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

if:

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

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.

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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
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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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