation normalizes the wave to DC
equivalence.

Not to put too fine a point on it, but since we're attempting to
be technically correct, the above is only true for a sine wave
in the case where the impedance is frequency dependent. For any
other waveform one must break the signal down into Fourier
components and use the appropriate frequency-dependent impedance
for each component, then add 'em up.

Nah - waveform is still irrelevant, as long as you have the
equation for how the impedance reacts with respect to
frequency... because even with a sin wave, the impedance is
frequency dependent.

What I'm saying is that at a given frequency the equation holds as
is for a sine wave. V(f) = Z(f) * I(f), and P = Re[I(f)^2 *
Z*(f)].

If your signal has harmonics in addition to the fundamental, things
change. If f1, f2, f3, etc. represent the signal and corresponding
harmonics, then a given Fourier component of voltage is
V(fn) = Z(fn) * I(fn) (with n = 1, 2, etc.) but the total power is

P = Re[I(f1)^2 * Z*(f1)] + Re[I(f2)^2 * Z*(f2)] + Re[I(f3)^2 *
Z*(f3)] + ...

That is, the impedances at the harmonic frequencies come into play,
in addition to the impedance at the fundamental.

Look it up.

I don't need to look it up - I'm well aware of it. The thing is that
the actual equation is:

P=sigma(1->X) Re[I(fx)^2 * Z*(fx)]

which is simply a condensed version of what I wrote...

The sine wave case is the special minimized case, just as the
Pythagorean Theorem is just a special case of the law of cosines.

So you're conceding then that you have to worry about not only the
impedance at the fundamental frequency but the impedances at the
harmonics as well. In other words, waveform *is* relevant in that it
dictates the magnitudes of the harmonics and you have to use a
different impedance with each harmonic.


You always have to worry about every frequency present - a sine wave
just happens to be the simple case. A true square wave falls at the
opposite side of the spectrum with having an infinite harmonic
series. It's not that waveform is irrelevant - it's that the equation
is exactly the same, regarldless of waveform.

The more general equation, yes.


.

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