Control Engineering Basics
Which of the following is the settling time, until the step response of the system G(s) = 25/(s² + 4s + 25) stays within ±2% of its final value, estimated with the approximation ts ≈ 4/(ζωn)?
- A0.5 s
- B0.8 s
- C2 s
Show answer and explanation
Answer: C. 2 s
This system has ωn = 5 rad/s and ζ = 0.4, so ζωn = 2 s⁻¹ and ts ≈ 4/2 = 2 s. The envelope of the oscillation shrinks as e^(−ζωn t), and e^(−4) ≈ 0.018 falls below 2%, which is where this approximation comes from. For a ±5% settling time, ts ≈ 3/(ζωn) is used. Because it is estimated from the envelope, it often comes out a little longer than the true value; computing this system numerically gives a ±2% settling time of about 1.7 s.
- A. This is 1/(ζωn), the time constant of the envelope. Staying within ±2% takes about four times that.
- B. This is 4/ωn, forgetting to multiply by ζ.
- C. Correct. The settling time is set by the real part ζωn of the poles.