Find the sequence whose z transform is [latex]\frac{1}{z^2(z-1)^2}[/latex].

The expression [latex]\frac{1}{z^2(z-1)^2}[/latex] does not appear in the table of transforms, but we observe that [latex]\frac{1}{z^2(z-1)^2}=\frac{1}{z^3} \frac{z}{(z-1)^2}[/latex] and [latex]\frac{z}{(z-1)^2}[/latex] does appear. It follows from Table 22.2 that [latex]\mathcal{Z}\{k\}=\frac{z}{(z-1)^2}[/latex] From the second shift property, [latex]z^{-3} \frac{z}{(z-1)^2}[/latex] is the z transform of (k — 3)u[k — 3].

Question 22.29: Find the sequence whose z transform is 1/z²(z-1)²....

Engineering Mathematics A Foundation for Electronic, Electrical, Communications and Systems Engineers [1040294]

Find the sequence whose z transform is $\frac{1}{z^2(z-1)^2}$ .

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The expression $\frac{1}{z^2(z-1)^2}$ does not appear in the table of transforms, but we observe that

\frac{1}{z^2(z-1)^2}=\frac{1}{z^3} \frac{z}{(z-1)^2}

and $\frac{z}{(z-1)^2}$ does appear. It follows from Table 22.2 that

\mathcal{Z}\{k\}=\frac{z}{(z-1)^2}

From the second shift property, $z^{-3} \frac{z}{(z-1)^2}$ is the z transform of (k — 3)u[k — 3].

Table 22.2
The z transforms of some common functions
$\begin{matrix}\hline f[k]&f(z)&f[k]&f(z)\\ \hline \delta[k]= \begin{cases}1 & k=0 \\ 0 & k \neq 0\end{cases}&1&k^2&\frac{z(z+1)}{(z-1)^3}\\ u[k]= \begin{cases}1 & k \geqslant 0 \\ 0 & k<0\end{cases}&\frac{z}{z-1}&k^3&\frac{z(z^2+4z+1)}{(z-1)^4}\\ k&\frac{z}{(z-1)^2}&\sin ak&\frac{z\sin a}{z^2-2z\cos a+1}\\ e^{-ak}&\frac{z}{z-e^{-a}}&\cos ak&\frac{z(z-\cos a)}{z^2-2z\cos a+1}\\a^k&\frac{z}{z-a}&e^{-ak}\sin bk&\frac{ze^{-a}\sin b}{z^2-2ze^{-a}\cos b+e^{-2a}}\\ ka^k&\frac{az}{(z-a)^2}&e^{-ak}\cos bk&\frac{z^2-ze^{-a}\cos b}{z^2-2ze^{-a}\cos b+e^{-2a}}\\k^2a^2&\frac{az(z+a)}{(z-a)^3}\\ \hline\end{matrix}$