Question 22.32: Find the sequence whose z transform is F(z) = 2z²-z/(z-5)(z......
Find the sequence whose z transform is
F(z)=\frac{2 z^2-z}{(z-5)(z+4)}The first stage is to divide the expression for F(z) by z. We have
\begin{aligned} & F(z)=\frac{2 z^2-z}{(z-5)(z+4)} \\ & F(z)=\frac{z(2 z-1)}{(z-5)(z+4)} \\ & \frac{F(z)}{z}=\frac{2 z-1}{(z-5)(z+4)} \end{aligned}We now split the r.h.s. expression into partial fractions using the standard techniques discussed in Section 1.7. This gives
\frac{F(z)}{z}=\frac{1}{z-5}+\frac{1}{z+4}Multiplying by z gives
F(z)=\frac{z}{z-5}+\frac{z}{z+4}It is now possible to invert this expression for F(z.) using the z transform table, Table 22.2. This gives
f[k]=5^k+(-4)^kTable 22.2
The z transforms ofsome common functions.
\begin{matrix}f[k]&F(z)&f[k]&f(z)\\\hline \delta[k]=\left\{\begin{matrix}1\quad k=0\\0\quad k\neq 0\end{matrix}\right.&1&k2&{\frac{z(z+1)}{(z-1)^{3}}}\\ u[k]=\left\{\begin{matrix}1\quad k\geq 0\\0\quad k<0\end{matrix}\right.&\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{z\mathbf{e}^{-a}\sin b}{z^{2}-2z\mathrm{e}^{-a}\cos b+\mathrm{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^k&\frac{az(z+a)}{(z-a)^3}\\ \hline\end{matrix}