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GATE Mock Test Electronics Engineering (ECE)- 6 - Electronics and Communication Engineering (ECE) MCQ


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65 Questions MCQ Test GATE ECE (Electronics) Mock Test Series 2025 - GATE Mock Test Electronics Engineering (ECE)- 6

GATE Mock Test Electronics Engineering (ECE)- 6 for Electronics and Communication Engineering (ECE) 2024 is part of GATE ECE (Electronics) Mock Test Series 2025 preparation. The GATE Mock Test Electronics Engineering (ECE)- 6 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The GATE Mock Test Electronics Engineering (ECE)- 6 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)- 6 below.
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GATE Mock Test Electronics Engineering (ECE)- 6 - Question 1

Direction: Study the following pie-chart and tables carefully and answer the questions given below.

Total Number of email received by the organization = 90000

Ratio of Read emails to Unread emails received by the organization

What is the ratio of the number of emails read in January to those unread in the month of April in the organization?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 1
Number of read emails in the month of January = (90000*17/100*8/15)

Number of unread emails in the month of April = (90000* 8/100 * 5/12)

Ratio = 68:25

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

The shape in the figure given below is folded to form a box. Choose from the alternatives (1), (2), (3) and (4), the boxes that are similar to the box formed from the shape below.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 2
The dot will lie opposite to one of the shaded surfaces.

Therefore, figure (2) cannot be formed.

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 3

In the following question, out of the given four alternatives, select the one which is opposite in the meaning of the given word.

Scurrilous

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 3
The meanings of the given words are as follows

Scurrilous— abusive

Coarse— rough or harsh in texture

Sophisticated— developed to a high degree of complexity.

Insolent— showing a rude and arrogant lack of respect

Complimentary— expressing a compliment; praising or approving

Clearly, option D is the antonym of the given word.

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 4

What is the chance that a leap year, selected at random, will contain 53 Sundays?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 4
There are 52 complete weeks in a calendar year i.e. 52 x 7 = 364 days

Number of days in a leap year = 366

∴ Probability of 53 Sundays = 2/7

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 5

Direction: Study the following information carefully and answer the questions given below:

P, Q, R, S, T, U, V and F are sitting around a circle facing the centre. U is third to the right of Q, who is third to the right of F. P is third to the left of F. R is fourth to the left of P. T is third to the right of S. S is not a neighbour of P.

Four of the following five are similar in a certain way based on their positions in the seating arrangement and so form a-group. Which of the following does not belong to that group?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 5

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

In a group, the ratio of the number of boys to the number of girls is P. The ratio of the number of girls to the number of boys is Q, then P + Q is always

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 6
We know that (x + 1/x) is always greater than or equal to 2.

Now, let x be the number of boys and y be the number of girls, then

x/y = P and y/x = Q

So, P + Q = (x/y) + (y/x) ≥ 2

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 7

In a school, 12th class consists of 30% male students of which 30% male students failed in the class. Total 82% students passed in 12th examination out of 900 students. Calculate the total number of female passed students?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 7
Short Trick

Total students = 900

Boys : Girls = 30 % : 70% = 3 : 7

So, exactly Boys : Girls = 270 : 630

Total failed students = 18% of 900 = 162

Failed boys = 30% of 270 = 81

Failed girls = 162 - 81 = 81

Passed female students = 630 - 81 = 549

Basic Method:

⇒ Let us consider total number of students be 100x

⇒ ∴ Males students = 30x , female students = 70x

⇒ According to condition given in the problem, of this 30x male students, 70 % male students pass the exam.

ie. Passed males = 30x × 70/100 = 21x

⇒ Total passed students = 100x × 82/100 = 82x

⇒ ∴ female students passed = 82x – 21x = 61x

⇒ ∴ Girls passed = 61% of total

∴ Answer is 61/100 x 900 = 549

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 8

A cube of side 10 cm is coloured red, with a 2 cm wide green strip along all the sides on all the faces. The cube is cut into 125 smaller cubes of equal sizes. How many cubes have at least two green faces each?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 8
There are 12 cubes in each of the layers 1 and 5 and 4 cubes in each of the layers 2, 3 and 4 which have two faces coloured green. Also, there are 4 cubes in each of the layers 1 and 5 which have three faces coloured green.

3 faces coloured cubes = 8

2 faces coloured cubes = 24 + 12 = 36

Total cubes = 44

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 9

Direction: In the following number series, only one number is wrong. Find out the wrong number.

17 25 34 98 121 339

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 9
The pattern followed here is,

⇒17 + 23 = 17 + 8 = 25

⇒25 + 32 = 25 + 9 = 34

⇒34 + 43 = 34 + 64 = 98

⇒98 + 52 = 98 + 25 = 123

⇒123 + 63 = 123 + 216 = 339

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 10

Directions: In the question given below, a statement is followed by three courses of action labelled (A), (B) and (C). A course of action is a step or administrative decision to be taken for improvement, follow-up or further action in regard to the problem, policy, etc. On the basis of the information given in the statement, you have to assume everything in the statement to be true and then decide which of the suggested courses of action logically follow(s) for pursuing.

Statement: Many political activists have decided to stage demonstrations and block traffic movement in the city during peak hours to protest against the steep rise in prices of essential commodities.

Courses of action:

(A) The Govt. should immediately ban all forms of agitations in the country.

(B) The police authority of the city should deploy additional forces all over the city to help traffic movement in the city.

(C) The state administration should carry out preventive arrests of the known criminals staying in the city.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 10
(A) is not feasible in a democracy. (C) is not followed because the problem is not concerned with criminals. (B) is the only course that authorities can resort to.
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 11

In the following figure, the J and K inputs of all the four Flip-Flops are made high. The frequency of the signal at output Y is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 11
Output of NAND is zero when Q3Q2Q1Q0 have state 1010=(10), Dec. Therefore, the given figure represents mod- 10 up counter. And, frequency of the signal will be

f = 10KHz/10 = 1KHz

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 12

For the function e-x, the linear approximation around x = 2 is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 12
We have f(x) = e-x = e-(x - 2) - 2 = e-(x - 2)e-2

= [1 - (x - 2)]e-2 Neglecting higher powers

= (3 - x)e-2

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 13

Consider the lossless transmission line circuit shown in the figure, The voltage standing wave ratio on 50 Ω line is given by

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 13
Zin at junction is given by

Zin = Zin(λ/4) ||Zin(150Ω)

Zin = Z02/ZL = 225 Ω

VSWR on 50 Ω is 225/50 = 4.5

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 14

The Laplace transform of a function is(S + 1)/S(S + 2) . The initial and final values, respectively, of the function are

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 14
For initial value, finding

For final value, Finding

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 15

Given x(t) = e−tu(t). Find the inverse laplace transform of e−3s X(2s).

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 15

x(t) = e−tu(t)

Taking laplace transfarm

X(s) = 1/s+1

X(2s) = 1/2s+1 = 1/2[1/(s+1/2)]

Taking inverse laplace transfarm

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 16

Under low level injection assumption, the injected minority carrier current for an extrinsic semiconductor is essentially the

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 16
Under low level injection, the minority current for an extrinsic semiconductor is by diffusion as there is concentration gradient due to low level injection.
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 17

Find the differential equation of the system described by the transfer function given as:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 17
We have,

Y(s)/X(s) = s + 3/s(2s + 5)

⇒Y(s) [2s2 + 5s] = X(s)⋅(s + 3)

⇒ 2s2⋅Y(s) + 5⋅Y(s) = s⋅X(s) + 3⋅X(s)

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 18

Choose proper substitutes for X and Y to make the following statement correct.

Tunnel diode and avalanche photo diode are operated in X bias and Y bias, respectively.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 18
Tunnel diode shows the negative characteristics in forward bias. It is used in forward bias. Avalanche photo diode is used in reverse bias.
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 19

The built-in potential of an abrupt p-n junction is 0.75 V. If its junction capacitance (Q) at a reverse bias (VR) of 1.25 V is 5 pF, the value of CJ (in pF) when VR = 7.25 V is________.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 19

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 20

Consider the state equation for a system given below.

Which of the following conditions is true for complete controllability?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 20

Controllability matrix = [B AB]

must not be equal to zero;

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 21

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

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 21
Given q = 1.6×10−19; kT/q = 25mV

μn = 1000cm2/v−s

From Einstein relation, Dnn = kT/q

⇒D = 25mV × 1000cm2/v−s

⇒ 25cm2/s

Diffusion current Density

J = qDndn/dx

= 1.6 x 10-19 x 25 x 1 x 1021

= 4000 A/cm2

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 22

Without any additional circuitry, an 8 : 1 MUX can be used to obtain

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 22
A 2n : 1 MUX can implement all logic functions of (n + 1) variable without any additional circuitry.

Here, n = 3.

Thus, an 8 : 1 MUX can implement all logic functions of 4 variables.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 23

The voltage gain of an amplifier is 100. A negative feedback is applied with β=0.04. The overall gain of the amplifier is:


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 23
Initially the voltage gain of the amplifier is 100 i.e. A=100.

Then a negative feedback is applied to the amplifier where the gain of the feedback loop is given as β = 0.04

Now the overall gain of the negative feedback amplifier is given as, = A/1+Aβ

= 100/[1+(100)(0.04)]

= 100/(1+4)

= 100/5

= 20

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 24

In the circuit shown, the device connected to Y5 can have address in the range

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 24
The simple decoder is shown below

Addressing is done based on the inputs and their respective logic circuit connected to the address decoders.

Calculation:

In the given question the chip select for Y5 is 101.

A10A9A8 are used for that chip select so A10A9A8 = 101.

A7 ⋯ A0 are initially zeroes and at the final address these will be all 1's.

To get the active high output for the NAND gate all the inputs must be 1 always.

The total addressing range is shown below:

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 25

Consider two resistors of 50 kΩ and 100 kΩ at room temperature of 27 OC. These resistors are passing a signal of bandwidth 50 kHz. Let N1 and N2 be the noise voltage generated while operating in series and parallel configuration, respectively. Determine N1 and N2.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 25
Given parameters,

T = 300 K

RS = 150 kΩ

RP = 100/3 kΩ.

B = 50 kHz.

We can find out the noise voltage generated through resistors using following formula,

Vn2 = 4RkTB, where k is Boltzmann constant.

Thus, N1(Series) would be 11.15 µV and N2(Parallel) would be 5.25 µV.

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 26

The transfer function of a phase-lead compensator is given by Gc = , where T > 0. What is the maximum phase shift of the compensator?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 26
The transfer function of given compensator is

Comparing with

The maximum phase shift is

or Φmax = π/6

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 27

The amplitude of a random signal is uniformly distributed between -5 V and 5 V.

If the signal to quantization noise ratio required in uniformly quantizing the signal is 43.5 dB, the step size (in V) of the quantization is approximately

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 27
(S/Nq)0dB = 1.76 + 6.02u

⇒43.5 = 17.6 + 6.024

⇒ n ≈ 7

Δ(Step size) = 2A/2′′

=10/27 = 0.07 Volts

Which is close to 0.0667 Volts

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 28

During transmission over a certain binary communication channel, bit errors occur independently with probability p. The probability of at most one bit in error in a block of n bits is given by

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 28
By Binomial distribution the probability of error is

Pe = nCr Pr (1- P) n - r

Probability of at most one error

= Probability of no error + Probability of one error

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 29

For a telephone line, β= 0.02rad/km. If frequency is 1 kHz. Calculate time taken to travel 3.14 km by the wave in µs


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 29
β = 0.02rad/km,f = 1kHz,d = 3.14km

Wavelength λ = 2π/β = 2π/0.02 = 314.159km

Velocity v = ω/β = 2πf/β = 2π∗1000/0.02 = 3.14 × 105km/sec

Time t = d/v = 3.14/(3.14 × 105) = 105Sec

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 30

The Laplace transform of i(t) is given by I(s) = 2/s(1 + s) . As t ⟶∞, the value of i(t) tends to be (Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 30
From the Final value theorem, we have

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 31

A 1MHz uniform plane wave propagates through fresh water for which σ = 0,μr = 1,εr =1 Calculate wavelength in meters.


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 31

As σ = 0, α will also be 0.

β = ω = √με

but μ = μ0μr and ε = ε0εr

Also Ω = 2π x 106

β = 0.185rad/m

λ = 2π/ β = 2π/0.185

λ = 33.963 m

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 32

The number of product terms in the minimised sum-of-product expression obtained through the following K-map is

(where "d" denotes do not care states)

(Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 32
As shown below, there are 2 terms in the minimised sum of product expression.

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 33

Which of the following coding scheme is most suitable for coding the successive positions in 8-position digital shaft encoder?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 33
In digital shaft encoder only one bit change in successive codes, so Gray Code is preferred for digital shaft encoder.

Hence option B is correct as it signify only 1-bit change between two successive codes.

000→001 →011→010→ 110→111→101→111

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 34

A transmission line is feeding 1 Watt of power to a horn antenna having a gain of 10 dB. The antenna is matched to the transmission line. The total power (in Watt) radiated by the horn antenna into the free space is (Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 34
We have,

10 log G = 10 dB

or G = 10

Now gain G = Prod/Pin

or 10 = Prod/(1 W)

or Prod = 10 Watt

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 35

ln INTEL 8085, suppose the peripheral mapped l/O has address length of M and memory mapped l/O address length of N. Then M+N =_______.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 35
Peripheral mapped l/O has address length

M = 8

Memory mapped l/O has address length

N = 16

M+N =8 +16 = 24

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 36

The feedback configuration and the pole-zero locations of

are shown below. The root locus for negative values of k i.e. for -∞ < k < 0 has breakaway/break-in points and angle of departure at pole P (with respect to the positive real axis) is equal to

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 36

The characteristic equation is

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

or s2 + 2s + 2 + K(s2 - 2s + 2) = 0

or

For break away & break in point differentiating above w.r.t. s we have

Thus (s2 - 2s + 2)(2s + 2) - (s2 + 2s + 2)(2s - 2) = 0

or s = ±√2

Let θd be the angle of departure at pole P, then

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 37

A 70 kHz bandwidth is to accommodate 7 AM broadcast signals simultaneously. The maximum modulation frequency must each station be limited to _____ kHz. Assume that a guard band of 5 kHz is also present.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 37
BW = n × 2fm + (n − 1)guard band

70 = 7 × 2 × fn + 6 × 5

40 = 14fm

fn = 40/14 = 2.857KHz

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 38

2 discrete time systems with impulse responses h1[n] = δ[n - 1] and h2[n] = δ[n - 2] are connected in cascade. The overall impulse response of the cascaded system is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 38
We have h1[n] = δ[n - 1] or H1 [Z] = Z-1

and h2[n] = δ[n - 2] or H2(Z) = Z-2

Response of cascaded system

H(z) = H1(z) · H2(z) = z-1 · z-2 = z-3

or, h[n] = δ[n - 3]

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 39

Consider two real valued signals, x(t) band-limited to [−500 Hz, 500Hz] and y(t) bandlimited to [−1kHz, 1kHz]. For z (t) = x(t). y(t), the Nyquist sampling frequency (in kHz) is __________

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 39
x(t ) is band limited to [−500Hz, 500Hz] y(t )is band limited to [−1000Hz, 1000Hz]

z(t ) = x (t).y(t)

Multiplication in time domain results convolution in frequency domain.

The range of convolution in frequency domain is [−1500Hz, 1500Hz]

So maximum frequency present in z(t) is 1500Hz Nyquist rate is 3000Hz or 3 kHz

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 40

Consider the following statements:

Any element connected in

  1. Series with an ideal current source is redundant.

  2. Parallel with an ideal current source is redundant.

  3. Series with an ideal voltage source is redundant.

  4. Parallel with an ideal voltage source is redundant.

Which of the above statements is/are correct?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 40

⟶ From figure (a), observe that if we keep or remove R1, the load voltage and current are the same. So, R1 is redundant here.

⟶ From figure (b), observe that if we keep or remove R1, VL & lL are the same and hence redundant.

⟶ Note that in figure (a), R1 can't be ∞ and in figure (b), R1 can't be 0, as it will result in violation of Kirchoff law.

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 41

A discrete time signal with input as x[n] has impulse response h[n] = kδ[n] where k is constant. The output y[n] of the system is given as:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 41
For a discrete time signal with input as x[n] has impulse response h[n] =kδ[n], the output y[n] of the system is given as

y[n] = x[n]*h[n]

y[n] = x[n]*kδ[n]

y[n] = k(x*δ)[n]

y[n] = kx[n]

Taking Fourier transform on both sides

Y(ω)=X(ω) x H(ω)

Therefore H(ω)= Y(ω)/X(ω)

This is known as transfer function of the system.

h[n]=F-1[H(ω)]

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 42

A second-order system has the transfer function C(s)/R(s) = 4/(s2 + 4s + 4) with r(t) as the unit-step function. The response c(t) of the system is represented by

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 42
The characteristics equation is s2 + 4s + 4 = 0

Comparing with s2 + 2εωn + ω2n = 0

we get 2εωn = 4 and ω2n = 4

Thus ε = 1

ts = 4/εωn = 4/1 x 2 = 2

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 43

For the given system, determine the overall response Y(s) of the system corresponding to both inputs.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 43
When multiple inputs are present in a linear system, output is computed by treating each input independently. The overall response of the system is the obtained by adding the outputs corresponding to each input.

Assuming X2(s) = 0,

The block diagram can be reduced to,

Reducing the inner feedback loop,

The block diagram is reduced to,

Thus,

YX1(s ) = Output due to X1(s) acting alone

Assuming X2(s) = 0,

The block diagram is reduced to,

Thus, YX2 (s) = Output due to X1(s) acting alone

Hence, the response of the system,

Y(s) = Yx1(s) + Yx2(s)

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 44

Consider the Schmidt trigger circuit shown below. A triangular wave which goes from -12 V to 12 V is applied to the inverting input of OPMAP. Assume that the output of the OPAMP swings from +15 V to -15 V. The voltage at the non-inverting input switches between

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 44
Let the voltage at non inverting terminal be V1, then after applying KCL at non inverting terminal side we have

or V1 = V0/3

If V0 swings from -15 to +15 V then V1 swings between -5 V to +5 V.

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 45

A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni= 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s= 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V )

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 45
We have the formula to find out Charge stored on either side on a junction given as follows,

We don’t have the value of W with us, it needs to be calculated. This can be easily calculated using the following formula,

where, VB = V0 − V, V is the applied Voltage and V0 is the barrier potential.

here V = 0 as no external bias is applied .

therefore,VB = V0 = 0.78V

Then the value of W is 2.01 x 10-6 cm.

So the value of QJ is found out to be 1.61 x 10-15 C

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 46

A linear, time-invariant and causal continuous time system has a rational transfer function with simple poles at s = -2 and s = -4 and one simple zero at s = -1. A unit step u(t) is applied at the input of the system. At steady state, the output has a constant value of 1. The impulse response of this system is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 46

At steady - state

Thus K/8 = 1 or K = 8 , s(∞) = 1

h(t) = L-1G(s) = (-4e-2t + 12e-4t)u(t)

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 47

Two ideal pn junction have exactly the same electrical and physical parameters except for the band gap of the semiconductor materials. The first has a bandgap energy of 0.525 eV and a forward-bias current of 10 mA with Va =0.255 V. The second pn junction diode is to be designed such that the diode current I = 10μA at a forward-bias voltage of Va = 0.32 V. The bandgap energy of second diode would be_____ (eV)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 47

Eg2 = 0.59 + 0.0259 In 103 = 0.769eV

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 48

Assuming all flip-flops are in reset condition initially, the count sequence observed at QA in the circuit shown is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 48
Let QA(n), QB(n), Qc(n) be the present states and QA(n + 1), QB(n + 1), QC(n + 1) be the next states of flop-flops.

In the circuit

QA(n + 1) = QB(n) Θ QC(n)

QB(n + 1) = QA(n)

QC(n + 1) = QB(n)

Initially all flip-flops are reset

1st clock pulse

QA = 0 Θ 0 = 1

QB = 0

QC = 0

2nd clock pulse

QA = 0 Θ 0 = 1

QB = 1

QC = 0

3rd clock pulse

QA=1 Θ 0 = 0

QB = 1

QC = 1

4th clock pulse

QA = 1 Θ 1 = 1

QB = 0

QC = 1

So, sequence QA = 01101…….

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 49

Calculate the hysteresis width (in Volts) for an inverting Schmitt trigger with feedback fraction 0.5. Assume the supply voltage to be 10V.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 49
For a inverting Schmitt trigger.

Given: β = 0.5, Vcc = 10V

Hysteresis Voltage VH = VUT-VLT

VUT = +β x Vcc

VLT = -β x Vcc

VH = 2 x β x Vcc

VH = 2 x 0.5 x 10

VH = 10V

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 50

Consider the below DTL NAND gate with A, B, C at logic 1 and the BJT Q is in saturation with = 0.8 V, = 0.2 V and hFE = 30. If all the diodes have forward conducting voltage of 0.7 V, then what will be the fan-out of the DTL NAND gate?

(Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 50
A, B, C are logic 1, output = Y = logic 0 i.e. transistor Q is in saturation

= (0.7 + 0.7 + 0.8) = 2.2 v

At the output node P, V(p) = 0.7 + V(y)

= (0.7 + 0.2)V = 0.9 V

I = (5 - 0.9)/5k = 0.82 mA

If ‘N’ be the fan out, then

Ic’ = 0.82 N + 2.18

= hFE x IB = 30 x 0.40

= 0.82 N + 2.18 = 30 x 0.40

N = 12

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 51

Let Q√γ be the BER of a BPSK system over an AWGN channel with two-sided noise power spectral density N0/2. The parameter γ is a function of bit energy and noise power spectral density.

A system with two independent and identical AWGN channels with noise power spectral density N0/2 is shown in the figure. The BPSK demodulator receives the sum of outputs of both the channels.

If the BER of this system is Q b√γ,then the value of b is _____________.


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 51
Bit error rate for BPSK

Function of bit energy and noise PSDN0/2

Counterllation diagram of BPSK

Channel is AWGN which implies noise sample as independent

Let 2x + n1 + n2 = x' + n'

where x' = 2x

n′ = n1 + n2

Now Bit error rate =

E1 is energy in x2

No1 is PSD of h1

E1 = 4E[ as amplitudes are getting doubled ]

No1 = N0[ independent and identical channel ]

⇒ Bit error rate =

⇒ b = √2 or 1.414

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 52

For the matrix , the Eigen value corresponding to the Eigen vector is (Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 52
We have

A =

Now [A - λI] [X] = 0

or (101) (4 - λ) + 2(101) = 0

or λ = 6

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 53

A biliary symmetric channel (BSC) has a transition probability of 1/8. If the binary transmit symbol X is such that P(X = 0) = 9/10, then the probability of error for an optimum receiver will be

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 53
Py = 1)/(x = 0P(x = 0)/Py = 1)/(x = 1P(x = 1)

P(x = 0) = 9/10

P(x = 0) + P(x = 1) = 1 SoP(x = 1) = 1/10

Transition Probability is nothing but Prabability af changing the value from 0 to 1 and 1 to 0

i.e.

P(0/1) = P(1/0) = 1/8

Pe = P(x = 0)Py = 1)/(x = 0 + P(x = 1)Py = 0)/(x = 1

= (9/10) × (1/8) + (1/10) × (1/8)

= 1/8

But this nat the Pe for optimum Receiver. So,Pe = 1 − Pc

where Pc is Probability of correct detection Use MAP detectionRule:

The received symbal m = m1 if P(y/x1)P(x1)/P(y/xj)P(xj) ≥ 1

This is abtained from following:

Pc(m = mi) = P(mi/y)⋅P(y) = P(xi/y)⋅P(y)

M is the estimated output signal Using Baye's thearem

P(xi/y)⋅P(y) = P(y/xi)⋅P(xi)

when o/py = 0 then Py = 0)/(x = 0P(x = 0)/Py = 0)/(x = 1P(x = 1)

= (7/8⋅9/10)/(1/8⋅1/10) = 63 ≥ 1

So when y = 0,m = 0

Similarly when y = 1 then = (1/8⋅9/10)/(7/8⋅1/10) = 9/7 ≥ 1

So when y = 0 then alsa m^ = 0

So if we choose the rule such that the message signal is 0 for y =0 ( or ) 1 then the Prabability of error is nothing but P(x = 1) = (1/10)

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 54

Consider the first order initial value problem

y' = y + 2x - x2, y(0) = 1, (0 ≤ x ≤ ∞ )

with exact solution y(x) = x2 + ex, For x = 0.1, the percentage difference between the exact solution and the solution obtained using a single iteration of the second-order Runge-Kutta method with step-size h = 0.1 is ______________.

(Answer up to two decimal places)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 54
dy/dx = y + 2x - x2

y(0) = 1, 0 ≤ x < ∞

Given f (x, y) = y + 2x - x2, xo = 0, yo = 1, h = 0.1

k1 = hf (xo, yo) = 0.1 (1 + 2(0) - 02) = 0.1

k2 = hg (x0 + h, y0 + k1) = 0.1 ((yo + k1) + 2(x0 + h) - (x0 + h)2)

= 0.1 ((1 + 0.1) + 2(0.1) - (0.1)2)

= 0.1 (1.1 + 0.2 - 0.01)

= 0.129

so, y1 = y0 + 1/2 (k1 + k2)

= 1 + 1/2 (0.1 + 0.129)

= 1 + 0.1145

= 1.1145

Exact solution, y (x) = x2 + ex

y (0.1) = (0.1)2 + e0.1

= 0.01 + 1.1052

= 1.1152

ERROR = 1.1152 - 1.1145

= 0.00069

Relative error = 0.00069/1.1152

= 0.00062

Percentage Error = 0.00062 × 100 = 0.06%

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 55

A 15 A source is operating at 100 MHz and feeds a Hertzian dipole of length 6mm situated at origin of space, determine electric field at P (4, 300, 900)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 55
since it is a current source driving a hertzian dipole, We need to get Magnetic field intensity and then Electric field intensity as,

Here, I0 = 15A,r = 4m,η = 377Ω,dl = 6mm,θ = 30 and β can be determined,

Thus, E = 706.875 mV/m

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 56

The open-loop transfer function of a unity feedback system is G(s) = k/s(s + 4) . If gain k is increased to infinity, then, the corresponding damping ratio will be

(Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 56
The characteristic equation is s2 + 4s + k = 0.

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 57

The divide by N counter is shown below. If initially Q0 = 0, Q1 = 1, Q2 = 0 the value of N is _______.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 57

State is repeating after 5 clock pulses.

N = 5

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 58

Consider a standard negative feedback configuration with G(s) = 1/(s + 1)(s + 2) and H(s) = (s + α)/s For the closed loop system to have a poles on the imaginary axis, the value of α should be equal to ____. (Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 58
Given G(s) = 1/(s + 1)(s + 2)

H(s) = (s + α)/s

C.E. = 1 + G(s) H(s) = 0;

s(s + 1) (s + 2) + (s + α ) = 0

s3 + 3s2 + 2s + s +α = 0

s3 + 3s2 + 3s + α = 0

If system is marginal stable

3 × 3 = α

α = 9

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 59

Consider following connection to the memory

The accessible range of address from memory is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 59
From the given data A00−A11 are 12 address lines while remaining 4 address lines are decided by OR gate If A12A13A14A15 all are zero then only chip select will active.

Hence A12−A15 will remain fixed at zero A15A14A13A12A11A10A9A8A7A6A5A4A3A2A1A0

Start address 0000000000000000

End address 0000111111111111

Accessible range is 0000− 0FFF H

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 60

When a plane wave travelling in free space is incident normally on a medium having εr = 4.0, then the fraction of power transmitted into the medium is given by x/9. Find the value of x.

(Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 60

Since η = μ/ε

Power transmitted into the medium can be calculated as,

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 61

An intrinsic semiconductor bar of Si is doped with donor type impurity to the extent of 1 atom per 108 silicon atoms, then the resistivity of silicon crystal will be (atomic density of Si crystals = 5×1022 atoms /cm3 and μn=1300cm2N−sec )


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 61

= 5 × 1014 donor /cm3

∵ It is an n -type semiconductor σ ≈ NDn

= 5 × 1014 × 1.6 × 10−19 × 1300

= 10.4 × 10−2

ρ = 1/σ

= 9.61(Ω−cm)

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 62

A 5-point sequence x(n) is given as x[–3] = 1, x[–2] = 1, x[–1] = 0, x[0] = 5, x[1] = 1. If X(e) denotes the discrete time Fourier transform of x [n], the value of ഽπX(e)dω is ____π (Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 62

For discrete time Fourier transform (DTFT) when N →∞

Putting n = 0 we get

or ഽπX(e)dω = 2πx[0] = 2π x 5 = 10π

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 63

(D2 − 5D + 6)y = excosx is a Partial differential equation whose solution is given by

y = C1e2x + C2e3x − Kex(3sin2x + cos2x)

The value of K is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 63

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 64

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

If the output terminal (2 –2') is short circuited then, input impedence (in Ω) of the network is

(Answer up to two decimal places)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 64
The ABCD parameters are defined as

V1 = AV2 - BI2 .... (i)

I1 = CV2 - DI2 .... (ii)

If the output terminal is short-circuited, then

V2 = 0

Substituting V2 = 0 into equation (i) and (ii), we get

V1 = - BI2

I1 = - DI2

Input impedance,

Zin = V1/I1 = B/D

=12/7 = 1.71 Ω

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 65

In a certain animal husbandry weight of animals had a distribution as shown in table below:

Mean weight of animals in the husbandry is______.


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 65

Mean=∑f(x)/∑f = 4305/75 = 57.4kg

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