# how to take a second order transfer function without any zeros and represent it using two first order transfer functions

does anybody know how to take a second order transfer function with no zeros (no s terms in the numerator):

$$g(s) = K\frac{\omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2}$$

and represent it using two first order transfer functions.

the layout for my second order transfer function is

     (number)/((number)s^2 + (number)s + (number))


this second order transfer function was previously in the form

(number)/((number) s + (number))^2 before i expanded the brackets

• Not clear, you want to represent G(s) by 2 1st-order transfer function having the same denominator ? – Ben Mar 22 at 19:12
• and generally, you simply need to factorize the polynomial. $$\frac{-b± \sqrt{b^2-4ac}}{2a}$$ – Ben Mar 22 at 19:13

For $$|\zeta| \le 1$$, let $$\zeta= \cos\theta$$, so $$\theta=\mathrm{arccos}\,\zeta$$
\begin{align*}g(s) &= K\frac{\omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2}\\ \\ &= K\frac{\omega_n^2}{s^2 + 2\omega_n s \cos\theta+ \omega_n^2(\cos^2\theta +\sin^2\theta)}\\ \\ &= K\frac{\omega_n^2}{(s + \omega_n \cos\theta)^2+ \omega_n^2\sin^2\theta}\\ \\ &= \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} \\ \\ &= \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) &= K\frac{\omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2}\\ \\ &= K\frac{\omega_n^2}{s^2 \pm 2\omega_n s \cosh\theta+ \omega_n^2(\cosh^2\theta -\sinh^2\theta)}\\ \\ &= K\frac{\omega_n^2}{(s \pm \omega_n \cosh\theta)^2- \omega_n^2\sinh^2\theta}\\ \\ &= \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} \\ \\ &= \sqrt{K}\frac{\omega_n}{s \pm \omega_n e^{\theta}} \cdot \sqrt{K}\frac{\omega_n}{s \pm \omega_n e^{-\theta}} \\ \\ \end{align*}
• @Ben Yeah, I made the assumption that $\xi \in [-1,1]$. $\xi = \pm \cosh \theta$ could be used for $\xi$ not in that range. – Andy Walls Apr 3 at 1:44
The roots of the quadratic equation $$s^2 + 2\xi \omega_n + \omega_n^2=0$$ are $$s_{1,2} = -\xi w_n \pm \omega_n \sqrt{\xi^2 - 1} = -\xi w_n \pm i\omega_n \sqrt{1 - \xi^2}$$ where we define the damping frequency $$\omega_d = \omega_n \sqrt{1 - \xi^2}$$ when $$\xi < 1$$. Therefore, the usual second order transfer function $$\frac{\omega_n^2}{s^2 + 2\xi \omega_n + \omega_n^2=0} = \frac{\omega_n}{s-(-\xi w_n + i \omega_d)} \cdot \frac{\omega_n}{s-(-\xi w_n - i \omega_d)}$$.