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I am trying to understand a piece of notation used in several papers, the simplest/shortest of which is this paper by Crochiere. The equation in question is Equation 7 on the second page:

$x_m(sR) = \tilde{x}_{((m-sR))_M}(sR)$

followed by the remark:

where ((n)) denotes n reduced modulo M

In the context of Fourier transforms, DFTs and FFTs, where $x$, $\tilde{x}$, etc are sampled signals, what does "n reduced modulo M" mean? I've seen that notation elsewhere with essentially the same non-explanation, so it must have been something very obvious in context (and maybe still is) but I am completely failing to grasp it.

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  • $\begingroup$ $$\begin{align} ((n))_N \ &\triangleq \quad n \mod N \\ \\ &= n - N\left\lfloor \frac{n}{N} \right\rfloor \\ \end{align}$$ where $\lfloor x \rfloor$ is the floor() operator, the largest integer that is no greater than the argument $x$. $\endgroup$ May 5, 2022 at 3:39

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It just means that the number in the double parentheses is changed to be between $0$ and $M-1$ by adding to or subtracting from it $M$ an integer number of times.


See equation (5) of the paper:

$$ \stackrel{\large x}{\sim}_m(sR) = \sum_{l=-\infty}^{+\infty} x(sR+lm+m)h(-lm-m) \tag{5} $$

so the modulo version is just the time-aliased copies of $x$ summed over the interval of interest, $M$.

The diagram below attempts to illustrate this with $M=4$ (though the image only shows four subsequences, all subsequences should be summed).

enter image description here

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  • $\begingroup$ Bear with me: So if I had some signal $x(n)$ with thousands of samples, but M was some small number like 16, then I'm just cycling through a set of 16 points even as the index increases well outside that range? $\endgroup$
    – Novak
    May 3, 2022 at 2:46
  • $\begingroup$ @Novak No worries! Please see my addition to attempt to explain the detail of what's happening. $\endgroup$
    – Peter K.
    May 3, 2022 at 14:33
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    $\begingroup$ Ah, I think I see-- that's just a shorthand for the combined operation of "Window, and slide window back to origin". Which, in context, makes sense. Thank you! $\endgroup$
    – Novak
    May 4, 2022 at 0:34

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