# Unique solution for the damping factor in second order system

Given an underdamped ($$\ 0 < \zeta < 1 \$$) second order harmonic oscillator system with canonical form $$G(s) = \frac{\omega_n^2}{s^2 + 2 \zeta \omega_n + \omega_n^2}$$.

we know that the maximum of the bode plot is given by $$|G(j\omega_d)| = \alpha = \frac{1}{2 \zeta \sqrt{1 - \zeta^2}}$$.

Rearranging we get $$\zeta^4 - \zeta^2 + \frac{1}{4 \alpha^2} = 0$$. Substituting $$\ x = \zeta^2 \$$ we can apply the quadratic formula $$x_{1/2} = \frac{1}{2} \pm \sqrt{ \frac{1}{4} - \frac{1}{4 \alpha^2} }$$ and consequently $$\zeta = \sqrt{ x_{1/2} }$$.

But of course, $$\ \zeta \$$ is unique, so how do we know whether to take $$\ + \$$ or $$\ - \$$ ?

I know that this is usually visible from the plot (if $$\ \zeta < \frac{1}{\sqrt{2}} \$$ then we have a peak), but is there a more mathematical reasoning?

• You've got something messed up somewhere mathematically. if alpha is a peak amplitude then it is greater than 1 hence $\sqrt{0.25 - \alpha}$ is nonsense. Nov 6, 2020 at 15:24
• @Andyaka fixed. thanks. Nov 6, 2020 at 15:39

If $$\\alpha\$$ is very large (a highly peaky 2nd order low pass filter) the $$\x\$$ formula tends towards this: -
$$x = \dfrac{1}{2} \pm\sqrt{\dfrac{1}{4} - \dfrac{1}{4\cdot\infty}} = \dfrac{1}{2} \pm\sqrt{\dfrac{1}{4} }$$
And, if $$\\alpha\$$ is very large, we know that $$\x\$$ (and $$\\zeta\$$) have to be very small hence, the sign cannot be positive because it would make $$\x\$$ be very nearly unity. And that would be incorrect.
$$\boxed{\text{So the sign has to be negative}}$$