# Discrete Fourier Transform of Sinc Function

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]$. But I notice the difference between Dirichlet kernel and Sinc function. I wonder whether there is a method to bridge them together. Thank you for your help! • The Fourier transform acts on several possible domains - the real line to itself, the circle to the integers and vice versa, or the integers mod$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$)? – jmerry Feb 5 at 8:13 • @jmerry Yes. I am sorry for the ambiguity and I have added some details. But I am not sure$s_0=1$or$s_{N/2}=1$if the length of the signal equals to N. – C.L. Liu Feb 5 at 8:21 • The length of the signal? You're not talking about the same domain I am here. That would put you on the integers mod$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\$. – jmerry Feb 5 at 8:28
• @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. – C.L. Liu Feb 5 at 8:54
• 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. – jmerry Feb 5 at 8:59