Modelling of mechanical, electrical and fluid systems

Why it matters

A controller can only be designed for a plant you can describe mathematically. Mechatronic systems mix masses, springs, motors, circuits, tanks and valves, and the same few lumped-element laws turn each of them into a linear differential equation and a transfer function. Once you see the analogies, a hydraulic tank, an RC filter and a damped spring all become "a first- or second-order system".

Key ideas

• Lumped-parameter model: each physical effect (inertia, stiffness, dissipation) is concentrated in one ideal element. Valid when the system is small compared with wavelengths of interest and the elements are linear over the operating range; otherwise linearise about an operating point.
• Mechanical translational elements (force f, displacement x):
  • Mass: f = m·d²x/dt² (stores kinetic energy).
  • Viscous damper: f = b·dx/dt (dissipates energy).
  • Spring: f = k·x (stores potential energy).
  • Build a free-body diagram for each mass and apply Newton's second law: Σ applied forces = m·ẍ.
• Mechanical rotational elements (torque T, angle θ): inertia T = J·θ̈, rotational damper T = B·θ̇, torsional spring T = K·θ. Gear trains reflect inertia and damping by the square of the gear ratio: J_eq = J_1 + J_2·(N_1/N_2)².
• Electrical elements: resistor v = R·i, inductor v = L·di/dt, capacitor i = C·dv/dt. Use Kirchhoff's voltage law (loops) and current law (nodes).
• Analogies:
  • Force–voltage (loop) analogy: m ↔ L, b ↔ R, k ↔ 1/C, x ↔ q, v (velocity) ↔ i.
  • Force–current (node) analogy: m ↔ C, b ↔ 1/R, k ↔ 1/L, v (velocity) ↔ voltage.
• Fluid (liquid-level) elements: mass conservation (continuity) for a tank of cross-section A: A·dh/dt = q_in − q_out. Fluid capacitance C = A; outlet resistance R = dh/dq (linearised about the operating point for turbulent flow, where q ∝ √h). Fluid inertance I = ρ·L/A_pipe comes from Newton's second law applied to the fluid slug in a pipe.
• Thermal elements: thermal resistance R_th = ΔT/Q̇ and capacitance C_th = m·c_p; give first-order models of heaters and ovens.
• Transfer function: Laplace transform with zero initial conditions, output over input. Order of the model = number of independent energy-storage elements.
• Electromechanical coupling (DC motor): back emf e_b = K_b·ω, torque T = K_t·i; in SI units K_t = K_b. This is the classic mechatronic plant.

Formulas

• m·ẍ + b·ẋ + k·x = f(t) → X(s)/F(s) = 1/(m·s² + b·s + k) — m in kg, b in N·s/m, k in N/m, x in m, f in N.
• ω_n = √(k/m) (rad/s), ζ = b / (2·√(k·m)) (dimensionless) — for the mass–spring–damper.
• J·θ̈ + B·θ̇ + K·θ = T(t) → Θ(s)/T(s) = 1/(J·s² + B·s + K) — J in kg·m², B in N·m·s/rad, K in N·m/rad.
• Series RLC driven by v: L·q̈ + R·q̇ + q/C = v → I(s)/V(s) = C·s/(L·C·s² + R·C·s + 1); V_C(s)/V(s) = 1/(L·C·s² + R·C·s + 1).
• RC low-pass: V_o(s)/V_i(s) = 1/(R·C·s + 1), τ = R·C (s).
• Liquid-level tank with linear outlet: H(s)/Q_in(s) = R/(R·C·s + 1), τ = R·C — h in m, q in m³/s, C = A in m², R in s/m².
• Turbulent outlet linearised: q = K·√h → R = dh/dq = 2·h_0/q_0.
• Armature-controlled DC motor (speed): Ω(s)/V(s) = K_t / [(L_a·s + R_a)(J·s + B) + K_t·K_b].

Worked examples

Example 1 (standard). A mass–spring–damper has m = 5 kg, b = 10 N·s/m, k = 200 N/m. Find the transfer function X(s)/F(s), the undamped natural frequency, damping ratio and the steady deflection under a constant 20 N force.

1. Newton: m·ẍ + b·ẋ + k·x = f → X(s)/F(s) = 1/(5·s² + 10·s + 200).
2. ω_n = √(k/m) = √(200/5) = √40 = 6.32 rad/s.
3. ζ = b/(2·√(k·m)) = 10/(2·√1000) = 10/63.25 = 0.158 (underdamped).
4. Steady deflection: s → 0, DC gain 1/k = 1/200 = 0.005 m/N; x_ss = 20 × 0.005 = 0.1 m.
5. Result: X/F = 1/(5s² + 10s + 200), ω_n = 6.32 rad/s, ζ = 0.158, x_ss = 0.1 m (100 mm).

Example 2 (GATE level). A water tank of cross-section A = 2 m² drains through a valve whose linearised resistance is R = 50 s/m². The system is at steady state when the inflow is suddenly increased by Δq = 0.01 m³/s. Find the transfer function, the time constant, the final rise in level, and the rise after 200 s.

1. Continuity: A·dh/dt = q_in − h/R → H(s)/Q_in(s) = R/(R·A·s + 1).
2. Time constant: τ = R·A = 50 × 2 = 100 s. So H(s)/Q_in(s) = 50/(100·s + 1) (m per m³/s).
3. Final rise (final-value theorem, step Δq/s): Δh_∞ = R·Δq = 50 × 0.01 = 0.5 m.
4. Response: Δh(t) = 0.5·(1 − e^(−t/100)) m.
5. At t = 200 s: Δh = 0.5·(1 − e^(−2)) = 0.5 × 0.8647 = 0.432 m.
6. Result: τ = 100 s, final rise 0.5 m, rise after 200 s ≈ 0.432 m. The electrical analogue is an RC circuit with R ↔ R, C ↔ A, current source ↔ q_in.

Common mistakes

• Sign errors in free-body diagrams: spring and damper forces oppose the motion of the mass relative to their other end; in two-mass systems use the relative displacement (x_1 − x_2).
• Mixing the analogies: in force–voltage, the spring maps to 1/C, not C.
• Forgetting the square of the gear ratio when reflecting inertia or damping through gears.
• Using the nonlinear turbulent orifice law directly in a transfer function; it must be linearised about an operating point first.
• Including non-zero initial conditions in a transfer function; transfer functions assume zero initial conditions.
• Unit slips: μF, mH and N/mm must be converted to SI before computing τ or ω_n.

For GATE ME

Questions ask you to write the equation of motion or transfer function of a spring–mass–damper, rotor or two-mass system, to identify ω_n and ζ from the model, to use force–voltage or force–current analogies, and to find time constants of RC, thermal or liquid-level systems. Practise free-body diagrams with relative displacements and quick DC-gain checks with s = 0.

Quick check

1. In the force–voltage analogy, what is the electrical analogue of a damper b?
2. Find ω_n for m = 2 kg, k = 50 N/m.
3. What is the time constant of an RC circuit with R = 10 kΩ, C = 10 μF?
4. What law gives the liquid-level tank equation?
5. Number of energy-storage elements in a series RLC circuit?

Answers: 1. Resistance R; 2. 5 rad/s; 3. 0.1 s; 4. Conservation of mass (continuity); 5. Two (L and C), so second order.