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 $x[n]$.

  • $\begingroup$ 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. $\endgroup$
    – Filip
    Oct 15 '18 at 0:25
  • $\begingroup$ 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. $\endgroup$
    – Fat32
    Oct 15 '18 at 0:30
  • 1
    $\begingroup$ or stated in other words: $$ \tilde{y}[n] = \sum_k x[n-kM] $$ $\endgroup$
    – Fat32
    Oct 15 '18 at 0:32
  • $\begingroup$ But what is the math behind this fact, to me it is not obvious why there is aliasing in the discrete time domain. $\endgroup$
    – Filip
    Oct 15 '18 at 0:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.