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Test: Time Domain Analysis of Control Systems- 2 - Electronics and Communication Engineering (ECE) MCQ


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20 Questions MCQ Test - Test: Time Domain Analysis of Control Systems- 2

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Test: Time Domain Analysis of Control Systems- 2 - Question 1

The system shown in figure below has unit step input.

In order that the steady state error is 0.1, the value of K required should be

Detailed Solution for Test: Time Domain Analysis of Control Systems- 2 - Question 1


Since Input is unit step, therefore steady state error is

Here, 


Test: Time Domain Analysis of Control Systems- 2 - Question 2

The steady-state error coefficients for a system are Kp = ∞ , Kv = finite constant and Ka = 0. The type of the system is

Detailed Solution for Test: Time Domain Analysis of Control Systems- 2 - Question 2

We have, 


= finite constant 

From above values of Kp, Kv and Ka we conclude that the type of the system should be one.

Test: Time Domain Analysis of Control Systems- 2 - Question 3

The steady-state error of a feedback control system with an acceleration input becomes finite in a

Detailed Solution for Test: Time Domain Analysis of Control Systems- 2 - Question 3

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.

Test: Time Domain Analysis of Control Systems- 2 - Question 4

Consider the unity feedback system shown below:​

The settling time of the resulting second order system for 2% tolerance band will be

Detailed Solution for Test: Time Domain Analysis of Control Systems- 2 - Question 4

The characteristic equation of the given closed loop system is

Comparing above equation with

Thus, setting time for 2% tolerance band is

or, 

Test: Time Domain Analysis of Control Systems- 2 - Question 5

For a control system, the Laplace transform of error signal e(t) is given by . The steady state value of the error will be

Detailed Solution for Test: Time Domain Analysis of Control Systems- 2 - Question 5

Given, 


(Using final value theorem) 
or, 

Test: Time Domain Analysis of Control Systems- 2 - Question 6

For a type one system, the steady-state error due to step input is equal to

Detailed Solution for Test: Time Domain Analysis of Control Systems- 2 - Question 6


where, 

(Here, A = magnitude of step input)
Since system is type -1, therefore Kp = ∞

Test: Time Domain Analysis of Control Systems- 2 - Question 7

For an unity feedback control system with  the value of K for damping ratio of 0.5 is

Detailed Solution for Test: Time Domain Analysis of Control Systems- 2 - Question 7

Characteristic equation is


or, K = 64

Test: Time Domain Analysis of Control Systems- 2 - Question 8

The damping ratio of a system having the characteristic equation s2 + 2s + 8 = 0 is

Detailed Solution for Test: Time Domain Analysis of Control Systems- 2 - Question 8

Given, s2 + 2s + 8 = 0
Here, ωn = √8 rad/s = 2√2 rad/sec

system is underdamped.

Test: Time Domain Analysis of Control Systems- 2 - Question 9

The closed-loop transfer function of a unity - feedback system is given by  The steady state error to a unit ramp input is:

Detailed Solution for Test: Time Domain Analysis of Control Systems- 2 - Question 9

Given, 

(Since H(s) = 1)

For a unit ramp input,

Test: Time Domain Analysis of Control Systems- 2 - Question 10

The peak overshoot of step-input response of an underdamped second-order system is an indication of

Detailed Solution for Test: Time Domain Analysis of Control Systems- 2 - Question 10

If damping of system increases, peak overshoot Mp decreases and vice-versa.

Test: Time Domain Analysis of Control Systems- 2 - Question 11

Consider the position control system shown below:

The value of K such that the steady state error is 20° for input θr = 300t rad/sec, is

Detailed Solution for Test: Time Domain Analysis of Control Systems- 2 - Question 11

Input = 300t rad/sec = ramp input of magnitude A = 300.
For ramp input, steady state error is

where, 

Now, from given block diagram, we have:


So, value of gain is K = 42.97

Test: Time Domain Analysis of Control Systems- 2 - Question 12

Assertion (A): With the increase in bandwidth of the system the response of the system becomes fast.
Reason (R): Damping ratio of the system decreases with the increase in bandwidth.

Detailed Solution for Test: Time Domain Analysis of Control Systems- 2 - Question 12

When BW is increased, the system response becomes fast due to fait in rise time (tr).
With the increase in bandwidth of the system, damping ratio (ξ) decreases.
Thus, both assertion and reason are true but, reason is not the correct explanation of assertion.

Test: Time Domain Analysis of Control Systems- 2 - Question 13

For the stable system described by the block diagram shown below, Match List - I (Static error coefficients) with List - II (Values) and select the correct answer using the codes given below the lists:

List-I
A. Ka
B. Kv
C. Kp

List-II
1. ∞
2. 0
3. 2
4. -1

Codes:

Detailed Solution for Test: Time Domain Analysis of Control Systems- 2 - Question 13

We have,



Also, 



and 

Test: Time Domain Analysis of Control Systems- 2 - Question 14

The unit step response of a system is given by c(t) - 1 + 0.25 e-50t - 1.25 e-10t 
The given system is

Detailed Solution for Test: Time Domain Analysis of Control Systems- 2 - Question 14

Given, 


Since, ξ > 1, therefore given system is overdamped.

Test: Time Domain Analysis of Control Systems- 2 - Question 15

Match List - I (Transfer function of systems) with List - II (Nature of damping) and select the correct answer using the codes given below the lists:

Codes:

Detailed Solution for Test: Time Domain Analysis of Control Systems- 2 - Question 15

Characteristic equations are:
• s2 + 8s + 12 = 0

• s2 + 8s + 16 = 0
Here, ωn = 4 and 2 ξωn = 8
or, ξ = 1 (critically damped)
• s2 + 8s + 20 = 0
Here, ω = √20
and 2ξω= 8
or, ξ = 0.894 (ξ < 1 ∴ underdamped)
• s2 + 4 = 0
or, ωn = 2
and ξ = 0 (∴ undamped)

Test: Time Domain Analysis of Control Systems- 2 - Question 16

The block diagram of an electronic pacemaker is given in figure below.

What is the value of K for which the steady-state error to a unit ramp input is 0.02?

Detailed Solution for Test: Time Domain Analysis of Control Systems- 2 - Question 16

Here, 

As the input is a unit ramp function, therefore

Test: Time Domain Analysis of Control Systems- 2 - Question 17

Which one of the following equations gives the steady-state error for a unity feedback system excited by r(t) = 2 + 5t + 2t2 ?

Detailed Solution for Test: Time Domain Analysis of Control Systems- 2 - Question 17

Given, r(t) = 2 + 5t + 2t2



Test: Time Domain Analysis of Control Systems- 2 - Question 18

An unity feedback control system with closed loon transfer function is given by

The steady state error due a unit ramp input response is

Detailed Solution for Test: Time Domain Analysis of Control Systems- 2 - Question 18


Test: Time Domain Analysis of Control Systems- 2 - Question 19

The second order approximation using dominant pole concept for the transfer function

Detailed Solution for Test: Time Domain Analysis of Control Systems- 2 - Question 19

Given, 

In time constant form,


Using dominant pole concept, the given transfer function can be approximated to

Test: Time Domain Analysis of Control Systems- 2 - Question 20

The unit step response of the system represented by the block diagram shown below is

Detailed Solution for Test: Time Domain Analysis of Control Systems- 2 - Question 20

From given block diagram, we have


∴ c(t) = (4e-t - 3e-2t - 1)
= Required step response

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