ec objective paper 2 1011


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Directors Message

UPSC has introduced the sectional cutoffs of each paper and screening cut off in

three objective papers (out of 600 marks). The conventional answer sheets of only

those students will be evaluated who will qualify the screening cut offs.

In my opinion the General Ability Paper was easier than last year but Civil

Engineering objective Paper-I and objective Paper-II both are little tougher/

lengthier. Hence the cut off may be less than last year. The objective papers of ME

and EE branches are average but E&T papers are easier than last year.

Category

Percentage

Marks

Expected Minimum Qualifying Marks in Each Paper (out of 200 Marks)OBJECTIVE

GEN

15%

30

OBC

15%

30

SC

15%

30

ST

15%

30

PH

10%

20

Category

Percentage

Marks

Expected Minimum Qualifying Marks in Each Paper (out of 200 Marks)CONVENTIONAL

GEN

15%

30

OBC

15%

30

SC

15%

30

ST

15%

30

PH

10%

20

Branch

CE

ME

EE

E&T

Expected Screening cut off out of 600 Marks (ESE 2016)

GEN

225

280

310

335

OBC

210

260

290

320

SC

160

220

260

290

ST

150

200

230

260

Note: These are expected screening cut offs for ESE 2016. MADE EASY does not

take guarantee if any variation is found in actual cutoffs.

B. Singh (Ex. IES)CMD , MADE EASY Group

MADE EASY team has tried to provide the best possible/closest answers, however if

you find any discrepancy then contest your answer at www.madeeasy.in or write your

query/doubts to MADE EASY at: [email protected]

MADE EASY owes no responsibility for any kind of error due to data insufficiency/misprint/human errors etc.


3/52

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members. MADE EASY announces exclusive batches for General Studies and Engineering Aptitude to cover the

syllabus of Paper-I of Preliminary exam. The classes will be conducted by experienced faculties of MADE EASY focusing

on new pattern of Engineering Services Examination, 2017. Latest and updated study material with effective

presentation will be provided to score well in Paper-I.

Roadmap forESE 2017 Prelims

Paper-IGeneral Studies &Engineering Aptitude

Indias Best Institute for IES, GATE PSUs

1. Current Affairs:Current National and International issues, bilateral issues, current economic affairs,

Defence, Science and Technology, Current Government Schemes, Persons in news, Awards &

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2. Reasoning and Aptitude :Algebra and Geometry, Reasoning and Data Interpretation, Arithmetic,

coding and decoding, Venn diagram, number system, ratio & proportion, percentage, profit & loss,

simple interest & compound interest, time & work, time & distance, blood relationship, direction

sense test, permutation & combinations etc.

3. Engineering Mathematics : Differential equations, complex functions, calculus, linear algebra,

numerical methods, Laplace transforms, Fourier series, Linear partial differential equations,

probability and statistics etc.

Paper-I : General Studies & Engineering Aptitude

P.T.O. (Page 1 of 3)

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4. General Principles of Design, Drawing, Importance of Safety : Engineering Drawing,

Drawing instruments, drawing standard, geometric construction and curves, orthographic

projections, methods of projection, profile planes side views, projection of points, projection of

straight lines, positions of a straight line with respect to HP and VP, determining true length and

true inclinations of a straight line, rotation methods, trace of a line, projection of planes,

importance of safety etc.

5. Standards and quality practices in production, construction, maintenance and services:

ISO Standards, ISO-9000 Quality Management, ISO-14000 other, BIS Codes, ECBC, IS, TQM ME,

TPM, PDCA, PDSA, Six Sigma, 5S System, 7 Quality Control Tools , ISHIKAWAS -7QC Tools, Kaizer

Tools-3m, TQM : Most Importance, Deming's: 14 Principles, Lean Manufacturing ME, Quality

Circles, Quality Control, Sampling.

6. Basics of Energy and Environment:Renewable and non renewable energy resources, energy

conservation, ecology, biodiversity, environmental degradation, environmental pollution,

climate change, conventions on climate change, evidences of climate change, global warming,greenhouse gases, environmental laws for controlling pollution, ozone depletion, acid rain,

biomagnification, carbon credit, benefits of EIA etc.

7. Basics of Project Management:Project characteristics and types, Project appraisal and project

cost estimations, project organization, project evaluation and post project evaluation, risk

analysis, project financing and financial appraisal, project cost control etc.

8. Basics of Material Science and Engineering:Introduction of material science, classification of

materials, Chemical bonding, electronic materials, insulators, polar molecules, semi conductor

materials, photo conductors, classification of magnetic materials, ceramics, polymers, ferrous

and non ferrous metals, crystallography, cubic crystal structures, miller indices, crystal

imperfections, hexagonal closed packing, dielectrics, hall effect, thermistors, plastics,

thermoplastic materials, thermosetting materials, compounding materials, fracture, cast iron,

wrought iron, steel, special alloys steels, aluminum, copper, titanium, tungsten etc.

9. Information and Communication Technologies : Introduction to ICT, Components of ICT,

Concept of System Software, Application of computer, origin and development of ICT, virtual

classroom, digital libraries, multimedia systems, e-learning, e-governance, network topologies,

ICT in networking, history and development of internet, electronic mail, GPS navigation system,smart classes, meaning of cloud computing, cloud computing architecture, need of ICT in

education, national mission on education through ICT, EDUSAT (Education satellite), network

configuration of EDUSAT, uses of EDUSAT, wireless transmission, fibre optic cable etc.

10. Ethics and values in engineering profession:ethics for engineers, Ethical dilemma, elements

of ethical dilemmas, indian ethics, ethics and sustainability, ethical theories, environmental

ethics, human values, safety, risks, accidents, human progress, professional codes,

responsibilities of engineers etc.

Page 2 of 3Scroll down

or Answer Key of ESE 2016

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Corporate Office: 44-A/1, Kalu Sarai, New Delhi-16 | Email : [email protected] | Visit: www.madeeasy.in

ESE-2016 : Electronics Engg.Solutions of Objective Paper-II | Set-A

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Paper-II (Electronics Engineering)

1.1.1.1.1. A single-stage amplifier has a voltage gain of 100. The load connected to the collector

is 500 and its input impedance is 1 k. Two such stages are connected in cascadethrough an R-C coupling. The overall gain is

(a) 10000 (b) 6666.66

(c) 5000 (d) 1666.66

Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)

A1

A2 VoVi

Voltage gain of A1decreases due to loading effect.

+

R01

A vv 1i Ri2V01

V01 =

=

+ +i

i

= 66.66

whereas AV2= 100

Overall gain, Av =AV1 AV2= 66.66 100 = 6666.66

2.2.2.2.2. Assuming VCE(Sat)= 0.3 V for a Silicon transistor at ambient temperature of 25C and

hFE= 50, the minimum base current IB required to drive the transistor into saturation

for the circuit shown is

+5 V

1 k

IB

Q

(a) 64 A (b) 78 A(c) 94 A (d) 140 A

Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)

IC sat =

=

IBmin =

= =

I


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1 2 3 4 5 6 7 8 9 0

Page4

3.3.3.3.3. Which of the following regions of operation are mainly responsible for heating of the

transistor under switching operation?

1. Saturation region

2. Cut-off region

3. Transition from saturation to cut-off

4. Transition from cut-off to saturation Select the correct answer using the codes given

below:

(a) 1, 2, and 4 only (b) 1, 3, and 4 only

(c) 2 and 3 only (d) 1 and 3 only


4.4.4.4.4. In a sinusoidal oscillator, sustained oscillations will be produced only if the loop gain

(at the oscillation frequency) is

(a) Less than unity but not zero (b) Zero

(c) Unity (d) Greater than unity


Oscillators have unity loop gain.

5.5.5.5.5. The Class-B push-pull amplifier is an efficient two-transistor circuit, in which the two

transistors operate in the following way :

(a) Both transistors operate in the active region throughout the negative ac cycle

(b) Both transistors operate in the active region for more than half-cycle but less than

a whole cycle

(c) One transistor conducts during the positive half-cycle and the other during the

negative half-cycle

(d) Full supply voltage appears across each of the transistors


6.6.6.6.6. Consider the following statements regarding Wien Bridge oscillator:

1. It has a larger bandwidth than the phase shift oscillator.

2. It has a smaller bandwidth than the phase shift oscillator.

3. It has 2 capacitors while the phase shift oscillator has 3 capacitors.

4. It has 3 capacitors while the phase shift oscillator has 2 capacitors.

Which of the above statements are correct?

(a) 1 and 3 only (b) 2 and 4 only


Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)


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1 2 3 4 5 6 7 8 9 0

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1 2 3 4 5 6 7 8 9 0

Page5

7.7.7.7.7. For normal operation of a transistor

(a) Forward bias the emitter diode and reverse bias the collector diode

(b) Forward bias the emitter diode as well as the collector diode

(c) Reverse bias the emitter diode as well as the collector diode

(d) Reverse bias the emitter diode and forward bias the collector diode


8.8.8.8.8. Consider the following statements regarding linear power supply:

1. It requires low frequency transformer.

2. It requires high frequency transformer.

3. The transistor works in active region. Which of the above statements is/are correct?

(a) 1 only (b) 2 and 3 only

(c) 1 and 3 only (d) 3 only


9.9.9.9.9. The capacitance of a full wave rectifier, with 60 Hz input signal, peak output voltage

Vp= 10 V, load resistance R= 10 k and input ripple voltage Vr = 0.2 V, is

(a) 22.7 F (b) 33.3 F(c) 41.7 F (d) 83.4 F


IDC

= =

Vr

=

I

C=

= =

I

10.10.10.10.10. A full wave rectifier connected to the output terminals of the mains transformer produces

an RMS voltage of 18 V across the secondary. The no-load voltage across the secondary

of the transformer is

(a) 1.62 V (b) 16.2 V

(c) 61.2 V (d) 6.12 V


VRMS =

=

Vm=


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Batches MADE EASY offers rank improvement batches for ESE 2017 & GATE 2017. These batches are designed for repeater students who havealready taken regular classroom coaching or prepared themselves and already attempted GATE/ESE Exams , but want to give next attempt

for better result. The content of Rank Improvement Batch is designed to give exposure for solving different types of questions within xed

time frame. The selection of questions will be such that the Ex. MADE EASY students are best benetted by new set of questions.

Eligibility :Features :

Syllabus Covered :Technical Syllabus of GATE-2017 & ESE-2017 Course Duration : Approximately 25 weeks (400 teaching hours)

Old students who have undergone classroom course from anycentre of MADE EASY or any other Institute

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Qualied in ESE written exam

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Ex. MADE EASY Students

Those students who were enrolled inPostal Study Course, Rank Improvement,

Conventional, G.S, Post GATE batches

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Students

Fee is inclusive of classes, study material and taxes

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Those students whowere enrolled in long termclassroom programs only

Batch

Rank Improvement Batch

Rank Improvement Batch

+ General Studies &

Engineering Aptitude Batch

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Documents required :M.Tech marksheet, PSUs/IES Interview call leer, GATE score card, MADE EASY I-card 2 photos + ID proof

Rank Improvement Batches will be conducted at Delhi Centre only.

Note: 1. These batches will be focusing on solving problems and doubt clearing sessions. Therefore if a student is weak in basic concepts& fundamentals then he/she is recommended to join regular classroom course.

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3. The course fee is designed without Study Material/Books, General Studies and Online Test Series (OTS) . However those subjects

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Page6

VDCNL

=

= =

11.11.11.11.11. An Op-Amp can be connected to provide

1. Voltage controlled current source

2. Current controlled voltage source

3. Current controlled current source Which of the above statements are correct?


(c) 2 and 3 only (d) 1, 2 and 3


12.12.12.12.12. In an Op-Amp, if the feedback voltage is reduced by connecting a voltage divider at

the output, which of the following will happen?

1. Input impedance increases

2. Output impedance reduces

3. Overall gain increases

Which of the above statements is/are correct?

(a) 1 only (b) 2 only

(c) 3 only (d) 1, 2 and 3


If feedback voltage is reduced then overall gain increases.

13.13.13.13.13. The transient response rise time (unity gain) of an Op-Amp is 0.05 s. The small signal

bandwidth is

(a) 7 kHz (b) 20 kHz

(c) 7 MHz (d) 20 MHz


Bandwidth =

=

14.14.14.14.14. A negative feedback of = 2.5 x 103is applied to an amplifier of open-loop gain 1000.What is the change in overall gain of the feedback amplifier, if the gain of the internal

amplifier is reduced by 20%?

(a) 295.7 (b) 286.7

(c) 275.7 (d) 266.7


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1 2 3 4 5 6 7 8 9 0

1 2 3 4 5 6 7 8 9 0

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Page7

Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)

A = 0.8 1000 = 800

New overall gain, Af =

+

Af =

=

+

15.15.15.15.15. If the quality factor of a single-stage single-tuned amplifier is doubled, the bandwidth

will

(a) Remain the same (b) Become half

(c) Become double (d) Become four times


Bandwidth =

Bandwidth becomes half.

16.16.16.16.16. Consider the following statements related to oscillator circuits :

1. The tank circuit of a Hartley oscillator is made up of a tapped capacitor and a common

inductor.

2. The tank circuit of a Colpitts oscillator is made up of a tapped capacitor and a

common oscillator.

3. The Wien Bridge oscillator is essentially a two-stage amplifier with an RC bridge

in the first stage, and, the second stage serving as an inverter.

4. Crystal oscillators are fixed frequency oscillators with a high Q-factor.


(a) 1, 2 and 3 only (b) 2, 3 and 4 only

(c) 1, 2 and 4 only (d) 1,3 and 4 only


17.17.17.17.17. The most commonly used transistor configuration for use as a switching device is

(a) Common-base configuration

(b) Common-collector configuration

(c) Collector-emitter shorted configuration

(d) Common-emitter configuration



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Page8

18.18.18.18.18. The value of hFE(the hybrid parameters) of a Common-Emitter (CE) connection of a Bipolar

Junction Transistor (BJT) is given as 250. What is the value of dc(ratio of collector currentto emitter current), for this BJT?

(a) 0.436 (b) 0.656

(c) 0.874 (d) 0.996


=

=

+

19.19.19.19.19. For realizing a binary half-subtractor having two inputs Aand B, the correct set of logical

expressions for the outputs D (Aminus B) and X (borrow) are

1. The difference output D=

+2. The borrow output B=



(c) Both 1 and 2 (d) Neither 1 nor 2


The truth table of Half subtractor is shown below

Difference = +

Borrow =

20.20.20.20.20. The simplified form of the Boolean expression AB + A(B + C) + B(B+ C) is given

by

(a) AB + AC (b) B + AC

(c) BC + AC (d) AB + C


f=AB + A(B + C) + B(B + C)

=AB + AB + AC + BB + BC

=AB + AC + B + BC

=AB + AC + B (1 + C)

=AB + AC + B

= B (A + 1) + AC

=B + AC


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Page9

21.21.21.21.21. Product of Maxterms representation for the Boolean function = + + is(a) M (1, 3, 5, 7) (b) M (0, 2, 4, 6)

(c) M (0,1, 2, 3) (d) M (4, 5, 6, 7)Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)

F= + +

13

5 7 61 1

11

BD

BD BD BD BD

A

A

A

0

4

2

F=m(1, 3, 5, 7)Converting the function from Minterms form to Maxterms form,

F=

M(0, 2, 4, 6)

22.22.22.22.22. Simplified form of the Boolean expression = + + + is

(a) + + + (b) + + + +

(c) + + (d) A (B+ C)


Y= + + +

= + + + +

= + + = + + +

=m(1, 2, 3, 4, 5, 6)=M(0, 7)

= (A + B + C) + +

23.23.23.23.23. What is the maximum frequency for a sine wave output voltage of 10 V peak with an

Op-Amp whose slew rate is 1 V/s?

(a) 15.92 kHz (b) 19.73 kHz(c) 23.54 kHz (d) 27.336 kHz


fo

fo

fo 15.92 kHz


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Page10

24.24.24.24.24. Which one of the following statements is correct?

(a) TTL logic cannot be used in digital circuits.

(b) Digital circuits are linear circuits.

(c) AND gate is a logic circuit whose output is equal to its highest input.

(d) In a four-input AND circuit, all inputs must be high for the output to be high.


In an AND gate, the output is high only when all inputs are high.

25.25.25.25.25. The slew rate is the rate of change of output. voltage of an operational amplifier when

a particular input is applied. What is that input?

(a) Sine wave input (b) Ramp input

(c) Pulse input (d) Step input


Slew rate of op-amp in measured by applying step input.

26.26.26.26.26. Except at high frequencies of switching, nearly all the power dissipated in the switch

mode operation of a BJT occurs, when the transistor is in the

(a) Active region (b) Blocking state

(c) Hard saturation region (d) Soft saturation region


27.27.27.27.27. Consider the following statements with respect to combinational circuit:

1. The output at any time depends only on the present combination of inputs.

2. It does not employ storage elements.

3. It performs an operation that can be specified logically by a set of Boolean functions.



(c) 2 and 3 only (d) 1, 2 and 3


A

CombiB

C

f

g

It has no storage present output depends only on present input.


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28.28.28.28.28. Consider the following statements : A multiplexer

1. selects one of the several inputs and transmits it to a single output.

2. routes the data from a single input to one of many outputs.

3. converts parallel data into serial data.

4. is a combinational circuit.



(c) 1,3 and 4 only (d) 2, 3 and 4 only


Data is routed from a single input to one of many outputs in a demultiplexer and not

in a multiplexer.

29.29.29.29.29. What are the two types of basic adder circuits?

(a) Half adder and full adder

(b) Half adder and parallel adder

(c) Asynchronous adder and synchronous adder

(d) One's complement adder and two's complement adder


A half adder adds 2 bits.

A full adder adds 3 bits.

These are the two types of basic adder circuits.

30.30.30.30.30. Consider the following statements :

1. An 8-input MUX can be used to implement any 4 variable functions.

2. A 3-line to 8-line DEMUX can be used to implement any 4 variable functions.

3. A 64-input MUX can be built using nine 8-input MUXs.

4. A 6-line to 64-line DEMUX can be built using nine 3-line to 8-line DEMUXs.


(a) 1, 2, 3 and 4 (b) 1, 2 and 4 only

(c) 3 and 4 only (d) 1, 2 and 3 only


An 8-input MUX can implement only some functions of 4 variables and not all functions

of 4 variables.

31.31.31.31.31. For ann-bit binary adder, what is the number of gates through which a carry has to

propagate from input to output?

(a) n (b) 2n

(c) n2 (d) n + 1


17/52


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EC

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Nishant Bansal

ME

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EE

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EE

3AIR

Piyush Kumar

CE

4AIR

Sumit Kumar

ME

8AIR

Stuti Arya

EE

10AIR

Brahmanand

CE

10AIR

Rahul Jalan

CE

10AIR

9 in Top 10 58 in Top 100

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CE

3AIR

Anas Feroz

EE

6AIR

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CE

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Arun Kumar

ME

10AIR

6 in Top 10 Total 47 selections

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In a single adder, the carry propagates through 2 gates.

in an n-bit binary adder, the carry would propagate through 2n gates.

32.32.32.32.32. The main disadvantage of DTL logic circuits is

(a) Medium speed (b) Very large power supply voltage

(c) High cost (d) Very large gate propagation delay


The propagation delay of DTL logic circuits is about 30 ns, which is relatively quite large.

33.33.33.33.33. Which one of the following statements best describes the operation of a negative-edge-triggered D flip-flop?

(a) The logic level at the Dinput is transferred to Qon NGT of CLK.

(b) The Qoutput is always identical to the CLK input if the D input is high.

(c) The Qoutput is always identical to the D input when CLK = PGT.

(d) The Qoutput is always identical to the D input.


D QInput

clk

Q+ =D

The logic level at the D input is transferred at negative edge.

D

Q

34.34.34.34.34. A 3-bit ripple counter is constructed using three Tflip-flops to do the binary counting.

The three flip-flops have T-inputs fixed at

(a) 0, 0 and 1 (b) 1, 0 and 1

(c) 0, 1 and 1 (d) 1, 1 and 1


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Page13


In a ripple counter, the T inputs of all T flip flops are always applied logic 1.

35.35.35.35.35. What is the function Y= A+ in Product-of-Sums (POS) form?

(a) M6 M5 M4 M3 (b) M3M2M4 M0(c) M0M2 M3 (d) M4M3M2 M1


Y= +

the function is plotted on K map below,

01

3 2

4 5 7 61 1

1

BC

BC BC BC BC A

A

A

1 1

Y=m(1, 4, 5, 6, 7)Converting the function from SOP to POS form

Y=M(0, 2, 3)=M

0 M

2 M

3

36.36.36.36.36. The initial content of a four-bit shift register is 1000. What is the register content after

it is shifted four times to the right, with the serial input being 111100?(a) 1111 (b) 1100

(c) 1000 (d) 0011


1 1 1 1

0

0

1

1

1 0

1

0

0

1

0 0

0

1

0

0

0 0

0

0

1

0

37.37.37.37.37. When a large number of analog signals is to be converted to digital form, an analog

multiplexer is used. The A-to-D converter most suitable in this case will be

(a) Forward counter type (b) Up-down counter type

(c) Successive approximation type (d) Dual slope type


Conversion time of successive approximation ADC is less than up counter type, up-

down counter and dual slope ADC.


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Page14

38.38.38.38.38. For Emitter-Coupled Logic (ECL), the switching speed is very high because

(a) Negative logic is used

(b) The transistors are not saturated when they are conducting

(c) Multi-emitter transistors are used

(d) Of low fan-out


In ECL, the transistors operate only in Active and Cut off regions.

39.39.39.39.39. A flip-flop is a

(a) Combinational logic circuit and edge sensitive

(b) Sequential logic circuit and edge sensitive

(c) Combinational logic circuit and level sensitive

(d) Sequential logic circuit and level sensitive


Flip-flop is sequential and edge triggered.

40.40.40.40.40. The transfer function

+will have

(a) dc gain 1 and high frequency gain 1

(b) dc gain 0 and high frequency gain (c) dc gain 1 and high frequency gain 0

(d) dc gain 0 and high frequency gain 1


T F=

+

M=

+

M at = 0 is 1 and at = is 0.

41.41.41.41.41. The closed-loop transfer function of a certain control system is given by

=

+ +. Then the settling time for a 2% tolerance band is given by

(a) 0.8 s (b) 1.2 s

(c) 1.5 s (d) 2.1 s


21/52

MADE EASY Students Top in GATE-2016

METop 10

9in

A I R Nishant Bansal Gaurav Sharma

AIR-2

Manpreet Singh

AIR-2

Sayeesh T. M.

AIR-4

Aakash Tayal

AIR-5

Tushar

AIR-7

Sumit Kumar

AIR-8

Amarjeet Kumar

AIR-6

Suman Dutta

AIR-10

EETop 10

10in

Arvind Biswal

AIR-2

Udit

AIR-3

Keshav

AIR-4

Tanuj Sharma

AIR-5

Arvind

AIR-6

Rajat Chaudhary

AIR-7

Sudarshan

AIR-8

Rohit Agarwal

AIR-9

Stuti Arya

AIR-10

A I R Anupam Samantaray

CETop 10

8in

Piyush Kumar Srivastav

AIR-4

Roopak Jain

AIR-6

Jatin Kumar Lakhmani

AIR-8

Vikas Bijarniya

AIR-8

Brahmanand

AIR-10

Rahul Jalan

AIR-10

Rahul SIngh

AIR-3

Kumar ChitranshA I R

ECTop 10

6in

K K Sri Nivas

AIR-2

Jayanta Kumar Deka

AIR-3

Amit Rawat

AIR-5

Pillai Muthuraj

AIR-6

Saurabh Chakraberty

AIR-10

A I R Agam Kumar Garg

INTop 10

9in

A I R Harshvardhan Sinha Avinash Kumar

AIR-2

Shobhit Mishra

AIR-3

Ali Zafar

AIR-4

Rajesh

AIR-7

Chaitanya

AIR-8

Palak Bansal

AIR-10

Saket Saurabh

AIR-10

Shubham Tiwari

AIR-8

53Selections in Top 101 Ranks inst

ME, EE, EC, CS, IN & PI 368 Selections in Top 10096 Selections in Top 20

METop 20 Top 100

17Selections Selections

68

CETop 20 Top 100


65

ECTop 20 Top 100


45

EETop 20 Top 100


76

CS & PITop 20 Top 100


53

INTop 20 Top 100


61

CSTop 10

6in

Himanshu Agarwal

AIR-4

Jain Ujjwal Omprakash

AIR-6

Nilesh Agrawal

AIR-10

Sreyans Nahata

AIR-9

Debangshu Chatterjee

AIR-2

A I R Ankita Jain

Akash Ghosh

AIR-4

Niklank Kumar Jain

AIR-7

Shree Namah Sharma

AIR-8

Agniwesh Pratap Maurya

AIR-8

PITop 10

5

A I R Gaurav Sharma

in

etailed results are available at

www m dee sy in

MADE EASY is the only institute which has consistently produced toppers in ESE & GATE


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Page15


T F=

+ +2n= 10 n= 5

ts (2%) =

= =

42.42.42.42.42. The unit step input response of a certain control system is given by

c(t) = 1 + 0.2 e60t 1.2 e10t. Then the undamped natural frequency nand damping

ratio are, respectively

(a) 24.5 and 1.27 (b) 33.5 and 1.27(c) 24.5 and 1.43 (d) 33.5 and 1.43


SR= 1 + 0.2 e60t 1.2 e10t

From SR the poles are at 10, 60

Hence q(s) = (S+ 10) (S + 60) = 0

q(s) =S2 + 70S + 600 = 0

Wn=

2n

= 70

= 1.43

43.43.43.43.43. In Force-Voltage Analogy

(a) Force is analogous to current

(b) Mass is analogous to capacitance

(c) Velocity is analogous to current

(d) Displacement is analogous to magnetic flux linkage


As per theory of analogy velocity is analogous to current.

44.44.44.44.44. For a unity feedback control system having an open-loop transfer function

=+

,

what is the time tp at which the peak of the step input response occurs?

(a) 0.52 s (b) 2.75 s

(c) 0.79 s (d) 1.57 s


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Page16


GH(s) =

+

q(s) = 1 + GH(s) = S2 + 6S + 25 = 0

n= 5 ; = 0.6 d= 4

tp=

=

45.45.45.45.45. The transfer function

=

+ represents

(a) Unstable system (b) Minimum phase system

(c) Non-minimum phase system (d) PID controller system


GH(s) =

+

If a system has at least a zero (or) a pole in right side of Splane then it is called Non-

minimum phase system.

46.46.46.46.46. What is the maximum input frequency limit of a 3-bit Ripple counter configured aroundflip-flops, with inherent propagation delay time tpd= 50 ns?

(a) 6670 MHz (b) 667 MHz

(c) 66.7 MHz (d) 6.67 MHz


fCLK =

= =

47.47.47.47.47. The characteristic equation of a certain feedback control system is given by

s4+ 4s3+ 13s2+ 36s + k= 0. The range of values of k for which the feedback system

is stable, is given by

(a) 0


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q(s) =s4 + 4s3 + 13S2 + 36S + K = 0

For stability K > 0 and K < 36

0 < K < 36

48.48.48.48.48. The closed-loop transfer function of a unity feedback control system is,

=

+ + . The velocity error constant of the system is

(a)

(b)

(c)

(d)


T F=

+ +

OLTFGH(s) =

+

Velocity error constant =Kv =

Kv=

=

+

49.49.49.49.49. The system described by the following state equations

[ ]

= + =

1. Completely controllable

2. Completely observable





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MADE EASY Students TOP in ESE-2015

MADE EASY selections in ESE-2015 : 82% of Total Vacancies

ME Selections in Top 10 10 18 83 99 83%Selections in Top 20 MADE EASY Selections Out of Vacancies MADE EASY Percentage

CE 10 20 120 151 79%Selections in Top 10 Selections in Top 20 MADE EASY Selections Out of Vacancies MADE EASY Percentage

EE 9 16 67 86 78%Selections in Top 10 Selections in Top 20 MADE EASY Selections Out of Vacancies MADE EASY Percentage

E T 9 19 82 98 84%& Selections in Top 10 Selections in Top 20 MADE EASY Selections Out of Vacancies MADE EASY Percentage

352Selections

out of total

434

73in

Top 20

38in

Top 10

4 StreamsToppers4

All 4 MADE EASY

Students

Ashok Bansal

AIR 2

Aarish Bansal

AIR 3

Vikalp Yadav

AIR 4

Naveen Kumar

AIR 5

Raju R N

AIR 6

Sudhir Jain

AIR 7

Bandi Sreenihar

AIR 8

Kotnana Krishna

AIR 9

Arun Kr. Maurya

AIR 10

AIR

PratapME

Partha Sarathi T.

AIR 2

Anas Feroz

AIR 6

Amal Sebastian

AIR 7

Vishal Rathi

AIR 10

Sudhakar Kumar

AIR 9AIR 8

Dharmini SachinNagendra Tiwari

AIR 5

Nikki Bansal

AIR 3

S. Siddhikh Hussain

AIR

EE

Siddharth S.

AIR 3

Piyush Vijay

AIR 4

Kumbhar Piyush

AIR 7

Shruti Kushwaha

AIR 9

Anurag Rawat

AIR 10

Nidhi

AIR 8

Saurabh Pratap

AIR 2

Manas Kumar Panda

AIR 5

Ijaz M Yousuf

AIR

E&T

Piyush Pathak

AIR 2

Amit Kumar Mishra

AIR 3

Amit Sharma

AIR 4

Dhiraj Agarwal

AIR 5

Pawan Jeph

AIR 6

Kirti Kaushik

AIR 7

Aman Gupta

AIR 8

Mangal Yadav

AIR 9

Aishwarya Alok

AIR 10

Palash Pagaria

AIR

CE

ME Top 10Selections in

10

EETop 10

Selections in

9

E&TTop 10

Selections in

9

CETop 10

Selections in

10

etailed results are available at

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MADE EASY is the only institute which has consistently produced toppers in ESE & GATE


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=

+

Y= [ ]

QC=

[ ]

= =

For controllability 0

Q0 =

=

= 4 0

For observability 0

50.50.50.50.50. Consider the system with

+=

+ +and H(s) = 1. The breakaway point(s) of

the root loci is/are at

(a) 0.265 only (b) 3.735 only

(c) 3.735 and 0.265 (d) There is no breakaway point


GH(s) =

++ +

Breakaway point is solution of

that lies on root locus.

Solution of

are

= 3.73 lies on root locus

51.51.51.51.51. How would a binary number 0010 be represented by a 4-bit binary word, if the range

of voltage is 0 to 10 V?

(a) 0.666 V (b) 1.333 V

(c) 0.333 V (d) 2000 V


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Page19


VFS= 10 V

n= 40010 D = 2

R=

= =

Analog output =R D= 0.66667 2= 1.333 volts

52.52.52.52.52. For a unity feedback system with open-loop transfer function

+, the resonant peak

at output Mmand the corresponding resonant frequency m, are respectively(a) 2.6 and 2.67 r/s (b) 1.04 and 2.67 r/s

(c) 2.6 and 4.8 r/s (d) 1.04 and 4.8 r/s


GH(s) =

+

q(s) = 1 + GH(s) = s2 + 6s + 25 = 0

n= 5 : = 0.6

Wr = =

Mr

=

=

53.53.53.53.53. The transfer function of a control system is said to be All Pass System, if it has

(a) Unit magnitude at all frequencies with anti-symmetric pole-zero pattern

(b) Unit magnitude at all frequencies with symmetric pole-zero pattern

(c) Magnitude varying with frequency and with anti-symmetric pole-zero pattern

(d) Unit magnitude at some frequencies with symmetric pole-zero pattern


All pass system has constant magnitude response for all frequencies. This can happen

only if poles and zeros are symmetric w.r.t. j-axis.


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Page20

54.54.54.54.54. Consider the following:

1. Bode plot 2. Nyquist plot 3. Nichols chart

Which of the above frequency response plots are commonly employed in the analysisof control systems?


(c) 2 and 3 only (d) 1, 2 and 3


All the mentioned plots are popular and commonly used in control analysis.

55.55.55.55.55. An A-to-D converter in which one sub-circuit is a D-to-A converter is

(a) Parallel A/D converter

(b) Dual slope A/D converter

(c) Successive approximation A/D converter

(d) Extended parallel type A/D converter


Successive approximation ADC

+

Controlcircuit

SAR

DAC

Start EOC

Va

56.56.56.56.56. Consider the transfer function :

+ +=

+ +The corner frequencies in Bode's plot for this transfer function are as

(a) 10 r/s and 10 r/s (b) 100 r/s and 10 r/s

(c) 10 r/s and 1 r/s (d) 100 r/s and 1 r/s


GH(s) =

++ +

Corner frequency in Bode plot is defined for finite poles and zeros, which are complex

in given system.

For complex pole, zeros corner frequency.

Hence 10, 1.


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57.57.57.57.57. Consider the transfer function (0.1 + 0.1s) for a PD controller. What is the frequency at

which the magnitude is 20 dB (by using asymptotic Bodes plot)?

(a) 2000 r/s (b) 1000 r/s

(c) 200 r/s (d) 100 r/s


GH(s) = [0.1 + 0.01s] = 0.1 [1 + 0.1s]

Bode plot is

20 dB

0 dB

20 dB

1 10 100 1000

+20 dB

decade

58.58.58.58.58. The main objectives of drawing the root-locus plot are

1. To obtain a clear picture of the open-loop poles and zeros of the system.

2. To obtain a clear picture of the transient response of the system for varying gain, K.

3. To find the range of K to make the system stable.


(a) 1, 2 and 3 (b) 1 and 2 only



Root locus obtained mainly to obtain response and stability of system.

59.59.59.59.59. A unity feedback system has open-loop poles at s= 2 j2, s= 1 and s= 0 and a

zero at s= 3. What are the angles made by the root-loci asymptotes with the real axis?

(a) 60, 180 and 60 (b) 30, 90 and 60

(c) 60, 120 and 30 (d) 30, 60 and 180


Total number of poles = 4

Total number of zeros = 1

Angles of asymptotes =

+

K= 0, 1, 2

Angles = 60, 180, 300 (or) 60


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60.60.60.60.60. The open-loop transfer function of a unity feedback system is

=

+. The

gain K that results in a phase margin of 45 is(a) 35 (b) 30

(c) 25 (d) 20


GH(s) =

+q(s) = 1 + GH(s) = S2 + 5S + K= 0

n= ; =

PM = 100 =

= K 35


The Gain margin and Phase margin of an unstable system may respectively be

1. Positive, negative 2. Negative, positive 3. Negative, negative


(a) 3 only (b) 1 and 2 only

(c) 2 and 3 only (d) 1, 2 and 3

Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)GM and PM of unstable system are always negative

62.62.62.62.62. A phase lead compensator has its transfer function,

+=

+. The maximum

phase lead and the corresponding frequency, respectively are nearly

(a) sin1 (0.9) and 6 r/s (b) sin1(0.82) and 4 r/s

(c) sin1 (0.9) and 4 r/s (d) sin1(0.82) and 6 r/s


Gc(s) =

++

Zero : S= 2; pole : S = 20 ; =

=

M= [ ]

= +

M= =


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1. Lead compensation decreases the bandwidth of the system.

2. Lag compensation increases the bandwidth of the system.





Lead compensator is high pass filter hence it increases bandwidth Lag compensator

is low pass filter hence it decreases bandwidth.

64.64.64.64.64. A proportional controller with transfer function, Kpis used with a first-order system having

its transfer function as

=+

, in unity feedback structure. For step inputs, an

increase in Kpwill

(a) Increase the time constant and decrease the steady state error

(b) Decrease the time constant and decrease the steady state error

(c) Decrease the time constant and increase the steady state error

(d) Increase the time constant and increase the steady state error


GH(s) =

=

+

CLTF =

+=

+ + +

time constant =

+

ess =

If KP is increased then both time constant and essdecreases.


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65.65.65.65.65. For a second-order differential equation, if the damping ratio , is unity, then(a) The poles are imaginary and complex conjugate

(b) The poles are in the right half of s-plane

(c) The poles are equal, negative and real

(d) Both the poles are unequal, negative and real


= 1 means critically damped.Hence, roots are real, equal, negative.

66.66.66.66.66. Consider the following statements associated with microstrip patch antenna :

1. The microstrip patch behaves more like a leaky cavity rather than like a radiator and

this is not a highly efficient antenna.

2. They can be adapted for radiation of circularly polarized waves.





67.67.67.67.67. A carrier waveform 10 cos ctand modulating signal 3 cos mthave fc= 100 kHz andfm

= 4 kHz. Given that sensitivity of FM is 4 kHz/V and FM spectra beyond J6 is

negligible, what are the channel bandwidth requirements forAMand FM, respectively?

(a) 12 kHz and 48 kHz (b) 8 kHz and 48 kHz

(c) 12 kHz and 24 kHz (d) 8 kHz and 24 kHz


AM BW = 2 fm

= 8 kHz

FM BW = 12 fm

(significant sidebands upto 6th order, so bandwidth 12 fm

)

= 48 kHz

68.68.68.68.68. When the modulating frequency is doubled, the modulation index is halved, and the

modulating voltage remains constant. The modulation system is

1. Amplitude modulation

2. Phase modulation

3. Frequency modulation

Select the correct answer from the codes given below :


(c) 3 only (d) 1, 2 and 3


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For FM; =

If modulating frequency doubled, then modulation index is half.

69.69.69.69.69. What is the modulation index of anFMsignal having a carrier swing of 100 kHz and

modulating frequency of 8 kHz?

(a) 4.75 (b) 5.50

(c) 6.25 (d) 7.50


Carrier swing

2f= 100 kHzf= 50 kHz

=

= =

70.70.70.70.70. In a Pulse Code Modulated system, the number of bits is increased from 7 to 8 bits.

The improvement in signal to quantization noise ratio will be

(a) 2 dB (b) 4 dB

(c) 6 dB (d) 8 dB


If k bits increased, then SQNR is increased by 6 k dB.

if one bit is increased, then SQNR is increased by 6 dB

71.71.71.71.71. In the process of modulation

(a) Some characteristics of a high frequency sine wave are varied in accordance with

the instantaneous value of a low frequency signal

(b) Parameters of carrier wave are held constant

(c) For proper and efficient radiation, the receiving antennas should have heights

comparable to half-wavelength of the signal received

(d) The signal is converted first within the range of 10 Hz to 20 Hz



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72.72.72.72.72. If the sampling is carried out at a rate higher than twice the highest frequency of the

original signal (fmax), then it is possible to receive the original signal from the sampled

signal by passing it through

(a) A high-pass filter with the cut-off frequency equal to fmax(b) A low-pass filter with the cut-off frequency equal to fmax(c) A high-pass filter with the cut-off frequency greater than fmax(d) A low-pass filter with the cut-off frequency greater than fmax


In sampling process with sampling frequency greater than nyquist frequency for

reconstruction of original signal, LPF is used.

73.73.73.73.73. The open-loop transfer function of a unity feedback system is

++ The

phase shift at = 0 and = , will be, respectively(a) 90 and 180 (b) 0 and 180

(c) 90 and 90 (d) 0 and 0


GH(s) = [ ]

++

type = 0 ; P Z = 0

Phase at = 0 is 90 type = 0phase at = is 90 (P Z) = 0

74.74.74.74.74. The conversion time for a 10-bit successive approximationA/Dconverter, for a clock

frequency of 1 MHz is

(a) 1 s (b) 5 s(c) 10 s (d) 15 s


fCLK = 1 MHz

TCLK

= 1 sIn successive approximation ADC, the conversion time is always constant, and is equal

to number of bits in ADC. Tconv.

= 10 1 s = 10 s.

75.75.75.75.75. The minimum bandwidth of the link needed for a guard band of 10 kHz frequency to

prevent interference between six channels, each with 100 kHz frequency, is

(a) 425 kHz (b) 575 kHz

(c) 650 kHz (d) 725 kHz


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6 signals multiplexed. So 5 guard bands will exist

BW = 6 100 kHz + 5 10 kHz= 650 kHz

76.76.76.76.76. The different access methods which permit many satellite users to operate in parallel

through a single transponder without interfering with each other are

1. Frequency Division Multiple Access (FDMA)

2. Time Division Multiple Access (TDMA)

3. Code Division Multiple Access (CDMA)

Which of the above are correct?


(c) 2 and 3 only (d) 1, 2 and 3


In TDMA, different users access the transponder in different time slots, not parallely.

77.77.77.77.77. In an optical fibre, the pulse dispersion effect is minimized by

1. Using a high frequency light source

2. Using plastic cladding

3. Minimizing the core diameter


(a) 1 only (b) 2 only(c) 3 only (d) 1, 2 and 3


78.78.78.78.78. Consider the following statements:

As compared to short-circuited stubs, open-circuited stubs are not preferred because

the latter

1. Are of different characteristic impedance

2. Have a tendency to radiate Which of the above statements is/are correct?

(a) 1 only (b) 2 only(c) Both 1 and 2 (d) Neither 1 nor 2


79.79.79.79.79. Consider the following statements for multiple access system in a satellite earth station:

1. Access to same repeater sub-systems and same RF channel is possible.

2. Frequency division multiple access is used.

3. Several carriers are not amplified by same TWT.


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(c) 1 and 2 only (d) 1, 2 and 3


80.80.80.80.80. The Bode plot of the open-loop transfer function of a system is described as follows:

Slope 40 dB/decade < 0.1 rad/s

Slope 20 dB/decade 0.1 < < 10 rad/s

Slope 0 > 10 rad/s

The system described will have

(a) 1 pole and 2 zeros (b) 2 poles and 2 zeros

(c) 2 poles and 1 zero (d) 1 pole and 1 zero


As per given details Bode plot is

40

200

Initial slope indicates 2 poles at origin.

Final slope indicates

P= Z P= Z = 2

81.81.81.81.81. From the Nichols chart, one can determine the following quantities pertaining to a closed-

loop system:

(a) Magnitude, bandwidth and phase (b) Bandwidth and phase only

(c) Magnitude and phase only (d) Bandwidth only


Nichols chart provides complete frequency response of a system.

82.82.82.82.82. In position control systems, the Tacho-generator feedback is used to

(a) Increase the effective damping in the system

(b) Decrease the effective damping in the system

(c) Decrease the steady state error

(d) Increase the steady state error


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Tacho generator (or) derivative controller mainly used to increase system damping

new =old+

KD

= tachometer constant


1. The pin diode consists of two narrow, but highly doped, semiconductor regions

separated by a thicker, lightly doped material called the intrinsic region.

2. Silicon is used most often for its power-handling capability and because it provides

a highly resistive intrinsic region.

3. The pin diode acts as an ordinary diode at frequencies above 100 MHz.



(c) 2 and 3 only (d) 1, 2 and 3



1. Additional cavities serve to velocity-modulate the electron beam and produce an

increase in the energy available at the output.

2. The addition of intermediate cavities between the input and output cavities of the

basic klystron greatly improves the amplification, power output, and efficiency ofthe klystron.





85.85.85.85.85. In a waveguide with perfectly conducting flat wall, the angle of reflection is equal to the

angle of

(a) Diffraction (b) Incidence(c) Refraction (d) Penetration



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86.86.86.86.86. In microwave system, waveguides have the advantages of

(a) High power-handling capability and low loss

(b) Thin dielectric substrate

(c) Low power-handling and adequate stability

(d) Positive phase shift


87.87.87.87.87. A straight dipole radiator fed in the centre will produce maximum radiation at

1. The plane parallel to its axis

2. The plane normal to its axis

3. Extreme ends





88.88.88.88.88. In communication systems, modulation is the process of

(a) Improving frequency stability of transmitter

(b) Combining signal and radio frequency waves

(c) Generating constant frequency radio waves

(d) Reducing distortion in RF waves


Modulation is the process of transmitting low frequency message signal by combining

with high frequency carrier signal.

89.89.89.89.89. Which one of the following statements is correct?

(a) Sampling and quantization operate in amplitude domain.

(b) Sampling and quantization operate in time domain.

(c) Sampling operates in time domain and quantization operates in amplitude domain.

(d) Sampling operates in amplitude domain and quantization operates in time domain.


Sampling is done in time domain and quantization is done is amplitude domain.

90.90.90.90.90. What is the voltage attenuation provided by a 25 cm length of waveguide having a= 1 cm

and b= 0.5 cm in which a 1 GHz signal is propagated in the dominant mode?

(a) 721 dB (b) 681 dB

(c) 521 dB (d) 481 dB


43