# Practice Question on Computing the Output of an LTI system by Convolution

The unit impulse response h[n] of a DT LTI system is

$h[n]= 5^n u[-n]. \$

Use convolution to compute the system's response to the input

$x[n]= u[n] \$

$y[n]=h[n]*x[n]=\sum_{k=-\infty}^\infty 5^{k}u[-k]u[n-k] = \sum_{k=-\infty}^0 5^{k}u[n-k] = \Bigg(\sum_{k=-\infty}^n 5^{k}\Bigg)u[-n]$

I'm not sure where to go with this sum. I tried convolving in the other order, but the result wasn't any more helpful (as far as I can tell).

$y[n]=x[n]*h[n]=\Bigg(\sum_{k=n}^\infty 5^{n-k}\Bigg)u[n]$

Am I making some kind of mistake? How do I solve this sum?

--Cmcmican 21:17, 31 January 2011 (UTC)

Starting from $(\sum_{k=-\infty}^n 5^{k})u[-n]$ It can be observed that $(\sum_{k=-\infty}^n 5^{k}) = (5^{-\infty} - 5^{n+1})/(1-5) = 5^{n+1}/4$ Therefore, $(\sum_{k=-\infty}^n 5^{k})u[-n] = 5^{n+1}/4*u[-n]$(Clarkjv 23:35, 31 January 2011 (UTC))