CHEN90032 · Process Simulation and Control
Time Delay, Integrating Processes, Poles & Zeros
This chapter of the CHEN90032 Process Simulation and Control exam guide (University of Melbourne) covers the process dynamics that break the tidy first- and second-order picture: pure time delay (dead time) e−θs, fitting a first-order-plus-time-delay (FOPTD) model, Padé approximants, integrating (non-self-regulating) processes 1/(As), and the poles & zeros that set speed, stability and inverse response.
These ideas feed the transfer-function, frequency-response and controller-tuning questions later in the written final, so getting the units and signs right here pays off across the paper.
What this chapter covers
- 01Write pure time delay as e^(−θs): AR = 1, phase −ωθ (rad), and why it is the most destabilising loop element
- 02Compute a transport delay θ = (pipe volume)/(volumetric flowrate) with units
- 03State the FOPTD model G(s) = K e^(−θs)/(τs+1) and identify K, τ, θ from a step reaction curve
- 04Fit τ and θ with the two-point rule: τ = (t₂ₓ₃ − t₁ₓ₃)/0.7 and θ = t₁ₓ₃ − 0.4τ
- 05Replace the delay with 1/1 and 2/2 Padé approximants so it can be inverted or Routh-tested
- 06Recognise an integrating process 1/(As): a pole at s = 0, a ramp response and no steady-state gain
- 07Locate poles and zeros; use m ≤ n; read stability (LHP) and speed (τ = 1/a) from pole positions
- 08Diagnose inverse response from a right-half-plane (positive) zero, including the lead-lag unit K(τₓs+1)/(τ₁s+1)
Fit a FOPTD model to step-test data
- +2Steady-state gain: Δu = 30 − 20 = 10 %, Δy_ss = 62.0 − 50.0 = 12.0 °C, so K = Δy_ss/Δu = 12.0/10 = 1.2 °C per %.
- +2Time constant from the two-point rule: τ = (t_2/3 − t_1/3)/0.70 = (21.0 − 12.0)/0.70 = 9.0/0.70 = 12.86 s.
- +1Dead time: θ = t_1/3 − 0.40τ = 12.0 − 0.40(12.86) = 12.0 − 5.14 = 6.86 s.
- +1Assemble the model (times in seconds): G(s) = 1.2 e^(−6.86s)/(12.86 s + 1).
Key terms
- Time delay (dead time) θ
- A pure transport lag that reproduces the input unchanged but shifted later by θ [time]; its transfer function is e^(−θs), with amplitude ratio 1 and phase −ωθ (radians).
- FOPTD model
- First-order-plus-time-delay transfer function G(s) = K e^(−θs)/(τs+1); three parameters (gain K, time constant τ, dead time θ) fitted from a step reaction curve for tuning.
- Padé approximant
- A rational approximation of e^(−θs) (1/1: (1−θs/2)/(1+θs/2)) that makes the transcendental delay algebraic so it can be inverted or tested with a Routh array.
- Integrating (non-self-regulating) process
- A process such as a pump-drained tank, H(s)/Q_i(s) = 1/(As), with a pole at s = 0; a step input gives a ramp that never levels off, so there is no steady-state gain.
- Pole
- A root of the denominator of a transfer function. Poles set speed (a real pole at s = −a gives τ = 1/a) and stability — any pole with positive real part makes the process unstable.
- Zero
- A root of the numerator of a transfer function. Zeros never change stability but shape the transient; a right-half-plane (positive) zero causes inverse response.
- Inverse response
- A response that first moves in the opposite direction to its final value, caused by a right-half-plane zero (e.g. two opposing first-order effects where the faster one dominates the initial slope).
- Lead-lag unit
- G(s) = K(τ_a s + 1)/(τ_1 s + 1): lead when τ_a > τ_1 (speeds up), lag when τ_a < τ_1 (slows down), and inverse response when τ_a < 0 (a RHP zero).
Time Delay, Integrating Processes, Poles & Zeros FAQ
What is the difference between dead time and inverse response?
Dead time means the output does nothing for a period θ and then moves the right way; inverse response means the output moves immediately but in the wrong direction first before turning around. Dead time comes from a transport delay e^(−θs); inverse response comes from a right-half-plane (positive) zero. Both reduce phase margin and make control harder, but only a right-half-plane pole (not a zero) causes actual instability.
Why does an integrating process have no steady-state gain?
For a pump-drained tank the model is H(s)/Q_i(s) = 1/(As), which has a pole at s = 0. A step in inlet flow produces a ramp h(t) = (Δq_i/A)t that grows without bound, so the Final-Value Theorem gives no finite steady state. Such non-self-regulating loops only settle under feedback control.
Can AI help me with time delay, FOPTD and poles & zeros in CHEN90032?
Yes. An AI study tool like Sia can explain each idea step by step — how to read θ and τ off a step reaction curve, when a zero sits in the right half-plane, or why an integrator ramps — and can walk you through practice problems until the method clicks. Use it to understand and check your own working; it is a tutor for building the skill, not a source of ready-made exam or assignment answers, and it never guarantees a mark or a pass.
Studying with AI? Sia — free AI chemical engineering tutor works through CHEN90032 step by step.
Exam move
Anchor everything on the four archetypes: dead time (e^(−θs), AR = 1, phase −ωθ), the FOPTD fit, the integrating pole at s = 0, and poles/zeros for stability and inverse response. Practise fitting K, τ and θ from step data until the two-point rule (τ = (t₂ₓ₃−t₁ₓ₃)/0.7, θ = t₁ₓ₃−0.4τ) is automatic, and always carry units on K. Rehearse the trap pairs — dead time versus inverse response, and a right-half-plane zero (still stable) versus a right-half-plane pole (unstable). The written final is 4 questions over 100 marks in 3 hours (plus 15 minutes reading), closed book with a provided formula sheet, worth 60% and a hurdle you must pass, so budget about 1.8 minutes per mark and confirm the exact date on Canvas for the ~June end-of-Semester-1 exam period.
