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GATE Mock Test Electronics Engineering (ECE)- 1 - Electronics and Communication Engineering (ECE) MCQ


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65 Questions MCQ Test GATE ECE (Electronics) 2024 Mock Test Series - GATE Mock Test Electronics Engineering (ECE)- 1

GATE Mock Test Electronics Engineering (ECE)- 1 for Electronics and Communication Engineering (ECE) 2024 is part of GATE ECE (Electronics) 2024 Mock Test Series preparation. The GATE Mock Test Electronics Engineering (ECE)- 1 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The GATE Mock Test Electronics Engineering (ECE)- 1 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)- 1 below.
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GATE Mock Test Electronics Engineering (ECE)- 1 - Question 1

A 300-meter-long train passes a 450-meter-long platform in 5 sec. If a man is walking at a speed of 4 m/sec along the track and the train is 100 m away from him, how much time will it take to reach the man?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 1
The train can cover (300 + 450) m distance in 5 sec.

The speed of the train = 150 m/sec

Relative speed of the man and the train is 154 m/sec or 146 m/sec

To cover the distance of 100 m, in either of the case, it will take less than 1 second.

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 2

Directions: Select the most appropriate word/phrase among the choices to complete the sentence.

You can make your quixotic experiments with someone else, I do not wish to be your ________.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 2
'Guinea pig' is term used to refer any person or thing used in an experiment. As first part of sentence refers the experiments as quixotic (impractical or daring), so perfect word suited to this blank is 'guinea pig', which is used in labs for testing on biological experiments.
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 3

Which of the following word is opposite to the word “connivance”?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 3
Connivance- willingness to allow or be secretly involved in an immoral or illegal act

Conspiracy- a secret plan by a group to do something unlawful or harmful.

Sufferance - capacity to endure pain, hardship, etc.; endurance

Complot - a plot involving several participants; conspiracy

Intolerance - unwillingness or refusal to tolerate or respect persons of a different social group, especially members of a minority group

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 4

If prices reduce by 20% and sales increase by 15%, what is the net effect on gross receipts?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 4
Let original price = p, and original sales = s. Therefore, original gross receipts = ps.

Let new price = 0.80p, and new sales = 1.15s. Therefore, new gross receipts = 0.92ps.

Gross receipts are only 92% of what they were. Hence, gross receipts decrease by 8%.

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 5

Direction: Each statement has a blank followed by four options. Select the most appropriate word for the blank.

Guru was always able to maintain a _______ face when he said something silly, and that contrast made everyone laugh.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 5
The statement means to state that Guru had an ability to maintain a serious face while saying something silly. The biggest hint in the question is the word ‘contrast’.
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 6

In a town, 48% people are educated, 51% people are young and 60% are servicemen. 24% are educated and young, 25% are young and servicemen, 27% are educated and servicemen and 5% have all the qualities. If the total number of persons in this town is 300, what is the ratio of those who have exactly two characteristics and those who have only one characteristic?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 6

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 7

In the following question, two/three statements are given followed by four conclusions. You have to consider the statements to be true even if they seem to be at variance from commonly known facts. You have to decide which of the given conclusions, if any, follow from the given statements.

Statements:

  1. Some cats are owls.

  2. Some owls are elephant.

Conclusion:

  1. Some cats are elephant.

  2. All owls are elephant.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 7
Here two possibilities emerge based on statement I and II.

Conclusion I:- It cannot be said with certainty that some cats are elephants.

Conclusion II: - It cannot be said with certainty that all owls are elephants.

Hence, neither conclusion I nor II follows.

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 8

Directions: Read the given context and answer the question that follows.

Most Reality TV shows centre on two common motivators: fame and money. The shows transform waitresses, hairdressers, investment bankers, counsellors and teachers, to name a few, from obscure figures to household names. A lucky few successfully parlay their fifteen minutes of fame into celebrity. The luckiest stars of Reality TV also reap huge financial rewards for acts, including eating large insects, marrying someone they barely know, and revealing their innermost thoughts to millions of people.

Which of the following options best supports the above paragraph?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 8
The correct answer is the first option as it is stated in the passage that most Reality TV shows centre on two common motivators: fame and money.

Choices 2 and 4 are not supported by the passage because passage is about Reality TV stars and not Reality TV.

Choice 3 is incorrect because the paragraph states that some Reality TV stars manage to parlay their fifteen minutes of fame into celebrity.

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 9

The sum of five consecutive integer is a and the sum of next five consecutive integer is b. Then, (b-a)/100 is equal to

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 9
Let the five consecutive integers be x,(x + 1),(x + 2),(x + 3) and (x + 4) respectively.

Then, x + x + 1 + x + 2 + x + 3 + x + 4 = a

5x + 10 = a

a = 5x + 10

Let the next five consecutive integers be (x + 5),(x + 6),(x + 7),(x + 8) and (x + 9) respectively.

Then, x + 5 + x + 6 + x + 7 + x + 8 + x + 9 = b

5x + 35 = b

b = 5x + 35

On subtracting equation (i) from equation (ii). Then, we get, b − a = (5x + 35)−(5x + 10)

(b − a)/100 = (5x + 35)−(5x + 10)/100

(b − a)/100 = 25/100

(b - a)/100 = 1/4

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 10

Directions: Insert the missing number.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 10

Middle number = (Difference between numbers of left column) × (Sum of numbers of right column)

i.e. 80 = (15 - 5) × (2 + 6) = (10) × (8) and 65 = (9 - 4) × (7 + 6) = (5) × (13)

∴ ? = (13 - 11) × (16 + 8) = 2 × 24 = 48

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 11

f(x) = x, g(x) = 1/x ; by using Cauchy mean value theorem mean value for the function in [a,b] will be

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 11
According to Cauchy mean value theorem, the mean value 'c' is given by

Putting x=c we have

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 12

The electric field of a uniform plane electromagnetic wave is:

The polarization of the wave is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 12

For finding polarization, put Z = 0 so that polarization can be seen in my plane. Let

Ez = cosωt

ω = 2π x 107 t

Ey = 4 cos(ω + π/2) = -4 sin ωt

As y direction component is multiplied by j, so it is π/2 shifted. Hence, it is left hand elliptical polarization.

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 13

If A2 - A + I = 0, then the inverse of A is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 13
A2 − A + I = 0

⇒I = A − A ⋅ A

Multiplying the equation with A-

⇒A-1 = I - A

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 14

In the design of a single mode step index optical fibre close to upper cut-off, the single-mode operation is not preserved if

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 14
The cut off frequency is given by f = 2πa sin α/wavelength, where a is the radius of core. For a single mode step index fiber, f must lie between 0 and 2.405. Thus to get f below 2.405, the radius must be doubled and wavelength must be halved.
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 15

If where x > 0, then dy/dx is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 15
If where x > 0, then dy/dx is

So we can write it as X = ey + x

Differentiate w.r. to x after taking logarithm both sides ⇒

∴ dy/dx = (1 - x)/x

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 16

The value of the integral is equal to

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 16

By the defining of L.T., we have

Putting s = 0,

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 17

For each of the positive edge-triggered J-K flip flop used in the following figure, the propagation delay is ΔT

Which of the following waveforms correctly represents the output at Q1?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 17
Since the input to both JK flip-flop is 11, the output will change every time with clock pulse. The input to clock is

The output Q0 of first FF occurs after time ΔT and it is as shown below

The output Q1 of second FF occurs after time ΔT when it gets input (i.e. after ΔT from t1) and it is as shown below

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 18

Let Gcc(s) be a PD controller. If f(t) = sin 2t, the amplitude of the frequency component of y(t) due to f(t) is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 18

If Gcc(jω) = 1 - jω, then Gcc(s) = 1 - s.

It can be understood as a PD controller.

Gcc(s) = 1 - s = KP + KDs

KP = 1 and KD = -1G(s) = (s2 + s + 1 - s) ÷ (s2 + s + 1)

G(jω) = (1 - ω2) ÷ [(1 - ω2) + jω]

If f(t) = 1 sin 2t, then by comparing it with frequency equation, we get

ω = 2 and modulus of G(j2) =

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 19

In an ideal silicon junction diode:

τn0 = τp0 = 10−7 sec

Dn = 25cm2/sec

Dp = 10cm2/sec

Then the ratia of Nd/N2 so that 80% of current in depletion region is carried by electrons is


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 19
As the given condition is

80% of current in depletion region is carried by electrons then,

Jn/(Jp+Jn)=0.8−−−−(1)

On putting these values in (1), we get

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 20

Consider a compensator defined as

Then, the maximum phase shift of the compensator is ____.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 20
Maximum phase shift

Gc(s) is a lead compensator

T = 0.3, βT = 0.9

Β = 1/α = 3

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 21

In the given circuit, the values of V1 and V2 respectively are

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 21
Perfarming Nadal analysis on the given circuit

(V2−V1)/4 = V1/4 + V1/4 + 2I

(V2 − V1)/4 = 5

since,

V1/4 = I,

I + I + 2I = 4I =5

V2 = 20 + V1

V1 = 4I = 5V

Hence, V2 = 20 + 5 = 25 V

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 22

A communication channel with AWGN operating at a signal to noise ratio of SNR >> 1 and bandwidth B has capacity C1. If the SNR is doubled keeping it constant, then the resulting capacity C2 is given by

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 22
We have C1 = Blog2(1 + S/N)

≈Blog2(S/N)

If we double the S/N ratio then

C2 ≈ Blog2(2S/N)

≈ Blog22 + Blog2(S/N)

≈ B + C1

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 23

A filter has filter response given by h(t) = (sin(0.25πt))/πt. An input signal x(t) is applied at the input. Determine which of the signal component would be visible at the output.

x(t) = cos(0.125πt) – 2sin(0.125πt) + 0.125sin(0.5πt)

  1. cos(0.125πt)

  2. sin(0.125πt)

  3. sin(0.5πt)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 23
Here, h(t) is a sinc() function whose transform would give a rect function ranging from [-0.25π, 0.25π].

Transform of x(t) would give impulses centered at -0.125π, 0.125π, -0.5π and 0.5π.

Since the filter window is from -0.25π to 0.25π, it would allow frequencies lying between these two extremes. i.e. only -0.125π and 0.125π components would pass whereas -0.5π and 0.5π would be blocked.

Thus we will get sin(0.125πt) and cos(0.125πt) components at the output.

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 24

The photo resist that is insoluble initially and becomes soluble after exposure to UV light is called

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 24
A Positive Photoresist is relatively developer insoluble until exposed to light at which point the exposed portion exhibits increased developer solubility. A positive replica of the mask pattern is obtained on a layer of positive PR upon exposure to UV light.
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 25

Refer the following figure to find the value of gain of the feedback circuit (feedback factor) and closed loop voltage gain respectively if R1=1kΩ and Rf = 10kΩ.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 25

This is a non-inverting feedback Amplifier. Given: R1 = 1kΩ and Rf = 10kΩ

Gain af the feedback circuit

B = 1/11

B = 0.09

Voltage Gain

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 26

Choose the correct match for input resistance of various amplifier configurations shown below.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 26
For the different combinations the table is as follows

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 27

A sailor goes 12 km downstream in 48 minutes and returns in 1 hour 20 minutes. The speed of the sailor in still water is:


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 27
Let the speed of sailor in still water be x kmph.

Speed of current = y kmph

∴ x + y =

=12/48 × 60 = 15 kmph

and,x − y =

=12/80 × 60 =9 kmph

Adding these equations,

2x = 15 + 9 = 24

x = 24/2 = 12 kmph

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 28

A proportional plus derivative controller

  1. Has high sensitivity.
  2. Increases the stability of the system.
  3. Improves the steady-state accuracy.

Which of the above statements are correct?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 28

The additive combination of proportional & Derivative control is known as P-D control.

The overall transfer function for a PD controller is given by:

Gc(s) =U(s)/E(s) = Kp+ sKD

PD controller is nothing but a differentiator (or) a High Pass Filter.

The frequency of noise is very high.

So this high pass filter will allow noise into the system which results in noise amplification.

Effects of Proportional Derivative (PD) controllers:

  • Decreases the type of the system by one
  • Reduces the rise time and settling time
  • It has high sensitivity.
  • Rise time and settling time decreases and Bandwidth increases
  • The speed of response is increased i.e. the transient response is improved
  • Improves gain margin, phase margin, and resonant peak
  • Increases the input noise
  • Improves the stability
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 29

A rectangular pulse of duration T is applied to a matched filter, output obtained out of this system would be most likely

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 29
We know that matched filter for a signal s(t) is given as

h(t) = s(T - t)

Thus, we would obtain convolution of rect(t), input, and rect(T-t), response of filter, at the output. This implies output would be convolution of two rectangular pulses of duration T, this would result into triangular pulse of duration 2T.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 30

If C is any closed path and U is a scalar function, then the value of contour integral ∮c(∇U).dl is (Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 30
According to the Stoke's theorem,

LA .dl = ∫(∇ x A).ds

So, ∮c(∇U).dl = ∫[∇ x (∇U)].ds

Since, ∇ x (∇U) = 0

(curl of the gradient of a scalar field is always zero)

So, the contour integral is zero.

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 31

The electrical path of a lossless transmission line of length 50 cm which is operated at 25 MHz in air is ______

(Given, L = 10 μH/m, C = 40 pF/m)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 31
To find electrical path length

LC = 10 × 10−6 × 40 × 10−12 = 4 × 10 − 16

∴ Electrical path length = 1/λ = 0.52 =¼

= 0.25

= 0.25 x 2 πrad = π/2 rad

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 32

If two semiconductor materials have exactly the same properties except that, material A has a band gap of 1.0 eV and material B has a band gap of 5.0 eV then, the ratio of the intrinsic concentration of material A to that of B will be ___ × 1033

(Answer up to two decimal places)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 32

ni(A)/ni(B) = 2.55 x 1033

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 33

Which one of the following is the correct perfect square (in base 10)?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 33
(29)11 = (?)10

(29)11 = 2 x 111 + 9 x 110

= 31

(17)7 = (?)10

(17)7 = 1 x 71 + 7 x 70

= 14

(23)3 = (?)10

(23)3 = 2 x 31 + 3 x 30

= 9

(46)6 = (?)10

(46)6 = 4 x 61 + 6 x 60

= 30

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 34

Two half-wave dipole antennas placed as shown in the figure are excited with sinusoidally varying currents of frequency 3 MHz and phase shift of π/2 between them (the element at the origin leads in phase). If the maximum radiated E-field at the point P in the x-y plane occurs at an azimuthal angle of 60°, the distance d (in metres) between the antennas is (Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 34
For maximum electric field, we have

ψ = βd cos θ + α = 0

where

θ = Azimuthal angle = 60o

α = Phase shift = -π/2

Substituting these value in equation (1), we get

= 50 m

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 35

A circuit is designed with 2- JK FF’s. If the output QAQB=10 at starting, what will be the output QAQB after 13th clock pulse will be

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 35

After 13th clock pulse the output will be 11

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 36

A transfer function G(s), with the degree of its numerator polynomial zero and the degree of its denominator polynomial as two, has a Nyquist plot shown in the figure. The transfer function represents

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 36
G(s) = 1/s(s - 1)

Phase Φ = -270o + tan-1(ω)

ω = 0 M = ∞, Φ = -270o

ω = ∞ M = 0 Φ = -270o + 90o = -180o

∴ our assumption of G(s) is true

The system is type-1 and is unstable.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 37

What will be the output Y, if the input number B3B2B1B0 = 0101 as shown in the below digital circuit?


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 37

Addend will be = 1010

S3S2S1S0 = 1010 + A2A2A1A0 + C11

= 1010 + 1001

= 0011

SE1SE0 = 00

and

E = 1

Y = 1

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 38

The diagonal clipping in Amplitude Demodulation (using envelope detector) can be avoided if RC time-constant of the envelope detector satisfies which of the following conditions?

(Here, W is message bandwidth and ωc is carrier frequency both in rad/sec)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 38
The diagonal clipping in AM using envelop detector can be avoided if

1/ ωc < rc="" />< />

We can say that RC depends on W, thus

RC < />

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 39

The distance between two spacecrafts is 3000 km. Both of them has a paraboidal reflector antenna of 0.85m diameter which operates at a frequency of 2 GHz, with an aperture efficiency of 64 %. If the spacecraft A’s receiver requires 1 pW for a 20 dB signal-to-noise ratio then what transmitted power (in Watts)is required on the spacecraft B to achieve this signal-to-noise ratio ?


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 39
R = 3000km = 3 × 106m

Diameter of antenna,

D = 0.85m λ = c/f = (3 × 1013)/(2 × 109) = 0.15m

Power required far receiving antenna

A, PA = 1pW = 10−12W

For SNR = 20dB = 100

Gain of the antenna B,GB =

Power density from B,W =

Where, PB is the power transmitted by B¯.

Aperture efficiency, eA = eB =64%

Operating frequency, f = 2GHz From (1) and (2)

Power received at the transmitting antenna,

From eq (4)

From E.q. (3) and E.q.(5)

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 40

Consider the following statements:

  1. A flip-flop is used to store 1 bit of information.

  2. Race around condition occurs in a J-K flip flop when both of its inputs are 1.

  3. Master slave configuration is used in flip-flops to store 2 bits of information.

  4. A transparent latch consists of D-type flip-flops.

Which of the above statements are correct?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 40
According to the definition of flip-flop, it consists of a latch used to store 1-bit of information. Race around condition occurs in J-K flip-flop, if the output continuously toggles. D-type flip-flop's output is equal to input.
*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 41

A Silicon PN junction diode under reverse bias has depletion region of with 10 µm. The relative permittivity of Silicon, εr = 11.7 and the permittivity of free space ε0 = 8.85 × 10-12 F/m. the depletion capacitance of the diode per square meter is


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

C = ε0εrA/d

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 42

For the linear network shown below, V-I characteristics are also given in the figure. The value of Norton equivalent current and resistance, respectively are

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 42
The circuit with Norton equivalent is shown below

So, IN + I = V/RN

(General form) I = V/RN - IN

From the given graph, the equation of line is

I = 2V − 6

Comparing with general form,

1/RN = 2 or RN = 0.5Ω

IN = 6 A

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 43

A message signal m(t) has been sent by imposing it on a carrier frequency c(t). There are two Modulations schemes under consideration, AM and FM. Peak Frequency deviation for FM is set to be 2 times that of bandwidth used in AM, whereas magnitude for spectral components at 10 MHz ± 4 kHz are same for both schemes. Given below are the signals. The modulations indices for AM and FM respectively under given constraints is m(t) = cos⁡[(8π ∗ 103)t],c (t) = 5 cos⁡[(2π ∗ 106)t]

[Values of Bessel function if required: J1(2) = 0.577, J1(4) = 0.066, J1(8) = 0.235, J1(16) = 0.094]

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 43
Here we have AM and FM schemes. BW for AM is 2fm = 8 kHz

For modulation index of FM,

β = Δf/fm, where Δf is 2 ∗ BWAM

Then, β=4

Now we have relation for spectral components that both are equal at 4 kHz, Component for AM is μAd2 at ± 4 kHz and component for FM is given by AcJ1(4).

μAc/2 = AJ1 (4)

Then,

μ = 0.13

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 44

For the arrangement shown below, the frequency spectrum of output y(t) will be

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 44

The output of AM modulator is

x(t) = (10 sin 200 πt) + 4 cos (60 πt) sin (200 πt)

= 10 {1+ 0.4 cos 60 πt} sin 200 πt

α = 0.4, AC = 10

x(t)sin 20 πt

10 sin2(200 πt) + 4 cos (60 πt) sin2 (200 πt)

= 5 – 5 cos (400 πt) + 2 cos 60 πt – {cos(340 πt) + cos(460 πt)}

The output of bandpass filter (80Hz – 180Hz) is

cos (340 πt) = cos(2π .170t)

fc = 170 Hz

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 45

Consider a signal x(t) = 8cos(4πt + π/4)

Its Autocorrelation and PSD would be

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 45
We have the function x(t) = 8 ∗ cos⁡(4πt + π/4)

since, x(t) is a power signal, its calculations would be simpler in f domain instead of ω domain. Autocorrelation in time domain is convolution on a signal with its time reversed signal, i.e. RX(τ) = Convolution [x(τ), x(−τ)]

So, in frequency domain, SX(f) = X(f) ∗ X(−f)

Thus, SX(f) = 16δ(f − 2) + 16δ(f + 2)

Then Inverse Fourier transform to get Autocorrelation function,

RX(τ) = 32 cos⁡(4πτ)

And its PSD is given as

(Here the signal is centered at 0Hz, so Bandwidth will be f instead of 2f as negative frequencies don't exist

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 46

Consider the system shown in the figure below. The transfer function of the system will be

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 46

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 47

Two RC coupled amplifiers are connected to form a 2-stage amplier. If the lower and upper cutoff frequencies of each individual amplifier respectively are 100 HZ and 20 kHZ, What these frequencies are for the 2-stage amplifier?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 47
For multi-stage amplifier consisting of n identical amplifying stages, the lower cut off f1(multi) is expressed as,

Where, f1 is lower cut-off of individual stages And the upper cutoff

f2( multi ) is expressed as,

Where f2 is upper cut-off of individual stages Therefore, the lower cut-off of 2-stages amplifier

Similarly, the upper cutoff of the 2 -stage amplifier is

thus the band width gets reduced in a multi-stage amplifier.

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 48

Consider a two-port network as shown in the figure below.

The ABCD-parameter matrix for this network is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 48
The given network is equivalent to the cascade connection of two-port network shown as-

For the above network, T-equivalent circuit is shown as-

z12 = z21 = 1

z11 – z12 = 1

z11 = 1 + z12 = 2

z22 = 1 + z12 = 2

For the cascade connection,

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 49

Find the value of integral is.....

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 49
As we know Inverse DTFT is given by

x(n) = (0.4)n + 3u(n + 3)

x(0) = (0.4)3

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 50

The open loop transfer function of a unity feedback control system is:

The phase margin for the system is ________ (in degrees). (Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 50

⇒ 3ω4 + 3ω2 - 1 = 0 ⇒ ω2 = 0.263

PM = 180o - 90o - 27.19o = 63o

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 51

Consider the circuit given below

The Thevenin equivalents of the circuit VTh, RTh between the terminals A and B are respectively

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 51
Redrawing the circuit

Applying nodal analysis

V1/R1 + V2/R2 = I0……..(i)

V2 − V1 = αIa…………..(ii)

V1/R1 = Ia

By equation

(i), (ii) and (iii)

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 52

A carrier signal AC cosωCt is frequency modulated using a message signal 3 cos 103πt. If the signal-to-noise ratio at input and output of the receiver are found to be 20 dB and 40 dB respectively, then what will be the sensitivity (in kHz/V) of frequency modulator? (Answer up to two decimal places)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 52
Si/Ni = 20 dB = 20 =10 log10 2

Si/Ni = 100, So/No = 40 dB ⇒ So/No

Kf = (8.16 x 0.5)/3 = 1.36 kHzN

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 53

In the figure shown, all elements used are ideal. For time t < 0.="" />1 remained closed and S2 open. At t = 0, S1 is opened and S2 is closed. If the voltage Vc2 across the capacitor C2 at t = 0 is zero, the voltage across the capacitor combination at t = 0+ will be ______Volts.


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 53
For t < 0,="" s1="" in="" closed="" and="" s2="" />

Qc1 = C1V = 3C

Qc2 = 0

At t = 0, S1 is opened and S2 is closed.

Now, let Q’c1 and Q’c2 is the change stored after redistribution

then, Q’c1 + Q’c2 = Qc1 + Qc2 = 3C …A.

Q’c1/c1 = Q’c2/c2

[Equal potential across C1 and C2 ] ⇒Q’c1/1=Q’c2/2

By equation A. and equation B. Q’c1 = 1C as Q’c2 = 2C

and voltage across capacitor combination is,

=Q’c1/C1 = 1 volt

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 54

If the transistor shown below is operating in the saturation region, the base current IB is greater than ___ μA. (Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 54
lesat = 10/5kΩ = 2 mA

IB sat 2 mA/100 = 20μA, IB > IB sat, IB > 20 μA

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 55

A silicon bar is doped with 1017 arsenic atoms /cm3. The equilibrium hole concentration at 300k and the location of the Fermi level (EF) of sample relative to intrinsic Fermi level (EFi) are

(Given, ni = 1.5 × 1010/cm3 )

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 55
As silicon bar is doped with 1017 arsenic atoms /cm3 so,

ND = 1017/cm3

therefore, n = 1017/cm3

Given, ni = 1.5 × 1010/cm3

Now, hole concentration, p = n2f/n

= 2.25 × 103/cm3

For n -type material,

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 56

X is a binary memory-less source with probability, p(x = 0) = 0.3. If the source is transmitted over a BSC with cross-over probability 0.1, then the capacity of the channel is

(Answer up to two decimal places)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 56
Capacity of the channel is,

C = 1 - H(0.1)

= 1 - 0.469 = 0.53

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 57

The parameter of the transistor in figure. VTN =1.2V , Kn= 0.5 mA/V2 and λ = 0.

The voltage VDS is ________(V)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 57
Given VTN = 1.2 V

Kn = 0.5 mA/v2

λ = 0

VDS = ?

From figure ID = 50μA

ID = Kn[VGs - Vr]2 [1 + λVDS]

⇒ 50 x 10-6 = 0.5 x 10-3[Vgs - 1.2]2

Therefore, VGS = 1.52 V

VDS = VGS = 1.52 V

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 58

If the Laurent series expansion of f(z) = is valid in the region | z | > 2, then what will be the absolute value of coefficient of 1/z2 ? (Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 58

Thus, the coefficient of 1/z2 = -1

Hence absolute value is 1.

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 59

The open loop transfer function of a system is G(s)H(s) = the root locus wil intersect the imaginary axis at

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 59
CE = s3 + 2s2 + s(2 + k) + 4k = 0

For finding critically damping frequency,

s1 row should be zero

4 – 2 k = 0

k = 2

For finding the interaction with imaginary axis is

2s2 + 4k = 0

Put s = jω , 2(jω)2 + 4k = 0

2 ω2 = 4 × 2

ω = 2 rad/sec.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 60

Consider a signal x(t) defined as

x(t) = sinc (100 πt) + 2 sinc2 (150 πt)

What will be the Nyquist sampling rate (in Hz) for this signal?

(Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 60
x(t) = sinc (100 πt)+ 2 sinc2(150 πt)

The sampling rate of sinc (100 πt)is

fs1 = 2 fm1 = 2 x 100 = 200 Hz

The sampling rate of sinc2 (150 πt) is

fs2 = 2 fm2 = 2 x 150 = 300 Hz

When both the signals are added, maximum frequency component will be 150 Hz. So, the sampling rate is 300 Hz.

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 61

The value of line integral along a path joining origin to the point (4,4,2)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 61
∫ F ⋅ dr

∇ × F = 0,

F is irratational, F is conservative

x = 2 x0.5y1.5

y = 2x1.5y0.5

z = 1

∅ = 2/1.5x1.5y1.5+z+c

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 62

The unit-step response of a unity feedback system with open loop transfer function G(s) = K/((s + 1)(s + 2)) is shown in the figure below. What is the value of K? (Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 62
From the figure we can see that steady state error for given system is

Ess = 1 - 0.75 = 0.25

Steady state error for unity feed back system is given by

So, ess = 2/2 + K = 0.25

2 = 0.5 + 0.25 K

K = 1.5/0.25 = 6

GATE Mock Test Electronics Engineering (ECE)- 1 - Question 63

The area of the region bounded by the curves y = |x – 1| and y = 3 - |x| is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 63

Required area

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 64

In a 8-bit SISO register, if 16-bit data 1011 0101 0010 1010 is applied at the input, the simplified form of the expression Z = is

(Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 64

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 1 - Question 65

Given

dy/dx = x2 − y2, y(0)

The value of y(0/1) using Runge-kutta Second order method is (Take h = 0.1)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 1 - Question 65
y’ = x2 - y2

y(0) = 1

y1p = y0 + h.f (xo,yo)

=1 + 0.1(-1) = 0.9

Modified value of y1 is

y1c= y0+ h/2 [f(x0,y0) + f(x1,y1p)]

=1+ 0.1/2[-1 + (0.01 - 0.81)]

=1 - 0.09

= 0.91

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