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 12th December 2010, 11:02 PM #11 DF96   diyAudio Member   Join Date: May 2007 If you can do calculus than you know enough maths to understand RMS for sine waves. Remember that power goes as V^2/R, so you can calculate the average power produced from Vsin(x) = (V^2/R) integral(sin^2(x)) - the answer is (V^2/2R) (by inspection or trig). What DC voltage would give you this power? Answer: V/sqrt(2). I think you are confusing average voltage with average power.
MichaelJHuman
diyAudio Member

Join Date: Aug 2005
Quote:
 Originally Posted by DF96 If you can do calculus than you know enough maths to understand RMS for sine waves. Remember that power goes as V^2/R, so you can calculate the average power produced from Vsin(x) = (V^2/R) integral(sin^2(x)) - the answer is (V^2/2R) (by inspection or trig). What DC voltage would give you this power? Answer: V/sqrt(2). I think you are confusing average voltage with average power.
You can calculate average value of abs(sin x) using calculus. Integrate from 0 to PI and then divide the result by PI, as memory serves.

2 / PI I think or .637. Isn't that what I said above?

The integral is -

(-cos PI - (- cos 0 ) / PI = 2 / PI, as memory serves...been awhile.

No, I was not confusing average voltage with power, least I don't think so...of course there's also a simple equation using integrals for RMS power, which is the integral of RMS voltage times RMS current divided by delta time I believe.

Last edited by MichaelJHuman; 13th December 2010 at 12:40 AM.

Elvee
diyAudio Member

Join Date: Sep 2006
Quote:
 Originally Posted by tomchr Note that for sine waves the average and the RMS values are the same. This is not the case for all waveforms. A square wave, for example, has Vavg = Vpeak * D and Vrms = Vpeak * sqrt(D), where D is the duty cycle. ~Tom
No they aren't (I assume you're speaking about the absolute value of the waveforms).
In this case, the rms is Vpeak/sqrt2, and the average Vpeak/0.5*Pi.
The difference is not huge, but significant.

 13th December 2010, 11:23 AM #14 DF96   diyAudio Member   Join Date: May 2007 The average of abs(sin(x)) is 2/pi, as you said. I wasn't disputing that, but it has absolutely nothing whatsoever to do with the RMS value of a sine wave! You were asking about RMS vs. peak for a sine wave so I explained where the factor of sqrt(2) comes from. It may be that your confusion is not about maths but physics.
MichaelJHuman
diyAudio Member

Join Date: Aug 2005
Quote:
 Originally Posted by DF96 The average of abs(sin(x)) is 2/pi, as you said. I wasn't disputing that, but it has absolutely nothing whatsoever to do with the RMS value of a sine wave! You were asking about RMS vs. peak for a sine wave so I explained where the factor of sqrt(2) comes from. It may be that your confusion is not about maths but physics.
I understand RMS.

What I did not understand, and someone explained above somewhere, was how the reservoir cap allows the DC voltage to be as high as it is. In my brain, the capacitor would have caused the DC signal (ignoring ripple,) to be some sort of average voltage value from the rectifier.

I understand, sort of, what he was saying.

 13th December 2010, 12:33 PM #16 DF96   diyAudio Member   Join Date: May 2007 Your post #7 suggested that you didn't understand RMS, as you spoke about waveform average. RMS is about waveform^2 average. Then you seemed to confirm this in post #10. Sorry if I misunderstood. The 2/pi figure appears in choke input supply theory, but you were asking about capacitor input supplies.
MichaelJHuman
diyAudio Member

Join Date: Aug 2005
Quote:
 Originally Posted by DF96 Your post #7 suggested that you didn't understand RMS, as you spoke about waveform average. RMS is about waveform^2 average. Then you seemed to confirm this in post #10. Sorry if I misunderstood. The 2/pi figure appears in choke input supply theory, but you were asking about capacitor input supplies.
I follow you. I was speaking imprecisely in post 7. RMS voltage != average voltage.

What always gets me is that average power is calculated from RMS voltage and RMS current not average voltage and average current. But the math works out, I believe.

 13th December 2010, 01:30 PM #18 DF96   diyAudio Member   Join Date: May 2007 The reason is that the average of a product is not necessarily equal to the product of the individual averages.

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