The function t u(t) is sampled at intervals T = 1 to give k u[k]. This sample is then shifted to the right by one sampling interval to give
(k-1) u[k-1]
Find its z transform.
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
\mathcal{Z}\{k\}=\frac{z}{(z-1)^2}and so, from the second shift theorem with i = 1, we have
\mathcal{Z}\{(k-1) u[k-1]\}=z^{-1} \frac{z}{(z-1)^2}=\frac{1}{(z-1)^2}Table 22.2
The z transforms of some common functions
\begin{matrix}\hline f[k]&f(z)&f[k]&f(z)\\ \hline \delta[k]= \begin{cases}1 & k=0 \\ 0 & k \neq 0\end{cases}&1&k^2&\frac{z(z+1)}{(z-1)^3}\\ u[k]= \begin{cases}1 & k \geqslant 0 \\ 0 & k<0\end{cases}&\frac{z}{z-1}&k^3&\frac{z(z^2+4z+1)}{(z-1)^4}\\ k&\frac{z}{(z-1)^2}&\sin ak&\frac{z\sin a}{z^2-2z\cos a+1}\\ e^{-ak}&\frac{z}{z-e^{-a}}&\cos ak&\frac{z(z-\cos a)}{z^2-2z\cos a+1}\\a^k&\frac{z}{z-a}&e^{-ak}\sin bk&\frac{ze^{-a}\sin b}{z^2-2ze^{-a}\cos b+e^{-2a}}\\ ka^k&\frac{az}{(z-a)^2}&e^{-ak}\cos bk&\frac{z^2-ze^{-a}\cos b}{z^2-2ze^{-a}\cos b+e^{-2a}}\\k^2a^2&\frac{az(z+a)}{(z-a)^3}\\ \hline\end{matrix}