I want to perform the convolution of the following discrete signals: $$h[n]=u[n-2] $$ and $$x[n] = (0.5)^nu[n+2]$$.

That's what I've done so far: $$\sum_{k=-\infty}^{\infty} (0.5)^ku[k+2]*u[n-2-k]$$

So: $$\sum_{k=0}^{\infty} (0.5)^ku[k+2]*u[n-2-k]$$

I know that the upper limit does not go to infinity, however, I cannot determine the correct upper limit. Any thoughts?


Well, you know that $u[n]=0$ for all $n<0$. So from the two terms in your sum, all terms will be zero where either $k+2 < 0$ (i.e., $k < -2$) or $n-2-k<0$ (i.e., $k>n-2$). From this your sum will need to run from $-2$ to $n-2$. Can you take it from here?

  • $\begingroup$ Yeah you're right. It's been a long day. :) I correct my reply. $\endgroup$
    – Florian
    Feb 25 at 19:07
  • $\begingroup$ Oh, of course, it's a geometric series! Thanks! $\endgroup$
    – July H.
    Feb 25 at 20:55

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.