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

Explanation:

Characteristics of Roots:

• Stable System: All roots lie in the left half of the s-plane (negative real parts). The system will return to equilibrium after a disturbance.
• Marginally Stable System: Roots lie on the imaginary axis, but no root has a positive real part. The system neither grows unbounded nor settles completely but oscillates perpetually.
• Unstable System: At least one root lies in the right half of the s-plane (positive real part) or there are repeated roots on the imaginary axis. The system grows unbounded in response to disturbances.

Analysis of Given Lists:

The characteristic equations and their respective roots provide insight into the system's behavior:

• List 1: Represents the location of roots of the characteristic equation.
• List 2: Represents the corresponding system characteristics.

Matching List 1 and List 2:

Option (A): (-1 + j), (-1 - j): The roots lie in the left half of the s-plane (negative real part), indicating a marginally stable system. Hence, A matches with D.

Option (B): (-2 + j), (-2 - j), (2), (-2j): Here, one root lies in the right half of the s-plane (positive real part, i.e., 2). This makes the system unstable. Hence, B matches with E.

Option (C): (-j), (j), (1), (-1): The root (1) lies in the right half of the s-plane (positive real part), making the system unstable. Hence, C matches with E. However, if (-j) and (j) were the only roots on the imaginary axis, the system would be marginally stable. But the presence of (1) shifts the system to instability.

The correct match from the options provided is: Option 3:

Explanation:

Routh Stability Criterion:

The Routh stability criterion is a mathematical technique used in control systems to determine the stability of a linear time-invariant system.

For the system to be stable, all the roots of the characteristic equation must lie in the left half of the s-plane (stable region).

The characteristic equation of a second-order polynomial can be expressed as:

G(s) = a2s2 + a1s + a0

Where:

• a2, a1, and a0 are the coefficients of the polynomial.
• s is the complex Laplace variable.

According to the Routh stability criterion, for the system to be stable, all the coefficients of the characteristic equation must be positive, and none of the coefficients should be zero. Additionally, there should be no sign changes in the first column of the Routh array.

For a second-order polynomial, the stability criterion simplifies to:

a₂ > 0, a₁ > 0, a₀ > 0

This ensures that all roots of the characteristic equation lie in the left-half of the s-plane, and the system is stable.

Explanation:

Feedback Control System Stability Analysis

Understanding the Problem: To determine the stability of the given feedback control system, we need to analyze the location of the system poles. The poles of the system are the roots of the characteristic equation, which is derived from the closed-loop transfer function. Stability is determined based on the following criteria:

• Left Half Plane (LHP) Poles: Poles in the LHP correspond to stable components of the system.
• Right Half Plane (RHP) Poles: Poles in the RHP indicate unstable components of the system.
• jω-Axis Poles: Poles on the imaginary axis may suggest marginal stability or oscillatory behavior, depending on their multiplicity.

Analysis of the Correct Option (Option 2):

Given: The system has:

• 2 LHP Poles
• 2 RHP Poles
• 0 Poles on the jω-Axis

Step-by-Step Analysis:

1. Determine the Characteristic Equation: The characteristic equation of the system is derived from the closed-loop transfer function, which is typically given as:

T(s) = G(s) / [1 + G(s)H(s)]

Where G(s) is the open-loop transfer function, and H(s) is the feedback transfer function. The characteristic equation is obtained by setting the denominator to zero:

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

2. Locate the Poles: Solving the characteristic equation provides the locations of the poles in the complex plane. Based on the problem statement, the poles are distributed as follows:

• 2 poles in the Left Half Plane (LHP): These poles indicate stable components of the system.
• 2 poles in the Right Half Plane (RHP): These poles indicate unstable components of the system.
• 0 poles on the jω-axis: There are no marginally stable or purely oscillatory components.

Note: The presence of RHP poles means the system is inherently unstable, regardless of the LHP poles.

3. Determine System Stability: A system is stable if all poles are located in the LHP. In this case, the presence of 2 RHP poles makes the system unstable. The 2 LHP poles do not compensate for the instability caused by the RHP poles.

Conclusion: The system is unstable because it has 2 poles in the RHP. This aligns with the correct answer, which is Option 2.

Additional Information

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

Option 1: 1 LHP pole, 3 RHP poles, 0 jω poles, system is unstable.

This option suggests that there is only 1 LHP pole and 3 RHP poles. While this configuration also makes the system unstable (due to the RHP poles), it does not match the given system's pole distribution (2 LHP poles and 2 RHP poles). Therefore, this option is incorrect.

Option 3: 2 LHP poles, 2 RHP poles, 0 jω poles, system is stable.

This option claims that the system is stable despite having 2 RHP poles. This is incorrect because any RHP poles render the system unstable. Stability requires all poles to be in the LHP. Hence, this option is incorrect.

Option 4: 1 LHP pole, 3 RHP poles, 0 jω poles, system is stable.

This option is inconsistent because it claims stability despite having 3 RHP poles. A stable system cannot have any RHP poles. Thus, this option is also incorrect.

Conclusion:

The correct analysis of the pole locations reveals that the system has 2 LHP poles, 2 RHP poles, and 0 jω-axis poles, making the system unstable. This matches Option 2. The other options misrepresent the pole distribution or incorrectly assess system stability, leading to their elimination.

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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<sub>n</sub>).

Step 1: Standard Form of the Characteristic Equation

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

s² + 2ζω<sub>n</sub>s + ω<sub>n</sub>² = 0

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

• 2ζω<sub>n</sub> = 2 → ζω<sub>n</sub> = 1
• ω<sub>n</sub>² = 2 → ω<sub>n</sub> = √2

Step 2: Calculate the Damping Ratio (ζ)

From the first equation, ζ can be calculated as:

ζ = (1 / ω<sub>n</sub>) = (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 Underdamped system
• ζ = 0: Undamped system

Since ζ = 0.707, which lies in the range 0 , 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

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

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

Incorrect. The system is stable for \( 0

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

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

Conclusion:

Based on the Routh-Hurwitz analysis, the correct statement is that the system is stable for \( 0

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,

s<sup>3</sup> + Ks<sup>2</sup> + (K + 2)s + 3 = 0

For system to be stable,

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

⇒ K > 0, K<sup>2</sup> + 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:

⇒ 2s<sup>5</sup> + 3s<sup>4</sup> + 2s<sup>3</sup> + 3s<sup>2</sup> + 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.

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: s<sup>4</sup> + 2ks<sup>3</sup> + s<sup>2</sup> + 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 > 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,

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

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.

The system to be stable,

k > 0 and  0\)

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:

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  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