ઓપન-લૂપ ટ્રાન્સફર ફંક્શન G(s) = 4/s(s + 5) ધરાવતી ફીડબેક સિસ્ટમનો સ્ટેપ ઇનપુટ માટે પ્રતિભાવ શું છે?
જવાબ અને પગલાવાર ઉકેલ
યુનિટી ફીડબેકમાં પહેલાં T(s)=G(s)/(1+G(s)) બનાવો, પછી સ્ટેપ ઇનપુટ 1/s વડે ગુણી આંશિક અપૂર્ણાંક કરો. G(s)=4/[s(s+5)] માટે ક્લોઝ્ડ-લૂપ છેદ s²+5s+4=(s+1)(s+4) છે, જેથી પ્રતિભાવ 1−(1/3)(4e⁻ᵗ−e⁻⁴ᵗ) મળે છે.
વિગતવાર ઉકેલ (મૂળ ભાષામાં)
Concept & Reasoning
The closed-loop transfer function is 4/[s(s+5)+4]=4/[(s+1)(s+4)]. For a unit step, C(s)=4/[s(s+1)(s+4)]. Partial-fraction expansion gives 1/s−(4/3)/(s+1)+(1/3)/(s+4). Taking the inverse Laplace transform produces the keyed result.
How to Solve It in the Exam
Unity feedback: close the loop before applying the input.
Important Exam Point
Check c(0⁺)=0 and c(∞)=1 as quick validation.
Quick Trick
Unity feedback: close the loop before applying the input.
Common Mistake
Using open-loop G(s) directly as the step-response transfer function.
What is the response of the feedback function with open loop transfer function G(s) = 4/s(s + 5) and input is input step?
Answer & step-by-step solution
For unity feedback, first form T(s)=G(s)/(1+G(s)), then multiply by the step input 1/s and perform partial fractions. For G(s)=4/[s(s+5)], the closed-loop denominator is s²+5s+4=(s+1)(s+4), giving the response 1−(1/3)(4e⁻ᵗ−e⁻⁴ᵗ).
Concept & Reasoning
The closed-loop transfer function is 4/[s(s+5)+4]=4/[(s+1)(s+4)]. For a unit step, C(s)=4/[s(s+1)(s+4)]. Partial-fraction expansion gives 1/s−(4/3)/(s+1)+(1/3)/(s+4). Taking the inverse Laplace transform produces the keyed result.
How to Solve It in the Exam
Unity feedback: close the loop before applying the input.
Important Exam Point
Check c(0⁺)=0 and c(∞)=1 as quick validation.
Quick Trick
Unity feedback: close the loop before applying the input.
Common Mistake
Using open-loop G(s) directly as the step-response transfer function.
ओपन-लूप ट्रांसफर फंक्शन G(s) = 4/s(s + 5) वाले फीडबैक सिस्टम की स्टेप इनपुट के लिए प्रतिक्रिया क्या है?
उत्तर और चरणबद्ध हल
यूनिटी फीडबैक में पहले T(s)=G(s)/(1+G(s)) बनाएँ, फिर स्टेप इनपुट 1/s से गुणा कर आंशिक भिन्न करें। G(s)=4/[s(s+5)] के लिए क्लोज़्ड-लूप हर s²+5s+4=(s+1)(s+4) है, जिससे प्रतिक्रिया 1−(1/3)(4e⁻ᵗ−e⁻⁴ᵗ) मिलती है।
विस्तृत हल (मूल भाषा में)
Concept & Reasoning
The closed-loop transfer function is 4/[s(s+5)+4]=4/[(s+1)(s+4)]. For a unit step, C(s)=4/[s(s+1)(s+4)]. Partial-fraction expansion gives 1/s−(4/3)/(s+1)+(1/3)/(s+4). Taking the inverse Laplace transform produces the keyed result.
How to Solve It in the Exam
Unity feedback: close the loop before applying the input.
Important Exam Point
Check c(0⁺)=0 and c(∞)=1 as quick validation.
Quick Trick
Unity feedback: close the loop before applying the input.
Common Mistake
Using open-loop G(s) directly as the step-response transfer function.
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