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?

• this is the kinda question that people doing watermarking ask. is that what you're doing? – robert bristow-johnson Jun 8 '16 at 4:10
• 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 8 '16 at 8:04
• @BrianP : Please merge your accounts: stackoverflow.com/help/merging-accounts – Peter K. Jun 8 '16 at 11:44

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: $$\frac{2 \sigma^4}{N-1}$$

• Thanks, this is useful. Can anything be said about the PSD of the new "variance of $x[t]$" signal? – Brian P Jun 8 '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 8 '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 8 '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 8 '16 at 18:08

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 $SNR_t = \frac{\bar{A}}{\sigma_A}$.

• Right, but I'm interested in a theoretical derivation of $\sigma_A$ to compare with such an empirical measurement. – Brian P Jun 8 '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 8 '16 at 20:17