Which classical stability assessment explicitly uses the open-loop transfer function to determine closed-loop stability margins and encirclements?

Difficulty: Easy

Correct Answer: Nyquist

Explanation:


Introduction / Context:
Different classical methods assess closed-loop stability from various viewpoints: algebraic criteria on the characteristic polynomial, frequency-domain loci, and graphical encirclements. Recognizing what each method requires as input helps select the right tool for a given problem.

Given Data / Assumptions:

  • Loop structure is unity or known; open-loop transfer function L(s) is available.
  • Plant and controller are linear time-invariant.
  • Interest is in stability and gain/phase margins.


Concept / Approach:
The Nyquist method maps the frequency response of the open-loop transfer function L(jω) around the Nyquist contour and counts encirclements of the critical point (−1, 0). This directly relates open-loop frequency response to closed-loop stability and margins. In contrast, Routh–Hurwitz is an algebraic criterion based on the characteristic polynomial, and Mikhailov uses the frequency response of that polynomial itself, not the open-loop L(jω) per se.


Step-by-Step Solution:

Compute L(jω) for ω from 0 to ∞ and reflect negative frequencies.Plot Nyquist locus and check encirclements of −1.Infer stability and margins from the open-loop plot.


Verification / Alternative check:
Bode plots (also based on L(jω)) can provide margins; Nyquist provides a full stability mapping including non-minimum-phase peculiarities.


Why Other Options Are Wrong:

Mikhailov: uses characteristic polynomial vector locus.Routh: tests signs of Routh array from polynomial coefficients.None: incorrect because Nyquist is correct.


Common Pitfalls:
Confusing “open-loop response” (Nyquist/Bode) with “closed-loop” plots; Nyquist uses open-loop to infer closed-loop stability.


Final Answer:
Nyquist

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