Show that the z-transform for the unit step sequence {xk}={1, a, a^2, a^3, a^4, …}={a^k} is give

step 1 So in this question we have to find Z transform for unit step sequence. So, so Z transform of x

Show that the z-transform for the unit step sequence...

Show that the z-transform for the unit step sequence $\left\{x_{k}\right\}=\left\{1, a, a^{2}, a^{3}, a^{4}, \ldots\right\}=\left\{a^{k}\right\}$ is given by $\frac{z}{z-a}$ The z-transform of $\left\{x_{k}\right\}$ is given by: $$ \begin{aligned} &z\left\{x_{k}\right\}=z\left\{a^{k}\right\}=\sum_{k=0}^{\infty} \frac{a^{k}}{z^{k}}=\sum_{k=0}^{\infty}\left(\frac{a}{z}\right)^{k} \\ &=\left(\frac{a}{z}\right)^{0}+\left(\frac{a}{z}\right)^{1}+\left(\frac{a}{z}\right)^{2}+\left(\frac{a}{z}\right)^{3}+\left(\frac{a}{z}\right)^{4}+\ldots \\ &=1+\frac{a}{z}+\left(\frac{a}{z}\right)^{2}+\left(\frac{a}{z}\right)^{3}+\left(\frac{a}{z}\right)^{4}+\ldots \end{aligned} $$ Comparing this with the series expansion of $\frac{1}{1-x}=1+x+x^{2}+x^{3}+\ldots$ which is valid for $|x|<1$ i.e. equation (2) above, shows that: $$ \begin{aligned} F(z) &=1+\frac{a}{z}+\left(\frac{a}{z}\right)^{2}+\left(\frac{a}{z}\right)^{3}+\left(\frac{a}{z}\right)^{4}+\ldots \\ &=\frac{1}{1-\frac{a}{z}} \text { provided }\left|\frac{a}{z}\right|<1 \end{aligned} $$ Hence, $\frac{1}{1-\frac{a}{z}}=\frac{1}{\frac{z-a}{z}}=\frac{z}{z-a}$ and $\mathbf{F}(\mathbf{z})=\frac{z}{z-a}$ provided $|z|>|a|$

Show that the z-transform for the unit step sequence $\left\{x_{k}\right\}=\left\{1, a, a^{2}, a^{3}, a^{4}, \ldots\right\}=\left\{a^{k}\right\}$ is given by $\frac{z}{z-a}$
The z-transform of $\left\{x_{k}\right\}$ is given by:
$$
\begin{aligned}
&z\left\{x_{k}\right\}=z\left\{a^{k}\right\}=\sum_{k=0}^{\infty} \frac{a^{k}}{z^{k}}=\sum_{k=0}^{\infty}\left(\frac{a}{z}\right)^{k} \\
&=\left(\frac{a}{z}\right)^{0}+\left(\frac{a}{z}\right)^{1}+\left(\frac{a}{z}\right)^{2}+\left(\frac{a}{z}\right)^{3}+\left(\frac{a}{z}\right)^{4}+\ldots \\
&=1+\frac{a}{z}+\left(\frac{a}{z}\right)^{2}+\left(\frac{a}{z}\right)^{3}+\left(\frac{a}{z}\right)^{4}+\ldots
\end{aligned}
$$
Comparing this with the series expansion of $\frac{1}{1-x}=1+x+x^{2}+x^{3}+\ldots$ which is valid for $|x|<1$ i.e. equation (2) above, shows that:
$$
\begin{aligned}
F(z) &=1+\frac{a}{z}+\left(\frac{a}{z}\right)^{2}+\left(\frac{a}{z}\right)^{3}+\left(\frac{a}{z}\right)^{4}+\ldots \\
&=\frac{1}{1-\frac{a}{z}} \text { provided }\left|\frac{a}{z}\right|<1
\end{aligned}
$$
Hence, $\frac{1}{1-\frac{a}{z}}=\frac{1}{\frac{z-a}{z}}=\frac{z}{z-a}$ and $\mathbf{F}(\mathbf{z})=\frac{z}{z-a}$ provided $|z|>|a|$

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