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.


1 Answer 1


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 $$ X_M[k] = X\left( e^{j 2 \pi\frac{k}{M}} \right) ~~~,~~ k=0,1,...,M-1. $$ Then the $M$-point inverse DFT of $X_M[k]$ will be: $$ y[n] = IDFT_M\Big\{ X_M[k] \Big\} = \begin{cases}{ x[n] ~~~,~~~0 \leq n < N \\ ~ \\ ~~~0~ ~~~~,~~~N \leq n < M } \end{cases} $$ $y[n]$ 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
    Commented Oct 15, 2018 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
    Commented Oct 15, 2018 at 0:30
  • 1
    $\begingroup$ or stated in other words: $$ \tilde{y}[n] = \sum_k x[n-kM] $$ $\endgroup$
    – Fat32
    Commented Oct 15, 2018 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
    Commented Oct 15, 2018 at 0:47

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