intensely_human , 1 month ago

The Norquist-Shannon rate sampling theorem only asserts that for a given maximum frequency, you only need another other given maximum frequency of sampling to represent it.

It does not say you can “perfectly” reproduce a signal. Only that you can reproduce all fourier components of the signal that are below half your sampling rate in frequency. It perfectly does that, yes.

But the signals that only contain a finite number of frequencies all below a certain maximum frequency are abstractions used in signal theory classes for teaching that theorem, and in engineering to hit a “good enough” target, not a “perfect” target.

Any frequencies bouncing around the room at over 22 kHz are lost at least to something using the 44 kHz sampling format.

TL;DR: Norquist-Shannon lets you completely reproduce signals with finite information in them. But real life sound doesn’t have finite information in it.