1

RF RC

VCC

vi

5V

5V

Detailed solution of IES 2103 (ECE) objective Paper – II: Set – ‘A’

1. (C)

Here output impedance is

feF0 C

0

hRR R || r

gm11A

Here fe0 C C

hA gm R .R

r

So 0

50A 4K 200

1k

So

F

0 C

R 40KR R || 4K ||

11 11 200200

0R 4K

2. (D) But it should be 2.84 mA

CEO CBOI 1 I 410 1 .5 81

Now C B CBOI I 1 I 81 30 410 2430 410 CI 2840 A 2.84 mA

3. (C) In general R. bias current of a diode depends upon temperature only but for Si due to

recombination it depends upon R. bias voltage also.

so in general it is indep of R. bias but depends in case of Si.

4. (D)

Let Transistor is in saturation region then using (B–E) Loop.

B B

4.25 100I 0.8 I mA

100

C C

4.85 5I 0.2 I mA

5

For saturation condition is CB

II

but here is not given so one can’t decide whether it is

in saturation or active region.

2

C

Q2 Q1

RRC2RC1

VCC

Trigger

RC

Cc

vo

Cz

Cb

vi

R2

R1

Re

VCC

FrequencyfL

Gain

5. (C)

It has one stable state and one quasi stable state. Other names are one shot, single step circuit,

gate circuit & delay circuit it is called a delay circuit because Monostable generates a fast

transition at predetermined time T after i/p trigger. It is called delay circuit it generates

rectangular waveform which can be used to gate other circuit so it is called gate circuit.

Main applications are gate and delay circuit. Since gating circuits generate rectangular wave

which can be used in counters. It is used for changing pulse width but does not mean in

regeneration of distorted output.

6. (D) 1

* nH Hf f 2 1 *

H Hn 2, f 0.64 f

7. (D)

Low frequency response depends upon coupling, by pass and blocking capacitor.

Value of Lf is inversely proportional to capacitance, so it can be improved or increased by

reducing any one of capacitors.

8. (C) 0fQ

B.W

For increasing band width Q factor must be decreased. So option (1) is not correct.

9. (C) if iZ Z 1 A 1K 1 0.99 100 100Kohm

0of

Z 100Z 1

1 A 1 0.99 100

3

R

C

–

+

vo

vi

C2

Q2 Q1

R2RC2

RC1

VCC

C1

R1

10. (A) Here ii/p v is given as

3

iv 5sin 2 10 t mV

Now

t

0 i

0

1v v dt

RC

t

t 03 3 3

0 5 6 3

0

1 10 5 1v 5sin 2 10 t cos 2 10 t cos 2 10 t 1

10 10 2 10 40

11. (C) Here Q is not 105 but 105

For series LCR 0

0

L 1Q &Q

R RC

5

12

5

150

10 400 C

1C 500 10 F

10 400 50

12. (B) with negative feedback

f

A 100A

1 A 1 0.03 100

f

100A 25

4

13. (B) It has resistive negative feedback so system will be stable at all frequency.

14. (C) For HWR 1.21& FWR 0.48

15. (A) In general differential amplifier is used at i/p stage to () high i/p impedance and this has

high CMRR also

16 . (B) HWR 40.6% & FWR 81.2%

17. (A) CMRR in dB 20log CMRR

1080dB 20log CMRR

So 4 d

c

ACMRR 10

A

4

c 4

2 10A 2

10

18. (C) ECL has the highest speed because it is nonsaturated logic and does not enter into Saturation

so time taken in removal of storage delay time is reduced or becomes zero.

19. (C) It is time taken in reacting from 10% to 90%

20. (A)

It is an astable multivibrator

4

01001000 0111

(8 7 4)10

00 01 11 10

0

1

BCA

1 1

1

H.S.B

H.S.B

x

y

D

Z D

D

1 1 2 2

3 8 4 8 5 4 8

5 3

5

5

T 0.69R C 0.69R C

T 0.69 2 10 10 0.69 2 10 5 10 0.69 2 10 2 10 5 10

T 0.69 2 10 10

1 1 10f 1420.85 Hz

T 0.69 102 10 102 0.69

21. (B) Transistor becomes heated when it enters into saturation or is in saturation or come back from

saturation to cut-off

D CE C D CP V .I P due to in I .

So power dissipation at collector () and which dissipates same heat.

22. (A)

23. (A) f A B A B AA AB BA B B A A B

24. (B) f ABD ABD AB D D = AB

25. (C) f A AB ABC

f A AB ABC A AB (A A)(A B) (A B)

For (A B) 4 NAND gates are required.

26. (A) f A,B,C,D A BC B CD AB ACD BC BCD

27. (None)

f A B A B B B A A

f AB AB BA

28. (A)

f AB BC

29. (B) FET has more temperature stability, high i/p impedance and less noisy than BJT.

30. (B) f AB ACD

f AB C C D D ACD B B

AB CD CD CD CD ABCD ABCD

ABCD ABCD ABCD ABCD ABCD ABCD

ABCD ABCD ABCD ABCD ABCD

31. (B) Full subtractor 2H.S 1 OR gate

5

32. (C) MUX is a Many to one switch and in which all parallel i/ps can be changed to serial

output.MUX is not used for chip select but decoder is used for chip select.

33. (C) & (D) are correct.

PROM Fixed AND + Programmable OR

PLA Programmable AND & OR

PAL Fixed OR + Programmable AND

34. (A)

n 1

n

n

J K Q

0 0 Q

0 1 0

1 0 1

1 1 Q

Here nQ 0 and n 1Q 1

it means J 1 & K X

35. (B) Output depends upon past output as well as present inputs.

36. (B) Dual slope ADC has the highest accuracy on cost of its speed.

37. (None)

2

s 1G s

s s 1

For D.C. gain put s = 0

Ps 0

G 0 1

K limG s 1

So ss

1e 0.5

1 1

38. (D)

2

2

d x t dx t5 7x t y t

dt dt

2s 5s 7 X s Y s

Characteristic equation is 2s 5s 7 0

2

n n

n

7 & 2 5

2.57 & 0.94

7

39. (B)

2

d1 d

1

C s s 4s 1 1

K .sR s 4s s 1 K1

s 4s 1

2

d

2

d

100 1

C s 4s s 1 K

100R s1

4s s 1 K

6

2

d

C s 100

R s 4s s 1 K 100

Characteristic equation is 2

d4s s 1 K 100 0

2

d

ss 1 K 25 0

4

d

n n

1 K5 2

4

d d40 0.5 1 K K 19

40. (C) Settling time s

n

4t

Here n 10 & n2 10

So s

4t 0.8sec

5

41. (D) 4 3 2s 20Ks 5s 10s 15 0

4

3

2

1

0

s 1 5 15

s 20K 10 0

100K 10s 15 0

20K

300K 20Ks 10 0 0

100K 10

s 15

For stable system check

300K 2K10

10K 1

2100K 10 600K

10K 1

2

2

100K 10 600K

600K 100K 10 0

100 10000 24000

K2 600

This equation will be always +ve, so it will be unstable for all values of K.

42. (C) 3 2s 4s s 6 K 0

3

2

1

0

s 1 1

s 4 K 6

4 K 6s 0

4

s K 6

For stable operation 10 K 0

K 10

7

K 6 0 K 6

43. (A)

44. (C) 2

K1 0

s s 6s 10

3 2s 6s 10s K 0

3

2

1

0

s 1 10

s 6 K

Ks 10 0

6

s K

For oscillations K

10 06

K 60

For frequency 26s K 0

26 K 0 10 3.16 Rad/ sec

45. (D)

C s G s

R s 1 G s H s

For closed loop poles

1 G s H s 0

2

2

11 0 s 4 4s 1 0

s 2

2 4 16 20 4 2js 5 4s 0 s s 2 1j

2 2

46. (A) Band width is Noise filtering characteristics higher is the B.W, higher is noise value.

47. (D) 21

pM e

48. (B) Since K = 10

So gain in dB at 1 will be 20 dB.

49. (C)

50. (C)

51. (C) G.M 20log40 20log20

20 log 2 6dB

G.M is value of additional gain to make a system unstable.

52. (D) For 0 system is undamped so marginally stable hence P.M = 0

So here option (4) is correct and option (3) is incorrect. Hence (D) is correct.

Alternatively formula for P.M. in terms of is:

8

1

2 4

2P.M. tan

2 1 4

53. (D)

1G s H s

s s 1

190º tan

For PCF 180º 1tan 90º tan90º

2

1G s H s 0

1

1G.M

G s H s

54. (A) Addition of zero means system becomes phase lead system.

55. (C) m 4 25 10 Rad/ sec

56. (A) 1 2 2 21 2

1 1H p log p logp p

2 2

1 1H log 2 log 64

2 64

1 6 38 19

H bits/outcome2 64 64 32

So Rate (in bps) 19 bits 16outcomes

32 outcome sec

19bits/ sec

2

57. (D) 2

SC Blog 1

N

S/N 15dB

So 1015 dB 10log S/N

S/N anti log 1.5 =31.6

4 4 3

2

log32.6C 10 log 1 31.6 10 50.27 10 bps

log 2

58. (A)

2

1I x log

p x bits/message

10

1I x log Hartley

p x

59. (B) 2 2 2

1 1 1H log 2 log 2 log 8

2 2 8

1 1 3 3 7H8 42 2 8

H 7/4

7/8L 2

9

60. (C) r 2 4 2 16KHz

2 2 2 2

1 1 3H log 8 log 8 log 8/3 3/8log 8/3

8 8 8

2 2

2 3 2H 3 log 8 log 3

8 8

3 3 0.4771 3

3 1 3 1.56 1.84 4 0.3010 4

H = 1.8 level/message=3.6 bits/message

Information rate= 33.6bits / message 4000message / sec 14.4 10 kbps

61. (A) 2

at c

MP P 1

2

If am 1 t cP 1.5 P

Modulating Power = 0.5 Pc

m

t

P1/3

P

62. (C) c mf f 555

m

c

2f 10

f 560KHz

63. (B) mB.W 2 f f

m

m

ff 2.4f

f

So B.W = 6.8 fm

64. (B) 2

at c

mP P 1

2

c

2

ac

P 1000W (1)

mP . 400W (2)

2

So 2

am 400 2

2 1000 5

2 2

a a a

4m m 0.8 m 0.894

5

65. (C) ma

c

A 15m 0.25

A 60

66. (C) sf 10B.W n. 7 35KHz

2 2

67. (A)

68. (B) t rd 4.12 h h

4.12 15 1.5 21Km

69. (A) Duration without collision = 310 10 10 0.1

70. (D) All are correct

10

b

a(a > b)

d

MicrostripSubstrate

Ground plane

d

Strip Conductor

Dielectric Substrate

Ground plane

71. (A) Kenedy book objective question

72. (A) Access time is sum of seek time and rotational latency

acuss

0.5T 30ms sec 30 16.67msec 47msec

30

73. (A) Kenedy book objective question

74. (D) For cf f signal passes there a W.G. so it is HPF.

75. (C) Mode excitation in rectangular W.G.:

Modes are excited by use of a typical coaxial line W.G. probe coupling. In order to launch a

particular mode probe is chosen which will produce lines of E & H that are roughly (||) to

lines of E & H of particular mode.

76. (B) Dominant mode of Rectangular W.G. is TE10.

77. (C) For a 3-port reciprocal N/W if all 3 ports are matched then it is physically not possible. So

one can terminate one port as short so other two ports can be matched.

3-port Non reciprocal matched

78. (D)

(Micro strip line uses high dielectric constant so that it reduces radiation losses because most

of the electromagnetic field is concentrated in a dielectric b/w conductive strip and ground

plane.

The real benefits in having a dielectric constant is that pekoes size () approx by square root

of dielectric const.

79. (D) Kenedy book quetsions

80. (A) Kenedy book question

81. (A) In LASER: SE is for Stimulated emission

82. (D)

83. (C) In OFC there has less interference

84. (B) Kenedy book question

85. (D) 2

10110.11 22.75

1 2

10110 22

0.11 1 2 1 2 3/4 0.75

86. (B)

87. (C) 1 2 3 4 5 6 7h h h h h h h

1 1 1 1 1 0 1

Hammg code

11

* 0 1 1 1× 1 1 1

0 1 1 1 0 1 1 1 ×0 1 1 1 × 0

44,5,6,7 1101 C 1 Parityodd.

22,3,6,7 1101 C 1 odd

11,3,5,7 1111 C 0 even

110 6th

position is correct

So correct code is 1111111

Hamming code is used for 1 bit correction so only option (C) is correct

88. (A) 2C 2Blog M

28000 1000 log M

So But M = 2N

Where N No. of bits

So N 8

Signal elements = 82 256

89. (D)

90. (D)

91. (B) In vectored address is not provided externally.

92. (A)

93. (B)

94. (C) DAD (Double Add) allows 16-bit addition b/w HL register pair and any one of BC, DE, HL

or SP register pairs so it is not necessary that register must be DE.

95. (A)

96. (C)

97. (C) TRAP works as both Level & edge trigged but can work as +ve edge

98. (A)

99. (B)

Need of 2, 4-bit address.

100. (A) These are not used in DMA controller.

101. (C) CMRR of OPAMP is high.

102. (B) Reason is not correct

103. (C) Transistor does not enter at all in saturation but can enter into cut-off

104. (A)

105. (D) Base width is decreased due to Early effect.

106. (A)

107. (A)

108. (C) PIN Photo diode is preferred to PN photo diode in high speed or at high frequency. It can n’t

be used as constant voltage generator because it detects light and increases current with

reference to light.

12

109. (D) Derivative control action depends upon how fast error is changing with respect to time. If

error is constant controller output is zero.

110. (B)

111. (A)

112. (C) Pole can be introduced any where order should be increased.

113. (A)

114. (C) Baud rate is not after modulation

115. (A)

116. (A)

117. (B)

118. (C) Gain of slot antenna is in range of 10-15dBi while that of Horn is in the range of 25 dBi

119. (B)

120. (A)