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I am trying to understand how the Nyquist-Shannon theorem applies to sampling in the time domain. Suppose I want to sample a function whose time constants I know. From what I understand, the bandwidth is determined by the shortest time constant. After that, I'm shaky. Wikipedia appears to suggest that the Nyquist rate is twice the bandwidth as I defined it previously, and that the Nyquist frequency should be between that and the bandwidth (https://en.wikipedia.org/wiki/Nyquist_frequency). On the other hand, a signal processing book that I consulted (Oppenheim and Schafer) suggests that twice the bandwidth is the Nyquist rate, and that the Nyquist frequency should be between that and the bandwidth. And it also shows the sampling frequency as being $2\pi$ times the bandwidth (I think) for reasons that I don't understand.

Based on the latter source, I would guess that my sampling rate in the time domain must fall within that interval in order not to produce artifacts in reproducing the function that I mentioned. From what I understand, aliasing is one such artifact.

Can somebody help clear this up for me? This is very muddy in my head. (My background is in an area of science that doesn't involve signal processing as part of the education, and I am trying to understand this because it seems quite fundamental.)

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This feels a bit like duplicate of your previous question How to sample a signal in time based on a set of relaxation times?

The answer is still the same: you cannot sample a first order decay process (with a single time constant) without aliasing.

The sampling criteria requires the signal to be bandlimited. That means the spectrum must be 0 above the Nyquist frequency. Bandwidth for a process with a time constant is typically the -3dB point which is something different.

You should sample at a rate substantially higher than the -3dB point. How much higher depends on how much aliasing your application can live with.

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  • $\begingroup$ Okay, but why the $2\pi$ in determining the Nyquist rate? For example, in the link posted above by @MBaz the sampling frequency is supposed to be at least twice the bandwidth, but I don't see $2\pi$. Also, if you sample too much, is that where aliasing comes in? $\endgroup$
    – user62149
    Commented Apr 7, 2022 at 2:40
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    $\begingroup$ The only place where $2\pi$ comes in is when you use a time constant to determine the $-3dB$ frequency $f_c$ . For a first order process we simply have $\omega_c = 1/\tau$ hence $f_c= 1/(2\pi\tau)$. This only loosely related to the Nyquist frequency. $\endgroup$
    – Hilmar
    Commented Apr 7, 2022 at 12:27
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    $\begingroup$ You cannot sample "too much". Aliasing happens only if you sample at too low a rate. $\endgroup$
    – Hilmar
    Commented Apr 7, 2022 at 12:28

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