Pretty much a newbie here, but would like to understand a pretty basic concept.

We all know that a quick and dirty way of calculating signal power of a part of a discrete sampled system (such as audio signals) is the age-old power formula for RMS power, where we simply break the signal into chunks of N samples and sum the squares of the amplitudes, then divide by the number of samples

$$ P_\mathrm{x} = \dfrac{1}{N}\sum_{n=1}^{N}|x[n]|^2 $$

So let's say I have 10ms samples of two completely uncorrelated signals at 48khz (480 samples), and I have calculated the RMS power of these samples using the above formula and have come up with P1 and P2, the power of my small segment of each signal.

Then if I sum the two signals (simple software mixer function), what is the resultant power of the two summed signals? Is it simply P1 + P2?

If so, do I simply keep going? So if I had four signals with P1, P2, P3 and P4, if I sum all four of them together, is the resultant RMS power P1+P2+P3+P4?


The formula you gave is an estimator for the "statistical power", i.e. the expectation of the square of a continuous underlying random process:

$$ P_X = E(X^2(t)) $$

Computing the latter theoretically would require to know the probabilistic behavior of $x(t)$, which you don't have access to in real life, especially since:

  • you get sampled data (not continuous),
  • you generally have a limited number of realizations.

This is even more complicated when you add variables, because you should know their paired probability behavior. A classical result is that, for variables $X$ and $Y$,

$$\mathrm{Var}(aX+bY) = a^2\mathrm{Var}(X)+ b^2\mathrm{Var}(Y)+ 2.a.b.\mathrm{Cov}(X,Y)$$

This quantity can take different values, depending on the last term: $\mathrm{Cov}(X,Y)$. For uncorrelated variables, the covariance vanishes, and the variables add as:

$$\mathrm{Var}(aX+bY) = a^2\mathrm{Var}(X)+ b^2\mathrm{Var}(Y)$$

So, for uncorrelated signals, powers add. And this generalizes to several signals.

Note that when you sum over a time frame to estimate such quantities, you somehow assumes that you can estimate quantities by averaging in time, with assume that the signals are correctly sampled for a stationary and ergodic process.

For $K$ signals of identical power $P$, the combined power can take value:

  • minimum: $0$ when signals sum to zero,
  • $KP$ when uncorrelated,
  • maximum: $K^2 P$ when signals are identical.

Additional pointers:


This one is easy:

Uncorrelated signals sum in power, so yes, it's just $P_1+P_2+P_3+...$

However, for most signals correlation is a statistical property, i.e. you need a large enough sample size to get accurate results. For short snippets, you can see significant variability of the results.


For two signals in general:

$ {1 \over N} \sum_{n=0}^{N-1} (x_1[n] + x_2[n])^2 = {1 \over N}(\sum_{n=0}^{N-1} x_1[n]^2 + \sum_{n=0}^{N-1} x_2[n]^2 + \sum_{n=0}^{N-1} 2x_1[n] x_2[n])$

For uncorrelated signals the last sum must be zero. This leads to:

$ {1 \over N} \sum_{n=0}^{N-1} ={1 \over N} ( \sum_{n=0}^{N-1} x_1[n]^2 + \sum_{n=0}^{N-1} x_2[n]^2)$

The same would work for more than two uncorrelated signals.

  • $\begingroup$ OP states that the signals are uncorrelated, i.e. for sine waves the phase angle would have to be 90 degrees $\endgroup$ – Hilmar Aug 22 '19 at 10:51
  • $\begingroup$ Yes you are absolutely right. I corrected my post. $\endgroup$ – chirp-rate Aug 22 '19 at 12:04

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