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

Explanation:

System Damping Analysis

The given characteristic equation of the closed-loop system is:

s² + 2s + 2 = 0

This is a second-order system, and its response is determined by analyzing its damping ratio (ζ) and natural frequency (ω_n).

Step 1: Standard Form of the Characteristic Equation

The standard form of a second-order system's characteristic equation is given by:

s² + 2ζω_ns + ω_n² = 0

Comparing the given equation (s² + 2s + 2 = 0) with the standard form:

• 2ζω_n = 2 → ζω_n = 1
• ω_n² = 2 → ω_n = √2

Step 2: Calculate the Damping Ratio (ζ)

From the first equation, ζ can be calculated as:

ζ = (1 / ω_n) = (1 / √2)

Therefore:

ζ = 0.707

Step 3: Determine the System's Damping Condition

The damping condition of a second-order system is determined by the value of the damping ratio (ζ):

• ζ > 1: Overdamped system
• ζ = 1: Critically damped system
• 0 < ζ < 1: Underdamped system
• ζ = 0: Undamped system

Since ζ = 0.707, which lies in the range 0 < ζ < 1, the system is underdamped.

Correct Option: Option 3 (Under damped)

Additional Information

To further analyze the other options, let's evaluate the damping conditions:

Option 1: Overdamped

In an overdamped system, the damping ratio (ζ) is greater than 1. This means that the system returns to its steady-state value without oscillating, but it does so slowly. Since ζ = 0.707 (less than 1) for the given characteristic equation, the system is not overdamped. Thus, this option is incorrect.

Option 2: Critically Damped

A critically damped system occurs when ζ = 1. In this case, the system returns to its steady-state value as quickly as possible without oscillating. However, for the given system, ζ = 0.707 (not equal to 1), so the system is not critically damped. Hence, this option is also incorrect.

Option 4: Undamped

An undamped system corresponds to ζ = 0. In this case, there is no damping force, and the system oscillates indefinitely at its natural frequency. Since ζ = 0.707 for the given system, it is not undamped. Therefore, this option is incorrect.

Option 5: Not Given in the Statement

This option does not apply to the given question as the characteristic equation directly provides enough information to determine the damping condition of the system.

Conclusion:

The damping ratio (ζ) is a critical parameter for determining the damping condition of a second-order system. Based on the given characteristic equation (s² + 2s + 2 = 0), we calculated ζ = 0.707, which classifies the system as underdamped. This means the system will exhibit oscillatory behavior with gradually decreasing amplitude over time.

Explanation:

Electromechanical Closed-Loop Control System

Problem Statement:

The transfer function of the electromechanical closed-loop control system is given as:

C(s)/R(s) = k / [s × (s² + s + 1) × (s + 4) + k]

We are tasked with determining the stability of the system for various values of the gain parameter \( k \). The correct answer is given as:

Option 1: The system is stable for all positive values of \( k \).

To verify this claim and analyze the other options, we will use the principles of control system stability analysis, specifically the location of the poles in the s-plane and the Routh-Hurwitz criterion.

Solution:

Step 1: Closed-Loop Characteristic Equation

The denominator of the transfer function represents the characteristic equation of the closed-loop system:

Characteristic Equation: \( s × (s² + s + 1) × (s + 4) + k = 0 \)

Expanding this equation:

• \( s × (s² + s + 1) = s³ + s² + s \)
• \( (s³ + s² + s) × (s + 4) = s⁴ + 4s³ + s³ + 4s² + s² + 4s = s⁴ + 5s³ + 5s² + 4s \)

Thus, the characteristic equation becomes:

Denominator: \( s⁴ + 5s³ + 5s² + 4s + k = 0 \)

Step 2: Stability Analysis using Routh-Hurwitz Criterion

The Routh-Hurwitz criterion provides a systematic method to determine the stability of a system by examining the signs of the coefficients in the Routh array. A system is stable if all the roots of the characteristic equation have negative real parts, which corresponds to all the elements in the first column of the Routh array being positive.

Let us construct the Routh array for the characteristic equation:

\( s⁴ + 5s³ + 5s² + 4s + k = 0 \)

Step 3: Construct the Routh Array

The key term in the second row of the Routh array is:

\( \dfrac{5 × 4 - k}{5} \)

For stability, this term must be positive. Thus:

\( 20 - k > 0 \)

\( k < 20 \)

Analysis of Options:

Option 1: The system is stable for all positive values of \( k \).

This is incorrect because the system is stable only for \( k < 20 \). Beyond \( k = 20 \), the system becomes unstable.

Option 2: The system is unstable for all values of \( k \).

Incorrect. The system is stable for \( 0 < k < 20 \).

Option 3: The system is stable for values of \( k \) between zero and 3.36.

Incorrect. The system is stable for a broader range of \( k \), specifically \( 0 < k < 20 \).

Option 4: The system is stable for values of \( k \) between 1.6 and 2.45.

Incorrect. This range is overly restrictive; the system is stable for \( 0 < k < 20 \).

Conclusion:

Based on the Routh-Hurwitz analysis, the correct statement is that the system is stable for \( 0 < k < 20 \), making none of the provided options entirely accurate. The correct answer should specify this range explicitly.

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 \( K \).

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