Test: Bode Plot - 1 - Electrical Engineering (EE) MCQ

# Test: Bode Plot - 1 - Electrical Engineering (EE) MCQ

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## 15 Questions MCQ Test GATE Electrical Engineering (EE) Mock Test Series 2025 - Test: Bode Plot - 1

Test: Bode Plot - 1 for Electrical Engineering (EE) 2024 is part of GATE Electrical Engineering (EE) Mock Test Series 2025 preparation. The Test: Bode Plot - 1 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Bode Plot - 1 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Bode Plot - 1 below.
Solutions of Test: Bode Plot - 1 questions in English are available as part of our GATE Electrical Engineering (EE) Mock Test Series 2025 for Electrical Engineering (EE) & Test: Bode Plot - 1 solutions in Hindi for GATE Electrical Engineering (EE) Mock Test Series 2025 course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Bode Plot - 1 | 15 questions in 45 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study GATE Electrical Engineering (EE) Mock Test Series 2025 for Electrical Engineering (EE) Exam | Download free PDF with solutions
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Test: Bode Plot - 1 - Question 1

### 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: Bode Plot - 1 - Question 1

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

Test: Bode Plot - 1 - Question 2

### The constant M-circle represented by the equation x2 + 2.25x + y2 = -1.25 has the value of M equal to:

Detailed Solution for Test: Bode Plot - 1 - Question 2

Comparing with the M circle equation we have the value of M = 3.

Test: Bode Plot - 1 - Question 3

### The constant N loci represented by the equation x2 + x + y2 = 0 is for the value of phase angle equal to:

Detailed Solution for Test: Bode Plot - 1 - Question 3

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

Test: Bode Plot - 1 - Question 4

Consider the following statements:
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: Bode Plot - 1 - Question 4

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: Bode Plot - 1 - Question 5

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: Bode Plot - 1 - Question 5

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: Bode Plot - 1 - Question 6

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

Detailed Solution for Test: Bode Plot - 1 - Question 6

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

Test: Bode Plot - 1 - Question 7

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

Detailed Solution for Test: Bode Plot - 1 - Question 7

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

Test: Bode Plot - 1 - Question 8

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: Bode Plot - 1 - Question 8

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

Test: Bode Plot - 1 - Question 9

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: Bode Plot - 1 - Question 9

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

Test: Bode Plot - 1 - Question 10

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

Detailed Solution for Test: Bode Plot - 1 - Question 10

Comparing with the M circle equation we have the value of M = 3.

Test: Bode Plot - 1 - Question 11

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

Detailed Solution for Test: Bode Plot - 1 - Question 11

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

Test: Bode Plot - 1 - Question 12

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

Detailed Solution for Test: Bode Plot - 1 - Question 12

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

Test: Bode Plot - 1 - Question 13

Frequency range of bode magnitude and phases are decided by :

Detailed Solution for Test: Bode Plot - 1 - Question 13

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

Test: Bode Plot - 1 - Question 14

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: Bode Plot - 1 - Question 14

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

Test: Bode Plot - 1 - Question 15

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

Detailed Solution for Test: Bode Plot - 1 - Question 15

Real part represents the damping and imaginary part damped frequency.

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