Difficulty: Easy
Correct Answer: its response grows linearly with time (ramp)
Explanation:
Introduction / Context: Pole locations dictate time-domain behaviour. A single pole at the origin gives an integrator; a double pole at the origin corresponds to a double integrator. Such systems cannot return to equilibrium after a disturbance and are considered unstable by BIBO criteria, even if the growth is not exponential.
Given Data / Assumptions:
Concept / Approach: An integrator (1/s) converts a step input into a ramp. A double integrator (1/s^2) converts a step into a parabolic growth and an impulse into a ramp. Even the simplest disturbances thus cause outputs that diverge with time, violating bounded-input bounded-output stability. The essential signature is unbounded growth (at least linear), not decay or bounded oscillation.
Step-by-Step Solution:
Model near the origin: G(s) ≈ K / s^2.Apply a bounded input (e.g., step R(s) = 1/s) → Y(s) = K / s^3.Time domain: y(t) ∝ t^2 (parabolic) for step; for impulse input y(t) ∝ t (ramp).Any of these grow unbounded → instability.Verification / Alternative check: BIBO stability requires all poles strictly in the left half-plane. Poles at the origin violate that requirement; a repeated pole at the origin is worse than a simple integrator and guarantees unbounded growth for common inputs.
Why Other Options Are Wrong:
Constant or decaying responses imply poles with negative real parts.Exponential growth suggests right-half-plane poles; while still unstable, it is not the correct hallmark for a pole at exactly s = 0.Bounded oscillation needs imaginary-axis conjugate poles without damping, not a double pole at zero.Common Pitfalls: Assuming “unstable” must mean exponential blow-up; integrators are unstable by unbounded growth even at polynomial rates.
Final Answer: its response grows linearly with time (ramp)
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