# convolution sum

Discussion in 'Homework Help' started by jut, Sep 20, 2009.

1. ### jut Thread Starter Senior Member

Aug 25, 2007
224
2
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?

Last edited: Sep 20, 2009
2. ### steinar96 Active Member

Apr 18, 2009
239
4
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.

Aug 25, 2007
224
2