Peak overshoot of an underdamped second-order system
BlinkNBuild practice problem · GATE standard, authored and verified in-house
A unity-feedback system has the closed-loop transfer function $$T(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2},$$ with $\zeta = 0.5$ and $\omega_n = 10\ \text{rad/s}$.
The peak overshoot for a unit-step input, in percent, is ________.
Show the step-by-step derivation
Step-by-step derivation
- For an underdamped second-order system ($0 < \zeta < 1$) the peak overshoot depends on the damping ratio alone: $$M_p = \exp\!\left(\frac{-\pi\zeta}{\sqrt{1-\zeta^2}}\right).$$ $\omega_n$ sets how fast the response is, not how far it overshoots.
- Compute the radical first, with $\zeta = 0.5$: $$\sqrt{1-\zeta^2} = \sqrt{1 - 0.25} = \sqrt{0.75} = 0.8660.$$
- Form the exponent: $$\frac{-\pi(0.5)}{0.8660} = \frac{-1.5708}{0.8660} = -1.8138.$$
- Exponentiate: $$M_p = e^{-1.8138} = 0.1630.$$
- Convert to a percentage: $$M_p = 0.1630 \times 100 = \mathbf{16.3\%}.$$
- Anchors worth memorising. $\zeta = 0.707$ gives $4.3\%$ overshoot (the maximally-flat design point), $\zeta = 0.5$ gives $16.3\%$, and $\zeta = 0.3$ gives $37.2\%$. If your answer falls outside the pattern, the arithmetic is wrong.
The idea behind this question
A standard second-order system is fully described by its damping ratio $\zeta$ and natural frequency $\omega_n$. The damping ratio alone decides how far the step response overshoots: $M_p = e^{-\pi\zeta/\sqrt{1-\zeta^2}}$. For example, $\zeta = 0.5$ gives 16.3% and $\zeta = 0.7$ about 4.6%. The natural frequency only sets how quickly it all happens.
Try a variation
What is the peak overshoot for $\zeta = 0.6$?
Show the answer
Answer: $9.48\%$
$e^{-\pi \times 0.6/\sqrt{1 - 0.36}} = 0.0948$.
Other mistakes to avoid
- Giving the overshoot as a fraction when the question asks for a percentage, or the reverse.
- Applying the formula to a system with extra poles or zeros, where it is only an approximation.