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I am working with a convolution sum of the form

$$ h(j) = \sum_{k=0}^2 f\!\big((j-k) \bmod 3\big)\, g(k), $$

where $f, g : \{0,1,2\} \to \mathbb{C}$. Because of the modulo $3$ structure in the index shift, this looks like a (discrete?) circular convolution of $f$ and $g$. How can I rigorously show that it is indeed a circular convolution? Also, is this sum generalizable to any modulo $n$?

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    $\begingroup$ (1) If you want to show that a function is a circular convolution, what is the precise definition of circular convolution you are using? (2) To me the sum seems immediately generalizable to any $n$. $\endgroup$ Commented Sep 30 at 3:14
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    $\begingroup$ People define convolution of functions over any Abelian group, and this is just the definition for $\mathbb Z_3$. $\endgroup$ Commented Sep 30 at 3:43
  • $\begingroup$ You can find the definition of circular convolution on Wikipedia. Also, you can write the circular convolution in terms of a matrix-vector multiplication over the finite field ${\Bbb F}_3$ and this can be generalized to any finite ring $\endgroup$ Commented Oct 10 at 8:18

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