082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

22.10 Properties of the z transform 713
Example22.21 The functiont u(t) is sampled at intervals T = 1 to give k u[k]. This sample is then
shifted tothe right by one sampling interval togive
(k −1)u[k −1]
Find itsztransform.
Solution Figure 22.23(a) shows t u(t) and Figure 22.23(b) shows the sampled function. Figures
22.23(c) and (d) show (t − 1)u(t − 1) and (k − 1)u[k − 1], respectively. From
Table 22.2, we have
z
Z{k} =
(z −1) 2
and so, fromthe second shift theorem withi = 1,wehave
Z{(k −1)u[k −1]} =z −1 z
(z −1) = 1
2 (z −1) 2
2
2
1
1
1
1
1
t
1 2 3 k
1 2 3 t
(a) (b) (c) (d)
1 2 3 k
Figure22.23
GraphsforExample 22.21:(a)t u(t);(b)ku[k];(c) (t −1)u(t −1);(d) (k −1)u[k −1].
Example22.22 Find theztransform of the unit step functionu(t) and the shifted unit stepu(t − 2T),
sampledatintervals ofT seconds.
Solution If the function u(t) is sampled at intervals T then we are concerned with finding the
z transform of the sequence u[k]. This has been derived earlier: Z{u[k]} =
z
z −1 . If
u(t −2T) issampled, we have
{
1 k=2,3,4,...
u[k−2]=
0 otherwise
Therefore, by the second shift theorem,
Z{u[k −2]} =z −2 Z{u[k]}
=z −2 z
z −1
=
1
z(z −1)