Relationship between the IDFT of a sampled DTFT and its discrete-time domain signal
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 &lt; 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 $M &lt; N$ 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.