$x[n]_M$ is a finite length sequence of length M.


$$ y = x[nN]_M \tag{1} $$

is called downsampling in the time-domain.

then what do you call the process of converting going from a M-point to an N-point DFT?

$$ y[n] = \Bigg[ \sum_{k=-\infty}^{\infty} x[n-Nk]_M \Bigg] R_N[n] \tag{2} $$

does that have a name? like frequency downsampling? frequency rebinning...frequency re-pie cutting or something like that? just guessing...

($R_N[n]$ = window function, which is 1 between 0 and N-1 and 0 otherwise.)

$$ Y[k]_N = X(e^{j\omega})\bigg|_{\omega=2\pi k/N} \tag{3} $$

rebinning $Y[k]_N$ to $Y[k]_M$ for example.



closed as unclear what you're asking by Marcus Müller, MBaz, lennon310, Peter K. Feb 5 at 14:07

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  • $\begingroup$ maybe call it "frequency rebinning" for lack of a better name? then its a choice of "up binning" or "down binning"? $\endgroup$ – MrCasuality Jan 28 at 18:49
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    $\begingroup$ I'd recommend removing the window function; it adds nothing to this question, but makes it harder to argue based on the "pure" DFT; anyways, the formula for $y[n]$ that you wrote has nothing to do with a DFT; not quite sure where you're taking use here. $\endgroup$ – Marcus Müller Jan 28 at 18:52
  • $\begingroup$ yeah... I know what you mean... the window function is driving me crazy also... connection with DFT is apparently because this is the time-domain representation of doing the same "rebinning" operation in the DFT domain.... of taking the unit circle apple pie and cutting it up for 3 people instead of 6 people, nevermind if we need to put two pieces of pie on top of the other because of aliasing... $\endgroup$ – MrCasuality Jan 28 at 18:56
  • $\begingroup$ I'm just going to call it rebinning in my notes… other people can say I don't know #$@#$ later.... $\endgroup$ – MrCasuality Jan 28 at 19:00
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    $\begingroup$ it's definitely not resampling. $\endgroup$ – Marcus Müller Jan 28 at 19:52

Your second equation suggests that $y[n]$ is obtained by performing an $N$-point inverse DFT of an $M$-point DFT $X[k]$ of an $M$-point sequence $x[n]$.

The result can be interpreted as shifted copies of $x[n]$ are superimposed, causing possible aliasing in time, which depends on whether $ N \geq M$ or $N < M$.

If $N \geq M$, there is no aliasing and $y[n]$ will be an $N$ point sequence whose first $M$ samples will be identical to $x[n]$ for $n=0,1,...,M-1$ and remaining $N-M$ samples will be zero-padded.

$$ y[n] = \begin{cases} { x[n] ~~~,~~~ n = 0,1,...,M-1 \\ ~~~~~0 ~~~,~~~ n= M,M+1,...N-1} \end{cases} $$

Otherwise if $N < M$, then $y[n]$ will be a time-aliased version of $x[n]$, as shifted copies of $x[n]$ would be overlapping each other. If you want any non-aliased samples in $y[n]$, then $N > M/2$ in general and in such a case $y[n]$ could be defined as:

$$ y[n] = \begin{cases} { x[n]+x[n+N] ~~~, n = 0,1,...,M-N \\ x[n] ~~~~~~~~~~~~~~~~~~~~~,~~~ n= M-N+1,...,N-1 } \end{cases} $$


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