While considering an input to be periodic of Period N, can the impulse response not be periodic of period greater than N ? If it can be, how can one compute it’s convolution?


Circular convolution assumes that all signals ($x[n]$, $h[n]$ and $y[n]$) are periodic in the same integer $L$. When any of the signals are shorther than $L$ then they are padded with enough zeros to make them periodic with $L$. When $x$ or $h$ has a period larger than $L$ then there will be aliasing in the computed output $y[n]$ which is still periodic with $L$.

| improve this answer | |
  • $\begingroup$ can you explain a little bit on the aliasing part please $\endgroup$ – Syed Mohammad Asjad Oct 22 '17 at 16:43
  • $\begingroup$ I mean would the alias be having a period longer than L or shorter ? @Fat32 $\endgroup$ – Syed Mohammad Asjad Oct 22 '17 at 16:52
  • $\begingroup$ result of circular convolution will always be periodic in L, whether there is aliasing or not. $\endgroup$ – Fat32 Oct 22 '17 at 17:21
  • $\begingroup$ I can understand that, but let's just say that we have a sine wave, periodic of period N, let the impulse response be an impulse as well, If I am to convolve the two for a period L, where L< N , I cannot understand how the output y[n] be an alias because it is not periodic of period N. $\endgroup$ – Syed Mohammad Asjad Oct 22 '17 at 17:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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