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We have a system $A$, and I measure that system's state, $s(t)$ using a very accurate probing tool that captures zero noise. The state can be whatever signal, like the voltage, current, etc. Assume that the highest frequency of $s(t)$ is 5K, and according to Nyquist's theory, I need at least 10K sampling frequency to recover the signal. I am trying to connect this theory with the shortest amount of time, $I$, where I see no change in $s(t)$ during $I$. For instance, if I magically have a sampling frequency of 5M, I would see no changes in every 5 consecutive points. So the sampled signal would be like this: $\dots 1,1,1,1,1,5,5,5,5,5,2,2,2,2,2,\dots$. If I have a sampling frequency of 1M, the signal would like $\dots 1,5,2,\dots$. So the redundancy has getting rid of.

At first glance, I thought Nyquist's theorem says that if you sample at least double the highest frequency of a signal, you are faster than the dynamics of the signal, so increasing the sampling rate higher than that (i.e., double the highest frequency) will give no more information, and you will end up seeing the same sample more than once. However, I think this is not true in practice (maybe because of the noise?)

Is there a theorem that connects the Nyquist rate with finding the interval $I$?

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  • $\begingroup$ Let $s(t) = \cos \left ( 2 \pi 1000 t \right)$, so it's perfectly bandlimited to well within $5 \mathrm{kHz}$. What is the shortest interval $I$ such that for any $\Delta t \le I$, $s(t)$ always shows no change? Figure that out, and you have your theorem -- or you've disproved your hypothesis. $\endgroup$
    – TimWescott
    Commented Dec 27, 2022 at 16:08

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so increasing the sampling rate higher than that will give no more information,

Correct

and you will end up seeing the same sample more than once.

Incorrect. "no more information" means that the additional samples can be calculated from the original ones (using sinc interpolation), NOT that samples get repeated.

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