Control Engineering Basics
A step is applied to a second-order system (standard form without zeros) with a damping ratio ζ = 0.5. Which of the following is closest to the overshoot (the maximum excursion beyond the final value, as a fraction of the final value)?
- AAbout 16%
- BAbout 21%
- C50%
Show answer and explanation
Answer: A. About 16%
The overshoot of a standard second-order system is exp(−πζ/√(1 − ζ²)) and does not depend on the natural angular frequency ωn. For ζ = 0.5 it is exp(−π × 0.5/√0.75) ≈ exp(−1.81) ≈ 0.163, about 16%. For ζ ≈ 0.707 (1/√2) it is about 4.3%, a value often used as a guideline for a response with little overshoot. The overshoot peaks at the time π/ωd, where ωd = ωn√(1 − ζ²).
- A. Correct. The overshoot is determined by ζ alone.
- B. This is exp(−πζ), forgetting to divide by √(1 − ζ²).
- C. Calculating 1 − ζ does not give the overshoot.