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


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10 Questions MCQ Test GATE Electrical Engineering (EE) Mock Test Series 2025 - Test: Routh-Hurwitz Stability Criteria

Test: Routh-Hurwitz Stability Criteria for Electrical Engineering (EE) 2024 is part of GATE Electrical Engineering (EE) Mock Test Series 2025 preparation. The Test: Routh-Hurwitz Stability Criteria questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Routh-Hurwitz Stability Criteria 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 Criteria below.
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Test: Routh-Hurwitz Stability Criteria - Question 1

The forward transfer function of a ufb system is  The system will be

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

Given transfer function 
Characteristic equation: 2s4 + 5s3 + s2 + 2s + 1 = 0

As there are sign changes, the system is unstable.

Test: Routh-Hurwitz Stability Criteria - Question 2

Consider the characteristic equation of a control system given by s3 + (K + 0.5)s2 + 4Ks + 50 = 0. Find the value of the frequency if the system has sustained oscillations for a given K.

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

Characteristic Equation 
s3 + (K + 0.5)s2 + 4Ks + 50 = 0
For cubic equation
Inner product = Outer product
(K+0.5)4K = 50
4K2 +2K-50 =0
K= 3.29
For Frequency of oscillation :
Auxillary Equation 
3.79 S2 + 50 =0
S= 3.63 rad/sec

Test: Routh-Hurwitz Stability Criteria - Question 3

Determine the stability range of k for a feedback control system having characteristic equation -
s4 + 2s3 + 10s2 + (k − 10)s + k = 0

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

Concept:
Routh-Hurwitz Stability Criterion
The necessary condition is that the coefficients of the characteristic polynomial should be positive. This implies that all the roots of the characteristic equation should have negative real parts.
The sufficient condition is that all the elements of the first column of the Routh array should have the same sign. This means that all the elements of the first column of the Routh array should be either positive or negative.
Calculation:
Given;
Characteristic equation: 
s4 + 2s3 + 10s2 + (k − 10)s + k = 0

For the system to be stable:
The necessary condition is that the coefficients of the characteristic polynomial should be positive.
(K-10) > 0 
K > 10
The sufficient condition is that all the elements of the first column of the Routh array should have the same sign. This means that all the elements of the first column of the Routh array should be either positive or negative.

Condition: (K > 10) ∩ (K < 13.10) ∩ (K < 22.89)
Pictorially 

In the shaded region all the conditions will be satisfy
Condition for stability: (10 < K < 13.10)

Test: Routh-Hurwitz Stability Criteria - Question 4

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 Criteria - Question 4

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 Criteria - Question 5

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

  • a pair of real roots with opposite sign
  • complex conjugate roots on the imaginary axis
  • a pair of complex conjugate roots with opposite real parts

Which of the above statements are correct?

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

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 Criteria - Question 6

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 Criteria - Question 6

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.

For all the values of k > 2.5,  gives negative values.
Therefore, the given system is unstable for all the values of k.

Test: Routh-Hurwitz Stability Criteria - Question 7

Which one of the following options correctly describes the locations of the roots of the equation s4 + s2 + 1 = 0 on the complex plane?

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

CE: s4 + 0s3 + 1s2 + 0s + 1
Routh array

We have row zero at s3 row
Solving the auxiliary equation, we get:
s4 + s2 + 1 = 0
By differentiating, we get:
4s3 + 2s = 0
The Routh array is modified as shown above.
Observations:
The row of zero indicates symmetric roots about the origin.
2 sign changes below row of zero indicate 2 poles in the right half of the s-plane.
∴ Two poles are on the right side and 2 poles symmetrically lying on left-half.

Test: Routh-Hurwitz Stability Criteria - Question 8

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 Criteria - Question 8

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 + 5s2 + 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 Criteria - Question 9

Consider the characteristic equation of a control system given by s3 + (K + 0.5)s2 + 4Ks + 50 = 0. Find the value of K for the system to have sustained oscillations.

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

Given characteristic equation of a control system is:
s3 + (K + 0.5)s2 + 4Ks + 50 = 0
By applying Routh Tabulation method:

The condition for the system to be marginally stable is:
4K(K + 0.5) − 50 = 0
4K+ 2K − 50 = 0
2K+ K − 25 = 0
On finding the roots for the above equation, K = 3.28 
Therefore the value of K for the system to have sustained oscillations is 3.3.

Test: Routh-Hurwitz Stability Criteria - Question 10

How many roots of characteristic equation P(s) = s4 + s3 + 2s2 + 2s + 3 have (+)ve real part?

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

NOTE:

  • If all the elements of a row is zero in a Routh Hurwitz table then we consider it as a row of zeros(ROZ) , for eliminating this ROZ we find the derivative of the above row and write its coefficient
  • Note that ROZ occurs in odd power of S rows only
  • If the first element is only zero  in any row then in that place we consider an arbitrary constant whose value is tending to zero
  • Order of the algebraic equation gives the number of poles 
  • Number of sign changes in first column of a row gives the number of poles at the right side of the origin, or on the positive side.

Now, forming Routh Hurwitz table fore the given equation 

so as we see at s1 we get negative infinite, which means we get two significant changes on the first column means two poles at the right-hand side of the origin.
 or We get two poles on the positive hand side

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