How it works
For a first-order system G(s) = K/(τs + 1) with unit step input, the output is c(t) = K(1 − e^(−t/τ))u(t). Rise time (10% to 90%) ≈ 2.2τ; settling time (2% criterion) ≈ 4τ. For a second-order underdamped system, the step response overshoot %Mp = e^(−πζ/√(1−ζ²)) × 100%, peak time t_p = π/ω_d where ω_d = ω_n√(1−ζ²), and settling time t_s ≈ 4/(ζω_n) for the 2% criterion. These formulas are the core of transient analysis.
Key points to remember
For ζ = 0.5 and ω_n = 10 rad/s: %Mp = e^(−π·0.5/√0.75) × 100% ≈ 16.3%, t_p = π/(10√0.75) ≈ 0.36 s, t_s ≈ 4/(0.5×10) = 0.8 s. Critically damped (ζ = 1) gives no overshoot and fastest settling without oscillation. Overdamped (ζ > 1) gives sluggish response; underdamped (ζ < 1) gives oscillatory response. Delay time t_d is the time to reach 50% of final value ≈ (1 + 0.7ζ)/ω_n. Rise time and settling time are inversely proportional to ω_n — increasing ω_n speeds up the response.
Exam tip
The examiner always asks you to calculate %Mp, t_p, and t_s for a given second-order transfer function — identify ζ and ω_n from the denominator coefficients first, because every formula flows from those two numbers.