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Test: Automatic Control Systems - 4


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25 Questions MCQ Test | Test: Automatic Control Systems - 4

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Test: Automatic Control Systems - 4 - Question 1

Lag compensation increases bandwidth of the system.

Detailed Solution for Test: Automatic Control Systems - 4 - Question 1

Lag compensation improves steady state behaviour.

Test: Automatic Control Systems - 4 - Question 2

Assertion (A): The time response of the system with will not have any overshoot for unit step input.

Reason (R): A critically damped system does not have any overshoot to a unit step input.

Detailed Solution for Test: Automatic Control Systems - 4 - Question 2

System in A is overdamped.

Test: Automatic Control Systems - 4 - Question 3

For the unity feed back system of the given figure, the closed loop transfer function is

Detailed Solution for Test: Automatic Control Systems - 4 - Question 3

.

Test: Automatic Control Systems - 4 - Question 4

The polar plot of the given figure is for the term

Detailed Solution for Test: Automatic Control Systems - 4 - Question 4

For the term , magnitude is 0 at ω = ∞ and infinite at ω = 0. Moreover phase angle is -180º.

Test: Automatic Control Systems - 4 - Question 5

The effect of negative feedback on distortion and bandwidth because

Detailed Solution for Test: Automatic Control Systems - 4 - Question 5

Negative feedback increase bandwidth and reduces distortion.

Test: Automatic Control Systems - 4 - Question 6

Which of the following rotors are used in two phase ac servomotor?

  1. Solid iron rotor
  2. Squirrel cage rotor
  3. Drag cup rotor
Of the above the correct ones are
Detailed Solution for Test: Automatic Control Systems - 4 - Question 6

Solid iron rotor would give rise to eddy currents.

Test: Automatic Control Systems - 4 - Question 7

For the system of given figure the closed loop transfer function

Detailed Solution for Test: Automatic Control Systems - 4 - Question 7

or .

Test: Automatic Control Systems - 4 - Question 8

Given figure shows a unity feedback system. The ratio of time constants for open loop and closed loop response are

Detailed Solution for Test: Automatic Control Systems - 4 - Question 8

For open loop time constant = 0.25

For closed loop and time constant =

Hence ratio of time constants is 3 : 2.

Test: Automatic Control Systems - 4 - Question 9

For a second order system damping factor is 0.1. If ωd is frequency of damped oscillations and ωn is natural frequency, then

Detailed Solution for Test: Automatic Control Systems - 4 - Question 9

ωd = ωn(1 - 0.1)2 ⋍ ωn

Test: Automatic Control Systems - 4 - Question 10

For the block diagram of in the given figure, the transfer function C(s)/R(s) is

Detailed Solution for Test: Automatic Control Systems - 4 - Question 10

Forward transfer function is G + H1.

Test: Automatic Control Systems - 4 - Question 11

The phase angle curve of G(jω) H(jω) can be drawn

Detailed Solution for Test: Automatic Control Systems - 4 - Question 11

When terms are multiplied, phase angles are added.

Test: Automatic Control Systems - 4 - Question 12

The most commonly used devices for differencing and amplification, in control systems, are

Detailed Solution for Test: Automatic Control Systems - 4 - Question 12

Op-amps are commonly used due to their several advantages.

Test: Automatic Control Systems - 4 - Question 13

For the system in the given figure, the ratio C(s)/R(s) =

Detailed Solution for Test: Automatic Control Systems - 4 - Question 13

Since positive feedback is used,

Test: Automatic Control Systems - 4 - Question 14

In Bode method of stability analysis, the error curve is not symmetric about corner frequency.

Detailed Solution for Test: Automatic Control Systems - 4 - Question 14

Error curve is symmetric.

Test: Automatic Control Systems - 4 - Question 15

The transfer function can be for

Detailed Solution for Test: Automatic Control Systems - 4 - Question 15

For a lag compensator β > 1

Therefore is less than .

Test: Automatic Control Systems - 4 - Question 16

For the feedback system with closed loop transfer function the settling time for 2% tolerance

Detailed Solution for Test: Automatic Control Systems - 4 - Question 16

and settling time for 2% tolerance is 4T.

Test: Automatic Control Systems - 4 - Question 17

In a second order system with a unit step input, the speed of response is high if system is

Detailed Solution for Test: Automatic Control Systems - 4 - Question 17

An underdamped system with damping factor slightly less than 1 gives high speed of response.

Test: Automatic Control Systems - 4 - Question 18

A thermometer requires 1 minute to indicate 98% of its final response to a step input. If it is a first order system the time constant is

Detailed Solution for Test: Automatic Control Systems - 4 - Question 18

0.98 = 1 - e-1/T or T = 0.25 minute.

Test: Automatic Control Systems - 4 - Question 19

Assertion (A): A slight change in pole-zero configuration can cause only minor changes in root locus configuration.

Reason (R): Asymptotes show the behaviour of root loci for |s| >> 1

Detailed Solution for Test: Automatic Control Systems - 4 - Question 19

Even a slight change in pole-zero position may affect root locus appreciably.

Test: Automatic Control Systems - 4 - Question 20

In a second order system ωn = 10 rad/sec and ζ = 0.1, then ωd =

Detailed Solution for Test: Automatic Control Systems - 4 - Question 20

Wd = ωn1 - ζ2 = 10 (1 - 0.01)05 = 9.95 rad/s.

Test: Automatic Control Systems - 4 - Question 21

A system has 12 poles and 2 zeros. Its high frequency asymptote in its magnitude plot has a slope of

Detailed Solution for Test: Automatic Control Systems - 4 - Question 21

-(12 - 2) 20 or - 200 dB/decade.

Test: Automatic Control Systems - 4 - Question 22

The open loop transfer function of a unity feedback system is The poles of closed loop system are at

Detailed Solution for Test: Automatic Control Systems - 4 - Question 22

for s2 + 4s + 5 = 0, the roots are - 2 j1.

Test: Automatic Control Systems - 4 - Question 23

For the system of the given figure the closed loop poles are located at

Detailed Solution for Test: Automatic Control Systems - 4 - Question 23

Transfer function = .

The roots of equation s2 + 2s + 4 = 0 are at -1 3.

Test: Automatic Control Systems - 4 - Question 24

The system in the given figure, x(0) = 0 and x (0) = 0, At t = 0 the unit impulse δ(t) is applied X(s)

Detailed Solution for Test: Automatic Control Systems - 4 - Question 24

Therefore displacement .

Test: Automatic Control Systems - 4 - Question 25

If the polar plot, as the frequency is varied from 0 to infinity is

Detailed Solution for Test: Automatic Control Systems - 4 - Question 25

At ω = 0, G(jω) = 1 ∠ 0. At ω = ∞, G(jω) = 0 ∠ -90º.

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