Convolution of discrete signals using convolution sum

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

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?

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