- #1

Xyius

- 508

- 4

## Homework Statement

Find ##x[n] \ast h[n]## when ##x[n] = 3 u[2-n]## and ##h[n] = 4\left( \frac{1}{2} \right)^{n+2}u[n+4]## where ##u[n-k]## is the unit step function.

## Homework Equations

None really

## The Attempt at a Solution

So I know this is probably simple but I am confused.

So the convolution is written as,

[tex]\sum_{k=-\infty}^{\infty} 4\left( \frac{1}{2} \right)^{k+2}u[k+4](3)u[2-n+k][/tex]

This simplifies to,

[tex]3\sum_{k=-\infty}^{\infty} \left( \frac{1}{2} \right)^{k}u[k+4]u[2-n+k][/tex]

The first step function of ##u[k+4]## means that the terms in the sum corresponding to ## k < -4 ## are zero, therefore this becomes the bottom limit.

[tex]3\sum_{k=-4}^{\infty} \left( \frac{1}{2} \right)^{k}u[2-n+k][/tex]

This form is confusing to me. I don't see an obvious limit on k except for that its LOWER limit is ##n-2##. What I tried to do is do some change of variables, such as ##k'=2-n+k##, this changes the lower limit to -2-n and the upper limit to infinity. So I have...

[tex]3\sum_{k'=-2-n}^{\infty} \left( \frac{1}{2} \right)^{k'-2+n}u[k'][/tex]

So this means that ##k'## must be larger than zero, which means ## n < -2##. So in this regime, the step function is 1 and I get..

[tex]3\sum_{k'=0}^{\infty} \left( \frac{1}{2} \right)^{k'-2+n}=3\left( \frac{1}{2} \right)^{n-1}[/tex]

Am I doing this correctly? I am just not sure. Thank you in advance for anyone that takes the time to read this!