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Nyquist–Shannon sampling theorem

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Alexandru Miculescu
Alexandru Miculescu on 22 Mar 2016
Answered: Star Strider on 22 Mar 2016
Hi. Can anyone make me understand how a signal can be reconstructed according to Nyquist–Shannon sampling theorem? How on earth can you reconstruct a signal just by sampling 2 times faster then the process itself? In the picture below I have sampled a 1hz sinewave at 4hz and it looks really really bad.

Answers (1)

Star Strider
Star Strider on 22 Mar 2016
No picture.
I can’t ‘make’ you understand it, but yours is a legitimate question and important to understanding the fundamentals of discrete signal processing.
The sampling frequency determines the highest resolvable frequency in a sampled signal. That is the Nyquist frequency, defined as half the sampling frequency.
If you sample at the frequency of the sine, you get a straight line, because you are sampling at the same point in the cycle over as many cycles as you want. If you sample between that frequency and twice the frequency, you get a slowly-varying sine that will appear in the sampled signal as a low-frequency signal. This is called ‘aliasing’, because it is a frequency that does not actually exist but ‘pretends’ to be a low-frequency component that appears to exist. (This is the reason that analog-digital converters have hardware low-pass filters — usually Bessel filters — that eliminate all frequencies above the Nyquist frequency before the signal is sampled.) If you sample at twice the sampling frequency, you will be able to uniquely resolve the sine, and if you sample more frequently, you will improve the frequency resolution of your sampled signal.
I’m not certain this resolves all of your uncertainties, but it’s the best I can do.

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