# EC GATE 2010 (Test 1)

Tag: ec gate 2010
Q.1
Which of the following options is closest in meaning to the world below:
A. Cyclic
B. Indirect
C. Confusing
D. Crooked
Explaination / Solution:

Circuitous means round about or not direct. Indirect is closest in meaning to this circuitous (A) Cyclic : Recurring in nature (B) Indirect : Not direct (C) Confusing : lacking clarity of meaning (D) Crooked : set at an angle; not straight

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Topic: Arithmetic Tag: EC GATE 2010
Q.2
25 persons are in a room, 15 of them play hockey, 17 of them football and 10 of them play both hockey and football. Then the number of persons playing neither hockey nor football is ;
A. 2
B. 17
C. 13
D. 3
Explaination / Solution:

Number of people who play hockey n(A) = 15 Number of people who play football n(B) = 17 Persons who play both hockey and football n(A∩B) = 10 Persons who play either hockey or football or both : n(AUB) = n(A) + n(B) - n(A∩B) = 15 + 17 - 10 = 22 Thus people who play neither hockey nor football = 25 - 22 = 3

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Q.3
Match the logic gates in Column A with their equivalents in Column B

A. P-2, Q-4, R-1, S-3
B. P-4, Q-2, R-1, S-3
C. P-2, Q-4, R-3, S-1
D. P-4, Q-2, R-3, S-1
Explaination / Solution:

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Q.4
For the 8085 assembly language program given below, the content of the accumulator after the execution of the program is

A. 00H
B. 45H
C. 67H
D. E7H
Explaination / Solution:

By executing instruction one by one

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Q.5
In the silicon BJT circuit shown below, assume that the emitter area of transistor Q1 is half that of transistor Q2

The value of current I0 is approximately
A. 0.5 mA
B. 2 mA
C. 9.3 mA
D. 15 mA
Explaination / Solution:

Since, emitter area of transistor Q1 is half of transistor Q2, so current

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Q.6
Consider the common emitter amplifier shown below with the following circuit parameters:

The lower cut-off frequency due to C2 is
A. 33.9 Hz
B. 27.1 Hz
C. 13.6 Hz
D. 16.9 Hz
Explaination / Solution:

Cut-off frequency due to C2

Lower cut-off frequency

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Q.7
The Nyquist sampling rate for the signal  is given by
A. 400 Hz
B. 600 Hz
C. 1200 Hz
D. 1400 Hz
Explaination / Solution:

S(t) = sinc(500t) sinc(700t)
S(f ) is convolution of two signals whose spectrum covers f1 = 250 Hz and f2 = 350 Hz. So convolution extends
f = 25 + 350 = 600 Hz
Nyquist sampling rate
N = 2f = 2#600 = 1200 Hz

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Q.8
The transfer function Y(s)/R(s) of the system shown is

A. 0
B. 1/s+1
C. 2/s+1
D. 2/s+3
Explaination / Solution:

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Q.9
A unity negative feedback closed loop system has a plant with the transfer function   and a controller G(s) in the
feed forward path. For a unit set input, the transfer function of the controller that gives minimum steady state error is
A.
B.
C.
D.
Explaination / Solution:

Steady state error is given as

ess will be minimum if  is maximum In option (D)

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Q.10
A continuous time LTI system is described by

Assuming zero initial conditions, the response y(t) of the above system for the input  is given by
A. (et - e3t)u(t)
B. (e-t - e-3t)u(t)
C. (e-t  + e-3t)u(t)
D. (et  + e3t)u(t)
Explaination / Solution:

System is described as

Taking laplace transform on both side of given equation

Transfer function of the system

By Partial fraction

Taking inverse laplace transform

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