Finding partial fractions of $\frac{z^3+2z^2-2z}{(z-2)(z^2+2}$ and/or using Cauchy's formula to solve

Question

I am trying to find the inverse z-transform of

\begin{equation} x(z)=\frac{z^3+2z^2-2z}{(z-2)(z^2+2)} \end{equation}

and for this we need to get partial fractions. I have tried multiple approaches, the first

$z^3+2z^2-2z:z^3-2z^2+2z-4=1+\frac{4z^2-4z-4}{(z-2)(z^2+2)}$ which can be further separated as:

$1+4\frac{z^2(z-1)}{(z-2)(z^2+2)}-\frac{4}{(z-2)(z^2+2)}$

But here , nothing cancels out, so with those denominators, nothing can be made out for an inverse z-transform.

Then I tried

\begin{equation} \frac{z^3+2z^2-2z}{(z-2)(z^2+2)}=\frac{z(z^2+2z-2)}{(z-2)(z^2+2)} \end{equation}

\begin{equation} \frac{z(z^2+2z-2)}{(z-2)(z^2+2)}=\frac{z^2(z+2)}{(z-2)(z^2+2)}-\frac{2z}{(z-2)(z^2+2)} \end{equation}

But here also, no part of the function can be simplified further.

I was tempted to use Cauchy integral formula on for higher order functions, on

\begin{equation} x(z)=\frac{z^3+2z^2-2z}{(z-2)(z^2+2)} \end{equation}

but can that be done?

Thanks

UPDATE;

taking the partial fractions from the decomposition:

$1+\frac{2}{z-2}+\frac{2z}{(z^2+2)}$, we get the inverse z.transform by the table:

Inverse z-transform of 1:

$1=\delta_n $ for $ \{\delta_0=1, \ \ \delta_{n\ne0}=0, \}$

$\frac{2}{z-2}$ For this, I couldn't find a inverse z-transform from the table, however I did a calculation of it using Cauchy's theorem

$\frac{2}{z-2}\rightarrow Res(2)=\lim_{z\rightarrow 2}\frac{2z^{n-1}}{z-2}=\frac{(z-2)2z^{n-1}}{z-2}=2^n$

Same for the third term, I couldn't find an inverse z-transform from the table, hence:

$\frac{2z}{(z^2+2)}\rightarrow Res(\sqrt{2}i)=\lim_{z\rightarrow \sqrt{2}i}\frac{2zz^{n-1}}{(z^2+2)}=\frac{2z}{(z+2i)}=\frac{2(\sqrt{2}i)^n}{(2\sqrt{2}i)}=( i)^{n-1}$

and the second pole:

$\frac{2z}{(z^2+2)}\rightarrow Res(-\sqrt{2}i)=\lim_{z\rightarrow -\sqrt{2}i}\frac{2zz^{n-1}}{(z^2+2)}=\frac{2z}{(z-2i)}=\frac{2(-\sqrt{2}i)^n}{(-2\sqrt{2}i)}= (-1)^n(i)^{n-1}$

but this last transform is not correct according to WA

https://www.wolframalpha.com/input?i=Z-transform+calculator&assumption=%7B%22F%22%2C+%22ZTransformCalculator%22%2C+%22transformfunction%22%7D+-%3E%222z%2F%28z%5E2%2B2%29%22&assumption=%7B%22F%22%2C+%22ZTransformCalculator%22%2C+%22variable1%22%7D+-%3E%22n%22&assumption=%7B%22F%22%2C+%22ZTransformCalculator%22%2C+%22variable2%22%7D+-%3E%22z%22

Answer 1

$$ \frac{z^3+2z^2-2z}{(z-2)(z^2+2)} = 1 + {2 \over z-2} + {2z \over z^2+2} $$ You know that the solution is of the form $$ a + {b \over z-2} + {cz+d \over z^2+2} $$ Reduce to the same denominator, identify with your initial fraction, deduce linear relations between a,b,c,d, solve.

You can be a bit faster, e.g. by noticing that a is the limit of the fraction when $z$ goes to infinity, and thus should be 1. Also you can get $b$ by multiplying the fraction by $z-2$ and letting $z$ goes to 2.

Note also that you can use a CAS such as maxima to check your result.

(%i1) partfrac((z^3+2*z^2-2*z)/(z-2)/(z^2+2), z);

$$1 + {2 \over z-2} + {2z \over z^2+2}$$