CHEN90032 · Process Simulation and Control
Laplace Transforms
In University of Melbourne CHEN90032 Process Simulation and Control, the Laplace transform is the mathematical engine of the whole subject: it turns a dynamic process model — a differential equation in time — into an algebraic equation in the complex variable s, so a hard ODE becomes a fraction you rearrange and look up. This chapter covers the transform-pair table, the derivative property, inverse transforms by partial-fraction expansion (distinct, repeated and complex factors), solving ODEs, and the initial- and final-value theorems. Master it and every later topic — transfer functions, stability, frequency response and controller tuning — falls out of the same algebra.
What this chapter covers
- 01The Laplace transform definition and its key transform pairs (Seborg Table 3.1)
- 02The derivative property: why every d/dt becomes a factor of s (and why deviation variables zero the initial-condition terms)
- 03The five-step routine: transform → rearrange for Y(s) → partial-fraction → invert → check
- 04Partial-fraction expansion with distinct real factors (the Heaviside cover-up rule)
- 05Repeated factors: one term at every power, coefficients by differentiation
- 06Complex (quadratic) factors: complete the square to reach damped-sinusoid pairs
- 07Reading a first-order or under-damped response back into the time domain, with units
- 08Initial- and Final-Value Theorems — and when the Final-Value Theorem is invalid
- 09How Laplace feeds directly into the transfer function G(s)=Y(s)/U(s) used from here on
Solve a first-order ODE by Laplace transform
- +1Transform every term. Using L{dy/dt}=sY(s)-y(0) with y(0)=0, and the step 3 ↔ 3/s: 2sY + Y = 3/s. Rearrange for the output transform: (2s+1)Y = 3/s, so Y(s) = 3 / [s(2s+1)].
- +1Partial fractions. Write 3/[s(2s+1)] = P/s + Q/(2s+1). Cover-up at s=0 gives P = 3/(2·0+1) = 3; matching the s-coefficient, 0 = 2P + Q, so Q = -6. Thus Y(s) = 3/s - 6/(2s+1) = 3/s - 3/(s+0.5).
- +1Invert term by term. With 1/s → 1 and 1/(s+a) → e^(-at) where a = 1/τ = 0.5 min^-1: y(t) = 3 - 3·e^(-t/2) = 3(1 - e^(-t/2)) mol/L.
- +1Check the steady state with the Final-Value Theorem: lim (t→∞) y = lim (s→0) sY(s) = lim 3/(2s+1) = 3 mol/L. It matches the e^(-t/2) term dying out, so y settles at 3 mol/L. The pole sits at s = -0.5 (left-half plane), so the theorem is valid here.
Key terms
- Laplace transform
- The operation F(s) = ∫₀∞ f(t) e^(-st) dt that maps a time function f(t) into a function F(s) of the complex variable s. It is linear, so a model can be transformed term by term.
- s (Laplace variable)
- The complex frequency variable, with units of inverse time (1/min or 1/s). Differentiation in time corresponds to multiplication by s in the transform domain.
- Transform pair
- A matched f(t) ↔ F(s) entry in the Laplace table (Seborg Table 3.1). Inverting a transform is mostly a matter of rewriting Y(s) as a sum of recognisable pairs and reading each one back.
- Derivative property
- L{df/dt} = sF(s) - f(0); L{d²f/dt²} = s²F(s) - s f(0) - f'(0). In deviation variables the initial-condition terms are zero, which is why transfer functions carry no initial-value information.
- Partial-fraction expansion
- Rewriting a ratio N(s)/D(s) as a sum of simple fractions, one per pole (root of D). The form of each term depends on whether the root is distinct real, repeated, or complex.
- Heaviside cover-up rule
- For a distinct real root p, the coefficient of 1/(s-p) is found by deleting the (s-p) factor from the denominator and evaluating what remains at s = p.
- Final-Value Theorem
- lim (t→∞) y(t) = lim (s→0) sY(s), giving the steady-state value without inverting. Valid only when sY(s) has all poles in the left-half plane (the limit must actually exist); it gives a wrong answer for an unstable system.
- Initial-Value Theorem
- lim (t→0⁺) y(t) = lim (s→∞) sY(s), giving the starting value of the response directly from its transform.
Laplace Transforms FAQ
Why bother with Laplace transforms instead of solving the ODE directly?
Because it converts calculus into algebra. Transforming a differential equation turns every derivative into a power of s, so the equation becomes a fraction you rearrange for Y(s) and invert with a table — no integrating factors or characteristic-equation gymnastics. It also produces the transfer function G(s)=Y(s)/U(s) that the rest of CHEN90032 (transfer functions, stability, frequency response, controller tuning) is built on.
What is examined, and what can I take in?
The end-of-semester exam is worth 60% of the subject and is a hurdle you must pass to pass the unit: four questions, 100 marks, 15 minutes reading plus 3 hours writing, closed-book with a PROVIDED formula sheet (which includes the full Laplace transform table) and a Casio FX82 calculator. That works out to about 1.8 minutes per mark. Laplace is rarely a whole question on its own — it is the tool inside the model-derivation and transfer-function questions — and marks are awarded for showing the transform, the split and the inversion, so write every step. Confirm the exact date for the roughly June 2027 end-of-Semester-1 exam period on Canvas.
Can AI help me with Laplace transforms in CHEN90032?
Yes — used as a study aid, not an answer service. Sia can explain the derivative property step by step, walk you through why a repeated root needs a term at every power, show how completing the square turns a complex factor into a damped-sinusoid pair, and check your reasoning on a practice inversion. Use it to understand the method and rehearse worked examples; do your own graded assignments and exam preparation, and follow the unit's stated policy on acceptable use of AI tools.
Studying with AI? Sia — free AI chemical engineering tutor works through CHEN90032 step by step.
Exam move
Treat this chapter as a set of reflexes, because in the exam it appears inside bigger dynamics and control questions rather than on its own. First, memorise the shape of the common transform pairs (step, ramp, exponential, sine/cosine, and the first-order step response) even though the table is provided — recognising them instantly saves time. Second, drill the five-step routine until it is automatic: transform in deviation form, rearrange for Y(s), partial-fraction by root type, invert, then check the ends with the value theorems. Third, practise all three partial-fraction cases separately — distinct roots with cover-up, repeated roots with differentiation, and complex roots by completing the square — since a single sign or factor error here propagates into every later part of a question. Budget your time at about 1.8 minutes per mark, and always write the intermediate steps: method earns marks even when one arithmetic slip spoils the final number.
