Inferring a process transfer function from a step test: A process is at steady state with output y = 1 for input u = 1. At t = 0, the input is stepped to 2. The measured output is y(t) = 1 + 2 t (a ramp with slope 2). What is the transfer function G(s) = Y(s)/U(s) of the process?

Introduction / Context:
Identifying simple process dynamics from step tests is fundamental in control engineering. An output ramp in response to a step input suggests integrating behavior. This question asks you to infer the transfer function from the given input–output relationship.

Given Data / Assumptions:

• Initial steady state: y(0−) = 1 at u(0−) = 1.
• At t = 0, input steps from 1 to 2, so Δu = 1 (unit step).
• Observed response: y(t) = 1 + 2 t for t ≥ 0 (a ramp with slope 2).
• Assume linear time-invariant SISO system.

Concept / Approach:
For an integrator G(s) = K/s, a unit step input U(s) = 1/s produces output Y(s) = (K/s)*(1/s) = K/s^2. In time domain, y(t) = K * t (a ramp) plus any initial condition. The ramp slope equals K times the step amplitude. With Δu = 1 and observed slope 2, the process gain K is 2, hence G(s) = 2/s.

Step-by-Step Solution:

Verification / Alternative check:
Differentiate y(t): dy/dt = 2 = K * Δu; consistent with integrator behavior. No exponential term appears, ruling out first-order lag forms like 2/(s+1).

Why Other Options Are Wrong:

• 1/s: Would yield slope 1, not 2.
• 2/(s + 1): Step response is asymptotic, not a ramp.
• 2 s: Noncausal differentiation-like behavior; step input would produce an impulse, not a ramp.
• 1/(s^2): Step would give a parabolic output (t^2/2), not a ramp.

Common Pitfalls:
Using steady-state gain logic for an integrating process; integrators have infinite steady-state gain and require rate-based interpretation.

Final Answer:
2/s