Time response of first- and second-order systems

Why it matters

When you command a servo to a new position or a heater to a new temperature, the specification is written in time-response terms: how fast it rises, how far it overshoots and how soon it settles. Most real loops behave like a first-order or a dominant second-order system, so the formulas here let you read performance straight off a transfer function — and choose a gain to meet a specification.

Key ideas

Test inputs. Responses are compared using the unit step (sudden change of set point), ramp (constant-velocity command), impulse (shock) and sinusoid. The step response is the standard for transient specifications. Total response = transient part (dies out if stable) + steady-state part.

First-order system G(s) = K/(τs + 1): one energy store (a capacitor charging through a resistor, a thermometer bulb, a tank with an outlet valve).

• Unit-step response y(t) = K(1 − e^(−t/τ)): no overshoot, monotonic rise.
• At t = τ the output reaches 63.2 % of the final value; at 4τ, 98.2 %.
• The initial slope is K/τ — if the response kept its starting slope it would reach the final value in exactly τ.
• Pole at s = −1/τ: the further left, the faster.
• For a unit ramp input with K = 1, the output follows the ramp with a constant lag of τ seconds (steady-state error τ).

Second-order system G(s) = ωn²/(s² + 2ζωn·s + ωn²): two energy stores (mass and spring, L and C, a motor with inertia in a position loop).

• ωn (natural frequency) sets the time scale; ζ (damping ratio, dimensionless) sets the shape.
• Poles: s = −ζωn ± jωn√(1 − ζ²). The real part σ = ζωn sets the decay rate; the imaginary part ωd sets the oscillation frequency; the angle θ = cos⁻¹ζ from the negative real axis fixes overshoot.
• ζ = 0 undamped: poles on the jω-axis, sustained oscillation at ωn (marginally stable).
• 0 < ζ < 1 underdamped: complex poles, decaying oscillation at ωd.
• ζ = 1 critically damped: repeated real pole at −ωn, fastest response with no overshoot.
• ζ > 1 overdamped: two real poles, sluggish, no overshoot.

Time-domain specifications (underdamped, unit step): delay time, rise time tr (0–100 % for underdamped; 10–90 % is used for overdamped), peak time tp, peak overshoot Mp and settling time ts (time to stay within ±2 % or ±5 %). Increasing ζ lowers Mp but lengthens tr and tp; increasing ωn speeds everything up without changing Mp.

Limits of validity. The formulas assume exactly the standard form with no zeros. Extra zeros (which add overshoot) or extra poles (which slow the response) break them unless the extra poles are at least about five times further from the imaginary axis than the dominant pair — then the dominant pair approximation holds.

Formulas

y(t) = K(1 − e^(−t/τ)) — first-order unit-step response

• K: steady-state gain (output units per input unit); τ: time constant (s).

ts ≈ 4τ (2 %), ts ≈ 3τ (5 %), tr(10–90 %) = 2.2τ — first order.

ωd = ωn·√(1 − ζ²) — damped natural frequency (rad/s).

θ = cos⁻¹ζ (rad) — angle of the pole from the negative real axis.

tr = (π − θ)/ωd — rise time 0–100 % (s).

tp = π/ωd — peak time (s).

Mp = exp(−πζ/√(1 − ζ²)) × 100 % — peak overshoot.

ts ≈ 4/(ζωn) (2 %), ts ≈ 3/(ζωn) (5 %) — settling time (s), envelope approximation, valid for 0 < ζ < about 0.8.

N ≈ ts/(2π/ωd) — number of oscillations before settling.

For a unity-feedback loop with G(s) = K/(s(s + a)): closed loop K/(s² + a·s + K), so ωn = √K and ζ = a/(2√K).

Worked examples

Example 1 (standard, first order). A temperature sensor has G(s) = 5/(2s + 1) (mV/°C). A step of 1 °C is applied at t = 0. Find τ, the output at t = 2 s and t = 4 s, and the 2 % settling time.

1. Standard form K/(τs + 1): K = 5 mV/°C, τ = 2 s.
2. y(t) = 5(1 − e^(−t/2)) mV.
3. t = 2 s: y = 5(1 − e^(−1)) = 5 × 0.632 = 3.16 mV.
4. t = 4 s: y = 5(1 − e^(−2)) = 5 × 0.865 = 4.32 mV.
5. ts = 4τ = 8 s.

Example 2 (GATE level, gain selection). A position servo has unity negative feedback with G(s) = K/(s(s + 6)). Find K for ζ = 0.6, then ωd, Mp, tr, tp and the 2 % settling time.

1. Closed loop: T(s) = K/(s² + 6s + K), so ωn² = K and 2ζωn = 6.
2. With ζ = 0.6: ωn = 6/(2 × 0.6) = 5 rad/s, so K = 25.
3. ωd = 5√(1 − 0.36) = 5 × 0.8 = 4.0 rad/s.
4. Mp = exp(−π × 0.6/0.8) = exp(−2.356) = 9.48 %.
5. θ = cos⁻¹(0.6) = 0.927 rad; tr = (π − 0.927)/4.0 = 0.554 s.
6. tp = π/4.0 = 0.785 s.
7. ts = 4/(0.6 × 5) = 1.33 s.

Common mistakes

• Using ωn instead of ωd in tp and tr. The oscillation you see is at ωd.
• Taking θ in degrees inside tr = (π − θ)/ωd. θ must be in radians.
• Calling ζ = 0 "unstable". It is marginally stable — bounded but never-decaying oscillation.
• Applying Mp and ts formulas to a system with a zero or a non-dominant third pole close to the dominant pair.
• Forgetting to put the transfer function in standard form first: in 5/(2s + 1), τ = 2 s and K = 5, not τ = 1/2.
• Mixing up the 2 % (4/ζωn) and 5 % (3/ζωn) settling criteria.
• Reading the closed-loop ωn and ζ from the open-loop denominator.

For GATE ME

Expect: finding ζ and ωn from a closed-loop transfer function or a characteristic equation; computing Mp, tp, ts or the response at a given time; choosing a gain to meet a damping or overshoot target; recognising response type from pole locations; and first-order problems on time constant and 63.2 % / 98 % points. Practise inverting the overshoot formula (ζ from a given Mp) and keeping angles in radians.

Quick check

1. A first-order system reaches 63.2 % of its final value in 3 s. What is τ?
2. ζ = 0.5, ωn = 4 rad/s. Find ωd.
3. For ζ = 0.7 and ωn = 10 rad/s, what is the 2 % settling time?
4. Which damping ratio gives the fastest response with no overshoot?
5. Does increasing ωn (same ζ) change the peak overshoot?

Answers: 1. 3 s. 2. 3.46 rad/s. 3. 0.571 s. 4. ζ = 1 (critical damping). 5. No — Mp depends only on ζ.