Practice Test: Electronics Engineering (ECE)- 2


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65 Questions MCQ Test GATE ECE (Electronics) 2023 Mock Test Series | Practice Test: Electronics Engineering (ECE)- 2

Practice Test: Electronics Engineering (ECE)- 2 for Electronics and Communication Engineering (ECE) 2023 is part of GATE ECE (Electronics) 2023 Mock Test Series preparation. The Practice Test: Electronics Engineering (ECE)- 2 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Practice Test: Electronics Engineering (ECE)- 2 MCQs are made for Electronics and Communication Engineering (ECE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Test: Electronics Engineering (ECE)- 2 below.
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Practice Test: Electronics Engineering (ECE)- 2 - Question 1

A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in total from both as interest. The rate of interest per annum is.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 1 Let the rate % = R

According to the question

100R + 120 R = 2200

2200 R = 2200

R = 10%

Hence required rate % = 10%

Practice Test: Electronics Engineering (ECE)- 2 - Question 2

Amit rows a boat 9 kilometres in 2 hours down-stream and returns upstream in 6 hours. The speed of the boat (in kmph) is:

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 2

Down-stream rate = 9/2 = 4.5 kmph

Upstream rate = 9/6= 1.5 kmph

The speed of the boat = (4.5 – 1.5) kmph = 3 kmph

Practice Test: Electronics Engineering (ECE)- 2 - Question 3

The sentences given with blanks are to be filled with an appropriate word(s). Four alternatives are suggested for each question. For each question, choose the correct alternative and click the button corresponding to it.

They abandoned their comrades ______the wolves.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 3 When something is left in between more than 2 things or persons, ‘among’ is used. There are a number of wolves in the sentence.

Hence the correct answer is option D.

Practice Test: Electronics Engineering (ECE)- 2 - Question 4

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

Turgid

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 4 Turgid = swollen and distended or congested; pompous

Bloated = inflated

Humble = plain, simple

Puffy = inflated

Tumescent = swollen

Hence, humble is the correct answer.

Practice Test: Electronics Engineering (ECE)- 2 - Question 5

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

The following pie chart shows the percentage distribution of total number of Apples (Dry + Wet) sold in different days of a week :

The pie chart 2 shows the percentage distribution of total number of Apples (Wet) sold in different days of a week.

Find the ratio of Apples (Dry + Wet) sold on Tuesday to that of the Apples (Wet) sold on Thursday?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 5 Required ratio = 13% of 7000: 23% of 4500 = 182:207
Practice Test: Electronics Engineering (ECE)- 2 - Question 6

Directions: In each of the questions below, some statements are given followed by some conclusions. You have to consider the statements to be true even if they seem to be at variance with commonly known facts. You have to decide which of the following conclusions logically follows from the given statements. Give answer.

Statements:

Some jacket is shirt.

Some shirts are trouser.

No shoes are t-shirt.

All trousers are shoes.

All pants are t-shirt.

Conclusions:

I. All shirts being t-shirt is a possibility.

II. Some shoes are trouser.

III. Some jackets are t-shirt.

IV. Some shirts being t-shirt is a possibility.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 6

Practice Test: Electronics Engineering (ECE)- 2 - Question 7

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: July 1969 was to see a transformed Indira Gandhi.

Q: Quite a few people contributed with ideas.

R: She sounded the bugle through her historic ‘Note on Economic Policy and Programme’ that was circulated among delegates at Bangalore on July 9, 1969.

S: But the pivot was P.N.Haksar who gave shape, structure and substance with the help of some of his colleagues in the Prime Minister’s Secretariat.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 7

P is clearly the first sentence as it introduces the main person of the paragraph, i.e. Indira Gandhi. The pronoun ‘she’ in R refers to Indira Gandhi mention in P. Thus, R forms the second statement. Now, in the sentence R, ‘delegates’ is mentioned which says about the people, who are given in the sentence Q. So, sentence Q must follow R.

Thus, the sequence after rearrangement is PRQS and option A is the correct answer.

Practice Test: Electronics Engineering (ECE)- 2 - Question 8

Direction: Which of the following is the MOST SIMILAR in meaning to the given word?

Remnant

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 8

‘Remnant’ means ‘remainder’ which is same as ‘residue’.

Practice Test: Electronics Engineering (ECE)- 2 - Question 9

Direction: Read the information carefully and give the answer of the following questions:

Total number of children = 2000

If two-ninths of the children who play football are females, then the number of male football children is approximately what percent of the total number of children who play cricket?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 9

57.4% (If 2/9 th of children play football are female , then male children = 1 - 2/9 = 7/9th of children play football;

Let's take z% of male football children equal to cricket children , then

⇒ z% of cricket children = 7/9 th of football;

⇒ z% of (23% of 2000) = 7/9 th of (17% of 2000);

⇒ z% of 23 = 7/9 th of 17

⇒ z = 7× 17 × 100 = 57.4%

Short-cut :

Required percentage = (7/9)×17×100/ 23 = 57.4%

Practice Test: Electronics Engineering (ECE)- 2 - Question 10

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

Seven people, P, Q, R, S, T, W and X sitting in a straight line facing north, not necessarily in the same order. R sits at one of the extreme ends of the line. T has as many people sitting on his right, as on his left. S sits third to the left of X. Q sits on the immediate left of W. Q does not sit at any of the extreme ends of the line.

If all the people are made to sit in alphabetical order from right to left, the positions of how many people will remain unchanged?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 10

From the given conditions, we can conclude

After arranging in alphabetical order,

Hence, only Q's position will remain unchanged.

Practice Test: Electronics Engineering (ECE)- 2 - Question 11

The figure shows the circuit of a gate in the resistor transistor logic (RTL) Then the circuit represent which gate?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 11

Hence gate is OR Gate.

Practice Test: Electronics Engineering (ECE)- 2 - Question 12

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 12

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 2 - Question 13

For the given p+ -n junction diode the minority carrier life time in the n -region is ΥP = 1.1 μsec. If the excess minority charges present in the n -region at any time t is given as QP(t) then QP(t) at t = 1.5 μsec is _____ μC?


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 13 For t < />

I0 = 2A

Diode in FB has diffusion capacitance, as current increases the charge inside increase. Hence voltage increase.

Now

Here

Y = life time of holes in N-region

= 1.1μsec

Multiplying equation (1) by C gives

And t = 1.5 \musec

Hence

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 2 - Question 14

For the given network, the power given to the load is ____ kW? [Assume R =1 Ω and ideal diode]

V = 100. sinωt


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 14 (3,4)

Using mirror symmetry

The resistive network between AB can be simplified as.

Further simplifying

Practice Test: Electronics Engineering (ECE)- 2 - Question 15

A current source with an internal resistance of RS, supplied power to load RL. The plot of power delivered as a function of load RL is?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 15

Hence option B is correct.

Practice Test: Electronics Engineering (ECE)- 2 - Question 16

where (x) denotes greatest integer

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 16

Practice Test: Electronics Engineering (ECE)- 2 - Question 17

For given system

Angle of arrival is?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 17 Zeros:- s= -3 ± j

Poles:- s = -1 ± j

ϕ’ = 180 - tan-11 = 135°

ϕA = 180° + ϕ

Where ϕ = SP - SZ

= 180 + 135 - 90

= 225°

∴ ϕA = 180° + ϕ = 405°

= 45°

Hence ϕA = ±45°

Practice Test: Electronics Engineering (ECE)- 2 - Question 18

A system is shown in figure as follows

Where, hk (n) = δ [n – k.(1/2)k]

Then value of overall impulse response of above system h(n) at h = 2 is ______?


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 18 We have properties

Therefore h(n) can be drawn as

∴ at n = 2h(n) = 1

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 2 - Question 19

A square wave with frequency = 5 kHz

This signal is passed through an ideal low-pass filter with 0 dB of passband gain and 10 kHz of cut-off frequency. The filtered signal is subsequently buried additively into a zero-mean noise processed with one-sided power spectral density (PSD) of 20 nW Hz–1. Upto a frequency of 2.5 MHz. The PSD of noise is zero beyond 2.5 MHz. The signal to noise ratio of the output is _______dB.


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 19 We have

Now, Noise power = Area

= One sided PSD × Frequency range

Practice Test: Electronics Engineering (ECE)- 2 - Question 20

The output of filter is a

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 20 x(t) can be drawn like that

Hence impulse response of matched filter h(t) = x(T – t) is

∴ output y(t) = x(t) * h(t)

∴ Hence option D.

Practice Test: Electronics Engineering (ECE)- 2 - Question 21

For a MOD -30 ripple counter, the maximum operating frequency is 8 MHz. The propagation delay of each flip flop must be _________nsec?


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 21 Mod-30 counter required

∴ Propagation delay pe flip flops

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 2 - Question 22

For a successive approximation type of ADC shown.

If Va = 8V and resolution of D/A = 0.25 volt

Then the error voltage between Va and SAR output at 5th clock is_______volt?


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 22 Resolution = 0.25

Va = 8v

VDAC = Resolution × Decimal equivalent of binary o/p of SAR

∴at 5th clock max o/p = 7.75 volt

∴ error = 8 – 7.75 = 0.25 volt.

Practice Test: Electronics Engineering (ECE)- 2 - Question 23

For the transfer function G(jω) = 6 + jω. The corresponding Nyquist plot for positive frequency has the form

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 23 As real part σ = 6

Only imaginary port increases with ω.

Hence option D is correct.

Practice Test: Electronics Engineering (ECE)- 2 - Question 24

Consider a random process as z(t) = x(t) + y(t)

Where x(t) and y(t) are two random processes with zero mean and they are individually WSS. The PSD of x(t) is Sx(f) and y(t) is Sy(f). Then PSD of z(t) is?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 24 We know

Rz(T) = E[z(t).z(t − T)]

Here z(t) = x(t) + y(t)

∴ Rz(T) = E[(x(t) + y(t))⋅(x(t − T) + y(t − T))]

= E[x(t)⋅x(t − T)] + E[y(t)⋅x(t − T)] + E[x(t)⋅y(t − T)] + E[y(t)⋅y(t − T)]

∴ Rz(T) = Rx(T) + Ry(T) + Rxy(T) + Ryx(T)

We know

Rz(T) ⟶ PSD Sz(f)

∴ S2(f) = Sx(f) + Sy(f) + S(f) + Sxy(f)

Hence option D is correct.

Practice Test: Electronics Engineering (ECE)- 2 - Question 25

A digital circuit is designed in a way that it accept 3-bit number and generate it’s square. The circuit is to be designed using ROM with minimum hardware. Then the size of ROM is _______?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 25 Truth table is

It can be seen that

D1 = 0 and D0 = z

Hence ROM size = 23 × 4 = 8 × 4

The design is

Practice Test: Electronics Engineering (ECE)- 2 - Question 26

Eight voice signals, each limited to 4 kHz and sampled at Nyquist rate, are converted into binary PCM signal using 256 quantization levels. The bit transmission rate for the time-division multiplexed signal will be (in kbps)____


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 26 Nyquist rate = 4×103×2 = 8×103 samples/sec

256 levels = 8 bits.

So one sample is represented by 8 -bit. So total = 8×8×103 bits/sec

For 8 voice signal = 8×8×8×103 bits/sec

= 512 kbps

Practice Test: Electronics Engineering (ECE)- 2 - Question 27

When a relatively small d.c. voltage is applied across a thin slice of Gallium Arsenide with thickness of order of few tenths of a micrometer and the voltage gradient across the slice is in excess of about 3300 V/cm, then the negative resistance will manifest itself. Which of the following devices operate based on the above phenomenon?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 27 The above-mentioned phenomenon occurs in Gunn diode and it is called the GUNN EFFECT. Gunn Effect is exhibited by semiconductor materials like gallium arsenide, indium phosphide, cadmium telluride and indium arsenide. Thus, option B is the correct answer.
Practice Test: Electronics Engineering (ECE)- 2 - Question 28

A signal of bandwidth 100 MHz and strength 10mW is transmitted through a transmitter through a cable that has 40 dB loss. If N0/2 = 0.5 × 10−20ω/Hz then the signal to noise ratio at the input of the receiver is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 28

Practice Test: Electronics Engineering (ECE)- 2 - Question 29

In a superheterodyne AM receiver, the image channel selectivity is determined by :

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 29 Image frequency rejection can be improved by placing more no of tank circuits, in between Antenna and the IF amplifier and by increasing their selectivity against image frequency.
Practice Test: Electronics Engineering (ECE)- 2 - Question 30

Consider a medium such that μ=μ0,ε = 81ε0,σ = 20S/m. The frequency at which the conduction current density is 10 times the displacement current density in magnitude is ____(GHz)


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 30

on putting all these values.

= 36 GHz

Practice Test: Electronics Engineering (ECE)- 2 - Question 31

A 78 Ω lossless planar line was designed but did not meet a requirement. The fraction of the widths of strip (in %)should be added or removed to get the characteristic impedance of 75Ω is ___


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 31

i.e. the width must be increased by 4%.

Practice Test: Electronics Engineering (ECE)- 2 - Question 32

A 16K × 8 ROM IC is interfaced to 8085 microprocessor. What is the address space?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 32

16K × 8 ⇒ 24 × 210 × 8 = 214 × 8

i.e. 14 address input lines and 8 data lines

With 14 address means there are 14 different address lines out of 16 address lines

Starting address should be 0000 H

Hence starting address will be

0000 0000 0000 0000 = 0000 H

And end address

0011 1111 1111 1111 = 3FFF H

Hence address space is 0000 H – 3FFF H

Practice Test: Electronics Engineering (ECE)- 2 - Question 33

A 16 bit computer can address 1M word and has 128K words installed from 64K 1 chips. Calculate the no. of chips required and no. of bits of address bus that allows to select the memory chips

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 33

As it is a 64K memory chips hence 16 bits are needed to address them

i.e. 64K = 216

As we want 128K × 16 from 64K × 1 chips

Now as we have two rows of 16 chips each, we will need 1 bit to difference one row from another. So A16 will be used to select the memory chips

Practice Test: Electronics Engineering (ECE)- 2 - Question 34

Maximum number of Self Dual possible for 4-variable function.


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 34 For n variable Boolean function maximum number of self-dual possible is

Practice Test: Electronics Engineering (ECE)- 2 - Question 35

A minimized Boolean expression is a combination of

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 35 Points to Remember

The prime implicant which contains at least one 1 which cannot be covered by any other prime implicant is called an essential prime implicant (EPI).

The prime implicant whose each 1 is covered by at least one EPI is called Redundant prime implicant (RPI).

The prime implicant which is neither EPI nor RPI is called selective prime implicant (SPI).

Practice Test: Electronics Engineering (ECE)- 2 - Question 36

If then which of the following holds

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 36

Partially differentiating with respect to y,

Partially differentiating with respect to x,

Hence,

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

Calculate the value of


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 37

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 2 - Question 38

If dt/dx = sin(t + x) + cos(t + x) then find out the value of t at x = π/3 correct upto 2 decimal places. It is given that t = π/2 at x = 0


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 38

Practice Test: Electronics Engineering (ECE)- 2 - Question 39

A poisson variate satisfies P(x = 1) = 0.5 P(x = 2). Then find the value of P(x = 4) is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 39

Practice Test: Electronics Engineering (ECE)- 2 - Question 40

Find the general solution of the equation

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 40

This is Cauchy’s Homogeneous Equation,

Therefore,

Practice Test: Electronics Engineering (ECE)- 2 - 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 Practice Test: Electronics Engineering (ECE)- 2 - 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)]

Practice Test: Electronics Engineering (ECE)- 2 - Question 42

Find the complete solution of the state equation given below

Given the initial condition as

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 42

Consider the state equation be,

X˙= AX+BU

We know that,

State transition matrix,

Practice Test: Electronics Engineering (ECE)- 2 - Question 43

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

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - 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,

The block diagram is reduced to,

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

Hence, the response of the system,

Practice Test: Electronics Engineering (ECE)- 2 - Question 44

An integral controller is used for speed control with a set-point of 120 rpm within a range of 100 to 150 rpm. The controller output is 22% initially. The constant KI for controller output is 0.15%/sec for unit percentage error. If the speed goes to 105 rpm, then the controller output after two seconds for a constant error ep is (in %).


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 44

Where r = set point value and b = measured value

The controller output for constant error will be

Practice Test: Electronics Engineering (ECE)- 2 - 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 Practice Test: Electronics Engineering (ECE)- 2 - 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,

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

Practice Test: Electronics Engineering (ECE)- 2 - Question 46

A thin P-type silicon sample is uniformly illuminated with light which generates excess carriers. The recombination rate is directly proportional to

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 46

Recombination rate, R = B(nn0 + nn)(Pn0 + Pn)Electron and hole concentrations respectively under thermal equilibrium nn0 & Pn0 Excess elements and hole concentrations respectively. Thus, option B is the correct answer.

Practice Test: Electronics Engineering (ECE)- 2 - 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 Practice Test: Electronics Engineering (ECE)- 2 - Question 47

Practice Test: Electronics Engineering (ECE)- 2 - Question 48

Consider the given figure and calculate the feedback factor of the wave generator if R1 = 5kΩ, R2 = 20kΩ, R = 500Ω and C = 10 µF. Also calculate the one cycle period?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 48

The given circuit is that of a square wave generator.

Given: R1 = 5kΩ, R2 = 20kΩ, R = 500Ω and C = 10 µF.

To calculate feedback factor:

The period T is given by,

T = 0.01 x 0.405

T = 4.05 x 10-3s

T = 4.05ms

Practice Test: Electronics Engineering (ECE)- 2 - 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 Practice Test: Electronics Engineering (ECE)- 2 - 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

Practice Test: Electronics Engineering (ECE)- 2 - Question 50

The combination of resistors is shown in figure below. If R1 is operated at T and R2 is operated at temperature (T+t) ,then the equivalent noise temperature if the circuit is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 50

The circuit can be redrawn as

Practice Test: Electronics Engineering (ECE)- 2 - Question 51

Decide the stability of the system whose characteristic equation is given by s5 + 2s4 + 5s3 + 10s2 + 4s + 8 = 0

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 51

s3 row elements are all zero, thus the even polynomial above is given by:

f(s) = 2s4 + 10s2 + 8

f′(s) = 8s3 + 20s

There is no sign change in the first column, thus no zeros of the polynomial are in

the RHP, with 4 poles on the imaginary axis and one in LHP. Hence the system response is converging oscillations, hence stable.

Practice Test: Electronics Engineering (ECE)- 2 - Question 52

Consider the electronic PID controller shown in figure below

If R1 = R2 = R and C1 = C2 = C2 = C, then find Kp, Kd and Ki

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 52

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 2 - Question 53

Two point charges of 10 µC at point M and 20 µC at point N exerts a force of 4.8 N and 10 N respectively on a point charge of 40 µC at point O. Determine the electric field intensity at point O.


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 53 The electric field intensity is given as the force per unit charge.

E = F/Q

The electric field intensity due to two point charges Q1 at point M and Q2 at point N is the sum of the forces on the test charge Qt at point O, caused by the point charges Q1 and Q2 acting alone. Thus,

Practice Test: Electronics Engineering (ECE)- 2 - Question 54

An op-amp is used in a feedback configuration having open loop forward gain value of 50 as shown. The feedback gain is β in the circuit, the feedback circuit is given below. For which of the combination of R1 and R2 respectively would the configuration work as an oscillator?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 54

Here, we can see from the figure that the feedback is connected in a negative feedback configuration. Then, overall gain is given by,

Where, Af is the open loop forward gain. β is the gain provided by the feedback component. According to Barkhausen Criteria, in this situation the given arrangement would work as an oscillator only if Af∗β would be ′−1′

Thus, here β should be -0.02. Among the given op-amp configurations for feedback element, given that gain for op-amp as inverting amplifier İs given as

Only C provides the required gain β.

Practice Test: Electronics Engineering (ECE)- 2 - Question 55

Find the h parameters of the following circuit

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 55

For parallel connection Y parameters gets added Y = Y1 + Y2

Practice Test: Electronics Engineering (ECE)- 2 - Question 56

The expression for unit tangent ∅ = e2x casyz at the or igin at the or igin in the direction of the tangent to curve x = a sin t, y = a cos⁡ t,z = 2α sin ⁡t, at t = T/4

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 56

Practice Test: Electronics Engineering (ECE)- 2 - Question 57

Root locus of 2s(s+4)+k(s+8) is a circle. The co-ordinates of the centre of the circle?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 57

Characteristic Equation is

Now for the centre of the root locus we have to find the breakaway point of the plot For breakaway point,

On solving for s, s1 = −1.1255 and s2 = -12.74

So, the centre will be at and the coordinates are (-7, 0)

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

In the following table, x is a discrete random variable and p(x) is the probability density. The difference between the variance & the standard deviation is


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 58

Difference of variance & standard deviation = 3.06 - 1.749 = 1.311

Practice Test: Electronics Engineering (ECE)- 2 - Question 59

Find the state transition diagram for the logic circuit shown below-

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 59

Hence option C is correct

Practice Test: Electronics Engineering (ECE)- 2 - Question 60

Consider the following circuit consisting of three D-flip flops.

If all the flip-flops were reset to 0 at power on, then the total number of different output states represented by the counter ABC is equal to-


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 60

So it requires 4 states

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 2 - Question 61

Consider a silicon PN junction at T = 300K with current density 3.6 10-11 A/cm2. If the photocurrent density is 10 mA/cm2. Then find the open-circuit voltage (in V) of a solar cell when the solar intensity is increased by a factor of

Assume thermal voltage VT = 26 mV


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 61 Given photo current density of laser is JL = 10mA/cm2

Current density JS = 3.6 × 10−11A/cm2since the solar intensity is increased by 10 times

JL = 10 × 10mA/cm2 = 100mA/cm2The open-circuit voltage of solar cell is

Practice Test: Electronics Engineering (ECE)- 2 - Question 62

DTFT of x(n) is X(ω). Hence x(n) = (0.5)nu(n). Then the time signal corresponding to X2(ω) is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 62

Practice Test: Electronics Engineering (ECE)- 2 - Question 63

A varying magnetic flux linking a coil is φ=1/3 λt3. If the value of λ = −1Wb/s2, then at what time the induced EMF is 9V?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 63

Practice Test: Electronics Engineering (ECE)- 2 - Question 64

Determine length of a cavity resonator(rectangular) if it has dimensions a = 3cm, b = 2cm, and is excited by TE101 mode at 30GHz frequency.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 64

Given: f = 30GHz = 30 × 109Hz

We have TE101 mode, which means: m = 1,n = 0 and p = 1 Also, given 1 = 3cm = 0.03m and b = 2cm = 0.02m.

Taking square on both sides, we get:

Now, we have a = 0.03m. So, on putting the values and solving for 'd' we get:

(1/d)2 = 38888.89 ⇒ d = 5.07 x 10-3 m

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 2 - Question 65

If directive gain of a half-wave dipole is 2.64 W/W, and ohmic losses is 8Ω, then determine its power gain(in dB).


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 65 We know that the radiation resistance for a half-wave dipole is given as

Rrad = 73Ω

Now, antenna efficiency is given as:

ƞr = (Rrad) / (Rrad + Rloss) = 73 / (73 + 8) = 0.901.

Also, we know that efficiency of antenna can also be written as:

ƞr = Gp/ Gd = Power gain / directive gain

Now, we have, ƞr = 0.901 and directive gain Gd = 2.64

So, power gain Gp can be given as:

0.901 = Gp/ (2.64)

So, Gp = 2.37

Gp(in db) can be give as : Gp(dB)= 10log10 Gp = 10log(2.37)= 3.74 dB.

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