I was calculating the Allan deviation (utilizing the Python module allantools.adev, which relies on eq. (6) in https://tsapps.nist.gov/publication/get_pdf.cfm?pub_id=50505). The expected behavior is a linear decrease as the system exhibits white frequency noise:

enter image description here

This expected linear decrease can be observed up to an averaging time of about 0.1 s. However, for longer averaging times the Allan Deviation starts to decay very rapidly. Has anyone observed such behavior and do you have an idea of how to interpret this/where it originates from?

I would be very grateful for your help and insights! :)

  • $\begingroup$ How large is your sample set and what is your sample rate ? $\endgroup$
    – Hilmar
    Commented Feb 7, 2022 at 19:22
  • $\begingroup$ I have 1400 samples with a sampling rate of 660 Hz. $\endgroup$ Commented Feb 7, 2022 at 20:09
  • $\begingroup$ Then you probably just run out of samples. You have about two seconds of data. If you try to average this over a 1/s of second, you are left with only 6 samples or so. $\endgroup$
    – Hilmar
    Commented Feb 8, 2022 at 14:35
  • $\begingroup$ Yeah that makes sense, but shouldn't the Allan Deviation still produce reasonable results (as the equation defining the Allan Deviation still works also for a few samples, even if there are only 2 samples left) or is there a problem at some point if the number of samples becomes too small? $\endgroup$ Commented Feb 8, 2022 at 17:28
  • $\begingroup$ Consider the simpler case of the computation of the mean and standard deviation from experimental data: it is an estimate of the true standard deviation of the underlying random process. The more samples we have, the better the estimate is. Confidence Intervals (en.wikipedia.org/wiki/Confidence_interval) tell us how likely the estimate is within a given range of the true underlying figure. With fewer samples the confidence interval increases significantly. $\endgroup$ Commented Feb 11, 2022 at 17:27

1 Answer 1


It is true that a white-FM process (or whatever units we are actually using for the waveform that allantools is expecting to be fractional frequency versus time) would go down at a rate of 1/root-tau consistent with averaging a white noise process over N samples with the standard deviation going down at the square-root of N. Note that the samples to the far right on the Allan Deviation have the biggest error confidence bar intervals since they have the lowest number of samples in their own solution, so can be least trusted in their own accuracy. However with sufficient samples we can still get this behaviour where the ADEV goes down faster than rate 1/root-tau when the spectrum of the noise process itself is not white and specifically the noise closer to the carrier is lower than the noise further away. Examples where this can occur would be the noise spectrum out of sigma-delta converters where the upper portion of the spectrum past the noise shaping corner can indeed be white, but below the noise shaping corner the noise rolls off rapidly. For that we would see a behavior consistent with the OP's plot of having noise going down at 1/root-tau up to a certain tau, and then dropping off faster than that consistent with the noise shaping slope.

Below is an example of such a sigma-delta spectrum where we could see this in the ADEV.

Sigma Delta Spectrum


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.