I'm trying to do the inverse Z-transform of: $$H(Z)=\frac{1+(2\cdot z^{-1})}{1-(0,5z{^-1})}$$
My first thought was to divide the expression into two additions:
$H1(Z) = \frac{1}{1-(0,5z^{-1})}$
$H2(Z) = \frac{2\cdot z^{-1}}{1-(0,5\cdot z^{-1})}$
$h1(n)$ would then be $(0,5)^{n}\cdot u(n)$
I cannot find $h2(n)$
Is this even the right way to go when I want to find $h(n)$? Or should I maybe use the fact that $H(z)=\frac{Y(z)}{X(Z)}$ and work from there? Thanks
You need the basic $\mathcal{Z}-$transform pair
$$\frac{1}{1-az^{-1}}=\frac{z}{z-a}\Longleftrightarrow a^nu(n)$$
(where $u(n)$ is the step function.)
Write $H(z)$ as you did:
$$H(z)=\frac{1}{1-0.5z^{-1}}+2\frac{z^{-1}}{1-0.5z^{-1}}$$
Multiplication by $z^{-1}$ is just a shift by one sample. So the inverse transform of $H(z)$ is given by
$$H(z)\Longleftrightarrow \left(\frac{1}{2}\right)^nu(n)+2\left(\frac{1}{2}\right)^{n-1}u(n-1)$$