Typically the Nyquist Shannon theorem is stated to say that you need to sample at twice the maximum frequency for complete reproducibility. Applying this to a digital signal, that does indeed work with a discrete Fourier transform. Applying this in the perspective of a sample rate (i.e. *.wav file of a recorded audio signal) does not hold true though in my opinion as the discrete signal has become a scalar, not a vector of the Fourier example.
If we take the highest reproducible frequency, two samples per wave, we find we could perfectly sample at the highest and lowest values of that wave, but we could have equally sampled the zero-crossing point, all depending on where in the phase the sample rate aligns with a given wave form. As the sampling has lost significant information, I believe a sample rate should be much higher than what Nyquist-Shannon would suggest for a high degree of reproducibility.
If your source is a digital signal, and you only need to reproduce that signal, of course 2x is ample.
Actually the Theorem states, that you must sample with a rate strictly GREATER THAN (not equal) than twice the highest frequency. Exactly because sampling only the zero crossings is not enough.
Ofcourse depending on how exactly you want to process your samples it might be convenient to have an even higher sampling rate. And if you know your signal does not contain low frequencies (=not using the full bandwidth) you might get away with even lower sampling rates.
But the general case is: you must sample with a rate strictly greater than twice the highest frequences.
Comments
Typically the Nyquist Shannon theorem is stated to say that you need to sample at twice the maximum frequency for complete reproducibility. Applying this to a digital signal, that does indeed work with a discrete Fourier transform. Applying this in the perspective of a sample rate (i.e. *.wav file of a recorded audio signal) does not hold true though in my opinion as the discrete signal has become a scalar, not a vector of the Fourier example.
If we take the highest reproducible frequency, two samples per wave, we find we could perfectly sample at the highest and lowest values of that wave, but we could have equally sampled the zero-crossing point, all depending on where in the phase the sample rate aligns with a given wave form. As the sampling has lost significant information, I believe a sample rate should be much higher than what Nyquist-Shannon would suggest for a high degree of reproducibility.
If your source is a digital signal, and you only need to reproduce that signal, of course 2x is ample.
Actually the Theorem states, that you must sample with a rate strictly GREATER THAN (not equal) than twice the highest frequency. Exactly because sampling only the zero crossings is not enough.
Ofcourse depending on how exactly you want to process your samples it might be convenient to have an even higher sampling rate. And if you know your signal does not contain low frequencies (=not using the full bandwidth) you might get away with even lower sampling rates.
But the general case is: you must sample with a rate strictly greater than twice the highest frequences.