CHEN90032 · Process Simulation and Control
Transfer Functions & First-Order Systems
This chapter of the University of Melbourne CHEN90032 Process Simulation and Control guide builds the transfer function G(s) = Y(s)/U(s) on deviation variables, the series and parallel rules for combining blocks, and the first-order model K/(τs+1) at the heart of process dynamics. You will learn what the steady-state gain K and time constant τ each mean (with units), the step, impulse and ramp responses, the 63.2% rule, and how an empirical step test identifies K and τ from real plant data — the workflow behind the lab practical and the exam's dynamics question.
What this chapter covers
- 01Define a transfer function G(s) = Y(s)/U(s) on deviation variables, with zero initial conditions
- 02Combine blocks: series (cascade) transfer functions multiply; parallel paths add
- 03Recognise and write the first-order model K/(τs+1) and its single pole at s = −1/τ
- 04Interpret the steady-state gain K (Δy_ss/Δu, with units and sign) versus the time constant τ (units of time)
- 05Derive and use the step, impulse and ramp responses of a first-order system
- 06Apply the 63.2% rule: a step reaches 63.2% of its final change in one time constant
- 07Read the standard fractions 63.2% (τ), 86.5% (2τ), 95.0% (3τ), 99.3% (5τ)
- 08Run an empirical step test to identify K and τ from a recorded response
- 09Spot when a dead time θ is needed, giving a first-order-plus-time-delay model
First-order step response: size and timing
- +1Ultimate change Δy_ss = K Δu = 0.4 (mol/L)/(kg/h) × 5 kg/h = 2.0 mol/L. The kg/h units cancel, leaving mol/L.
- +1State the step response: ŷ(t) = K Δu (1 − e^(−t/τ)), which rises toward the final value 2.0 mol/L.
- +1At t = τ the 63.2% rule gives ŷ(τ) = 0.632 × 2.0 mol/L = 1.26 mol/L above the starting concentration.
- +1For 95%: 0.95 = 1 − e^(−t/τ), so e^(−t/τ) = 0.05.
- +1Solve: t = −τ ln(0.05) = 6 min × 2.996 = 17.97 min ≈ 18 min, which equals 3τ (the standard 95.0% point).
- +1Sanity check: 18 min = 3τ matches the tabulated 95.0% fraction, and the answer is independent of K — only τ sets the timing.
Key terms
- Transfer function G(s)
- The ratio of the Laplace transform of the output deviation to that of the input deviation, with zero initial conditions: G(s) = Y(s)/U(s). It describes how the output moves when the input moves, not the absolute operating point.
- Deviation variable
- A signal measured as its departure from steady state, y-hat(t) = y(t) − y-bar. Transfer functions act only on deviations, which is why every initial condition is zero.
- Steady-state gain K
- The ultimate output change per unit sustained input change, K = Δy_ss/Δu. It carries the ratio of output-to-input units, and its sign shows whether the output rises or falls with the input.
- Time constant τ
- The parameter that sets response speed in a first-order system, with units of time. The pole sits at s = −1/τ; a larger τ means a slower response. For a first-order lag, τ is the time to reach 63.2% of a step change.
- First-order transfer function
- The model K/(τs+1): one gain, one time constant and a single real pole at s = −1/τ. It models many sensors, valves and single tanks.
- 63.2% rule
- Because 1 − e^(−1) = 0.632, a first-order step response reaches 63.2% of its total change in exactly one time constant — the fastest way to read τ off a curve.
- Empirical step testing
- Identifying a model from data: step the input, record the output, then take K = Δy_ss/Δu and τ = the time to reach 63.2% of the total change (or the initial-slope tangent intercept).
- First-order-plus-time-delay (FOPTD)
- The extended model K e^(−θs)/(τs+1), used when a step response shows a dead time θ before it starts to move — an S-shaped reaction curve.
Transfer Functions & First-Order Systems FAQ
What is the difference between the gain K and the time constant τ?
K answers 'how far': it is the ultimate output change per unit input change (Δy_ss/Δu), carrying the ratio of the two engineering units and a sign. τ answers 'how fast': it has units of time and fixes the response speed. A first-order fit must report both, each with its correct unit — they are not interchangeable.
Why does a first-order step response reach 63.2% at one time constant?
Setting t = τ in ŷ(t) = KA(1 − e^(−t/τ)) gives the factor 1 − e^(−1) = 0.632, independent of K and the step size A. So in one time constant the output always covers 63.2% of its total change, which is why a step test reads τ at the 63.2% point.
Can AI help me with transfer functions and first-order systems in CHEN90032?
Yes — used as a study aid, not an answer service. Sia can explain, step by step, how G(s) = Y/U is formed on deviation variables, why series blocks multiply, and how to read K and τ from a step test, and it can check your reasoning against your own worked attempt. It will not sit your closed-book exam or lab for you, and no tool can promise a specific grade; the aim is to help you understand the method so you can reproduce it under exam conditions.
Studying with AI? Sia — free AI chemical engineering tutor works through CHEN90032 step by step.
Exam move
Anchor everything on the two questions a first-order model answers: 'how far' (the gain K) and 'how fast' (the time constant τ). Practise moving both ways — given K and τ, predict the size and timing of a response; and given a recorded step response, read K = Δy_ss/Δu and τ at the 63.2% point. Memorise the standard fractions (63.2% at τ, 86.5% at 2τ, 95.0% at 3τ, 99.3% at 5τ) so you can sanity-check any timing answer in seconds. The written final is closed-book with a provided formula sheet, so drill the method rather than the formulae, and always attach units and a sign to K and a time unit to τ — method and units carry the marks. Pace at about 1.8 minutes per mark (3 hours over 100 marks), and treat this chapter as the foundation for the closed-loop, stability, frequency-response and tuning chapters that follow. Confirm the exam date for the ~June 2027 end-of-Semester-1 exam period on Canvas.
