Talk:Audio power

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i suggest removing all of the RMSes that are misnomers and using mean power. yeah it's a valid abbreviation, but it's still misleading. just mention it at the beginning and then use the real thing. - Omegatron 19:22, Dec 31, 2004 (UTC)

like it or lump it RMS is the term everyone in the buisness uses despite the fact it is a misnomer. Therefore i think it should stay. It is clearly stated at the begining of the article why the term is a misnomer and it is in quotes later on to try and reinforce this i think thats enough. Plugwash 19:37, 31 Dec 2004 (UTC)
Just because everyone's in the habit of doing something wrong doesn't mean we should encourage them. Besides, lots of manufacturers don't use it, because they know it's wrong. I believe the normal wikipedia response would be to explain the incorrect usage and then use the proper term everywhere else. Since the article is basically about "watts rms" i will change what i think should be changed and you can revert it if you want. - Omegatron 22:47, Dec 31, 2004 (UTC)
Eh, you're right. It's totally appropriate in this article because it's about marketing literature. - Omegatron 22:57, Dec 31, 2004 (UTC)
RMS is not a misnomer if you consider that (power)amps are normally rated for power output using a sine-wave at a nice middle-of-the-bandwidth frequency, in this case RMS means RMS, unless you want to nit pick over distortion and such caused by the amp itself. Secondly does anyone object to a section being added on peak-to-peak power (just a UK thing?) and PMPO being expanded to include peak music program/power output?--Pypex 23:01, 14 September 2005 (UTC)
No, that is simply mistaken, without any recourse to nit picking. If you were to actually measure the RMS power of an amplifier on a sinewave signal you would get a figure higher than the true power figure (the one you think you would get) which is actually the mean of the instantaneous power at all points over one cycle. Squaring the power is wrong, and gives more weight to the power at peak voltage than is correct. We square the voltage, to arrive at power, but that is a separate issue. A correct term, if you want to refer to RMS would be 'power based on RMS voltage measurement', but this is silly because there is no other valid way to measure power on an AC signal. The only correct term is 'mean sinewave power', since without a waveform specification you can chose any figure you want. I would object to any other method being described here, other than as in terms of explaining why it is not valid, unless it is backed by a standard. We have standards, notably an IEC one which specifies precise times for test duration and method. These should take priority over 'folklore'. Unless you can show me otherwise, I maintain that the term PMPO has nowhere ever been defined, and certainly not by a standards body. UK thing? As a professional audio measurement expert in the UK I can only say that the term PMPO is hated and despised by all fellow professionals. It's marketing speak, based only on a desire for high numbers by people trying to sell, should be exposed as such in any article on audio. --Lindosland 11:43, 21 March 2006 (UTC)
I do not like the statement that peak music power is usually twice the sinewave power. It confuses three issues - power supply sag, peak to mean ratio, and thermal limitation. Most power amplifiers will only give a little more true power short term, perhaps 10%, by virtue of having the full power supply voltage. Using peak power, which is twice the normal, is simply cheating on figures, and to be rejected. Thermal limitation is a complex business. Few modern amps are thermally limited in practice. Integrated amps chips have thermal protectin built in, but they are usually provided with sufficient heat sinking to prevent them cutting out (a sudden muting). Some high power amps do have thermal protection coupled with gain reduction. The common supposition that testing power amps at full output on sinewaves overstresses them is mistaken, since calculations show that maximum dissipation in the output devices of a class B amp occurs at about 60% of full output on sinewave drive, where the voltage across the devices is high. At 100% output the amp will run cooler, and at 100% flat out and clipping it will run cooler still (because the devices are almost switching, with no voltage across them for mush of the time). I believe there is actually a case for designing an amp with real music power capability, with say 10 times the steady sinewave power available on peaks, and sophisticated protection so that this was only available for handling brief music peaks (see Programme levels). Only with such a large ratio does 'peak music power' take on a useful meaning, but to my knowledge this has never been done. On modern compressed material it would have no advantage, but on properly recorded material it would be a revelation. --Lindosland 11:43, 21 March 2006 (UTC)

Contents

[edit] Incorrect equation at the end.

The last section claims that P_\mathrm{avg} = V_\mathrm{rms} \cdot I_\mathrm{rms} = \sqrt{\frac{1}{T}\int_{0}^{T} v^2(t)\, dt} \cdot \sqrt{\frac{1}{T}\int_{0}^{T} i^2(t)\, dt}

but it seems to me that it should be something like

P_\mathrm{avg} = \frac{1}{T}\int_{0}^{T} v(t)i(t)\, dt

Could someone show how this is derived? It looks like the magniture of a phasor, but that is only true of a sinusoidal system, not power in general. --njh 11:48, 16 May 2006 (UTC)

Well, the P_\mathrm{avg} = V_\mathrm{rms} \cdot I_\mathrm{rms} is correct (for a resistive load only), and the square root integrals are the equations for RMS, so it looks correct. Your version agrees with Power_(physics)#Average_electrical_power_for_AC and [1], though. Are these really just the same equation for the case of the resistive load? — Omegatron 14:27, 16 May 2006 (UTC)
I think that is correct, that purely resistive loads have ave power = rms voltage * rms current, but speakers are not very resistive (they have lots of inertia). --njh 02:45, 17 May 2006 (UTC)
I've updated it. — Omegatron 11:25, 18 May 2006 (UTC)

[edit] Example

"An amplifier labeled "500 W PMPO" but fitted with a 5-amp fuse can therefore deliver an average power of 5 A × 14.4 V × 60%, or about 43 watts."

Why are they assuming a 14.4 V supply? Is this a car amplifier? Cars have 12 V supplies. I don't know which type of amp this is assuming. — Omegatron 13:41, 17 May 2006 (UTC)
Car batteries are only nominally "12 volt". At rest, a healthy car battery is actually 12.6 volts. While the engine is running, the alternator always maintains the cars electrical system at a higher voltage (in order to recharge the battery as necessary), ranging from about 13.5 to 15 volts. 14.4 volts is very common - much lower and the charging doesn't work very well, much higher and devices on board may be damaged. human 00:06, 29 March 2007 (UTC)

[edit] H-bridge

I don't understand NathanHurst's edit. An H-bridge seems to be a switching network for running DC motors. I assume he means a bridged amplifier, which is a very similar idea, but which I have never heard called an "H bridge". Also, if car amps are commonly bridged, then you could get 12 Vpeak out of them instead of 6 Vpeak, as the example in the article describes. Also, I don't know what any of this has to do with his removal of the fused plug example. — Omegatron 13:05, 18 May 2006 (UTC)

H-bridge is a topology to get full supply p-p voltage. most car amps provide this. But the example was using 6v p-p and so was presumably not using h-bridge topology, instead a single class ab amp. I just clarified this. You could talk about a bridged amp, and then multiply everything by 2 or 4, but that would distract from the point. I removed the remark about the fuse as it is simply a non-sequetar - a fuse blows based on average power, pmpo, whatever it means, is presumably a short duration thing. Perhaps this particular model has a flywheel it charges up in quiet sections. (Yes, obviously it doesn't, but it is not _proof_ that PMPO is wrong) --njh 22:33, 18 May 2006 (UTC)

[edit] To RMS or not to RMS, that is the question!

Dear engineer types,

I appreciate your discussion and follow most of it. But I believe this academic level is useful only to BSEE types (design engineers, most often) and not nearly as useful to we mere hands on technicians and especially the laity of the general public who might read this. For most just comparing the RMS power number is useful and real. The ins & outs of analyzing complex waveforms goes beyond the need for this article. Feel free to publish all you want in your engineering forums or in formal technical papers, but I request that we keep Wikipedia articles like this one user friendly for the common man. Don't get me wrong, your discussions certainly have their place, but blowing away people with the sophistication of your knowledge does not help them understand the basic question being asked.

(In other words, RMS power is plenty good enough in practice if you are not a design engineer)

sjb

137.164.224.92 23:35, 12 January 2007 (UTC)

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