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?


1 Answer 1


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, 2021 at 19:07
  • $\begingroup$ Oh, of course, it's a geometric series! Thanks! $\endgroup$
    – carraro
    Feb 25, 2021 at 20:55

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