This question was asked inGPSC Deputy Executive Engineer (Electrical) Paper 2024Question 8 of 33 published · Gujarat Public Service CommissionOpen paper
Q9ModerateControllers & Feedback

Read the following statements in the context of negative feedback system.
Statement 1: Negative feedback always makes the overall system response stable.
Statement 2: Negative feedback system may improve the system’s robustness towards parametric uncertainties.
Choose the correct option from the following:

Answer & step-by-step solution
Correct answer: C. Statement 2 is correct and statement 1 is incorrect.
Quick explanation

Statement 1 is false because negative feedback does not always guarantee stable closed-loop poles. Statement 2 is true because negative feedback can reduce parameter sensitivity; for a standard loop, S = 1/[1 + G(s)H(s)]. Therefore option C is correct.

T(s) = G(s)/[1 + G(s)H(s)] Characteristic equation: 1 + G(s)H(s) = 0 Sensitivity: S = 1/[1 + G(s)H(s)] T(s) = G(s)/[1 + G(s)H(s)] Characteristic equation: 1 + G(s)H(s) = 0 S = 1/[1 + G(s)H(s)]

What This Question Is Really Testing

Two different ideas are being tested: closed-loop stability and robustness to parameter variation . Negative feedback can improve both in many practical designs, but it guarantees neither simply because the feedback sign is negative.

Concept Foundation

For a standard negative-feedback loop:

T(s) = G(s)/[1 + G(s)H(s)]

Characteristic equation: 1 + G(s)H(s) = 0

The roots of the characteristic equation are the closed-loop poles. Their locations decide stability.

Step 1 — Evaluate Statement 1

Statement 1 is false. Negative feedback does not automatically force all closed-loop poles into the left half-plane. If the loop gain and phase cause the characteristic equation to have unstable roots, the closed-loop system is unstable even though the feedback connection is negative.

Step 2 — Evaluate Statement 2

Statement 2 is true. One important benefit of negative feedback is reduced sensitivity to changes in forward-path parameters.

S = 1/[1 + G(s)H(s)]

When the loop gain is appropriately large and the loop remains stable, the sensitivity magnitude can become much smaller than one.

Step 3 — Select the Statement Combination

Statement 1 is false and Statement 2 is true. Hence the correct option is C .

Why This Method Works

The denominator 1 + G(s)H(s) simultaneously explains both ideas: its roots govern stability, while the same loop-gain term appears in the sensitivity reduction factor.

Independent Verification / Cross-Check

The logic is internally consistent: a system can be robust to moderate parameter changes yet still require an independent stability check from poles, Routh, Nyquist or Bode margins.

Concept Extension

  • Negative feedback commonly reduces sensitivity and steady-state error.
  • It may increase bandwidth.
  • Improper loop phase or excessive gain can still create instability.

Exam Strategy

In statement questions, watch absolute words such as always . Separate “feedback benefit” from “guaranteed stability,” then judge each statement independently.

Quick Trick

“Always stable” is too strong; “may improve robustness” is valid.

Why Other Options Are Wrong

A and B incorrectly accept Statement 1. D incorrectly rejects the valid robustness statement. Only C matches “Statement 1 false, Statement 2 true.”

Common Mistake

Equating “negative feedback usually improves stability” with “negative feedback always guarantees stability.”

Solution diagram – Read the following statements in the context of negative feedback system.…
Exam tipCheck feedback statements separately: poles decide stability; sensitivity function explains robustness.
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