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Test: Polar And Bode Plots - Electrical Engineering (EE) MCQ


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29 Questions MCQ Test - Test: Polar And Bode Plots

Test: Polar And Bode Plots for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Test: Polar And Bode Plots questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Polar And Bode Plots MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Polar And Bode Plots below.
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Test: Polar And Bode Plots - Question 1

The constant M circle for M=1 is the

Detailed Solution for Test: Polar And Bode Plots - Question 1

Answer: a
Explanation: For M =1 the constant M circle is a straight line at x=-1/2.

Test: Polar And Bode Plots - Question 2

The polar plot of a transfer function passes through the critical point (-1,0). Gain margin is

Detailed Solution for Test: Polar And Bode Plots - Question 2

Answer: a
Explanation: Gain margin of a polar plot passing through the critical point is zero.

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Test: Polar And Bode Plots - Question 3

Consider the following statements:
1. The effect of feedback is to reduce the system error
2. Feedback increases the gain of the system in one frequency range but decreases in another
3. Feedback can cause a system that is originally stable to become unstable
Which of these statements are correct.

Detailed Solution for Test: Polar And Bode Plots - Question 3

Answer: c
Explanation: Feedback can cause the increase in gain and also can cause stable system to become unstable.

Test: Polar And Bode Plots - Question 4

The open loop transfer function of a system is G(s) H(s)= K / (1+s)(1+2s)(1+3s)

The phase cross over frequency ωc is

Detailed Solution for Test: Polar And Bode Plots - Question 4

Answer: b
Explanation: Phase crossover frequency is calculated as by calculating the magnitude of the transfer function and equating it to 1 and the frequency calculated at this magnitude is phase cross over frequency.

Test: Polar And Bode Plots - Question 5

If the gain of the open-loop system is doubled, the gain margin

Detailed Solution for Test: Polar And Bode Plots - Question 5

Answer: a
Explanation: If the gain of the open-loop system is doubled, the gain margin gets doubled.

Test: Polar And Bode Plots - Question 6

The unit circle of the Nyquist plot transforms into 0dB line of the amplitude plot of the Bode diagram at

Detailed Solution for Test: Polar And Bode Plots - Question 6

Answer: d
Explanation: The unit circle of the Nyquist plot transforms into 0dB line of the amplitude plot of the Bode diagram at any frequency.

Test: Polar And Bode Plots - Question 7

Consider the following statements:
The gain margin and phase margin of an unstable system may respectively be
1. Positive, positive
2. Positive, negative
3. Negative, positive
4. Negative, negative
Of these statements

Detailed Solution for Test: Polar And Bode Plots - Question 7

Answer: d
Explanation: For unstable system the signs of gain margin and phase margin are always different or they can both be negative.

Test: Polar And Bode Plots - Question 8

If a system has an open loop transfer function1-s / 1+s, then the gain of the system at frequency of 1 rad/s will be

Detailed Solution for Test: Polar And Bode Plots - Question 8

Answer: d
Explanation: The system is all pass system and the gain of the system at frequency of 1 rad/sec.

Test: Polar And Bode Plots - Question 9

The polar plot of the open loop transfer function of a feedback control system intersects the real axis at -2. The gain margin of the system is

Detailed Solution for Test: Polar And Bode Plots - Question 9

Answer: c
Explanation: Gain margin of the system is inverse of the intersect on the real axis and calculated in decibels.
G(s) = 1+s / s(1+0.5s).

Test: Polar And Bode Plots - Question 10

The corner frequencies are

Detailed Solution for Test: Polar And Bode Plots - Question 10

Answer: d
Explanation: Corner frequency can be calculated by time constant form of the transfer function and here the corner frequencies are 1 and 2.

Test: Polar And Bode Plots - Question 11

For the transfer function

G(s) H(s) = 1 / s(s+1) (s+0.5), the phase cross-over frequency is

Detailed Solution for Test: Polar And Bode Plots - Question 11

Answer: b
Explanation: Phase cross over frequency is calculated at the point where magnitude of the polar plot is 1.

Test: Polar And Bode Plots - Question 12

The gain margin (in dB) of a system having the loop transfer function

G(s) H(s) = 2 / s(s+1) is

Detailed Solution for Test: Polar And Bode Plots - Question 12

Answer: d
Explanation: Gain margin of a system is calculated at the phase cross over frequency and expressed in decibels.

Test: Polar And Bode Plots - Question 13

The gain margin for the system with open loop transfer function

G(s) H(s) = G(s) =2(1+s) / s2 is

Detailed Solution for Test: Polar And Bode Plots - Question 13

Answer: 0
Explanation: Gain margin of a system is calculated at the phase cross over frequency and expressed in decibels.

Test: Polar And Bode Plots - Question 14

Statement 1: In constant M circles, as M increases from 1 to 8 radius of circle increases from 0 to 8 and Centre shifts from (-1,0) to (-8,0)
Statement 2: The circle intersects real axis at point (-1/2, 0)

Detailed Solution for Test: Polar And Bode Plots - Question 14

Answer: d
Explanation: All the circles pass through the points (0,0) and (-1,0).

Test: Polar And Bode Plots - Question 15

Assertion (A): Relative stability of the system reduces due to the presence of transportation lag.
Reason (R): Transportation lag can be conveniently handled by Bode plot.

Detailed Solution for Test: Polar And Bode Plots - Question 15

Answer: b
Explanation: Transportation lag can be conveniently handled on Bode plot as well without the need to make any approximation.

Test: Polar And Bode Plots - Question 16

Assertion (A): The phase angle plot in Bode diagram is not affected by the variation in the gain of the system.
Reason(R): The variation in the gain of the system has no effect on the phase margin of the system.

Detailed Solution for Test: Polar And Bode Plots - Question 16

Answer: c
Explanation: The variation in the gain of the system has effect on the phase margin but phase plot is not affected.

Test: Polar And Bode Plots - Question 17

A system has poles at 0.01 Hz, 1 Hz and 80Hz, zeroes at 5Hz, 100Hz and 200Hz. The approximate phase of the system response at 20 Hz is : 

Detailed Solution for Test: Polar And Bode Plots - Question 17

Answer: a
Explanation: Pole at 0.01 Hz gives -180° phase. Zero at 5Hz gives 90° phase therefore at 20Hz -90° phase shift is provided.

Test: Polar And Bode Plots - Question 18

The constant M-circle represented by the equation x^2+2.25x+y^2=-1.25 has the value of M equal to: 

Detailed Solution for Test: Polar And Bode Plots - Question 18

Answer: c
Explanation: Comparing with the M circle equation we have the value of M =3.

Test: Polar And Bode Plots - Question 19

What is the value of M for the constant M circle represented by the equation 8x2+18x+8y2+9=0? 

Detailed Solution for Test: Polar And Bode Plots - Question 19

Answer: c
Explanation: Comparing with the M circle equation we have the value of M =3.

Test: Polar And Bode Plots - Question 20

The constant N loci represented by the equation x^2+x+y^2=0 is for the value of phase angle equal to: 

Detailed Solution for Test: Polar And Bode Plots - Question 20

Answer: d
Explanation: Centre = (-0.5, 0)
Centre of N circle is (-1/2, 1/2N)
N =tanα
α =90°.

Test: Polar And Bode Plots - Question 21

All the constant N-circles in G planes cross the real axis at the fixed points. Which are these points?

Detailed Solution for Test: Polar And Bode Plots - Question 21

Answer: a
Explanation: Centre of N circle is (-1/2, 1/2N)
N =tanα
Constant –N circles always pass through (-1, 0) and (0, 0).

Test: Polar And Bode Plots - Question 22

Consider the following statements:
Nichol’s chart gives information about.
i. Closed loop frequency response.
ii. The value of the peak magnitude of the closed loop frequency response Mp.
iii. The frequency at which Mp occurs.
Which of the above statements are correct?

Detailed Solution for Test: Polar And Bode Plots - Question 22

Answer: d
Explanation: Nichol’s chart gives information about closed loop frequency response, value of the peak magnitude of the closed loop frequency response Mp and the frequency at which Mp occurs.

Test: Polar And Bode Plots - Question 23

Which one of the following statements is correct? Nichol’s chart is useful for the detailed study analysis of:

Detailed Solution for Test: Polar And Bode Plots - Question 23

Answer: a
Explanation: Nichol’s chart is useful for the detailed study analysis of closed loop frequency response.

Test: Polar And Bode Plots - Question 24

In a bode magnitude plot, which one of the following slopes would be exhibited at high frequencies by a 4th order all-pole system?

Detailed Solution for Test: Polar And Bode Plots - Question 24

Answer: a
Explanation: A 4th order all pole system means that the system must be having no zero or s-term in numerator and s4 terms in denominator. One pole exhibits -20dB/decade slope, so 4 pole exhibits a slope of -80 dB /decade.

Test: Polar And Bode Plots - Question 25

Frequency range of bode magnitude and phases are decided by :

Detailed Solution for Test: Polar And Bode Plots - Question 25

Answer: d
Explanation: T. F. = Kp (1+Tds)
There is only one zero which will give slope of +20dB/decade.

Test: Polar And Bode Plots - Question 26

OLTF contains one zero in right half of s-plane then

Detailed Solution for Test: Polar And Bode Plots - Question 26

Answer: c
Explanation: OLTF contains one zero in right half of s-plane then Close loop system is unstable for higher gain.

Test: Polar And Bode Plots - Question 27

The critical value of gain for a system is 40 and gain margin is 6dB. The system is operating at a gain of:

Detailed Solution for Test: Polar And Bode Plots - Question 27

Answer: a
Explanation: Gm (dB) = 20log⁡GM
GM =2
As we know, GM =K (marginal)/K (desired)
K desired =40/2 =20.

Test: Polar And Bode Plots - Question 28

Nichol’s chart is useful for the detailed study and analysis of:

Detailed Solution for Test: Polar And Bode Plots - Question 28

Answer: a
Explanation: Nichol’s chart is useful for the detailed study and analysis of closed loop frequency response.

Test: Polar And Bode Plots - Question 29

The roots of the characteristic equation of the second order system in which real and imaginary part represents the :

Detailed Solution for Test: Polar And Bode Plots - Question 29

Answer: b
Explanation: Real part represents the damping and imaginary part damped frequency.

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