I measure noise in different systems by looking at voltage fluctuations. I wanted to use a Keithley SMU for that purpose, by storing the data in the buffer and then reading them out. However, I found that the data points are not measured at equally spaced points. While the difference is small, it is still visible. For example, the time intervals for a given trial where: 3.0 ms, 3.01 ms, 2.99 ms, etc, instead of a fixed 3 ms time interval.

I am wondering whether I can still overlook this error or not. If not, what is the main reason? Also, is there anything else I can do?


2 Answers 2


You still can do a lot. There is no perfect sampler, jitter always exists. Moreover, traditional systems include quantization which prevents (theoretically) the satisfaction of Shannon-Nyquist-Kotelnikov-Raabe... conditions. If your system includes low-pass filtering, or/and the phenomenon you look at is very weak above 166 Hz, the variations are likely to be small enough for your purposes. I would propose to look at an harmonic behavior with the following three approaches:

  1. Suppose the periods are even, then perform Fourier analysis,
  2. Resample in a simple fashion at even samples (linear interpolation), then perform Fourier analysis,
  3. Use a Fourier tool adapted to irregular sampling, like the Lomb-Scargle periodogram.

If you detect no meaningful difference in the features you extract, then you are fine (for the time being). You even can, routinely, use these methods in parallel to detect a weird behavior when they differ.

If you see meaningful difference then you might want to invest into irregular sampling processing methods. Luckily, in theory, when you have irregular samples, and a sufficient precision in the timing, you can extract information beyond Nyquist, at the cost of more involved methods. There is a handful of methods for Fourier or time-frequency methods, FIR results, etc.

  • $\begingroup$ I have a question. Usually I do oversampling for filter design reasons, and then downsample (decimate) the signal again. This is simply done by taking every nth sample in case of uniform sampling. However, with non-uniform sampling, it does not seem to be a simple way...what do you think? $\endgroup$
    – student1
    Commented Sep 2, 2015 at 1:47

A couple of intuitions. Small sample jitter over most of the band is equivalent to additive noise, (maybe) correlated to the signal. For non-ideal sampling, where out of band energy is non-zero, jitter has the beneficial effect of randomizing the sampling with the phase of repetitive signals, so rather than classical aliasing, the out of bands signals can appear as random, wideband noise. So as long as your jitter is small, and your signal energy is primarily well away from Nyquist, you can treat the sampling as ideal, with additive white noise.

Things get more tricky if the jitter is cyclic, nonzero mean, or otherwise oddly distributed.

Ref Introduction to Shannon Sampling and Interpolation Theory, Robert Marks


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.