I'm wondering why noise sources are added such a way their rms values are added in square root of sum of squares fashion.

For instance if the rms noise values from three sources are X,Y and Z, then the total noise is given as \$N = \sqrt{X^2+Y^2+Z^2}\$.

Is there a way to show this relation mathematically?


If the noise sources are uncorrelated, then the power (variance) of its sum equals the sum of their powers (variances), because all the co-variances between them are zero, and only the variances remain.

An example with two noise sources (with zero mean):

\$\sigma_N^2 = E[N^2] = E[(X+Y)^2] = E[X^2]+E[Y^2]+2E[XY] = \sigma_X^2 + \sigma_Y^2 +2\sigma_{XY}\$

If \$X\$ and \$Y\$ are uncorrelated then \$\sigma_{XY}=0\$ and \$\sigma_N^2 = \sigma_X^2 + \sigma_Y^2 \$, thus:

\$\sigma_N = \sqrt{\sigma_X^2 + \sigma_Y^2} \$, where \$\sigma_N\$, \$\sigma_X\$ and \$\sigma_Y\$ are the RMS values of \$N\$, \$X\$ and \$Y\$.

It can be demonstrated for any number of sources, if required. The result ends up being the same: power of the sum equals the sum of the powers.


If you take two waveforms having RMS values x and y, to get the total RMS you square (x+y) to get: -

$$x^2 + 2xy +y^2$$

Then you take the mean and the 2xy will have a mean of zero because x and y are uncorrelated noise sources. If they correlated then xy would not have a zero mean value.

So, the mean of the squares is \$x^2+y^2\$ and, to get the total RMS you take the square root to get: -


RMS means root of the mean of the square.


You could turn this on its head and say that a defining feature of noise is that it adds as power to any signal, that is, as RMS.

Two uncorrelated signals add as power. You can determine the degree of correlation by using the amount the power of their sum differs from the sum of their powers.

  • \$\begingroup\$ Your last sentence is very interesting.... So let X and Y are rms voltage noises. Are you saying that the equation: |(X^2+Y^2) - (X+Y)^2| will determine the degree of correlation? The bigger this subtraction outcome the higher the correlation? \$\endgroup\$ – cm64 Sep 10 '19 at 12:39
  • \$\begingroup\$ The word 'using' means I am not specifying the equation, but there is an equation. I don't know it off the top of my head, and can't be bothered to look it up. A few minutes with google and 'correlation' in the search ought to get you the answer. \$\endgroup\$ – Neil_UK Sep 10 '19 at 12:41

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