# Discrete Time Convolution Integral

• Apr 11th 2011, 09:53 AM
rebghb
Discrete Time Convolution Integral
Hello Everyone!

I've been stuck on this DT convolution sum for quite sum time, it's because I have not much experience in calculating DT sums. Well here goes:

$\displaystyle x[n] = (1/2)^{n}.u[n+2]$
$\displaystyle h[n] = (-3)^{n}.u[-n-1]$

Now, $\displaystyle x[n] \ast h[n] = \displaystyle \sum ^{\infty} _{k = -\infty} (1/2)^k .u[k+2] . (-3)^{n-k}.u[-n-1+k]$

Now here, I saw that whenevr we have step functions we can remove them and change the limits of the sum, so I modified the sum to be from n+1 to infinity as:
$\displaystyle \displaystyle \sum ^{\infty} _{k=n+1}(1/2)^k . (-3)^{n-k}$.
But I think I am very wrong.

Any help would be greatly appreciated!
• Apr 20th 2011, 01:00 PM
Mondreus
If you mirror and shift h[n], the limits become: http://i.imgur.com/irLG2.png

u[-n-1] is 1 for -n-1 >= 0 => n<=-1. If you mirror and shift that you end up with k-n<=-1 => k <= n-1, hence the upper limit.