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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.

Dajuvaja
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    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Apr 28 '23 at 03:27
  • Perhaps a link to "$z$-transform", since it is unknown to most mathematicians (and there is more than one possibility). Is this the question... Find coefficients $a_n$ so that $$\frac{1}{\left(z-\frac{1}{2}\right)^3} =\sum_{n=-\infty}^{+\infty} a_n z^{-n}$$ – GEdgar Apr 28 '23 at 07:06
  • https://signalprocessingsystems.netlify.app/disciplines/discrete/discretesignalprocessing_transforms_ztransform_inverse/ – Claude Leibovici Apr 28 '23 at 10:30

1 Answers1

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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:

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

gpmath
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