Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
179 views
in Electronics by (237k points)
closed by
A unit feedback system has the open loop transfer function \(G(s) = \frac {10^4}{s(s+10)^2}\). The closed loop system is
1. stable
2. marginally stable
3. unstable
4. stable at an angular frequency of 10 rad / s

1 Answer

0 votes
by (239k points)
selected by
 
Best answer
Correct Answer - Option 3 : unstable

Routh-Hurwitz stability criterion: 

  • Used to find whether the system is stable or unstable.
  • Used to find the how many roots of the given system lies in left and right half of the s-plane. 

 

Calculation:

Given:

Open-loop transfer function ⇒ \(G\left( s \right) = \frac{{{{10}^4}}}{{s{{\left( {s + 10} \right)}^2}}}\)

For unity feedback ⇒ H(s) = 1

Characteristic equation ⇒ 1 + G(s)H(s) = 0

⇒ \(1 + \frac{{{{10}^4}}}{{s{{\left( {s + 10} \right)}^2}}} = 0\)

⇒ s(s+10)2 + 104 = 0

⇒ s3 + 20s2 +100s + 10000 = 0

Routh Hurwitz table:

s3

1

100

s2

20

10000

s1

\(\frac{{20 \times 100 - 1 \times 10000}}{{20}} = - 400\)

 

s0

\(\frac{{ - 400 \times 1000 - 0}}{{ - 400}} = 10000\)

 

 

There are 2 sign change

First from 20 to -400

Second from -400 to 10000

So, two roots will lie on the right half of the s-plane.

System is unstable.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...