નેગેટિવ ફીડબેક સિસ્ટમના સંદર્ભમાં નીચેના વિધાનો વાંચો.
વિધાન 1: નેગેટિવ ફીડબેક હંમેશા સમગ્ર સિસ્ટમ પ્રતિભાવને સ્થિર બનાવે છે.
વિધાન 2: નેગેટિવ ફીડબેક સિસ્ટમ પેરામેટ્રિક અનિશ્ચિતતાઓ સામે સિસ્ટમની મજબૂતાઈ (robustness) વધારી શકે છે.
નીચેનામાંથી સાચો વિકલ્પ પસંદ કરો:
જવાબ અને પગલાવાર ઉકેલ
વિધાન 1 ખોટું છે કારણ કે નેગેટિવ ફીડબેક હંમેશા સ્થિર ક્લોઝ્ડ-લૂપ પોલની ખાતરી આપતું નથી. વિધાન 2 સાચું છે કારણ કે નેગેટિવ ફીડબેક પેરામીટર સંવેદનશીલતા ઘટાડે છે; પ્રમાણભૂત લૂપ માટે S = 1/[1 + G(s)H(s)]. તેથી વિકલ્પ C સાચો છે.
વિગતવાર ઉકેલ (મૂળ ભાષામાં)
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.”

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
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.
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.”

नेगेटिव फीडबैक सिस्टम के संदर्भ में निम्नलिखित कथन पढ़िए।
कथन 1: नेगेटिव फीडबैक हमेशा समग्र सिस्टम प्रतिक्रिया को स्थिर बनाता है।
कथन 2: नेगेटिव फीडबैक सिस्टम पैरामीट्रिक अनिश्चितताओं के प्रति सिस्टम की मज़बूती (robustness) बढ़ा सकता है।
निम्नलिखित में से सही विकल्प चुनिए:
उत्तर और चरणबद्ध हल
कथन 1 गलत है क्योंकि नेगेटिव फीडबैक हमेशा स्थिर क्लोज़्ड-लूप पोल की गारंटी नहीं देता। कथन 2 सही है क्योंकि नेगेटिव फीडबैक पैरामीटर संवेदनशीलता घटाता है; मानक लूप के लिए S = 1/[1 + G(s)H(s)]। अतः विकल्प C सही है।
विस्तृत हल (मूल भाषा में)
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.”
