how to take a second order transfer function without any zeros and represent it using two first order transfer functions
For $|\zeta| \le 1$, let $\zeta= \cos\theta$, so $\theta=\mathrm{arccos}\,\zeta$
$$\begin{align*}g(s) &amp;= K\frac{\omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2}\\
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&amp;= K\frac{\omega_n^2}{s^2 + 2\omega_n s \cos\theta+ \omega_n^2(\cos^2\theta +\sin^2\theta)}\\
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&amp;= K\frac{\omega_n^2}{(s + \omega_n \cos\theta)^2+ \omega_n^2\sin^2\theta}\\
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&amp;= \sqrt{K}\frac{\omega_n}{s + \omega_n \cos\theta + j\omega_n\sin\theta} \cdot \sqrt{K}\frac{\omega_n}{s + \omega_n \cos\theta  -j\omega_n\sin\theta} \\
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&amp;= \sqrt{K}\frac{\omega_n}{s + \omega_n e^{j\theta}} \cdot \sqrt{K}\frac{\omega_n}{s + \omega_n e^{-j\theta}} \\
\\
\end{align*}$$
For $|\zeta| \ge 1$, since $\cosh\theta \ge 1$, let $\zeta= \pm\cosh\theta$, so $\theta=\mathrm{acosh}\,\left(\pm\zeta\right)$
$$\begin{align*}g(s) &amp;= K\frac{\omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2}\\
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&amp;= K\frac{\omega_n^2}{s^2 \pm 2\omega_n s \cosh\theta+ \omega_n^2(\cosh^2\theta -\sinh^2\theta)}\\
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&amp;= K\frac{\omega_n^2}{(s \pm \omega_n \cosh\theta)^2- \omega_n^2\sinh^2\theta}\\
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&amp;= \sqrt{K}\frac{\omega_n}{s \pm \omega_n \cosh\theta + \omega_n\sinh\theta} \cdot \sqrt{K}\frac{\omega_n}{s \pm \omega_n \cosh\theta  -\omega_n\sinh\theta} \\
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&amp;= \sqrt{K}\frac{\omega_n}{s \pm \omega_n e^{\theta}} \cdot \sqrt{K}\frac{\omega_n}{s \pm \omega_n e^{-\theta}} \\
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\end{align*}$$