Module 2 : Modeling Discrete Time Systems by Pulse Transfer Function

Lecture 1 : Motivation for using Z-transform

The initial conditions are y(0) = 0, y(1) = 0 .
We have to find the solution y(k) for k > 0.

Taking z-transform on both sides of the above equation:

Using partial fraction expansion:

$\displaystyle Y(z)$
 $\displaystyle =$ $\displaystyle -\frac {0.8937 z}{z-0.5}+\frac{7.123 z}{z+0.2}-\frac{6.25 z}{z+0.3}$  
Taking Inverse Laplace:$\displaystyle \;\;\; y(k)$
 $\displaystyle =$ $\displaystyle -0.893(0.5)^k+7.143(-0.2)^{k}-6.25(-0.3)^{k}$  

To emphasize the fact that y(k) = 0 for k < 0 , it is a common practice to write the solution as:

$\displaystyle y(k)=-0.893(0.5)^ku_s(k)+7.143(-0.2)^{k}u_s(k)-6.25(-0.3)^{k}u_s(k) $

where $ u_s(k)$ is the unit step sequence.