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?
Best Answer
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.
$$\boxed{\text{So the sign has to be negative}}$$