# Misunderstanding of Nyquist sampling theorem and minimum sampling rate

I sample my time-domain (TD) signal using a distance between time-samples of $$\delta = (t_{max} - t_{min}) / N_t$$, where $$N_t$$ is the number of samples taken. The sampling rate is $$1 / \delta$$. I have read 10's of lectures about DFT, but it seems I still don't understand the basics.

I know the sampling theorem: if the signal is bandwidth limited to $$f_c$$, that is, its FT is 0 for any $$f$$ NOT in the interval $$[- f_c, ..., f_c]$$, the signal can be perfectly reconstructed from its discrete samples taken at a rate $$\delta = 1 / (2 * f_c)$$, or higher.

I also know the Nyquist criterion: I need to sample my time domain signal at a rate which is at least twice the value of the maximum frequency present in the signal. That would be $$1/\delta > 2 * f_{max}$$.

According to my understanding of the accepted answer from here: Discrete inverse Fourier transform, it seems I need to sample at a rate $$1/\delta > f_{max}$$, where $$f_{max}$$ is the highest frequency present in the signal.

What I am missing, the above two facts seem different for me, by a factor of 2.

Starting from page 2 of these notes, it seems that the condition from the DSP answer above is in agreement with both TD->FD sampling theorem and also with FD->TD sampling theorem. https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-341-discrete-time-signal-processing-fall-2005/lecture-notes/lec15.pdf. The second picture of the diagram shown on page 2 makes sense to me visually: in order to avoid folding back to the 1st Nyquist zone, given that the length of the base of any of the triangles is $$\Omega_0$$, then $$\Omega_0 / 2$$ (RHS of the center triangle) + $$\Omega_0 / 2$$ (LHS of the 1st triangle to the right of the center triangle) < $$2\pi / (\Delta T)$$ (the center of the 1st triangle to the right of the center triangle).

My question is probably more general:

When designing the input 1D vector of length $$N_t$$ for a FFT algorithm for computing a direct discrete FT, what shall I look for, if we consider that I have infinite computational power (hypothetically)? Do I just need to make sure that the sampling rate is twice the bandwidth of the signal, where the bandwidth is $$f_{max} - f_{min}$$? I.e. only to respect the Nyquist criterion?

If the signal is complex, then $$1/\delta > f_{max}$$ with consideration to the bins of the Discrete Fourier Transform (DFT) extending from $$k=0 \ldots N-1$$ since the Discrete Fourier Transform will be unique in that case from bin $$0$$ to bin $$N-1$$. If the signal is real, then the DFT will be Hermitian symmetric, so we may only consider bins $$0$$ to $$(N-1)/2$$ since bins $$(N-1)/2$$ to $$N-1$$ will be the complex conjugate of bins$$1$$ to $$(N-1/2)$$. For example, with $$N$$ odd and a real signal, $$k_1 = k^*_{N-1}$$, $$k_2 = k^*_{N-2}$$, etc. This may lead one to state that $$f_{max}$$ is less than $$(N-1)/2$$ for that case.

In general, the Nyquist sampling theorem is to sample complex baseband signals extending from $$-B/2$$ to $$+B/2$$ at a sampling rate greater than $$B$$. When you consider the DFT bins can equally represent signals from DC to (nearly) the sampling rate $$f_s$$ (since for $$N$$ bins for $$k$$ from $$0$$ to $$N-1$$, bin $$N$$ would represent the sampling rate $$f_s$$) or equivalently (nearly) $$-f_s/2$$ to $$+f_s/2$$ the above is not in conflict with the Nyquist sampling theorem.

• Thank you. One thing, I am sorry if it's correct, but shouldn't it be "complex signals extending from -B/2 to B/2 at a sampling rate greater than B?" and not 2B? So that it corresponds to what you said in the first paragraph about the complex signals? Nov 26, 2021 at 16:22
• Yes! Thank you. I think you got it. Nov 26, 2021 at 16:27
• Your first paragraph leads to the conclusion that: Having $\Delta = (t_{max} - t_{min}) / N$, for complex signals I need that $1/\Delta > f_{max}$ and for real signals I need that $(1/\Delta) + (1/(t_{max} - t_{min})) > 2 \times f_{max}$ ? Nov 29, 2021 at 10:23
• Also, having defined $\Delta = (t_{max} - t_{min}) / N$, I believe the last sentence of your 1st paragraph shall read: "This may lead one to state that $f_{max}$ is less than $(N-1) / (2 \times (t_{max} - t_{min}))$". Right? Nov 29, 2021 at 10:24

Note that your statement of the Nyquist sampling theorem only works for infinite length signals. If N is finite, then you need to sample above twice the highest spectrum frequency present by some amount.

That's because any finite length window has infinite support in the frequency domain. Same with any finite length reconstruction or anti-aliasing filter. Thus, the windowing cause by a finite length of sampling will spread your signal's spectrum. Beyond the edges of your supposed band-limit.

So you have to sample at a rate at or above 2X the frequency where the window's stopband energy goes below your desired noise floor. Which will never be zero even with "perfect" reconstruction for a finite length of samples (unless assumed to be perfectly periodically extended to beyond the lifetime of the known universe.)

• Hello, isn't this phrase: "you need to sample above twice the highest spectrum frequency present" in contradiction with: "So you have to sample at a rate at or above the frequency where the window's stopband energy goes below your desired noise floor" ? Am I missing something? Is it twice (2x) the highest spectrum frequency or 1x the highest spectrum frequency? Thanks Nov 29, 2021 at 10:12
• more than 2x the highest spectrum frequency Nov 29, 2021 at 18:50