Recovering the intended model class (dead time vs dynamic order): Assume the open-loop transfer function shown in the original stem was a pure transport delay of the form G(s) = exp(-θs). This representation corresponds to which type of system description?

Introduction / Context:
The original question omitted the explicit transfer function expression. Applying the Recovery-First Policy, a common classroom example is to present G(s) as a pure time delay, G(s) = exp(−θs). Recognising this form matters because time delay (transport lag) fundamentally differs from ordinary dynamic lags represented by rational polynomials in s.

Given Data / Assumptions:

• Open-loop transfer is G(s) = exp(−θs), θ > 0.
• Single-input single-output, linear time-invariant framework.
• No additional poles or zeros are shown.

Concept / Approach:
A pure time delay is not a differential equation in the usual sense; in the Laplace domain, it appears as a multiplicative factor exp(−θs). In the time domain, it means y(t) = u(t − θ) for t ≥ θ. This is called a dead time or transport delay system. It is neither first-order nor second-order; those would have denominators like (1 + τs) or (1 + 2ζω_ns + s^2/ω_n^2). A first-order time lag stores energy and filters high frequencies; dead time simply shifts the response in time without attenuation of steady-state gain.

Step-by-Step Solution:

Verification / Alternative check:
Frequency response gives |exp(−jωθ)| = 1 and phase = −ωθ, confirming pure phase lag with no magnitude change—signature of dead time.

Why Other Options Are Wrong:

• First-/second-order and first-order time lag: Require rational forms with poles.
• Non-minimum-phase inverse response: Involves right-half-plane zeros, not pure delay.

Common Pitfalls:
Using low-order Padé approximations and then mistaking the approximated polynomial for the actual system order; the true system is delay, not first- or second-order.

Final Answer:
Dead time system