Time response of first and second-order systems

Why it matters

Most sensors and loops in instrumentation behave, at least approximately, like first- or second-order systems: a thermometer in a well is first order, a pressure gauge or a galvanometer is second order, and a tuned closed loop is usually designed to look like a second-order system with chosen damping. Rise time, overshoot and settling time are the numbers on a specification sheet, and they come straight from τ, ζ and ωn.

Key ideas

Standard test inputs: step (sudden change of set point), ramp (constant-rate change), impulse (short shock), and sinusoid (frequency response). The step response is the most used because it shows both speed and damping.

First-order system G(s) = K/(τs + 1)

• One energy store, one pole at s = −1/τ.
• Step response rises exponentially with no overshoot. After τ it has reached 63.2 % of the final change; after 4τ, 98.2 %; after 5τ, 99.3 %.
• The initial slope of the step response is K/τ per unit step; if it kept that slope it would reach the final value in exactly τ.
• For a ramp input the output lags the input by τ in steady state (for K = 1).

Second-order system G(s) = ωn² / (s² + 2ζωn s + ωn²) (unity DC gain)

• ωn is the undamped natural frequency; ζ is the damping ratio.
• Poles: s = −ζωn ± ωn√(ζ² − 1).
• Cases: 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, fastest response without overshoot); ζ > 1 overdamped (two real poles, sluggish).
• For underdamped poles, ζωn is the distance of the poles from the jω-axis (sets the decay rate) and ωd is their height (sets the ringing frequency). The angle θ between the pole and the negative real axis satisfies cos θ = ζ.
• Peak overshoot depends only on ζ. Settling time depends on ζωn.
• Typical design target: ζ between 0.4 and 0.8, giving about 25 % to 1.5 % overshoot.

Higher-order systems: if one pair of poles is much closer to the jω-axis than the others (roughly five times or more), that dominant pair sets the response and the second-order formulas apply approximately.

Formulas

First order, unit step input:

• c(t) = K·(1 − e^(−t/τ))
• t_s (2 %) ≈ 4τ; t_r (10–90 %) = τ·ln 9 ≈ 2.2τ Second order, underdamped, unit step input:
• ωd = ωn·√(1 − ζ²)
• c(t) = 1 − [e^(−ζωn t)/√(1 − ζ²)]·sin(ωd t + θ), with θ = cos⁻¹ ζ
• t_r = (π − θ)/ωd (0 to 100 % rise time)
• t_p = π/ωd
• Mp = e^(−πζ/√(1 − ζ²)) (fraction; multiply by 100 for %)
• ζ = −ln(Mp) / √(π² + ln²(Mp)) (inverse, Mp as a fraction)
• t_s ≈ 4/(ζωn) (2 % band), t_s ≈ 3/(ζωn) (5 % band)
• Number of oscillations before settling ≈ t_s·ωd/(2π)

Symbols: K steady-state gain (output unit/input unit), τ time constant (s), ωn, ωd rad/s, ζ dimensionless, θ rad, times in s. Settling-time formulas use the exponential envelope and are approximate; the rest are exact for the standard form with no zeros.

Worked examples

Example 1 — first-order sensor Given: G(s) = 5/(s + 2), unit step input.

1. Standard form: G(s) = 2.5/(0.5s + 1), so K = 2.5 and τ = 0.5 s.
2. Final value: c(∞) = K = 2.5.
3. At t = 1 s: c(1) = 2.5·(1 − e^(−1/0.5)) = 2.5 × (1 − 0.1353) = 2.162.
4. Settling time (2 %): 4τ = 2.0 s.

Result: τ = 0.5 s, c(∞) = 2.5, c(1 s) = 2.16, t_s ≈ 2.0 s. Note that τ is not the reciprocal of the numerator; it comes from writing the denominator as τs + 1.

Example 2 — unity-feedback position loop (GATE level) Given: G(s) = 25/[s(s + 6)], unity feedback, unit step input.

1. Closed loop: T(s) = 25/(s² + 6s + 25), so ωn² = 25 → ωn = 5 rad/s and 2ζωn = 6 → ζ = 0.6.
2. ωd = 5·√(1 − 0.36) = 5 × 0.8 = 4 rad/s; θ = cos⁻¹ 0.6 = 0.927 rad.
3. Rise time: t_r = (π − 0.927)/4 = 2.214/4 = 0.554 s.
4. Peak time: t_p = π/4 = 0.785 s.
5. Overshoot: Mp = e^(−π × 0.6/0.8) = e^(−2.356) = 0.0948, i.e. 9.48 %.
6. Settling time (2 %): t_s = 4/(ζωn) = 4/3 = 1.33 s.

Result: ωn = 5 rad/s, ζ = 0.6, t_r = 0.554 s, t_p = 0.785 s, Mp = 9.5 %, t_s ≈ 1.33 s.

Common mistakes

• Reading τ = 1/2 directly from 5/(s + 2) but then using K = 5 instead of 5/2 = 2.5.
• Using the open-loop transfer function to find ζ and ωn when the question asks about the closed-loop response.
• Forgetting the phase θ inside the sine of the underdamped response, which makes the response start at the wrong value.
• Taking θ in degrees inside t_r = (π − θ)/ωd.
• Believing a higher ωn reduces overshoot: overshoot depends only on ζ.
• Applying second-order formulas to a system with a nearby zero or a non-dominant pole, where overshoot changes noticeably.

For GATE IN

Expect questions that give G(s) (often in a unity-feedback loop) and ask for ωn, ζ, peak time, overshoot or settling time; questions that give the overshoot or peak time and ask for a gain or ζ; first-order sensor questions on time to reach a percentage of final value or on ramp lag; and pole-location MCQs (which pole pattern gives which response). Practise moving between pole positions, ζ/ωn, and the time-domain numbers.

Quick check

1. A first-order system with τ = 3 s gets a step. How long until it reaches 95 % of its final value?
2. For ζ = 0.5, what is the percentage overshoot?
3. T(s) = 16/(s² + 4s + 16). Find ζ and ωn.
4. Which two pole parameters set the settling time and the overshoot respectively?

Answers: 1. −3·ln 0.05 = 8.99 s (about 3τ); 2. 16.3 %; 3. ωn = 4 rad/s, ζ = 0.5; 4. Settling time is set by ζωn (real part), overshoot by ζ (pole angle).