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
Answer
16.3 % (accept 16.0 to 16.6) % (accept 16.0 to 16.6)
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 trap this question is built around
Using $\omega_n = 10$ somewhere in the overshoot formula. It appears in the peak time $t_p = \pi/(\omega_n\sqrt{1-\zeta^2})$ and in the settling time, but never in $M_p$. Two systems with the same $\zeta$ overshoot identically no matter how different their speeds.