Convolution sum. Compute $y[n]=x[n]\ast h[n]$

Question

Compute $y[n]=x[n]\ast h[n]$

$x[n]=(-\frac{1}{2})^2u[n-4]$

$h[n]=4^nu[2-n]$

In this question, when I try to calculate the convolution sum. I face with: $$\sum_{k=-\infty}^{+\infty}(-\frac{1}{2})^ku[k-4]4^{n-k}u[n+k+2]$$

Then I tried to rearrange the boundaries according to $u[n]$'s: $$\sum_{k=4}^{+\infty}(-\frac{1}{2})^k4^{n-k}u[n+k+2]$$

I'm stuck here. The only idea I have now is, to choose values of $n$ before taking convolution and go from different sums at the beginning. Would this be correct or any other suggestions ?

Answer 1

The convolution sum is

$$y[n]=\sum_{k=-\infty}^{\infty}x[k]h[n-k]\tag{1}$$

To find $h[n-k]$ you need to replace the argument $n$ in $h[n]$ with $n-k$, which gives

$$h[n-k]=4^{n-k}u[2-(n-k)]=4^{n-k}u[2-n+k]\tag{2}$$

Combining (1) and (2) gives

$$\begin{align}y[n]&=\sum_{k=-\infty}^{\infty}\left(-\frac12\right)^ku[k-4]4^{n-k}u[2-n+k]\\&=4^n\sum_{k=4}^{\infty}\left(-\frac18\right)^ku[2-n+k]\\ &=4^n\sum_{k=\max\{4,n-2\}}^{\infty}\left(-\frac18\right)^k\end{align}\tag{3}$$

From (3) you have to deal with two cases: $4\ge n-2$, i.e. $n\le 6$, where you need to use $k=4$ as the lower summation limit, and the case $n>6$, where the lower summation limit becomes $n-2$. The rest is a straightforward application of the formula for the geometric sum.

Answer 2

The problem here I have is about flipping and time-shifting. First I was trying to $h[k] => h[-k] => h[n-k]$ but then I tried replacing $n$ with $n-k$. This led to correct answer.