Suppose we are given an input signal s[m,n] with DTFT $S(\omega_1, \omega_2)$.
We sample it at $\omega_1 = \frac{2 \pi k}{256}$ and $\omega_2 = \frac{2 \pi l}{256}$ to get a 256 point DFT S[k,l]. Now suppose we take the IDFT of S[k,l] to get $\tilde{s}[m,n]$
I am trying to understand the relationship between $\tilde{s}[m,n]$ and s[m,n]. My understanding is that I should be able to express $\tilde{s}[m,n]$ as a summation of every 256th sample of the original signal.
So far I have been advised to try expressing the Sampled DTFT as a dirac comb, i.e.:
$S[k,l] = \sum_{k = -\infty}^{\infty} \sum_{l = -\infty}^{\infty} S( \frac{2 \pi k}{256}, \frac{2 \pi l}{256}) \delta(\omega_1 - \frac{2 \pi k}{256}, \omega_2-\frac{2 \pi l}{256}) $
and use the definition of the IDFT:
$\tilde{s}[m,n] = \frac{1}{256^2}\sum_{k=0}^{255} \sum_{l=0}^{255} S[k,l] e^{j(\frac{2 \pi km}{256}+\frac{2 \pi ln}{256})}$
Using these two pieces of information I can get the expression:
$\tilde{s}[m,n] = \frac{1}{256^2}\sum_{k=0}^{255} \sum_{l=0}^{255} [\sum_{k = -\infty}^{\infty} \sum_{l = -\infty}^{\infty} S( \frac{2 \pi k}{256}, \frac{2 \pi l}{256}) \delta(\omega_1 - \frac{2 \pi k}{256}, \omega_2-\frac{2 \pi l}{256})] e^{j(\frac{2 \pi km}{256}+\frac{2 \pi ln}{256})}$
but I am not sure how to proceed from here. Any advice would be helpful I have been racking my brain over this for a couple hours now.