0
$\begingroup$

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$n$?

Rodrigo de Azevedo's user avatar
Rodrigo de Azevedo
23.6k7 gold badges49 silver badges117 bronze badges
askedSep 30 at 1:28
AmB's user avatar
$\endgroup$
3
  • 1
    $\begingroup$(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 $n$.$\endgroup$CommentedSep 30 at 3:14
  • 2
    $\begingroup$People define convolution of functions over any Abelian group, and this is just the definition for $\mathbb Z_3$.$\endgroup$CommentedSep 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$\endgroup$CommentedOct 10 at 8:18

0

You mustlog in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.