Nyquist frequency - Sampling theory

Question

If a signal contains frequencies from $500$ to $1000\textrm{ Hz}$, what is the Nyquist sampling frequency, $2$ times $500$ or $2$ times $1000$?

Twice the highest frequency of the signal, or twice the bandwidth of the signal?

Answer 1

Both. For baseband sampling (thus using only low-pass filtering), a sampling rate over over 2x the highest frequency is necessary. For band-pass under-sampling, sampling at a bit over twice the bandwidth will work, as long as you remember the original band and don't fold (alias) the given band onto itself. 1000 Hz sampling will fold and invert a (slightly narrower than) 500 to 1000 Hz signal down to 500 to 0 Hz, and thus the signal won't be aliased with itself, but with an empty frequency band (after band-pass filtering stuff out, if needed). Thus, no information will be lost (aliased), as long as you remember the original frequency band (e.g. it was not baseband). Reconstruction may require a band-pass anti-aliasing filter, rather than a low-pass anti-aliasing filter. Using knowledge of the original band, you can de-invert and upconvert the samples back to the original lower side-band of a 1000 Hz carrier, for example.

Answer 2

The "Nyquist rate" is often defined as:

the minimum rate at which a signal can be sampled without introducing errors

The most conservative option is twice the maximum frequency. A cleverer option is to take advantage of the actual bandwidth or the signal. This resorts to generalized sampling theorems, by Papoulis and Gerchberg, notably. And using the most recent findings, you can go below, using the fact that inside $[500,1000]$ there may be holes.

My take is to go for $1000$ $\text{Hz}$, for rhetorical reasons, as the question is about Nyquist. Would the question have been about "the minimum sampling frequency", I would have said "something" below $500$ $\text{Hz}$, since we do not have enough information.

In the special case of $M$-band multi-band wavelets, or multi-rate filter banks, the context is more specific. $M$ filters, quite often FIR, are designed to allow perfect reconstruction, even if they are not ideal. So each channel, that span $1/M$th of the frequency range, can be subsampled at an additional $1/M$ rate. In the $2$-band case, subsampling both the $[0,500]$ and the $[500,1000]$ bands at the same rate is perfectly valid, as long as the low- and high-pass filters are designed correctly. However, note that after filtering and downsampling, there still exists aliasing in the subbands (esp. due to FIR)), but it is cancelled at the synthesis.