Originally Posted by

**jut** Sorry, let me restate the original problem and be more explicit.

Find the convolution of x(n) and h(n).

$\displaystyle x(n) = (-1/2)^n u(n-4)$

$\displaystyle h(n) = 4^n u(2-n)$

where $\displaystyle u(n)=\begin{cases}1,n\geq0\\0,n<0\end{cases}$

where the convolution is $\displaystyle y(n)=x(n)\ast y(n)=\sum_{k=-\infty}^{\infty}x(k)h(n-k)$

note the $\displaystyle \ast$ symbol is not multiply, it is the convolution operator. Also y(n) is causal, which means y(n) is only defined for $\displaystyle n\geq 0$, where n=0,1,2,3....

Now,

$\displaystyle x(k) = (-1/2)^k u(k-4)$

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

Plugging in,

$\displaystyle x(n)\ast y(n)=\sum_{k=-\infty}^{\infty} (-1/2)^k u(k-4) 4^{n-k} u(2-n+k)$

Simplifying,

$\displaystyle x(n)\ast y(n)=4^n\sum_{k=-\infty}^{\infty}(-1/2)^k 4^{-k} u(k-4) u(2-n+k)$

The argument passed into the unit step must be $\displaystyle \geq 1$ or else the value of the unit step function is 0. So if I need to change my limits of summation so that the two unit step functions resolve to 1. That is,

if $\displaystyle k-4\geq 0$ then $\displaystyle u(k-4)=1$.

if $\displaystyle k+2-n\geq 0$ then $\displaystyle u(k+2-n)=1$.

So the limits on k would be,

if $\displaystyle k\geq 4$

if $\displaystyle k\geq n-2$

I am bamboozled by the two lower limits on k. How can I do a summation with two lower limits? Would the summation diverge?