Determine the z-transfom for the. unit step sequence $\left\{x_{k}\right\}=\{0,1,2,3,4, \ldots,\}=(k$, and show that it is equivalent to $\frac{2}{(z-1)^{2}}$
The $z$-transform of $\left\{u_{k}\right\}$ is given by:
$$
\begin{aligned}
Z\left\{x_{k} \mid\right.&=F(z)=\sum_{k=0}^{\infty} \frac{x_{\lambda}}{z^{2}}=\sum_{k=0}^{\infty} \frac{k}{z^{k}} \\
&=\frac{0}{z^{0}}+\frac{1}{z^{1}}+\frac{2}{z^{2}}+\frac{3}{z^{3}}+\frac{4}{z^{4}}+\ldots \\
&=0+\frac{1}{z}+\frac{2}{z^{2}}+\frac{3}{z^{3}}+\frac{4}{z^{4}}+\ldots
\end{aligned}
$$
It was shown earlier, equation (2), that
$$
\frac{1}{1-x}=(1-x)^{-1}=1+x+x^{2}+x^{3}+\ldots
$$
Now $\quad \frac{d}{d x}\left(1+x+x^{2}+x^{3}+x^{4}+\ldots\right)$
$$
=1+2 x+3 x^{2}+4 x^{3}+\ldots
$$
and $\frac{d}{d x}\left[(1-x)^{-1}\right]=-(1-x)^{-2}(-1)=\frac{1}{(1-x)^{2}}$ using the function of a function rule,
i.e. $\quad 1+2 x+3 x^{2}+4 x^{3}+\ldots=\frac{1}{(1-x)^{2}}$
Comparing equations (3) and $(4)$ shows that by multiplying $F(z)$ by $z$
then $z F(z)=z\left(0+\frac{1}{z}+\frac{2}{z^{2}}+\frac{3}{z^{1}}+\frac{4}{z^{4}}+\ldots\right)$ $\quad=1+\frac{2}{z}+\frac{3}{z^{2}}+\frac{4}{z^{3}}+\ldots=\frac{1}{\left(1-\frac{1}{z}\right)^{2}}$
$$
\begin{gathered}
\begin{aligned}
F(z)=\frac{1}{z\left(1-\frac{1}{z}\right)^{2}}=& \frac{1}{z\left(\frac{z-1}{z}\right)^{2}}=\frac{1}{z \frac{(z-1)^{2}}{z^{2}}} \\
=& \frac{1}{\frac{(z-1)^{2}}{z}}=\frac{z}{(z-1)^{2}} \\
\text { Hence, } \mathbf{Z}\left(x_{2}\right)=F(z) &=0+\frac{1}{z}+\frac{2}{z^{2}}+\frac{3}{z^{3}}+\frac{4}{z^{4}}+\ldots \\
&=\frac{z}{(z-1)^{2}}
\end{aligned}
\end{gathered}
$$
From the results obtained in Problems 1 to 4 , together with some additional results, a summary of some 2 : transforms is shown in Table $79.1$, on page 828 , which may now be accepted and used.
Here are some further problems using Table 79.1.