चरण पश्च क्षतिपूरक का स्थानांतरण फंलन क्या है ∝ और τ के मान ∝ > 1 और τ > 0 के रूप में दिए गए हैं?

This question was previously asked in
ESE Electronics 2011 Paper 2: Official Paper
View all UPSC IES Papers >
  1. \(\frac{1}{ \propto }\frac{{\left( {s + \frac{1}{\tau }} \right)}}{{\left( {s + \frac{1}{{ \propto \tau }}} \right)}}\)
  2. \(\frac{1}{ \propto }\frac{{\left( {s - \frac{1}{\tau }} \right)}}{{\left( {s -\frac{1}{{ \propto \tau }}} \right)}}\)
  3. \(\frac{1}{ \propto }\frac{{\left( {s + \frac{1}{\tau }} \right)}}{{\left( {s - \frac{1}{{ \propto \tau }}} \right)}}\)
  4. \(\frac{1}{ \propto }\frac{{\left( {s - \frac{1}{\tau }} \right)}}{{\left( {s + \frac{1}{{ \propto \tau }}} \right)}}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{1}{ \propto }\frac{{\left( {s + \frac{1}{\tau }} \right)}}{{\left( {s + \frac{1}{{ \propto \tau }}} \right)}}\)
Free
ST 1: UPSC ESE (IES) Civil - Building Materials
6.3 K Users
20 Questions 40 Marks 24 Mins

Detailed Solution

Download Solution PDF

संकल्पना:

चरण पश्च क्षतिपूरक का स्थानांतरण फंलन निम्न द्वारा दिया जाता है

\(G\left( s \right) = \frac{{1 + \tau s}}{{1 + a\tau s}}\)

> 1 और > 0

\(G\left( s \right) = \frac{\tau ({s + \frac{1}{\tau}})}{\alpha \tau ({s + \frac{1}{\alpha \tau}})}\)

\(G\left( s \right) = \frac{ ({s + \frac{1}{\tau}})}{\alpha ({s + \frac{1}{\alpha \tau}})}\)

एक पश्च क्षतिपूरक के लिए ध्रुव शून्य की तुलना में मूल के अधिक करीब होते हैं।

महत्वपूर्ण बिंदु

चरण पश्च क्षतिपूरक का स्थानांतरण फंलन निम्न द्वारा दिया जाता है

\(G\left( s \right) = \frac{{1 + Ts}}{{1 + aTs}}\)

\(G\left( {j\omega } \right) = \frac{{1 + j\omega T}}{{1 + j\omega aT}}\)

चरण कोण, \(\angle G\;\left( {j\omega } \right) = {\tan ^{ - 1}}\omega T - {\tan ^{ - 1}}a\omega T\)

ϕ = tan-1 ωT – tan-1 aωT

अधिकतम चरण ωm पर पश्च होता है जैसे कि \(\frac{{d\phi }}{{d\omega }} = 0\)

\(\Rightarrow {\omega _m} = \frac{1}{{T\sqrt a }}\)

यह इसकी दो कोनों की आवृत्तियों का एक ज्यामितीय माध्य है

\({\omega _m} = \sqrt {\frac{1}{T} \times \frac{1}{{aT}}} = \frac{1}{{T\sqrt a }}\)

अधिकतम चरण पश्च,

\({\phi _m} = {\tan ^{ - 1}}\omega T - {\tan ^{ - 1}}a\omega T\)

\(= {\tan ^{ - 1}}\left( {\frac{1}{{T\sqrt \alpha }}} \right)T - {\tan ^{ - 1}}\left( {\frac{1}{{T\sqrt \alpha }}T} \right)\)

\(= {\tan ^{ - 1}}\frac{1}{{\sqrt a }} - {\tan ^{ - 1}}\sqrt a\)

\({\phi _m} = {\tan ^{ - 1}}\left( {\frac{{\frac{1}{{\sqrt a }} - \sqrt a }}{{1 + \sqrt a \cdot \frac{1}{{\sqrt a }}}}} \right)\)

\({\phi _m} = {\tan ^{ - 1}}\left( {\frac{{1 - a}}{{2\sqrt a }}} \right)\)

\(= {\sin ^{ - 1}}\left( {\frac{{1 - a}}{{a + 1}}} \right) = {\cos ^{ - 1}}\left( {\frac{{2\sqrt a }}{{a + 1}}} \right)\)
Latest UPSC IES Updates

Last updated on May 28, 2025

->  UPSC ESE admit card 2025 for the prelims exam has been released. 

-> The UPSC IES Prelims 2025 will be held on 8th June 2025.

-> The selection process includes a Prelims and a Mains Examination, followed by a Personality Test/Interview.

-> Candidates should attempt the UPSC IES mock tests to increase their efficiency. The UPSC IES previous year papers can be downloaded here.

More Compensators Questions

More Controllers and Compensators Questions

Get Free Access Now
Hot Links: online teen patti real money teen patti diya teen patti master 2024 teen patti royal teen patti master real cash