# Can I sample at Nyquist rate if I know that my samples are taken only at the signal's maxima or minima?

I know that in general the sampling rate, $$f_s$$, must be greater than twice the highest frequency of the signal, $$f$$.

If I sample at the Nyquist rate, it can lead to the following: However, if the signal is sampled only at maxima and minima, then we get this: My signal is not pure sinusoidal, it is something like this: Will I loose any frequency components if I sample it at the Nyquist rate?

• Plot the spectrum of your signal to see firstly how bandlimited your signal is. To me it looks like samples of square wavish signal at first. Which is not band limited strictly Sep 20 '20 at 5:53
• Like DspGuySam states, you should make sure your signal is "enough" bandlimited. If it is not, you won't (just) loose frequency components, but they will be aliased! Additionally, in theory (please correct me if I am wrong), if you could use a sinc function to perform interpolation between the samples you would be able to completely reconstruct your original signal (provided it was bandlimited and you sampled after a brick-wall filter, which corresponds to the sinc function you are supposed to use for the reconstruction). Sep 20 '20 at 9:26
• @VladislavGladkikh yes, but only by assuming an interpolation function, and that inherently assumes a bandwidth. So, yes, without knowing what your original signal is, you can't preclude losing spectral components through them being indistinguishable aliased to other frequencies. Corollary: How do you know these samples are taken exactly at the minimum / maximum? Sep 20 '20 at 10:53
• well, the problem that I have here is that I don't understand how you're certain that the values you observe (which seem interestingly quantized, but that might actually be a quantum thing?) are actually maxima of a continuous and continuously diff'able function; you say "if we only sample at the maxima, can we...", but I don't see how you're actually sampling at the maxima; you might just be seeing exactly two or three different values, because your digitization process only allows for these discrete steps, telling you very little about the functions in between. Sep 20 '20 at 11:51
• But: you know these are electron densities. Do you have a mathematical model for these (I guess that might be what you're investigating)? Something like them being a superposition of Gaussians with different expectation and variance, or something like them being the squared magnitude of a some Fourier transform of a known impulse space function with a few unknown parameters? Anything as a mathematical basis that says "If you get a couple of values observed, you can make more general statements beyond only the exact observation"? Sep 20 '20 at 11:54

Thus if we sample exactly at Nyquist ($$f_s/2$$ where $$f_s$$ is the sampling rate, there is no realizable filter that will pass every signal of interest up to $$f_s/2$$ while reject everything immediately after that which would otherwise fold in. It is for this reason that we need to sample at some frequency above Nyquist and the decision is based on the analog filter design prior sampling. This "filter" may be inherent in the signal we are filtering, and our knowledge of the signal and noise content and how much we care about the higher frequency noise that would otherwise fold into band and be indistinguishable.
when you have a quantized signal (like in your case) you also have a noise floor in the frequency spectrum. This noise floor is proportional also to the sampling rate. If you sample a signal at twice its maximum frequency you can distinguish all the frequency components above the noise floor. In your case you are sampling at $$F_p = 2$$ because there are 2 samples every second therefore you are assuming your signal has a maximum period of 1. There is one case when you are interested on sampling at the same frequency of the signal. For example if you know that your signal has an unwanted ripple at the frequency $$f_{ripple}$$ you can sample at this frequency $$F_p = f_{ripple}$$ preserving the lower frequencies while the ripple is transformed into a constant error which basically depends on the ripple waveform.