Some Basic Audio Questions

MrChips

Joined Oct 2, 2009
30,708
Let us assume a sine wave input signal. The way the amplifier power is measured is to use a nominated load, and then to wind up the volume control until the output is just below the point at which it begins to distort significantly (according to some agreed and stated distortion measure, though this is not critical in practice). The average output power over many sine wave cycles is then measured. This is defined as the time integral of the instantaneous power over the measured time, divided by that time (which is always taken to be very large).

In simpler words, it is a straight average or mean of the output power, measured over a long time, and has a real technical significance (e.g. it measures the heating power of the amplifier).

By contrast, RMS (root mean square) power, would have to be defined as the square root of the time average of the square of the instantaneous power, since this is what 'RMS' means. This could be done, but it is not the power as measured, and furthermore, it would have no technical significance (e.g. it doesn't measure heating power).

© 2003 by Stephen Dawson
I think this writer does not know his mathematics.

Average power and RMS power of a sine wave are mathematically identical.
 

endolith

Joined Jun 21, 2010
27
I think this writer does not know his mathematics.

Average power and RMS power of a sine wave are mathematically identical.
Are you just trolling? The RMS power of a sine wave is 1.225 times the average power of a sine wave. Do the math.
 
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#12

Joined Nov 30, 2010
18,224
I really shouldn't have presented post #16. I know better than to argue math with an audiophile. Please forgive my lapse of judgment.
 

endolith

Joined Jun 21, 2010
27
Show me your math.
V = sin(t)
P = V^2/R ∝ sin(t)^2
mean(sin(t)^2) = mean(1/2-1/2cos(t)) = mean(1/2-0) = 1/2
rms(sin(t)^2) = sqrt(mean((sin(t)^2)^2) = sqrt(mean(1/8 (-4 cos(2 t)+cos(4 t)+3))= sqrt(mean(1/8 (0+0+3)) = sqrt(3/8) = 0.612

0.612/0.5 = 1.224
 

endolith

Joined Jun 21, 2010
27
Do you know how to do integrals?
Sure.

\(P = sin(t)^2\) (assuming R = 1 to make it simple)

mean(P) = \(\frac{1}{T} \int_0^T sin(t)^2 dt\) = \(\frac{1}{T} \int_0^T \frac{1-cos(2t)}{2} dt\) = \(\frac{1}{T} (\frac{1}{2} \int_0^T dt - \frac{1}{2}\int_0^T cos(2t) dt)\) = \(\frac{1}{T} (\frac{1}{2} T - \frac{1}{2}\frac{1}{2} sin(2t)|_0^{2\pi} )\) = \(\frac{1}{T} (\frac{1}{2} T - 0 )\) = \(\frac{1}{T} (\frac{1}{2} T )\) = 1/2


rms(P) = \(\sqrt{\frac{1}{T} \int_0^T (sin(t)^2)^2 \ dt}\) = \(\sqrt{\frac{1}{T} \int_0^T ( \frac{1-cos(2t)}{2})^2 \ dt}\) = \(\sqrt{\frac{1}{T} \int_0^T \frac{1-2cos(2t)+cos(2t)^2}{4} \ dt}\) = \(\sqrt{\frac{1}{T} (\frac{1}{4} \int_0^T dt - \frac{1}{4} \int_0^T 2cos(2t) dt + \frac{1}{4} \int_0^T cos(2t)^2 \ dt})\) = \(\sqrt{\frac{1}{T} (\frac{1}{4}T - \frac{1}{2} \frac{1}{2} sin(2t)|_0^{2\pi} + \frac{1}{4} \int_0^T \frac{1+cos(4t)}{2} \ dt})\) = \(\sqrt{\frac{1}{T} (\frac{1}{4}T - 0 + \frac{1}{4} \frac{1}{2} \int_0^T dt + \frac{1}{8} \int_0^T cos(4t) \ dt})\) = \(\sqrt{\frac{1}{T} (\frac{1}{4}T + \frac{1}{8}T + \frac{1}{8} \frac{1}{4} sin(4t)|_0^{2\pi})\) = \(\sqrt{\frac{1}{T} (\frac{2}{8}T + \frac{1}{8}T + 0)\) = \(\sqrt{\frac{1}{T} (\frac{3}{8}T)\) = \(\sqrt{\frac{3}{8}}\)

sqrt(3/8) / (1/2) = 1.22474

My math tells me that the RMS value of a 1V sine wave is 0.707V.
We're talking about power, not voltage.

Incidentally, what does your math tell you is the average voltage of a 1V sine wave? (It's not 0.707V.)
 

rogs

Joined Aug 28, 2009
279
If you go along with the comment in wayneh's post :

Common usage - opinion - defines what words mean, not "facts".
Then this thread will probably run and run, in ever decreasing circles, until it disappears up its own...

If, however, you go along with the view that scientific terms are specific, then the definition of the Watt as...

"the SI unit of power, equivalent to one joule per second, corresponding to the rate of consumption of energy in an electric circuit where the potential difference is one volt and the current one ampere"

..does not allow for any reference to 'RMS Watts' - 'peak power Watts' - 'music Watts' - 'average Watts' -- or indeed any further qualification of the 'Watt' at all.....

So, 'RMS Watts' maybe a phrase in common use.

That doesn't make it correct........
 

studiot

Joined Nov 9, 2007
4,998
Isn't it strange that someone has started a thread on the root cause (pun intended) of this dispute?

http://www.scienceforums.net/topic/83438-precision-in-english/

There are rights and wrongs on both sides of the RMS argument.

First a definition.

The Root Mean Square function or operator is the square root of average of the squares of a set of numbers.

Now it just so happens that when this definition is applied to the alternating current or voltage in a pure resistance the resultant RMS current or voltage is numerically equal to the the value of direct current or voltage that would produce the same heating effect in the same pure resistance.

But a loudspeaker is far from a pure resistance.

Original usage was to call this the effective current or voltage.

Note the term effective was introduced quite deliberately because the mathematical definition of average makes the average value of the alternating current of voltage over a whole number of cycles zero. (not 0.707 times the peak)

Note also that the term effective can be applied to a non integer number of cycles, the 'average' yields a different number.

As regards to the use of RMS instead of effective,

I use the convention "Watts, RMS" not RMS watts to distinguish the product of the RMS voltage and the RMS current, since this may or may not be the RMS value of the power, depending upon whether the load is complex or real.
I have already noted that loudspeakers present complex loads.

Finally it has been the custom and practice of many manufacturers to specify the output watts of an amplifier that can never be achieved. This is because the real power supply in the unit is 'replaced for the purpose of specification' by a perfect one.
 
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endolith

Joined Jun 21, 2010
27
This site also doesn't know how to figure out the (lack of) difference between RMS and average.

http://www.electronics-tutorials.ws/accircuits/average-voltage.html
Nah, they understand: "The result is that the average or mean value of a symmetrical alternating quantity is zero because the area above the horizontal axis (the positive half cycle) is the same as the area below the axis (the negative half cycle) and cancel each other out in the sum of the two areas as a negative cancels a positive producing zero average voltage."

So they measure average rectified value instead, which is the same method that multimeters use to estimate the RMS value. "True RMS" multimeters do not use this method, in other words.

Average rectified power P_ARV is equal to average power P_avg, though, because the power waveform is always positive already.

Here are some of the nobodies that use RMS watts for their ratings:
Well if they're literally using RMS power and not average power, then they're inflating their specs by 22%, but I doubt their marketing departments would be happy with that; it's too small of an exaggeration. ;)

..does not allow for any reference to 'RMS Watts' - 'peak power Watts' - 'music Watts' - 'average Watts' -- or indeed any further qualification of the 'Watt' at all.....
Well, you can say "average power = 100 W" and "peak power = 200 W". Those are meaningful and correct. But yes, the units are just "watts".
 
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rogs

Joined Aug 28, 2009
279
Well, you can say "average power = 100 W" and "peak power = 200 W". Those are meaningful and correct. But yes, the units are just "watts".
The most common use of the phrase would seem to be '100W RMS' or '200W RMS'..etc.... There have been (and continue to be) literally thousands of such examples quoted in audio amplifier specifications over the years.
it's a phrase that's commonly 'understood'......it just happens to be wrong..
 
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