Derive maximum overshoot of second order system

Maximum (peak) overshoot is the largest amount by which the response exceeds its final value, usually given as a percentage of the final value. To find it, derive $c(t)$, find the time of the first peak, and substitute.

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 $\cos\omega_d t\,\sin\theta + \sin\omega_d t\,\cos\theta = \sin(\omega_d t + \theta)$:

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

Peak time

At a peak, $\dfrac{dc}{dt} = 0$. Differentiating $c(t)$:

$$\frac{dc}{dt} = \frac{\omega_n}{\sqrt{1-\xi^2}}e^{-\xi\omega_n t}\sin\omega_d t = 0$$

So $\sin\omega_d t = 0$, which gives $\omega_d t = n\pi$. The first peak (maximum overshoot) is at $n = 1$:

$$t_p = \frac{\pi}{\omega_d} = \frac{\pi}{\omega_n\sqrt{1-\xi^2}}$$

Maximum overshoot

Substitute $t_p$ in $c(t)$. Since $\omega_d t_p = \pi$, $\sin(\pi + \theta) = -\sin\theta = -\sqrt{1-\xi^2}$:

$$\begin{aligned} c(t_p) &= 1 - \frac{e^{-\xi\omega_n\pi/\omega_d}}{\sqrt{1-\xi^2}}\left(-\sqrt{1-\xi^2}\right) \\ &= 1 + e^{-\xi\pi/\sqrt{1-\xi^2}} \end{aligned}$$

With final value $c(\infty) = 1$:

$$M_p = c(t_p) - c(\infty) = e^{-\xi\pi/\sqrt{1-\xi^2}}$$ $$\%M_p = 100\,e^{-\xi\pi/\sqrt{1-\xi^2}}$$

Remarks

• $M_p$ depends only on $\xi$, not on $\omega_n$.
• When $\xi$ increases, $M_p$ decreases: e.g. $\xi = 0.5$ gives 16.3%, $\xi = 0.707$ gives 4.3%, and $\xi \ge 1$ gives no overshoot.