z transform of $n^2 (1/3)^n$

Question

How can i find the z transform of $n^2(1/3)^n$ using the properties? I refer to the two sided z transfrom. There is a property where $nx[n]$ becomes $-zdX(z)/z$ after z-transform is applied.But how can i find then the $x[n]=(1/3)^n$ without using the definition of z transform and using only the properties?

Answer 1

You can look at it from the following perspective:

$$f[n]=n^2\left(\frac{1}{3}\right)^n=n\left(n\left(\frac{1}{3}\right)^n\right)$$

Applying the $Z$-Transform yields:

$$\mathcal{Z}\left\{n\left(n\left(\frac{1}{3}\right)^n\right)\right\}=-z\frac{d}{dz}\mathcal{Z}\left\{n\left(\frac{1}{3}\right)^n\right\}=-z\frac{d}{dz}\left[-z\frac{d}{dz}\mathcal{Z}\left\{\left(\frac{1}{3}\right)^n\right\}\right]$$

Basically every $Z$-Transform table that you take will present the following $a^n=\frac{z}{z-a}$, hence:

\begin{align} -z\frac{d}{dz}\left[-z\frac{d}{dz}\mathcal{Z}\left\{\left(\frac{1}{3}\right)^n\right\}\right]&=-z\frac{d}{dz}\left[-z\frac{d}{dz}\left(\frac{z}{z-\frac{1}{3}}\right)\right]\\ &=-z\frac{d}{dz}\left[\frac{\frac{1}{3}z}{\left(z-\frac{1}{3}\right)^2}\right]\\ &=\frac{z}{3}\frac{z+\frac{1}{3}}{\left(z-\frac{1}{3}\right)^3} \end{align}

Or, if you prefer:

$$\mathcal{Z}\left\{f[n]\right\}=\frac{3z\left(3z+1\right)}{\left(3z-1\right)^3}$$