First-order plus time delay (FOPDT) model Select the correct transfer function form for a first-order process with pure time delay (dead time) θ, process gain K, and time constant τ.

Difficulty: Easy

Correct Answer: K * e^{-θ s} / (τ s + 1)

Explanation:


Introduction / Context:
The first-order plus dead time (FOPDT) model is the most widely used low-order approximation in process control, capturing dominant lag and transport delay for many thermal, flow, and composition loops. Recognizing its canonical transfer function form is essential for tuning (e.g., Ziegler–Nichols, Cohen–Coon) and for controller design heuristics.

Given Data / Assumptions:

  • Process gain: K.
  • Time constant: τ.
  • Pure time delay (dead time): θ.
  • Linearity and small-signal behavior around an operating point.


Concept / Approach:
A dead time element multiplies any transfer function by e^{−θ s} (unity magnitude, frequency-dependent phase). A first-order lag is 1/(τ s + 1). Therefore, the FOPDT model is K * e^{−θ s} / (τ s + 1). Higher-order denominators or extra 1/s factors indicate different dynamics (second order or integrating behavior) and are not correct for a simple FOPDT.

Step-by-Step Solution:

Identify first-order lag → denominator (τ s + 1).Incorporate pure delay → multiplicative factor e^{−θ s} in the numerator.Multiply by steady-state gain K → K * e^{−θ s} / (τ s + 1).


Verification / Alternative check:
Frequency response of e^{−θ s} has |G| = 1 and phase = −ω θ, while 1/(τ s + 1) gives −tan^{-1}(ω τ) phase and 1/√(1 + (ω τ)^2) magnitude roll-off, matching FOPDT characteristics.


Why Other Options Are Wrong:

  • K/(τ s + 1): No dead time included.
  • K * e^{−θ s} /(τ s + 1)^2: Second-order lag, not first order.
  • K /(s(τ s + 1)): Integrating plus first-order, not a simple FOPDT.


Common Pitfalls:
Modeling delay as extra lag terms; true transport delay is exponential in s.


Final Answer:
K * e^{-θ s} / (τ s + 1)

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