If I have understood correctly, averaging $N$ noisy independent segments or signals increases the signal-to-noise $\sqrt{N}$-fold.

How does one derive this result?

  • $\begingroup$ The noiseless part of the signal needs to be the same for all segments/signals for that to be the case. If both the noise and the signal are independent between the signals/segments then you get no such signal-to-noise ratio (SNR) improvement. $\endgroup$ Oct 11, 2015 at 15:43

3 Answers 3


I will show how to calculate the SNR for the case of $N=2$ measurements; it is easy to extend the result to general $N$. Assume a signal $s(t)$ has power $S$, and the noise $n(t)$ has variance $\sigma^2$ and zero mean. Then, the signal $s(t)+n(t)$ has SNR equal to $S/\sigma^2$.

Now assume you observe $s(t)$ twice, each time with different, uncorrelated noise, $n_1(t)$ and $n_2(t)$, each with zero mean and variance $\sigma^2$. You average the two observations to get $$\frac{2s(t)+n_1(t)+n_2(t)}2 = s(t)+\frac{n_1(t)}2+\frac{n_2(t)}2.$$

The variance of both $n_1(t)/2$ and $n_2(t)/2$ is $\sigma^2/4$, so the total noise variance is $\sigma^2/2$ and the SNR is $2S/\sigma^2$, for an improvement of 2.

For some reason I don't understand, the wikipedia page on signal averaging defines the SNR as $S/\sigma$. Under that definition, the improvement will indeed be $\sqrt{N}$.

  • 1
    $\begingroup$ You also need to assume that $n_1$ and $n_2$ are independent or, at least, uncorrelated. Otherwise $E[n_1(t)n_2(t)]$ will be non-zero and contribute to the overall noise in the averaged signal. $\endgroup$
    – Peter K.
    Oct 11, 2015 at 20:44
  • $\begingroup$ Standard deviation versus variance: note that your form would not scale correctly. Like in physics we have "symmetries" in our expectations. In this case if we amplify a reading by x we expect that what we call signal-to-noise to remain the same. $\frac{S}{\sigma}$ is and $\frac{S}{\sigma^2}$ isn't. $\endgroup$
    – rrogers
    Oct 14, 2015 at 13:49
  • 1
    $\begingroup$ @rrogers The way I see it, if $s(t)$ has average power $S$ and $n(t)$ has variance $\sigma^2$ (so the SNR is $S/\sigma^2$), then the power of $as(t)$ is $a^2S$, and the variance of $an(t)$ is $a^2\sigma^2$, so the SNR is still $S/\sigma^2$. $\endgroup$
    – MBaz
    Oct 14, 2015 at 17:51
  • $\begingroup$ @MBaz agreeing with you, Wikipedia's Signal Averaging really uses a non-canonical, and even more importantly, non-useful definition of SNR. $\endgroup$ Oct 15, 2015 at 8:33
  • 1
    $\begingroup$ As far as I'm concerned the standard deviation definition (as opposed to variance) is incorrect, because it's not power. I believe also that the square of the mean definition only applies if there's zero variance. Good answer though $\endgroup$ Nov 16, 2021 at 14:44

First, let's properly define the problem. Given $N$ observations $\left \{ x_i \right \}^{N}_{i=1}$ whereas $x_i \overset{i.i.d.}{\sim} {\mathcal{N}}(0, \sigma^2), \forall i \in \left [ 1,2,\dots, N \right ]$ we wish to find the variance of the sample mean estimator compared to a single sample variance.

Obviously, a single sample variance is $\sigma^2$, now let's see how averaging the samples will decrease this variance:

\begin{equation} {\mathrm{Var}}\left [ \frac{1}{N} \sum_{i=1}^{N} x_i\right ]\overset{i.i.d}{=} \frac{1}{N^2} \sum_{i=1}^{N} {\mathrm{Var}}\left ( x_i \right ) = \frac{1}{N^2}N\sigma^2 = \frac{\sigma^2}{N}. \end{equation}

So the error is inversely proportional to the sample size $N$. In fact, just by adding one more sample you increase your SNR by 3dB.

Note that $\sigma^2/N$ is also the Cramer Rao (lower) Bound (CRB) for the sample mean in a linear Gaussian problem, so this is in fact the best unbiased estimator for this problem.


A slightly different calculation is: $\frac{S}{N}=\frac{s+s}{\sqrt{\sigma^{2}+\sigma^{2}}}=\sqrt{2}\cdot\frac{s}{\sigma}$

Where $\frac{s}{\sigma}$ is the signal/noise for each reading

Two signals add: $s_{1}+s_{2}$

The two noises add and normalize: $\sqrt{\sigma_{1}^{2}+\sigma_{2}^{2}}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.