# using z transform pairs to solve the question.

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?

• @LordSharktheUnknown can you help me on this one? Jan 6 '19 at 16:15

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}