# 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, I prefer a 1D notation without losing generality.

The IDFT associated with the properly taken uniform (in frequency) samples of a given valid DTFT can be shown to be the following.

Case-1: Assume a finite length signal $$x[n]$$ of length $$N$$, and consider its DTFT $$X(\omega)$$. Now if you take $$M \geq N$$ uniform (frequency) samples of $$X(\omega)$$ according to $$X[k] = X( \frac{2\pi}{M}k) ~~~, k=0,1,...,M-1$$ then the M-point inverse DFT of $$X[k]$$ will exactly be $$x[n]$$ padded with $$M-N$$ zeros; i.e., $$y[n] = IDFT_M\{ DFT_M\{ x[n] \} \} = x_M[n]$$ So the sequence $$x[n]$$ and $$y[n]$$ have the same values for the first $$N$$ samples (i.e., all samples of $$x[n]$$ are retained in $$y[n]$$), but $$y[n]$$ have further (effectively padded) zero samples.

If $$M < N$$ , then time-aliasing in $$x[n]$$ will happen and the first $$N-M$$ samples of $$y[n]$$ will be in error. But the remaning $$2M-N$$ samples of $$y[n]$$ will be identical to the original $$x[n]$$ at the corresponding index positions.

Case-2: $$x[n]$$ is infinite length (and stable) sequence and $$X(\omega)$$ is its DTFT. Then according to $$M < N$$ subcondition of case-1, all samples of $$y[n]$$ will be aliased. No samples of $$y[n]$$ will be equal to $$x[n]$$, except other than (pure) coincidence.

• 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. – Filip 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. $y[0] = x[0] + x[M]$, $y[1] = x[1]+x[M+1]$,...,$y[N-M-1] = x[N-M+1]+x[N-1]$. You can see those indices by a simple plot of the periodic extension of the circularly alised inverse DFT. – Fat32 Oct 15 '18 at 0:30
• or stated in other words: $$\tilde{y}[n] = \sum_k x[n-kM]$$ – Fat32 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. – Filip Oct 15 '18 at 0:47