Talk:Signal averaging

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According to the top paragraph, by averaging several signals "...S/N, will be increased, ideally in proportion to the square root of the number of measurements", which is indeed a common conclusion.

However, according to the calculations below and using the given definition of SNR as the power of the signal over the noise variance, averaging n signals increases the S/N by n, not the square root of n: "...averaging n realizations of the same, uncorrelated noise reduces noise power by a factor of n."

I believe the statement that the SNR is increased by the square root of n arises from using the RMS of the signal over the RMS of the noise as the definition of SNR, rather than the power. Perhaps this could be clarified.

2600:1702:2DD0:C4C0:97C:E50:637E:B5EA (talk) 23:44, 19 August 2018 (UTC)Reply

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Talk:Signal averaging