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GATE Mock Test Electronics Engineering (ECE)- 2 - Electronics and Communication Engineering (ECE) MCQ


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65 Questions MCQ Test - GATE Mock Test Electronics Engineering (ECE)- 2

GATE Mock Test Electronics Engineering (ECE)- 2 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The GATE Mock Test Electronics Engineering (ECE)- 2 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The GATE Mock Test Electronics Engineering (ECE)- 2 MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for GATE Mock Test Electronics Engineering (ECE)- 2 below.
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GATE Mock Test Electronics Engineering (ECE)- 2 - Question 1

Direction: In the following question, there are three sentences. Each sentence has a pair of words which is underlined. From the underlined words, select the most appropriate word (1) or (2) to form correct sentences. Choose the correct options to form correct sentences.

The report submitted by the committee does not quite jibe(1)/gibe(2) with the observations.

Ryan tried to flesh(1)/flush (2) out his novel with incidents borrowed from other writers.

Anger is always a bad thing, even when it is a condign (1)/ condescend(2) response to events.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 1
The word ‘jibe’ means ‘be in harmony’ and ‘gibe’ means ‘to mock’. Thus 1 fits here best.

The word ‘flesh out’ means ‘to expand’ and ‘flush out’ means ‘to clean’. Thus 1 conveys the correct meaning.

The word ‘condign’ means ‘appropriate’ and ‘condescend’ means ‘to lower to a less dignified level’. Thus 1 fits here correctly.

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 2

Directions: In the following question, a set of three figures X, Y and Z shows a sequence in which a paper is folded and finally cut from a particular section. Below these figures, a set of answer figures (a), (b), (c) and (d) shows the design which the paper actually acquires when it is unfolded. You have to select the answer figure which most closely resembles the unfolded piece of paper.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 2

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 3

If the rate of change in the circumference of a circle is 0.5 cm/s, then find the rate of change in the area of the circle when the radius is 8 cm.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 3
C = 2πr

⇒ dC/dt = 2πdr/dt

⇒0.5/2π = dr/dt

∴ A = πr2

⇒dA/dt = 2πrdr/dt = 0.5 x 8 = 4 sq.units

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 4

Which of the following is an antonym of the word PROFESSIONAL?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 4
Professional is the person who does something as a part of his job. Amateur is a person who does something because he loves doing it.
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 5

Find the wrong number in the series

2, 9,5, 36, 125, 648

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 5

5 is wrong number, starting new series with 5, we get

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 6

The ratio of cost price and marked price of a cell phone is 2 : 3 and ratio of profit percentage and discount percentage is 3 : 2. What is the discount percentage?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 6

CP:MP = 2:3

Let CP = 200, MP = 300

(%) profit : (%) discount = 3:2

⇒ 3x/100 x 200 + 2x/100 x 300 = 100

⇒ x = 8.33%

Discount = 2x = 16.66%

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 7

The line graph shows Net exports in million USD of a certain country (Net exports = Exports - Imports). Study the diagram and answer the following questions. 

Cumulative net exports (in million USD) from the beginning of 2011 to the end of 2014 was?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 7

Cumulative net exports (in million USD) from the beginning of 2011 to the end of 2014 
= 20 - 10 - 20 + 30 = 20

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 8

Directions: Choose the sentence that best combines the given sentences.

The airport is called the Glynco Jetport. The airline reservations and travel systems refer to the location as Brunswick, Georgia.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 8
Option 3 is the most logical subordinating phrase, showing a contrast. The other choices are not only illogical but also grammatically incorrect.
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 9

The question below consists of a set of labelled sentences. These sentences, when properly sequenced form a coherent paragraph. Select the most logical order of sentences from among the options.

P. This partial eclipse began at 7:18 IST on July 13, reached its maximum magnitude of 0.3367 at 8:31 IST and ended at 09:43 IST.

Q. The year's second solar eclipse occurred on July 13, 2018.

R. The partial eclipse was seen only in parts of Australia and New Zealand as it took place almost entirely over open waters of Antartica.

S. Solar eclipse is a natural phenomenon that occurs when the moon comes in between the sun and the earth.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 9
The given passage talks about the second solar eclipse of this year. Hence, Q should be the first sentence as it introduces the topic. R should come next as it tells about the eclipse is that it was a partial eclipse. P should follow it as it uses the term ‘this partial eclipse’ (another hint that partial eclipse introduced in earlier sentence) gives further information about timing of the eclipse. In the end, sentence S gives general information about solar eclipse. Hence the correct sequence is QRPS
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 10

Leo participated in a race of motor bikes. Flags of different colours are lowered to give different signals. How many different signals can be made here, using any number of flags from 5 flags of different colours?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 10
Signals can be made by using at a time one, two, three, four and five flags.

∴ Total number of signals that can be made = 5P1 + 5P2 + 5P3 + 5P4 + 5P5

= 5!/4! + 5!/3! + 5!/2! + 5!/1! + 5!/1

= 5 + (5 × 4) + (5 × 4 × 3) + (5 × 4 × 3 × 2 × 1) + (5 × 4 × 3 × 2 × 1)

= 5 + 20 + 60 + 120 + 120

= 325

So, 325 signals can be made using five flags.

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 11

If A is 3 × 3 matrix, whose elements are given by aij = i2 – j2 where 1 ≤ i, j ≤ 3 then A–1 = _____

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 11
aij = i2 – j2∀i, j

∴A3×3is singular

∴|A| = 0

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 12

The load curve is useful in deciding:

  1. The operating schedule of generating units

  2. The total installed capacity

Which of the above statements is/are correct?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 12
The load curve indicates average (KWh) energy consumption during a given period and hence, it is useful in deciding both the operating schedule of generating units and the total installed capacity.
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 13

What is the behavior of following one i/p Flipflop ‘X’ ?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 13

→ Qn+1 = X

Therefore it is D-Flipflop.

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 14

Consider the circuit diagram shown below. The circuit represents

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 14

In the case of transistor AND gate, when the inputs X, Y = 0 V or when X = 0 V and Y = +5 V or when X = +5 V and Y = 0 V, both the transistors Q1 and Q2 are at OFF state. At the same time, transistor Q3 gets enough base drive from the supply through resistor R3, so transistor Q3 will be ON. Therefore, output voltage Z = Vce(sat) corresponds to 0 V. When both the inputs are equal to +5 V, transistors Q1 and Q2 will be ON; therefore, the voltage at the collector of transistor Q1 will drop. As a result, transistor Q3 does not get enough base drive and turns OFF. Therefore, no current flows through the collector resistor of Q3, hence no voltage drop across it. So, the final output voltage corresponds to +5 V.

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 15

In the given figure, Zin at the generator is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 15

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 16

For static electric and magnetic fields in an homogeneous source-free medium, which of the following represents the correct form of Maxwell equations?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 16
Maxwell equations

For static electric and magnetic fields

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 17

Consider the circuit below with initial state Q0 = 1, Q1 = Q2 = 0. The state of the circuit is given by the value of 4Q2 + 2Q1 + Q0

Which one of the following is the correct state sequence of the circuit?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 17

State= 4Q2 + 2Q1 + Q0

This State Equation is Decimal Equivalent of binary State.

Ex: Qo Q1 Q2 = 100, state = 4 × 0 + 2 × 0 + 1 = 1

Hence the Answer is 1,2,5,3,7,6,4 (See the table)

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 18

Consider the following circuits:

  1. Oscillator

  2. Emitter follower

  3. Power amplifier

Which of the above circuits employ feedback?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 18
Oscillator

With the use of positive feedback or regenerative feedback, the oscillations will take place in a circuit and the circuit acts as an oscillator.

Emitter follower:

It has negative current feedback as shown in the figure below.

Power Amplifier:

A power amplifier has negative feedback. The negative feedback is used to reduce the distortion and noise and to increase the bandwidth.

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 19

A sinusoidal message signal of amplitude Am and frequency fm is passed through a delta modulator whose step size in 0.628 V and sampling rate is given by 40K sample/ sec then which of the following delta modulator will be slope overloaded.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 19
Step size (s)=0.628 V

Condn for slope overload Δ/Ts < />mAm

0.628 x fs < />mAm

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 20

The switch in the figure below was closed for a long time. It is opened at t = 0. The current in the inductor of 2 H for t ≥ 0, is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 20
At t = 0-, inductor acts as short circuit,

Now, the circuit is reduced to,

Equivalent resistance Req = 10 Ω

Current, I =50/10 = 5 A

By using current division rule,

IL (0-) = 5/2 = 2.5 A

At t = 0+,

IL (0+) = IL (0-) = 2.5 A

At steady state (t =∞ )

Time constant of the circuit (τ) = L/Req

Req is the equivalent resistance across inductor.

Req = 16||32||32 = 8 Ω

Time constant:

(τ) = L/Req = 2/8 = 0.25 sec

= 0.25 sec

= 0 + (2.5 - 0) e-4t = 2.5 e-4t A

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 21

In the question a circuit and a waveform for the input voltage is given. The diode in circuit has cut-in voltageVγ=0 Choose the option for the waveform of output voltage V0.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 21
During positive cycle when vs < 8V, both diode is OFF vo = vs⋅ For vs > 8,v0 = 8V,D1 is ON.

During negative cycle when |vs|<6, both diode is OFF, vo = vs. For |vs| > 6V,D2 is on = -6 V

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 22

If a, b, c are 3-input variables, then Boolean function y = ab + bc + ca represents:

  1. A 3-input majority gate

  2. A 3-input minority gate

  3. Carry output of a full adder

  4. Product circuit for a, b, and c

Which of the above statements are correct?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 22
In the given input combination, if the majority of inputs are 1, then the Boolean (SOP) function gives a majority gate. In the given 3-input a, b, c if at least two inputs are '1' then the Boolean function y = ab + bc + ca represents majority gate. Consider the truth table of full adder

By using K-map method, the carry output is obtained as follows:

Thus the given Boolean function represents carry o/p of a full adder.

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 23

A gallium arsenide pn junction is operating in reverse-bias voltage VB = 5V. The doping profile are Na = Nd = 1016cm−3. The minority carrier life- time are τp0 = τn0 = τ0 = 10−8s. The reverse-biased generation current density is s (εr = 13.1, ni = 1.8 x 106)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 23

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 24

Using the residue theorem, the value of the integral (counterclockwise) around a circle with center at z = 0 and radius = 8 (where z is a complex number and i = √-1 ), is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 24

circle having centre z = 0

Radius = 8

z = 4 is a simple pole.

= -20

Using Residue theorem,

= 2πi (-20) = -40πi

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 25

Under which of the following conditions, the N-MOS transistor operates in depletion mode?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 25

For an N-MOSFET, when Vt < 0, the transistor operates in depletion mode and when Vt > 0, the transistor operates in enhancement mode.

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 26

In the following circuit, a network and its Thevenin and Norton equivalent are given.

The value of the parameters are

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 26
Thevenin voltage: (Open circuit voltage)

VTh = 4 + (2 x 2) = 4 + 4 = 8 V

Thevenin resistance:

RTh = 2 + 3 = 5Ω = RN

Norton current:

IN = VTh/RTh = 8/5A

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 27

Which of the following meters is based on the principle of Hall Effect?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 27
Voltmeter and ammeter works on the principle of electrostatic principle. Gauss meter works on the principle of Half effect.
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 28

In the Op-Amp circuit shown, assume that the diode current follows the equation I = Is exp (V/VT). For Vi = 2V, V0 = V01; and for Vi = 4 V, V0 = V02.

The relationship between V01 and V02 is:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 28
Here the inverting terminal is at virtual ground and the current in resistor and diode current is equal i.e.

IR = ID

or VD = VT In Vi/IsR

For the first condition

VD = 0 - V01 = VT In 2/IsR

For the second condition

VD = 0 - V02 = VT In 4/IsR

Subtracting above equation

V01 - V02 = VT In 4/IsR - VT In 2/IsR

or V01 - V02 = VT In 4/2 = VT In 2

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 29

Let f(x) be a real, periodic function adequate the condition f (-x) = -f(x). The general form of its Fourier series representation would be

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 29

Given signal is odd function so a0=0

f (-x) = -f(x)

And the Fourier series will content sine term only

So

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 30

A white noise process X(t) with two-sided power spectral density 1 x 10-10 W/Hz is input to a filter whose magnitude squared response is shown below.

The power (in W) of the output process Y(t) is given by ___× 10-6

(Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 30
Power of the output process Y(t),

Y(t) = |H(t)|2 . X(t)

X(t) = N0/2 = 1 x 10-10W/Hz

|H(t)|2 = area under curve |u(t)|2

= 2(1/2 x 10 x 103 x 1) = 104 Hz

Then, Y(t) = 10-6 W

or Y(t) = 1 x 10-6 W

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 31

For the periodic signal x(t) shown below with period T = 10sec, the power in the 20th harmonic will be-


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 31
The given signal is odd and half wave symmetry so it has only odd harmonics

Hence power in 20th harmonic will be zero.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 32

The divergence of the vector field V = x2i + 2y3j + z4k at x = 1, y = 2, z = 3 is ____ (Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 32

= 2x + 6y2 + 4z3

At x = 1, y = 2 and z = 3

[Divergence (V)] = 2 x 1 + 6 x 22 + 4 x 33 = 134

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 33

Assume electronic charge q = 1.6×1019C,AT/q = 25mV and electron mobility μn = 1000cm2/V−s. If the concentration gradient of electrons injected into a P-type silicon sample is 1×1021cm4, the magnitude of electron diffusion current density {in A/cm ?


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 33
Given

q = 1.6 × 10−19; kT/q = 2.5mV

μn = 1000cm2/v−s

From Eindtein relation,

D/μn = kT/q ⇒ D = 25mV × 1000cm2/v−s

⇒ 25em2/s

Diffusion current Density

J = qDndn/dx =1.6 × 10−19 × 25 × 1 × 1021

= 4000.A/cm2

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 34

Newton-Raphson method is to be used to find root of equation 3x - ex + sinx = 0. If the initial trial value for the root is taken as 0.333, the next approximation for the root would be ______________.

(Answer up to two decimal places)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 34
Let f(x) = 3x - ex + sinx and x0 = 0.333 ≈ 1/3

⇒ f'(x) = 3 - ex + cos x

f(x0) = -0.069 and f'(x0) = 2.55

∴ x1 = x0 - [f(xo)]/[f’(x0)] (Using Newton-Rapshon method)

= 0.333 + 0.069/2.55 = 0.36 is the required next approximation

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 35

Consider the circuit given below

The Leq for the circuit is _______H.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 35
Redrawing the circuit

Applying KVL

By equation (i) and (ii)

Vi = 8.625di/dt Vi = Legdi/dt

Leg = 8.625 H

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 36

The forward path transfer function of a unity feedback system is given by

The system is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 36
Characteristic equation is

s4 + 10s3 + 35s2 + 50s + 264 = 0

According to Routh's stability criteria, a system is stable only when all the values in the first column are positive and the system does not satisfy this condition. Hence, the system is unstable.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 37

If the median of (x-1), (x-1), (x-4), (x+4), (x-3) is zero then value of x is ________?


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 37

The median of a set of numbers is the middle value when the numbers are arranged in ascending or descending order. If we have an odd number of numbers, the median is the middle number. If we have an even number of numbers, the median is the average of the two middle numbers.

In this case, we have five numbers: (x-1), (x-1), (x-4), (x+4), (x-3). To find the median, we need to arrange these in ascending order:

(x-4), (x-3), (x-1), (x-1), (x+4)

Since we have an odd number of numbers, the median is the middle number, which is (x-1).

According to the problem, the median is zero. So:

x-1 = 0

Solving for x, we find:

x = 1

So, the value of x that makes the median of these numbers zero is 1.

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 38

Consider an analytic function f(z) = u + iv, where z = x + iy. If the real part of analytic function is u = x3 − 3xy2, then the analytic function f(z) is (where 'c' is a constant)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 38

= 3z2 - i(-6z x 0) = 3z2

f(z) = z3 + c

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 39

By using Simpson's 1/3rd rule find ∫06sin⁡xdx by taking a width of 1


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 39
Divide the internal (0.6) into six parts each of width h = 1

Width h = 1

y = f(x) = sin x

Put calculator in radian mode

By using Simpson's 1/3rd rule, we have

06sin⁡x dx = h/3[(y0 + y6) + 4(y1 + y3 + y5) + 2(y2 + y4)]

= 1/3[(0 − 0.279) + 4(0.841 + 0.141 − 0.958) + 2(0.909 − 0.756)]

= 0.041

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 40

The schematic shown below generates FM signal as a wideband frequency modulator using the indirect method. The output FM signal is characterized by fc = 100 MHz, Δf = 75 kHz (min).

If the phase modulator has a modulation index of β = 0.3 radians, then the values of n1, n2 will be respectively

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 40
The lowest modulation frequency i.e., tx = 100 Hz

Produces a frequency deviation f1 = β x fme = (0.2 100) = 20 Hz

The highest modulation frequency i.e. fm = 15 kHz

Produces a frequency deviation = Δf1 = β x fmh = (0.2 x 15) = 3 kHz

Let us take the case of 100 H(fmi) as the worst case.

n1n2 = Δf/Δf1 = 75000/20 = 3750… (i)

And f1 - n1f1 = fc/n2

Hence, with f1 = 0.1 MHz, f2 = 9.5 MHz, fc = 100 MHz

We have, 9.5 – 0.1 n1 = 100/n2 … (ii)

Solving (i) and (ii), we get

n1 = 75, n2 = 50

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 41

Find the minimum number of NAND gates required to realise the given boolean expression: F = A'B + CD' + B'C


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 41
Step 1.

Check if the expression is minimized using K-Map. If it isn't minimized minimize it.(If we're trying to find minimum no. of NAND Gates obtain SOP form, else obtain POS form).

In this particular Example, the expression is already minimized.

Step 2.

We try to realize the expression using only AND and NOT by applying Demorgan's Law.

A'B+CD'+B'C

= C(B'+D')+A'B

= C(BD)'+A'B

= ((C(BD)')'.(A'B)')'

So number of NAND Gates = 1 [for (BD)'] + 1 [for (C(BD)')'] + 1 [for (A'B)'] + 1 [for ((C(BD)')'.(A'B)')'] + 1 [for A'] = 5

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 42

An 8085 executes the following instructions:

2710 LXI H, 30A0 H

2713 DAD H

2414 PCHL

All addresses and constants are in Hex. Let PC be the contents of the program counter and HL be the contents of the HL register pair just after executing PCHL.

Which of the following statements is correct?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 42
2710H LXI H, 30A0H ; Load 16 bit data 30A0 in HL pair

2713H DAD H ; 6140H → HL

2714H PCHL ; Copy the contents 6140H of HL in PC

Thus, after execution, the above instruction contents of PC and HL are same, that is 6140H.

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 43

A new type of synchronous flip-flop has the following characteristic table.

Choose the correct corresponding excitation table entry for Q(t) = 0 and Q(t+1) = 1. Here * denotes don’t care.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 43
Now we will draw extended characteristic table of the flip – flop

Now we will see X and Y in table corresponding to Q(t) = 0 and Q(t+1) = 1 (marked in yellow)

So there are two set of values X = 0, Y = 0 and X = 0 ,Y = 1

So the final answer will be X = 0 , Y = * .

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 44

The transfer function of a first-order controller is given as Gc (s) = K(s +a)/s + b, where K, a and b are positive real numbers.

The condition for this controller to act as a phase lead compensator is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 44
The pole zero plot of lead compensator is shown below

By comparing it to the given problem, we get

1/τ = a

1/ατ = b

So, a/b = α

For lead compensator,

α < />

a/b < />

a < />

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 45

A car radio antenna 1m long operates in the AM frequency of 1.5 MHz. How much current is required to transmit 4W of power in Ampere?


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 45

This is a monopole antenna

λ = c/f = 3×108/1.5×106 = 200

l<<λ ,hence it is a Hertizian monopole.

= 810.54

I0 = 28.47A

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 46

Consider a causal LTI system described by the following differential equation.

The correct block diagram representation of the system is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 46
Option (2) is the correct representation of the system.

Differentiating, we get

4x(t) - 2y(t) = dy(t)/dt

Or dy(t)/dt + 2y(t) = 4x(t)

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 47

In the given figure

region y < 0="" consist="" of="" a="" perfect="" conductor="" while="" region="" y="" /> 0 is a dielectric medium. If surface charge on conductor is 2nC/m2, then value of E at B is ______ ay (v/m)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 47
Point B (–4, 1, 5) is in the dielectric medium since y = 1 > 0 at B.

Dn = ρs = 2 nC/m2

Hence, D = 2ay nC/m2

and

E = D/ε0εr = 2 x 10-9 x 36π/2 x 109ay

= 36πay

= 113.1ayV/m

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 48

Consider the following equations

Where p and q are constants. V(x, y) that satisfies the above equations is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 48
By total differential of V, we have

Since ∂v/∂x = px2 + y2 + 2xy & ∂v/∂x = x2 + qy2 + 2xy.

Le M = Px2 + y2 + 2xy & N = x2 + qy2 + 2xy

⇒ ∂M/∂y = 2y + 2x; ⇒ ∂N/∂x = 2x + 2y

∴ ∂M/∂y = ∂N/∂x

∴ Eq(1) is exact.

∴ The solution of eq(1)

[Treat 'y' terms as constants] [Integrate w.r.t 'y' only for 'y' terms]

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 49

If transmitted power,ST is 200m;

WL is 90 dB, n is 10-15 W/Hz and Pe is 10-4

Then maximum allowable bit rate using non coherent FSK will be

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 49
sT/L = SR = Ebrb = ηVbrb

rb = sT/ηLVb = 2 x 105/Vb

Also

Bit rate ≤ 2 x 105/17 = 11.8 bps

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 50

The input-output characteristics of a Schmitt trigger is shown in the figure below. The noise margin (in V) of the Schmitt trigger is ____.(Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 50

Noise margin is tolerable voltage range for which we interpret the inverter output.

NM = VO - VI

NM = 3 V − 1 V

NM = 2 V

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 51

Consider the signal

The Fourier transform of signal x(t) is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 51
Expressing a signal y(t)

x(t) = ∫t−∞y(t)dt

So y(t) F.T .⟷ 2sin⁡(ω/2)/ω

Using intergration property

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 52

In the circuit given below if V1 = 4 V and V2 = 6 V, then what is the value of Vo? (in Volts)

(Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 52

We choose the super node which includes source V1 as shown.

Writing KCL for super node,

V0 + 4 + 4 V0 – 8 + 2 V0 – 12 + V0 = 0

8 V0 = 16

V0 = 2 volt

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 53

Consider the following statements:

  1. The transfer function of a system is defined as the pseudo-static relationship between the input and output variable.

  2. We can compute the frequency response of an interconnected system of components with the help of Transfer functions.

  3. While using the Laplace transform on a transfer function it is possible to get information related to the qualitative behaviour of the response by solving for the dynamic response of the system

  4. A transfer function is defined only for LTI systems when all initial conditions are assumed to be zero.

Which of the following are true?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 53
All are true except c as it doesn’t solve for dynamic response.
*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 54

What is the size (in bits) of the ROM required for 3 bit binary multiplier? (Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 54
Size of ROM = 22n × 2n bits

n = 3

Size of ROM = 22×3 × 2(3) bits

Size of ROM = 26 × 6 bits

Size of ROM = 64 × 6 bits

Size of ROM = 384 bits

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 55

Parameter in the base region of an npn bipolar transistor are as follows:-

Dn = 20cm2/s

nBO = 104cm− 3, XB = 1μm and ABE = 104cm2

If VBE = 0.5V then the collector current IC in μAis:-


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 55

IC = 3.2 x 10-14e(0.5/0.0259)

= 7.75μA

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 56

The waveform of the current drawn by a semi-converter from a sinusoidal AC voltage source is shown in the figure. If I0 = 20 A, the rms value of fundamental component of the current is ___________ A (Answer up to two decimal places)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 56

Fourier series representation of io(t) is

RMS value of nth harmonic is n =

RMS value of fundamental is

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 57

Feedback control system is given by, G(s) = 1/(s+9) and H(s) = 1/(s+5) How many poles are in right half of the s-plane?


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 57
Characteristic equation is given by

1+G(s)H(s) = 0

≫1 + (1/s+9)(1/s+5) = 0

1 + 1/(s2+9s+5s+45) = 0

≫ s2 + 9s + 5s + 45 + 1 = 0

≫ s2 + 14s + 46 = 0

Now, Routh’ array of this equation is,

As, all the elements in the first column is of same sign so there are no poles in the right half of the s-plane.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 58

Consider a discrete memoryless source having a set of 5 letters each occurring with equal probability. If three letters are encoded at a time into a binary sequence, then what will be the efficiency of the binary code? (Answer up to three decimal place)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 58
The source entropy is

In case of encoding three letters at a time, we have 53 = 125 possible sequences.

We need [log2 125] + 1 = [6.96] + 1 = 6 + 1 = 7 bits/3 letter

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 59

For the network shown below, find the transfer function of the system

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 59

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 60

In a lossy medium(εr = 8, αr = 0.5, σ = 0.01S/m), a plane wave is travelling in + az direction that has the electric field intensity E = 0.5 cos(109πt + π/3 ) ax at z = 0.

What will be the distance (in mm) travelled by the wave to have a phase shift of 10°? (Answer up to two decimal places)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 60
From the field intensity, we get

w = 109π

And it is given that, αr = 0.5, σ = 0.01 S/m, εr = 8.

So, the phase constant,

= 20.95

Let the distance travelled by the wave be z to have a phase shift of 10o.

So, βz = 10o = 10π/180 rad

z = π/18 x(20.95) = 8.33 mm

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 61

A stable real linear time-invariant system with single pole at p, has a transfer function

H(s) = (s2+100)/s−p with a dc gain of 5. The smallest positive frequency, in rad/s, at unity gain is closest to:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 61
H(s) = (s2+100)/s−p

D.C. gain = 5

D.C. gain = H{s}|s = 0

= (s2+100)/s−p|s=0 = 5

100/−p = 5 ⇒ p= −20

H(s) = (s2+100)/(s+20)

For frequency domain s = jω

Frequency at unity gain

(100 − w2)2 = (400 + w2)

Solving this equation

ω=8.84 rad/sec and 11.08 radisec.

Smallest positive frequency = 8.84 rad/sec.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 62

In the circuit below, two series connected capacitors of 3 μF and 6 μF having initial voltage of 4 V and 6 V are connected across a black box at t = 0.

For t ≥ 0, it is given that i(t) = 20e−t μA.

The total energy delivered to the black box will be ___μJ

(Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 62

i(t) = 20e-t μA, v(0) = 10V

Power, p(t) = v(t) i(t)

= (10e-t)(20e-t) x 10-6 = (200 x 10-6)e-2t

Energy delivered to the black box in the interval 0 < t="" />< />

= 100 x 10-6 = 100 μJ

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 63

Consider the two port network given below

For the given network admittance parameter Y12 is ______ mƱ.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 63

Applying KVL

25I1 + 100(I2 - I1) = 0

100I2 = 75I1

V2 = 100(I2 - I1)

V2 = 100I2 - 100I1

V2 = 75I1 - 100I1

V2 = 25I1

I1/V2 = -1/25 = 140 mho

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 64

Let u(x, t) be the bounded solution of ∂u/∂t - ∂2u/∂x2 with u(x,0) = (e2x - 1)/(e2x + 1)

What will be the value of u(0, t)?

(Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 64

We have, ∂u/∂t - ∂2u/∂x2 = 0 ……..(i)

Let u = X(x)T(t), this implies that ∂u/∂t = XT' and ∂2u/∂x2 = X''T

Putting in (i), we get

XT' = X"T

or X”/X = T’/T = P2 (say)

Thus X" - p2 X = 0 & T' − P2T = 0

A.E. D12 - P2 = 0 and D2 – P2 = 0

D1 = ±P and D2 = P2

u(x,0) = (c1ePx + c2e-Px)c3

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 65

The infinite series

1 + x + x2/2! + x3/3! + x4/4! +… corresponds to

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 65
Letf(x) = ex ⇒ f(0) = e0 = 1

f′(x) = ex ⇒ f′(0) = e0 = 1

f′′(x) = ex ⇒ f′′(0) = e0 = 1

f′′(x) = ex ⇒ f′′′(0) = e0 =1

………………………………

fn(x) = ex ⇒ fn(0) = e0 = 1

Now, by Maclaurin’s series

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