Characteristic equation in feedback control:\nFor a closed-loop transfer function written as G1 / (1 + G2), the characteristic equation is 1 + G2 = 0. What does this imply about dependence, stability assessment, and applicability to set-point vs load changes?

Difficulty: Easy

Correct Answer: All (a), (b) & (c)

Explanation:


Introduction / Context:
In classical control theory, the characteristic equation captures the closed-loop poles, which entirely determine stability and transient behaviour. This question probes your understanding of how the characteristic equation relates to the open-loop dynamics and whether it changes with the type of excitation (set-point versus load disturbance).


Given Data / Assumptions:

  • Closed-loop form is written so that the denominator is 1 + G2.
  • Linear time-invariant system with negative feedback.
  • Set-point and load disturbances enter through standard summing junctions.


Concept / Approach:
The characteristic equation is obtained by setting the closed-loop denominator to zero. It depends only on the open-loop transfer function combination that appears in the denominator (typically the product of controller and plant). Because closed-loop poles are independent of the input path (set-point or load), the same characteristic equation governs stability for any excitation. Hence, it both depends only on open-loop dynamics and determines stability, and it is common to all input cases.


Step-by-Step Solution:

Write closed-loop denominator: 1 + G2.Set 1 + G2 = 0 to obtain the characteristic equation.Note that poles (roots) depend on open-loop elements only.Conclude: same characteristic equation applies to set-point and load analyses.


Verification / Alternative check:
Block-diagram manipulations or Mason’s gain formula show that closed-loop poles do not depend on input placement, confirming invariance across set-point and disturbance cases.


Why Other Options Are Wrong:

  • Choosing only one of (a), (b), or (c) ignores the full role of the characteristic equation.
  • “None of these” contradicts standard control results.


Common Pitfalls:
Confusing the characteristic equation (poles) with the full closed-loop transfer function (which includes zeros and may differ for set-point vs load paths).


Final Answer:
All (a), (b) & (c)

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