One of the earliest extensions of this theorem was stated by Shannon himself in his 1949 paper, which says that if $x(t)$ and its first $(M - 1)$ derivatives are available, then uniformly spaced samples of these, taken at the reduced rate of $M$ times, are sufficient to reconstruct $x(t)$.

This result will be referred to as the derivative sampling theorem in this paper. I'm working with $M=3$ and having trouble designing a synthesis filter with an impulse response of the following form:


Can you help?

  • $\begingroup$ I have no idea why you want those strange-looking impulse responses. But if you have them already defined in the time domain, you're nearly done. You do have to window them, with a decent window (like Kaiser), to a finite length and to sample them with a high sample rate and make tables of coefficients outa that. $\endgroup$ Commented Apr 18 at 17:10
  • $\begingroup$ My problem is applying these 3 filters in speech signal processing. The input is a speech signal, which is then sampled including the signal, the first derivative of the signal, and the second derivative of the signal. Can using the above 3 filters restore the original signal? $\endgroup$ Commented Apr 21 at 17:46
  • $\begingroup$ I dunno. Appears to me that $$ t^2 \mathrm{sinc}^3(t) $$ is the inverse Fourier Transform of a the second-derivative of the piecewise quadratic function that is continuous to the second-derivative. $\endgroup$ Commented Apr 22 at 3:48

1 Answer 1


You may find the paper Filterbank reconstruction of bandlimited signals from nonuniform and generalized samples by Y. Eldar and Al Oppenheim, which discusses some approach to synthesize the filters, useful.

  • $\begingroup$ Thank you very much. I have a question that you can answer for me: multichannel data acquisition (my problem is 3 channels including: signal sample, first derivative sample of the signal, second derivative sample of the signal) is considered uniform or non-uniform sampling? $\endgroup$ Commented Apr 20 at 9:41
  • $\begingroup$ Being uniform or non-uniform depends on how the sampling points are distributed. Let's say that the sample instances are $t_{n-1},t_{n},t_{n+1}$. If $t_{n}-t_{n-1} = T_{s} = \text{const},~\forall n$, then the function (signal) is uniformly sampled. Otherwise, the resampling is done in a non-uniform fashion. Recall that reconstruction from the derivatives allows to resample the signal at sub-Nyquist. In the problem at your hand, you can reconstruct a signal sampled at rate $\Omega_{N}/3$. $\endgroup$
    – AHT
    Commented Apr 21 at 15:05
  • $\begingroup$ You're right. I read the paper but didn't understand much. Can you explain more clearly to me? If possible, to make it easier to understand, can you simulate it on a program like matlab? $\endgroup$ Commented Apr 22 at 8:25

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