I have $n^2 3^n$, and want to find the z-transform of it.
I take
$$n^2 3^n= n\cdot n 3^n$$
and we can see from a z-transform table that
$$3^n \rightarrow\frac{z}{z-3}$$
$$n3^{n-1} \rightarrow\frac{z}{(z-3)^2}$$
$$n \rightarrow\frac{z}{(z-1)^2}$$
and these are awfully familiar with the correct result:
\begin{equation} n^2 3^n \rightarrow \frac{3 z (3 + z)}{(z-3)^3} \end{equation}
An alternative is to use the definition \begin{equation} V(z)=\sum_{n=-\infty}^\infty v(n)z^{-n}=\sum_{n=-\infty}^\infty n^2 3^n z^{-n}=\sum_{n=-\infty}^\infty \frac{n^2 3^n}{ z^{n}} \end{equation}
But this is not the same as the answer by WA.
How is it calculated?
Thanks
I found a solution, thanks to some help. The table of transforms states that $nx_n\rightarrow -zX'(z)$.
Putting $x_n=n3^{n}$, which is the function to transform specifically, we use the rule
$na^{n-1}\rightarrow \frac{z}{(z-1)^2}$ on it.
Re-writing:
$x_n=n3^{n}=3\cdot n3^{n-1}$, we use the z-transform rule given above, and obtain:
$3\cdot n3^{n-1}\rightarrow X(z)=\frac{3z}{(z-3)^2}$.
We then need to find $X'(z)$. So,
$\frac{3z}{(z-3)^2}'= -3\frac{3 + z}{(x-3)^3}$
The first z-transform rule at the top gives then:
\begin{equation} nx_n=n\cdot n3^{n}\rightarrow -zX'(z)\rightarrow z\bigg(\frac{9 + 3z)}{(z-3)^3}\bigg) \end{equation}