Z-Transform of $x(n) = 3^n$

Question

First of all, thank you all for your answers. I know the z transform for

$$ x(n)=3^n \space ; \space n\geqslant 3 $$

or rather

$$ x(n)= 3^n u(n-3) $$$$\begin{align}X(z)&=\sum_{n=-\infty}^{\infty}x[n]z^{-n}\\&=\sum_{n=-\infty}^\infty3^nu(n-3)z^{-n}\\& \text{change of variables: } n'=n-3 \\& =\sum_{n=-\infty}^\infty 3^{(n'+3)} u(n') z^{(-n'-3)}\\& \text{recall: } n'-> n \\&=3^3z^{-3}\sum_{n=0}^{\infty}(3 z^{-1})^{n}\\& X(z) =3^3 z^{-3}\frac{1}{1-3{z^{-1}}},\quad |z|>3 \tag{1}\end{align} $$\begin{align} \\& \text{instead, for the sequence: } x(n)=3^{n} \ ; n\leqslant -3 \\&=\sum_{n=-\infty}^\infty3^nu(-n-3)z^{-n}\\& \text{change of variables: } n'=-n-3 \\& =\sum_{n=-\infty}^\infty 3^{(-n'-3)} u(n') z^{-(-n'-3)}\\& \text{recall: } n'-> n \\&=3^{-3} z^{3}\sum_{n=0}^{\infty}(3^{-1} z)^{n}\\& X(Z)= 3^{-3}z^{3}\frac{1}{1- \frac{z}{3}} \quad |z|<3\tag{2}\end{align} Is it correct? And for @Matt L. I don't understand your mathematical step initial when you write: $$\begin {align} \sum_{n=-\infty}^3 3^nz^{-n} =\sum_{n=-\infty}^0 3^{(n+3)} u(n') z^{-(n+3)} \end {align} $$

Can you write me the generic formula?

Answer 1

The signal $x[n]=3^n$ for $n\le -3$ (and zero otherwise, I assume) can be written as

$$x[n]=3^nu[-n-3]\tag{1}$$

To compute its $\mathcal{Z}$-transform you simply need to use the formula:

$$\begin{align}X(z)&=\sum_{n=-\infty}^{\infty}x[n]z^{-n}\\&=\sum_{n=-\infty}^{-3}3^nz^{-n}\\&=\sum_{n=-\infty}^03^{(n-3)}z^{-(n-3)}\\&=3^{-3}z^{3}\sum_{n=0}^{\infty}\left(\frac{z}{3}\right)^n\\&=3^{-3}z^{3}\frac{1}{1-\frac13 z},\quad |z|<3\tag{2}\end{align}$$

It's important to specify the region of convergence ($|z|<3$), otherwise the $\mathcal{Z}$-transform does not uniquely identify the corresponding sequence $x[n]$.