Bill Turner wrote:
On Thu, 07 Oct 2004 01:28:13 -0700, Roy Lewallen wrote:


You can use the basic definition of RMS to calculate an RMS value of
power from the instantaneous power, but it's not useful for anything. A
resistor dissipating 10 watts of average power gets exactly as hot if
that average power is supplied by DC, a sine wave, or any other
waveform. That's not true of the RMS power -- different waveforms
producing the same average power and causing the same amount of heat
will produce different RMS powers. So average power is a very useful
value, while RMS power is not.


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That goes against everything I've ever read about RMS power, at least
for sine waves. I have always heard that a certain value of RMS power
produces the same heating as the same value of DC power.

I'd be interested in where you've read this. It's wrong. I suspect that
if you go back and read carefully, you'll find that it's the RMS value
of the voltage or current that causes the same heating as the same DC
value, not the RMS value of the power.

Incidentally, I should mention that "RMS power" is used by the makers of
audio amplifiers. About four years ago, this topic came up on
rec.radio.amateur.antenna, and Jim Kelley commented about it:

Jim Kelley wrote:

I found out the answer to this usage of RMS power. The audio folks
have co-opted the RMS idea to mean the following: If the signal is a
single sine wave, then RMS power is understood to mean the average
power output due to that sine wave. It is deceptive to the following
extent: An amplifier with a certain "RMS power rating" may go
completely flat on peaks that are only a little greater than the RMS
rating. Anyway, the use of the term is pervasive.

So the term is sometimes used in the consumer audio world, although
incorrectly. One shouldn't look to the audio consumer world for accurate
technical information about anything.

In your
statement above, you say that's true only for average power, not RMS,
and is true for *any* waveform, including sine waves.

Let me demonstrate my statement about the average power being the square
of the RMS voltage divided by R for any waveform.

Call the voltage time waveform v(t). This was V * sin(wt) for my sine
wave example, but let it be any periodic waveform. The RMS value of the
voltage Vrms is, by definition, sqrt(avg(v(t)^2)). Applied to a
resistor, the power time waveform p(t) is v(t)^2/R, so the average power
is avg(p(t)) = avg(v(t)^2/R) = avg(v(t)^2)/R. (The 1/R term can be moved
out of the average since it doesn't vary with time -- the time average
of 1/R = 1/R.) From the definition of RMS voltage, you can see that
avg(v(t)^2)/R is simply Vrms^2/R. This was demonstrated without any
assumption about the nature of v(t) except that it's periodic.

The RMS power caused by that v(t) waveform would be sqrt(avg(p(t)^2)) =
sqrt(avg(v(t)^4/R^2)) = sqrt(avg(v(t)^4))/R. This is different from the
average, with the size of the difference depending on the shape of the
waveform.

Is there a new world order?

No, but hopefully there's some new knowledge being gained.

Roy Lewallen, W7EL