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Time Response Analysis MCQs for Electrical Engineering (EE) Exam
It covers all Important Questions with answers on Time Response Analysis for the Electrical Engineering (EE) exam. The questions are based on important topics. Details about the questions:- Topic: Time Response Analysis
- Type of Questions: MCQs with solutions
- Number of Questions: 50
- You can attempt them on EduRev to score high in Electrical Engineering (EE) exam.
A ufb control system has a forward path transfer function
If the system is subjected to an input r(t) = 1 + t + 1/2 t2 , t > 0the steady state error of the system will be- a)0
- b)0.1
- c)10
- d)∞
Correct answer is option 'C'. Can you explain this answer?
A ufb control system has a forward path transfer function
If the system is subjected to an input r(t) = 1 + t + 1/2 t2 , t > 0the steady state error of the system will be
a)
0
b)
0.1
c)
10
d)
∞
 | Lekshmi Kulkarni answered |
System is type 2. Therefore error due to 1+t would be zero and due to 
would be 
Note that you may calculate error from the formula
would be Note that you may calculate error from the formula
The steady-state error of a feedback control system with an acceleration input becomes finite in a- a)type-3 system
- b)type-2 system
- c)type-1 system
- d)type-0 system
Correct answer is option 'B'. Can you explain this answer?
The steady-state error of a feedback control system with an acceleration input becomes finite in a
a)
type-3 system
b)
type-2 system
c)
type-1 system
d)
type-0 system
 | Raj Choudhary answered |
Steady state error with an acceleration input having an amplitude of A is given by
where,
Hence, if the type of the system = 2, then Ka = some non-zero value or finite value due to which we will get some finite vaiue of Ka.
where,
Hence, if the type of the system = 2, then Ka = some non-zero value or finite value due to which we will get some finite vaiue of Ka.
The time required for the response curve to reach and stay within the specified 2-5% of final value is referred to as :- a)Peak time
- b)Rise time
- c)Settling time
- d)Peak overshoot time
Correct answer is option 'C'. Can you explain this answer?
The time required for the response curve to reach and stay within the specified 2-5% of final value is referred to as :
a)
Peak time
b)
Rise time
c)
Settling time
d)
Peak overshoot time
 | Imtiaz Ahmad answered |
The correct answer is option 3): Settling time
Concept:
- The time that is required for the response to reach and stay within the specified range (2% to 5%) of its final value is called the settling time.
- Peak Time. It is the time required for the response to reach the peak value for the first time. It is denoted by tp.
- The time is defined as "the time required for the response to rise from x% to y% of its final value", with 0% to 100% rise time common for underdamped second order systems, 5% to 95% for critically damped and 10% to 90% for overdamped ones.
- Peak time (tp) is simply the time required by response to reach its first peak i.e. the peak of first cycle of oscillation, or first overshoot. On differentiating the expression of c(t)
Which of the following is not true regarding addition of a pole to the forward-path transfer function?- a)Maximum overshoot of the closed-loop system is increased.
- b)Rise time of the step response is increased.
- c)Stability of the system is increased.
- d)Bandwidth of the system is reduced.
Correct answer is option 'C'. Can you explain this answer?
Which of the following is not true regarding addition of a pole to the forward-path transfer function?
a)
Maximum overshoot of the closed-loop system is increased.
b)
Rise time of the step response is increased.
c)
Stability of the system is increased.
d)
Bandwidth of the system is reduced.
 | Sandeep Chatterjee answered |
With addition of a pole to the forward-path transfer function, bandwidth will reduce.
Since
therefore rise time will increase.
Since rise time
therefore stability will decrease.
Since
therefore rise time will increase.Since rise time
therefore stability will decrease.The peak overshoot of step-input response of an underdamped second-order system is an indication of- a)rise time of the system
- b)settling time of the system
- c)damping ratio of the system
- d)natural frequency of the system
Correct answer is option 'C'. Can you explain this answer?
The peak overshoot of step-input response of an underdamped second-order system is an indication of
a)
rise time of the system
b)
settling time of the system
c)
damping ratio of the system
d)
natural frequency of the system
 | Avik Iyer answered |
If damping of system increases, peak overshoot Mp decreases and vice-versa.
For type-2 system, the steady state error due to parabolic input is- a)finite
- b)zero
- c)infinite
- d)indeterminate
Correct answer is option 'A'. Can you explain this answer?
For type-2 system, the steady state error due to parabolic input is
a)
finite
b)
zero
c)
infinite
d)
indeterminate
 | Aman Jain answered |
For parabolic input, 
(A = constant)
= Finite for type-2 system
So, ess = Some finite value
(A = constant)= Finite for type-2 system
So, ess = Some finite value
How can the steady state error can be reduced?- a)By decreasing the type of the system
- b)By increasing the input
- c)By decreasing the static error constant
- d)By increasing system gain
Correct answer is option 'D'. Can you explain this answer?
How can the steady state error can be reduced?
a)
By decreasing the type of the system
b)
By increasing the input
c)
By decreasing the static error constant
d)
By increasing system gain
 | Prisha Sengupta answered |
From the above table, it is clear that for type – 1 system, a system shows zero steady-state error for step-input, finite steady-state error for Ramp-input and ∞ steady-state error for parabolic-input.
As the type of the system increases, the steady-state error decreases.
The steady-state error is inversely proportional to the gain. Therefore, it can be reduced by increasing the system gain.
The unit step response of a system is given by c(t) - 1 + 0,25 e-50t - 1.25 er-10t The given system is- a)underdamped
- b)critically damped
- c)overdamped
- d)undamped
Correct answer is option 'C'. Can you explain this answer?
The unit step response of a system is given by c(t) - 1 + 0,25 e-50t - 1.25 er-10t The given system is
a)
underdamped
b)
critically damped
c)
overdamped
d)
undamped
 | Manoj Mehra answered |
Given, 
∴
or,
∴
or,
Thus,
and
Thus,
Since, ξ > 1, therefore given system is overdamped.
∴
or,
∴
or,
Thus,
and
Thus,
Since, ξ > 1, therefore given system is overdamped.
A system has position error constant Kp = 3. The steady state error for input of 8tu(t) is- a)2.67
- b)2
- c)∞
- d)0
Correct answer is option 'C'. Can you explain this answer?
A system has position error constant Kp = 3. The steady state error for input of 8tu(t) is
a)
2.67
b)
2
c)
∞
d)
0
 | Aryan Mathur answered |
Steady state error is a measure of how well a control system is able to track a desired input signal. It represents the difference between the desired output and the actual output of the system when the input signal has reached a steady state.
In this system, the position error constant Kp is given as 3. The position error constant relates the steady state error to the input signal. It is defined as the ratio of the steady state error to the input signal. Mathematically, Kp = 1 / steady state error.
Given that the input signal is 8tu(t), where u(t) is the unit step function, we can determine the steady state error using the position error constant.
Let's calculate the steady state error step by step:
1. First, we need to determine the Laplace transform of the input signal.
The Laplace transform of tu(t) is 1/s^2.
Therefore, the Laplace transform of 8tu(t) is 8/s^2.
2. Next, we determine the Laplace transform of the steady state error.
Let the Laplace transform of the steady state error be E(s).
Then, E(s) = Kp / (s * 8/s^2) = Kp * s^2 / 8.
3. To find the steady state error, we need to take the inverse Laplace transform of E(s).
Using inverse Laplace transform tables or software, we can determine that the inverse Laplace transform of Kp * s^2 / 8 is Kp/8 * t^2.
4. Since the input signal is a unit step function, the steady state error is the value of the inverse Laplace transform at t = infinity.
Therefore, the steady state error is Kp/8 * infinity^2 = Kp/8 * ∞ = 3/8 * ∞ = 0.
Hence, the correct answer is option 'C', which states that the steady state error is 0.
In this system, the position error constant Kp is given as 3. The position error constant relates the steady state error to the input signal. It is defined as the ratio of the steady state error to the input signal. Mathematically, Kp = 1 / steady state error.
Given that the input signal is 8tu(t), where u(t) is the unit step function, we can determine the steady state error using the position error constant.
Let's calculate the steady state error step by step:
1. First, we need to determine the Laplace transform of the input signal.
The Laplace transform of tu(t) is 1/s^2.
Therefore, the Laplace transform of 8tu(t) is 8/s^2.
2. Next, we determine the Laplace transform of the steady state error.
Let the Laplace transform of the steady state error be E(s).
Then, E(s) = Kp / (s * 8/s^2) = Kp * s^2 / 8.
3. To find the steady state error, we need to take the inverse Laplace transform of E(s).
Using inverse Laplace transform tables or software, we can determine that the inverse Laplace transform of Kp * s^2 / 8 is Kp/8 * t^2.
4. Since the input signal is a unit step function, the steady state error is the value of the inverse Laplace transform at t = infinity.
Therefore, the steady state error is Kp/8 * infinity^2 = Kp/8 * ∞ = 3/8 * ∞ = 0.
Hence, the correct answer is option 'C', which states that the steady state error is 0.
In time domain specification, the time delay is the time required for the response to reach- a)25% of final value
- b)50% of the final value
- c)10% of the final value
- d)63.3% of the final value
Correct answer is option 'B'. Can you explain this answer?
In time domain specification, the time delay is the time required for the response to reach
a)
25% of final value
b)
50% of the final value
c)
10% of the final value
d)
63.3% of the final value
 | Srestha Chavan answered |
Delay time (td) is the time required for the response curve to reach 50% of the final value.
Assertion (A): The "dominant pole" concept is used to control the dynamic performance of the system, whereas the “insignificant poles” are used for the purpose of. ensuring that the controller transfer function can be realized by physical components.
Reason (R): if the magnitude of the real part of a pole is atleast 5 to 10 times that of a dominant pole or a pair of complex dominant poles, the pole may be regarded as insignificant in so far as the transient response is concerned.- a)Both A and R are true and R is a correct explanation of A.
- b)Both A and R are true but R is not a correct explanation of A.
- c)A is true but R is false.
- d)A is false but R is true.
Correct answer is option 'B'. Can you explain this answer?
Assertion (A): The "dominant pole" concept is used to control the dynamic performance of the system, whereas the “insignificant poles” are used for the purpose of. ensuring that the controller transfer function can be realized by physical components.
Reason (R): if the magnitude of the real part of a pole is atleast 5 to 10 times that of a dominant pole or a pair of complex dominant poles, the pole may be regarded as insignificant in so far as the transient response is concerned.
Reason (R): if the magnitude of the real part of a pole is atleast 5 to 10 times that of a dominant pole or a pair of complex dominant poles, the pole may be regarded as insignificant in so far as the transient response is concerned.
a)
Both A and R are true and R is a correct explanation of A.
b)
Both A and R are true but R is not a correct explanation of A.
c)
A is true but R is false.
d)
A is false but R is true.
| | Devansh Das answered |
Both assertion and reason are true. However, the correct explanation of assertion is that the poles having real part magnitude 5 to 10 times of the magnitude of real part of dominant poles doesn't affect the transient response (dynamic performance) of the system to a great extent due to which they are termed as insignificant poies and hence are neglected for transient stability study purpose.
When the period of the observation is large, the type of the error will be:- a)Transient error
- b)Steady state error
- c)Half-power error
- d)Position error constant
Correct answer is option 'B'. Can you explain this answer?
When the period of the observation is large, the type of the error will be:
a)
Transient error
b)
Steady state error
c)
Half-power error
d)
Position error constant
| | Maulik Chatterjee answered |
Answer: b
Explanation: The error will be the steady state error if the period of observation is large as the time if large then the final value theorem can be directly applied.
Explanation: The error will be the steady state error if the period of observation is large as the time if large then the final value theorem can be directly applied.
The steady state error of a control system can be reduced by- a)increasing time constant of the system only.
- b)increasing gain constant of the system only.
- c)increasing both the time constant and gain of the system.
- d)none of the above.
Correct answer is option 'C'. Can you explain this answer?
The steady state error of a control system can be reduced by
a)
increasing time constant of the system only.
b)
increasing gain constant of the system only.
c)
increasing both the time constant and gain of the system.
d)
none of the above.
 | Anuj Kapoor answered |
Time constant of the system,
When
is increased means damping is reduced, thus steady state error reduces.
Also, when gain of the system increases, then damping reduces and therefore, steady state error reduces.
When
is increased means damping is reduced, thus steady state error reduces.Also, when gain of the system increases, then damping reduces and therefore, steady state error reduces.
With negative feedback in a closed loop control system, the system sensitivity to parameter variation:- a)Increases
- b)Decreases
- c)Becomes zero
- d)Becomes infinite
Correct answer is option 'B'. Can you explain this answer?
With negative feedback in a closed loop control system, the system sensitivity to parameter variation:
a)
Increases
b)
Decreases
c)
Becomes zero
d)
Becomes infinite
| | Mahi Bose answered |
Answer: b
Explanation: Sensitivity is defined as the change in the output with respect to the change in the input and due to negative feedback reduces by a factor of 1/ (1+GH).
Explanation: Sensitivity is defined as the change in the output with respect to the change in the input and due to negative feedback reduces by a factor of 1/ (1+GH).
A third order system is approximated to an equivalent second order system. The rise time of this approximated lower order system will be:- a)Same as the original system for any input
- b)Smaller than the original system for any input
- c)Larger than the original system for any input
- d)Larger or smaller depending on the input
Correct answer is option 'B'. Can you explain this answer?
A third order system is approximated to an equivalent second order system. The rise time of this approximated lower order system will be:
a)
Same as the original system for any input
b)
Smaller than the original system for any input
c)
Larger than the original system for any input
d)
Larger or smaller depending on the input
 | Mansi Choudhury answered |
Answer: b
Explanation: As order of the system increases the system approaches more towards the ideal characteristics and if the third order system is approximated to an equivalent second order system then the rise time of this will be smaller than the original system for any input.
Explanation: As order of the system increases the system approaches more towards the ideal characteristics and if the third order system is approximated to an equivalent second order system then the rise time of this will be smaller than the original system for any input.
Consider the following statements related to second order control system with unit step input:
1. The right-half s-plane corresponds to negative damping.
2. Negative damping gives a response that doesn’t grows in magnitude without bound with time.
3. The imaginary axis corresponds to zero damping.
4. The maximum overshoot is often used to measure the relative stability of a control system.
5. Higher the damping in the system larger is the peak overshoot in the transient response of the system.
Which of the above statements is/are correct?- a)1 and 3
- b)1, 4 and 5
- c)2, 3 and 4
- d)1,3 and 4
Correct answer is option 'D'. Can you explain this answer?
Consider the following statements related to second order control system with unit step input:
1. The right-half s-plane corresponds to negative damping.
2. Negative damping gives a response that doesn’t grows in magnitude without bound with time.
3. The imaginary axis corresponds to zero damping.
4. The maximum overshoot is often used to measure the relative stability of a control system.
5. Higher the damping in the system larger is the peak overshoot in the transient response of the system.
Which of the above statements is/are correct?
1. The right-half s-plane corresponds to negative damping.
2. Negative damping gives a response that doesn’t grows in magnitude without bound with time.
3. The imaginary axis corresponds to zero damping.
4. The maximum overshoot is often used to measure the relative stability of a control system.
5. Higher the damping in the system larger is the peak overshoot in the transient response of the system.
Which of the above statements is/are correct?
a)
1 and 3
b)
1, 4 and 5
c)
2, 3 and 4
d)
1,3 and 4
| | Vaibhav Joshi answered |
Negative damping gives a response that grows in magnitude without bound in time. Hence, statement-2 is false.
Since
therefore higher is the damping
in the system, less is the peak overshoot of the system. Hence, statement-5 is not correct
Since
therefore higher is the dampingin the system, less is the peak overshoot of the system. Hence, statement-5 is not correct
For a type one system, the steady-state error due to step input is equal to- a)zero
- b)0.75
- c)infinite
- d)0.50
Correct answer is option 'A'. Can you explain this answer?
For a type one system, the steady-state error due to step input is equal to
a)
zero
b)
0.75
c)
infinite
d)
0.50
 | Mahesh Datta answered |
where,
(Here, A = magnitude of step input) Since system is type -1, therefore Kp = ∞
∴
The unit impulse response of a linear time invariant second order system is
g(t) = 10e-8t sin 6t (t > C)
The natural frequency and the damping ratio of the system are respectively - a)10 rad/sec and 0.6
- b)10 rad/sec and 0.8
- c)5 rad/sec and 0.6
- d)6 rad/sec and 0.8
Correct answer is option 'B'. Can you explain this answer?
The unit impulse response of a linear time invariant second order system is
g(t) = 10e-8t sin 6t (t > C)
The natural frequency and the damping ratio of the system are respectively
g(t) = 10e-8t sin 6t (t > C)
The natural frequency and the damping ratio of the system are respectively
a)
10 rad/sec and 0.6
b)
10 rad/sec and 0.8
c)
5 rad/sec and 0.6
d)
6 rad/sec and 0.8
| | Bijoy Nair answered |
Given, c(t) = g(t) = 10e-8t sin6t
Here,
and ξωn = 8
or,
or,
= 10 rad/sec
So,
Here,
and ξωn = 8
or,
or,
= 10 rad/sec
So,
A second order system is said to be critically damped if the damping factor (ξ) is - a)ξ= 0.707
- b)ξ<1
- c)ξ=1
- d)ξ>1
Correct answer is option 'C'. Can you explain this answer?
A second order system is said to be critically damped if the damping factor (ξ) is
a)
ξ= 0.707
b)
ξ<1
c)
ξ=1
d)
ξ>1
 | Priyanka Ahuja answered |
If ξ < 1, system is underdamped.
If ξ > 1, system is overdamped.
If ξ= 1, system is critically damped.
If ξ > 1, system is overdamped.
If ξ= 1, system is critically damped.
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