Characteristic equation in feedback control: For 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?
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ADepends only upon the open-loop transfer function
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BDetermines the stability of the system
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CIs the same for set-point or load changes
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DAll (a), (b) & (c)
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ENone of these
Answer
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)