|
|
|
|
|
|
by GlenTheMachine
581 days ago
|
|
|
If you look at the Wikipedia entry on the Nyquist Sampling Theorem, you should note that the summations to reconstruct the original signal go from negative infinity to positive infinity. In other words, that sum requires an infinite number of samples. There are many signals of practical interest that can be approximately reconstructed with a finite truncation of the series. Note, however, that any signal that has only a finite length, eg has a uniformly zero amplitude after some time t_final, does not have a finite bandwidth, and cannot be exactly reconstructed by any sampling scheme. This is the case whenever you stop sampling a signal, eg it is always the case whenever you step outside the mathematical abstraction and start running real code on a real computer. So any signal reconstructed from samples is always approximate, except for some relatively trivial special cases. |
|
|
What if you take a function that has bounded support in the time domain, and then turn it into a periodic function? Might the resulting function have bounded support in the frequency domain even though the original function did not? I suppose doing this would force the Fourier transform to have discrete support? But under what conditions would it have bounded support?…
I guess technically a low-pass filter applied to a signal with finite support in the time domain, would result in a function which has infinite support in the time domain.
I suppose sinc(f t + c) doesn’t have bounded support, and it is unsurprising that a non-trivial linear combination of finitely many terms of this form would also not have finite support.
Still, such a linear combination could decay rather quickly, I imagine. (Idk if asymptotically faster than (1/t) , but (1/(f t)) is still pretty fast I think, for large f.)
Soon enough the decay should be enough that the amplitude should be smaller than the smallest that the speaker hardware is capable of producing, I suppose.