A second order control theory function looks like:
$$\text{H}_{(s)}=\frac{\text{K}_p}{\frac{1}{\omega_0^2}\cdot s^2+\frac{2\beta}{\omega_0}\cdot s+1}$$
Now I've got the function, with $a,r,k\in\mathbb{R^+}$ and $k$ is the variabele:
$$\text{H}_{(s)}=\frac{kr}{(ars+1)^2+kr}$$
And I alrady now that $\text{K}_p=\frac{1}{1+\frac{1}{kr}}$
But how do I get my function in the normal form? (So I would like to find $\omega_0$ and $\beta$)
This is plain algebra.
$$H(s)=\frac{kr}{(ars+1)^2+kr}$$
First $(ars+1)^2 = a^2r^2s^2 + 2ars + 1$. We get:
$$H(s) = \frac{kr}{a^2r^2s^2 + 2ars + (1 + kr)}$$
Multiply the nominator and denominator with $\frac{1}{1 + kr}$. We get:
$$H(s) = \frac{\frac{kr}{1 + kr}}{\frac{a^2r^2}{1 + kr}s^2 + 2\frac{ar}{1 + kr}s + 1}$$
As such,
$$K_p = \frac{kr}{1 + kr} = \frac{1}{1 + \frac{1}{kr}}$$
$$\frac{1}{\omega_0^2} = \frac{a^2r^2}{1 + kr} \Leftrightarrow \omega_o = \sqrt{\frac{1 + kr}{a^2r^2}}$$
$$2\frac{\beta}{\omega_0} = 2\frac{ar}{1 + kr} \Leftrightarrow\beta = \frac{ar}{(1 + kr)}\omega_0$$