This is a follow-up question to Allan deviation to determine averaging time

I was under the impression that with DC signals I couldn't capture or average a signal forever to get better SNR, because of 1/f or pink noise. I thought the Allan deviation was a good way to determine averaging time.

I have now convinced myself the Allan deviation isn't useful for DC signals like this. It seems like averaging the signal will overcome any extra noise power obtained from a longer capture window. (speaking of white and pink noise)

Here is an example. This is a mixture of white and pink noise.

enter image description here

Here is the associated power spectral density.

enter image description here

I split the signal into 1 second chunks. In each of those chunks, I varied the capture window. On one extreme, I'm taking the first sample and throwing away the rest of the chunk. On the other extreme, I'm taking every sample and averaging them together. This is repeated over many seconds, and the standard deviation taken. Here is the result.

enter image description here

This makes me think I should always make my capture windows as long as possible, and because of averaging I will always get the best noise numbers.

Is this true? Is it because finite capture is a high pass with a cutoff at 1/T, but averaging is also a low pass with cutoff at 1/T?

I repeated the plot for pure pink noise, and it looks like averaging kills the noise.

enter image description here

Thanks for the help.


1 Answer 1


A necessary detail to consider is if the signal has a true long term mean or is it in the long term a non-stationary process (does the mean drift with time)? The utility of ADEV is with non-stationary processes that tend to drift in the long term as it indicates the time duration over which such stationarity can be assumed- and thus measures the stability of the signal.

That said, where the ADEV curve flattens out and starts to turn up indicates where stationarity can no longer be assumed. We can continue to average the signal which will of course reduce the total noise in the given capture (as a low pass filter), but the resulting mean will not be meaningful (consider in general a drifting process- what is the significance of the mean when it is changing?).

In many applications, the signal we are observing (and using ADEV to characterize) is a “reference” we want to use and for this purpose we want it to be stable and not changing in the mean. That is what ADEV tells us: how accurate it can be as a reference and for how long. If the mean isn’t changing; then yes we can continue to average and improve our estimate of that mean. The success in being able to do this and over which time intervals is exactly what the ADEV curve tells us.

With regards to SNR, it may help to think in terms of noise density rather than total noise since the averaging operation keeps reducing the bandwidth, but not the SNR when considered as a ratio of signal to noise density within the passband of the filter. For many (if not most) applications, there is a minimum measurement bandwidth that our measurement system will need based on the actual bandwidth of the signal we are observing (this is the case whenever what we are observing or measuring is ultimately changing with time - as is typically the case with anything that has information content). This sets an upper limit on how much averaging we can do before we start removing information. The other consideration is how long we can capture for-- this sets the lowest observable frequency effectively as a high pass filter. It is under this consideration where the SNR for the lower frequency components of what we are observing would be increasingly reduced by the effects of 1/f noise, importantly under a consistent noise density metric (this is exactly what the power spectral density shows us as we see the noise floor increasing as we approach DC). So in this regard, ADEV is telling us how long of a capture we can make without increasing the noise density for the lower frequency components.


Your Answer

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

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