एक अतिअवमंदित दूसरे क्रम प्रणाली की आवृत्ति प्रतिक्रिया का परिमाण 0 रेड/सेकंड पर 5 है और 5 √2 रेड/सेकंड पर \(\frac{10}{\sqrt{3}}\) पर शीर्ष है। प्रणाली का स्थानांतरण फलन है

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UPSC ESE (Prelims) Electronics and Telecommunication Engineering 19 Feb 2023 Official Paper
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  1. \(\frac{{245}}{{\left( {{s^2} + 1.21s + 49} \right)}}\)
  2. \(\frac{375}{s^2+5 s+75}\)
  3. \(\frac{500}{s^2+12 s+100}\)
  4. \(\frac{1125}{s^2+25 s+225}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{{245}}{{\left( {{s^2} + 1.21s + 49} \right)}}\)
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ST 1: UPSC ESE (IES) Civil - Building Materials
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20 Questions 40 Marks 24 Mins

Detailed Solution

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अवधारणा:

DC लब्धि = 5

\({M_r} = \frac{10}{\sqrt 3} = \frac{1}{{2\xi \sqrt {1 - {\xi ^2}} }}\)

100 [4ε2 (1 – ε2)] = 3

400 ε2– 400 ε4 = 3

400x2 – 400 x + 3 = 0  

x2 – x + 0.0075 = 0

ξ2 = 0.99244 , ξ2 = 0.00755

ξ = 0.99 , ξ = 0.08689

यदि अनुनाद शिखर > 1 तब

 \(\zeta \le \frac{1}{{\sqrt 2 }} = 0.707\)

∴ ξ = 0.08689

\(5\sqrt 2 = {\omega _n}\sqrt {1 - 2{\xi ^2}} \)

\(5\sqrt 2 = {\omega _n}\sqrt {1 - 2 \times 0.00755} \)

ωn = 7 rad/s

\({s^2} + 2\xi {\omega _n}s + \omega _n^2\)

= s2+ 1.21 s + 49

अब \({\left. {\frac{K}{{{s^2} + 1.21 s + 49}}} \right|_{s = 0}} = 5\) 

K = 49 × 5 = 245

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