Solve the difference equation y[k+1]-3 y[k]=0, y[0]=4.
Taking the z transform of both sides of the equation we have
\mathcal{Z}\{y[k+1]-3 y[k]\}=\mathcal{Z}\{0\}=0since \mathcal{Z}\{0\}=0. Using the properties of linearity we find
\mathcal{Z}\{y[k+1]\}-3 \mathcal{Z}\{y[k]\}=0Using the first shift theorem on the first of the terms on the l.h.s. we obtain
z \mathcal{Z}\{y[k]\}-4 z-3 \mathcal{Z}\{y[k]\}=0Writing \mathcal{Z}\{y[k]\}=Y(z), this becomes
(z-3) Y(z)=4 zso that
Y(z)=\frac{4 z}{z-3}The function on the r.h.s. is the z transform of the required solution. Inverting this, from Table 22.2 we find y[k]=4(3)^k
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}