How can I find the inverse z-transform of $\frac{1}{\left(z-\frac{1}{2}\right)^3}$?

Question

I'm trying to find the inverse z-transform of $Y(z) = \frac{1}{\left(z-\frac{1}{2}\right)^3}$ but I just don't find how. I was trying to adapt the function so I could find it directly by using properties but it was useless. Btw it is my first question on this site so I'm sorry if there are any mistakes with the format.

Answer 1

Let $Y(z)=\frac{1}{\left(z-\frac{1}{2}\right)^3}=\frac{z^{-3}}{\left(1-\frac{1}{2}z^{-1}\right)^3}$

We want to use the known z-transform reference ($\theta(k)$ is the unit step function): $$\mathcal{Z}_z^{-1}\left[\frac{1}{\left(1-\alpha \cdot z^{-1}\right)^m}\right](k)=(-\alpha )^k \theta (k) \binom{-m}{k}$$

With $\alpha=\frac{1}{2}$ and $m=3$ we then get

$$g(k)\mapsto\left(-\frac{1}{2}\right)^k \binom{-3}{k} \theta (k)$$

Finally with time delay $z^{-3}$ we get

$$y(k)=\mathcal{Z}_z^{-1}[Y(z)](k)=g(k-3)=\left(-\frac{1}{2}\right)^{k-3} \binom{-3}{k-3} \theta (k-3)$$

Values for $k=1..10$: $\left\{0,0,1,\frac{3}{2},\frac{3}{2},\frac{5}{4},\frac{15}{16},\frac{21}{32},\frac{7}{16},\frac{9}{32}\right\}$

These values also appear in the Laurent series:

$$Y(z)\simeq\frac{1}{z^{3}}+\frac{3}{2 z^4}+\frac{3}{2 z^5}+\frac{5}{4 z^6}+\frac{15}{16 z^7}+\frac{21}{32 z^8}+\frac{7}{16 z^9}+\frac{9}{32 z^{10}}+O\left(\left(\frac{1}{z}\right)^{11}\right)$$