First-order systems and their dynamic response

Why it matters

A surprising number of process elements behave as first-order systems: a thermometer in a bath, a liquid tank draining through a valve, a stirred mixing tank, a sensor with a thermowell. Knowing how they respond to steps, ramps and sinusoids tells you how long a reading takes to settle, how much a measurement lags a changing process, and how fast a loop can possibly be.

Key ideas

Definition. A first-order system obeys τ·dy/dt + y = K·u, with y and u as deviation variables. Its transfer function is G(s) = Y(s)/U(s) = K/(τs + 1), with a single pole at s = −1/τ.

• Steady-state gain K: the final change in output per unit change in input (units of output/input).
• Time constant τ: a measure of speed; it is the product of a capacitance and a resistance (storage ÷ throughput).

Typical derivations.

• Mercury thermometer: energy balance m·Cp·dT/dt = h·A·(T_bath − T) gives τ = m·Cp/(h·A) and K = 1.
• Liquid-level tank with linear outlet resistance: A·dh/dt = q_in − h/R gives H(s)/Q_in(s) = R/(A·R·s + 1), so τ = A·R and K = R.
• Stirred mixing tank: V·dc/dt = q·(c_in − c) gives τ = V/q (the residence time) and K = 1.
• A non-linear tank (outflow ∝ √h) is first order only after linearisation about the operating point (see the linearisation topic).

Step response. For an input step of size M, y(t) = K·M·(1 − e^(−t/τ)). The response reaches 63.2 % of its final change at t = τ, 86.5 % at 2τ, 95.0 % at 3τ, 98.2 % at 4τ and 99.3 % at 5τ. The initial slope is K·M/τ; if the response kept that slope it would reach the final value at t = τ. There is no overshoot and no inflection.

Impulse response. For an impulse of area M, y(t) = (K·M/τ)·e^(−t/τ): a jump followed by exponential decay.

Ramp response. For u = a·t, y(t) = K·a·(t − τ + τ·e^(−t/τ)). After the transient, the output follows a ramp delayed by τ; with K = 1 the measurement lags the true value by a·τ.

Sinusoidal response. For u = A·sin(ωt), after transients the output is sinusoidal at the same frequency with amplitude ratio K/√(1 + ω²τ²) and phase lag φ = tan⁻¹(ωτ) (between 0 and 90°).

Pure capacity (integrating) process. A tank whose outflow is fixed by a pump gives G(s) = 1/(A·s): the level ramps without limit after a step. It is self-regulating only if the outflow depends on the level.

Formulas

G(s) = K / (τ·s + 1)

• K steady-state gain (output unit/input unit); τ time constant (s or min).

y(t) = K·M·(1 − e^(−t/τ))

• Step response to an input step of size M.

t = τ·ln(1/(1 − f))

• Time to reach fraction f of the final change.

y(t) = K·a·(t − τ + τ·e^(−t/τ))

• Ramp response, a = input ramp rate (input unit/s). Long-time lag = K·a·τ (for K = 1).

AR = Â_out / (K·Â_in) = 1/√(1 + ω²·τ²), φ = −tan⁻¹(ω·τ)

• Normalised amplitude ratio and phase angle for a sinusoid of angular frequency ω (rad/s). Time lag of the output wave = |φ|/ω (φ in rad).

τ = m·Cp/(h·A) (thermometer), τ = A·R and K = R (linear tank), τ = V/q (mixing tank)

• m mass (kg), Cp heat capacity (J/(kg·K)), h film coefficient (W/(m²·K)), A area (m²), R outlet resistance (s/m²), V volume (m³), q flow (m³/s).

Worked examples

Example 1 (standard): tank with linear outlet resistance. A tank of cross-section A = 2 m² drains through a linear resistance R = 0.5 min/m² (so h = R·q_out with h in m, q in m³/min). The inlet flow steps up by 0.4 m³/min. Find the final level change and the level change after 2 min.

1. τ = A·R = 2·0.5 = 1.0 min; K = R = 0.5 m per (m³/min).
2. Final change: Δh(∞) = K·M = 0.5·0.4 = 0.20 m.
3. Δh(t) = 0.20·(1 − e^(−t/1.0)).
4. At t = 2 min: Δh = 0.20·(1 − e^(−2)) = 0.20·0.8647 = 0.173 m.

Δh(∞) = 0.20 m; Δh(2 min) ≈ 0.173 m

Example 2 (GATE level): thermometer in a ramping bath. A thermometer with τ = 0.5 min is at steady state in a bath at 50 °C. The bath temperature then rises linearly at 2 °C/min. Find the reading and the true bath temperature at t = 2 min, and the long-time lag.

1. Deviation input u = a·t with a = 2 °C/min; K = 1.
2. y(t) = a·(t − τ + τ·e^(−t/τ)) = 2·(2 − 0.5 + 0.5·e^(−4)) = 2·(1.5 + 0.00916) = 3.018 °C.
3. Reading = 50 + 3.018 = 53.02 °C; bath = 50 + 2·2 = 54.00 °C.
4. Long-time lag = a·τ = 2·0.5 = 1.0 °C (the reading lags by τ = 0.5 min in time).

Reading ≈ 53.02 °C while the bath is at 54.00 °C; the lag tends to 1.0 °C

Example 3 (GATE level): sinusoidal bath temperature. The same thermometer (τ = 0.5 min) sits in a bath oscillating ±2 °C with a period of 2 min.

1. ω = 2π/P = 2π/2 = 3.1416 rad/min; ωτ = 1.571.
2. AR = 1/√(1 + 1.571²) = 1/√3.467 = 0.537.
3. Reading amplitude = 0.537·2 = ±1.07 °C.
4. φ = −tan⁻¹(1.571) = −57.5°; time lag = 1.004 rad / 3.1416 rad/min = 0.32 min.

Common mistakes

• Reading τ as the time to reach the final value; at t = τ only 63.2 % of the change has occurred.
• Forgetting to multiply by the input step size M or by the gain K.
• Using absolute instead of deviation variables, so the initial steady value is lost or double-counted.
• Using the period P instead of ω = 2π/P in the amplitude ratio.
• Treating a tank with a pumped outlet as first-order; it is a pure integrator with no steady state.
• Calling a first-order system "overdamped": damping ratio applies to second-order systems.

For GATE CH

This is one of the most frequently examined topics. Expect: time to reach a given fraction of the final value, the reading of a thermometer at a given time, the time constant from a test, ramp lag, and amplitude ratio or phase lag of a sinusoidal input. Derivations of τ and K for tanks and thermometers also appear, often combined with linearisation. Memorise the 63.2/86.5/95/98.2 % markers.

Quick check

1. A first-order system with τ = 5 s receives a step. How long to reach 90 % of the final change?
2. What is the gain of G(s) = 6/(3s + 1), and its time constant?
3. A thermometer (τ = 0.2 min) follows a bath rising at 5 °C/min. What is the long-time lag?
4. At ωτ = 1, what are the amplitude ratio and phase lag?

Answers: 1. 5·ln 10 = 11.5 s. 2. K = 6, τ = 3 time units. 3. 1.0 °C. 4. AR = 0.707, phase lag 45°.