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We know that the discrete Fourier transform (DFT) of a discrete rectangular function is related to Dirichlet kernel: $D_n$(x)=$\frac{sin[(n+1/2)x]}{sin(x/2)}$, and the Fourier transform of a continuous rectangular function is Sinc function. But now I wonder what is the (inverse) discrete Fourier transform of a discretely sampled Sinc function? In my current work, I have two finite discrete signal $f_1$ and $f_2$ both in frequency domain, and now I want to simplify the equation:

$f_1[x]$$f_2[x]$$\frac{sin[kx]}{kx}$.

I want to first fix $f_1[x]$ and use the convolution theorem, getting the result:

=$f_1[x]$$\hat{f_2}[x]$,

where $\hat{f_2}[x]$ is the DFT of average smoothed inverse DFT of $$f_2[x]$. But I notice the difference between Dirichlet kernel and Sinc function. I wonder whether there is a method to bridge them together.

Thank you for your help!

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  • $\begingroup$ The Fourier transform acts on several possible domains - the real line to itself, the circle to the integers and vice versa, or the integers mod $n$ to themselves. By its nature, sinc should be on one of the infinite domains; as you said "discrete", do you mean that to be the (double-ended) sequence $s_n = \frac{\sin cn}{cn}$ (and $s_0=1$)? $\endgroup$ – jmerry Feb 5 at 8:13
  • $\begingroup$ @jmerry Yes. I am sorry for the ambiguity and I have added some details. But I am not sure $s_0=1$ or $s_{N/2}=1$ if the length of the signal equals to N. $\endgroup$ – C.L. Liu Feb 5 at 8:21
  • $\begingroup$ The length of the signal? You're not talking about the same domain I am here. That would put you on the integers mod $N$ - and the sinc function is problematic there, both because we have no reason for the $\sin$ in the numerator to have period $N$ and because dividing by $n$ isn't well-defined when $n$ is interpreted mod $N$. $\endgroup$ – jmerry Feb 5 at 8:28
  • $\begingroup$ @jmerry. Thanks for your comment. I found it indeed makes thing complicate and not necessary to use DFT here. I will try to rethink the problem using discrete time Fourier transform. $\endgroup$ – C.L. Liu Feb 5 at 8:54
  • $\begingroup$ And then, there's a general principle for moving between domains: the operations "wrap" and "sample" are taken to each other by the Fourier transform. Our sampled sinc is the transform of a box, so we'll wrap that box around the circle. The result should be a step function, the exact details of which depend on the parameter involved. $\endgroup$ – jmerry Feb 5 at 8:59

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