answers - ies masteriesmaster.org/public/archive/2016/im-1482668626.pdf · 2017. 5. 26. · r...

1. (b)

2. (c)

3. (c)

4. (b)

5. (b)

6. (c)

7. (d)

8. (b)

9. (a)

10. (d)

11. (d)

12. (c)

13. (d)

14. (b)

15. (c)

16. (d)

17. (d)

18. (a)

19. (d)

20. (a)

21. (c)

22. (a)

23. (c)

24. (d)

25. (d)

26. (d)

27. (b)

28. (b)

29. (d)

30. (a)

31. (a)

32. (c)

33. (a)

34. (a)

35. (d)

36. (d)

37. (b)

38. (c)

39. (c)

40. (c)

41. (d)

42. (a)

43. (b)

44. (b)

45. (a)

46. (b)

47. (c)

48. (b)

49. (d)

50. (b)

51. (b)

52. (c)

53. (a)

54. (d)

55. (a)

56. (b)

57. (c)

58. (b)

59. (b)

60. (c)

61. (c)

62. (a)

63. (a)

64. (c)

65. (b)

66. (c)

67. (c)

68. (d)

69. (b)

70. (a)

71. (c)

72. (a)

73. (b)

74. (c)

75. (b)

76. (b)

77. (b)

78. (d)

79. (b)

80. (b)

81. (b)

82. (b)

83. (c)

84. (d)

85. (c)

86. (b)

87. (a)

88. (c)

89. (d)

90. (a)

91. (a)

92. (a)

93. (d)

94. (b)

95. (b)

96. (d)

97. (a)

98. (c)

99. (d)

100. (c)

101. (d)

102. (b)

103. (a)

104. (b)

105. (b)

106. (c)

107. (d)

108. (d)

109. (c)

110. (c)

111. (b)

112. (c)

113. (b)

114. (a)

115. (c)

116. (c)

117. (c)

118. (b)

119. (b)

120. (c)

ESE-2017 PRELIMS TEST SERIESDate: 25th December, 2016

ANSWERS

121. (c)

122. (c)

123. (b)

124. (b)

125. (d)

126. (c)

127. (d)

128. (a)

129. (c)

130. (a)

131. (b)

132. (c)

133. (b)

134. (b)

135. (c)

136. (a)

137. (b)

138. (b)

139. (d)

140. (a)

141. (a)

142. (a)

143. (a)

144. (d)

145. (a)

146. (b)

147. (b)

148. (d)

149. (d)

150. (c)

151. (c)

152. (d)

153. (a)

154. (d)

155. (c)

156. (d)

157. (c)

158. (c)

159. (a)

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-18) 25th December 2016 (3)

1. (b)

Since, p = Ewhere p is dipole moment is polarizabilityE is electric field intensity

= p q.dE V d

= 2q .dV

= C.d2

= (Farad) .(meter)2

2. (c)Ferroelectric materials exhibits thephenomenon of hysteresis.

3. (c)Ferromagnetic involves an electrostaticinteraction between adjacent atoms thataffects the alignment of the resultantelectrons spin of the atoms. If the interactionare positive, the spin are aligned parallel toone other and their magnetic moments areadditive and ferromagnetism results.

4. (b)

Q , C11

V1

Q , C22

V2

After putting in contact and then separatedthe voltage across each sphere will be same.Since, Q = CV

i.e. V =QC

So, V = 1 2

1 2

Q QC C

i.e. 1

2

QQ = 1

2

CC

5. (b)For parallel plate capacitor,

• Charge in the plate will remain same as it isgiven to the plate.

• Since, Electric field E

for a plate is 0 ,(constant) and potential difference betweenplates,

V = dE

i.e. V depends on d

• Energy of the capacitor, 21E CV2

willchange with change in ‘d’ as C and V bothare changing.

• Energy density 2E 0

1W E2

, as E

isconstant, WE will be constant.So, only V and energy of capacitor will changewith change in separation.

6. (c)

2

–Q

R2

R1

1+Q

Electric field intensity between two cylinders.

E.2 xL = Q

[at a distance x from the centre]

[According to Gauss’s Law]

E =Q

2 xL

So, V =1

2

R1

2 R

Q 1E.dx dx2 L x

= 2

1

RR

Qln x2 L

=2

1

RQ n R2 L

l

Since, C = 2 2

1 1

Q Q 2 LR RV Q n nR R2 L

l l

7. (d)In electrostatics, a perfect conductor hasfollowing characteristics :

1. No electric field exist within a conductor i.e.electric field will be only external

2. Electric field will be normal to its surface.

IES M

ASTER



(4) (Test-18) 25th December 2016

3. Since E V 0 ,there will be no potentialdifference between any points on its surface.

8. (b)Ampere’s Law :

I = lC

HdBiot-Savart Law :

H = l r

2Id a4 r

Coulomb’s Law :

F = 1 2 1 221 122 2

0 0

Q Q Q Qa or a4 R 4 R

where a12 and a21 are unit vectors along theline joining Q1 and Q2 .Gauss’s Law :

Q =s

D.ds

9. (a)

• Propagation constant, = j j

For highly conducting medium, >>

= j

= j j j2 2

= 2

• Radiation intensity 2U r P,

=2 2r E2

• Wave propagation, P = H

10. (d) = impedance of free space = 120

hence2E

2 = 3 21.2 10 W m

E = 31.2 10 2 120

= 951 V/m

11. (d)2a r = 2 6 231.25 10 4.91 10 m

Bair =5

60.6 10 1.22T

a 4.19 10

Hair = air7

0

B 1.224 10

= 59.71 10 AT / m

12. (c)Poynting vector, P E H

So, the direction of P

will be perpendicularto both E

and H

.

13. (d)

14. (b)

Given, 6 yˆE 10cos a V m10 t 50x

D = 0 rE

= 0E

[In free space, r 1 ]= 60 yˆ10 cos a10 t 50x

Now, the displacement current density,

dJ = dDdt

= 60d 10 cos 10 t 50xdt

= 6 6010 10 sin 10 t 50x

= 7 260 yˆ10 sin a A m10 t 50x

15. (c)

x

y

z

According to Ampere’s Law.

H =

I a2

where is the perpendicular distance ofpoint from conductor.

y

x

(–3, 4) 4

30I

a

IES M

ASTER




Hence, 2 2 20 53 4

So, H = 10 a

2 5

= ˆ1.a

= x yˆ ˆsin a cos a

= x y4 3ˆ ˆa a A m5 5

16. (d)

V0

1K

V10K 20K

1K

Vi

iV V1 10

= 0

V = –Vi

0V VV V10 1 20

=0

1.15V = 0V20

0

i

VV = –230

17. (d)

Z =0 R R

R 0 R

[Z] =R RR R

Z =R R

0R R

Z = 0

Hence Y does not exist.

18. (a)

(100 ) (0.5A 0.25A) 25V

19. (d)Using thevenin theorem

+–

+ –

+–

20V

20V

5K 4K

5K 15K

5K

10V

+–

+ –

+–

20V

15V

2.5K 4K

3.75K

5V

i1

Current (i1) =15 5 20

R

= 0

20. (a)Source transformation :

R RVR

VR

R2

2VR

R2V

21. (c)

Z12 =1

1

2 I 0

VV

When I1 = 0 , V2 =

12 V

3

I2 = 1 122V V3

I2 = 14 V3

1

2

VI =

34

22. (a)Two port network,

I1 = 11 1 12 2Y V Y V

I2 = 21 1 22 2Y V Y V

Applying KCL,

I1 = l 1 2V V Vs 1

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ASTER




I1 = 1 21V V1s

Y11 = 1211 , Y 1s

I2 = 2 2 1sV V V

= 1 2V V 1 s Y22 = 1 + s, Y21 = –1

23. (c)

SCi 5A [Given], we can find Rth

8 4 4

8 4Rth

Rth = 8 8 4

Norton equivalent = SC thI R & 4

Max power = 2 thSC RI 25 4 25W4 4

24. (d)a Pole on positive real axisb Poles and zeros do not interlacec Poles and zeros do not interlaced is valid immitance.

25. (d)Cut-off frequency is given by: fc =

2 2

2 20 0

1 m na b2 E

For TE02, fc = 12 GHz

12 × 109 =83 10 2

2 b

b =8

93 10 1

4012 10

a = 2b a = 120

For TM11, fc =8

8 2 22 2

3 10 1 1 1.5 10 20 402 a b

fc =8 81.5 10 400 1600 1.5 10 10 2 5 3 5 GHz

26. d

wavelength =

8

6c 3 10 300m7 10

For quarter wave monopole length =

300 75m4 4

27. (b)

Pavg = 2in1 P

=211 10

4

=11 10

16

=15 1016

= 9.38 W

28. (b)

Given l4

, and LZ (i.e. open circuit).

Since, input impedance

Zin =ll

L 00

0 L

Z jZ tanZ Z jZ tan

Here, l 2tan tan tan. 24

So, Zin = l

l

ll

L0

00

L

Z jZtantanZ

Z jZtantan

Zin =20

L

ZZ

But as LZ So, Zin= 0

29. (d)• As the line is perfectly matched, so

according to maximum power theorem,there will be maximum transfer of powerto the load.

• Reflection coefficient,

= L 0 0 0

L 0 0 0

Z Z Z Z 0Z Z Z Z

so, there will be no reflection.

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• VSWR =1 1 0 11 1 0

so, no standing wave pattern.

30. (a)The input impedance,

inZ = ll

L 00

0 L

Z jZ tanZZ jZ tan

Here, ltan = lP

tan V

=6

82 100 10tan 1

2 10

= tan= 0

So, inZ = L0 L

0

ZZ Z 30 j40Z

31. (a)Given,

ZOC = 12 40

and, ZSC = 48 16

then, characteristic impedance

Z0 = OC SCZ Z

= 12 40 48 16

= 12 48 56

= 24 28

32. (c)Perfectly matching takes place using aquarter wave transformer when the loadimpedance and the input impedance areresistive in nature.

33. (a)Phasor diagrams for leading and laggingpower factors are shown below:

I

V

Ip

Leading pf

I

V

Ip

Lagging pf

It is clear from the phasor diagrams that thewattmeter will read low when load powerfactor is leading as in that case the effect ofpressure coil inductance is the increase thephase angle between load current andpressure coil current lagging loads thewattmeter will read high as the effect of theinductance of the pressure coil circuit is tobring the pressure coil current more nearlyinto phase with the load current than wouldbe the case if this inductance were zero.

34. (a)The correction factor is a factor by which theactual wattmeter reading is multiplied to getthe true power. For leading power factor,

Correction factor = cos

cos cos( )

35. (d)A ballistic galvanometer can be used as fluxmeter by eliminating the control springs. Inthe fluxmeter the controlling torque is verysmall and electromagnetic damping is heavy.The current is led into the coil with the helpof very loose helices of very thin annealedsilver strips. The controlling torque is thusreduced to minimum.

36. (d)Shunts are used for extending the range ofammeters.Multipliers are used for extending the rangeof voltmeters.Current transformers and potentialtransformers are alos used for rangeextension purposes. When instruments areused in conjuction with instrumenttransformers, their readings do not dependupon their constants (R, L, C) as is the casewith shunts and multipliers.

IES M

ASTER




37. (b)The deflecting torque exerted,

Td = p

V dMIcosR d

ddMT cosd

…(i)

Here rate of change of flux linkage = dMd

Mutual inductance M = maxM cosMmax = Maximum mutual inductance = Angle between the axes of two coils

dMd = maxM sin [sign disregarded] …(ii)

d maxT cos M sin …(iii)[from equation (i) and (ii)

Case (i) : = 60°, cos 0.866

Td1 max0.866M sin60 [from equation(iii)]

Case(ii): = 90°, cos 0.5

2d maxT 0.5M sin90 [from equation(iii)]

1

2

d

d

T 0.866 sin 60 1.499 1.5T 0.5 sin 90

38. (c)

Power factor angle 1 1 2

1 2

3(P P )tanP P

P2 = 0

= 1tan 3

= 60°

Power factor = cos cos60 0.5

39. (c)

Resolution on 10V range = 41010 = 0.001V

Therefore on 10V range reading can bedisplayed only upto 3rd place of decimal.Hence, 0.8624 V will be displayed as 00.862on 10V scale.

40. (c)The synchros and circular potentimeters areused to measure angular position.But LVDT is used to measure lineardisplacement.

41. (d)

(330)8 = 2011 011 000

= 20000 1101 1000

= (D8)H

(330)8 = 2 13 8 3 8 0

= 3 × 64 + 24= 192 + 24= 216

42. (a)• The excess-3 code for each decimal digit is

obtained by adding decimal 3 to the naturalBCD code of the digit.

• It is a self-complementing code• It is not a weighted code.43. (b)

44. (b)

45. (a)

46. (b)

–42 recur (12,3)

–36

–24

0

recur (6,6) –2×3

recur (3,12) –12

recur (1,24) –24

return 0

47. (c)

48. (b)

49. (d)

50. (b)

51. (b)

52. (c)

53. (a)

54. (d)

55. (a)

56. (b)

IES M

ASTER




57. (c)

58. (b)

Y = AB A.B

= A BA B

= A B

59. (b)• Base width modulation (Early effect) occurs

in common-base Transistor (BJT)• MOS-capacitor works as constant voltage

source (flat band voltage) during saturationregion.

• LASER diode works on the principle ofpopulation inversion.

• Pinch-off of channel occurs in JFET whendrain current becomes zero.

60. (c)• Emitter injects the charge carrier into BJT

therefor it has very high carrierconcentration due to which highconductivity.

• Collector has very large size in BJT as itworks as a collector of charge carriers dueto which temperature increases so it needslarger surface to work as heat sink.

61. (c)

20V

EI CI5K15K

30V

BIBEV

Applying KVL at base emitter section, wegetVEE + IE15K + VBE = 0

EI = BE EEV V15K

= 30 2 mA15K

E C CBI I , V = CC CV I 5K

= 10V

62. (a)

IS would be Vm45 5 for positive cycle and zero

for negative cycle

dc avgI I = mV50

= mV50

63. (a)

Avg speed of electron V = E = MobilityE = Electric field.

= 2cmV sec

cm / secV / cm

64. (c)

12V

11.3K

oV20V 100

300B

C E

BEV

Loop1

Applying KVL in Loop-1, we getVz – VBE – V0 = 0

oV = z BEV V

= 12 0.7= 11.3 V

IE = 0V 11.3 1mA11.3K 11.3K

Thus, IC = EI1

=100 1mA

1 100

IC = 0.99 mAFrom the circuit,

CEV = i oV V 20 11.3

= 8.7V Power dissipated in the transistor

= VCE×IC

= 8.7V 0.99 mA

= 8.6 mW

65. (b)

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ASTER




10k

10k

10k

10k

5V10V

Di

DV– +AK

From the circuit,

VA = 10k 5

k10 10

= 2.5 V

and VK = 10k 10 5V

k10 10

Thus, VD = A KV V 2.5V

Since, VA < VK the diode is not conducting iD = 0

66. (c)V0 is maximum for +ve cycle of input. In the+ve half cycle, diode D2 is ON for Vi>2V.

0, maxV10V2V

500

250

0, maxV = 2502 10 2750

= 4.67V

V0 is minimum for –ve cycle of input. In the–ve half cycle, diode D1 is ON.

0, minV10V

500

1250

0, minV = 125010 7.14V1750

67. (c)

Since, XY 1 i.e. X 1 X 0 and Y = 1

Both are NOT gate

XY

X

Y Y

X Y 0 1 0 Y

68. (d)

Y = AB B= B BA B

= A B B B 1

A

BY

69. (b)1. Immediate Addressing Mode: - An

immediate is transferred directly to theregister.Eg: - MVI A, 30H (30H is copied into theregister A)MVI B,40H(40H is copied into the registerB).

2. Register Addressing Mode: - Data iscopied from one register to anotherregister.Eg: - MOV B, A (the content of A is copiedinto the register B)

MOV A, C (the content of C is copied intothe register A).

3. Direct Addressing Mode: - Data is directlycopied from the given address to theregister.Eg: - LDA 3000H (The content at thelocation 3000H is copied to the registerA).

4. Indirect Addressing Mode: - The data istransferred from the address pointed bythe data in a register to other register.Eg: - MOV A, M (data is transferred fromthe memory location pointed by theregiser to the accumulator).

5. Implied Addressing Mode: - This modedoesn’t require any operand. The data isspecified by opcode itself.Eg: - RAL, CMP

70. (a)The address start from 0000 to 1FFFH. Thechip select pins are A13, A14, A15. They haveto be zero to address memory. So we have tofix A13, A14, A15 as zero. Other address linescan be zero or one accordingly we can access0000 to 1FFF memory.

71. (c)HL register pair is known as primary data

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pointed as used in register indirectaddressing.

72. (a)DMA transfers data form memory to memorywithout consuming microprocessor time. Forharddisk where lot of memory has to betransferred, DMA is most suited.

73. (b)Output, F is given by

F = ABC ABC ABC ABC

= BA C C AB C C

= BA AB

= B A

74. (c)Counter, flip-flop & shift registers have amemory storage element. Therefore, they aresequential circuits. Only adder is acombinational circuit.

75. (b)

y = x x x x = 0 + 0= 0

76. (b)In asynchronous counter, as CLK is notapplied at all the flip-flops simultaneously,so there is always a propagation delay atevery stage of flip-flop which results into alarge delay for output i.e. the effect of aninput clock pulse “ripples” through thecounter, taking some time, due to propagationdelays, to reach the last flip-flop.This cummulative delay of an asynchronouscounter is a major disadvantage for high-frequency application because it limits therate at which the counter can be clocked andcreates decoding problem.

77. (b)As the ring counter utilizes one flip-flop foreach state in its sequence. It has theadvantage that decoding gates are notrequired. For this, initially, a 1 is presentinto the first flip-flop, and the rest of flip-flops are cleared.

0 1 2 3CLK Q Q Q Q0 1 0 0 01 0 1 0 02 0 0 1 03 0 0 0 1

78. (d)Cases should be replaced by ceases PIPOmode is not done by a shift register.

79. (b)

Maximum counting speed = Pd

1N t FF

= 1

4 50ns= 5MHz

80. (b)

D-flip flop with a feedback at Q acts as atoggle mode.

In toggle mode fout = inf2

= 10KHz2

= 5KHz

81. (b)Mnemonics are techniques for rememberinginformation that is otherwise quite difficultto recall. It is used in microprocessor for ashort abbreviation for the operation to beperformed.eg ADD – for additionANI – for AND operation

82. (b)The 8085 has six general purpose registersto store 8-bit data. These are B, C,D,E, H, L.They can be combined as register pairs BC,DE and HL, to perform 16-bit operations.

83. (c)The INTR pin is checked by themicroprocessor in the last state of the lastmachine cycle of each instruction. WhenINTR is HIGH, the microprocessor saves thecontents of the program counter on the stackand then sends an interrupt acknowledgesignal, INTA to the external hardware.

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84. (d)

Modulation index, m = max min

max min

V VV V

= 7 37 3

= 410

= 0.4

85. (c)fm = 100 Hzmf = 15

Bandwidth = mf2 fm 1

= 2 10015 1

= 3.2 KHz

86. (b)FM requires more bandwidth than AM andone more advantage of FM is that in FM,less modulating power is required comparedto AM.

87. (a)

x(t) = 6cos V10 2sin400 t 2 10 t

= 610 cos V1 0.2sin400 t 2 10 t

c0.2, A 10

2C

10P 50W2

Total side band power = 2

cP2

= 250 0.22

= 1 W

88. (c)A parallel encoder ADC or Flash type ADCrequires 2n–1 comparators for an n-bitconversion.Here, n = 8 Number of comperators required

= 82 1= 256 – 1= 255

89. (d)

• Simultaneous A/D converter or Flash typeA/D converter is a fastest converter

• But for a higher bit output, it requireslarge number of comparators. If ‘n’ is thenumber of output bit, the number ofcomparator required is

N = 2n – 1

90. (a)Output voltage of a R/2R ladder DAC,V0 = (Resolution) × (binary equivalent ofinput) × (gain)

= 0 1 2 34

205 2 1 2 0 2 1 2 0 102

= 5

5 216

= 5016

= – 3.12 V

91. (a)

2Z = 22

2

R sLR sLR sL

Z1 = 11

sR C 11RsC sC

0

i

V sV s = 2 1

2

sR L sR C 1sL R sC

= 2

2

2 1

s R LCR sL sR C 1

By observing this equation, we get it as anexpression of high pass filter.

92. (a)

20k20k

2k

240k

24k24k

Star delta

transform

5kV0

24k 24k

240k

–

+Vix

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Gain = 0

1

V 240KV 5K

= – 48

93. (d)

H(x) =

4

2i 1

1P logi P i

= 2 2 2 21 1 1 1log 2 log 8 log 4 log 82 8 4 8

= 1.75 bits per symbol

94. (b)For equiprobable symbols, the averageinformation

H(x) = 2log mm = number of symbols

H(x) = 2log 50

= 5.6 bits/symbolThis is between (5 and 6) bits/symbol.

95. (b)

From the property of entropy, we know thatthe maximum value of H is log2 M bits/symbolwhen M symbols are equal probable andminimum value of H is 0, hence,0

2H log M.

96. (d)

An optimal code for a source has minimumaverage code length and uniquedecipherability of encoded sequence.

97. (a)

A prefix code which is uniquely decodablesatisfies Kraft-Millman inequality.

98. (c)

Modulation used in GSM is GuassianMinimum-Shift Keying (GMSK).

99. (d)

Cell splitting increases handoff rate persubscriber and the cost of the system.

100. (c)

Due to their high operating power LASERhave shorter life span than LEDs.

101. (d)

PIN photodiode and avalanche photodiode aregenerally used as optical detector.

102. (b)

Frequency modulation index = mf

= 310 10 Hz

fm = 200 Hz

Modulation index = 310 10 50

200

103. (a)

Power in a frequency modulation doesnotchange with modulation. The power remainssame as before modulation.

104. (b)

Mixed signal y(t) =

m c cA cos2 f t.cos 2 f tt

output of filter = A m cost2

Pout =2 2

mP A cos4

[Pm = Power of m(t)]

PU =2

mA P2

out

U

PP = 2 21 cos 0.5cos

2

105. (b)

iV t = C mcos t 0.5cos t

i(t) = C m10 2 cos t 0.5cos t

2C m0.2 cos t 0.5cos t

The AM signal is

= c m2cos t 0.2cos t cos t

= cm cos t2 0.2cos t

Modulation = 0.2 0.1 or 10%2

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106. (c)

PC = 10 KW

= 0.7

Pt =2

CP 12

=20.710 1

2

= 12.45 W

107. (d)

x 20cos400 t 4cos360 t 4cos440 tt

= cos400 t20 8cos40 t

= modulation = 8 0.420

2C1P 200W202

2 2Sb

11P 16W4 42 2

Efficiency = 16 0.074200 16

108. (d)Nyquist rate = 2 × 30 K = 60 KHz6 level PCM, so we need at least n = 33 2 6 .

Bit rate = 60 3 180KHz

= 180 Kbps

109. (c)

The band width of x(t) = 410

2 Hz

Nyquist rate = 2 × BW

=4102

2

= 104

= 10 KHz

110. (c)

H(x) = 2 21 3 82 2log 8 log8 8 3

= 1.8 bits/symbol

Nyquist rate = 2 Hz

Information rate = 2 1.8

= 3.6 bits sec

111. (b)

As the sampling rate is 1600 samples/sec.The minimum recoverable signal would be

1600 samples sec 800Hz2

112. (c)

Since one pole lie in right half of s plane sofinal value theorem is not applicable givenimpulse response is causal then ROC 2

Img

–1 1 Real2

H(s) =s – 1

(s 1) (s – 2)

Using partial fraction

H(s) =A B

s 1 (s – 2)

By solving

A = 23 and B = 1

3

H(s) =2 1 1 13 s 1 3 (s – 2)

By taking inverse laplace transform, we get

h(t) = – t 2 t2 1e e u(t)3 3

H( ) = 0 +

h( )

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113. (b)

We know

x(s) = –st

–x(t) e dt

By putting the value of x(t), we get

x(s) = –2t – t –st

–3e u(t) – 2e u(t) e dt

On solving and simplifying, we get

x(s) = 23 2 s – 1–

s 2 s 1 s 3s 2

to determine ROC

e–t u(t) 1

s 1 ; ROC > –1

e–2 u(t) 1s 2

; > –2

for which both terms converge is > –1

114. (a)

Signal x(t) is Real X( j ) is conjugatesymmetric.

X( j ) is conjugate symmetric then it’s Realpart should be even and its imaginary partshould be odd

X( j ) = Re [ X( j ) ] + img. [ X( j ) ]

given Re [ X( j ) ] = 0

then X( j ) = img.[ X( j ) ] is odd

if X( j ) is odd and imaginary then x(t)should be Real and Odd.

115. (c)

Givenx(t) = e–b| t |

x(t) = e–bt u(t) + ebt u(–t) ...(1)

e–bt u(t) LT1 – b

s b

...(2)

ebt u(–t) LT1 b

s – b ...(3)

So laplace transform of x(t) is

X(s) = 2 2–2b ; – b b

s – b

116. (c)

• For a finite length signal, the ROC is theentire plane, therefore, there can be no polesin the finite z-plane for a finite lengthsignal.

• Since the signal is absolutely summable,the ROC must include unit circle for left

sided signal 0 < |z| < 12

does not include

unit circle so signal is right sided signaland stable because it in clude unit circle.

117. (c)

Let g(t)= dx(t)

dt

t1–1

1

+t

–1

+1–1

+1

Taking Fourier transform of given signal

G(j ) = j j2 sin( .1) ( 1)e ( 1)e

j X( j ) = j j2sin e e

22sin 2cosx( )

jj

118. (b)

Given

F.Th(t) H(j ) = 22

(1 j ) …(1)

x(t)=

F.Tcost ( 1) ( 1) X( j )

…(2)

y(t) = h(t) * x(t)

Y(j ) = X(j ) H(j )

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1x(t) (t t ) = 1 1x(t ) (1 t )

Putting the values from eqn. (1) and eqn.(2)

=2 2

( 1) ( 1)2(1 j( 1) (1 j)

= 2 2( 1) ( 1)2

(1 j) (1 j)

= 2 2 j ( 1) j ( 1)4

= IFTj{ ( 1) ( 1)} sint

119. (b)

Given

x(t) = 0 0 0 0

0 0

j t j t j t j t

j(2 t /4) j(2 t /4)

e e e e1 22j 2

e e2

Since we know

x(t) = 0jn tn

n 0C e

x(t) = 0 0j t j t0 1 1C C e C e

0 02j t j2 t2 2C e C e

on comparing eqn (1) and eqn. (2)

C–2 = j /41 e2

…(3)

C2 = j /41 e2

…(4)

C2 + C–2 = j /4 j /41 e e2

= 1 2cos / 42

= cos / 4

=12

120. (c)

2G R s 1G

C(s)

2

2

G1 G R s 1G

C(s)

Transfer function =

1 2

2

G GC sR 1 Gs

121. (c)

• Except for certain special design, all ACservomotors are primarily of 2 inductionmotor.

• The rotor of the servomotor is built withhigh resistance, so that its X/R ratio is smalland the torque-speed characteristic is linear.

• The diameter of the rotor is kept small inorder to reduce inertia for betteraccelerating characteristics.

122. (c)

123. (b)

R

V1 V2C

2

1

VV = 1

1 RCs

This has a finite pole at s = 1 RC andzero at infinity

124. (b)

T.F. =

2 1

1 1 2

V R sLsV R sL sLs

=

3 3

3 3 3100 10 s 100 10

100 10 s 100 10 100 10

=5

510 0.1s10 0.2s

=6

6s 102s 10

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125. (d)

A system is stable if it is BIBO stable andwhen system output donot increase withinitial conditions when no input is given.These two conditions must be satisfied forstable system.

126. (c)

When the gain is increased, the poles tendto go towards right side, so became underdamped.

Under damped

127. (d)

A state-space model can be applied to linearand non-linear, time invariant and variantsystems.

128. (a)

Given open loop transfer function

G(s) + H(s) = 2K

s(s 2)(s 2s 2)

(a)angle made by pole at s = 0 is

2

(-1+j1) -90°

1-45°-2Real

-135°(-1- j1)

Imag

1

tan = 1

1 = 45°

angle = –180 + 45°= –135°

(b)angle amde by pole at (–1+j1) is –90°(c)angle made by by pole at s = –2 is – 45°

Angle of departure = 180 – ( anglesmade by poles – angles made by zero)

= 180 – (–135 – 90 – 45)

= 180 – (–270)

= 450

= 90°

129. (c)The point where asymptotes of a root locusmeet is called centroid hence statement-I isfalse

Root locus always starts from open loop pole(k=0) and end at open loop zero (k = ).Hence statement-II is true.

130. (a)

A system is controllable if the matrix

D = 2B AB A B............. is non

singular and Rank of A = Rank of D = orderof the matrix. Solving from the options

(a) B = 21

AB = 1 0 20 1 1

= 21

D = 2 21 1

D 0 hence controllable.

option (a) is controllable.

131. (b)

Writing the loop equation & capacitorequation

cc i

dvdiL Ri v v ; C idt dt

L 1 H, R = 10 , C 1 F

1 cc i

dvdi 10i v v ; idt dt

cc i

dvdi 10i v v ; idt dt

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icc

dii10 1 1dt vvdv 1 0 0

dt

A = 10 11 0

132. (c)

• For parallel connection of transformers,primary windings of the Transformersare connected to source bus-bars andsecondary windings are connected tothe load bus-bars.

• Various conditions that must befulfilled for the successful paralleloperation of transformers:

1. Same voltage Ratio & Turns Ratio(both primary and secondary VoltageRating is same).

2. Same Percentage Impedance and X/Rratio.

3. Identical Position of Tap changer.

4. Same KVA ratings.

5. Same Phase angle shift (vector groupare same).

6. Same Frequency rating.

7. Same Polarity.

8. Same Phase sequence.

Some of these conditions are convenientand some are mandatory.

• The convenient are: Same voltage Ratio& Turns Ratio, Same PercentageImpedance, Same KVA Rating, SamePosition of Tap changer.

• The mandatory conditions are: SamePhase Angle Shift, Same Polarity, SamePhase Sequence and Same Frequency.

• When the convenient conditions are notmet parallel operation is possible butnot optimal.

For 3 phase transformers, conditions forparallel operations are

1. Same polarity

2. Zero relative phase displacement

3. Same phase sequence

4. Same voltage ratio133. (b)

If one of the single phase transformenr in3 phase connection is taken out ofservice, the connection will reduced toopen delta connection i.e. V/V.

V/VS =1 S3

= 0.577 × (3 × 100) KVA

= 173.2 KVA

134. (b)For lap wound machine = No. of poles

= No.of parallel pathsNo. of armature turns = 200 × 2 = 400Nol. of armature turns per path

400 1004

Resistance of each path = 100 × 0.04 = 4

Resistance of armature circuit 4 14

135. (c)

When DC machine runs as motor,

1bE = V – IaRa

= 240 – 20 × 2

= 200 V

When DC machine runs as generator,

2bE = a aV I R

= 240 + 20 × 2= 280 V

2 1b bE E = 280 – 200 = 80 V

136. (a)

In lap winding, the number of parallel pathsare equal to number of poles and hence itcan support larger current rating.

In wave winding the paths are 2 and coilsare connected in series in both the parallelpath and hence it can support large voltagerating.

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137. (b)

Rotor power factor = 2´2 ´

2

RcosZ

If slip increases, then rotor frequencyincreases. This will increase ´

2Z and sinceR2 is constant, p.f. will decrease.

138. (b)

Mechanical power developed = (1 – s) ×power

input to rotor

= 10001 0.02

= 980 W

139. (d)

This is kind of motor used depending onthe requirement of load. Universal motor isused in vacuum clearner. In traction ACseries motor is used to get high startingtorque.

140. (a)

A Klystron is a vacuum tube used forgeneration/amplification of microwaves.

An electron beam is produced by oxide coatedindirectly heated cathode and is focussed andaccelerated by focussing electrode.

This beam is transmitted through a glass tube.The input cavity where the beam enters theglass tube is called buncher.

As electrons move ahead they see anaccelerating field for half cycle and retardingfield for the other half cycle.

Therefore, some electrons are acceleratedand some are retarded. This process is calledvelocity modulation.

The velocity modulation causes bunching ofelectrons. This bunching effect convertsvelocity modulation into density modulation ofbeam.

The input is fed at buncher cavity and outputis taken at catcher cavity.

In a two cavity klystron only buncher andcatcher cavity are used. In multi cavity klystronone or more intermediate cavities are alsoused.

The features of a multicavity klystron are :

1. Frequency range - 0.25 GHz to 100 GHz

2. Power output - 10 kW to several hundredkW

3. Power gain - 60 dB (nominal value)

4. Efficiency - about 40%.

A multicavity klystron is used in UHF TVtransmitters, Radar transmitter and satellitecommunication.

141. (a)

It is somewhat similar to TWT and can delivermicrowave power over a wide frequency band.

It has an electron gun and a helix structure.However the interaction between electronbeam and RF wave is different than in TWT.

The growing RF wave travels in oppositedirection to the electron beam.

The frequency of wave can be changed bychanging the voltage which controls the beamvelocity.

Moreover the amplitude of oscillations can bedecreased continuously to zero by changingthe beam current.

It features are:

1. Frequency range - 1 GHz to 1000 GHz.

2. Power output - 10 mV to 150 mW(continuous wave) 250kW (pulsed).

It is used as signal source in transmitters andinstruments.

142. (a)

Applegate diagram is distance time plot.

143. (a)

144. (d)

145. (a)

146. (b)

147. (b)

148. (d)

149. (d)

The desired output of an FIR filter with aninput h(n) and using a window of length M isgiven as

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M –jωnn=0 h n e H

150. (c)

There is a potential problem for frequencysampling realization of the FIR linear phasefilter. The frequency sampling realization ofthe FIR filter introduces poles and zeros atequally spaced points on the unit circle.

151. (c)

152. (d)Forbidden energy gap of Si and Ge are givenas 1.12 eV and 0.72 eV and as we know

2in = gE /KT3

0A T e

ni gE /KTehigher the band gap lower the intrinsicconcentration, which results into lowconductivity

= i nn q = niTherefore, Si has higher forbidden energygap but low conductivity.

153. (a)At absolute zero temperature T 0K allthe electrons in semiconductor remains invalence band (VB) and conduction band (CB)remains empty as no thermal energy isavailable for electrons to move intoconduction band from valence band.

154. (d)Energy gap in diamond (Eg) = 5.47 eVEnergy gap in silicon (Eg) = 1.12 eV

155. (c)Addition of small quantity of silicon ironimproves several of qualities of iron eg. theresistivity increases, the coecive forcedecreases etc. So, the iron with addition ofsilicon is used as soft magnetic material.

156. (d)In case the moving coil instruemnt is usedas a voltmeter a large series resistance ofnegligible temperature coefficient (made ofmaterial like manganin) is used. Thiseliminates the error due to temperature.This is because the copper coil forms a verysmall fraction of the total resistance of theinstrument circuit and thus any change inits resistance has negligible effect on thetotal resistance.

157. (c)The phasor diagram of a wattmeter for alagging power factor is given below

V

IP

I

Here,V Voltage applied to pressure coil circuitIP Current in pressure coil circuitI Current in current coil circuit Angle by which IP lags V

Angle by which I lags V(load angle)

As the know,True power

Actual wattmeter reading

=cos

cos cos( )

× Actual wattmeter reading

[For lagging loads]

158. (c)

Clipper circuit is used to remove the certainportion of a waveform. A clipper can bemade of resistor and diode e.g. half wavediode can be used to remove one half(positive or negative) of sinusoidalwaveform.

159. (a)Tappings in a transformer are provided onhv side due to following reasons:1. No. of turns on hv side is large. Hence

a fine voltage regulation can beobtained

2. The hv winding is placed on the outerside

3. The current in the hv winding is low.

160. (b)

Armature reaction Shifts the MagneticNeutral Axis (MNA) axis of dc axis. Henceshifting the brushes would result in smoothcommutation but it is the effect of shiftingof brushes that flux per pole gets reduced.

answers - ies masteriesmaster.org/public/archive/2016/im-1482668626.pdf · 2017. 5. 26. · r...

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