ANSWERS - IES Masteriesmaster.org/public/archive/2016/IM-1481464255.pdfH = 1 ne where n is the carrier-concentration, A B R R = 21 B 21 A n 4 10 4 :1 n 1 10 50. (d) The conductivity

ESE-2017 PRELIMS TEST SERIESDate: 11th December, 2016

1. (b)

2. (a)

3. (a)

4. (c)

5. (d)

6. (b)

7. (c)

8. (d)

9. (d)

10. (c)

11. (a)

12. (b)

13. (d)

14. (a)

15. (c)

16. (b)

17. (c)

18. (c)

19. (a)

20. (a)

21. (d)

22. (c)

23. (d)

24. (b)

25. (b)

26. (b)

27. (c)

28. (c)

29. (a)

30. (c)

31. (a)

32. (a)

33. (a)

34. (b)

35. (d)

36. (c)

37. (b)

38. (d)

39. (d)

40. (a)

41. (b)

42. (a)

43. (b)

44. (a)

45. (b)

46. (a)

47. (d)

48. (c)

49. (d)

50. (d)

51. (c)

52. (b)

53. (a)

54. (b)

55. (c)

56. (d)

57. (c)

58. (b)

59. (d)

60. (a)

61. (c)

62. (d)

63. (b)

64. (b)

65. (c)

66. (b)

67. (c)

68. (b)

69. (c)

70. (b)

71. (c)

72. (c)

73. (a)

74. (d)

75. (a)

76. (b)

77. (d)

78. (a)

79. (a)

80. (b)

81. (c)

82. (d)

83. (c)

84. (b)

85. (a)

86. (d)

87. (b)

88. (b)

89. (c)

90. (b)

91. (b)

92. (a)

93. (d)

94. (a)

95. (c)

96. (c)

97. (b)

98. (c)

99. (b)

100. (b)

101. (b)

102. (d)

103. (b)

104. (a)

105. (d)

106. (c)

107. (c)

108. (c)

109. (c)

110. (d)

111. (b)

112. (c)

113. (c)

114. (b)

115. (c)

116. (d)

117. (a)

118. (b)

119. (d)

120. (b)

ANSWERS

121. (a)

122. (c)

123. (a)

124. (d)

125. (a)

126. (b)

127. (b)

128. (a)

129. (b)

130. (c)

131. (d)

132. (c)

133. (a)

134. (d)

135. (a)

136. (a)

137. (b)

138. (c)

139. (b)

140. (a)

141. (c)

142. (a)

143. (b)

144. (c)

145. (c)

146. (d)

147. (b)

148. (d)

149. (d)

150. (d)

151. (d)

152. (a)

153. (a)

154. (b)

155. (c)

156. (a)

157. (a)

158. (b)

159. (b)

160. (b)

IES M

ASTER

Office : Phone : F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 011-26522064

8130909220, 9711853908 [email protected], [email protected]. : E-mail:

(Test-15 Solution) 11th December 2016 (3)

1. (b)2. (a)3. (a)

4. (c)5. (d)6. (b)7. (c)

8. (d)9. (d)10. (c)

11. (a)12. (b)13. (d)14. (a)

15. (c)16. (b)17. (c)

There are five interrupt sources for the 8051,which means that they can recognize 5different events that can interrupt regularprogram execution. Each interrupt can beenabled or disabled by setting bits of the IEregister. Likewise, the whole interrupt systemcan be disabled by clearing the EA bit ofthe same register.

18. (c)19. (a)

20. (a)21. (d)22. (c)23. (d)

24. (b)25. (b)26. (b)

27. (c)The ring counter utilizes one flip-flop for eachstate in its sequence. So, for modulus-12ring counter i.e. for 12 different output stateit needs 12 flip-flops.

28. (c)

In a Johnson counter, the complement ofthe output of the last flip-flop is connectedback to the input of the first flip-flop.

CLKQ0 Q1 Q2 Q3

0 1 1 1CLK5

0 0 1 16

i.e. 0011

29. (a)

Case 1: When Qn = 0

D Qn T Q n+1 (Required for D FF)

0

1

0

0

0

1

0

1

(i.e. Q )n

(i.e. Q )–n

Case 2: When Qn = 1

D Qn T Q n+1

0

1

1

1

1

0

(i.e. Q )–n0

1 (i.e. Q )n

So, T = n nDQ DQ

= nD Q

i.e. required gate is EXOR gate.

30. (c)

• A ripple counter counts in a straight binarysequence. So n FFs will have 2n outputconditions.

• Modulus or, Mod of a counter is the totalnumber of states through which thecounter can progress.

• For ripple counter, Mode-N = 2n

i.e. 16 = 2n

n = 4• But for ring counter, number of different

states is equal to number of Flip-flopsMode-N = n = 16

Where, n is the number of FFs required formode-N counter.

31. (a)

Wired Logic: If the output of the gates areconnected together as shown in figure below,additional logic is performed withoutadditional hardware. This type of connectionis referred to as wired logic e.g. wired -AND

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(4) (Test-15 Solution) 11th December 2016

AB

CD

Y1

Y2

Y = Y Y1 2

TTL logic gates with open-collector outputcan be used for wired-AND operation.

32. (a)

ECL logic family has lowest propagation delaytime because the transistors are used indifference amplifier configuration, in whichthey are never driven into saturation andthereby the storage time is eliminated.

33. (a)

Since, VNH = VOH(min) – VIH(min)

= 2.4 – 2= 0.4 V

and VNL = IL(max) OL maxV V

= 0.8 – 0.4= 0.4 V

3.3V

2V

0.8V

0VInput

VIH(min)

VIL(max)

Output

3.3V2.4V

0.4V0V

VOH(min)

VOL(max)

34. (b)

A BA B T T Y0 0 OFF OFF 10 1 OFF ON 01 0 ON OFF 01 1 ON ON 0

So NOR gate

35. (d)

Direct memory access (DMA) is a feature ofcomputerized systems that allows certainhardware subsystems to access main systemmemory independently of the centralprocessing unit (CPU).Without DMA, whenthe CPU is using programmed input/output,it is typically fully occupied for the entireduration of the read or write operation, andis thus unavailable to perform other work.With DMA, the CPU first initiates the transfer,then it does other operations while the

transfer is in progress, and it finally receivesan interrupt from the DMA controller whenthe operation is done. This feature is usefulat any time that the CPU cannot keep upwith the rate of data transfer, or when theCPU needs to perform useful work whilewaiting for a relatively slow I/O data transfer.DMA is the fastest mode of data transfer.

36. (c)

The advantage of immediate addressing likeMVI B, 2EH is immediate execution ofinstruction. 8085 microprocessor has to doother operations like decoding and fetchingin immediate addressing also.

37. (b)

Handshake and interruption method I/Ooperations can be done. DMA is for memoryoperations.

38. (d)LDA address = 3 bytes

ORI bytes = 2 bytesMOV A, B = 1 bytes

39. (d)

In a 8085 micro processor, On receiving aninterrupt from an I/O device, the CPUBranches off to the interrupt service routineafter completion of current instruction.

40. (a)clock frequency = 10MHz

Time required to execute 1T-state is t = 1 fTime required to execute 18 T-states is

T = 18 t

= 618 1

10 10

= 1.8 s

41. (b)

16-bit register pairs BC, DE, HL formedfrom 8 bit registers.

42. (a)

Interrupt type Trigger Priority Maskable Vector

addressTRAP

RST 7.5

RST 6.5

RST 5.5

INTR

Edge

Level

Level

Level

1st

2nd

3rd

4th

5th

No

Yes

Yes

Yes

Yes

0024H

003CH

0034H

002CH

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43. (b)

The total number of memory locations for an-bit memory address register is 2n

216 = 64 K bytes

44. (a)

XRA A instruction execution resets contentsof accumulator and hence sets zero flag.

45. (b)Insulator has Eg 5 eVSemiconductor has Eg 1 eV

46. (a)Conductor: Valence band and conductionband overlaps to each other.Semiconductor:Small gap between valanceband and conduction band.Insulator: Wider gap between valence bandand conduction band.

47. (d)• At absolute zero temperature (i.e. –

273°C), no free electrons is available inthe intrinsic semiconductor and so itbehaves like a perfect insulator. But asthe temperature is increased, electronsbreaks away from the atom and movesfrom the valence band to conductionband. Hence conductivity increases.

• As the semiconductor is doped, thematerials gets either free electrons or,holes for conduction and henceconductivity increases.

• As the mobility of electrons is higher thanthat of holes, the conductivity of n-typeof material is higher. Because n-type ofmaterials have electrons as a charge-carrier.

48. (c)

For any intrinsic semiconductor, the intrinsicconcentration is directly proportional toT3/2. As conductiv ity of the instrinsicsemiconductor is directly proportional to

intrinsic concentration. So, 3/2T .

49. (d)

Half-coefficient of semiconductor materials

RH =1ne

where n is the carrier-concentration,

A

B

RR =

21B

21A

n 4 10 4 :1n 1 10

50. (d)The conductivity at any given temperature(except at absolute zero) is due to bothelectrons and holes, and it is given by

i = i e hn e

=e h2

ie h

n e m m

where

i = conductivity of intrinsicmaterial

e h, = relaxation time forelectrons and holes

e hm ,m = mass of electron andholes

e h, = mobility of electron andholes

51. (c)

• The semiconductor in its purest form iscalled intrinsic semiconductor e.g. Si, Ge

• When some impurity is added to theintrinsic semiconductor, i t is calledextrinsic semiconductor.

• N-type semiconductor and P-typesemiconductor are classified on the basisof type of impurity added.

• If impurity is a donor (or Penta valent)e.g. P, Ar, Sb, etc. it is called N-typesemiconductor.

• If impurity is a acceptor (or tri valent)e.g. B, Al, Ga etc.

52. (b)

Let x is the displacement of electron and Eis applied electric field across conductor.

F =2

e 2d xm eEdt

dxdt =

e

eE drift velocitym

As, mobility, =drift velocity

Electric field intensity

= e

eEm

E

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=e

em

53. (a)

Band diagram for extrinsic semiconductor :

N-type P-type

Conduction bandFermi Energy

level

Valence band

Donor energy level

acceptor energy levelFermi level

• In N-type material, there are electronenergy levels near the top of the bandgap so tht they can be easily excited intothe conduction band.

• In P-type material, extra holes in the bandgap allow excitation of valence bandelectrons, leaving mobile holes in thevalence band.

54. (b)

Net doping will be of trivalent nature astrivalent doping concentration is higher thanpentavalent dopin concentration. So thematerial is P-type.

Net concentration = trivalent penavalentn n

= 16 162 10 10

= 16 310 cm55. (c)

56. (d)

57. (c)

58. (b)

59. (d)

60. (a)

61. (c)

62. (d)

63. (b)

64. (b)

65. (c)

66. (b)

67. (c)

68. (b)

69. (c)

70. (b)

Reflex Klystron is used as an oscillator.

71. (c)

In multicav ity Klystron, each of theintermediate cavities act as a buncher withthe passing electron beam inducing anenhanced RF voltage than the previouscavity.

72. (c)

Helix is used in TWT to reduce velocity ofwave in certain direction to enable interactionwith e– beam.

73. (a)

Both M-type and O-type BWOs are tunableover a wide range of frequencies by varyingthe accelerating voltage.

74. (d)

For a distortionless transmission line d 0dL

Which gives R G LGor RL C C .

75. (a)

The linear velocity of orbital motion of

satellite is given as v = GM,

r where r is

total height of orbiting satellite

As r decreases, v increases.

76. (b)

CE amplifier provides high voltage gain inthe audio frequency range.

CB amplifier has a very good high frequencyresponse. Thus, it is referred to as Radiofrequency amplifier.

JFET amplif ier has very high inputimpedance, in the order of megaohms. JFETamplifiers have very low bandwidth.

Common collector amplifier is known asBuffer amplifier as it has unity voltage gainand helps in impedance matching betweenhigh impedance source and low impedanceload.

77. (d)

IES M

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Frequency response of R-C coupledamplifier:

fGain at low frequency drops due to thepresence of coupling and bypass capacitorsand gain at the high frequency falls off dueto the presence of parasitic capacitances ofthe BJT.Frequency response of Transformer coupledamplifier:

fFrequency response of direct-coupledamplifier:

fDirect-coupled amplif ier can amplifyextremely low frequency signals

Frequency response of Tuned amplifier:

ffo

78. (a)

The frequency of oscillation of BJT phaseshift oscillator is given by

fo =1

2 RC 6 4k , where

cRk

R

Here, R = 8k , C = 1500 pF and Rc =6k . Thus,

k = cR 6k 3R 8k 4

fo = 1

32 8k 1500p 6 44

=1

2 12 3

= 4.42 kHz

79. (a)

Voltage regulator ICs with their outputvoltages, are given below:

1. 7809 : + 9V

2. 7805 : + 5V

3. 7815 : +15V

4. 7812 : +12V

80. (b)

The given circuit is an instrumentationamplifier, providing an output based on thedifference between two inputs. It is adifferential amplifier optimized for high inputimpedance and high CMRR. It is typicallyused in applications in which a smalldifference voltage and a large common modevoltage are the inputs.

Analysis of the above circuit yields followingequations:

V3 = 1 11 CM 2

p p

R R1 V V V

R R

V4 = 2 22 CM 2

p p

R R1 V V V

R R

Output voltage,

5 4 3 2 1p

2RV V V 1 V V ,R

(for R3 = R4 = R5 = R6 and R1 = R2 = R)81. (c)

–+

v0vin

R1R2Vf

+

–When ov = + Vsat.

Vf =2

ut sat1 2

RV V

R R

= satVWhere Vut is the upper trip pointWhen ov = – Vsat.

IES M

ASTER



(8) (Test-15 Solut ion) 11th December 2016

Vf = 2lt sat

1 2

RV VR R

= satV

Where l tV is the lower trip point

Hysteresis voltage (VH) is given by

VH = ut ltV V

= sat satV V

= sat2 V

82. (d)

+–

Rf

R1

vin vout

1

2

An ideal Op-amp has a perfect balancebecause of the infinite voltage gain. Thus,V1 = V2.

Applying KCL at node ‘2’, we get

2 out 2

f 1

V v VR R

= 0

out

2

vV =

f

1

R1

R

out

in

vv =

f

1

R1

R 2 1 inV V v

+–

voutvin

This is voltage buffer circuit i.e. out inv v

out

in

vv = 1

+– vout

vin

Rf

R1 V1

Due to perfect balance, V1 = 0 (virtualground) Applying KCL at node ‘1’, we get

in 1

1

v VR

=1 out

f

V vR

in

1

v 0R

=out

f

0 vR

out

in

vv =

f

1

RR

83. (c)• For underdamping 1 , response is

y t

• For overdamped 1 , response is

y t

• For undamped 0 , response is

y t

• Exponential decaying function y(t) = ate

y tate

84. (b)

85. (a)

Given,

G(s) = s 2 1, H ss 6s a s 4

For step input, steady state error,

ess =P

11 K

KP = s 0LimG Hs s

IES M

ASTER




= 2s 0

s 2 1Lim .s 4s 6s a

=2 1

a 4 2a

ess =1 2a 01 2a 112a

sse 0 given

a = 0

86. (d)

Transfer function =

C s T GR s

Sensitivity of open transfer T with respect toG is given by

TGS =

dT T dT G G. 1. 1dG G dG T G

...(i)

Transfer function=

C Gs TR 1 GHs

Sensitivity of closed loop transfer T withrespect to G is given by

TGS =

dT T dT G.dG G dG T

= 21 G 1

G 1 GH1 GH1 GH

...(ii)

Hence equation (i) and (ii), clear that with anegative feedback in the closed loop controlsystem, the system sensitivity to parametervariation decreases.

87. (b)

For a system,

Gain × Bandwidth = Constant

OL OLA BW = CL CLA BW

where AOL = open loop gain = G

OLBW = Open loop bandwidth

ACL = Closed loop gain

=G

1 GH

CLBW = Closed loop bandwidth

OLGBW = CLG BW

1 GH

CLBW = OLBW1 GH

88. (b)

In electric switch does not consists anyfeedback element, hence it work as a openloop control system.

89. (c)

• If a system is sensitive to parameters(i.e temperature, pressure etc.), it will givedifferent output for same input at differenttime, temperature or pressure.

• If a system is insensitive to inputcommand, it will give same output atdifferent inputs.

Hence for a good control system, i tinsensitive to parameters and sensitive toinput command.

90. (b)

The transfer function is applicable to lineartime-invariant (LTI) system with initialconditions are zero.

91. (b)

R G

B

R G

B GC C

92. (a)

G OutputInput

H

Given, Input = 3V, G = 2, H = 1

OutputInput =

G1 GH

Output

3V =22

1 2 3

Output =2 3V 2V3

93. (d)

B

C

AR

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B CAR

Transfer function = G

1 GH

Here G = A(B–C), H = 1

OR =

A B C1 A B C

=AB AC

1 AB AC

94. (a)

For R(s) = 0, the above system can beredrawn as.

G (s)2 C(s)N(s)

H(s)

G (s)n G (s)1

C sN s =

2

n 12 1

G s 1 G s G s1 G s H s G s

i.e.C(s) =

2 n 1 2

1 2

G s G s G s G s N s1 G s G s H s

To eliminate the effect of Noise N(s) on C(s)

2 n 1 2G s G s G s G s = 0

nG s = 1

1G s

95. (c)

V0(s) = iR sLI (s)R sL

0

i

V sI s =

R sLsLR 1R

=

sL11 sLR

So, the block diagram will be,

sLI (s)i V (s)0

1R

96. (c)

–H2–H1

2R 1/2 CG1

Forward Path,

P1 = 112×G × 2 = 1G,Δ =1

Individual loops :

L1 = –H1, L2 = –H2,

Non-touching loops,

L1L2 = –H1×–H2 = H1H2

Using Mason’s gain formula

CR =

1 1

1 2 1 2

P1 L L L L

= 1

1 2 1 2

G1 H H H H

CR = 1

1 2 1 2

G1 H H H H

= 1

1 2

G1 H 1 H

97. (b)

s 1

R(s) E(s) C(s)

0

s 1 E s

s 1

E ss 1

R(s) C(s)1s 1

C sR s =

1s 1

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98. (c)

x1

e f g

a x2 b x3 c x4 d x5

Applying Mason’s gain formula

5

1

xx =

n

K KK 1

P

Forwards path :

P1 = abcd, 1 1

Loops : L1 = be, L2 = cf, L3 = dg

Non-touching Loop :

L1L3 = bedg

5

1

xx =

1 1

1 31 2 3

P1 L LL L L

5

1

xx =

abcd1 bedgbc cf dg

99. (b)

If a change of polarisation direction isrequired, a twist section may be usedextending over two or more wavelengths.

Tapers are used to couple waveguides ofdifferent dimensions.

Bends & corners are used to changedirection.

100. (b)

Average power is given by d rad

ave 2

GP

4 r

Here, d 10 dG 10dB 10log G

dG 10

3rad 20 10 w and 3r 10 m

3

ave 610 20 10

4 10

= 1 W

20

101. (b)

Horn antenna, an example of an apertureantenna is a tapered section of a waveguide

providing a transection between a waveguideand the surroundings. It finds its applicationin aircraft.

Parabolic dish Reflector reflects EM wavesby conducitng sheet. Wirid antennas are ledby transmission lines.

102. (d)

The wave is propagating in +z direction. Theexpression of electric field for TM mode isgiven by :

i zZ 0

m x n yE E sin sin ea b

comparing with

i zZE 30sin 50 x sin 60 y e

i zsin 60 y e

m 50a

2m 50a 50 8 10 4

n 60b

2n 60 5 10 3

Mode is 43TM

103. (b)

A waveguide behaves as a high pass filter.There is heavy attenuation for frequenciesbelow cut-off although the waveguide itselfis virtually lossless. Such attenuation is dueto reflection at the mouth of the guide andsome take place in evanescent modes.

104. (a)

The wavelength of the wave propagating ina waveguide is given by

2m12a

where Free space wavelength.

Hence is always greater than A.

IES M

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105. (d)

Phase velocity is the velocity with which thewavechanges phase at a plan foundary andnot the velocity with which it travels alongthe boundary this apparent velocity which isgreater than velocity of light in space is calledphase velocity.

106. (c)

A parasitic element shorter than the drivenone or radiator tends to increase radiation inits own direction and behaves as convergentconvex lense which is called a Director.

Radiator

Director

Reflector

A parasitic element longer than the drivenone and which reduces signal strength in itsown direction and increases in oppositedirection is called a reflector.

107. (c)

TEM (Transverse Electromagnetic Mode)does not have any cut-off frequency andhence it will be supported at any frequency.

Cut-off frequency is given lay: 0 = m c

nd(for mode m)

108. (c)

A waveguide may consist of one or moreconductors, or no conductors at all and willsupport waveguide modes.

109. (c)

For a Transverse magnetic wave propagatingin a waveguide, the electric field componentalong the wave propagation is given by:

Ez =

z0

m x n yE sin sin ea b

For m = 0, n 0 , Ez = 0

For n = 0, m 0 , Ez = 0

TM01 and TM10 will not exist and TEMmode cannot propagate in a rectangularwaveguide.

Only TM11 can propagate.

110. (d)

Loop Antenna consists of a wire with one ormore turns. They are also called wireAntennas and used in Buildings, ships,Automobiles, etc.

Horn Antenna is an example of ApertureAntenna is a tapered section of a wave guide.It is used in Aircraft as it can be convenientlyflush mounted.

Parabolic dish reflector reflects EM wave bya conducting sheet. It is used incommunication, Radar & Astronomy. YagiUda Array is used for short wavetransmissions by Radio Amateurs.

111. (b)

After a wafer has been coated withphotoresist, the correct sequence of stepscarried out in photolithography is

(i) Exposure to UV radiation:

(ii) Developing : The wafer is developedusing a suitable chemical l iketrichloroethylene. This results in theremoval of photoresist f i lm wherewindows are required.

(iii) Etching : Wafer is dipped in an etchingsolution to remove the layers to openwindows for impurity diffusion or makingcontacts.

(iv) Stripping : Stripping is used to removethe photoresist mask with sulphuric acidand by means of mechanical abrasionprocess.

112. (c)

The main purpose of the buried layer is tominimize the series resistance of thecollector.

113. (c)

The diffusion profile of the dopant atoms isdependent on the initial and boundaryconditions. The two important cases are

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(i) Constant-surface-concentrationdiffusion: In this case, impurity atoms aretransported from a vapour source onto thesemiconductor surface and diffused into thesemiconductor wafers. The vapour sourcemaintains a constant level of surfaceconcentration during the entire diffusionperiod. The doping profile in this case is acomplementary error function (erfc)distribution.

(ii) Constant-total-dopant diffusion: In thiscase, a fixed amount of dopant is depositedonto the semiconductor surface and issubsequently diffused into the wafers. Thedoping profile for this case is the Gaussiandistribution.

114. (b)

LED works on the principle of spontaneousemission.

LASER works on the principle of stimulatedemission of radiation.

Solar cel ls work on the principle ofphotovoltaic effect i.e. voltage is developedacross it when illuminated by light.

Photodiode works on the principle ofphotoconductive effect or photoconductivityi.e. conductivity of the photodiode changesdepending on the light intensity.

115. (c)

The magnitude of current under large reversebias is given by

TV Vs oI I I 1 e ...(i)

Where Io is the reverse saturation current,

Is is the short-circuit current which isproportional to the light intensity

and V is voltage across the cell.

Photovoltaic potential is the voltage at whichzero resultant current is obtained under open-circuit conditions.

From Equation (i),

0 = max TV Vs oI I 1 e

Vmax =s

To

IV n 1

I

l

116. (d)

In optocouplers, the wavelength response ofthe each device (light emitters and lightdetectors) is made as identical as possible.This is done so as to permit the highestmeasure of coupling possible.

117. (a)

118. (b)

119. (d)

120. (b)

121. (a)

122. (c)

123. (a)

124. (d)

The stalls are a type of hazards that affect apipelined system.

125. (a)

Data hazards are generally caused when thedata is not ready on the destination side.

126. (b)

127. (b)

128. (a)

129. (b)

Mutually exclusive events cannot occur

simultaneously, hence, A B

130. (c)

In case of statistically independent eventsoccurance of one event doesn’t affectprobability of other event, ie.

APB = P (A)

Using Bayes’ Rule P (A/B) =

P(A) P B A P(AB)

P(B) P(B)

or P (AB) = P (A) P(B)

131. (d)

Probability density function is defined as

Fx(x) = xd F (x)

dx

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and xF (x) 0

132. (c)The Guassian probability density function is

defined as

22x 2x 2

1f (x) e2

133. (a)

As per central limit theorem as N increases,probability density of a sum of N independentrandom variables tends to approach aGuassian density.

134. (d)Given, receiver noise resistance =

eq100 R

Noise figure F = 3

We have, F = eq

a

R1

R

3 = a

1001R

Ra= 50 input resistance of antenna

135. (a)

Given to = 273 + 27 = 300 K

Effective noise temperature of the receiver

Te = (F–1) To

10F 10log 10

Te = (10–1) × 300 = 3000K

The overall effective noise temperature

TIN = Ta+Te

= 0 + 3000 = 3000K

Overall noise figure

F =IN

o

T1

T

=30001 11300

FdB = 10 log1011 = 10dB

136. (a)

For a band-pass system, minimum samplingfrequency

fs = H2fK

where K = floor

H

H L

f 4floor 8f f 4 3.5

fs =2 4 1MHz

8

137. (b)

In Mid-rise quantizer, input value between 0and 1 is mapped to an output value of 0.5.

138. (c)

For X(t) to be stationary µx(t) must beindependent of t i.e.

x(t) = E[X(t)] = E[P] sin t + E[Q] cos tmust be independent of t which is possibleonly if x(t) = 0 E[P] = E[Q] = 0.

139. (b)

Given h(t) = e–t u(t)

h(f) = F·T· [h(t)]

= 1

1 j2 f

so, Sy (f) = 2xS f h f

=

0

2

N2

1 2 f

Autocorrelation function yR = Inverse F.T..[Sy(f)]

yR =

01

2

N2F

1 2 f

= 0N e4

140. (a)

We have variance 2 = 22E X E X

For a uniformaly distributed random variable

Mean µx = E[X] =

x f x dx

=

b

a1 a bx dx

b a 2

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and

2E X =

b3b 2a

a

1 1 xx dxb a b a 3

So 2 = 22E X E X

= 2b a

,12

here b = 5 and

a = 2

So 2 =

25 20.75

12141. (c)

Lempel-ziv coding scheme doesn’t dependupon source statistics, hence it is universalcoding scheme.

142. (a)

Differential entropy of a continuous randomvariable is given as

h(x) =

x 2x

1f (x)log dxf (x)

Here fx(x) =

1 , K x K2k0, otherwise

h(x) =k

2k

11 log dx12k 2k

=k

2

k

log 2k dx2k

= 2log 2k 2k2k

= log2 2k143. (b)

Mutual information of a BSC is

I(X, Y) = H (Y) + p log2 p + (1–p) log2 (1–p)

which is maximum when H(Y) is maximum.Maximum H(Y) is achieved when each outputhas probability of 0.5 and H(Y)max = 1

So channel capacity of BSC

Cs = P(x)

max I (X, Y) = 1 + p log2 p + (1 – p)

log2 (1 – p)

144. (c)

In a block code dmin, the minimum distancebetween any two code words of the code is(2t + 1).

145. (c)

code tree of an (n, k, L) linear convolutioncode repeats after (L + 1)th stage.

146. (d)Fro a raised-cosine spectrum Bandwidth B =

W 1

Where W = b

b

R12T 2

Given, Rb = 28 Kbps

W = 28 14 KHz2

Bandwidth = 14 1 0.5

0.5= 14 × 1.5 = 21 KHz

147. (b)

Given 2 i 2i i

1 1Log n log 1P P

It can be written as

2 i i 2 iLog P n log P 1Multiplying by Pi and summing over i yields

l

m m m

i 2 i i i i 2 ii i 1 i 1

P log P n P P log P 1 ...(1)

m m m

i 2 i 2 i ii 1 i 1 i 1

P log P 1 Pi log P P H X 1

Thus eqn. (1) reduces to

H(x) L H X 1148. (d)

Thermal noise is also called as resistor noiseor Johnson noise.

149. (d)

Sequential circuits :• It gives outputs depending on the present

input and past input also.

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• It contains atleast one feedback path.• It contains some memory.• It exhibit cyclic nature i.e. its output

sequence repeats after some clock cycle.

150. (d)

In intrinsic semiconductor, the total electricalconductivity.

e h e hne ne Although the number of electrons inconduction band and holes in valence bandare equal in intrinsic semiconductor, theconductivity due to electrons is more thanthe conductivity due to holes. Because themobility of electrons is higher than that ofholes.

151. (d)

With increase in temperature mobility ofcharge carriers decreases as vibration ofatom’s in molecule increases but at the sametime carrier concentration increases.

2in = gE /KT3

0A T e

in increases with temperature increase. Now conductiv ity of intrinsicsemiconductor.

= i nn qAs with temperature intrinsic concentrationni increases and mobility decreases, butincrease in carrier concentration is muchmore than decrease in carrier mobilitytherefore conductivity effectively increases.Hence resistivity decreases.

152. (a)Doping materials are called impurities whichare used to increase the carrier concentrationin intrinsic (pure) semiconductor, when theyadded to intrinsic semiconductor they alteredthe crystal structure of semiconductor bymaking the covalent bonds with other atomsand release/accept free charge carriers.

153. (a)

555 timers, operating as multivibrator, arepreferred over those with BJT and digital ICsbecause they provide variation in duty cyclefrom 0% to 100% and its output can drive aTTL load.

154. (b)

The E field, H f ield and direction ofpropagation in space are mutuallyperpendicular to each other. Hence it is TEMwave in space.

TEM wave will not propagate in a waveguidebecause electric field would be short circuitedby the walls since walls are perfectconductors and potential cannot exist acrossthem.

155. (c)

The mode with lowest possible frequency thatcan be propagated in a waveguide is calledDominant mode.

cf

wavelength is highest

156. (a)

MOS ICs based on MOSFET structure findwide application in digital field. This isbecause MOS ICs have small size and easyto fabricate.

157. (a)

An LED is a light emitting diode that givesout spontaneous emissions, when it isenergized.

The recombination of holes and electrons insemiconductor materials gives off energy(equal to the Band gap energy) in the formof light and heat.

158. (b)

159. (b)

160. (b)

Documents

ANSWERS - IES Masteriesmaster.org/public/archive/2016/IM-1481464255.pdfH = 1 ne where n is the carrier-concentration, A B R R = 21 B 21 A n 4 10 4 :1 n 1 10 50. (d) The conductivity