Relationship between the IDFT of a sampled DTFT and its discrete-time domain signal

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.

For simplicity, 1D notation can be used without losing generality.

IDFT associated with uniform (frequency) samples of a (valid) DTFT $$X(e^{j \omega})$$ of $$x[n]$$:

Case-1: finite length $$x[n]$$ of lenght $$N$$ :

If $$M \geq N$$ uniform samples of $$X(e^{j \omega})$$ is taken to obtain $$X_M[k] = X( \frac{2\pi}{M}k) ~~~,~~ k=0,1,...,M-1.$$ Then the $$M$$-point inverse DFT of $$X_M[k]$$ will be: $$y[n] = IDFT_M\{ X_M[k] \} = \begin{cases}{ x[n] ~~~,~~~0 \leq n < N \\ ~ \\ ~~~0~ ~~~~,~~~N \leq n < M } \end{cases}$$ $$y[n]$$ is actually $$x[n]$$ padded with $$M-N$$ zeros.

If $$M < N$$ , then time-aliasing in $$x[n]$$ happens: $$N-M$$ samples of $$y[n]$$ in the range $$0\leq n < N-M$$ will be corrupted, while the remaining $$2M-N$$ samples of $$y[n]$$ in the range $$N-M \leq n < M$$ will be identical to $$x[n]$$, provided $$M > N-M$$ holds true.

Case-2: $$x[n]$$ is of infinite length ($$N \to \infty$$) and $$X(e^{j \omega})$$ exists:

Then it falls into second subcondition of case-1, and thus all samples of $$y[n]$$ will be aliased, since no finite $$M$$ can be larger than $$N$$. No samples of $$y[n]$$ will be equal to $$x[n]$$.

• Correct, I am expecting aliasing in y[n], but I am trying to figure out how to prove derive the expression which relates the aliased $\tilde{s}[m,n]$ values to those of the original signal. Oct 15 '18 at 0:25
• aliased values of y[n] are addition of last $N-M$ samples of $x[n]$ to first $N-M$ samples of $x[n]$ ; i.e. $y = x + x[M]$, $y = x+x[M+1]$,...,$y[N-M-1] = x[N-M+1]+x[N-1]$. You can see those indices by a simple plot of the periodic extension of the circularly alised inverse DFT. Oct 15 '18 at 0:30
• or stated in other words: $$\tilde{y}[n] = \sum_k x[n-kM]$$ Oct 15 '18 at 0:32
• But what is the math behind this fact, to me it is not obvious why there is aliasing in the discrete time domain. Oct 15 '18 at 0:47