Servo differential equation: ζ, frequency and overshoot

Take the Laplace transform with zero initial conditions and compare with the standard second order form $\omega_n^2/(s^2 + 2\zeta\omega_n s + \omega_n^2)$.

Transfer function

$$\frac{d^2y}{dt^2} + 8\frac{dy}{dt} = 64(u - y) \;\Rightarrow\; (s^2 + 8s + 64)Y(s) = 64U(s)$$ $$\frac{Y(s)}{U(s)} = \frac{64}{s^2 + 8s + 64}$$

Parameters

$$\begin{aligned} \omega_n^2 &= 64 \;\Rightarrow\; \omega_n = 8\ \text{rad/s} \\ 2\zeta\omega_n &= 8 \;\Rightarrow\; \zeta = \frac{8}{16} = 0.5 \end{aligned}$$

Frequency of sustained oscillation

Sustained (non-decaying) oscillation would occur with zero damping, at the undamped natural frequency:

$$\omega_n = 8\ \text{rad/s} \quad (f_n = \omega_n/2\pi = 1.27\ \text{Hz})$$

With the actual damping ($\zeta = 0.5$), the step response oscillates and decays at the damped frequency:

$$\omega_d = \omega_n\sqrt{1 - \zeta^2} = 8\sqrt{0.75} = 6.928\ \text{rad/s}$$

Maximum overshoot

$$M_p = e^{-\pi\zeta/\sqrt{1-\zeta^2}} = e^{-\pi(0.5)/0.866} = e^{-1.814} = 0.1630 = 16.30\%$$

Answer: frequency of sustained oscillation $\omega_n = 8$ rad/s (damped frequency $\omega_d = 6.93$ rad/s), $\zeta = 0.5$, $M_p = 16.3\%$.