Second-order systems: underdamped and overdamped response

Why it matters

Second-order behaviour appears whenever two lags act in series, in inherently oscillatory elements such as a U-tube manometer or a control valve with its actuator, and, most importantly, in almost every feedback loop under PI control. The terms overshoot, decay ratio and period are how plant engineers describe loop performance, and the classic tuning target of a quarter decay ratio is a second-order idea.

Key ideas

Standard form. In process control a second-order system is written G(s) = K / (τ²s² + 2ζτs + 1), where K is the steady-state gain, τ the natural period of oscillation (time) and ζ the damping factor (dimensionless). The equivalent mechanical form uses the natural frequency ω_n = 1/τ: for a mass–spring–damper, ω_n = √(k/m) and ζ = c/(2√(k·m)).

Sources of second-order dynamics.

• Multicapacity processes: two first-order lags in series give τ² = τ₁τ₂ and 2ζτ = τ₁ + τ₂, always with ζ ≥ 1.
• Inherently second-order elements: a U-tube manometer, a pneumatic valve actuator, a spring-loaded instrument.
• Processes with controllers: a first-order process under PI control gives a closed loop that can be underdamped.

Poles and damping. The poles are s = (−ζ ± √(ζ² − 1))/τ.

• ζ > 1, overdamped: two distinct real poles; sluggish, no overshoot. The system factors into two first-order lags with time constants τ/(ζ − √(ζ² − 1)) and τ/(ζ + √(ζ² − 1)).
• ζ = 1, critically damped: a repeated real pole; the fastest response without overshoot.
• 0 < ζ < 1, underdamped: complex poles; the response overshoots and oscillates with decaying amplitude.
• ζ = 0, undamped: sustained oscillation at the natural frequency 1/τ rad per unit time.
• ζ < 0: growing oscillation (unstable).

Underdamped step response (unit step, K = 1). y(t) = 1 − (1/√(1 − ζ²))·e^(−ζt/τ)·sin(√(1 − ζ²)·t/τ + φ), with φ = tan⁻¹(√(1 − ζ²)/ζ). Its characteristics depend only on ζ (and τ for the time scale):

• Overshoot (fraction of the final change by which the first peak exceeds it).
• Decay ratio (ratio of successive peak heights above the final value), equal to the square of the overshoot.
• Period of oscillation and time to first peak.
• Rise time (first time the final value is reached) and response (settling) time (time to stay within ±5 % or ±2 %). Smaller ζ means faster rise but more overshoot and longer settling.

Process-control targets. A quarter decay ratio (DR = 0.25) corresponds to ζ ≈ 0.215 and an overshoot of 50 %; it was the classic Ziegler–Nichols target. Many modern tunings aim at ζ around 0.5–0.7 for less overshoot and more robustness.

Formulas

G(s) = K / (τ²·s² + 2·ζ·τ·s + 1)

• K gain; τ natural period (s); ζ damping factor (–).

Overshoot OS = exp(−π·ζ / √(1 − ζ²))

• Valid for 0 < ζ < 1, as a fraction of the final change.

Decay ratio DR = exp(−2·π·ζ / √(1 − ζ²)) = OS²

Period T = 2·π·τ / √(1 − ζ²), ω_d = √(1 − ζ²)/τ

• T in s; ω_d damped angular frequency (rad/s).

Time to first peak t_p = π·τ / √(1 − ζ²)

Settling time t_s ≈ 4·τ/ζ (±2 %), ≈ 3·τ/ζ (±5 %)

• Approximation based on the decay envelope.

ζ = −ln(OS) / √(π² + (ln OS)²)

• Damping factor from a measured overshoot.

τ₁, τ₂ = τ / (ζ ∓ √(ζ² − 1))

• Overdamped system split into two first-order time constants.

ω_n = 1/τ = √(k/m), ζ = c / (2·√(k·m))

• Mass–spring–damper analogue: k stiffness (N/m), m mass (kg), c damping coefficient (N·s/m).

Worked examples

Example 1 (standard): characteristics of an underdamped system. G(s) = 2/(s² + 0.8s + 1), time in minutes, unit step input. Find ζ, τ, overshoot, decay ratio, period, time to first peak and the peak output.

1. Compare with τ²s² + 2ζτs + 1: τ² = 1, so τ = 1 min; 2ζτ = 0.8, so ζ = 0.4.
2. √(1 − ζ²) = √0.84 = 0.9165.
3. OS = exp(−π·0.4/0.9165) = exp(−1.371) = 0.254.
4. DR = OS² = 0.0644.
5. T = 2π·1/0.9165 = 6.86 min; t_p = T/2 = 3.43 min.
6. Peak output = K·(1 + OS) = 2·1.254 = 2.51.
7. Settling (±2 %) ≈ 4τ/ζ = 4·1/0.4 = 10 min.

OS = 25.4 %, DR = 0.064, T = 6.86 min, t_p = 3.43 min, y_max ≈ 2.51

Example 2 (GATE level): identifying ζ and τ from a test. A closed-loop step test shows an overshoot of 30 % and a period of oscillation of 6 s. Find ζ, τ and the decay ratio.

1. ln(OS) = ln(0.30) = −1.204.
2. ζ = 1.204/√(π² + 1.204²) = 1.204/√(9.870 + 1.450) = 1.204/3.365 = 0.358.
3. √(1 − ζ²) = √(1 − 0.128) = 0.934.
4. τ = T·√(1 − ζ²)/(2π) = 6·0.934/6.283 = 0.892 s.
5. DR = OS² = 0.30² = 0.09.

ζ ≈ 0.358, τ ≈ 0.89 s, DR = 0.09

Example 3 (short): overdamped factoring. G(s) = 1/(4s² + 5s + 1): τ = 2, ζ = 5/(2·2) = 1.25. Time constants τ/(ζ ∓ √(ζ² − 1)) = 2/(1.25 ∓ 0.75) = 4 and 1, so G = 1/((4s + 1)(s + 1)).

Common mistakes

• Confusing τ (natural period) with a first-order time constant; for an underdamped system the oscillation period is 2πτ/√(1 − ζ²), not τ.
• Taking the decay ratio equal to the overshoot; DR = OS².
• Reading 2ζτ as ζτ when matching coefficients, or forgetting to normalise the denominator so the constant term is 1.
• Using the overshoot formula for ζ ≥ 1 (there is no overshoot).
• Using a mass–spring formula with ω_n in Hz instead of rad/s.

For GATE CH

Expect: extracting K, τ and ζ from a transfer function; computing overshoot, decay ratio, period or time to peak; working backwards from a step test to ζ; deciding whether a response oscillates from the characteristic equation; and finding the closed-loop ζ of a first-order process under PI control or of two lags under P control. Practise normalising the denominator before matching coefficients.

Quick check

1. For G = 1/(9s² + 3s + 1), find τ and ζ.
2. What decay ratio corresponds to an overshoot of 0.5?
3. Can two non-interacting first-order lags in series give ζ < 1?
4. For ζ = 0, what does the step response look like?

Answers: 1. τ = 3, ζ = 3/(2·3) = 0.5. 2. 0.25. 3. No, ζ ≥ 1. 4. A sustained oscillation of constant amplitude about the final value.