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Consider the following contrived situation. Imagine a Gaussian white noise process $x[t]$, with bandwidth $Δf$, with PSD equal to some quantity $A$ which you would like to measure.

So the way to measure this seems to measure the variance of the process $x[t]$, which by Parseval's theorem will be $AΔf$.

So you measure points with some frequency $f_m$, probably $2Δf$. And at a lower rate, say $f_v$, you compute the variance of the preceding block of $f_m/f_v$ points and take that to be a measurement. What will the noise/variance be in this measurement of the variance of $x[t]$? How can I approach this question?

Brian P
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  • this is the kinda question that people doing watermarking ask. is that what you're doing? – robert bristow-johnson Jun 08 '16 at 04:10
  • Robert, I posted this as a guest but have lost my cookie and appear to have already had an account; this account does not have enough reputation to comment and I do not know a way to get the guest cookie back. This is actually an infrared detector application. – Brian P Jun 08 '16 at 08:04
  • @BrianP : Please merge your accounts: http://stackoverflow.com/help/merging-accounts – Peter K. Jun 08 '16 at 11:44

2 Answers2

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There is a derivation over on the math.SE site that might be useful.

Provided your data is Gaussian with variance $\sigma^2$.

If you use the unbiased estimator of the population variance: $$ S^2_{N-1} = \frac{1}{N-1} \sum_{n=0}^{N-1} (x[n]- \bar{x})^2 $$ where $N$ is the number of points in your batch and $\bar{x}$ is the sample mean, then the variance of this estimate will be: $$ \frac{2 \sigma^4}{N-1} $$

Community
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Peter K.
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  • Thanks, this is useful. Can anything be said about the PSD of the new "variance of $x[t]$" signal? – Brian P Jun 08 '16 at 17:48
  • Provided there are no overlapping entries in each calculation of the variance estimate, I expect all the estimates will be uncorrelated from each other... so the sequence of variance estimates will be white, so the PSD should be a constant. – Peter K. Jun 08 '16 at 17:50
  • Thanks. And, one more question: suppose the noise were not white or $x[t]$ were not Gaussian. Am I correct that this would still approximately hold for $f_m/f_v \gg 1$, by the Central Limit Theorem? – Brian P Jun 08 '16 at 17:56
  • Well, the original variance estimate would need to be revised, as it assumes whiteness. And then you'd have to see how that coloration carried through. – Peter K. Jun 08 '16 at 18:08
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Is SNR really the quantity that you desire? In your situation, I would assume you are more interested in how accurately you can measure $A$. Why don't you simply take, let's say, 100 measurements, and compute the mean and standard deviation of all $A_i$ from your this data? If you need to compare different measurement strategies, you could then use a "temporal SNR" given by the mean and standard deviation of $A$ as $SNR_t = \frac{\bar{A}}{\sigma_A}$.

M529
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  • Right, but I'm interested in a theoretical derivation of $\sigma_A$ to compare with such an empirical measurement. – Brian P Jun 08 '16 at 17:51
  • I understand. If this is just an empirical measure... no, I wouldn't say that. Compare it to medical imaging (specifically MRI): The SNR derivation there from ROIs within signal and noise areas is just an easy substitute for doing the measurement a few hundred times and calculating the SNR pixel-wise by this formula. It is also known in fMRI as *temporal SNR* (however, imho this is a bit doubtful there). Maybe you can find an appropriate theoretical framework with this *temporal* approach. If your measurement itself does not exhibit any noise, it might be hard to link it to common formula. – M529 Jun 08 '16 at 20:17