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Latest Condition For Stability MCQ Objective Questions

Top Condition For Stability MCQ Objective Questions

Condition For Stability Question 1:

Which stability criteria is used to check the stability of a non linear control system?

  1. Hough Transform
  2. Routh stability criterion
  3. Lyapunov's stability criterion
  4. Root locus

Answer (Detailed Solution Below)

Option 3 : Lyapunov's stability criterion

Condition For Stability Question 1 Detailed Solution

Explanation:

Lyapunov's stability criterion is used to check the stability of a nonlinear control system.

Lyapunov's stability criterion: 

1 Lyapunov stability is named after Aleksandr Mikhailovich Lyapunov, a Russian mathematician who defended the thesis The General Problem of Stability of Motion regarding the asymptotic stability of nonlinear systems.

2 In simple terms, if the solutions that start out near an equilibrium point xe stay near xforever, then xis Lyapunov stable. More strongly, if xis Lyapunov stable and all solutions that start out near xconverge to xe, then xis asymptotically stable.

Hence option (3) is the correct answer.

Important Points

Hough Transform:

Hough Transform is a popular technique to detect any shape if you can represent that shape in mathematical form. It can detect the shape even if it is broken or distorted a little bit.

Routh stability criterion and Root locus are used to check the stability of the linear control system.

Condition For Stability Question 2:

Which of the following statement is incorrect regarding stability of control systems?

  1. For a stable system only the natural response remains and the forced response dies out. 
  2. A linear time invariant system is unstable if the natural response grows without bound as time approaches infinity. 
  3. A linear time invariant system is unstable if the forced response grows without bound as time approaches infinity. 
  4. A system is unstable if any bounded input yields unbounded output. 

Answer (Detailed Solution Below)

Option 3 : A linear time invariant system is unstable if the forced response grows without bound as time approaches infinity. 

Condition For Stability Question 2 Detailed Solution

Concept:

The output response of a system is equal to the sum of natural response and forced response.

Forced response: The response generated due to the pole of the input function is called the forced response.

Natural response: The response generated due to the pole of system function is called the natural response.

For linear systems, the stability is based on the natural response which leads to zero for a stable system, however, for an unstable system the natural response grows without bound and the forced response is overpowered by the natural response and the control is lost over the system. 

 

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The response reaches a steady state after almost 5-time constants and then the difference in the input and output is termed as the steady-state error. 

The forced response remains and the natural response dies out for a stable system. 

Any system is called a stable system if the output of the system is bounded for any bounded input. If any of the bounded input yields unbounded output then the system is not stable. 

Condition For Stability Question 3:

For the system shown in the figure, the range of K for which the system is stable is

F1 Uday 21.10.20 Pallavi D 6

  1. -0.5 < K < 2
  2. K > -0.5
  3. – 1 < K < 1
  4. The system is unstable for all the values of K

Answer (Detailed Solution Below)

Option 3 : – 1 < K < 1

Condition For Stability Question 3 Detailed Solution

Characteristic equation:

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

\( \Rightarrow 1 + K\left( {{s^2} - 2s + 2} \right)\frac{1}{{\left( {{s^2} + 2s + 4} \right)}} = 0\)

⇒ s2 + 2s + 4 + Ks2 – 2Ks + 2K = 0

⇒ (K + 1) s2 + 2 (1 - K)s + (2K + 4) = 0

Since all co-efficient must be positive for stability in a second order polynomial,

-1 < K ; -∞ < K < 1 ; - 4 < 2 K < ∞

2K + 4 > 0; 2K > - 4; K > - 2

With all the conditions in hand we can write:

\( \Rightarrow -1 < K < 1\)

Condition For Stability Question 4:

A linear time-invariant (LTI) system has a transfer function, \(H(s) = \dfrac{1}{(s - 5)}\). To make this system stable it needs to be cascaded with another LTI system having a transfer function H1(s). The correct choice of H1(s) can be _________

  1. (s2 - 3s - 10)
  2. (s2 + 2s - 35)
  3. (s2 + s - 6)
  4. (s2 - 2s + 12)

Answer (Detailed Solution Below)

Option :

Condition For Stability Question 4 Detailed Solution

Concept: 

All the poles must have the negative real part.

Calculation:

Option 1:

(s2 - 3s - 10) → (s + 2) (s - 5) Cascading with (s - 5), we get

⇒ \(\dfrac{1}{(s - 5)} (s + 2) ( s - 5)\)

⇒ (s + 2)

Option 2:

(s2 + 2s - 35) → (s + 7) (s - 5)

Cascading with \(\dfrac{1}{(s - 5)}\), we get

⇒ \(\dfrac{1}{(s - 5)} (s + 7) ( s - 5)\)

⇒ s + 7

Option 3:

(s2 + s - 6) → (s - 2) (s + 3)

does not give a stable system on cascading.

Option 4:

(s2 - 2s + 12) → (s - 2)(s - 6)

does not give a stable system on cascading.

Condition For Stability Question 5:

To increase the stability of a feedback system.

  1. reduce the gain and increase the phase of open loop transfer function 
  2. reduce the gain and reduce the phase of open loop transfer function 
  3. increase the gain and increase the phase of open loop transfer function 
  4. increase the gain and reduce the phase of open loop transfer function 

Answer (Detailed Solution Below)

Option 1 : reduce the gain and increase the phase of open loop transfer function 

Condition For Stability Question 5 Detailed Solution

To increase the stability of a feedback system reduce the gain and increase the phase of open loop transfer function.

Condition For Stability Question 6:

If the characteristics equation of a closed loop system is s2+2s+2=0, then the system is 

  1. Over damped
  2. Critically damped
  3. Under damped 
  4. Undamped 

Answer (Detailed Solution Below)

Option 1 : Over damped

Condition For Stability Question 6 Detailed Solution

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 (ωn).

Step 1: Standard Form of the Characteristic Equation

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

s² + 2ζωns + ωn² = 0

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

  • 2ζωn = 2 → ζωn = 1
  • ωn² = 2 → ωn = √2

Step 2: Calculate the Damping Ratio (ζ)

From the first equation, ζ can be calculated as:

ζ = (1 / ωn) = (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.

Condition For Stability Question 7:

The following item consists of two statements, one labeled as ‘Statement (I)’ and the other as ‘Statement (II)’. Examine these two statements carefully and select the answers to these items using the code given below.

Statement (I): The standard definition of stability precludes the sin ω0t term in the impulse response.

Statement (II): sin ω0t is a periodic function.

Code:

  1. Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I)
  2. Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I)
  3. Statement (I) is true but Statement (II) is false
  4. Statement (I) is false but Statement (II) is true

Answer (Detailed Solution Below)

Option 1 : Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I)

Condition For Stability Question 7 Detailed Solution

Explanation:

sin ωot introduces poles on the jω axis. This will make the system marginally stable, and hence the stability will be ruined.

Also, sin ωot is also a periodic function, not letting to converge the impulse response since it exists from -∞ to ∞ time duration.

∴ The standard definition of stability precludes the sin ω0t term in the impulse response.

Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I)

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