# Why use RMS for averages?

1. ### iScience

408
consider a sin wave

to find the average of this function over interval 2pi, why not just do...

$$\frac{1}{2\pi}\int |sinx|dx$$

this turns out to be..

$$\frac{4}{2\pi}= 0.63619$$

the whole purpose of squaring and then sqrt'ing at the end is to make sure there are no negative values. this would be fine if its value came out to be the same as the average of the absolute value (which is what i thought we were trying to find in the first place), but the value is different.

so.. i guess i have two questions:

1.) is 0.636 more correct to use as an avg than 0.702?

2.)why do we always use RMS value in science as opposed to the ACTUAL avg (of the absolute)?

2. ### HallsofIvy

40,369
Staff Emeritus
That is a perfectly valid "average" with slightly different properties than "rms". Which you choose to use depends upon the data and what you plan to do with the data. One can show, for example, that if your data is from a "normal distribution" the rms will, on average, lie closer to the true mean of the distribution than the absolute value mean. Also, since the absolute value function is not differentiable, it can be harder to work with, analytically, than rms.

### Staff: Mentor

One of the key uses of the RMS of a sinusoid is in alternating current. Suppose you have a light bulb and you want to make it shine equally brightly using direct current as opposed to a 120 volt RMS AC supply? The answer is 120 volts. The RMS voltage (or current) gives the equivalent DC voltage (or current).

No.

What makes you think your average is the "ACTUAL" one?

There are many ways of computing a "norm" or average. Your's is but one, RMS is another. Yet another is the maximum absolute deviation. There are others as well. Which one is "right"? That's the wrong question. They all are, in their own way.

4. ### jbunniii

3,356
One reason that RMS is a natural measurement to use is that you can think of it as the extension to infinite dimensions of standard euclidean distance.

If we have some point, say, (3,4,5) in 3-dimensional euclidean space, then the norm (the distance from this point to the origin) is ##\sqrt{3^2 + 4^2 + 5^2}##. We can think of a function as a "point" in infinite-dimensional space, and its "distance" from the origin (the zero function) is ##\sqrt{\int |f(x)|^2 dx}##.

But even in euclidean space, there are many other norms we can use, for example ##(3^p + 4^p + 5^p)^{1/p}## where ##p## is any real number ##\geq 1##. The special case ##p=2## gives euclidean distance. Similarly, ##(\int |f(x)|^p dx)^{1/p}## is a perfectly valid norm to use for functions.

A couple of advantages of the ##p=2## case (RMS):

1. The pythagorean theorem: if ##f## and ##g## are orthogonal (meaning ##\int f(x)g(x) = 0## in the case of functions), then ##\int |f(x) + g(x)|^2 dx = \int |f(x)|^2 dx + \int |g(x)|^2 dx##. This makes it easy to calculate the RMS of the sum of certain kinds of functions, such as sinusoids or noise.

2. The Cauchy-Schwarz inequality: ##|\int f(x) g(x) dx| \leq \sqrt{\int |f(x)|^2 dx}\sqrt{\int |g(x)|^2 dx}##

5. ### George Jones

6,384
Staff Emeritus
Related to DH's post:

RMS voltage multiplied by (in phase) RMS current gives average power. Some references call this product RMS power, but this is not correct, it is average power, hence the equivalent brightness of a lightbulb.