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Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  Test: Time Response Of Second Order Systems - 4 - Electrical Engineering (EE) MCQ

Test: Time Response Of Second Order Systems - 4 - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test - Test: Time Response Of Second Order Systems - 4

Test: Time Response Of Second Order Systems - 4 for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Test: Time Response Of Second Order Systems - 4 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Time Response Of Second Order Systems - 4 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Time Response Of Second Order Systems - 4 below.
Solutions of Test: Time Response Of Second Order Systems - 4 questions in English are available as part of our course for Electrical Engineering (EE) & Test: Time Response Of Second Order Systems - 4 solutions in Hindi for Electrical Engineering (EE) course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Time Response Of Second Order Systems - 4 | 10 questions in 10 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study for Electrical Engineering (EE) Exam | Download free PDF with solutions
Test: Time Response Of Second Order Systems - 4 - Question 1
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The standard second order system to a unit step input shows the 0.36 as the first peak undershoot, hence its second overshoot is:

Detailed Solution for Test: Time Response Of Second Order Systems - 4 - Question 1

Answer: b
Explanation: Overshoot and undershoot are calculated from the formula of peak time as odd peaks denote the overshoot and even denotes the under shoot.

Test: Time Response Of Second Order Systems - 4 - Question 2
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Consider the input with the inputs 4 u(t) and the impulse response 5 +7 , the time constants of the output are,

Detailed Solution for Test: Time Response Of Second Order Systems - 4 - Question 2

Answer: a
Explanation: Output response so calculated is the inverse Laplace transform of the input and impulse response and hence converting the resultant form in the time constant form time constant can be directly determined.

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Test: Time Response Of Second Order Systems - 4 - Question 3
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In a second order feedback control system natural frequency and damping

Detailed Solution for Test: Time Response Of Second Order Systems - 4 - Question 3

Answer: a
Explanation: Natural frequency and damping can be designed by changing the gain of the individual system.

Test: Time Response Of Second Order Systems - 4 - Question 4
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Undamped natural frequency of a second order system has the following influence on the response due to various excitations:
Detailed Solution for Test: Time Response Of Second Order Systems - 4 - Question 4

Answer: a
Explanation: Undamped natural frequency is the frequency that has suffered damping and gets affected by the increase in the speed of response and decrease in sensitivity.

Test: Time Response Of Second Order Systems - 4 - Question 5
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Normalized response of a dynamic system refers to:
Detailed Solution for Test: Time Response Of Second Order Systems - 4 - Question 5

Answer: a
Explanation: Normalization refers to the desired to the reference value and normalized response of the dynamic system refers to the characteristic feature of a response due to specific excitation irrespective of its amplitude.

Test: Time Response Of Second Order Systems - 4 - Question 6
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The transfer function of a system is G(s) = 100/(s+1) (s+100). For a unit step input to the system the approximate settling time for 2% criterion is:
Detailed Solution for Test: Time Response Of Second Order Systems - 4 - Question 6

Answer: b
Explanation: Comparing the equation with the characteristic equation and then finding the value of G and w and calculating the value of settling time as 4 sec from 4/Gw.

Test: Time Response Of Second Order Systems - 4 - Question 7
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The characteristic equation of a control system is s(s2+ 6s+13)+K=0. The value of k such that the characteristic equation has a pair of complex roots with real part -1 will be : 
Detailed Solution for Test: Time Response Of Second Order Systems - 4 - Question 7

Answer: b
Explanation: The characteristic equation is considered and the values of G and w are calculated and further the value of k can be calculated.

Test: Time Response Of Second Order Systems - 4 - Question 8
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Normalized difference between the time response peak and steady state output is ______________
Detailed Solution for Test: Time Response Of Second Order Systems - 4 - Question 8

Answer: a
Explanation: Maximum peak overshoot is the normalized difference between the time response peak and steady state output.

Test: Time Response Of Second Order Systems - 4 - Question 9
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Rise time, Peak time, maximum peak overshoot, settling time, and steady state error are mutually dependent.
Detailed Solution for Test: Time Response Of Second Order Systems - 4 - Question 9

Answer: a
Explanation: Rise time, peak time, settling time and maximum peak overshoot are the prime factors of the time domain analysis and they must be specified in a consistent manner but they are mutually dependent.

Test: Time Response Of Second Order Systems - 4 - Question 10
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Control system are normally designed to be:
Detailed Solution for Test: Time Response Of Second Order Systems - 4 - Question 10

Answer: b
Explanation: Practically there are some non-linearity present in the system as friction but in mathematical model these are considered by considering high gain and lower damping.

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