Transportation lag and linearisation

Why it matters

Dead time is the single biggest enemy of good control: during it the controller acts blind, and it adds phase lag without limit as frequency rises. Linearisation is what lets us use transfer functions at all, because real tanks, valves, reactions and heat exchangers are non-linear. Both appear in nearly every realistic process model, typically as the first-order-plus-dead-time (FOPDT) model used for tuning.

Key ideas

Transportation lag (dead time). When material must travel from where a change happens to where it is felt or measured, the output repeats the input exactly, only later: y(t) = u(t − θ). Typical causes are flow through pipes (plug flow), conveyor belts, sample lines to analysers and sampled measurements.

• For plug flow, θ = L/v = V/q (pipe length over velocity, or pipe volume over volumetric flow).
• In the Laplace domain the dead time is e^(−θs). It is not a ratio of polynomials, so it cannot be handled by partial fractions or the Routh test directly; it is replaced by an approximation when needed.
• Padé (1/1) approximation: e^(−θs) ≈ (1 − θs/2)/(1 + θs/2). It has a right-half-plane zero and reproduces the phase lag well at low frequency.
• Taylor approximation: e^(−θs) ≈ 1/(θs + 1) (a first-order lag) or 1 − θs.
• Frequency response: amplitude ratio 1 at all frequencies, phase φ = −ωθ (rad), which grows without bound. This is why dead time limits the gain at which a loop stays stable.
• Many lags in series look like dead time, so higher-order processes are often fitted as K·e^(−θs)/(τs + 1). The ratio θ/τ indicates how hard the process is to control.

Compensation. Dead time cannot be removed by feedback. Remedies are to reduce it physically (relocate sensors, faster sample loops), detune the controller, use a Smith predictor (an internal model that lets the controller act on a delay-free prediction), or use feedforward when the disturbance can be measured early.

Linearisation. A non-linear term f(x) is replaced near a steady operating point x_s by the first two terms of its Taylor series: f(x) ≈ f(x_s) + (df/dx)_s·(x − x_s). For several variables, add a partial-derivative term for each. Subtracting the steady-state balance then leaves an equation in deviation variables, which can be Laplace transformed.

• The linear model is accurate only for small deviations; its gain and time constant change with the operating point.
• Tank with square-root outflow: q = C·√h linearises to q ≈ q_s + (h − h_s)/R with R = 2√h_s/C = 2h_s/q_s, giving H(s)/Q_in(s) = R/(A·R·s + 1).
• Arrhenius: k = k₀·e^(−E/(R_g·T)) linearises to k ≈ k_s·[1 + E/(R_g·T_s²)·(T − T_s)].
• Radiation: T⁴ ≈ T_s⁴ + 4T_s³·(T − T_s).

Formulas

L{u(t − θ)} = e^(−θ·s)·U(s)

• θ dead time (s).

θ = L/v = V/q

• L pipe length (m), v mean velocity (m/s), V pipe volume (m³), q flow (m³/s). Plug flow only.

e^(−θ·s) ≈ (1 − θ·s/2)/(1 + θ·s/2) and e^(−θ·s) ≈ 1/(θ·s + 1)

• First-order Padé and Taylor approximations, good for ωθ well below 1.

AR = 1, φ = −ω·θ (rad)

• Frequency response of pure dead time; ω in rad/s.

G(s) = K·e^(−θ·s)/(τ·s + 1)

• FOPDT model.

f(x) ≈ f(x_s) + (df/dx)_s·(x − x_s)

• First-order Taylor linearisation about steady state x_s.

R = 2·√h_s / C = 2·h_s/q_s, τ = A·R

• Linearised resistance and time constant of a tank with q = C·√h; A area (m²), h level (m), q flow (m³/s), R in s/m².

Worked examples

Example 1 (standard): dead time of a pipe. Water flows at 0.01 m³/s through a 50 m pipe of 0.1 m inside diameter to a temperature sensor. Find the dead time, and the phase lag it adds to a sinusoid of 0.05 rad/s.

1. Area A = π·d²/4 = π·0.1²/4 = 7.854 × 10⁻³ m².
2. Velocity v = q/A = 0.01/7.854 × 10⁻³ = 1.273 m/s.
3. θ = L/v = 50/1.273 = 39.3 s (same as V/q = 0.3927/0.01).
4. Phase lag = ω·θ = 0.05·39.3 = 1.96 rad = 112.5°.

θ ≈ 39.3 s; phase lag ≈ 112° at 0.05 rad/s

Example 2 (GATE level): linearising a tank. A tank (A = 1.5 m²) drains as q = 0.6·√h, with q in m³/min and h in m. It operates at h_s = 4 m. Find the linearised transfer function H(s)/Q_in(s), and compare the predicted steady-state level rise for an inlet step of +0.1 m³/min with the exact non-linear result.

1. Steady state: q_s = 0.6·√4 = 1.2 m³/min.
2. Linearise: dq/dh = 0.6/(2√h_s) = 0.6/4 = 0.15 m²/min, so R = 1/0.15 = 6.667 min/m² (check: 2h_s/q_s = 8/1.2 = 6.667).
3. τ = A·R = 1.5·6.667 = 10.0 min; H(s)/Q_in(s) = 6.667/(10s + 1).
4. Linear prediction: Δh = R·Δq = 6.667·0.1 = 0.667 m.
5. Exact: new steady state 1.3 = 0.6·√h, so h = (1.3/0.6)² = 4.694 m, Δh = 0.694 m.

H(s)/Q_in(s) = 6.667/(10s + 1) (min, m, m³/min); linear Δh = 0.667 m vs exact 0.694 m, an error of about 4 % for an 8 % flow change.

Common mistakes

• Writing dead time as e^(+θs), or as a first-order lag in a stability test where the exact phase −ωθ is needed.
• Taking R = h_s/q_s for a square-root tank; the linearised resistance is twice that.
• Linearising in absolute variables and forgetting to subtract the steady-state equation, which leaves constant terms in the transfer function.
• Using θ = L/v with the wrong velocity (for example, the velocity in a different pipe size) or forgetting unit conversion of flow from m³/h.
• Believing derivative action or higher gain can cancel dead time.

For GATE CH

Expect: dead time from pipe data; the phase angle of K·e^(−θs)/(τs + 1) at a given frequency (−ωθ − tan⁻¹ωτ); Padé or Taylor approximations; and linearisation of tanks, Arrhenius rates or valve equations to obtain τ and K. Dead time is central to crossover-frequency and gain-margin questions, so be fluent with radians versus degrees.

Quick check

1. What is the amplitude ratio of a pure dead time?
2. Liquid at 2 L/s fills a 40 L pipe section in plug flow. What is the dead time?
3. For q = C√h at h_s = 1 m and q_s = 0.5 m³/min, what is the linearised R?
4. Give the first-order Padé approximation of e^(−2s).

Answers: 1. 1 at all frequencies. 2. 20 s. 3. 2·1/0.5 = 4 min/m². 4. (1 − s)/(1 + s).