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I usually do the DFT using the fft in Matlab. After the Nyquist frequency I don't see any result. Is it possible to perform a dft looking after the Nyquist frequency. I am asking this because I have seen a paper where an author show the result of a spectrum after the Nyquist frequency

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  • $\begingroup$ Do you really mean to say Nyquist Frequency (which is half the sampling rate) or Nyquist Rate (which is the sampling rate)? $\endgroup$ Jun 9 at 10:57
  • $\begingroup$ I meant Nyquist Frequency. Yes, So I guess, that I was right and that there is something that should be investigated in the paper $\endgroup$ Jun 9 at 12:15
  • $\begingroup$ @LucaMirtanini It would help us answer if you can give more context. Perhaps a link to or screenshot of the paper in question? $\endgroup$
    – Peter K.
    Jun 9 at 12:38
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Is it possible to perform a dft looking after the Nyquist frequency.

Of course. Every DFT will do this. Here is why: Applying a DFT requires the signal to be time discrete, which means that it is periodic in the frequency domain with the sample rate being the period.

Let's say you are sample rate is 40 kHz. Then the value of the DFT at 1kHz will be the same 41kHz, 81 kHz, 121 kHz, -39kHz, -79kHz, etc. A single DFT will produce the values for ALL frequencies, not just the ones below Nyquist. It's just a matter if which period you want to look at.

It's easy enough to proof

$$X(k) = \sum_{n=0}^{N-1} e^{-j2\pi\frac{nk}{N}}$$

$$X(k+N) = \sum_{n=0}^{N-1} e^{-j2\pi\frac{n\cdot (k+N)}{N}}=\sum_{n=0}^{N-1} e^{-j2\pi\frac{n\cdot k}{N}} \cdot e^{-j2\pi\frac{n\cdot N}{N}}= X(k)$$

since

$$e^{-j2\pi\frac{n\cdot N}{N}} = e^{-j2\pi \cdot n} = 1$$

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  • $\begingroup$ Moreover, if you're taking the DFT of a real-valued signal, the value of the DFT at -1kHz will be the complex conjugate of the 1kHz value, and it will repeat at -41kHz, +39kHz, etc. $\endgroup$
    – TimWescott
    Jun 9 at 16:07
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At Nyquist the signal goes $[1, -1, 1, -1,...]$ - it's the fastest possible discrete variation for any input length. Zero padding won't help: it'll only lower the lowest possible (non-zero) frequency.

Going beyond Nyquist thus necessarily implies increasing the physical sampling rate, or "rate of information", such that the same discrete variation $[1, -1, ...]$ now represents a higher physical frequency.

A workaround is "imputation" or interpolation, but FFT interpolation here is meaningless, uniquely for Nyquist and dc bins.

Note: I understand the question as going beyond Nyquist without aliasing, i.e. attaining actually greater physical frequencies. That aside it's trivial to do, the DFT at Nyquist + 1 equals DFT at -(Nyquist - 1), and so on - "DFT periodicity". (and if input is real, spectrum's magnitude at Nyquist + 1 equals that at Nyquist - 1)

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  • $\begingroup$ @DanBoschen Thanks again. $\endgroup$ Jun 14 at 12:10
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In a digitized signal, the minimum time gap between two samples would be 1/(sampling frequency). Hence the maximum frequency component can't exceed sampling frequency (which can be Nyquist rate or greater than that)

For a real signal, the maximum frequency component with unique information will not exceed $\frac{sampling frequency}{2}$ $ $ ( because of symmetry in the spectrum about Nyquist frequency)

Nyquist rate is double of max. frequency component in a frequency-limited signal.

Nyquist frequency is half of the sampling rate (which can be equal or greater than the Nyquist rate)

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    $\begingroup$ That's technically not correct. The DFT is defined for all frequencies and that's often used in bandpass sampling. The range of frequency cannot be larger than the sample rate, but you do not need to include 0 Hz in your range but can shift around as you like $\endgroup$
    – Hilmar
    Jun 9 at 13:50
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If the OP's reference is showing a unique spectrum after the Nyquist frequency corresponding to $f_s/2$ where $f_s$ is the sampling rate (and perhaps this was the motivation for the question?), then this would imply the time domain waveform is complex as in that case the entire spectrum spanning over the width of the sampling rate (in any frequency interval) would map uniquely to the DFT bins. (Hence the need for an anti-aliasing filter to select such an interval prior to sampling an analog signal otherwise we wouldn't know which range of frequencies we are really looking at, or if many are mapping over).

For real signals, the frequency spectrum is complex conjugate symmetric (hence we can ignore half the DFT bins in this case as the information is redundant).

But for complex signals, there is no relationship between the positive and negative spectrum, so the spectrum is unique.

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