Underdamped second order unit step response derivation

For $\xi < 1$ (underdamped) the unit step response is a damped sinusoid that settles at 1.

Closed-loop transfer function

$$\frac{C(s)}{R(s)} = \frac{G}{1+G} = \frac{\omega_n^2}{s^2 + 2\xi\omega_n s + \omega_n^2}$$

For $\xi < 1$ the poles are complex:

$$s = -\xi\omega_n \pm j\omega_n\sqrt{1-\xi^2} = -\xi\omega_n \pm j\omega_d, \qquad \omega_d = \omega_n\sqrt{1-\xi^2}$$

Output for a unit step

With $R(s) = 1/s$:

$$C(s) = \frac{\omega_n^2}{s(s^2 + 2\xi\omega_n s + \omega_n^2)}$$

Partial fractions:

$$\begin{aligned} C(s) &= \frac{1}{s} - \frac{s + 2\xi\omega_n}{s^2 + 2\xi\omega_n s + \omega_n^2} \\ &= \frac{1}{s} - \frac{s + \xi\omega_n}{(s + \xi\omega_n)^2 + \omega_d^2} - \frac{\xi\omega_n}{(s + \xi\omega_n)^2 + \omega_d^2} \end{aligned}$$

Using $\mathcal{L}^{-1}\left\{\frac{s+a}{(s+a)^2+\omega^2}\right\} = e^{-at}\cos\omega t$ and $\mathcal{L}^{-1}\left\{\frac{\omega}{(s+a)^2+\omega^2}\right\} = e^{-at}\sin\omega t$:

$$\begin{aligned} c(t) &= 1 - e^{-\xi\omega_n t}\cos\omega_d t - \frac{\xi\omega_n}{\omega_d}e^{-\xi\omega_n t}\sin\omega_d t \\ &= 1 - e^{-\xi\omega_n t}\left[\cos\omega_d t + \frac{\xi}{\sqrt{1-\xi^2}}\sin\omega_d t\right] \end{aligned}$$

Put $\cos\theta = \xi$ and $\sin\theta = \sqrt{1-\xi^2}$, so that $\cos\omega_d t\,\sin\theta + \sin\omega_d t\,\cos\theta = \sin(\omega_d t + \theta)$:

$$c(t) = 1 - \frac{e^{-\xi\omega_n t}}{\sqrt{1-\xi^2}}\sin(\omega_d t + \theta), \qquad \theta = \tan^{-1}\frac{\sqrt{1-\xi^2}}{\xi}$$

The response is a sine wave at $\omega_d$ that decays inside the envelopes $1 \pm e^{-\xi\omega_n t}/\sqrt{1-\xi^2}$ and settles at the final value 1, so the steady state error is zero.

 c(t)
  |    _
  |   / \     _
 1|--/---\---/-\--------- final value
  | /     \_/
  |/
  +------------------------ t