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Sorry if I ask a basic question but I am a bit lost.

I have a data set in frequential domain, I need to go to time domain to perform time-gating and then going back to frequencies. In order to do this, I compute an inverse Discrete Fourier Transform.

Shannon theorem tells us that two times the difference between the greatest and the least frequency, respectively noted as $f_N$ and $f_0 ( \ne 0)$, contained in the signal must be less then the sample-rate in time domain ($F_t$). Do I need to do some zero padding to avoid aliasing ?

Without zero padding on frequential domain, what is the relation between $F_t$ and $f_N, f_0$ ? My guess is that $F_t=f_N-f_0$. Is that correct ?

If I must do some zero padding, should I do it this way :

$$(0,\dots,0,f_0,\dots,f_N,0 \dots,0)$$

With enough zeros such that the first frequency contained in the padded signal is $0$ and the last is $2 f_N$ ? If my guess above is correct, then the new sample rate is high enough to avoid aliasing, because it will be $F_t=2f_N>2(f_N-f_0)$.

Thanks for your help.

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    $\begingroup$ Can you define what "time-gating" means, mathematically, for the time-sample vector that you get out of your idft? $\endgroup$ Jul 30, 2018 at 9:39
  • $\begingroup$ Sure. Let $(x_0,\dots,x_N)$ the corresponding time vector (without zero padding), what I am doing is the Hadamard product (or termwise product) with another vector $p=(p_0,\dots,p_N)$. That vector $p$ has a lot of zeros at the beginning and at the end, and where it is non-null, I compute a Kaiser-Bessel window (accordingly to papers I have read, this is the most efficient window to use in my situation). And finally I obtain something like this : $(0,\dots,0,p_i x_i, \dots, p_{i+l} x_{i+l},0\dots,0)$ @MarcusMüller $\endgroup$
    – nicomezi
    Jul 30, 2018 at 10:22
  • $\begingroup$ did your process work correctly when you simply adhere to using powers of 2 number of samples so you can avoid needing to do any padding ? only after you have this simple approach working would I worry about using non powers of 2 $\endgroup$ Jul 30, 2018 at 13:35
  • $\begingroup$ @ScottStensland you're conflating things! this has nothing to do with zero-padding to achieve an efficient-to-calculate transform length (by the way, FFTs exist for far more sizes than just powers of 2). . $\endgroup$ Jul 30, 2018 at 13:38

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You'll not experience aliasing – in the end, the IDFT result per definition only contains frequency components that can be represented by the original frequency domain.

If you will, just stop worrying about aliasing: the DFT and inverse DFT are just base change matrices in $\mathbb C^N$, and are invertible.

Aliasing can only happen when something is non-invertible. This is not the case here; so, no problem.

What you might see is leakage, but you're already taking care of that by windowing.

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  • $\begingroup$ Sorry but after reflection I am having another problem. You said that aliasing can only happen if something is non invertible, and windowing is not invertible. So I am unsure that I am not introducing new frequencies that will produce aliasing when going back to frequency domain. Am I mistaken ? @MarcusMüller $\endgroup$
    – nicomezi
    Jul 31, 2018 at 7:00
  • $\begingroup$ yes, you are. Aliasing doesn't happen. $\endgroup$ Jul 31, 2018 at 7:24

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