Question 22.28: Find the sequence whose z transform is 1/z-1....
Find the sequence whose z transform is \frac{1}{z-1}.
From Table 22.2 we have
\mathcal{Z}\{u[k]\}=\frac{z}{z-1}So from the second shift property, we have
\mathcal{Z}\{u[k-1]\}=z^{-1} \frac{z}{z-1}The required sequence is therefore u[k – 1].
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}