Assessing stability from transfer-function denominators:\nWhich of the following single-input single-output systems is (asymptotically) stable based on its denominator polynomial?

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:

  • Stability means responses decay to zero for bounded inputs after transients.
  • We assess the sign of real parts of the roots of each denominator.
  • Time delays (e.g., exp(−20s)) do not move poles by themselves but can worsen margins; denominator roots still dictate asymptotic stability.


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:

  • (a): Sustained oscillations violate asymptotic stability.
  • (b), (d), (e): Each has at least one pole with positive real part.


Common Pitfalls:
Calling imaginary-axis poles “stable”; they are only marginally stable and sensitive to disturbances.


Final Answer:
1 / (s^2 + 2s + 2)

More Questions from Process Control and Instrumentation

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion