Average vs RMS

[Nigel Goodwin]

Huh? It has everything to do with power measurement, unless, I guess, you are an audiophile...

Which is what the thread is about

 

[MrAl]

Hello,

I dont see what this has to do with "audiophile".
If anyone, whoever they are, audiophile or not, wants to be able to read a data sheet and understand it enough to know what they are getting with the choice of amplifier chip then they absolutely must understand what "average power" is.

Excerpt from an audio power amplifier data sheet:

Features

• 68W Cont. Avg. Output Power into 4Ω at VCC = ±28V
• 38W Cont. Avg. Output Power into 8Ω at VCC = ±28V
• 50W Cont. Avg. Output Power into 8Ω at VCC = ±35V

 

[Endolith]

To try and get a clarification on the original question (as I understand it, at least) am I correct in saying we are all agreed that "Average Power" and "RMS Power" are the same thing. Even if the nomenclature is not correct, these two terms *don't* represent different measurements that might produce different results when carried out on the same piece of equipment?

Yes, that's the problem. The two terms represent different measurements that are different for the same signal in the same piece of equipment. One measurement is meaningful, while the other is nonsense, but they can both be calculated and are not the same number.

If you have a sine wave generator that goes from -14.1 V to +14.1 V peaks, and you apply it to a 10 ohm load:

• The average voltage is 0.0 V, since it spends just as much time positive as it does negative.
• The RMS voltage is 10 V (= peak value / √2).
• The average power is 10 W, since it varies from 0 W to 20 W during each cycle (avg(P) = avg(V^2/R) = RMS(V)^2 / R = heat dissipation)
  
• The RMS power, literally, is 12.2 W. (RMS(P) = RMS(V^2/R) = useless measurement)

However, you 'could' use the 'average' from it (as you've done below), which is why 'average' is a confusing term, and has no bearing on the power output measurement of an amplifier.

If you run a 100 W lightbulb for 12 hours per day and leave it off for 12 hours per day, the average power usage over a day is 50 W. Is that confusing?

Audio power is exactly the same:
If you play a square pulse train from your power amp into a speaker that produces 100 W for half a cycle and 0 W for half a cycle, the average power for a complete cycle is 50 W.
If you play a sine wave that reaches peaks of 100 W and troughs of 0 W, the average power over a complete cycle is 50 W.

 

[crutschow]

Here's an interesting question:
If you have a periodic positive voltage feeding a constant current load, do you use the RMS value of the voltage or the average value of the voltage to calculate the power dissipated in the load?

 

[Nigel Goodwin]

If you run a 100 W lightbulb for 12 hours per day and leave it off for 12 hours per day, the average power usage over a day is 50 W. Is that confusing?

Not at all - but it has nothing to do with audio power specifications.

By your 'measurements' my amplifier that is 60W+ RMS per channel is only going to rated very pathetically, as it's only used occasionally, so 'average' power would be close to zero.

BTW, perhaps you might like to 'grow up' a bit, and stop acting like a petty child clicking 'dislike' on every single post.

 

[MrAl]

Here's an interesting question:
If you have a periodic positive voltage feeding a constant current load, do you use the RMS value of the voltage or the average value of the voltage to calculate the power dissipated in the load?

Hi there,

Yes that is interesting

The general form of the voltage would be:
V=A*sin(w*t)+B
and the general form of the current would be a constant:
I=K
and so the instantaneous power would be:
p=V*I=K*(A*sin(w*t)+B)

and the average power is:
P=(1/T)*integral V*I dt

which expanded is:
P=(1/T)*integral [K*(A*sin(w*t)+B)] dt

and since K is a constant we can bring that out of the integral and get:
P=K*(1/T)*integral [(A*sin(w*t)+B)] dt

and since everything to the right of K is just the average voltage, we can write:
P=K*Vavg

and since K was the 'constant' current we can write:
P=I*Vavg

The RMS rule is only good for sine waves or special cases then because what does NOT work here is:
P = Irms*Vrms

so P != Irms*Vrms in this case, where "!=" is the symbol for "not equal to".

One thing that never fails is the integral form of calculating the average power:
P=integral [v*i] dt

and since v and i are more generally functions of time the more accurate writing of that is:
P=integral[v(t)*i(t)] dt

and this should always work as that is the definition of average power.

Just to note, in the past i have gotten good results with that last form even using a numerical integrator where each wave v(t) and i(t) are approximated as piecewise functions with straight lines connecting the measured points, where the points are read off the oscilloscope. With a little care the results are very reasonable.

Many things in electronics are dependent on the situation. For example, the average power is usually important, but in a battery run project the more important thing is the current drain because that affects the run time more than anything else.

 

[The Electrician]

Here's an interesting question:
If you have a periodic positive voltage feeding a constant current load, do you use the RMS value of the voltage or the average value of the voltage to calculate the power dissipated in the load?

Your question is ill-posed. Do you want to know how to calculate the RMS power, or the average power? MrAl showed how to calculate the average power, assuming that's what you wanted, but you didn't say that was what you wanted.

 

[MrAl]

Your question is ill-posed. Do you want to know how to calculate the RMS power, or the average power? MrAl showed how to calculate the average power, assuming that's what you wanted, but you didn't say that was what you wanted.

Hi,

How are you defining RMS Power and Average Power?
I guess you are defining Average Power as usual, but what about RMS Power how are you defining that?

 

[crutschow]

Your question is ill-posed. Do you want to know how to calculate the RMS power, or the average power? MrAl showed how to calculate the average power, assuming that's what you wanted, but you didn't say that was what you wanted.

I am talking about real (average) power since here are no reactive components mentioned.
As has been noted, RMS power is a misnomer.
There is RMS voltage and RMS current but power is either real or reactive.
Of course you can talk about "peak" power but that's not of particular interest here.

 

[The Electrician]

Hi,

How are you defining RMS Power and Average Power?
I guess you are defining Average Power as usual, but what about RMS Power how are you defining that?

Since this thread is all about the controversy surrounding the two terms "RMS power" and "average power", it seemed to me that crutschow's question asking how to calculate unspecified "power" was ambiguous. I see that he has clarified his question.