Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  Control Systems  >  Test: Routh-Hurwitz Stability - Electrical Engineering (EE) MCQ

Test: Routh-Hurwitz Stability - Electrical Engineering (EE) MCQ


Test Description

15 Questions MCQ Test Control Systems - Test: Routh-Hurwitz Stability

Test: Routh-Hurwitz Stability for Electrical Engineering (EE) 2024 is part of Control Systems preparation. The Test: Routh-Hurwitz Stability questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Routh-Hurwitz Stability MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Routh-Hurwitz Stability below.
Solutions of Test: Routh-Hurwitz Stability questions in English are available as part of our Control Systems for Electrical Engineering (EE) & Test: Routh-Hurwitz Stability solutions in Hindi for Control Systems course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Routh-Hurwitz Stability | 15 questions in 30 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study Control Systems for Electrical Engineering (EE) Exam | Download free PDF with solutions
Test: Routh-Hurwitz Stability - Question 1

The characteristic equation of given system is 6s + K = 0. Determined the range of K for which the system to be stable.

Detailed Solution for Test: Routh-Hurwitz Stability - Question 1

Routh-Hurwitz Stability Criterion: 

  • It is used to test the stability of an LTI system.
  • According to the Routh tabulation method, the system is said to be stable if there are no sign changes in the first column of the Routh array.
  • The number of poles lies on the right half of s plane = number of sign changes.
  • If there is a change in sign then the number of sign changes in the first column is equal to the number of roots of the characteristic equation in the right half of the s-plane i.e. equals to the number of roots with positive real parts.

Application:

F(s) = 6s + K

By applying the Routh tabulation method, we get:

System stable when

K > 0

Important Point

A row of zeros in a Routh table:

This situation occurs when the characteristic equation has

  • a pair of real roots with opposite sign (±a)
  • complex conjugate roots on the imaginary axis (± jω)
  • a pair of complex conjugate roots with opposite real parts (-a ± jb, a ± jb)
Test: Routh-Hurwitz Stability - Question 2

The characteristic equation of a linear time-invariant (LTI) system is given by Δ(s) = s4 + 3s3 + 3s2 + s + k = 0. The system is BIBO stable if

Detailed Solution for Test: Routh-Hurwitz Stability - Question 2

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Routh-Hurwitz Stability - Question 3

The characteristic equation of a feedback system is s3 + Ks2 + 5s + 10 = 0. For a stable system, the value of K should not be less than

Detailed Solution for Test: Routh-Hurwitz Stability - Question 3

Concept:

The characteristic equation for a given open-loop transfer function G(s) is

1 + G(s) H(s) = 0

To find the closed system stability by using RH criteria we require a  characteristic equation. Whereas in remaining all stability techniques we require open-loop transfer function.

The nth order general form of CE is

RH table shown below:

Necessary condition: All the coefficients of the characteristic equation should be positive and real.

Sufficient Conditions for stability:

1. All the coefficients in the first column should have the same sign and no coefficient should be zero.

2. If any sign changes in the first column, the system is unstable.

And the number of sign changes = Number of poles in right of s-plane.

Calculation:

Characteristic equation: s3 + Ks2 + 5s + 10 = 0

By applying Routh tabulation method,

The system to become stable, the sign changes in the first column of the Routh table must be zero.

5K – 10 > 0 and K > 0

⇒ K > 2

Test: Routh-Hurwitz Stability - Question 4

A closed loop system has the characteristic equation given by s3 + Ks2 + (K + 2)s + 3 = 0. For this system to be stable, which one of the following conditions should be satisfied?

Detailed Solution for Test: Routh-Hurwitz Stability - Question 4

Given that characteristic equation is,

s3 + Ks2 + (K + 2)s + 3 = 0

For system to be stable,

K > 0, K (K + 2) - 3 > 0

⇒ K > 0, K2 + 2K - 3 > 0

⇒ K > 0, (K + 3) (K - 1) > 0

⇒ K > 0, K > -3, K > 1 ⇒ K > 1

Test: Routh-Hurwitz Stability - Question 5

Routh Hurwitz criterion is used to determine

Detailed Solution for Test: Routh-Hurwitz Stability - Question 5

Routh-Hurwitz criterion:

  • Using the Routh-Hurwitz method, the stability information can be obtained without the need to solve the closed-loop system poles. This can be achieved by determining the number of poles that are in the left-half or right-half plane and on the imaginary axis.
  • This involves checking the roots of the characteristic polynomial of a linear system to determine its stability.
  • It is used to determine the absolute stability of a system.

Important points

Other methods of determining stability include:

Root locus:

  • This method gives the position of the roots of the characteristic equation as the gain K is varied.
  • With Root locus (unlike the case with Routh-Hurwitz criterion), we can do both analysis (i.e., for each gain value we know where the closed-loop poles are) and design (i.e., on the curve we can search for a gain value that results in the desired closed-loop poles).

Nyquist plot:

  • This method is mainly used for assessing the stability of a system with feedback.
  • While Nyquist is a graphical technique, it only provides a limited amount of intuition for why a system is stable or unstable, or how to modify an unstable system to be stable.

Techniques like Bode plots, while less general, are sometimes a more useful design tool.

Test: Routh-Hurwitz Stability - Question 6

What is the stability of the system s3 + s2 + s + 4 = 0 using Hurwitz criteria?

Detailed Solution for Test: Routh-Hurwitz Stability - Question 6

Concept:

To find the closed system stability by using RH criteria we require a characteristic equation.

Whereas in remaining all stability techniques we require open-loop transfer function.

The nth order general form of CE is a0 sn + a1 sn-1 + a2sn-2 + __________an-1 s1 +

an RH table shown below:

Necessary condition

All the coefficients of the characteristic equation should be positive and real.

Sufficient Conditions for stability:

1. All the coefficients in the first column should have the same sign and no coefficient should be zero.

2. If any sign changes in the first column, the system is unstable. And the number of sign changes = Number of poles in right of s-plane.

Calculation:

With the help of the

Routh table explained in the concept we can extend the calculation.

So in the first column, there are two sign changes, the system is unstable

Test: Routh-Hurwitz Stability - Question 7

The characteristic polynomial of a linear system is given as s4 + 3s3 + 5s+ 6s + K + 10 = 0. What should be the condition on K so that the system is stable?

Detailed Solution for Test: Routh-Hurwitz Stability - Question 7

Concept:

The characteristic equation for a given open-loop transfer function G(s) is

1 + G(s) H(s) = 0

To find the closed system stability by using RH criteria we require a  characteristic equation. Whereas in remaining all stability techniques we require open-loop transfer function.

The nth order general form of CE is

RH table shown below:

Necessary condition: All the coefficients of the characteristic equation should be positive and real.

Sufficient Conditions for stability:

1. All the coefficients in the first column should have the same sign and no coefficient should be zero.

2. If any sign changes in the first column, the system is unstable.

And the number of sign changes = Number of poles in right of s-plane.

Calculation:

Characteristic equation: s4 + 3s3 + 5s+ 6s + K + 10 = 0

By applying Routh tabulation method,

The system to become stable, the sign changes in the first column of the Routh table must be zero.

- 4 - K > 0 and K + 10 > 0

4 + K < 0 and K + 10 > 0

K < - 4 and K > - 10 

⇒ - 10 < K < - 4

*Answer can only contain numeric values
Test: Routh-Hurwitz Stability - Question 8

Given the following polynomial equation

s+ 5.5s2 + 8.5s + 3 = 0

the number of roots of the polynomial, which have real parts strictly less than −1, is ________ 


Detailed Solution for Test: Routh-Hurwitz Stability - Question 8

s+ 5.5s2 + 8.5s + 3 = 0

Putting s = z -1

z3 + 2.5z2 + 0.5z - 1 = 1

Taking routh criteria for the given characteristic equation

As there is one sign change hence, two roots of given polynomial will lie to the left of s = -1.

Test: Routh-Hurwitz Stability - Question 9

The number of roots of s3 + 5s2 + 7s + 3 = 0 in the left half of the s-plane is

Detailed Solution for Test: Routh-Hurwitz Stability - Question 9

Concept:

The characteristic equation for a given open-loop transfer function G(s) is

1 + G(s) H(s) = 0

According to the Routh tabulation method,

The system is said to be stable if there are no sign changes in the first column of the Routh array

The number of poles lies on the right half of s plane = number of sign changes

Calculation:

Characteristic equation: s3 + 5s+ 7s + 3 = 0

By applying the Routh tabulation method,

Number of roots lie on the right side of the s-plane = Number of sign changes = 0

Number of roots lie on the left side of the s-plane = Total number of roots of the characteristic equation - number of roots lie on the right side of the s-plane

As the characteristic equation is a cubic equation 

∴ Total number of roots of the characteristic equation = 3

Hence, the number of roots lie on the left side of the s-plane = 3 - 0 

Number of roots that lie on the left side of the s-plane = 3

Test: Routh-Hurwitz Stability - Question 10

In the formation of Routh-Hurwitz array for a polynomial, all the elements of a row have zero values. This premature termination of the array indicates the presence of

1. a pair of real roots with opposite sign

2. complex conjugate roots on the imaginary axis

3. a pair of complex conjugate roots with opposite real parts

Which of the above statements are correct?

Detailed Solution for Test: Routh-Hurwitz Stability - Question 10

Routh-Hurwitz Stability Criterion: It is used to test the stability of an LTI system.

The characteristic equation for a given open-loop transfer function G(s) is

1 + G(s) H(s) = 0

According to the Routh tabulation method,

The system is said to be stable if there are no sign changes in the first column of the Routh array

The number of poles lies on the right half of s plane = number of sign changes

A row of zeros in a Routh table:

This situation occurs when the characteristic equation has

  • a pair of real roots with opposite sign (±a)
  • complex conjugate roots on the imaginary axis (± jω)
  • a pair of complex conjugate roots with opposite real parts (-a ± jb, a ± jb)

The procedure to overcome this as follows:

  • Form the auxiliary equation from the preceding row to the row of zeros
  • Complete Routh array by replacing the zero row with the coefficients obtained by differentiating the auxiliary equation.
  • The roots of the auxiliary equation are also the roots of the characteristic equation.
  • The roots of the auxiliary equation occur in pairs and are of the opposite sign of each other.
  • The auxiliary equation is always even in order.
Test: Routh-Hurwitz Stability - Question 11

Which of the following is the correct comment on stability based on unknown k for the feedback system with characteristic s4 + 2ks3 + s2 + 5s + 5 = 0?

Detailed Solution for Test: Routh-Hurwitz Stability - Question 11

Concept:

The characteristic equation for a given open-loop transfer function G(s) is

1 + G(s) H(s) = 0

According to the Routh tabulation method,

The system is said to be stable if there are no sign changes in the first column of Routh array

The number of poles lie on the right half of s plane = number of sign changes

Calculation:

Characteristic equation: s4 + 2ks3 + s2 + 5s + 5 = 0

By applying the Routh tabulation method,

The system to become stable, the sign changes in the first column of Routh table must be zero.

⇒ k > 2.5

For all the values of  gives negative values.

Therefore, the given system is unstable for all the values of k.

Test: Routh-Hurwitz Stability - Question 12

Find the number of poles in the right-half plane (RHP) for the system as shown. Is the system stable?

Detailed Solution for Test: Routh-Hurwitz Stability - Question 12

Concept:

The characteristic equation for a given open-loop transfer function G(s) is

1 + G(s) H(s) = 0

According to the Routh tabulation method,

The system is said to be stable if there are no sign changes in the first column of the Routh array

The number of poles lie on the right half of s plane = number of sign changes

Calculation:

Characteristic equation: 

⇒ 2s5 + 3s4 + 2s3 + 3s2 + 2s + 1 = 0

By applying Routh tabulation method,

As ε is very small value, (3 – 4/ε) is a negative value and hence there are two sign changes.

Therefore, the number of right-half poles = 2

The system is unstable.

Test: Routh-Hurwitz Stability - Question 13

Match List I with List II:

Choose the correct answer from the options given below:

Detailed Solution for Test: Routh-Hurwitz Stability - Question 13

The locus of the open loop poles of the root locus is represented as:

The solution of s2+a1s+a2 = 0 is given by:

Nature of roots

Case 1: When roots are negative, real, and equal

The imaginary part must be zero

Case 2: When roots are conjugate imaginary

The imaginary part must exist and the real part should be zero.

a1 = 0 and a2 ≠ 0

Case 3: Negative Real and Unequal

Case 4: Conjugate Complex

Both real and imaginary parts must exist

Hence, the correct answer is option c.

Test: Routh-Hurwitz Stability - Question 14

Find the range of k for stable operation if H(s) = 1 and G(s) 

Detailed Solution for Test: Routh-Hurwitz Stability - Question 14

Concept:

  • The characteristic equation for a given open-loop transfer function
  • G(s) is 1 + G(s) H(s) = 0
  • According to the Routh tabulation method, The system is said to be stable if there are no sign changes in the first column of the Routh array The number of poles lies on the right half of s plane = number of sign changes

Calculation:

Characteristic equation: s3 + 3s2 + 2s + k = 0 By applying the Routh tabulation method,

For the system to become stable, the sign changes in the first column of the Routh table must be zero.

6 - K > 0 and K > 0, 

The range of k for stable operation 0 < K < 6

Test: Routh-Hurwitz Stability - Question 15

According to Hurwitz criterion, the characteristic equation

s4 + 8s3 + 18s2 + 16s + 5 = 0 is

Detailed Solution for Test: Routh-Hurwitz Stability - Question 15

Concept:

The Routh Stability Criterion is used to test the stability of an LTI system. The conditions for stability are:

  •  All the coefficients of the characteristic Equation must be present and must have the same sign.
  • It is necessary and sufficient that each term of the first column of the Routh Array of the Characteristic Equation is positive for the system to be stable, i.e. there should not be any sign changes in the first column of each row.
  • The number of sign changes represents the number of roots on the right side of the s-plane.
  • If the first term in any row of Routh Array is zero while the rest of the row has at least one non-zero term. Because of this term, the terms in the next row will become infinite.
  • When all elements in any row of the Routh array are zero. This condition indicates that there are symmetrical/imaginary roots in the s-plane.

Application:

A(s) = s4 + 8s3 + 18s2 + 16s + 5

Forming the Routh array, we get:

Since there are no sign changes in the first column of the Routh array, we conclude that the system is stable.

53 videos|73 docs|40 tests
Information about Test: Routh-Hurwitz Stability Page
In this test you can find the Exam questions for Test: Routh-Hurwitz Stability solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Routh-Hurwitz Stability, EduRev gives you an ample number of Online tests for practice

Top Courses for Electrical Engineering (EE)

53 videos|73 docs|40 tests
Download as PDF

Top Courses for Electrical Engineering (EE)