I have a question regarding discrete inverse Fourier transform, and no answer I found on the internet seem to be satisfying. This might be because I do not fully get some of them, so please excuse my ignorance.

Suppose you have full knowledge of a frequency-domain function $\hat{f}(\omega)$. It is a continuous, well-behaved bump function. You want to compute its inverse Fourier transform $f(t) = \Re \int_{-\infty}^{+\infty} \hat{f}(\omega) e^{j\omega t} d\omega$, but this cannot be done analytically, so you compute a discrete inverse Fourier transform. You know the interval $ \left[ \omega_{\min}, \omega_{\max} \right]$ on which it is reasonable to numerically compute the IDFT, so you now have to sample $\hat{f}(\omega)$ on this interval.

My question is the following: is there an equivalent of the sampling theorem for the inverse Fourier transform ? Is there a generic rule to sample my frequency-domain signal, as there is for time-domain ? I noticed that I obtain something similar to aliasing when I undersample $\hat{f}(\omega)$.

Side note : all of this is used in a finite element code to compute a complex source. As such the IDFT will be performed at least hundreds of thousand of times, so the question to limit its cost.

Thanks in advance !

  • $\begingroup$ Do the answers to this question answer your question? In short: sampling in one domain causes aliasing in the other domain. $\endgroup$ – Matt L. Nov 6 '18 at 16:29
  • $\begingroup$ Hi! is it about the boundary conditions ? $\endgroup$ – Fat32 Nov 6 '18 at 22:58
  • $\begingroup$ Matt L : it might, but I do not get the specific jargon (I am not really into signal processing in my everyday life) Fat32 : no it is not related to BC $\endgroup$ – Scrimbibete Nov 7 '18 at 6:40

As you already know, sampling in the frequency domain generally causes aliasing in the time domain. If you have a perfectly time-limited signal, which is zero outside the interval $[0,T]$, then sampling its Fourier transform with a distance of $\Delta\omega$ between the samples will not result in any aliasing error if


is satisfied. Eq. $(1)$ is of course just a specific version the sampling theorem.

  • $\begingroup$ Exactly what I needed, many thanks. As a matter of fact I finally found a document in the meantime that was saying that in a more complex way. I am still amazed of how it was difficult to find that clearly stated anywhere in the many courses and books I went through, given the importance of the "reciprocal version" of it. The statement might seem obvious for anyone in the field of signal processing, but it is not necessarily when you come from another field of mathematics. $\endgroup$ – Scrimbibete Nov 7 '18 at 21:00
  • $\begingroup$ @Scrimbibete: Yes, it's usually not stated explicitly in that way. But if you understand how the sampling theorem is derived, it's rather straightforward to exchange the roles of time and frequency and come up with the result you were looking for. But I understand that it's nice to just be able to look up the formula :) $\endgroup$ – Matt L. Nov 7 '18 at 21:05

The Fourier Transform is (mostly) symmetric in both directions so the same sampling rules apply.

The minimum frequency resolution is determined by the length in the time domain, the same way that the minimum time resolution is determined by the "length" in the frequency domain.

That creates a bit of theoretical problem: in order for the required frequency resolution to be finite, the length in the time domain has to be finite as well. However, a signal that has finite length in one domain has infinite length in the other. So you can't actually sample a signal in both domains at the same time.

In practice, you need to work around this by replacing "must be zero outside the supported interval" by "must be small enough outside the supported interval". So you accept some residual aliasing in either domain, provided you can make it small enough for your specific application

  • $\begingroup$ Thank you for answering. It is fairly clear to me that a residual aliasing is unavoidable, and this is ok with my application. The Shannon theorem is also clear to me, but I still have trouble stating its reciprocal properly (and I haven't found any clear statement of that kind anywhere, hence my question). $\endgroup$ – Scrimbibete Nov 6 '18 at 20:29
  • $\begingroup$ It's quite simple: if you your length in the time domain is, for example, 5 seconds, then then required minimum frequency resolution is simply the inverse: 0.2 Hz. $\endgroup$ – Hilmar Nov 7 '18 at 14:14

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