Mathematical modelling of physical systems

Why it matters

You cannot design a controller for a plant you cannot describe. Before tuning a temperature loop or choosing a compensator, an instrumentation engineer writes the governing equations of the process — electrical, mechanical, thermal or fluid — and turns them into a transfer function. A good model tells you the gain, the time constants and the order of the system, which is most of what the rest of control theory needs.

Key ideas

From physical law to transfer function

1. Write the balance law for each energy-storing element: Kirchhoff's laws (electrical), Newton's second law (mechanical), energy balance (thermal), mass balance (fluid).
2. Assume the system is linear and time-invariant (LTI), or linearise it about an operating point.
3. Take the Laplace transform with zero initial conditions.
4. The transfer function is G(s) = Output(s)/Input(s). It is a property of the system alone, not of the input.

Order of the model equals the number of independent energy-storing elements: an RC network is first order, a series RLC circuit or a mass-spring-damper is second order.

Element relations (time domain → s-domain impedance)

• Electrical: resistor v = R·i, inductor v = L·di/dt (Ls), capacitor i = C·dv/dt (1/(Cs)).
• Mechanical translational: mass f = M·d²x/dt², viscous damper f = B·dx/dt, spring f = K·x.
• Mechanical rotational: inertia T = J·d²θ/dt², damper T = B·dθ/dt, torsional spring T = K·θ.
• Thermal: thermal resistance R_th = Δθ/q (K/W), capacitance C_th = m·c_p (J/K).
• Liquid level: hydraulic resistance R = dh/dq (s/m²), capacitance C = A (tank cross-section, m²).

Electrical analogies let you build one circuit for any mechanical system.

• Force–voltage analogy (series): force ↔ voltage, mass ↔ inductance, damper B ↔ resistance, spring K ↔ 1/C, velocity ↔ current.
• Force–current analogy (parallel): force ↔ current, mass ↔ capacitance, damper B ↔ conductance 1/R, spring K ↔ 1/L, velocity ↔ voltage.

Linearisation Real processes are nonlinear (valve characteristics, outflow ∝ √h, radiation ∝ T⁴). For small deviations about a steady operating point (x0, y0), replace y = f(x) by Δy ≈ (df/dx)|x0 · Δx. The resulting model is valid only near that point; time constants and gains change when the operating point moves.

Transfer function properties

• Defined only for LTI systems with zero initial conditions.
• It is the Laplace transform of the impulse response.
• Poles are roots of the denominator (they set the natural modes); zeros are roots of the numerator.
• Proper: degree of numerator ≤ degree of denominator — every physical system is proper.

Formulas

• Vc(s)/Vi(s) = 1 / (LC·s² + RC·s + 1) — series RLC, output across C.
• ωn = 1/√(LC), ζ = (R/2)·√(C/L) — for the series RLC above.
• X(s)/F(s) = 1 / (M·s² + B·s + K) — mass-spring-damper.
• ωn = √(K/M), ζ = B / (2·√(K·M)) — for the mass-spring-damper.
• θ(s)/T(s) = 1 / (J·s² + B·s + K) — rotational analogue.
• H(s)/Qi(s) = R / (A·R·s + 1) — liquid-level tank, τ = A·R.
• R = dh/dq = 2·h0/q0 — linearised resistance for outflow q = k·√h.
• Θ(s)/Q(s) = R_th / (R_th·C_th·s + 1) — thermal first-order system.

Symbols and units: R Ω, L H, C F; M kg, B N·s/m, K N/m; J kg·m², rotational B N·m·s/rad, K N·m/rad; h m, q m³/s, A m²; R_th K/W, C_th J/K; ωn rad/s; ζ dimensionless.

Worked examples

Example 1 — series RLC network Given: R = 10 Ω, L = 0.1 H, C = 100 µF; output is the capacitor voltage.

1. KVL: vi = R·i + L·di/dt + vc, with i = C·dvc/dt.
2. Laplace transform (zero initial conditions): Vi = (LCs² + RCs + 1)·Vc.
3. LC = 0.1 × 100×10⁻⁶ = 1×10⁻⁵ s²; RC = 10 × 100×10⁻⁶ = 1×10⁻³ s.
4. G(s) = 1 / (1×10⁻⁵ s² + 1×10⁻³ s + 1).
5. ωn = 1/√(1×10⁻⁵) = 316.2 rad/s; ζ = (10/2)·√(100×10⁻⁶/0.1) = 5 × 0.03162 = 0.158.

Result: G(s) = 1/(10⁻⁵ s² + 10⁻³ s + 1), with ωn = 316 rad/s and ζ = 0.158 (lightly damped).

Example 2 — linearised liquid-level process (GATE level) Given: tank area A = 2 m²; outflow through a valve q = k·√h; steady state h0 = 4 m, q0 = 0.02 m³/s. Find the transfer function H(s)/Qi(s) for small deviations.

1. Mass balance: A·dh/dt = qi − k·√h.
2. Valve constant: k = q0/√h0 = 0.02/2 = 0.01 m^2.5/s.
3. Linearise: dq/dh = k/(2√h0) = q0/(2h0), so R = dh/dq = 2h0/q0 = 2 × 4 / 0.02 = 400 s/m².
4. Deviation equation: A·dΔh/dt = Δqi − Δh/R, giving ΔH(s)/ΔQi(s) = R/(ARs + 1).
5. τ = A·R = 2 × 400 = 800 s.

Result: ΔH(s)/ΔQi(s) = 400/(800s + 1) m per (m³/s), a first-order lag with τ = 800 s. Note that R = 2h0/q0, not h0/q0: the square-root valve is twice as "stiff" in small-signal terms as its large-signal ratio suggests.

Common mistakes

• Forgetting that the transfer function assumes zero initial conditions, then trying to include them.
• Mixing up the two analogies — in force–current, mass maps to capacitance, not inductance.
• Linearising with R = h0/q0 instead of the slope dh/dq.
• Dropping the sign of reaction forces when two masses are coupled through a spring: the spring force depends on (x1 − x2).
• Mixing grams and kilograms, or µF and F, in ωn and ζ calculations.

For GATE IN

Expect derivations of transfer functions for RLC networks, op-amp circuits, mechanical (translational and rotational) systems and thermal or liquid-level processes; identification of ωn and ζ from a model; force–voltage and force–current analogy matching; and linearisation of a nonlinear element about an operating point. Practise writing node or mass equations quickly and checking units at the end.

Quick check

1. In the force–voltage analogy, what is the electrical analogue of a spring of stiffness K?
2. A mass-spring-damper has M = 2 kg, B = 3 N·s/m, K = 5 N/m. What is its static gain X/F at s = 0?
3. For the system in Q2, find ωn.
4. Why must a linearised model be used only near its operating point?

Answers: 1. A capacitor with C = 1/K; 2. 1/K = 0.2 m/N; 3. √(5/2) = 1.58 rad/s; 4. Because the slope (gain) of the nonlinearity changes as the operating point changes.