Signal Processing Stack Exchange is a question and answer site for practitioners of the art and science of signal, image and video processing. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'd like to understand how to calculate the cyclic convolution as well as understand what that means exactly. How should I go about finding the output for various periods for a system?

I have an example:

$x(n) = {n\ \textrm{for} \ 1 <= n <= 3,\ 0\ \textrm{otherwise}}$

$h(n) = {n\ \textrm{for} \ 1 <= n <= 2,\ 0\ \textrm{otherwise}}$

If I perform the convolution, then I get the following values for $y(n)$:

$y(2) = 1$; $y(3) = 4$; $y(4) = 7$; $y(5) = 6$

Now, if I want a period = 3, then:

$x(n) = x(n+3k)$ and $h(n) = h(n+3k)$

At this point, I'm unsure of what to do to get values that correspond to a period.

share|improve this question
up vote 0 down vote accepted

People generally define the cyclic convolution of periodic sequences $x$ and $h$ of period $N$ as $$y[n] = \sum_{m=0}^{N-1}x[m]h[n-m], n = 0, 1, \ldots, N-1.\tag{1}$$ Note that the above expression consists of $N$ different sums that you have to compute, and if while computing any particular sum, the value of $n-m$ is not in the range of numbers for which you are given $h[\cdot]$, then you use the periodicity ($h[n-m] = h[n-m+N]$ or $h[n-m] = h[n-m-N]$) to get the argument into the range for which you know the value of $h[\cdot]$. Also, note that $(1)$ holds for all integers $n$, but we don't need to calculate more than $N$ sums like $(1)$ because $y[n]$ is also a periodic sequence of period $N$ and so we have for any integer $M$ that $y[M] = y[M \bmod N]$ where, of course, $0 \leq M \bmod N \leq N-1$.

Exercise: write out the above formula explicitly, meaning no summations, for $n = 0, 1, 2$ and proceed from there. Go on; you can do it. There are only three sums of three terms each.

share|improve this answer
Alright, I'll try it out. – Chris Harris Feb 25 '13 at 16:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.