Difficulty: Easy
Correct Answer: All poles of the closed-loop transfer function must lie in the left half of the complex plane.
Explanation:
Introduction / Context: In linear time-invariant control, stability is fundamentally tied to the locations of the closed-loop poles. Whether you analyse from time domain (characteristic equation) or frequency domain (Nyquist, Bode), the conclusion is the same: stable systems have decaying natural responses, which correspond to poles with negative real parts for continuous-time systems.
Given Data / Assumptions:
Concept / Approach: The closed-loop transfer function’s denominator roots (closed-loop poles) govern stability. Poles in the left half-plane (Re{s} < 0) yield exponentially decaying modes. Complex conjugate poles are permitted; they need not be real. Frequency-domain criteria (Nyquist, Bode gain/phase margins) are equivalent ways to ensure that all closed-loop poles reside in the left half-plane.
Step-by-Step Solution:
Relate stability to closed-loop pole locations.Note that “real” is not required; complex poles are common.Reject statements implying right-half-plane poles or misleading Bode monotonicity requirements.Verification / Alternative check: Routh–Hurwitz gives algebraic tests ensuring left-half-plane poles; Nyquist ensures encirclement conditions around −1 that imply the same.
Why Other Options Are Wrong:
Real roots — unnecessary; damping can arise with complex poles.Monotonic Bode magnitude — not a stability requirement.Right-half-plane poles — by definition unstable.Origin crossing in Nyquist — not a criterion.Common Pitfalls: Confusing necessary with sufficient conditions and misinterpreting frequency-response shapes as guarantees of stability.
Final Answer: All poles of the closed-loop transfer function must lie in the left half of the complex plane.
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