Second-order step response and damping

The second-order model in standard form is

$$\frac{1}{\omega_n^2}\ddot{q}_o + \frac{2\zeta}{\omega_n}\dot{q}_o + q_o = K q_i$$

For a step input $q_i = q_s$, $q_o(0) = \dot q_o(0) = 0$. Let $y = q_o/(Kq_s)$, so $y$ tends to 1.

Underdamped case ($0 < \zeta < 1$)

The characteristic roots are $s = -\zeta\omega_n \pm j\omega_n\sqrt{1-\zeta^2}$. With damped frequency $\omega_d = \omega_n\sqrt{1-\zeta^2}$, the solution is $y = 1 + e^{-\zeta\omega_n t}(A\cos\omega_d t + B\sin\omega_d t)$. The conditions $y(0) = 0$ and $\dot y(0) = 0$ give $A = -1$ and $B = -\zeta\omega_n/\omega_d$:

$$y(t) = 1 - \frac{e^{-\zeta\omega_n t}}{\sqrt{1-\zeta^2}}\sin\left(\omega_d t + \phi\right), \qquad \phi = \cos^{-1}\zeta$$

Setting $\dot y = 0$ gives peak time $t_p = \pi/\omega_d$, and the peak overshoot

$$M_p = \exp\left(\frac{-\pi\zeta}{\sqrt{1-\zeta^2}}\right)$$

Other cases

• Critically damped ($\zeta = 1$): $y = 1 - (1 + \omega_n t)e^{-\omega_n t}$. Fastest approach to the final value without overshoot.
• Overdamped ($\zeta > 1$): $y = 1 - \dfrac{e^{-\zeta\omega_n t}}{2\sqrt{\zeta^2-1}}\left[\dfrac{e^{\omega_n\sqrt{\zeta^2-1}\,t}}{\zeta - \sqrt{\zeta^2-1}} - \dfrac{e^{-\omega_n\sqrt{\zeta^2-1}\,t}}{\zeta + \sqrt{\zeta^2-1}}\right]$. No overshoot but slow.

 y
 1.4|  .-.            zeta = 0.2
 1.0|-/---`-.__.---- zeta = 0.7
    | /  _.--------- zeta = 1
    |/ .'  ........  zeta = 2 (slow)
    +----------------> w_n t

Why $\zeta \approx 0.7$

At $\zeta = 0.7$ the overshoot is only 4.6%, the 2% settling time is near its minimum, and the amplitude response is flat within 1% up to about $0.4\,\omega_n$ with a phase lag nearly proportional to frequency (little distortion). Hence most instruments use $\zeta = 0.6$ to $0.7$.