The magnitude of frequency response of an underdamped second-order system is 5 at 0 rad/sec and peaks at \(\frac{10}{\sqrt{3}}\) at 5√2 rad/sec. The transfer function of the system is

This question was previously asked in
UPSC ESE (Prelims) Electronics and Telecommunication Engineering 19 Feb 2023 Official Paper
View all UPSC IES Papers >
  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 : \(\frac{{245}}{{\left( {{s^2} + 1.21s + 49} \right)}}\)
Free
ST 1: UPSC ESE (IES) Civil - Building Materials
6.2 K Users
20 Questions 40 Marks 24 Mins

Detailed Solution

Download Solution PDF

Concept:

DC gain = 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

If resonant peak > 1 then

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

∴ 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

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

K = 49 × 5 = 245

So the required transfer function = \(\frac{{245}}{{\left( {{s^2} + 1.21s + 49} \right)}}\)

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 Frequency Response Analysis Questions

Get Free Access Now
Hot Links: teen patti flush teen patti yes teen patti gold apk download lotus teen patti