
Using the method of partial fraction expansion presented in Section 4.4.6, the inverse z-transform
of X(z) is (details are left as an exercise)
x
k
¼ 0:05
sin 0:1p
1 cos 0:1p
(1 cos 0:1kp) sin 0:1kp
, k ¼ 0, 1, 2, 3, ... (4:468)
(b) The continuous-time integrator output is obtained by integration of the input u(t),
x(t) ¼
ð
t
0
u(l)dl ¼
ð
t
0
sin pl dl ¼
1
p
(1 cos pt)(4:469)
The discrete-time signal x
k
and the continuous-time integrator output x( t) are plotted in Figure 4.44
for one cycle of the input.
4.7.1 NONZERO INITIAL CONDITIONS
Using the z-transform to solve a difference equation with nonzero initial conditions requires
additional terms to account