Stability of a system with a double pole at the origin A linear system has a double pole at s = 0. Which time-domain behaviour explains why such a system is unstable?

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:

• Transfer function has a repeated pole at s = 0.
• Consider response to bounded inputs such as steps and impulses.
• We use standard LTI system theory.

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:

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:

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)