Stability Analysis MCQ Quiz - Objective Question with Answer for Stability Analysis - Download Free PDF

Let's analyze the given Routh array and determine how to replace the row of zeros.

Understanding the Routh Array and Row of Zeros:

• The Routh array is used to determine the stability of a linear time-invariant system.
• A row of zeros in the Routh array indicates the presence of roots on the imaginary axis (jω-axis) or roots that are symmetrical about the origin.
• To continue the Routh array and determine stability, we need to form an auxiliary polynomial from the row above the row of zeros.

Forming the Auxiliary Polynomial:

• The row above the row of zeros is s⁴.
• The coefficients in this row are used to form the auxiliary polynomial.
• The auxiliary polynomial is formed using only even powers of 's'.

In this case, the row s⁴ has coefficients 2, 12, and 16.

Therefore, the auxiliary polynomial is:

A(s) = 2s⁴ + 12s² + 16

Finding the Derivative of the Auxiliary Polynomial:

To replace the row of zeros, we need to find the derivative of the auxiliary polynomial with respect to 's'.

dA(s)/ds = 8s³ + 24s

Replacing the Row of Zeros:

The coefficients of the derivative will replace the row of zeros in the Routh array.

The coefficients of the derivative are 8 and 24.

We can simplify these coefficients by dividing by 8:

8/8 = 1 24/8 = 3

So, the simplified coefficients are 1 and 3.

The row of zeros will be replaced by the coefficients of s³ + 3s.

Therefore, the correct answer is option 2.

Explanation:

In control system analysis, determining the stability and behavior of a system involves understanding the location of the roots (or poles) of its characteristic equation in the complex plane. The characteristic equation is typically derived from the system's transfer function or differential equations describing the system dynamics.

Given the characteristic equation:

s³ + 5s² + 7s + 3 = 0

We need to determine the number of roots (poles) that lie in the left half of the s-plane. This is crucial because the location of these roots directly affects the stability of the system. Roots in the left half-plane (with negative real parts) indicate a stable system, whereas roots in the right half-plane (with positive real parts) indicate instability.

Routh-Hurwitz Criterion:

One common method to determine the number of roots in the left half-plane is the Routh-Hurwitz criterion. This criterion uses the coefficients of the characteristic polynomial to form the Routh array, which can then be analyzed to determine the number of roots with positive real parts (i.e., in the right half-plane).

Formation of the Routh Array:

For the given characteristic equation:

s³ + 5s² + 7s + 3 = 0

The Routh array is formed as follows:

The elements in the first column are calculated as follows:

Thus, the Routh array is:

Analysis of the Routh Array:

To determine the number of roots in the left half-plane, we need to count the number of sign changes in the first column of the Routh array. The first column is:

1, 5, 6.4, 3

There are no sign changes in this sequence, indicating that all the roots have negative real parts (i.e., they are all in the left half-plane). Therefore, there are zero roots in the right half-plane, and all roots are in the left half-plane.

Conclusion:

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

Option 4: Three

This is the correct option. All roots are in the left half-plane, indicating the system is stable

Explanation:

To determine the value of \( K \) for the system to be stable, we first need to analyze the open-loop transfer function of the system. The given transfer function of the system is:

\[ G(s) = \frac{K}{s(s+2)(s+4)} \] and the feedback transfer function is \( H(s) = 1 \).

The closed-loop transfer function for a unity feedback system is given by:

\[ T(s) = \frac{G(s)}{1 + G(s)H(s)} = \frac{\frac{K}{s(s+2)(s+4)}}{1 + \frac{K}{s(s+2)(s+4)}} \]

To simplify, multiply the numerator and the denominator by the denominator of \( G(s) \):

\[ T(s) = \frac{K}{s(s+2)(s+4) + K} \]

For the system to be stable, all the poles of the closed-loop transfer function must lie in the left half of the s-plane (i.e., all the poles must have negative real parts). The poles of the closed-loop transfer function are the roots of the characteristic equation:

\[ 1 + G(s)H(s) = 0 \rightarrow 1 + \frac{K}{s(s+2)(s+4)} = 0 \]

Multiplying through by \( s(s+2)(s+4) \):

\[ s(s+2)(s+4) + K = 0 \]

Expanding the polynomial:

\[ s^3 + 6s^2 + 8s + K = 0 \]

To determine the range of \( K \) for stability, we can use the Routh-Hurwitz criterion. The Routh-Hurwitz criterion provides a systematic way to determine the number of roots of the characteristic equation that lie in the right half-plane, left half-plane, and on the imaginary axis. The characteristic equation for the system is:

\[ s^3 + 6s^2 + 8s + K = 0 \]

The Routh array for this characteristic equation is constructed as follows:

For the system to be stable, all the elements in the first column of the Routh array must be positive:

1. The coefficient of \( s^3 \): \( 1 \) (which is positive)
2. The coefficient of \( s^2 \): \( 6 \) (which is positive)
3. The coefficient of \( s^1 \): \(\frac{48 - K}{6} > 0 \rightarrow 48 - K > 0 \rightarrow K < 48\)
4. The coefficient of \( s^0 \): \( K > 0 \)

So, for the system to be stable:

\[ 0 < K < 48 \]

The correct option is therefore option 3: \( K < 48 \).

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: \( K > 48 \)

This option is incorrect because if \( K > 48 \), the coefficient of \( s^1 \) would become negative (\(\frac{48 - K}{6} < 0\)), indicating that the system would be unstable.

Option 2: \( K < 24 \)

While \( K < 24 \) is within the stability range, it is not the complete condition for stability. \( K \) must be less than 48 for the system to be stable, so this option is incomplete and not the best representation of the stability condition.

Option 4: \( K > 24 \)

This option is also incorrect because \( K > 24 \) does not guarantee stability. The critical upper limit for stability is \( K < 48 \).

Conclusion:

The stability of a control system is crucial for its correct operation. By applying the Routh-Hurwitz criterion to the characteristic equation, we determined that the system is stable for \( 0 < K < 48 \). This comprehensive analysis helps ensure that the system operates correctly within the specified range of \( K \).

Explanation:

The given problem is to identify the method that provides only the absolute stability of the system from the given options. The correct answer is option 3, R-H criteria. Let’s delve into a detailed explanation of why R-H criteria are the correct answer and analyze the other options to understand their roles in system stability analysis.

Routh-Hurwitz (R-H) Criteria:

Definition: The Routh-Hurwitz criterion is a mathematical test used to determine the absolute stability of a linear time-invariant (LTI) system. This criterion provides a systematic method to assess whether all the roots of the characteristic equation of the system have negative real parts, which is a necessary and sufficient condition for the system to be stable.

Working Principle: The R-H criterion involves constructing the Routh array from the coefficients of the characteristic polynomial of the system. The characteristic polynomial is usually derived from the denominator of the transfer function of the system, expressed as:

P(s) = a_nsⁿ + a_n-1s^n-1 + ... + a₁s + a₀

To apply the R-H criterion, the coefficients of the polynomial are arranged in a tabular form known as the Routh array. The stability of the system is then determined by examining the first column of the Routh array. The system is stable if and only if all the elements in the first column of the Routh array are positive. If any element in the first column is zero or negative, the system is unstable.

Advantages:

• Provides a clear and straightforward method to determine absolute stability without solving for the roots of the characteristic equation.
• Efficient for polynomials with high degrees, where finding roots analytically might be challenging.

Disadvantages:

• Does not provide information about the relative stability or the location of the poles in the complex plane.
• Cannot be used directly for systems with characteristic equations having coefficients with large numerical values or varying widely in magnitude.

Applications: The Routh-Hurwitz criterion is widely used in control system design and analysis to ensure that the system will remain stable under various operating conditions.

Correct Option Analysis:

The correct option is:

Option 3: R-H criteria

This option correctly identifies the method that gives only absolute stability of the system. The Routh-Hurwitz criterion is specifically designed to determine whether a system is stable (all poles have negative real parts) or unstable (at least one pole has a positive real part or zero).

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Bode plot

A Bode plot is a graphical representation of a system's frequency response. It consists of two plots: one showing the magnitude (gain) of the system as a function of frequency and the other showing the phase angle as a function of frequency. Bode plots are used to analyze the relative stability of a system and to design compensators to achieve desired performance specifications. However, Bode plots do not directly provide absolute stability information.

Option 2: Root locus

The root locus technique is a graphical method for examining how the roots of a system's characteristic equation change with variation in a certain system parameter, typically the gain. It provides insights into the relative stability and transient response of the system as the parameter changes. While root locus can help determine stability margins and the location of poles in the complex plane, it does not directly provide absolute stability information like the R-H criterion.

Option 4: Nyquist plot

A Nyquist plot is a graphical representation of a system's frequency response used to assess the stability of a feedback system. It plots the complex values of the system's transfer function as a function of frequency. The Nyquist criterion can determine both absolute and relative stability by examining the encirclements of the critical point (-1,0) in the complex plane. However, the Nyquist plot involves more complex analysis compared to the R-H criterion.

Conclusion:

The Routh-Hurwitz criterion is a powerful and efficient method for determining the absolute stability of a linear time-invariant system. It provides a clear indication of stability without requiring the computation of the roots of the characteristic equation. While other methods like Bode plots, root locus, and Nyquist plots offer valuable insights into the relative stability and dynamic behavior of systems, the R-H criterion remains the most straightforward tool for assessing absolute stability.

• Routh-Hurwitz Criterion: This criterion determines the stability of a linear time-invariant (LTI) system.
• Stability: A system is considered stable if all its poles (roots of the characteristic equation) lie in the left half of the s-plane.

• The Routh array is a tabular method used to analyze the stability of a system.
• The first column of the Routh array contains coefficients derived from the characteristic polynomial of the system.
• A necessary condition for stability is that all the elements in the first column of the Routh array must be positive. The system is unstable if any element in the first column is negative or zero

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.

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.

Given that characteristic equation is,

s³ + Ks² + (K + 2)s + 3 = 0

\(\left. {\begin{array}{*{20}{c}} {{s^3}}\\ {{s^2}}\\ {{s^1}}\\ {{s^0}} \end{array}} \right|\begin{array}{*{20}{c}} 1&{\left( {K + 2} \right)}\\ k&3\\ {\frac{{K\left( {K + 2} \right) - 3}}{K}}&0\\ 3&{} \end{array}\)

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

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: \(1 + \frac{1}{{s\left( {2{s^4} + 3{s^3} + 2{s^2} + 3s + 2} \right)}} = 0\)

⇒ 2s⁵ + 3s⁴ + 2s³ + 3s² + 2s + 1 = 0

By applying Routh tabulation method,

\(\begin{array}{*{20}{c}} {{s^5}}\\ {{s^4}}\\ {{s^3}}\\ {{s^2}}\\ {{s^1}}\\ {{s^0}} \end{array}\left| {\begin{array}{*{20}{c}} 2&2&2\\ 3&3&1\\ {0\left( \varepsilon \right)}&{\frac{4}{3}}&{}\\ {\left( {3 - \frac{4}{\varepsilon }} \right)}&1&{}\\ {\frac{4}{3}}&{}&{}\\ 1&{}&{} \end{array}} \right.\)

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.

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,

\(\begin{array}{*{20}{c}} {{s^4}}\\ {{s^3}}\\ {{s^2}}\\ {{s^1}}\\ {{s^0}} \end{array}\left| {\begin{array}{*{20}{c}} 1&1&5\\ {2k}&5&0\\ {1 - \frac{5}{{2k}}}&5&{}\\ {5 - \frac{{20{k^2}}}{{2k - 5}}}&0&{}\\ 5&{}&{} \end{array}} \right.\)

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

k > 0, \(1 - \frac{5}{{2k}} > 0\)

⇒ k > 2.5

For all the values of k > 2.5, \(5 - \frac{{20{k^2}}}{{2k - 5}}\) 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

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:

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

By applying Routh tabulation method,

\(\begin{array}{*{20}{c}} {{s^3}}\\ {{s^2}}\\ {{s^1}}\\ {{s^0}} \end{array}\left| {\begin{array}{*{20}{c}} 1&5\\ K&{10}\\ {\frac{{5K - 10}}{K}}&0\\ {10}&{} \end{array}} \right.\)

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

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.

\({\rm{\Delta }}\left( s \right) = {s^4} + 3{s^3} + 3{s^2} + s + k = 0\)

\(\left. {\begin{array}{*{20}{c}} {{s^4}}\\ {{s^3}}\\ {\begin{array}{*{20}{c}} {{s^2}}\\ {\begin{array}{*{20}{c}} {{s^1}}\\ {{s^0}} \end{array}} \end{array}} \end{array}} \right|\begin{array}{*{20}{c}} 1&3&k\\ 3&1&0\\ {8/3}&k&{}\\ {\left( {\frac{{8/3 - 3k}}{{8/3}}} \right)}&0&{}\\ k&0&{} \end{array}\)

The system to be stable,

k > 0 and \(\left( {\frac{8}{3} - 3k} \right) > 0\)

\(\Rightarrow 3k < \frac{8}{3} \Rightarrow k < \frac{8}{9}\)

\(\Rightarrow 0 < k < \frac{8}{9}\)

Stability:

System(1):

• From the response of system 1, we can observe that it is absolutely stable and following BIBO stable condition.
• And the poles of the system given are imaginary and lies in the left half of the s-plane, so it is a stable system.

System(2):

• From the response of the system, we can observe that the response of the system is infinite when 't' tends to infinity. So we are getting unbounded output, the system is unstable.
• Repeating poles at the origin (more than one) makes the system unstable.

System(3):

• Single pole at origin or on the imaginary axis makes the system marginally stable or just stable.

System(4):

• From the response of the above system(4), we can observe that the response has sustain oscillations, this represents a pair of poles on the imaginary axis.
• A pair of poles on the imaginary axis makes the system marginally stable or just stable.
• If more than one pair of poles on the imaginary axis then the system is Unstable.

The stability of a linear closed-loop system can be determined from the locations of closed-loop poles in the s-plane.

Stable System:

If all the roots of the characteristic equation lie on the left half of the 's' plane, i.e. if all the roots have negative real parts, then the system is said to be a stable system.

Ex: For a system with Closed Loop Transfer function as:

\(C.L.T.F.=\frac{1}{s+4}\)

It has a pole at s = - 4. The pole has a negative real part. This makes the system absolutely integrable and therefore, stable.

Marginally Stable System:

• A linear time-invariant system is said to be critically or marginally stable if for a bounded input its output oscillates with constant frequency & Amplitude.
• Such oscillation of output is called Undamped or Sustained oscillations. For such a system, one or more pairs of non-repeated roots are located on the imaginary axis.

Unstable System:

• If any or all the roots of the system lie on the left half of the 'S' plane then the system is said to be an unstable system.
• Also, if there are repeated poles located purely on the imaginary axis, then the system is said to be unstable.

The stability of the system for different pole positions is as shown:

NOTE:

1: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

2:Note that ROZ occurs in odd power of S rows only

3: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

 4:Order of the algebraic equation gives the number of poles 

 5: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 \(s^1\) 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