Minimum number of possibly non-contiguous samples to distinguish two periods

Abstract

Given that a sequence x(n) is periodic with period P belonging to a known integer set &#x007B;P<inf>1</inf>, P<inf>2</inf>, &#x2026; P<inf>L</inf>&#x007D;, what is the minimum number of samples of x(n) required to find the period? For the special case where the samples of x(n) are constrained to be contiguous in time, this problem has recently been solved. More generally, when the samples are allowed to be non-contiguous, the problem is quite difficult. This paper provides the answer for the restricted situation where P &#x2208; &#x007B;P<inf>1</inf>, P<inf>2</inf>&#x007D;. With P<inf>1</inf> &lt; P<inf>2</inf>, the necessary and sufficient number of (possibly noncontiguous) samples for period estimation turns out to be (a) P<inf>1</inf>, if P<inf>1</inf> is not a divisor of P<inf>2</inf>, and (b) P<inf>2</inf> otherwise. While the proof is quite involved even in this restricted case, it is likely to form the basis for addressing the more general situation where P &#x2208; &#x007B;P<inf>1</inf>, P<inf>2</inf>, &#x2026; P<inf>L</inf>&#x007D;.

DOI: 10.1109/ICASSP.2017.7952872

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Cite this paper

@article{Tenneti2017MinimumNO, title={Minimum number of possibly non-contiguous samples to distinguish two periods}, author={Srikanth V. Tenneti and P. P. Vaidyanathan}, journal={2017 IEEE International Conference on Acoustics, Speech and Signal Processing (ICASSP)}, year={2017}, pages={3824-3828} }