MEGR 3171 — Mathematical Modeling & Laplace Transforms

Learning Objectives

Write differential equations for mechanical, electrical, thermal, and fluid systems using analogies.
Apply Laplace transform properties to convert ODEs to algebraic equations in the s-domain.
Derive the transfer function from a differential equation or impedance model.
Simplify block diagrams using series, parallel, and feedback reduction rules.
Convert a transfer function to state-space form and vice versa.

1. Physical System Analogies

Mechanical, electrical, thermal, and fluid systems share mathematical structure. Identifying the analogs allows transfer of analysis methods across domains.

Quantity	Mechanical (Trans.)	Electrical	Thermal	Fluid
Through variable	Force F (N)	Current i (A)	Heat flow Q (W)	Flow rate q (m³/s)
Across variable	Velocity v (m/s)	Voltage V (V)	Temperature T (°C)	Pressure P (Pa)
Energy storage 1	Mass m (kg)	Inductance L (H)	—	Inertance I
Energy storage 2	Spring 1/k	Capacitance C (F)	Thermal cap. C_th	Fluid cap.
Dissipation	Damper b	Resistance R (Ω)	Thermal resist. R_th	Fluid resist.

2. Laplace Transform Properties

Key Laplace Transform Pairs and Properties

Unit step:         L{u(t)} = 1/s
Exponential:       L{e^(-at)} = 1/(s+a)
Sine:              L{sin(ωt)} = ω/(s²+ω²)
Cosine:            L{cos(ωt)} = s/(s²+ω²)

Differentiation:   L{dx/dt} = s·X(s) - x(0)
Integration:       L{∫x dt} = X(s)/s

Initial Value:     lim_{t→0+} x(t) = lim_{s→∞} s·X(s)
Final Value:       lim_{t→∞} x(t) = lim_{s→0} s·X(s)  [if stable]

The Laplace transform converts a differential equation into an algebraic equation in the complex frequency variable s = σ + jω. Solving the algebraic equation and inverse-transforming gives the time-domain response.

3. Transfer Functions

The transfer function G(s) = Y(s)/U(s) is the ratio of output to input in the Laplace domain, assuming zero initial conditions. It completely characterizes the input-output behavior of a linear, time-invariant (LTI) system.

Mass-Spring-Damper Example

ODE: m&ddot;x + b˙x + kx = F(t)

Laplace (zero ICs): ms²X(s) + bsX(s) + kX(s) = F(s)

Transfer function: G(s) = X(s)/F(s) = 1 / (ms² + bs + k)

4. Block Diagram Algebra

Series (Cascade)

G_total = G_1 · G_2 · ... · G_n. Multiply transfer functions of blocks in series.

Parallel

G_total = G_1 + G_2 + ... + G_n. Add transfer functions of blocks in parallel.

Negative Feedback

G_closed = G_forward / (1 + G_forward · H_feedback). The fundamental closed-loop formula.

Positive Feedback

G_closed = G_forward / (1 − G_forward · H_feedback). Uncommon; can be unstable.

Practice Problems

Problem 1 — Transfer Function Derivation A series RC circuit has input voltage V_in and output voltage across the capacitor V_out. Derive the transfer function G(s) = V_out(s) / V_in(s).

Using voltage divider in the s-domain: V_out = V_in · Z_C / (Z_R + Z_C) = V_in · (1/sC) / (R + 1/sC)

Multiply numerator and denominator by sC: G(s) = 1 / (1 + sRC) = (1/RC) / (s + 1/RC)

G(s) = 1 / (1 + τs) where τ = RC (first-order low-pass filter)

Problem 2 — Final Value Theorem A system has transfer function G(s) = 5 / (s(s+2)(s+5)). A unit step input U(s) = 1/s is applied. Find the steady-state output.

Y(s) = G(s) · U(s) = 5 / (s²(s+2)(s+5))

Final value: lim_{s→0} s · Y(s) = lim_{s→0} s · 5 / (s²(s+2)(s+5)) = lim_{s→0} 5 / (s(s+2)(s+5))

= 5 / (0 · 2 · 5) → ∞ (output grows without bound — system has a double integrator, non-converging step response)

The final value theorem gives ∞. This system has a pole at s = 0 (integrator) in the plant; a ramp input would be needed for bounded steady-state error analysis.

Problem 3 — Thermal Model A lumped thermal mass m·c_p = 500 J/°C is connected to ambient through a thermal resistance R_th = 0.4 °C/W. Write the transfer function T(s) / Q(s) where Q is heat input in Watts.

ODE: (m·c_p)·dT/dt + T/R_th = Q(t) → (m·c_p)·s·T(s) + T(s)/R_th = Q(s)

T(s)[(m·c_p)·s + 1/R_th] = Q(s)

G(s) = T(s)/Q(s) = R_th / (1 + (m·c_p·R_th)·s) = 0.4 / (1 + 200s)

G(s) = 0.4 / (1 + 200s) (τ = 200 s, DC gain = 0.4 °C/W)