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?


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.

| improve this answer | |
  • $\begingroup$ 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 Sep 2 '15 at 1:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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