With stationary noise we have constant mean and variance (let's assume it is Gaussian noise). My first question is, how is the noise power calculated and how it is related to the variance?

Now, I have a non-stationary noise which its variance is not constant and changes periodically over time. My second question is, how can I calculate the noise power for this noise and how it is related to the time-variant variance?

Any help and remark on the topic would be appreciated.

Thank you all.


1 Answer 1


If your signal is just $n(t)$ distributed as $N(\mu, \sigma^2)$, then the expected signal power is then: $$ p_{\rm ensemble} = E\left [ \left| n(t) \right|^2 \right] = \mu^2 + \sigma^2 $$

For the time-varying case, I'd suggest that the answer is: $$ p_{\rm time\ varying} = \mu^2(t) + \sigma^2(t) $$

  • $\begingroup$ I think you are right. Thank you very much. $\endgroup$
    – Jennifer
    Commented Sep 30, 2015 at 12:20
  • $\begingroup$ One question though, why they usually put $N_0$ as the noise power for a stationary WGN? Is that $N_0$ equal to the variance if the mean is zero? $\endgroup$
    – Jennifer
    Commented Sep 30, 2015 at 12:26
  • $\begingroup$ $N_0$ is not the noise power. It is the noise spectral density. See Dilip's answer here for more details. $\endgroup$
    – Peter K.
    Commented Sep 30, 2015 at 12:48
  • $\begingroup$ Thanks. Now everything is clear. I didn't know the relationship between variance and PSD. You were very helpful, thanks again. $\endgroup$
    – Jennifer
    Commented Sep 30, 2015 at 13:04

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