Following is a question about the z transform with: $$ (-\frac{1}{3})^n \, u(-n-2). $$ Now, I know that a transform pair similar to this is: $$ -\alpha^n \, u(-n-1) = \frac{1}{1-\alpha z^{-1}}. $$ Now using that property what I get is: $$ \frac{1}{1-(1/3)z^{-1}}. $$ Since the shift is of $-1$ so using that property my final answer is: $$ (z^{-1})(\frac{1}{1-(1/3)z^{-1}}). $$
Is it correct?
As written the result is incorrect. Using the definition of the unit step function then it is seen that $$ u(-n -2) = \begin{cases} 1 && n\leq -2 \\ 0 && n > -2 \end{cases} $$ and yields $$\sum_{n=0}^{\infty} a^n \, u(-n-2) \, z^{-n} = 0.$$
If the unit step function is $u(n-2)$, $$ u(n -2) = \begin{cases} 1 && n \geq 2 \\ 0 && n < 2 \end{cases}, $$ then the transform is as follows: \begin{align} \sum_{n=0}^{\infty} a^n \, u(n-2) \, z^{-n} &= \sum_{n=2}^{\infty} \left(\frac{a}{z} \right)^n = \frac{a^2}{z^2} \, \frac{1}{1 - \frac{a}{z}} = \frac{a^2}{z(z-a)}. \end{align}