What does RMS stand for???

[TechSpec]

I know that RMS is the rating for the maximum continious power, I was wondering what the acronym actually stood for. if you know please post it. please only post if you know what it means not what you think it means... like random magnitic signal, and yes that is what a salesman at a audio store tried to tell me it was. Thank you

[Mighty Favog]

I'm at work and I did find in the McGraw-Hill Dictionary of Sientific and Technical Terms-Fifth Edition the only refernce it makes to it is "root-mean-square". But I'm still looking...

O.k., here we go..."The square root of the time average of the square of a quantity; for a periodic quantity the average is taken over one complete cycle.

Looks like that's just the physics definition.

------------------

Tom

KLF-20 Mahogany

McIntosh C33 (In the Hospital--being discharged on June 29th)

Rotel RB-1080

Yamaha PF-800 Turntable/ Sure V15 Type V Cartridge

Ortofon VMS-30 mkII Cartridge

Stanton 999SS Cartridge

Carver TL-3100 CD

Yamaha K-1020 Cassette

dbx 1231 EQ

H.H. Scott 830z Analyzer

Monster Interlink 400mk II

Monster Interlink 300 mk II

Studio Tech U-48RW Cabinet

Original 12ga. Monster Cable

Enough boxes for a fire hazard!

This message has been edited by tblasing on 06-28-2001 at 01:03 PM

[J M O N]

Root Mean Square is the correct answer.

[TechSpec]

Thanks guys, I need to get a copy of that dictonary. What a salesman... I also had someone else tell me something like Recomended Magnitude Sample, and that just sounds like someone who has no clue what they are saying, but spew out words they have heard and don't know what they mean. Thanks Again.

[Ray Garrison]

see attachment for actual formula for calculating RMS values...

When you have a thing that fluctuates from one moment to the next, and you want to know the "average" value of that thing, the definition of what "average" means is somewhat arbitrary. In this case, we have an alternating current delivering power to the speakers. Keeping things as simple as possible for a moment, suppose that signal is a pure sine wave at 1kHz. Let's say the peak voltage is four volts, so you have a signal alternating smoothly between plus and minus four volts. Assume the speakers are a perfect 8 ohm load at 1kHz. How much power is going to the speakers? Well, if it was direct current, you'd have four volts into eight ohms, which is two watts (power equals voltage squared divided by resistance), and a constant current of a half an amp (voltage times current equals power). However, it's not direct current, it's alternating current. One thousand times per second, the voltage swings from plus four to minus four. The current swings back and forth, flowing first this way then that way. For some finite period during each cycle, the voltage is zero, and at some other point, the current is zero. How do you define the amount of power over a period of time? You can't simply average the voltage, the current or some combination... half the time it's positive, the other half the time it's negative, so your average would be zero. However, if you square the value present at any instant in time, you eliminate the negative numbers. If you take the average of these squared values over some period of time, then take the square root of that averaged value, you get a number which everyone has agreed can be used as the measurement of "power".

Hope that helps.

Ray

------------------

Music is art

Audio is engineering

[Colin]

it is a measurement of amp power now, but I remember the days when amps were measured by IHF standards (Institute of High Fidelity ?) but the new style of amps with flat little chips did mot measure so well by those standards, after much debate, the new standard was adopted ....

[WMcD]

Of course Ray is correct, as always.

The problem is that the sine wave is changing in level (amplitude) over it's cycle. What point of the changing amplitude of a continuous sine wave do we take as the "voltage". What dimension measured is the best number to tell us what we want to know?

By comparison, consider a circle. We could measure radius, diameter, or circumfrence. They are all inter-related. One or the other is not more true or less true.

Also, what do we want to know for a calculation? If the circle is the cross section of a pipe, we might want to know the area, because that is related to how much water will flow. The area can be calculated using the square of all the above. For example: Pi (or 3.141) x radius x radius = Area. Therefore, when talking about a circle, we might measure it by area, rather than a linear dimension.

Going back to the sine wave. Suppose it goes from + 4 volts to - 4 volts. That would be 4 volts "peak" or 8 volts "peak-to-peak". (Yup, the valley is called a peak.)

Sounds easy. And "peak" or "peak-to-peak" are commonly used means of expressing the level, particularly the level of a voltage. You can see these on an oscilloscope.

However, these expressions don't easily tell us one thing which we might want to know. That is, how much power will the sine wave voltage create when the given level of the sine wave is applied to a load.

Let's assume the load is an 8 ohm resistor. Let's also assume an 8 ohm speaker is really a resistor. (Which is not true, but it is a simplifying assumption.)

We know that for direct current:

Power = Volt x Amps.

Also by Ohm's law:

Amps = Volts / Resistance.

Putting those two equations together to eliminate Amps we find

Power = Volts x Volts / Resistance.

Note, the Volts are multiplied by itself, or squared! Also, the power equation gets us a bit closer to what we want to know: power. Also, we don't have to worry about the Amps of current, it has dropped out of the picture.

If we're dealing with D.C., like from a battery, the voltage is constant. But not so with alternating current, meaning our sine wave.

One thing we realize is that power is being delivered to the resistor whether the voltage is plus or minus. (Here at least, the negatives don't suck out power.) So, conceptually, we can flip over the bottom half of the sine to the top. Instead of mountains (+) and valleys (-), we have just mountains going up from zero.

Couldn't we just average out this whole string of mountains to get the equivalent of the D.C. for our power equation? Nope. The power equation requires Volts x Volts. So, we really have to square the voltage at every single point on the "partially flipped" sine wave. Then we can average that (or actually, integrate the continuous function).

When the math is done, we find that the effective value is 0.707 of the measured peak. This is to say, the RMS value is 0.707 of the peak of the sine.

In truth, you know some of this already in a vague way.

We all know we have 120 volts A.C. at our home power outlets in the U.S. (Tony, take note, 240 volts in Sunny Sal?) That is actually the RMS volt value of sine wave coming from the power company. If you looked at the "signal" on the hot wire of the outlet using an oscillosope, you'd see it has peaks of 170 volts, or 340 volts peak-to-peak. 170 x 0.707 is 120 volts RMS. (This is measured relative to the neutral wire, which should be at zero volts).

In a way, it is a means of saying, "What is the equivalent D.C. voltage which would create the same amount of power to a light bulb." For that matter, the amount of power delivered to our speaker. The equivalent is the RMS A.C. voltage. In almost all cases, A.C. voltmeters are calibrated to read out in "RMS" volts.

Let review that again, because it is difficult first time around, and very important:

When you measure your A.C. line sine voltage with an A.C. volt meter, it says 120 volts. But that is neither the peak, nor the peak-to-peak. It is 0.707 times the peak. We're asking the A.C. volt meter to tell us the number (level) of the D.C. voltage which would deliver the same amount of power, even though the meter it is not measuring D.C. at all. Wacky, but true.

On the other hand, maybe not so wacky. It saves a lot of math as set out in Ray's link.

Now you say, what does this have to do with audio and amplifiers?

Sales people sometimes incorrectly speak of RMS "power". I believe they are trying to say the amp will put out a certain amount of voltage over a long term. This is really continuous power. Amps can often put out higher power for short periods such as the peaks in music. Music is not a continuous sine wave. The "peaks" rise and fall.

In a way, the hi-fi salesmen are not too far off base even if nomenclature is sloppy. If the amp can put out a given voltage (RMS) for a long term, that will allow the calculation of the continuous power capability of the amp.

(The Federal Trade Commission stepped on the hi-fi industry decades ago. Power outputs were overstated, because they were short term peak power. Hence there was a demand to standardize measurement to continuous power, which is always lower.)

While we're here, I might as well throw in something about speaker testing. Near and dear to our hearts. We've come so far. It would be a shame to not make a final step.

You'll see that speaker outputs measured in, say, "98 dB at one meter distance" are often measured with a certain electrical input to the speaker. It will say something like, "2.83 volts input to 8 ohms, 1 watt".

Question: Why was 2.83 volts selected? Answer: It is related to 1 watt input.

First of all, that is 2.83 volts as measured by RMS, as I harp upon.

Looking at our power equation again:

Power = Volts x Volts / Resistance.

The resistance is assumed to be 8 ohms.

To get 1 watt of power input we need:

1 Watt = Volts x Volts / 8 Ohms.

What is the numerical value of the "Volts" needed above to make it come out to one Watt? Obviously it is the quantity which when multiplied by itself equals 8. (1 = 8 / 8)

We can make some guesses in our head: 2 x 2 = 4. 3 x 3 = 9. So, the answer is in there someplace bigger than 2 and smaller than 3.

Obviously, the answer is the square root of 8. That is 2.83 volts, according to the calculator.

What the testing engineer has done is measure and adjust the output of the amplifier (using an A.C. volt meter) so it is putting out 2.83 volts A.C. RMS. That creates one watt of input to the assumed 8 ohm load.

Sorry this missive has been so long, I didn't have time to make it short. Too simple for some; too complicated for others; too long for all. =8^o. However, I did want to touch all the bases.

Now you know!

Regards,

Gil

[Mr. Blorry]

Huh! And I thought it meant "Really Mean Speaker"

This message has been edited by Mr. Blorry on 06-29-2001 at 06:38 PM

[JohnA]

Gil and Ray did a great job! I'd like to add that the 2.83 volts is a further standardization, because no speaker is a uniform 8 ohm load. Using 2.83V RMS is an easily measured, precise calibration point.

The common statement is "104 dB @ 1 watt @ 1 meter". However, the Heritabe Klipsch typically vary from 4 ohms in the bass to 32 ohms in the midrange. Since power is volts x volts/resistance, if the resistance changes, the power input changes. Measuring only voltage (rather than power that really has 2 components, volts & resistance) is a simple, reliable calibration standard.

John

[buylow]

I'm impressed. Where else in all the internet can you discuss audio issues and learn so much at the same time? I continue to be amazed.

------------------

Mains: RF-3

Center: RC-3

Rears: RS-3

Sub: KSW-12

A/V Receiver: Yamaha HTR-5250

DVD Player: Yamaha DV-6280

CD Player: Yamaha CDC-506

VCR: Toshiba

TV: Toshiba 55" Projection

[mace]

Gil wrote:

Sales people sometimes incorrectly speak of RMS "power". I believe they are trying to say the amp will put out a certain amount of voltage over a long term.

Thanks for the nice refresher. Do amplifier manufacturers also incorrectly refer to "RMS power"??? In the literature for my amplifier it clearly states "continuous output power into 8 ohms" and it does not mention anything about voltage. I noticed that you left out amperes out of your equations but it was my understanding that actual amperes delivered to the speaker is very important, especially for controlling the voice coil of low impedence amplifiers.

Can't one just plot amperage as a function or voltage (power = current * voltage) and get a sine wave (assume constant current)? Thus, if my amp is delivering 56 W peak to peak, this is 0.7*56 = 40 W RMS, no??

Isn't it fun to talk about amps from amps?

While we're on the subject.... what exactly is subject to damage when impedences are incorrectly mismatched?? My amplifier has a switch for 4 and 8 ohms settings and my KLF 30's are rated at 8 ohms, as are my Boston's (which run off of my B channel). In another forum BobG recommended trying both ohm settings and seeing which one I liked better. However, can I do damage to the amp or speakers or both by having the amp set to 4 ohms when the speakers are rated at 8?? (although i suspect the KLF 30's tend to run lower than 8 when driven hard...)

Any comments would be appreciated.

I'm loving my new KLF 30's, BTW... awesome... I've had the whole weekend to myself (after travelling for 2 weeks) and am finally getting a chance to really give a good listen. Quite amazing...

Mace

[Klewless]

OK, guys.

Here is a place that explains RMS.

http://www.chipcenter.com/circuitcellar/january00/c10r24.htm

------------------

John P

St Paul, MN

[djk]

First off, there is no such thing as "watts RMS".This was a term coined by Marantz in the early '70s.RMS voltage times RMS amperage equals AVERAGE POWER.The RMS value assigned to an AC signal is the amount of DC required to produce an equivalent amount of heat in the same load.Marantz coined the term "watts RMS" as a marketing technique to distance themselves from manufacturers advertising "peak" and "IHF" figures."Dynamic headroom" is just another way to indicate how poorly your power supply is regulated.Unfortunately NONE of these numbers is going to give any indication as to how loud it will play on your speakers(or how good it will sound).I can give examples (if requested) of amplifiers that had differences of over 6dB in rated power (a 4:1 ratio)where the smaller amp not only played louder but sounded better too.

[mace]

djk,

Thanks for the clarification of "Watts RMS" and what it really means.

About the dynamic headroom issue... you make it sound like something like +6dB would only be found on a 'poorly regulated power supply' amplifier. At least with my experience, I've found dynamic headroom (at least 6 dB) to be a good thing in an amplifier.

My take on the dynamic headroom issue is that it is what it is... dynamically allocated umpff.. the amp goes to 11 if you will, albeit for only about 200 ms or so...

I agree, this has nothing to do with how "loud" an amplifier is but has everything to do with how dynamic an amplifier is. I wholeheartedly agree with you that dynamic headroom does not indicate how good an amp sounds. However, when done right, dynamic headroom can really make a smaller amplifier (like 40 VA RMS (is that better??)) sound very alive, especially when coupled with a speaker with 102 dB-W-m sensitivety.

If I read your post correctly, you indicated that rated RMS power had nothing to do with the sound volume an amp can produce. Take a 40 VA RMS amp and compare it to a 200 VA RMS amplifier... how can the 40 play louder than the 200??? Does it have something to do with peak current delivered?? How can RMS power not relate to overall sound volume the amplifier produces??? I'm just curious how this would be physically possible.

Always curious,

Mace

[djk]

I own a couple of Electrocompaniet AW 50 amplifiers.They are rated at 50 watts at 4 ohms.Because they have a regulated supply they have 0dB of dynamic headroom and only put out 25 watts at 8 ohms.On Klipschorns they have such an effortless sound like seem like they have 500 watts, up until the point they clip.Behavior at clipping is very important.Even on efficent speakers like Klipschorns.Some amps fall completly apart the instant they clip.Others can be driven 10dB into clipping and just become a bit "bright".A Heresy will sound better with a "well behaved" 100 watt amp than a 400 watt amp that will never come close to clipping.Tom Holman's APT model 1 amp had a Baker clamp so the output stage could never be driven into clipping.It had a "distortion" led that came on only after the amp had been clipping for more than 40mS.On dynamic program material you could drive that amp about 10dB into clipping before the led flashed.