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We record a signal between times $t_0 < t < t_0 + T$. We sample our recording $N$ times in this time interval, with each reading taken time $\delta t$ apart (so $N\delta t = T$). The DFT tells us that

$$X_k = \Sigma^{N-1}_{n=0}x_n\exp\left(-\frac{i2\pi kn}{N}\right),$$ $$x_n = \Sigma^{N-1}_{k=0}X_k\exp\left(\frac{i2\pi kn}{N}\right),$$

where $x_n$ is the recording at time n$\delta t$ and $X_k$ is the Fourier coefficient of the $k^{th}$ frequency component.

From these equations it is evident that the $k^{th}$ angular frequency in the Fourier series is $2\pi k/N\delta t$, and the $k^{th}$ (non-angular) frequency is $k/N\delta t$.

This means that the maximum frequency is $\nu_{max} = (N-1)/N\delta t$.

According to the Nyquist-Shannon sampling theorem, the maximum frequency should be half of the sampling rate, so $\nu_{nyquist} = 1/2{\delta t}$.

But $1-1/N \geq 1/2$ as long as we have more than one measurement, therefore $\nu_{max}\geq\nu_{nyquist}$, which appears to violate the Nyquist-Shannon theorem. According to the theorem, there shouldn't be a term in the Fourier decomposition which corresponds to a frequency higher than the Nyquist frequency. How can this contradiction be reconciled?

Pancake_Senpai
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  • The reconciliation is that the frequencies are taken modulo N so that $X_{-k} = X_{N-k}$. That is, the other half of frequencies are negative. – Somos Dec 16 '21 at 12:41
  • Could you please write a more detailed answer or provide a reference? All $k$ values in the summation are less than $N$, so interpreting them as modulo $N$ doesn't change anything. Are we simply attempting to make $k$ symmetric about the origin? – Pancake_Senpai Dec 16 '21 at 14:12
  • I've done some working and it's easy to show that if $N$ is odd, there are an even number of non-zero $k$ values and if we make the higher half of them negative then the Nyquist theorem is obeyed. However, in the case where $N$ is even, the $k$ values have to be distributed asymmetrically about the origin and the highest frequency in the Fourier series is then exactly the Nyquist frequency, which still violates the theorem since $\nu_{max}$ should strictly be less than $\nu_{nyquist}$. What are your thoughts on that? – Pancake_Senpai Dec 16 '21 at 14:15
  • For even $N$ then $\nu_{max} = \nu_{nyquist}$ but this is allowed. Read Wikipedia article about Shannon's version. – Somos Dec 16 '21 at 15:28
  • Could you provide a link? I can't find what you're referring to. – Pancake_Senpai Dec 16 '21 at 15:50
  • With regards to negative frequencies, I've looked at this a lot more and I can see now that making half the frequencies negative is equivalent to using the full set of positive ones and doesn't change the summation at all. However, if both are equivalent, why do we consider the set of positive and negative frequencies as correct and not the full set of positive frequencies? In other words, why do we take them modulo $N$? What's the reasoning? – Pancake_Senpai Dec 16 '21 at 15:52
  • Search for "Nyquist-Shannon sampling theorem" Wikipedia article. In the introduction, the 2nd sentence begins "Shannon's version of the theorem states:" – Somos Dec 16 '21 at 15:52
  • Although later in the article "Critical frequency" section, it states "That sort of ambiguity is the reason for the strict inequality of the sampling theorem's condition." – Somos Dec 16 '21 at 16:03

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