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Test: Transfer Function - 1 - Electrical Engineering (EE) MCQ


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20 Questions MCQ Test GATE Electrical Engineering (EE) Mock Test Series 2025 - Test: Transfer Function - 1

Test: Transfer Function - 1 for Electrical Engineering (EE) 2024 is part of GATE Electrical Engineering (EE) Mock Test Series 2025 preparation. The Test: Transfer Function - 1 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Transfer Function - 1 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Transfer Function - 1 below.
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Test: Transfer Function - 1 - Question 1

Detailed Solution for Test: Transfer Function - 1 - Question 1

Test: Transfer Function - 1 - Question 2

The output of the feedback control system must be a function of:

Detailed Solution for Test: Transfer Function - 1 - Question 2

Explanation: In a feedback control system, the output is determined by a combination of the input signal and the feedback signal. The input signal is the desired output or reference, while the feedback signal is a portion of the actual output that is fed back into the system to compare with the input. This comparison helps the system to adjust its output to minimize the error between the input and the actual output. By considering both the input and feedback signals, the system can continuously adapt and achieve the desired output.

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Test: Transfer Function - 1 - Question 3

The unit impulse response of a certain system is found to be e-8t. Its transfer function is _______.

Detailed Solution for Test: Transfer Function - 1 - Question 3

The impulse response is defined as the output of an LTI system due to a unit impulse signal input being applied at time t = 0.

y(t) = h(t) x(t) = h(t) δ(t)

where δ(t) is the unit impulse function and h(t) is the unit impulse response of a continuous-time LTI system.

Calculations:-

Given-

y(t)  = e-8t 

x(t) = δ(t)

For calculating the transfer function convert the time domain response into Laplace or S domain.

*Answer can only contain numeric values
Test: Transfer Function - 1 - Question 4

For the system shown in the figure, Y(s)/X(s) = _________. (Answer in integer )


Detailed Solution for Test: Transfer Function - 1 - Question 4

The circuit is redrawn as shown:

[X(s) – Y(s)] G(s) + X(s) = Y(s)

Given G(s) = 2

(X(s) – Y(s)) 2 + X(s) = Y(s)

= 2 X(s) + X(s) = Y(s) + 2Y(s)

= 3 X(s) = 3 Y(s)

Test: Transfer Function - 1 - Question 5

Assuming zero initial condition, the response y(t) of the system given below to a unit step input u(t) is

Detailed Solution for Test: Transfer Function - 1 - Question 5

Laplace transform of u(t) is given by 

Taking Inverse Laplace transform

y(t) = t.u(t)

Test: Transfer Function - 1 - Question 6

The sum of the gains of the feedback paths in the signal flow graph shown in fig.  is

Detailed Solution for Test: Transfer Function - 1 - Question 6
  • Any feedback path is considered as a valid feedback path if it does not touches any single node twice in its path.
  • Feedback paths in the above diagram are af, be, cd, abef, bcde, and abcdef.
  • abef is not a valid feedback path because node 'Q' is repeated twice in the path.
  • bcde is not a valid feedback path because node 'R' is repeated twice in the path.
  • abcdef is not a valid feedback path because node 'Q' and 'R' is repeated twice in the path.

So, af, be and cd are the only valid feedback paths. The sum of the gains of the feedback paths in the signal flow graph is af + be + cd.

Test: Transfer Function - 1 - Question 7

A linear system with H(s) = 1/s is excited by a unit step function input. The output for t > 0 is given by

Detailed Solution for Test: Transfer Function - 1 - Question 7

TF = H(s) = 1/s,

Input is unit step, R(s) = 1/s

Now output for t > 0 can be calculated as

C(s) = H(s) x R(s)

C(s) = (1/s) x (1/s)

C(s) = 1 / s2

L-1[C(s)] = t

C(t) = t = ramp function

Test: Transfer Function - 1 - Question 8

The overall transfer function C/R of the system shown in fig. will be:

Detailed Solution for Test: Transfer Function - 1 - Question 8


Test: Transfer Function - 1 - Question 9

Consider the List I and List II

The correct match is

Detailed Solution for Test: Transfer Function - 1 - Question 9
  • P. P1 = ab, Δ = 1, L = 0 ,T = ab
  • Q. P1 = a, P2 = 6 , Δ = 1, L = Δk = 0,T  = a+b
  • R. P1 = a, L1 = b, Δ = 1 - b, Δ1 =1,
  • S. P1 = a, L1 = ab, Δ = 1 - ab, Δ1 = 1, 
Test: Transfer Function - 1 - Question 10

For the signal flow graph shown in fig. an equivalent graph is

Detailed Solution for Test: Transfer Function - 1 - Question 10

While writing the transfer function of this signal flow graph,

e2= tae1 + tbe= (ta+ tb) e1 

Then, signal flow graph will lokk like this:

Test: Transfer Function - 1 - Question 11

The block diagram of a system is shown in fig. The closed loop transfer function of this system is

Detailed Solution for Test: Transfer Function - 1 - Question 11

Consider the block diagram as SFG. There are two feedback loop -G1G2H1 and -G2G3H2 and one forward path G1G2 G3 . So (D) is correct option.

Test: Transfer Function - 1 - Question 12

For the system shown in fig. transfer function C(s) R(s) is

Detailed Solution for Test: Transfer Function - 1 - Question 12

Consider the block diagram as a SFG. Two forward path G1G2 and G3 and three loops -G1G2 H2, -G2H1, -G3 H2
There are no nontouching loop. So (B) is correct.

Test: Transfer Function - 1 - Question 13

In the signal flow graph shown in fig. the transfer function is

Detailed Solution for Test: Transfer Function - 1 - Question 13

P1 = 5 x 3 x 2 = 30, Δ = 1 - (3x - 3) = 10
Δ1 = 1, 

Test: Transfer Function - 1 - Question 14

In the signal flow graph shown in fig. the gain C/R is

Detailed Solution for Test: Transfer Function - 1 - Question 14

P1 = 2 x 3 x 4 = 24 , P2 = 1 x 5 x 1 = 5
L1 = -2, L2 = -3, L3 = -4, L4 = -5,
L1L3 = 8, Δ = 1 -(-2 - 3 - 4 - 5) + 8 = 23, Δ1 = 1, Δ2 = 1 - (-3) = 4, 

Test: Transfer Function - 1 - Question 15

The gain C(s)/R(s) of the signal flow graph shown in fig.

Detailed Solution for Test: Transfer Function - 1 - Question 15

X1 = 1 - H1

X2 = (G1 + G3) X1 - X3H1

X= X2G2 + G4

Test: Transfer Function - 1 - Question 16

The negative feedback closed-loop system was subjected to 15V. The system has a forward gain of 2 and a feedback gain of 0.5. Determine the output voltage and the error voltage.

Detailed Solution for Test: Transfer Function - 1 - Question 16

Given: 
G(s) = 2 
H(s) = 0.5 and R(s) = 10V 
Output voltage:

= (2/1 + 2 x 0.5) x 15 = 15V 
Error voltage:

= (1/1 + 2 x 0.5) x 15 = 7.5V

Test: Transfer Function - 1 - Question 17

For the block diagram shown in fig. transfer function C(s)/R(s) is

Detailed Solution for Test: Transfer Function - 1 - Question 17

Four loops -G1G4, -G1G2G5, -G1,G2G5G7 and -G1G2G3G3G7.

There is no nontouching loop. So (B) is correct.

Test: Transfer Function - 1 - Question 18

For the block diagram shown in fig. the numerator of transfer function is

Detailed Solution for Test: Transfer Function - 1 - Question 18

SFG

P1 = G2G5G6 , P2 = G3G5G6, P3 = G3G6 , P4 = G4G6
If any path is deleted, there would not be any loop.
Hence Δ1 = Δ2 = Δ3 = Δ4 = 1 

Test: Transfer Function - 1 - Question 19

For the block diagram shown in fig.  the transfer function C(s)/R(s) is

Detailed Solution for Test: Transfer Function - 1 - Question 19

Test: Transfer Function - 1 - Question 20

In the signal flow graph of figure y/x equals

Detailed Solution for Test: Transfer Function - 1 - Question 20

Transfer function

PK = 5 x 2 x 1 = 10
ΔK = 1
Δ = 1 - (-4) = 5

= 2

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