# time downsampling vs. frequency downsampling [closed]

$$x[n]_M$$ 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

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• maybe call it "frequency rebinning" for lack of a better name? then its a choice of "up binning" or "down binning"? – MrCasuality Jan 28 at 18:49
• 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. – Marcus Müller Jan 28 at 18:52
• 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... – MrCasuality Jan 28 at 18:56
• I'm just going to call it rebinning in my notes… other people can say I don't know #$@#$ later.... – MrCasuality Jan 28 at 19:00
• it's definitely not resampling. – 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:
$$y[n] = \begin{cases} { x[n]+x[n+N] ~~~, n = 0,1,...,M-N \\ x[n] ~~~~~~~~~~~~~~~~~~~~~,~~~ n= M-N+1,...,N-1 } \end{cases}$$