Cannot find the inverse of Z-transform.

Question

Find the invers of $Z$-transform of $X(z)=\dfrac{z}{(z-2)(z^2+6z+9)}$.

I try as below.

Let \begin{align*} X(z)&=\dfrac{z}{(z-2)(z^2+6z+9)} =\dfrac{z}{(z-2)(z+3)^2}\\ &=\dfrac{A}{z-2}+\dfrac{Bz+C}{(z+3)^2}\\ &=\dfrac{A(z+3)^2+(Bz+C)(z-2)}{(z-2)(z+3)^2}\\ &=\dfrac{A(z^2+6z+9)+Bz^2-2Bz+Cz-2C}{(z-2)(z+3)^2}\\ &=\dfrac{(A+B)z^2+(6A-2B+C)z+(9A-2C)}{(z-2)(z+3)^2} \end{align*}

Now we have system of linear equation \begin{align*} A+B&=0\\ 6A-2B+C&=1\\ 9A-2C&=0 \end{align*} and now we have $A= \dfrac{2}{25}$, $B= -\dfrac{2}{25}$, and $C= \dfrac{9}{25}$.

We have \begin{align*} X(z)&=\dfrac{2}{25}\cdot \dfrac{1}{z-2}+\dfrac{-\dfrac{2}{25}z+\dfrac{9}{25}}{(z+3)^2}\\ &= \dfrac{2}{25}\cdot \dfrac{1}{z-2}-\dfrac{2}{25} \dfrac{z}{(z+3)^2}+\dfrac{9}{25} \dfrac{1}{(z+3)^2}. \end{align*}

Now I don't know what the inverse Z-transform of $\dfrac{1}{z-2}$, $\dfrac{z}{(z+3)^2}$, and $\dfrac{1}{(z+3)^2}$.

I just have this Z-transform table

Any one can give me hint for this problem?

Answer 1

Hints: $$\frac{1}{z-2} = z^{-1}\cdot\frac{z}{z-2}$$

$$\frac{z}{(z+3)^2} = \frac{-1}3\cdot\frac{(-3z)}{(z-(-3))^2}$$

$$\frac{1}{(z+3)^2} = \frac{-z^{-1}}{3}\cdot\frac{(-3z)}{(z-(-3))^2}$$

Then use shifting property

$$\mathcal{Z}\{x(n-m)\} = z^{-m}X(z) $$