Damping ratio, overshoot and settling time of a second-order system

Compare the closed-loop characteristic equation with s² + 2ζωₙs + ωₙ².

Worked answer.

Step 1: Closed-loop transfer function. With unity feedback, T(s) = G/(1 + G) = 25 / (s² + 6s + 25).

Step 2: Compare with the standard form. The standard second-order form is ωₙ² / (s² + 2ζωₙs + ωₙ²).

• ωₙ² = 25, so ωₙ = 5 rad/s.
• 2ζωₙ = 6, so ζ = 6 / (2 × 5) = 0.6 (underdamped).

Step 3: Damped frequency. ω_d = ωₙ√(1 − ζ²) = 5 × √(1 − 0.36) = 5 × 0.8 = 4 rad/s.

Step 4: Peak overshoot.
M_p = exp(−πζ / √(1 − ζ²)) = exp(−π × 0.6 / 0.8) = exp(−2.356) ≈ 0.0948, i.e. about 9.5%.

Step 5: Peak time. t_p = π / ω_d = π / 4 ≈ 0.785 s.

Step 6: Settling time (2% criterion). t_s ≈ 4 / (ζωₙ) = 4 / (0.6 × 5) = 4/3 ≈ 1.33 s.

Check. The closed-loop poles are the roots of s² + 6s + 25 = 0, which are s = −3 ± j4. The real part −3 equals −ζωₙ and the imaginary part 4 equals ω_d, both consistent with the values above. Also, because G(s) has one pole at the origin (type 1), the steady-state error to a step is zero, so the output settles at exactly 1.

Common mistake. Reading ωₙ and ζ from the open-loop G(s) instead of the closed-loop denominator. The "s + 6" factor is not 2ζωₙ until you form 1 + G(s).

Likely follow-up. "What gain K in G(s) = K/[s(s+6)] gives ζ = 0.707?" Then 2ζωₙ = 6 gives ωₙ = 4.243 rad/s, so K = ωₙ² = 18.