Question 22.31: If F(z) = z+3/z-2, find f[k]....
If F(z)=\frac{z+3}{z-2}, find f[k].
Note that we can write F(z) as
\frac{z+3}{z-2}=\frac{z}{z-2}+\frac{3}{z-2}The reason for this choice is that quantities like the first on the r.h.s. appear in Table 22.2. From this table we find
\mathcal{Z}\left\{2^k\right\}=\frac{z}{z-2}and write
\mathcal{Z}^{-1}\left\{\frac{z}{z-2}\right\}=2^kwhere \mathcal{Z}^{-1} denotes the inverse z transform. Also,
\begin{aligned} \frac{3}{z-2} & =\frac{3}{z} \frac{z}{z-2} \\ & =3 z^{-1} \frac{z}{z-2} \end{aligned}Using the second shift property we see that
\mathcal{Z}\left\{3(2)^{k-1} u[k-1]\right\}=\frac{3}{z} \frac{z}{z-2}so that
\begin{aligned} \mathcal{Z}^{-1}\left\{\frac{z+3}{z-2}\right\} & =\mathcal{Z}^{-1}\left\{\frac{z}{z-2}\right\}+\mathcal{Z}^{-1}\left\{3 z^{-1} \frac{z}{z-2}\right\} \\ & =2^k+3(2)^{k-1} u[k-1] \\ & = \begin{cases}1 & k=0 \\ 2^k+3(2)^{k-1} & k=1,2, \ldots\end{cases} \\ & = \begin{cases}1 & k=0 \\ 2(2)^{k-1}+3(2)^{k-1} & k=1,2, \ldots\end{cases} \\ & = \begin{cases}1 & k=0 \\ 5(2)^{k-1} & k=1,2, \ldots\end{cases} \end{aligned}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}