RMS Power?

@kijanki Al, I was right (he is not going to get it). RMS value of ANY sinusoidal waveform, having peak at 200 (of any unit) is 141 (of the same unit). 100W would be an average power value corresponding to VrmsxIrms and equal to half of peak power for sinewave (and equivalent to amount of DC power producing the same amount of heat). Guys, please, this is EE101.

Lets be a little careful here for those trying to understand this conversation. "RMS value of ANY sinusoidal waveform, having peak at 200 (of any unit) is 141 (of the same unit). Does that apply to watts? Are 200 peak watts the same heating value as 141 watts. They are the same units? I know you dont think so but one could easily interpret any to mean any. Granted going on its fine. Whats all this about average power? Whats the definition. I saw you 3 step calculus but, sorry I dont get it and I did fine in Calculus.

Yes RMS power is half the peak power, but you call it average power without informing us what average power is to you.  Al, who you appear to agree with. thinks 141 watts is the average power of 200 watts peak. He has said so. Still what is this average power? We are talking about a sine wave going on and on. 

Average power as used by most in audio means the average over a long time, playing music and not letting the voice coil get too hot. This is the definition I find most often for average power. I dont see how it applies to amplifiers except for the heat sinking.

RMS power is continuous sine wave in this discussion. Why say it is wrong. Whats wrong with it?

Come on kijanki, lets get this ironed out for everyone else who by now doesnt know what we are talking about. It is important. The early replies to you OP had no idea what to say.

I do think we agree yet use of the term average to describe heating is generally used as RMS. Do you really want to say this. " Prms = 0.61Ppeak "? not 0.5 . P rms= V rms x I rms. does it not. is so .7 x .7 =.5

I am writing a paper on how the FTC got involved, its not the way most people think and its not bad.

Al, who you appear to agree with. thinks 141 watts is the average power of 200 watts peak. He has said so.

No, I have not said that. I have said that 141 watts is the RMS value of a sinusoidal power waveform having a peak of 200 watts. "RMS" in the sense of a mathematically calculated root-mean-square. And in saying that I certainly recognize that the heating which occurs in that example corresponds to 100 watts, not to 141 watts.

On the other hand, though, in citing the 141 watt figure I overlooked the fact that the product of a sinusoidal voltage and a sinusoidal current is not sinusoidal, since it never goes negative in the case of a resistive load. And correspondingly positive power is being delivered to the load at all times, other than at the zero crossings. So I believe the 141 watt figure should be, per one of Kijanki’s posts early in the thread, 0.61 x 200 = 122 watts.

Also, Roger, a **much** better paper on the subject than the Wikipedia writeup Kijanki referred to is the one Imhififan linked to in a post early in the thread:

http://eznec.com/Amateur/RMS_Power.pdf

That author’s conclusions:

It should be noted that the term “RMS power” is (mis)used in the consumer audio industry. In that context, it means the average power when reproducing a single tone, but it’s not actually the RMS value of the power.

Summary:

I’ve shown that:

-- The equivalent heating power of a waveform is the average power.

-- The RMS power is different than the average power, and therefore isn’t the equivalent heating power. In fact, the RMS value of the power doesn’t represent anything useful.

--The RMS values of voltage and current are useful because they can be used to calculate the average power.

Imhififan also provided the following reference early in the thread, which again is highly supportive of Kijanki’s position:

http://www.n4lcd.com/RMS.pdf

In any event, I agree with PTSS that Ralph’s (Atmasphere’s) advocacy of a pragmatic outlook on this issue (see his post dated 6-30-2017) is well stated and appropriate. But in going forward on that basis it seems to me that at the very least we should also acknowledge the legitimacy of Kijanki’s point, that based on a strict interpretation "RMS power" does not equal the product of RMS voltage and RMS current, and therefore does not correspond to the heating effect of a given amount of power.

Regards,
-- Al

@kijanki @almarg That author’s conclusions:

It should be noted that the term “RMS power” is (mis)used in the consumer audio industry. In that context, it means the average power when reproducing a single tone, but it’s not actually the RMS value of the power.

Summary:

I’ve shown that:

-- The equivalent heating power of a waveform is the average power.

-- The RMS power is different than the average power, and therefore isn’t the equivalent heating power. In fact, the RMS value of the power doesn’t represent anything useful.

--The RMS values of voltage and current are useful because they can be used to calculate the average power.

Why do you quote a paper that is all about square waves when we are talking about sine waves? This is most unscientific.

http://eznec.com/Amateur/RMS_Power.pdf

I encourage readers, if there are any left over this foolishness, to note all the waves in the picture are square waves. The value of Vp= 1.41 x Vrms applies to sine waves only. Not to square waves where the average is 0.5 and so is the RMS.

In this amateur, by its own name, paper, which is highly flawed. In the first step he aready has the average, is correct and done. However he wants to prove something odd. So he applies 1.414 to the already correct answer and gets a new answer which is incorrect.

If one stops for a moment and looks at a square wave with at flat top the average and the DC value are both 1/2 the peak. Just cut the wave in half and fill in the hole. Then you get a straight DC line. No problem. But with a sine wave as the voltage peaks and current peaks there is a lot of energy at the top. The use of root 2 or 1/ root 2 ONLY APPLIES TO SINE WAVES, not triangular, not square, not you mothers fancy stiching.

TRUE RMS meters actually measure the heating value of nonperodic waves and can even tell you the RMS or heating value of music. That heating value is important to your woofer.

So far the author has supported his position with a flawed page from Wickipedia (flawed in their estimation also) and this paper on square waves. There is no point in going further with this.

Stating there is no such thing as RMS power is a bold statemtent that has uncountable support for the fact that RMS is real, useful and applies to amplifiers.

I did find today a mistake in the authors early math and will present the correct math, I dont know why this author wants to press this most unreasonable theory.

As to the authors commment Al (he wont get it) and( he didnt get it.) Rudely said but true. I dont get what you said and I dont know what engineer would.

http://www.n4lcd.com/RMS.pdf

This is interesting and if all we are actually arguing about is the term "RMS" then we have made a mess of things. Putting the term "RMS" in front of watts is a misnomer. Once you have watts you just have watts. There are no other kind of watts for continuous waves so AVERAGE watts does not apply either. Its just WATTS, AC, DC, any periodic constant value. The RMS I believe is to show that the watts were measured by RMS methods, not peak or peak to peak methods. 

Perhaps we have argued over nothing but 3 letters of the alphabet, however the OP has cited papers that are incorrect and certainly have muddied the water. Heres you out guys:)

We still have to agree on one thing. The heating watts of a 100 watt amplifier is 100 watts. It is measured by V rms sq/R load.

The Peak watts is 200 and there are no other meaningful numbers to be stated. Using RMS to mean  'Hey anything with RMS in front of it gets to be multiplied by 1.4"... is a no no.

Average watts is generally applied to a signal that is non constant and thus an average is needed. Average is not appplicable to measuring sine waves for power. In fact the average squared comes up low. 

@imhififan https://en.wikipedia.org/wiki/Root_mean_square

Thanks, this is most interesting. I note that the coefficient (mulitplier) for peak to peak is 2.8 for the sine, 3.5 for the sawtooth and 1.0 for the square. (divide those numbers in half for the peak, of course) Those are all RMS. So readers might like to know there is not just one RMS in this world. It depends on the waveform. Sines are the best because we have equipment to null out the fundamental and then we can measure and SEE the distortion products on a scope.

Do you have a comment to add to our lovely conversation? 

Recently I have come to the possibiity that the OP decided to apply RMS to power, which I cant imagine anyone doing. There is no RMS of power nor is there average of constant power it is just power. Power determined by RMS voltage. We dont RMS it again. Why go on to use mathmatical arguments to prove or disprove a simple misuse of terms. Thats just grandstanding.

If we want to be perfectly clear we should say:

100 Watts (measured by RMS voltage of an undistorted sinewave into a resistive load) .Then then there is no abiguity. Thats all he had to say!

At first no one was interested, look at 2nd post. They Ralph properly answered how its done. Then the OP objected to Ralphs answer which happend to be exactly how we do it. Though most of us read it off the Sound Tech which has a watts scale. Just Watts, my hands are steady, the meter is steady, the ship is on course, there is nothing to average.

The OP states that the word "average" is appied in ’every textbook". Every is a dangerous word. But im gonna look at a few I have here.