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

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)

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

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

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.

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 s^n-1 + a₂s^n-2 + __________a_n-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

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 + 5s² + 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

s³ + 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.

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

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.

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.

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.

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.

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

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.