Find the convolution of x(n) and h(n).
\(x(n) = (-1/2)^n u(n-4)\)
\(h(n) = 4^n u(2-n)\)

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

where the convolution is \(y(n)=x(n)\ast y(n)=\sum_{k=-\infty}^{\infty}x(k)h(n-k)\)
note the \(\ast\) symbol is not multiply, it is the convolution operator. Also y(n) is causal, which means y(n) is only defined for \(n\geq 0\), where n=0,1,2,3....

Now,
\(x(k) = (-1/2)^k u(k-4)\)
\(h(k) = 4^{n-k} u(2-n+k)\)

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


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

The argument passed into the unit step must be \(\geq 1\) 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 \(k-4\geq 0\) then \(u(k-4)=1\).
if \(k+2-n\geq 0\) then \(u(k+2-n)=1\).

So the limits on k would be,
if \(k\geq 4\)
if \(k\geq n-2\)

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

 

Signals and systems eh. Chapter 2 proplem . Funny thing is i got stuck on the same thing yesterday. The summation limits change with each n chosen which makes calculating the summation to a closed form a bit more tedious.

Most likely it needs to be solved by looking at different scenarios when you plug in different "ranges" of n.
K needs to be larger or equal to 4 so that u[k-4] be nonzero. From this we could say that we could start summing from k = 4.

however let's say we have y[7]. Which means that u[2-n+k] won't be nonzero till k = 5.
So with increasing n we can't start summing till k = n-2.

So for n between 0 and 5 you can start summing from k = 4. While for n >= 6 you have to begin the summation at K = n-2.

 

Hi, thanks for the reply.

I was able to solve it shortly after posting. It's like you said, the limits depend on values of n. So the ultimate solution is piecewise.