Practice Test: Electronics Engineering (ECE)- 15 - Electronics and Communication Engineering (ECE) MCQ

Practice Test: Electronics Engineering (ECE)- 15 - Electronics and Communication Engineering (ECE) MCQ

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65 Questions MCQ Test - Practice Test: Electronics Engineering (ECE)- 15

Practice Test: Electronics Engineering (ECE)- 15 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Practice Test: Electronics Engineering (ECE)- 15 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Practice Test: Electronics Engineering (ECE)- 15 MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Test: Electronics Engineering (ECE)- 15 below.
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Practice Test: Electronics Engineering (ECE)- 15 - Question 1

There are two lines made by joining points A, B,C. B lies between the line joining A and C. Is the distance between the A and C passes through the B more then 7 Km.I. The distance between A and B is 6 KmII. Distance between B to C is 7 Km long,

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 1
From Statement I

We know the distance between Points B and C but we do not know about the distance between A and B

Statement II

This statement tells us about the distance between A and B and by using Both statement we can calculate the distance between A and C that is longer than the 7 Km.

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

Direction: In the given question, a word/phrase is given followed by three statements; I, II and III. Choose the pair of sentences which can be combined using the given word/ phrase when used at the beginning of the new sentence.AndI: When Lionel Messi, Ronaldo and Neymar kick and dribble the football, they don’t remain confined to Argentina, Portugal or Brazil, respectively.II: Let the entire world realise that through sports, especially football, we all can cement our bonds, wash away our bitterness and prejudices.III: Live with a greater sense of closeness and joy with each other.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 2
‘And’ is used as a conjunction here which is used to introduce an additional comment or interjection.

Here sentences II and III are displaying same sense that is realizing the benefit of football in our lives whereas sentence I is providing entirely different aspect of football as a sport.

Hence option (B) will be the most appropriate choice.

New sentence: Let the entire world realise that through sports, especially football, we all can cement our bonds, wash away our bitterness and prejudices, and live with a greater sense of closeness and joy with each other.

Practice Test: Electronics Engineering (ECE)- 15 - Question 3

Which of the following is the MOST SIMILAR in meaning to Accreditation?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 3
Accreditation = an acknowledgement of a person's responsibility for or achievement of something.

Certification = an official document attesting to a status or level of achievement.

Meticulous = showing great attention to detail; very careful and precise.

Lurid = very vividly shocking.

Suppressive = tending or acting to suppress.

Agreement = harmony or accordance in opinion or feeling.

Thus, option B is the correct answer.

Practice Test: Electronics Engineering (ECE)- 15 - Question 4

In how many ways can you place 2 white bishops on an empty chess board?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 4
One white bishop can be placed on any of the 32 white boxes and the other white bishop can be placed on any of the 32 black boxes.

Ways of doing that = 32C1 * 32C1

= 32 × 32 = 1024

Practice Test: Electronics Engineering (ECE)- 15 - Question 5

Direction: In the question below are given statements followed by some conclusions. You have to take the given statements to be true even if they seem to be at variance with commonly known facts. Read all the conclusions and then decide which of the given conclusions logically follows from the given statements disregarding commonly known facts.

Statement:

No physics is maths.

Some chemistry is maths.

All sciences are chemistry.

Conclusion:

I. No science is physics.

II. Some physics are science.

III. Some physics are chemistry.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 5
The least possible Venn diagram for the given statements is as follows.

Conclusions:

I. No science is physics → False (It is possible but not definite).

II. Some physics are science → False (It is possible but not definite)

III. Some physics are chemistry → False (It is possible but not definite)

Conclusion I and II form complementary pair.

Hence, either conclusion I or II follows.

Practice Test: Electronics Engineering (ECE)- 15 - Question 6

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

(This pie chart shows the percentage of students appear in six different exams in 2016)

(This bar graph shows the percentage of failed students which appear in these six different exams in 2016)

If in 2016, total number of failed students in exam F was 4080, then how many passed students appears in the exam B?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 6

14280 (Let total number of students be = 100z.

Total number of students appears in exam F

= 16% of 100z

= 16z;

Total number of failed students in exam F

= 30% of 16z

= 4.8z = 4080;

⇒ z = 850;

Total number of passed students appears in the exam B

= (100 – 30)% of 24% of 100z

= 16.8z = 16.8 × 850 = 14280

Practice Test: Electronics Engineering (ECE)- 15 - Question 7

Direction: Two sentences with two blanks in each, followed by five alternatives with two words in each, are given. Choose that option as the answer which can fill both the blanks of both the sentences.

i. A sinking feeling of panic ________ over them and a temporary paralyzing fear engulfed them ________.

ii. Being a cleanliness freak, she ________ the floor and went down to the market only after the house was _________ clean.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 7

‘Sweep’ is to clean (an area) by brushing away dirt or litter. It fits the first blank of the second sentence, as the subject is cleaning the floor. Further, ‘sweep over’ means to overcome or overwhelm, thus fits in the first blank of the first sentence. The second blank of both the sentences can be filled by "completely", thus conveying an appropriate sense. Thus, option D is the correct answer.

Practice Test: Electronics Engineering (ECE)- 15 - Question 8

Direction: Five statements are given below, labelled 1, 2, 3, 4 and 5 which are supposed to be in a logical order. A statement labelled P is given thereafter. P can replace one of the five statements such that the four statements along with P would make a coherent paragraph. You have to identify which statement should P replace and then the find out the correct sequence from the options. If the five options are in logical order and form a coherent paragraph/passage, choose the fifth option “12345”.

1) The future of Germany’s coalition government is hanging in the balance after the country’s interior minister reportedly announced his intention to resign over a migration showdown with Angela Merkel.

2) Horst Seehofer, who is also leader of the Christian Social Union, on Sunday night offered to step down from his ministerial role and party leadership in a closed-door meeting in which he and fellow CSU leaders had debated the merits of the migration deal Merkel hammered out with fellow European Union leaders in Brussels.

3) But with CSU hardliners believed to have tried to talk the combative interior minister into staying, a press conference was postponed until Monday, with Seehofer seeking to go back to Merkel in search of a final compromise.

4) At a 2am media conference, Seehofer said he had agreed to meet again with Merkel’s party before he made his decision final.

5) “We’ll have more talks today with the CDU in Berlin with the hope that we can come to an agreement,” Seehofer said.

P. Merkel said on Sunday she wanted the CDU and its Bavarian allies to continue working together.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 8

The paragraph talks about the issues pertaining to the state of coalition government in Germany. The first sentence introduces the issue to us. The second sentence explains the problem. The third sentence then goes on to elaborate upon the meeting, and the fourth and fifth sentence talk about the current state and measures being taken. Thus, they are in perfect order. Sentence P is completely out of place here because CDU is not the subject of this paragraph and it does not matter what Merkel thinks they should do. Thus, D is the correct answer.

Practice Test: Electronics Engineering (ECE)- 15 - Question 9

4 identical solid spheres are melted and re-formed into a solid hemisphere. Then, the ratio of the curved surface area of the hemisphere to half of the surface area of a single sphere is -

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 9

Let radius of sphere be ‘r’ and radius of hemisphere be ‘R’.

Then,

ATQ,

Volume of Spheres = Volume of Hemisphere

4 x (4/3)πr3 = (2/3)πR3

Or, R / r = (8)1/3 = 2 …(1)

C.S.A of hemisphere = 2πR2

And, Surface area of sphere = 4πr2

ATQ,

2πR2 :( ½) of 4πr2

= 2πR2 :2πr2 = R2 :r2 {using (1)}

= 4 :1

Practice Test: Electronics Engineering (ECE)- 15 - Question 10

A study of people who reduced the calories they consumed has found the strongest evidence yet that such restrictions slow down metabolism, raising hopes that a low calories lifestyle or treatments stimulating biological effects of restricted eating, could prolong health in old age. The report provides the most robust evidence to date that everything we have learnt in other animals can be applied to humans.

Which of the following argument will prove that the above conclusion is flawed?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 10

The given passage gives the benefits of reducing calories to attain a healthier lifestyle and out of the given options, option B seems to give a contradictory fact stating that the lack of required calories may cause problems to the body. Hence, option B is correct.

Practice Test: Electronics Engineering (ECE)- 15 - Question 11

The value of k such that the system of equation

x + ky + z = 0, x + 3y + kz = 0, 2x + y + z = 0 has non-trivial solution.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 11

To find the value of K, we have to make rank of matrix less than unknown of the equation.

Apply R2 = R2 - R1

R3 = R3 - 2R1

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 12

The frequency at the output of the following cascaded circuit is …………. Hz

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 12
Modulus of 10 -bit ring counter = 10

Modulus of 4 -bit ripple counter = 2N = 24 = 16

Modulus of 5-bit Johnson counter = 2N = 2 × 5 = 10

Frequency at the output

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 13

A PCM (Pulse Code Modulation) system uses 8-bit encoder and uniform quantization. What is the maximum bandwidth(in MHz) of the low-pass input message signal for which the bit rate of the system is 20Mbps?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 13
Given system is PCM, n = 8bits, Rb = 20. Mbps

We have,

Again, by Nyquist Criterion, fz ≥ 2f

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 14

A super heterodyne receiver having RF amplifier is tuned to 1200 kHz, the intermediate frequency is 450 kHz. The quality factor of the tuned circuit at RF amplifier and at the mixer are same and is equal to 65. The image rejection ratio is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 14

a = Image rejection ratio

[Turned circuit of same quality factor used ar two stages]

Therefore,

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

Simplify the given Boolean expression

A+AB'+AB'C'+AB'C'D'+AB'C'D'E'+…………………

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 15
We have A (1 + B'+B'C'+B'C'D'+B'C'D'E'+…………………)

We know A + 1 = 1 (Rule)

Now A.1 = A

Practice Test: Electronics Engineering (ECE)- 15 - Question 16

If the skin depth is 60 µm at 3 MHz in a certain conducting medium then what will be the skin depth if the frequency is changed to 27 MHz?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 16

From the expression we find that δ is inversely proportional to square root of frequency,

Practice Test: Electronics Engineering (ECE)- 15 - Question 17

Find the Standing Wave Ratio of a transmission line of characteristic impedance 80Ω is connected to a load of 100Ω.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 17
Given, Z0 = 80Ω, ZL = 100Ω

Voltage reflection ratio,

Standing Wave Ratio, SWR

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 18

Consider the following continuous time signal x(t)

If Ck is the complex Fourier series coefficient of signal x(t). Then the value of |C2| will be

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 18

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 19

The circuit shown in the figure below is in steady state when switch is closed at t = 0, then the value of

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 19
At t = 0, the circuit is in steady state condition during this condition inductor will acts as short circuit

iL(0-) = 10/(9 + 1) = 10/10 = 1A

After t = 0+

By Applying KVL in the loop

VL + 1(1) = 0

VL = -1

Practice Test: Electronics Engineering (ECE)- 15 - Question 20

Characteristic equation of a system having unity feedback is

(s + 1)(s - 2)+ k(s - 3)(s + 4). The system is then which of the following?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 20
Transfer Function

For minimum phase system, all finite poles and zeroes are located in the left half of the s-plane. For a non-minimum phase system, one or more zeroes are located in the right half of the splane and remaining all zeroes and poles are located on the left half of the s-plane. For All-pass system, zeroes lies on the right half of the s-plane and poles lies on the left half of the s-plane and are symmetrical about the imaginary axis. Also, magnitude should be 1 and phase should be -180

Practice Test: Electronics Engineering (ECE)- 15 - Question 21

Consider a spatial curve in 3D space given on parameters form by

X = t

Y = 1

Z = 3t/π in 0 ≤ t ≤ π. The length of the curve is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 21

Practice Test: Electronics Engineering (ECE)- 15 - Question 22

The modified work function of N-channel MOSFET is -0.8V. The effective positive charge at the Semiconductor oxide layer interface is 9.8 x 10-5 C/m2 and the oxide capacitance is 300 μF/ m2, then calculate the flat band voltage.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 22
φms = -0.8V, Q0x = 9.8 x 10-8C/m2, C0x = 300μF/m2

The flat band voltage is given by

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 23

In the voltage controller circuit shown in the figure (assume that op-amp is ideal). The BJT has VBE = 0.7V & β is very large and V2 = 4.7V. For a regulated output of 9.4V, the value of R (in kΩ) is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 23
Out voltage for given circuit

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 24

Consider a lossless antenna with a directive gain of +6db. If 1mW of power is fed to it the total power radiated by the antenna will be

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 24
Lossless antenna with directive gain of +6dB = 4 (In linear)

Input power to antenna = 1 mW

Power radiated by antenna = 4 mW

Thus,

Radiated power = 4 × 1 mW = 4 mW

Hence option (a) is correct.

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 25

The output voltage Vo(in mV) of the circuit shown in figure

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 25

Practice Test: Electronics Engineering (ECE)- 15 - Question 26

For a given P+ - N Si junction diode with ND = 1015 cm-3 and ni = 1010 cm-3 and e = 8.85 × 10-14 C2/N-cm3. Find the electric field at the midpoint of the depletion region on the n-side. (let reverse bias voltage = 100V)

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 26

We have

Practice Test: Electronics Engineering (ECE)- 15 - Question 27

The divergence of the vector

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 27
We have

Divergence is,

Practice Test: Electronics Engineering (ECE)- 15 - Question 28

For the voltage amplifier circuit shown in the figure, Av is very high. Also RS = 1mΩ, RL = 1kΩ R1 = 1 kΩ and R2 = 1mΩ

Then what is the value of |Vo/Vs| = ?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 28
This is a negative feedback amplifier

Hence option A is correct.

Practice Test: Electronics Engineering (ECE)- 15 - Question 29

The value of d/dt x(t) at t = 1.5 for x(t) = u(t) + r(t) - 2r(t - 1) + r(t - 2) - u(t - 2) is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 29
Plotting each signal one by one

By adding all above signals we get x(t) as

by differentiation of x(t)

Practice Test: Electronics Engineering (ECE)- 15 - Question 30

Consider an angle modulated signal X(t) = 10.cos [108 πt + Sin 2π 103t]

The maximum frequency deviation is ________kHz?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 30
Given x(t) = 10. Cos [108 πt + 5. Sin 2π103 t]

θ(t) = 108 πt + 5. Sin 2π103t

= ωct + ϕ (t)

∴ ϕ(t) = 5. sin 2π103t

∴ ϕ’ (t) = 104.π.cos 2π 103 t

Maximum frequency deviation is

Δω = | ϕ’(t)|max

∴ ∆fmax = 5 kHz.

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 31

With G(s) = k/(s+3)5. Find kmax for stability.

k =

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 31
Here

Poles = 5

Zeros = 0

angle of asymptotes

For k = 0, 1, 2,…, (P − Z − 1)

∴ k = 0, 1, 2, 3, 4

∴ angles = 36º, 108º, 180º, 252º, 324º

Hence root locus is:

Practice Test: Electronics Engineering (ECE)- 15 - Question 32

For the circuit using op-amp -741 with unity gain frequency of 1 MHz. The overall 3 -dB frequency is ____?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 32
Unity gain frequency fT = f3dB x Gain

Hence option C is correct

Practice Test: Electronics Engineering (ECE)- 15 - Question 33

Given that D = z⋅ρ cos2⁡ϕa2c/m2, Calculate the charge density at (1, π/4, 3) and then the total charge enclosed by the cylinder of radius 1m with −2 ≤ z ≤ 2m is?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 33
Charge density ρv = ∇.D

=∂Dz/∂z =ρ⋅cos2⁡ϕ

Now charge enclosed is

Hence option C is correct.

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 34

Let c be triangle path connecting the points (0, 0), (4, 4) and (0, 4) in the counter clock wise direction then,

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 34

Acc. to Green theorem,

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 35

For the transfer function

The resonance peak magnitude is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 35
(5, 6)

For given system

∴ Also 0C gain k = 5

Resonance peak can be given as

Practice Test: Electronics Engineering (ECE)- 15 - Question 36

The residue of the function

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 36

Res f(z)z = a = 1/(n – 1)! dn-1/dzn-1 [(z – a)n f(z)]z=a

Here we have n = 2 and a = 2

Thus

Res f(z)z=2 = 1/(2-1)! d/dz [(z-2)2 1/(z-2)2(z+2)2]z=a

Res f(z)z=2 = d/dz [1/(z+2)2]z=a

Res f(z)z=2 = [-2/(z+2)3]z=a

Res f(z)z=2 = -2/64

Res f(z)z=2 = -1/32

Practice Test: Electronics Engineering (ECE)- 15 - Question 37

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 37

This is the line integral in the form

Comparing with the given function.

since path along which integration is to be done is not given, F may or may not be exact.

Curl F = 0, Hence this line integral is path independent.

ϕ = exz + y

ϕ1 at (2, 3, 0) is = e + 3 = 4

ϕ2 at (0, 1, 2) is = e + 1 = 2

Practice Test: Electronics Engineering (ECE)- 15 - Question 38

For the lattice circuit shown in figure, Za = 20Ω and Zb = 10Ω. The value of open circuit impedance parameter Z12 is ____ Ω ?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 38
(5, 5)

For lattice network

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 39

For given circuit, with R = 1 kΩ and C = 1 μF and also R1 = R2. The output frequency is ____ kHz. (Assume ideal opamp)

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 39

This is a Astable multivibrator.

Hence

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 40

A photo-detector circuit shown in figure. The photo-diode has an active area of 10+2 cm2 and sensitivity of 0.55 A/W. When light of intensity 10mW/cm2 falls on the photo-diode, the output voltage VO is ___ volt.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 40
The photo diode current is

I = light intensity x sensitivity x Area

= 10 x 10-3 x 0.55 x 102 x 10-4

= 0.55 x 10-4 A

Hence Vo1 = (-100k) x I = -5.5 volt

Now for 2nd op-amp,

Current through 20 kΩ resistor is

Practice Test: Electronics Engineering (ECE)- 15 - Question 41

Study the standard negative feedback configuration with G(s) Find the number of clockwise encirclements of (-1, 0) in the Nyquist of the Loop transfer function G(s) H(s) will be

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 41

Characteristics equation

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

3s2-0.2s+ 300 = 0

The given system is unstable with two roots in right hand side of s-plane

So Z = 2

Also P = 2 since OLTF has two poles in right hand side of s-plane

So N = P - Z = 2 – 2 = 0

N = 0

Hence the number of clockwise encirclements of (-1, 0) in the Nyquist of the Loop transfer function G(s) H(s) is zero.

Practice Test: Electronics Engineering (ECE)- 15 - Question 42

A second order process having transfer function is cascaded with a standard PD controller in unity feedback configuration. If the closed loop system have velocity error constant KV = 300 and damping ratio ζ = 0.6, then the value of controller gain (kP) is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 42
Standard TF of PD Controller is given by

Gc(s) = KP (1 + KDs)

Velocity error constant Kv=lim(s→0) s Gc(s) G(s)H(s)

300 = Kp (60/8)

Kp = 40

Practice Test: Electronics Engineering (ECE)- 15 - Question 43

Which of the following signal is obtained on integrating ramp signal?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 43

On integrating t, we get (1/2)t2 which is the amplitude of parabolic signal. Thus, option C is the correct answer.

Practice Test: Electronics Engineering (ECE)- 15 - Question 44

A unity feedback control system has an open loop transfer function.

The value of system gain K for which all the roots are equal is _______.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 44
Break point can be calculated as

dK/ds = 0

The characteristic equation for unity feedback system is

∴ The break point =−4

Equal roots are at =- 4

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 45

For a Si, one sided abrupt junction with NA = 2 × 1019/cm3 & ND = 8 × 1015/cm3 & ni = 9.65 × 109/cm3 The junction capacitance at zero bias is ______(in mF/cm2). Given,ϵr = 11.7

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 45

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 46

In an n-channel MOSFET, drain is shorted to the gate so that VGS = VDS. And

VT = 1V. If the drain current ID is 1 mA for VGS = 2V then for VGS = 3V, ID (in mA) is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 46
Given VT = 1V

ID1 = 1 mA, VGS1 = 2V

VGS2 = 3V, ID2 = ?

Practice Test: Electronics Engineering (ECE)- 15 - Question 47

A semiconductor of thickness 0.5μm is illuminated with monochromatic light of hϑ = 2eV, and the absorption coefficient of semiconductor α = −5 × 10−4cm−1. If the incident power is 10mW (Assume semiconductor has perfect quantum efficiency) then number of photons given from recombination events per second is photons/sec

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 47

I = I0e-αl

= 10-2 exp(-5 x 10-4 x 0.5 x 10-4)

= 0.82m W

Total energy absorbed = 10 - 0.82 = 9.18mW

Number of photons emitted

= 2.8687 x 1016 photons/sec

Practice Test: Electronics Engineering (ECE)- 15 - Question 48

In a 8085 microprocessor, the register which holds address of the next instruction to be fetched is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 48

A program counter is a register in a computer processor that contains the address (location) of the instruction being executed at the current time. As each instruction gets fetched, the program counter increases its stored value by 1.

After each instruction is fetched, the program counter points to the next instruction in the sequence. When the computer restarts or is reset, the program counter normally reverts to 0.

In computing, a program is a specific set of ordered operations for a computer to perform. An instruction is an order given to a computer processor by a program. Within a computer, an address is a specific location in memory or storage.

A register is one of a small set of data holding places that the processor uses. Thus, option B is the correct answer.

Practice Test: Electronics Engineering (ECE)- 15 - Question 49

Calculate Zth across the terminals A and B:

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 49

To calculate Zth consider the diagram

Current source has been open circuited.

XL = jωL = sL =10s since L = 10H

XC = 1/jωC = 1/sC = 1/10s since C = 10F

Therefore Zth= (XL+10)||(XC + 10)

= {(XL+10)* (XC + 10)}/(20 + XL + XC)

= {(10 +10s)*(10 +1/10s)}/(20 + 10s +1/10s)

Practice Test: Electronics Engineering (ECE)- 15 - Question 50

The response of a system at zero initial conditions and non-zero input is called ________.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 50

The response of a system at zero initial conditions and non-zero input is called zero state response and the response of a system at non-zero initial conditions and zero input is called zero input response.

The sum of zero state response and zero input response is called total response.

Thus, option A is the correct answer.

Practice Test: Electronics Engineering (ECE)- 15 - Question 51

Following the state diagram shown clocked sequential circuit:

How many states the sequential circuit has?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 51

Drawing a state table: (Let input be X)

State e and g are equivalent because the next states and outputs are same. We can eliminate any one them, say ‘g’. Thus ‘g’ is replaced by ‘e’ wherever it occurs. Now sated d and f are equivalent. Thus one of them say d, can be eliminated. Thus the reduced state table-

Practice Test: Electronics Engineering (ECE)- 15 - Question 52

Consider the following circuit. A = a2a1a0 and B = b2b1b0 are three bit binary numbers input to the circuit. The output is Z = z3z2z1z0. R0, R1 and R2 are registers with loading clock shown. The registers are loaded with their input data with the falling edge of a clock pulse (signal CLOCK shown) and appears as shown. The bits of input number A, B and the full adders are as shown in the circuit. Assume Clock period is greater than the settling time of all circuits.

For 8 clocks pulses on the CLOCK terminal and the inputs A, B as shown, obtain the output Z (sequence of 4-bit values of Z). Assume initial contents of R0, R1 and R2 as all zeros.

A = 110 011 111 101 000 000 000 000

B = 101 101 011 110 000 000 000 000

Clock No 1 2 3 4 5 6 7 8

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 52

A clock signal is a particular type of signal that oscillates between a high and a low state and is utilized like a metronome to coordinate actions of digital circuits. A clock signal is produced by a clock generator. Although more complex arrangements are used, the most Common clock signal is in the form of a square wave with a 50% duty cycle, usually with a fixed, constant frequency. Circuits using the clock signal for synchronization may become active at either the rising edge, falling edge, or, in the case of double data rate, both in the rising and in the falling edges of the clock cycle. Most integrated circuits (ICs) of sufficient complexity use a clock signal in order to synchronize different parts of the circuit, cycling at a rate less than the worst-case internal propagation delays. In some cases, more than one clock cycle is required to perform a predictable action. As ICs become more complex, the problem of supplying accurate and synchronized clocks to all the circuits becomes increasingly difficult.

Practice Test: Electronics Engineering (ECE)- 15 - Question 53

Find by double integration the volume of the solid generated by revolving the ellipse about the x-axis.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 53

Here we shall consider the revolution of the only upper half of the ellipse about the x-axis because volumes generated by upper and lower halves overlap. If V is the required volume, then

Practice Test: Electronics Engineering (ECE)- 15 - Question 54

Calculate the value of

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 54

Put 3θ = x

3dθ = dx

(use gamma function formula)

Practice Test: Electronics Engineering (ECE)- 15 - Question 55

Consider the following differential equation dy/dx = -x+1

If y (2) = 1 then the value of y (2.2) using Runge- Kutta third order method is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 55

here, f(x) = −x + 1

h=0.2,x0 = 2,y0 = 1

According to Runge-Kutta third order method

Where,

K1 = hf(x0, y0) = hf(2,1)

= 0.2 × (−2 + 1) = −0.2

= 0.2 × f(2.1,1.1)

= 0.2 × (−2.1+1)

= −0.22

K3 = hf(x0 + h,y0 + 2K2 − K1)

= hf(2+0.2,1+2×(−0.22)+0.2)

= 0.2 × f(2.2,0.76)

= 0.2 × (−2.2+1)

= -0.24

Practice Test: Electronics Engineering (ECE)- 15 - Question 56

The value of integral c∮(yzdx + zxdy + xydz) where c is the curve x2 + y2 = 1 and z = y2 is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 56

Practice Test: Electronics Engineering (ECE)- 15 - Question 57

Consider the two port network shown below

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 57

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 58

Consider the 2 port circuit given below

Hybrid parameter h12 for the circuit is _____.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 58

Redrawing the circuit by keeping I1 = 0

V2 = 10(I2 - Ix) = 4ix + 6ix + 10ix

10I2 = 30ix

I2 = 3ix

V1 = 10ix

V2 = 10(I2 - Ix)

V2 = 20Ix

V1/V2 = 10ix/20ix = 0.5

Practice Test: Electronics Engineering (ECE)- 15 - Question 59

For the opamp biasing shown in figure, it was found experimentally even at zero bias the current flowing into positive terminal is 15nA and current flowing into negative terminal is 10nA. If the output voltage of op-amp is 8mV. The value of feedback resistance Rf is__________.

Assume (Rf ≫ R). R = 0.5 kΩ

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 59

Assuming ideal op-amp and Rf ≫ R current can be modelled as voltage source

I+ = current at positive non-inverting terminal

I- = current at the negative inverting terminal

Vo+ = output voltage due to non-inverting current flow

Vo- = output voltage due to non-inverting current flow

(Taking inverting node ground)

V0 (off set) = V0+ + V0− (By superposition theorem)

= I+R+I+Rf−IRf

= I+R+Rf(l+−I)

8mV = 15 × 0.5 × 10−16 + Rf(15 − 10) × 10−9

Practice Test: Electronics Engineering (ECE)- 15 - Question 60

The input-output relationship of a causal stable LTI system is given as y[n] = αy[n - 1] + βx[n].

If the impulse response h[n] of this system satisfies the condition the relationship between α cond⁡ β is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 60

Given system equation as

Practice Test: Electronics Engineering (ECE)- 15 - Question 61

The Fourier transform of continuous time signal X(t) = 3cos(10t) + 4sin(10t) is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 61

x(t) = 3cos(10t) + 4sin(10t)

Practice Test: Electronics Engineering (ECE)- 15 - Question 62

If the Nyquist rate for x(t) is ω0, what will be the Nyquist rate for x(3t) and x(t/3)?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 62

Considering x(t)↔X(ω), then if y(t) = x(at)

Then for y(t)=x(3t):

As the Nyquist rate for x(t) is ω0, ω0,X(ω) = 0 for |ω| > ω0/2

Considering y(t), Y(ω) = 0 for |ω/3| > ω0/2 =>Y(ω) = 0 for| ω| > 3ω0/2

Therefore Nyquist rate for y(t) is 2×3ω0/2=3ω0

for y(t) = x(t/3): Y(ω) = 3X(3ω)

As the Nyquist rate for x(t) is ω0,X(ω) = 0 for |ω| > ω0/2

Considering y(t), Y(ω) = 0 for |3ω| > ω0/2 => Y(ω) = 0 for |ω| > ω0/6

Therefore Nyquist rate for y(t) is 2 x ω0/6 = ω0/3

Practice Test: Electronics Engineering (ECE)- 15 - Question 63

Find the systems response to the input x[n] = u[n]| if the unit impulse response h[n] of a DT LTI system is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 63

Practice Test: Electronics Engineering (ECE)- 15 - Question 64

Consider the speed control system of figure below where in the inner loop corresponds to motor back emf. The controller is an integrator with gain K observe that the load is inertia only.

Determine the value of k for which steady state error to unit ramp input (Vr = 1/s2) is less than 0.01 rad/sec.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 64

Reducing the inner loop

Practice Test: Electronics Engineering (ECE)- 15 - Question 65

Find the capacitor voltage VC at t = 1msec if switch is closed as t = 0. Assume capacitor initially uncharged and op-amp is not saturated.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 65

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