When studying solutions of DE's in the time domain we noted that the output from a system could be written as the convolution sum :
This convolution statement is often written using * as the convolution operator :
where x(n) is the driving input and h(n) is the system response to the unit sequence 
.
We couldn't take convolution very far for lack of a powerful method of computing h(n) .
Convolution by z-transform
In general if we have a DE and zero initial conditions and take its z-transform :
So we see that convolution in the time domain becomes multiplication in the z-domain.
But multiplication is a more straightforward operation to carry out than convolution.
Of course we must to the inverse z-transform to get y(n).
When we write 
we are assuming zero initial conditions. So we see that
this simple convolution method is the same as the more general DE solution by z-transform we
discussed already but assuming zero initial conditions. The convolution method will work in
general but it might be necessary to think a bit more about Z[h(n)].
Example of convolution by z-transform
A lowpass digital filter has a unit sample response :
It is subject to a unit step input x(n)=u(n) . Use z-transforms to find the output
y(n)=x(n)*h(n) .
Solution
The lookup table of z-transforms gives :
MATLAB functions to plot the solution to this problem - filter or dstep .