Suppose I have a digital signal measured with sampling time, $T_s=1$ sec. If I take it's derivative, it will, naturally have $T_s=1$ sec. But what are the implications if I re-sample this derivative to have a lower sampling time, such as $T_s=0.1$ sec?

For example, how is the accuracy or representatives of the derivative influenced by re-sampling?

  • $\begingroup$ Isn't this discrete system you are talking about? Thus giving you a difference not derivative. Indeed a difference can be characterized as the impulse response: 1, -1. The frequency response which is (simplified) a static flanger. As the sampling rate is lowered the bandwidth will decrease as usual. $\endgroup$
    – Dole
    Feb 13, 2016 at 18:11

2 Answers 2


As you start from a discrete signal, you are likely to interpolate the derivative at a finer step. The result depends on the combination of your interpolation scheme and your discrete derivative estimation.

Your numerical derivative will probably be more convenient, perhaps smoothed, but not "accurate" with respect to what you would have obtained from a $10$ Hz signal, unless you make assumptions on your original signal.

For instance, you can fit (locally) the signal by continuous functions, differentiate them (with penalities on thier variations), and sample the result.

  • 1
    $\begingroup$ Indeed. That was my reasoning also. Thank you for answering my question in such a clear way. :) $\endgroup$
    – Aquila
    Feb 13, 2016 at 10:30
  • $\begingroup$ Two additional comments: if your interpolation and derivative schemes are both linear, interpolating the signal before differenciation will be the same as the controverse (interpolating the derivative). Sample quantization also plays a role (as quantization is not linear). $\endgroup$ Feb 13, 2016 at 11:23

As Laurent said, no amount of interpolation will restore the information that's not in the original $f_s=1\,\text{Hz}$ signal.

It's been discussed in other posts' comments what a suitable digital differentiator looks like; that's not really easy to answer universally. However, it's certainly a high pass filter. So, I'd say: Try to understand your derivate as a filter operation, if possible. Maybe this gives you a suitable approach on how to implement a "good" (by your application's needs) differentiation.

Also, as Laurent said, when you make assumptions on your original signal, you might find a suitable derivative estimator. The typical approach for DSP (and where Nyquist's theorem mentioned above comes form) is to assume that your signal is a finite sum of complex sinusoids, and hence, every digital signal is equivalently representable by its Discrete Fourier Transform. You might find a derivative according to the differentiation theorem of the Fourier transform, if you want to represent your signal as a sum of orthogonal sinusoids, but you might also represent it according to any other base – and derive a differentiation algorithm that fits your needs best.

  • $\begingroup$ Great complement. May I suggest Numerical differentiation formulae as a starting point? $\endgroup$ Feb 13, 2016 at 12:00
  • $\begingroup$ I generated some results and conclusions based on the original signal + its derivative at $T_s=1$. Then, I had a discussion with somebody that asked 'Why not have the derivative signal with a lower $T_s$?' and re-analyze the results. But that requires a lot of work and I wanted to be sure in which directions to go with my motivation.. In fact, for my case, if I want the derivative accurate enough, I should measure it. Think for instance of velocity and acceleration. I hope I make some sense. Thank you both for your help! :) $\endgroup$
    – Aquila
    Feb 13, 2016 at 12:37
  • $\begingroup$ @laurentduval: that's pretty much the slide set I was looking for without knowing I was looking for it :) $\endgroup$ Feb 14, 2016 at 13:25

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