Question 3

Using the definition of z-transform, show that

(a) $\gamma^{k-1} u \left [k-1 \right ] {\Longrightarrow} \frac{\large 1}{\large z- \gamma }$

(b) $u \left [ k-m \right ] {\Longrightarrow} \frac{\large z}{\large z^{m}(\large z-1)}$

Solution.

(a) Let $x\left [k \right] = \gamma^{k-1} u \left [k-1 \right ],$ then from the definition of z-transform

$$\begin{eqnarray*}X \left (z \right) & = & \sum_{ k= -\infty }^{ \infty } \gamma^{k-1}u[k-1]z^{-k}\\ \end{eqnarray*}$$

Now let k - 1 = m, then the above equation is equivalent to

$$\begin{eqnarray*}X \left (z \right) & = & \sum_{ m= -\infty }^{ \infty } \gamma^... ...= & \frac{z^{-1}}{1-\gamma z^{-1}}\\ & = & \frac{1}{z-\gamma} \end{eqnarray*}$$

(b) From the definition of z-transform,

$$\begin{eqnarray*}X \left (z \right) & = & \sum_{ k= -\infty }^{ \infty } u[k-m]z^{-k}\\ \end{eqnarray*}$$

Let k - m = n, the above equation is equivalent to

$$\begin{eqnarray*}X \left (z \right) & = & \sum_{ n= -\infty }^{ \infty } u[n]z^{... ...ft(1-z^{-1} \right)}\\ & = & \frac{z}{z^m \left( z-1 \right)} \end{eqnarray*}$$