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In dealing with audio signals processing and talking with some engineers about general signals processing I've encountered the Root Mean Square which is some sort of mean. Here's a summary of the RMS of discrete data points and of a function on an interval:

$$\text{RMS}(\textbf{x})=\sqrt{\frac{1}{n}\sum_{i=1}^n x_i^2}=\frac{\|\textbf{x}\|_2}{\sqrt{n}}$$

$$\text{RMS}(f)=\sqrt{\frac{1}{b-a}\int_{a}^b [f(x)]^2dx}=\frac{\|f\|_2}{\sqrt{b-a}}$$

Geometrically this makes sense to me because it's essentially the euclidean norm. What I'm wondering is why is the "averaging term" under the radical? It seems to me that the squaring then the square root put the data in the same scale as the original function, wouldn't it make more sense to define it the following way?

$$\text{M}(\textbf{x})=\frac{1}{n}\sqrt{\sum_{i=1}^n x_i^2}=\frac{\|\textbf{x}\|_2}{n}$$

$$\text{M}(f)=\frac{1}{b-a}\sqrt{\int_{a}^b [f(x)]^2dx}=\frac{\|f\|_2}{b-a}$$

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I think $f(x)$ under the radical in the continuous definition should be squared as well, right? –  Bitrex Jul 31 '12 at 23:49
    
Bitrex: yes I've fixed it –  Nate Iverson Jul 31 '12 at 23:54
    
Yours would the the mean-root-square, not the root-mean-square. –  Pedro Tamaroff Jul 31 '12 at 23:56
    
@Peter, I would interpret mean root square as $\operatorname{mean}\circ\operatorname{root}\circ\operatorname{square}$, i.e. $\frac1n\sum_i\sqrt{x_i^2}$ instead :) This is more of a, I don't know, divide-by-$n$ root sum square. –  Rahul Aug 1 '12 at 0:08
    
@Rahul isn't that just the absolute mean for real valued x ? –  Nate Iverson Aug 1 '12 at 0:11

4 Answers 4

up vote 8 down vote accepted

Any sensible average, when given $n$ identical values, should result in the same value. The root mean square has this property, while your proposed definition does not.

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Rahul, I think that hits the nail on the head. –  Nate Iverson Jul 31 '12 at 23:55
    
...constant functions should return the constant under a sensible continuous mean as well. –  Nate Iverson Aug 1 '12 at 0:07

It's under the radical because when you're doing something which is "kind of like average" such as RMS, you want the result of $n$ identical data points to be that same value. There are other reasons, but I think that has to be a big one.

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Here's a physics exercise. Recall from your high-school physics course that volts times amps equals watts. That's with direct current. Show that that also works with sinusoidal alternating current precisely if "volts" is take to mean the root-mean-square voltage as conventionally defined, i.e. the square root of the mean of the square of the voltage.

Now remember that "standard deviation" is root-mean square deviation. When applying the central limit theorem to find the probability that the number of heads you get when you toss a coin 1600 times is between 1590 and 1630, show that everything works out neatly precisely if you use the conventional definition of root-mean-square.

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Just an interesting aside (hey! my main hangout is on EE stack exchange!) With sinusoidal AC, it's only called "watts" if both the voltage and current are in phase. If there's a phase difference, the product of RMS voltage and current is called "volt-amps reactive" and forms a right triangle with "real" power (watts) on the real axis and "reactive" power on the complex axis. In general, for an arbitrary waveform, "real" power is only consumed if the voltage and current harmonics are non-orthogonal. –  Bitrex Aug 1 '12 at 0:07
    
Sometimes I think I should go back to school and study electrical engineering. Some of the math involved seems intriguing. –  Michael Hardy Aug 1 '12 at 15:19
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I'd like to go back to school to study math! :P –  Bitrex Aug 1 '12 at 17:13
    
Oops! Just noticed a mistake - should be just "volt-amperes" for the RMS product. –  Bitrex Aug 4 '12 at 7:39

Electrical engineers use RMS values when dealing with AC. The instantaneous power delivered to a resistive load is $p(t) = R \, i^2(t)$ (or the equivalent formulation with voltage), where $R$ is the resistance and $i$ the instantaneous current.

For most AC (as in industrial & domestic power) considerations, the average power over a cycle is of more interest than the instantaneous power. If one does the calculation (with $i(t) = i_\max \sin(2 \pi 60 t)$), then this is expressed as $p_\mathrm{average} = R \, i_\mathrm{rms}^2$, where $i_\mathrm{rms} = \frac{1}{\sqrt{2}} i_\max$, thus retaining the same form.

AC voltage/current is typically quoted as an RMS value, since this usually reflects the actual power or losses of interest involved. So a 110-120V domestic supply actually reflects a maximum voltage of about 156-170V.

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I changed i_{max} to i_\max and i_{rms} to i_\mathrm{rms}. They look like this: $i_{max}$ versus $i_\max$. –  Michael Hardy Aug 1 '12 at 15:13
    
@MichaelHardy: Thanks! –  copper.hat Aug 1 '12 at 16:32

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