Process control – Transportation (time) delay: What is the transfer function representation of a pure transportation lag with delay time T?

Introduction / Context:
A transportation lag (also called a time delay or dead time) models the pure delay between an input signal change and the moment it is observed at the output. Time delays are common in chemical processes due to material holdup and transport in pipes, sampling lines, or sensor response paths.

Given Data / Assumptions:

• Pure delay of magnitude T seconds.
• Linear time-invariant framework with Laplace transforms.
• No dynamics besides the delay (i.e., no gain or storage in the path).

Concept / Approach:
In the Laplace domain, a time shift t → t − T translates into an exponential factor multiplying the transform. Specifically, a pure time delay T is represented by the factor exp(-Ts) multiplying the transfer function without delay. This captures the phase lag that increases linearly with frequency while keeping the magnitude at unity.

Step-by-Step Solution:
Let y(t) = x(t − T).Take Laplace transforms: Y(s) = exp(-Ts) * X(s).Therefore the transfer function G(s) = Y(s)/X(s) = exp(-Ts).Hence the correct compact representation is exp(-Ts).

Verification / Alternative check:
Frequency response of exp(-Tjω) has magnitude 1 and phase = −ωT radians, matching the expected behavior of a pure delay (no amplitude change, only phase lag).

Why Other Options Are Wrong:

• exp(+Ts): would imply a time advance, which is non-causal for real processes.
• 1/(1 + Ts): this is a first-order lag, not a pure delay.
• 1/s: represents an integrator, not a time delay.
• None of these: incorrect because exp(-Ts) is correct.

Common Pitfalls:
Confusing pure delay with first-order lag; a delay has unit magnitude across all frequencies but adds phase lag that grows with frequency.

Final Answer:
exp(-Ts)