Difficulty: Medium
Correct Answer: 1 / (s^2 + 2s + 2)
Explanation:
Introduction / Context:Continuous-time linear systems are asymptotically stable if and only if all poles lie strictly in the left-half complex plane (negative real parts). Recognising stability by inspection of the characteristic polynomial is a core skill for controls and process dynamics.
Given Data / Assumptions:
Concept / Approach:For a quadratic s^2 + 2ζω_ns + ω_n^2, stability requires ζ > 0 and ω_n > 0. Alternatively, compute roots directly and check real parts. Imaginary-axis poles indicate marginal stability (undamped oscillation) rather than asymptotic stability. Right-half-plane poles imply instability.
Step-by-Step Solution:
(a) s^2 + 2 = 0 ⇒ s = ± j√2 → marginal, not asymptotically stable.(b) s^2 − 2s + 3 = 0 ⇒ s = 1 ± j√2 → positive real part → unstable.(c) s^2 + 2s + 2 = 0 ⇒ s = −1 ± j → negative real part → stable.(d) s^2 + 2s − 1 = 0 ⇒ s = −1 ± √2 → one root at 0.414… > 0 → unstable (delay does not fix this).(e) s^2 − 1 = 0 ⇒ s = ±1 → right-half-plane pole → unstable.Verification / Alternative check:Routh–Hurwitz for (c): coefficients [1, 2, 2] are positive with no sign changes, confirming stability.
Why Other Options Are Wrong:
Common Pitfalls:Calling imaginary-axis poles “stable”; they are only marginally stable and sensitive to disturbances.
Final Answer:1 / (s^2 + 2s + 2)
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