Solutions All MOCK Tests

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GATE Kanodia Mock Test Solutions

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ANSWER KEY Practice Papers EC
Practice Paper-A
1. D 16. C 31. B 46. C 61. B 2. B 17. C 32. C 47. C 62. C 3. A 18. B 33. D 48. B 63. C 4. D 19. D 34. C 49. B 64. D 5, A 20. D 35. C 50. D 65. A 6. B 21. C 36. B 51. C 7. B 22. A 37. C 52. B 8. A 23. D 38. B 53. D 9. B 24. C 39. B 54. D . 10. D 25. C 40. A 55. B 11. C 26. A 41. B 56. B 12. B 27. B 42. A 57. C 13. B 28. A 43. B 58. C 14. B 29. B 44. B 59. C . 15. B 30. A 45. B 60. B

Practice Paper-B
1. C 16. B 31. C 46. B 61. A 2. B 17. A 32. C 47. C 62. A 3. B 18. B 33. B 48. B 63. C 4. A 19. A 34. B 49. A 64. A 5. C 20. B 35. D 50. A 65. D 6. C 21. A 36. C 51. D 7. D 22. B 37. A 52. B 8. C 23. C 38. C 53. C 9. A 24. A 39. B 54. A . 10. B 25. D 40. A 55. D 11. D 26. D 41. C 56. B 12. B 27. D 42. A 57. B 13. B 28. A 43. D 58. B 14. B 29. C 44. A 59. C . 15. B 30. C 45. A 60. C

Practice Paper-C
1. D 16. D 31. C 46. D 61. C 2. B 17. C 32. C 47. A 62. C 3. D 18. D 33. D 48. B 63. D 4. C 19. B 34. A 49. A 64. D 5, A 20. B 35. D 50. B 65. A 6. A 21. A 36. C 51. D 7. A 22. C 37. A 52. B 8. C 23. D 38. B 53. B 9. A 24. A 39. C 54. B . 10. C 25. A 40. A 55. B 11. D 26. C 41. C 56. A 12. D 27. D 42. C 57. D 13. C 28. C 43. C 58. B 14. D 29. A 44. C 59. D . 15. D 30. B 45. B 60. B

Practice Paper-D
1. A 16. B 31. C 46. C 61. C 2. A 17. B 32. B 47. D 62. A 3. A 18. B 33. A 48. B 63. D 4. B 19. B 34. C 49. D 64. D 5, A 20. B 35. A 50. D 65. B 6. D 21. C 36. A 51. B 7. D 22. A 37. B 52. A 8. A 23. B 38. C 53. D 9. C 24. A 39. A 54. B . 10. D 25. D 40. A 55. C 11. A 26. B 41. C 56. C 12. A 27. B 42. D 57. D 13. B 28. C 43. B 58. B 14. D 29. C 44. B 59. D . 15. B 30. C 45. D 60. C

Practice Paper-E
1. A 16. B 31. B 46. C 61. C 2. D 17. B 32. A 47. A 62. D 3. C 18. B 33. C 48. B 63. B 4. C 19. B 34. C 49. D 64. D 5. B 20. B 35. C 50. A 65. D 6. C 21. C 36. D 51. B 7. B 22. D 37. A 52. C 8. B 23. B 38. C 53. A 9. C 24. D 39. B 54. A . 10. B 25. C 40. D 55. C 11. C 26. C 41. C 56. B 12. B 27. B 42. B 57. A 13. C 28. D 43. B 58. A 14. A 29. B 44. D 59. A . 15. A 30. C 45. B 60. D

SOLUTIONS EC_Practice Paper A

SOL 1.1

Option (C) is correct. We have Z = [A + BC + D + EF ] [A + BC + D (E + F ) = [A + BC + EF + D] [A + BC + D + EF ] S S S 14 2 43 X Y X Y = (X + Y) (X + Y ) = X = A = BC For A = 0, Z = BC

SOL 1.2

Option (B) is correct. p (heart) = 13 = 1 , 52 4 I = log 2 4 = 2 bits Option (B) is correct. The processing gain is given as W = 500 or 27 dB R The εb /J 0 required to obtain an error probability of 10−5 for binary PSK is 9.5 dB. Hence, the jamming margin is εb Jav W b Pav l = b R l − a J 0 k = 27.95 or 17.5 dB dB dB dB

SOL 1.3

SOL 1.4

Option (A) is correct.

SOL 1.5

75 + j 25 − 60 Γ = ZL − Z 0 = = 0.212+48.55c 75 + j 25 + 60 ZL + Z 0 1+ Γ s = = 1 + 0.212 = 1.538 1 − 0.212 1− Γ Option (C) is correct. Gd = U = 4πr 2 Pave Uave Prad 103.6 # 15 # 103 = 1.32 mW/m 2 rad Pave = Gd # P = 4πr 2 4π (60 # 103) 2

SOL 1.6

Option (A) is correct.

Page 2

EC_Practice Paper A

Chapter 1

We know that adj(adj A ) = A n − 2 : A. Here n = 3 , and A = 3 So, adj (adjA) = 3(3 − 2) : A = 3A .
SOL 1.7

Option (D) is correct. Let I = I = I =

#
0

π 2

log (tan x) dx log tan a π − x k dx 2 log (cot x)

...(i)

# 0 # 0 # 0 # 0 # 0

π 2

π 2

...(ii)

Adding (1) and (2), we get 2I = = =
SOL 1.8
π 2

[log (tan x) + log (cot x) dx log (tan x. cot x) dx log 1dx = 0 & I = 0

π 2

π 2

Option (C) is correct. Let p1 = 0.4 , p2 = 0.3 , p 3 = 0.2 and p 4 = 0.1 P (the gun hits the plane) = P (the plane is hit at least once) = 1 − P (the plane is hit in none of the shots) = 1 − (1 − p1) (1 − p2) (1 − p 3) (1 − p 4) = 1 − (0.6 # 0.7 # 0.8 # 0.9) = (1 − 0.3024) = 0.6976 Option (A) is correct. 4: P = 0 , 4# P =−2Pz u y ! 0 2x P is a possible EM field 4: Q = 0 , 4# Q = 1 2 [10 cos (ωt − 2ρ)] u z ! 0 ρ 2ρ Q is a possible EM field sin φ 4: R = 1 2 (3ρ2 cot φ) ! 0, ρ 2ρ ρ R is not an EM field. 2 (sin2 φ) 1 !0 4: S = 2 sin (ωt − 6r) 2r r sin θ S is not an EM field.

SOL 1.9

SOL 1.10

Option (B) is correct. PXX = 1 2π

#

3

−3

6ω2 (1 + ω2) 3

Page 3

EC_Practice Paper A
3 ω = 6 ' −ω 2 2 + + 1 tan−1 (ω)1 2 π 4 (1 + ω ) 8 (1 + ω ) 8 0

Chapter 1

SOL 1.11

=3 8 Option (D) is correct. Initially plot has slope of 40 dB/decade so there must be two zero at origin. At ω = 10 slope changes to -60 dB/decade from 40 dB/decade. So there must be 5 Ks2 . The 40 pole at ω = 10 . Thus transfer function must be of the form = (1 + 0.1s) 5 dB/decade line will cross ω = 1 at 0 dB, 20 log K = 60 & K = 1000 , Option (B) is correct. Closed loop transfer function is T (s) = G (s) K (s2 + 1) = 1 + G (s) (1 + K) s2 + 3s + (2 + K)

SOL 1.12

SOL 1.13

1+K > 0 & K >−1 & K >−1 2 + K > 0 & K > − 24 Option (B) is correct. 1 2 1 + s 1 T (s) = 2 = s α s + αs + 1 1 + s2 + 1 Thus an equivalent system has G (s) = 2 1 and H (s) = αs . For this function s +1 root-locus is (B).

SOL 1.14

Option (C) is correct. X [k ] = 1 T x (t) e−jkω t dt = 1 # Ae−jkω t dt # T −T/4 −T/2
o o

T/2

T/4

−jkω t 4 = A; e = A sin b πk l 2 T − jkωo E−T πk
o

T

4

SOL 1.15

Option (B) is correct. X (z) = =
n =− 3 1 3 1 3

z / (3z−1) n = / b 1 3 l
n=1

−1

3

n

SOL 1.16

z , z <3 1− z = z 3−z Option (B) is correct. y [n] = {1, 4, 2, − 4, 1}

Page 4

EC_Practice Paper A

Chapter 1

SOL 1.17

Option (A) is correct. x (t) = 1, E3 =

#

3

−3

x (t) 2 dt = 3

SOL 1.18

So this is a power signal not a energy. 2 T P3 = lim 1 x (t) dt = 1 T " 32T −T Option (B) is correct. Step size = 9.99 = 10 mV 999

#

&
SOL 1.19

0110 1001 0101 6 9 5 m 6.95 # 10 = 6.95 V

Option (A) is correct. The truth table is as shown below A 1 1 B 1 0 X 0 0 Y 1 1

SOL 1.20

Option (B) is correct. Y = A (B + C (AB + AC )) = AB + AC [(A + B ) (A + C )] = AB + AC (A + AC + AB + BC ) = AB Option (A) is correct. In the circuit VGS = VG − VS − 0.5 = 1 V VDS = VD − VS = 0.5 − 0.5 = 0 V VDS (sat) = VGS − VTH = 1 − 0.4 = 0.6 V Here, VDS < VDS (sat) and VGS > VTH So, M1 is in linear (triode) region Option (B) is correct. When vi > 2 V , output is positive. When vi < 2 V , output is negative. The waveform is as shown below

SOL 1.21

SOL 1.22

Page 5

EC_Practice Paper A

Chapter 1

SOL 1.23

5π −π 6 Duty cycle = tON = 6 =1 t 2π 3 Option (C) is correct. During positive cycle when vi < 8 V, both diode are OFF vo = vi . For vi > 8 V, vo = 8 V, D1 is ON. During negative cycle when vi < 6 V, both diode are OFF, vo = vi . For vi > 6 V, D2 is on vo =− 6 V.

SOL 1.24

Option (D) is correct. Power P = vi 2 = 2ix # ix = 2i x ix = 4 A, P = 32 W (absorb) Option (A) is correct. Option (D) is correct. 2ε (V + VR) 1 1 2 W = ) s bi :Na + Na D3 e WA = (Vbia + VR) (NaA + NdA) (NaB NdB) 2 =(V + V ) (N + N ) N N G WB aA dB bib R aB dB d Vbi = Vi ln c Na N n i2 m 18 15 VbiA = 0.0259ln c 10 # 10 20 m = 0.754 V 2.25 # 10 18 16 VbiB = 0.0259ln c 10 # 10 20 m = 0.814 V 2.25 # 10 WA = 5.754 1018 + 1015 1016 2 = 3.13 =b 5.814 lc 1018 + 1016 mc 1015 mG WB
1 1 1

SOL 1.25 SOL 1.26

SOL 1.27

Option (A) is correct. The small-signal equivalent circuit is shown in fig.

Page 6

EC_Practice Paper A

Chapter 1

vo =− gm vgs (RD RL), vi =− vgs Av = vo = gm (RD RL) = (2m) (5k 4k) = 4.44 vi
SOL 1.28

SOL 1.29

SOL 1.30

Option (C) is correct. VSD = VSG , k' ID = p W (VGS + VTP ) 2 2 L 10−4 = b 25 lbW l (2.5 − 1.5) 2 & W = 32 μm 2 4 Option (C) is correct. Full wave rectifier vs = vs = 120 sin 2π60t V v max = 120 − 0.7 = 119.3 V vrip = 119.3 − 100 = 19.3 V 119.3 = 20.6 μF C = v max = 2fkvrip 2 (60) 2.5 # 103 # 14.4 Option (C) is correct. For each successive gate, that has a transistor in saturation, the current required is I V − VCE (sat) IB (sat) = C (sat) = CC β βRC = 5 − 0.2 = 0.15 mA 50 (640) For n attached gate Io = nIB (sat) To assure no logic error Vo = VCC − Io RC > VH = 3.5 V n # VCC − 3.5 = 5 − 3.5 = 15.6 & n # 15 RC IB (sat) 640 (0.15m)

SOL 1.31

Option (C) is correct. LHLD 3000H MOV E, M INX H MOV D, M LDAX D MOV L, A INX D LDAX D

;(3000A) " HL = 3002H ; (3002H) " E = 00 ; HL + 1 " HL = 3003H ; M " D = (3003H) = 02H ; (DE) " A = (3000H) = 02H ; A " L = 02H ; DE+1 " DE = 3001H ; (DE) " A = (3001) = 30H

Page 7

EC_Practice Paper A

Chapter 1

MOV H, A Hence HL pair contain 3002H.
SOL 1.32

; A " H = 30H

Option (C) is correct. Let Where I =

# 0

π

cosm x sinn xdx =

# 0

π

f (x) dx

f (x) = cosm x sinn x f (π − x) = cosm x (π − x) sinn (π − x) = (− cos x) m (sin x) n =− cosm x sinn x , if m is odd I =

# 0

π

cosm x sinn xdx = 0 , if m is odd

SOL 1.33

Option (B) is correct. Let f (z) = cos πz then f (z) is analytic within and on z = 3 , now by Cauchy’s integral formula f (z) f (z) dz = 2πif (z 0) f (z 0) = 1 # dz & # 2πi z − z 0 z − z0 take f (z) = cos πz, z 0 = 1, we have πz dz = 2πif (1) = 2πi cos π =− 2πi # cos z−1
z =3 c c

SOL 1.34

Option (B) is correct. Here f (x, y) = x + y2, x 0 = 0, y 0 = 0 We have, by Picard’s method y = y0 + y(1) = y 0 + = 0+ y(1) = y0 + = 0+

#
x0

x

f (x, yo) dx

The first approximation to y is given by

#
x0

x

f (x, y 0) dx = 0 +

#
0

x

f (x, 0) dx

#
0

x

2 xdx = x 2

The second approximation to y is given by

# x # 0

x

f (x, y0) dx = 0 +

0

# 0

x

f (x, 0) dx

x

2 xdx = x 2

The second approximation to y is given by y(2) = y 0 + =

#
x0

x

f (x, y(1)) dx = 0 +
4 2 5

#
0

x

f bx, x l dx 2

2

#
0

x

x x x bx + 4 l dx = 2 + 50

The third approximation is given by y(3) = y 0 +

#
x0

x

f (x, y(2)) dx

Page 8

EC_Practice Paper A

Chapter 1

= 0+ =

#
0

x

2 5 f bx, x + x l dx 2 20

#
0

x

x4 x10 2x7 bx + 4 + 400 + 40 l dx

SOL 1.35

2 5 8 11 =x +x + x + x 2 20 160 4400 Option (C) is correct.

1 s (s + 10s + 29) 1 T (s) = = 10 (s + α) (s + 3) α10 (s + 3) 1+ 1+ 2 s (s − 1) s (s2 + 10s + 29) α10 (s + 3) H (s) G (s) = s (s2 + 10s + 29) (A) is not a valid root-loci because all root-loci ends on a zero. (B) is also not a valid root-loci because loci exist to the left of even number of pole. Characteristic equation is s3 + 10s2 + (29 + 10α) s + 30α = 0
2

s3 s2 s1 s0

1 10 29 + 7α 30α

29 + 10α 30α

For α # 0 root-loci does not intersect the imaginary axis.
SOL 1.36 SOL 1.37

Option (D) is correct. Option (A) is correct. Routh table is as shown below

P (s) =− 6s 4 − 6 ,

Page 9

EC_Practice Paper A

Chapter 1

dP (s) =− 24s3 , ds There is two sign change from the s 4 row down to the s0 row. So two roots are on RHS. Because of symmetry rest two roots must be in LHP. From s6 to s 4 there is 1 sign change so 1 on RHP and 1 on LHP. Total LHP 3 root, RHP 3 root .
SOL 1.38

Option (C) is correct. x1 (t) = 6 sin c (100t) cos (200πt) X1 (f ) is convolution of two signals whose spectrum covers f = ! 50 Hz and ! 100 Hz. So convolution extends over f = 150 Hz and sampling rate N1 = 2f = 300 Hz x 2 (t) = 10 sin c2 (100t) Taking Fourier transform f x 2 (f ) = 0.1tri b 100 l B = 100 Hz Sampling rate N2 = 2B = 200 Hz

N1 = 300 = 3 200 2 N2
SOL 1.39

Option (B) is correct. Exponential series of g (t) is g (t) =

/ Xm e−jπω t
0

3

n =− 3

Fundamental period T0 = 2 ω0 = 2π = 2π = π 2 T0 Coefficient Xm is given by Xm = 1 T0 For m = 0 For m > 0 Xm = 1 2

# g (t) e

−jπω0 t

=1 2

#
0

2

te−jπωt dt

T0

2 X 0 = 1 # tdt = 1 20

#0

2

te−jmπt dt

2 1 = 1 =e−jmπt ; t − 2 2 − jmπ (− jmπ) E G 0

Page 10

EC_Practice Paper A

Chapter 1

= 1 );e−j 2mπ b 2 + 21 2 lE − (1) b 1 − jmπ π m 2 m2 π2 l3 e−j 2mπ = cos (− 2πm) + j sin (− 2πm) = 1 + j.0 So, Xm = 1 :(1) 2 + 21 2 − 21 2 D − jmπ π m 2 πm j j Xm = = e mπ mπ Option (A) is correct. x (t) = 5rect a t k * [δ (t + 1) + δ (t)] 2
π 2

SOL 1.40

by convolution property g (t) * δ (t − t 0) = g (t − t 0) so x (t) = 5rect a t k * δ (t + 1) + 5rect a t k * δ (t) 2 2 x (t) = 5rect b t + 1 l + 5rect a t k 2 2 x b 1 l = 5rect b 3 l + 5rect b 1 l 2 4 4 = 5 (0) + 5 (1) = 5
SOL 1.41

Option (C) is correct. Let Thevenin equivalent of both network

2 vTH R RTH + R k 2 vTH Pl = R f R + RTH p 2 2 = 4 a vTH R k 2R + RTH

P =a

Thus P < Pl < 4P
SOL 1.42

Option (A) is correct. Thevenin equivalent across the inductor for t > 0 can be obtain as

Page 11

EC_Practice Paper A

Chapter 1

by applying node equation is = 0.01vx +

(vs − va + vin) 100

100is = 1.va + vs − va + vin vs =− vin + 100is For thevenin’s equivalent circuit

vs = vth + is Rth So, vth =− vin =− 10 V, (for t > 0 ) Rth = 100 Ω − for t < 0, iL (0 ) = 0 iL (3) = − 10 =− 0.1 A 100 iL (t) = 0.1 61 − e τ @, where τ " time constant −3 τ = L = 20 # 10 100 Rth
−t

So,
SOL 1.43

iL (t) = 0.1 [1 − e−5000t] = (− 0.1 + 0.1e−5000t) A

Option (D) is correct. The π equivalent circuit of coupled coil is shown in figure below

L1 L 2 − M 2 = M

L1 L2 (1 − k 2) k

Page 12

EC_Practice Paper A

Chapter 1

= Output is zero if

0.12 # 0.27 (1.05) 2 = 0.27 0.5

−j + jCω = 0 0.27ω C = 1 2 = 33.33 μF 0.27ω

SOL 1.44

Option (A) is correct. V1 = 0.5V1 + I1 + 2 (I1 + I2) + aI1 & V1 = (6 + 2a) I1 + 4I2 V2 = 2 (I1 + I2) + aI1 & V2 = (2 + a) Ia + 2I2 For reciprocal network z12 = z21 , 4 = 2+a & a = 2 Option (A) is correct. P (miss/or more aircraft) = 1 − P (miss 0) = 1 − P (0 arrive) (λt) 0 e− λt = 1− 0! = 1 − e− λt = 1 − e− 60 (2) = 1 − e0.4 = 0.33
12

...(i) ...(ii)

SOL 1.45

SOL 1.46

Option (B) is correct. x (t) can be written as x (t) = (40 + 8 cos 40πt) cos 400πt Modulation index α = 8 = 0.2 40 Pc = 1 (40) 2 = 800 W 2 The components at 180 Hz and 220 Hz are side band Psb = 1 (4) 2 + 1 (4) 2 = 16 W, 2 2 16 Eeff = Psb = Pc + Psb 800 + 16

SOL 1.47

Option (C) is correct.

# F : dL = e # + # + # + #
L 1 2 3 4

o F : dL

The figure is as shown below

Page 13

EC_Practice Paper A

Chapter 1

For segment 1, y = 0 = z, ,

F = x2 ux , dL = dxu x
0 3 x2 dx = x 3 0 1

# F : dL = #
1

1

=− 1 3

For segment 2, x = 0 = z,
2

# F : dL = 0

F =− y2 u z , dL = dy u y

For segment 3, z = x,
3

dx = dz
1 0 1 3 (x2 − 1) dx = :x − x D =− 2 3 3 0

# F : dL = #

For segment 4, x = 1 F = u x − zu y − y2 u z , dL = dy u y + dz u z

# F : dL = # (− zdy − y dz) : z = y
2

4

&

# F : dL = #
4 L

0

1

(− y − y2) dy = ;− =− 1 6

y2 y 3 0 − E =5 2 3 1 6

+0−2+5 # F : dL =− 1 3 3 6
SOL 1.48

Option (B) is correct. o1 =− x1 + x2 , x o2 =− x2 + u , x o3 =− 2x 3 + u x R0V R− 1 1 0V W S W S o = S 0 − 1 0W x + S1W u, x S S W W S1W S 0 0 − 2W TR XV T R− 1 1 0V X S0W S W A = S 0 − 1 0W, B = S1W S S S1W W S 0 0 2W W T X T X

Page 14

EC_Practice Paper A

Chapter 1

SOL 1.49

R− 1 1 0VR0V R 1V WS W S W S AB = S 0 − 1 0WS1W = S− 1W S W W S WS S− 2W S1W S 0 0 − 2W XV X X T T T R− 1 1 0VR 1V R− 2W S W S W S 2 A B = S 0 − 1 0WS− 1W = S 1W S S 0 0 − 2W WS S− 2W W S S 4W W T XT X R0 T 1 X − 2V W S Cm = 9B AB A2 BC = S1 − 1 1W S W S1 − 2 4W X T Option (B) is correct. y = 10x1 − 10x2 + 10x 3 , y = 810 − 10 10B x R− 1 1 0V W S A = S 0 − 1 0W, C = 810 − 10 10B S W S 0 0 − 2W XR T V S− 1 1 0W CA = 810 − 10 10BS 0 − 1 0W = 810 0 − 20B S W S 0 0 − 2W T R− 1 1 0V X W S CA2 = 810 0 − 20BS 0 − 1 0W = 810 − 10 40B S W S 0 0 − 2W X T R C V R 10 − 10 10V W W S S OM = SCA W = S− 10 0 − 20W S W W S S 10 − 10 40W SCA2W X X T T Option (A) is correct. For λ line tan βl = 3 4 2 1002 = 32 − j 24 Zin1 = Z 0 = 200 + j150 ZL For λ line, tan βl = 1 8 0 + jZ 0 = jZ 0 = j100 Ω Zin2 = Z 0 c Z0 + 0 m At the 7λ line Zin1 Zin2 will be load. 8 (32 − j 24) (j100) = 47.06 − j11.76 ZL = 32 − j 24 + j100 tan βl = tan b 2π : 7λ l =− 1 8 λ 47.06 − j11.76 + j100 (− 1) Zin = 100 e = 94.11 − j 76.45 100 + j (47.06 − j11.76) (− 1) o

SOL 1.50

Page 15 SOL 1.51

EC_Practice Paper A

Chapter 1

Option (D) is correct.

If ZL = 3, at the input end of λ line 8 Z + jZ 0 =− jZ 0 =− j100 Zin2 = Z 0 c L Z 0 + jZL m ZL = (32 − j 24) (− j100) = 19.5 − j 24.4 Zin = 100 e
SOL 1.52

19.5 − j 24.4 + j100 (− 1) = 64 − j148 Ω 100 + j (19.5 − j 24.4) (− 1) o

Option (B) is correct. The total SNR for three hops is 20 + 13 dB. Therefore the SNR per hop is 20/3. The probability of a chip error with non-coherent detection is ε P = 1 e 2N 2 where εc = 20 . The probability of a bit error is 3 N0
c 0

Pb = 1 − (1 − p) 2 = 1 − (1 − 2p + p2) = 2p − p2 ε ε = e− 2N − 1 e− 2N = 0.0013 2
c c 0 0

SOL 1.53

Option (C) is correct. In the case of one hop per bit, the SNR per bit is 20, Hence, ε Pb = 1 e− 2N = 1 e−10 = 2.27 # 10−5 2 2
c 0

SOL 1.54

Option (B) is correct. VGS 2 = VGS 3 = 2.76 V ID 4 = Kn 4 (VGS 4 − VTN ) 2 = Kn 3 (VGS 3 − VTN ) 2 = 100 # 10−6 (2.76 − 1) 2 = 0.31 mA Option (C) is correct. VGS = 4.2 V, VDS = 0.1 V VDS < VGS − VTN , Thus transistor is in non saturation. ID = 5 − 0.1 = 0.49 mA 10k
2 ID = k n W {2 (VGS − VTN ) VDS − V DS } 2 L 0.49 = 0.015 bW l {2 (4.2 − 0.8) (0.1) − (0.1) 2} L 0.49 = bW l (0.67) & W = 0.731 L L '

SOL 1.55

& &

SOLUTIONS EC_Practice Paper B

SOL 1.1

Option (D) is correct. A is Hermitian then AQ = A Now, (iA)Q = i AQ =− iAQ =− iA, & (iA)Q =− (iA) Thus iA is Skew-Hermitian. Option (B) is correct. u = log eu = x2 + y2 , x+y

SOL 1.2

x2 + y2 = f (say) x+y

f is a homogeneous function of degree one u u 2f 2f x +y = f & x2e + y2e = eu 2x 2y 2x 2y or xeu 2u + yeu 2u = eu 2x 2y or, x2u + y2u = 1 2x 2y
SOL 1.3

SOL 1.4

Option (D) is correct. P (none dies) = (1 − p) (1 − p) ...n times = (1 − p) n P (at least one dies) = 1 − (1 − p) n P ( A1 dies) = 1 {1 − (1 − p) n} n Option (C) is correct. In a minimum phase system, all the poles as well as zeros are on the left half of the s -plane. In given system as there is right half zero ^s = 5h, the system is a nonminimum phase system. Option (A) is correct. In open loop system change will be 10% in C1 also but in closed loop system change will be less C2 = 10 = 10 , 10 + 1 11

SOL 1.5

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Page 2
' = C2

EC_Practice Paper B

Chapter 1

9 = 9 C is reduced by 1% 9 + 1 10 2 10 =1− 1 s s + 10 s (s + 10)

SOL 1.6

Option (A) is correct. C (s) = &

SOL 1.7

c (t) = 1 − e−10t a = 10 , Rise time Tr = 2.2 = 2.2 = 0.22 s a 10 Setting time Ts = 4 = 0.4 s a Option (A) is correct. F= = IdL # B # 0
1

# 0

1

100dzu z #

SOL 1.8

Option (D) is correct.

− 100μ0 u y = 0.4u x N/m 2π (5 # 10−3)

SOL 1.9

118.43 sin (120πt) i = vab = = 0.47 sin (120πt) A 250 R Option (C) is correct. Since antenna is installed at conducting ground, 2 Rrad = 80π2 b dl l λ = 80π2 b 50 4π2 3l = 5 Ω 0.5 # 10

SOL 1.10

Option (C) is correct. 2 2 2 2 α2 = αa 1 + α2 = 0.3 + 0.4 = 0.5 or α = 0.5 Option (D) is correct. Modulation Index = δ fm

SOL 1.11

δ = 50 # 500 = 25000 Hz
SOL 1.12

Option (D) is correct. The processing gain is PG = 4 # 4 = 16 Hence, in decibels, PG = 10 log 10 16 = 12 dB Option (C) is correct. 1, sgn (t) = * − 1, X (ω) = =e
−jωt 0 0

SOL 1.13

0#t<3 −3 < t < 0

3 (− 1) e−jωt dt + # (1) e−jωt dt # −3 0

SOL 1.14

−jωt − 3 +e = 2 jω − 3 − jω 0 jω Option (D) is correct.

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Page 3

EC_Practice Paper B

Chapter 1

N = 7,

Ω0 = 2π , 7 x [n] =

/ X [k] e jb 7 lkn
n =− 3

3



= 2e j (− 1)b 7 ln − 1 + 2e j (1)b 7 ln





SOL 1.15

= 4 cos b 2π n l − 1 7 Option (D) is correct. Not causal because h (t) ! 0 for t < 0 . Unstable because
3 h (t) dt = 3 # −3

SOL 1.16

Option (D) is correct. ABC + ABC + ABC + ABC = C (AB + AB) + C (AB + AB ) = C (AB + AB ) + C (AB + AB ) = A5B5C Option (C) is correct. Resolution = n 5 = 10 m 2 −1 & & 500 = 2n − 1 2n = 501, n > 8 , n = 9

SOL 1.17

SOL 1.18

Option (D) is correct. Converting in decimal we have (80) 16 = 8 # 16 = (128) 10 (128) 10 + (− 64) 10 = (64) 10 (64) 10 = (01000000) 2 Option (B) is correct. The circuit is as follows

SOL 1.19

&
SOL 1.20 SOL 1.21

v (4) v+ = 2.5 V = v− , 0 = 2.5 8+4 vo = 7.5 V

Option (C) is correct. Option (B) is correct. After killing all source equivalent resistance is R Open circuit voltage = v1

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Page 4 SOL 1.22

EC_Practice Paper B

Chapter 1

Option (A) is correct. Leq = L1 + L2 − 2M = 4 + 2 − 2 # 2 = 2 H Option (D) is correct. The output voltage cannot exceed the positive power supply voltage and cannot be lower than the negative power supply voltage. Option (D) is correct. M 2 is in saturation because VGS 2 = VDS2 > VGS2 − VTN M1 is in non saturation because VGS 1 = Vi = 5 V, VDS 1 = VD = 0 V VDS1 < VGS1 − VTN , ID1 = ID2 W W 2 2 b L l [2 (VGS 1 − VTN 1) VDS 1 − V DS 2] = b L l (VGS 2 − VTN 2) 1 1 W 2 & b l [2 (5 − 0.8) (0.1) − (0.1) ] = (1) (5 − 0.1 − 0.8) 2 L 1 W W b L l (0.83) = 16.81 & b L l = 20.3 1 1 Option (D) is correct. Option (C) is correct. Required incidence matrix is R− 1 1 1 0 S A = S 0 −1 0 1 S S 0 0 −1 −1 T The graph is as shown below

SOL 1.23

SOL 1.24

SOL 1.25 SOL 1.26

0 1 0

0V W 0W W 1W X

R− 1 1 1 0 S AAT = S 0 − 1 0 1 S S 0 0 −1 −1 T

0 1 0

SOL 1.27

R 3 − 1 − 1V R 3 S W S = S− 1 3 − 1W, detS− 1 S S S− 1 − 1 3W W S− 1 T T X Option (D) is correct.

R V S− 1 0 0W S 1 − 1 0W 0V WS 1 0 − 1W W 0WS S 0 1 − 1W W 1WS W XS 0 1 0W S 0 0 1W X − 1T − 1V W 3 − 1W = 16 W − 1 3W X

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Page 5

EC_Practice Paper B

Chapter 1

Rearranging the given expression x (t) =− 2u (t + 2) + u (t + 1) + u (t) At t =− 2 , the function steps down from 0 to − 2 for a change in magnitude of − 2 ; At t =− 1, the function steps up from − 2 − 1 for a change in magnitude of 1; At t = 0 , the function steps from − 1 to 0, for a change in magnitude of 1; Alternatively : For − 2 < t < 1, x (t) =− 2 For − 1 < t < 0 , x (t) =− 1 For 0 < t , x (t) = 0
SOL 1.28

Option (C) is correct. 2710H LXI H, 30A0H ; Load 16 bit data 30A0 in HL pair 2713H DAD H ; 6140H " HL 2714H PCHL ; Copy the contents 6140H of HL in PC Thus after execution above instruction contents of PC and HL are same and that is 6140H Option (A) is correct. A (A + B ) (A + B + C ) (AA + AB ) (A + B + C ) = A (A + B + C ) = A Therefore, no gate is required to implement this function. Option (B) is correct. From the combinational logic Let D is input, Qn is present state, Qn + 1 is next state, then R = D 5 Q, S = D 5 Q Characteristic equation of R-S flip-flop is given by Qn + 1 = S + RQn So, Qn + 1 = (D + Qn) + (D 5 Qn) Qn = (D 5 Qn) + (D 5 Qn) Qn = (D 5 Qn) (1 5 Qn) = (D 5 Qn) = DQ n + DQn For D = 0 , Qn + 1 = Qn D = 1, Qn + 1 = Qn So, the circuit function as a T-flip flop. Option (C) is correct. Z0 = L = C μd d =d ω ωd εω μ ε

SOL 1.29

SOL 1.30

SOL 1.31

Z 0 = d η0 = 81 Ω , ω Z0 ' = d = 75 Ω ω' η 0 w ' = 81 & w ' = 1.08w w 75

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Page 6 SOL 1.32

EC_Practice Paper B

Chapter 1

Option (C) is correct. Since a > b , the dominant mode is TE10 . In free space 8 fc = c = 3 # 10 = 3 GHz 2a 2 # 0.05 η0 377 η1 = = = 406.7 Ω 2 fc 2 3 1 −c m 1 −b l f 8 In dielectric medium 3 # 108 = = 2 GHz fc = c 2a εr 2 # 0.005 2.25 η0 = 377 = 251.33 Ω , εr 2.25 η2 = 251.33 = 259.23 Ω 2 1 −b2l 8 η2 − η1 Γ = η + η = 259.23 − 406.7 =− 0.22 259.23 + 406.7 2 1 1+ Γ s = = 1 + 0.22 = 1.564 1 − 0.22 1− Γ Option (D) is correct. v (0) # B = 2 (u x − 3u y − 4u z ) 105 # (− 3u x + 2u y − u z ) 10−3 = 1100u x + 1400u y − 500u z F (0) = q [E + v # B] = 2 # 10−16 [1200u x + 1200u y − 200u z ] = 4 # 10−14 [6u x + 6u y − u z ] −14 F = ma & a = F = 4 # 10−26 [6u x + 6u y − u z ] m 5 # 10 = 800 [6u x + 6u y − u z ] 109 m/s2 η = Option (A) is correct. The circuit is as shown below

SOL 1.33

SOL 1.34

At t = 3,

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Page 7

EC_Practice Paper B

Chapter 1

SOL 1.35

5 + 1 v 10 20 a = 2 + va vb = 1 + 1 + 1 5 10 10 20 vb 1 + 5 va = 20 10 = vb + 10 1 + 1 3 3 20 10 & va = 2 + va + 10 3 15 3 & va = 30 V 7 Option (D) is correct. 1 1 Y = j ω600 # 10−12 + 2 # 103 1.8 + j ω10−5 1.8 − j ω10−5 = j ω6 # 10−10 + 45.45 + 3.24 + ω 2 10−50 At resonance Im {Y } = 0 −10 ω0 6 # 10−10 (3.24 + ω2 ) − ω0 10−5 = 0 0 10
−10 = 16.67 # 103 ω0 = 12.9 M rad/s 3.24 + ω2 0 10 f 0 = ω0 = 2.05 MHz 2π Option (C) is correct. vC (0) = 0, iL (0) = 4 # 6 = 3 6+2 dv (0) = iL (0) = 3 0.02 C dt dvC (0) & = 150 dt α = 6 + 14 = 5 , 2#2 1 ω0 = =5 2 # 0.02 α = ω0 critically damped v (t) = 12 + (A + Bt) e−5t 0 = 12 + A, 150 =− 5A + B & A =− 12 , B = 90 v (t) = 12 + (90t − 12) e−5t iL (t) = 0.02 (− 5) e−5t (90t − 12) + 0.02 (90) e−5t = (3 − 9t) e−5t

SOL 1.36

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Page 8 SOL 1.37

EC_Practice Paper B

Chapter 1

Option (A) is correct. The circuit is as shown in figure below

Ca = 30 # 60 = 20 mF, 30 + 60 30 (20 + 40) Cb = = 20 mF 30 + 20 + 0 We can say Cd = 20 mF, and Ceq = 20 + 40 = 60 mF vC = 1 idt C = 1 b− 300 cos 20t l # 10−3 60m 20

#

=− 0.25 cos 20t V
SOL 1.38

Option (B) is correct. p 0 = N v e− At 300 K,
(EF − Ev) kT

& EF − Ev = kT1n b Nv l p0

Nv = 1.0 # 1019 cm−3 1019 = 0.239 eV EF − Ev = 0.02591n c 1.04 # m 1015 n0 Nc Ec − EF n0 = Nc e kT = 2.8 # 1019 cm−3 = 1.12 − 0.239 = 0.881 eV = 4.4 # 10 4 cm−3
BE t BC t BE t

(Ec − EF )

At 300 K,

SOL 1.39

Option (C) is correct.

&

V V V IE = Is _ec V m − ec V mi + Is _ec V m − 1i = 0 βF V V βR ec V m ec V m = 1 + 1 + βF 1 + BF V V V IC = Is 8ec V m − ec V mB − Is 8ec V m − 1B βR V V ICBO = Is 81 − ec V mB − Is 8ec V m − 1B 1 + βF βR V V ICBO = Is 81 − ec V mB − Is 8ec V m − 1B 1 + βF βR
BE t BC t BE t BC t BC t BC t BC t BC t BC t

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Page 9

EC_Practice Paper B

Chapter 1

SOL 1.40

VBC =− 5 V , Vt = 0.0259 V ICBO = Is (1 − 0) − Is (0 − 1) = 1.01Is = 1.01 # 10−15 A 101 1 Option (A) is correct. X (jω) = π [δ (ω + 1) + δ (ω − 1)] H1 (jω) = 1 + πδ (ω) jω Y1 (jω) = b 1 + πδ (ω)l π [δ (ω + 1) + δ (ω − 1)] jω = jπ [δ (ω + 1) − δ (ω − 1)], y1 (t) = sin t 1 − 2jω H2 (jω) =− 2 + 5 = 2 + jω 2 + jω 1 − 2jω π [δ (ω + 1) + δ (ω − 1)] Y2 (jω) = c 2 + jω m 1 + 2j 1 − 2j =c πδ (ω + 1) + c πδ (ω − 1) 2−j m 2+j m = jπ {δ (ω + 1) − δ (ω − 1) + δ (ω − 1)} = sin t H 3 (jω) = 2 (1 + jω) 2 2 Y3 (jω) = π {δ (ω + 1) + δ (ω − 1)} (1 + jω) 2 = 2π 2 δ (ω + 1) + 2π 2 δ (ω − 1) (1 − j) (1 + j) = jπ {δ (ω + 1) − δ (ω − 1)}, y 3 (t) = sin t
FS k jb 2π lb N ln N 2

SOL 1.41

Option (C) is correct. Let x [n] Then

a

(− 1) n x [n] = e
FS

x [n]

FS

ak − N 2

N = 8 , so, (− 1) n x [n] Given that ak So x [n] This implies that x [0] We are given that x [1]
SOL 1.42

ak − 4

=− ak − 4 =− (− 1) n x [n] = x [! 2] = x [! 4] = ... = 0 = x [5] = ... = 1 and x [3] = x [7] =− 1

Option (C) is correct. Initially plot has − 20 dB/decade slop hence pole at origin. At ω1 slope is changed − 20 dB/decade to 0 dB/decade hence zero at ω1 . At ω2 and 10 there will be pole. K (s + ω1) G (s) = s (s + ω 2) (s + 10)

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Page 10

EC_Practice Paper B

Chapter 1

From the plot 32 − 6 =− 20 (log 0.1 − log ω1) log ω1 = 0.3 , ω1 = 2 rad/sec 20 log K = 6 + 20 log 2, K = 4 Between ω2 and 10 there is a fall of 6 dB on a slope of − 20 dB/decade 6 =− 20 (log ω2 − log 10), ω2 = 5 rad/sec 4 (s + 2) G (s) = s (s + 5) (s + 10)
SOL 1.43

Option (C) is correct. vo = a 0 [A c' cos (2πf c' t) + m (t)] + a1 [A c' cos (2πf c' t) + m (t)] 3 = a 0 Ac' cos (2πf c' t) + a 0 m (t) + a1 [(A c' cos 2πf c' t) 3 + (A c' cos (2πf c' t)) 2 m (t) + 3A c' cos (2πf c' t) m2 (t) + m3 ^ t h] = a 0 A c' cos (2πf c' t) + a 0 m (t) + a1 (A c' cos 2πf c' t) 3 1 + cos (4πf c' t) + 3a1 A c'2 ; E m (t) 2 = 3a1 A c' cos (2πf c' t) m2 (t) + m3 (t) cos 4πf c' t The term 3a1 A c' ; E m (t) is a DSB-SC signal having carrier 2 frequency 1 MHz. Thus 2f c' = 1 MHz or f c' = 0.5 MHz Option (C) is correct. Given that z = z (u, v), u = x2 − 2xy − y2, v = a 2z = 2z .2u + 2z .2v 2x 2u 2x 2v 2x 2z = 2z .2u + 2z .2v and 2y 2u 2y 2v 2y From (i), 2u 2x 2u 2y 2v 2x 2v 2y = 2x − 2y , =− 2x − 2y , = 0, =0

SOL 1.44

...(i) ...(ii) ...(iii)

Substituting these values in (ii) and (iii) 2z = 2z (2x − 2y) + 2z .0 2v 2x 2u 2z = 2z . (− 2x − 2y) + 2z .0 and 2v 2y 2u

...(iv) ...(v)

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Page 11

EC_Practice Paper B

Chapter 1

From (iv) and (v), we get (x + y)2z = (y − x)2z 2x 2y
SOL 1.45

Option (B) is correct. dy x −y = dx

x2 + y2

Put y = vx and differential with respect to x , we get dy = v + x dv dx dx x bv + x dv l − vx = x 1 + v 2 , dx x dv = 1 + v 2 dx dv = dx 2 x 1+v or, log (v + 1 + v 2 ) = log x + log c , v + 1 + v 2 = cx y y2 or, + 1 + 2 = cx x x

#

#

or,
SOL 1.46

y+

x2 + y2 = cx2

Option (D) is correct. Since, 3 1 − 2z = 1 + 1 − 2z z − 1 2 (z − 2) z (z − 1) (z − 2) 1 − 2z dz # z (z − 1) (z − 2) = 1 I1 + I 2 − 3 I 3 2 2 ...(i)

c

Since, z = 0 is the only singularity for I1 = # 1 dz and it lies inside z = 15, z c therefore by Cauchy’s integral Fourmula ...(ii) I1 = # 1 dz = 2πi z c f (z) dz 1 >f (z 0) = 2πi # z − z 0 H [Here f (z) = 1 = f (z 0) and z 0 = 0 ]
c

Similarly, for I2 = # 1 dz , the singular point z = 1 lies inside z = 1.5 , so the z−1 c function f (z) is analytic everywhere in c i.e. z = 1.5 , hence by Cauchy’s integral theorem ...(iv) I 3 = # 1 dz = 0 z−2
c

using equations (ii), (iii), (iv) in (i) we get 1 − 2z dz = 1 (2πi) + 2πi − 3 (0) = 3πi # z (z − 2 2 − 1 ) ( z 2 ) c

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Page 12 SOL 1.47

EC_Practice Paper B

Chapter 1

Option (A) is correct. Here n = 5 and np + npq = 4.8 np (1 + q) = 4.8 & np (2 − p) = 4.8 & p (2 − p) = 0.96 & p2 − 2p + 0.96 = 0 & 25p2 − 50p + 24 = 0 & (5p − 5) (5p − 4) = 0 & p = 6 or p = 4 5 5 But, the value of p can not exceed 1. p = 4 and q = b1 − 4 l = 1 5 5 5
5 The distribution is b 1 + 4 l 5 5 Option (B) is correct. 3 f = (3 f1) n1 n2 = (25) (64) (48) = 76.8 kHz

SOL 1.48

SOL 1.49

Option (A) is correct. f2 = n 1 f1 = (64) (200) (103) = 12.8 MHz f 3 = f2 ! fLO = (12.8 ! 10.8) = 23.6 MHz, 2.0 MHz When f 3 = 23.6 MHz, then f c = n2 f 3 = (48) (23.6) = 1132.8 MHz When f 3 = 2 MHz, then f c = n 2 f3 = (48) (2) = 96 MHz Option (B) is correct.
6 4.2 = 21π βl = ωl = 2π # 150 # 10 # v 4 0.8 # 3 # 108

SOL 1.50

tan βl = 1

Zin = Z 0 e

' + jZ 0 tan βl ZL 80 − j 60 + j 50 o = 50 c 50 + 60 + j 80 m ' Z 0 + jZ L tan βl = 21.6 − j 20.3 Ω

SOL 1.51

Option (D) is correct.
' − Z 0 = 80 − j 60 − 50 = 0.468+ − 38.66c Γ' = ZL ' 80 − j 60 + 50 Z L + Z0 Γ = Γ ' = 0.468, But θΓ = θΓ ' + 2 # π =− 38.66c + 90c = 51.34c 4

Γ = 0.468+51.34c
SOL 1.52

Option (B) is correct. Closed loop gain of op-amp is given by ACL (3) ACL = A (3) 1 + CL AOL Where, ACL (3) " ideal closed loop gain AOL " open loop gain

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Page 13

EC_Practice Paper B

Chapter 1

SOL 1.53

SOL 1.54

ACL (3) = 1 + R2 = 1 + 495 = 100 5 R1 So, ACL = 10 = 99.90 1 + 100 105 Option (B) is correct. Change in closed loop gain is dACL = dAOL ACL (3) AOL c AOL m ACL = 10 # 100 = 0.01% 105 Option (B) is correct. o2 =− 5s2 − 21 x2 + u , x 4 o1 = x2 , x y = 5x1 + 4x2 0 1 x1 o1 x1 0 x 21 = > o H >− 5 − H>x H + >1H u , y = 85 4B>x H x2 2 4 2 Option (B) is correct. C 5 4 OM = > H = > CA − 20 1H det OM = 0 . Thus system is not observable 0 1 CM = 8B ABB = >1 − 21H 4 det CM =− 1. Thus system is controllable.

SOL 1.55

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SOLUTIONS EC_Practice Paper C

SOL 1.1

Option (A) is correct. 2v = 2y = h (x, y),2v = 2x = g (x, y) 2x 2y by Milne’s Method f l (z) = g (z, 0) + ih (z, 0) = 2z + i0 = 2z On integrating f (z) = z2 + c

SOL 1.2

Option (A) is correct. ρ (X, Y) = cov (X, Y) = var (X) var (Y) 10 =5 7 6.25 # 31.36

SOL 1.3

Option (A) is correct. We know that adj(adj A ) = A n − 2 : A. Here n = 3 , and A = 3 So, adj (adjA) = 3(3 − 2) : A = 3A . Option (B) is correct. E [X] = X =

SOL 1.4

/ x P (x )
i i i=1

4

= 1.0 (0.4) + 4 (0.25) + 9 (0.15) + 16 (0.1) = 4.35
SOL 1.5

Option (A) is correct. If modulation index α is 0, then 2 2 2 Pt1 = A c b1 + 0 l = A c 2 2 2 If modulation index is 1 then 2 2 2 , Pt 2 = A c b1 + 1 l = 3 A c 2 2 4 Pt 2 = 3 2 Pt 1 Thus Pt 2 = 1.5Pt 1 and Pt 2 is increases by 50%

SOL 1.6

Option (D) is correct. H (x) =

/ p log
i i=1

6

2

1 pi

Page 2

EC_Practice Paper C

Chapter 1

=− (0.1 log 2 0.1 + 0.2 log 2 0.2) + 0.3 log 2 0.3 + 0.05 log 2 0.05 + 0.15 log 2 0.15 + 0.2 log 2 0.2 = 2.4087 bits/symbol
SOL 1.7

Option (D) is correct. X [k ] = 1 T A = 10 , T = 5 , X [k] = 2 Option (A) is correct. X (jω) = = Aδ (t) e−jkω dt = A , # T −T/2
T/2

SOL 1.8

3 −4 t e e # −3 0

−jωt

dt

SOL 1.9

Option (C) is correct. Anti causal signal, x [0] = lim X (z) = 4
z"3

3 8 e 4t e−jωt + # e−4t e−jωt dt = # −3 0 16 + ω2

SOL 1.10

Option (D) is correct. For vi < 4 V the diode is ON and output vo = 4 V. For vi > 4 V diode is off and output vo = vi . Option (A) is correct. This is inverting amplifier Av =− RF =− 400 =− 10 40 R1 Option (A) is correct. Input impedance of instrumentation amplifier is very large Option (B) is correct. Option (B) is correct. Following figure shown that a d f is a tree.

SOL 1.11

SOL 1.12

SOL 1.13 SOL 1.14

SOL 1.15

Option (A) is correct. The circuit is as shown below

Page 3

EC_Practice Paper C

Chapter 1

SOL 1.16

RN = 2 4 + 2 = 10 Ω , 3 15 2 =6 V v1 = 1+1+1 2 2 4 isc = iN = v1 = 3 A 2 Option (B) is correct. For series critically damped circuit R 0 = 4L = 4 # 4 = 40 Ω 10m C R 120 = 40 Ω , R = 60 Ω

SOL 1.17

Option (B) is correct. ωc = 2πf c = 1 RC 1 & R = = 1.06 kΩ 2π # 15 # 10 # 10−6 Option (B) is correct. XZ + X Z = X (XY + XY) + X (XY + XY) = X (XY + X Y ) + XY = XY + XY = Y Option (B) is correct. If either or both of V1 and V2 are logic high, then Vo is high otherwise Vo = 0 . Thus OR logic. Option (B) is correct. In characteristic equation s3 + 2s2 + K , the term s is missing. Option (C) is correct. P1 = 5 # 3 # 2 = 30 , 3 = 1 − (3 # − 3) = 10 31 = 1, C = 30 = 3 R 10 Option (A) is correct. (r − 4) r2 = c x 2 + y 2 + z 2 m (x2 + y2 + z2) 2x 2y 2z = x (2x) + y (2y) + z (2z) = 2 (x2 + y2 + z2) = 2r2 Option (B) is correct. E =−4 V , − 3xy2 z3 satisfy this equation. Option (A) is correct. Jb = 4# M = 12 u z a Option (D) is correct.

SOL 1.18

SOL 1.19

SOL 1.20

SOL 1.21

SOL 1.22

SOL 1.23

SOL 1.24

SOL 1.25

Page 4

EC_Practice Paper C

Chapter 1

f l (x) = 3c2 − 12x + k f l (c) = 0 & 3c2 − 12c + k = 0 & &
SOL 1.26

c = 12 ! 144 − 12k & 144 − 12k = 1 6 36 3 144 − 12k = 12 & k = 11. R = V = 20 = 200 Ω 100m I
−3 = 0.01 (Ω − cm) −1 σ = L = 2 # 10 RA (200) (0.001)

Option (B) is correct.

σ . eμn n 0 , For Si, μn = 1350 & n 0 >> ni &
SOL 1.27

0.01 = (1.6 # 10−19) (1350) n 0 n 0 = 4.6 # 1013 cm−3 n 0 = Nd
d Vbi = Vt ln c Na N 2 m ni

Option (B) is correct.

= 2 (0.0259) ln c

1016 = 1.16 V 1.8 # 106 m
1

2 2ε (V + VR) 1 W = ) s bi + 1 D3 : e Na Na

Jgen

2 (13.1 # 8.85 # 10−14) (6.16) 2 2 == # b 1016 lG 1.6 # 10−19 = 1.34 # 10−4 cm = eni W 2τ0 = 1.6 # 10 # 1.8 # 10 # 13.4 # 10 2 # 10−8 = 1.93 # 10−9 A/cm2
2 −19 6 −4

1

SOL 1.28

Option (C) is correct. Pinch off voltage Vp = eW ND εs Let Now or Vp = Vp1 2 Vp1 W 12 = 2= W 2 Vp2 W 2 (2W ) 4Vp1 = Vp2

Initial trans-conductance gm = Kn ;1 − For first condition

Vbi − VGS E Vp

Page 5

EC_Practice Paper C

Chapter 1

gm1 = Kn =1 − Dividing Here
SOL 1.29

0 − (− 2) G = Kn ;1 − Vp1

2 Vp1 E

1 − 2/Vp1 gm1 =f p gm2 1 − 1/ (2Vp1) Vp = Vp1

Option (C) is correct. Output of gate 1 is Xo X1 Output of gate 2 is X 0 X1 + X2 Output of gate 3 is (X 0 X1 + X2) X 3 = X 0 X1 X 3 + X2 X 3 Output of gate 4 would be X 0 X1 X 3 + X2 X 3 + X 4 Output of gate 5 would be X 0 X1 X 3 X5 + X2 X 3 X5 + X 4 X5 So output of gate n would be Xo X1 X 3 X5 ...Xn + X2 X 3 X5 ...Xn + X 4 X5 X7 ...Xn + Xn − 1 Xn Option (C) is correct. The Input and output is as shown below

SOL 1.30

SOL 1.31

Option (C) is correct. f = 1 0.693 (RA + 2RB) C

&

RA = R1 = 10 kΩ RB = R2 + xR 3 10k < RB # 110k f min =

SOL 1.32

1 = 627 Hz 0.693 {10k + 2 (110k)} 0.01μ 1 f max = = 4.81 kHz 0.693 {10k + 2 (10k)} 0.01μ Option (A) is correct. Let X 3 X2 X1 X 0 be 1001 then Y3 Y2 Y1 Y0 will be 1111. Let X 3 X2 X1 X 0 be 1000 then Y3 Y2 Y1 Y0 will be 1110 Let X 3 X2 X1 X 0 be 0110 then Y3 Y2 Y1 Y0 will be 1100

Page 6

EC_Practice Paper C

Chapter 1

So this converts 2 − 4 − 2 − 1 BCD numbers.
SOL 1.33

Option (B) is correct. MVI B, 87H MOV A, B START: JMP NEXT XRA B JP START

JMP NEXT NEXT : XRA B JP START OUT PORT2
SOL 1.34

; ; ; ; ; ; ; ; ; ; ; ; ;

B = 87 A = B = 87 Jump to next A 5 B " A, A = 00, B = 87 Since A = 00 is positive so jump to START Jump to NEXT unconditionally A 5 B " A, A = 87 B = 87H will not jump as D7 of A is 1 A = 87 " PORT2

Option (A) is correct. 1. Over damped response (a, b) Poles : Two real and different on negative real axis. 2. Under damped response (c) Poles : Two complex in left half plane Poles : Two complex in left half plane 3. Undamped response (d) Poles : Two imaginary 4. Critically damped (e) Poles : Two real and same on negative real axis. Option (C) is correct. Any point on real axis of s -is part of root locus if number of OL poles and zeros to right of that point is even. Thus (b) and (c) are possible option. The characteristic equation is 1 + G (s) H (s) = 0 K (1 − s) or =0 1+ s (s + 3) or K = s + 3s 1−s
2

SOL 1.35

For break away & break in point dK = (1 − s) (2s + 3) + s2 + 3s = 0 ds or − s2 + 2s + 3 = 0 Which gives = 3, − 1 Here − 1 must be the break away point and 3 must be the break in point. Thus (C) is correct.

Page 7 SOL 1.36

EC_Practice Paper C

Chapter 1

Option (A) is correct. Here given x0 = 0 y0 = 1, h = 0.2 f (x, y) = x + y2 To find y1 = y(0.2), k1 = hf (x0, y0) = (0.2) f (0, 1) = (0.2) # 1 = 0.2 k2 = hf cx0 + h , y0 + k1 m 2 2 = (0.2) f (0.1, 1.1) = 0.2 (1.31) = 0.262 k3 = hf cx0 + h , y0 + k2 m 2 2 = 0.2f (0.1, 1.131) = 0.2758 k4 = hf (x0 + h, y0 + k3) = (0.2) f (0.2, 1.2758) = 0.3655 k = 1 [k1 + 2k2 + 2k 3 + 2k 4] 6 k = 1 [0.2 + 2 (0.262) + 2 (0.2758) + 0.3655] = 0.2735 6 Here, y1 = y^0.2h = y 0 + k = 1 + 0.2735 = 1.2735

SOL 1.37

Option (A) is correct. The given equation is d 2 y dy − − 2y = 10 cos x dx 2 dx or, (D 2 − D − 2) y = 10 cos x C.F. is given by y = c1 e−x + c2 e2x P.I. = 2 1 : 10 cos x = 10 2 1 cos x −1 − D − 2 D −D−2 =− 10 1 cos x =− 10 D2− 3 cos x D+3 D −9 − 3 cos x = (D − 3) cos x =− sin x − 3 cos x =− 10 D 2 −1 − 9 The required solution is y = c1 e−x + c2 e2x − sin x − 3 cos x Option (B) is correct. Let I = = Here, I1 =

SOL 1.38

# 0 # 0 # 0

π 2

ex sec2 x + 2 tan x dx 2a 2 2k 1 ex sec2 x dx + 2 2 1 ex sec2 x dx 2 2

π 2

# 0

π 2 x

e tan x dx = I1 + I2 2

π 2

Page 8

EC_Practice Paper C
π π 2 x

Chapter 1

= :1 ex .2 tan x D 2 − 1 2 2 0 2 = ae 2 tan π − 0k − 4
π π

# 0

e .2 tan x dx 2

# 0

π 2 x π

e tan x dx 2

= e 2 − I2, I1 + I2 = e 2 I = I1 + I 2 = e 2
SOL 1.39
π

Option (C) is correct. Let E = event that A speaks the truth. F = event that B speaks the truth Then, P (E) = 75 = 3 , 100 4 P (F) = 80 = 4 100 5 P (E ) = b1 − 3 l = 1 , 4 4 P (F ) = b1 − 4 l = 1 5 5 P ( A and B contradict each other) = P [( A speaks truth and B tells a lie) or ( A tells a lie and B speaks the truth)] = P ( E and F ) + P ( E and F ) = P (E) .P (F ) + P (E ) .P (F) = 3#1+1#4 = 3 +1 = 7 5 4 5 20 5 20 4 = b 7 # 100 l % = 35% 20 Option (B) is correct. If β >> 1 and transistor are identical Iref . IC 1 = IS e V , Io = IC 2 = IS e V I VBE 1 = Vt ln c ref m, VBE 2 = Vt ln b Io l IS IS I VBE 1 VBE 2 = Vt ln c ref m Io
t t

SOL 1.40

VBE 1

VBE 2

From the given circuit VBE 1 − VBE 2 = IE 2 RE . Io RE I Io RE = Vt ln c ref m Io RE = 0.026 ln 1 # 10−3 = 9.58 kΩ 12 # 10−6 c 12 # 10−6 m + − 5 − 7 − (− 5) R1 = V − VBE 1 − V = = 9.3 kΩ 1m Iref

SOL 1.41

Option (A) is correct.

Page 9

EC_Practice Paper C

Chapter 1

VD = ID RD − 5 = (0.4) (5) − 5 =− 3 V VSD = VS − VD = 2.21 − (− 3) = 5.21 V
SOL 1.42

Option (D) is correct. (VGS ) 1 = (VGS ) 2 =− V3 Assume transistors are in saturation ' are W are same for M1 and M2 ) ID1 = ID 2 = 200 = 100 μA ( K n 2 L For saturation region
' ID = K n bW l (VGS − VTH ) 2 2 L

100 # 10−6 = 100 # 10 # 20 (VGS − 1) 2 2 (VGS − 1) 2 = 0.1 & VGS = 1.32 V VD1 = 5 − ID1 (40 # 103) = 1 V = 5 − 100 # 10−6 (40 # 103) = 1 V (VDS ) sat = VGS − VTH = 1.32 − 1 = 0.32 V VGS = VG − VS = 1.32 0 − VS = 1.32 V VS =− 1.32 V So, VDS = VD − VS = 1 − (1.32) = 2.32 V (VDS ) > (VDS ) sat So, the assumption is true In the circuit, V3 =− VGS =− 1.32 V V1 = VD1 = 1 V, V2 = VD 2 = 1 V
SOL 1.43

−6

a VG = 0

Option (B) is correct. ICQ = b 100 l (0.35) = 0.347 mA 1001 The small-signal equivalent circuit is as shown below

vπ = vo vo vs gm

rπ 10k (vs), 500 + rπ 10k =− gm v π (ro 7k) rπ 10k =− gm f p (ro 7k) 500 + rπ 10k I = CQ = 0.347m = 13.13 mA/V 0.0259 Vt

Page 10

EC_Practice Paper C

Chapter 1

SOL 1.44

β βV = t = 100 = 7.6 kΩ gm 13.13m ICQ β = 100 = 288 kΩ ro = 0.347m ICQ ro 7k = 288 # 7 = 6.83 kΩ 288 + 7 rπ 10k = 7.6 # 10 = 4.32 kΩ 7.6 + 10 Av =− 13.13m b 4.32k l (6.83k) =− 80 500 + 4.32k Option (B) is correct. rπ = RXY = = E [X] = E [Y] =

# # xyf
−3 −3

3

3

X, Y

(x, y) dxdy

# # x 9y
0 3 0 2 2 0 3 0 2 2

3

2

2 2

dxdy = 8 3

4 # # x9y dxdy = 3

# # x9y dxdy = 2
0 0

Since RXY = 8 = E [X] E [Y] = 2 b 4 l = 8 , we have X and Y uncorrelated 3 3 3 form From marginal densities fX (x) = fY (y) = dy = x , 0 < x < 2 # xy 9 2
0 3

2y dy = , 0 < y < 3 # xy 9 9
0

2

xy we have fX (x) fY (y) = , 0 < x < 2 and 0 < y < 3 . 9 Thus
SOL 1.45

fX, Y (x, y) = fX (x) fY (y) and X and Y are statistically independent. 30 1k = 25 vC (0 ) = 1 + 1 + 1 1k 6.25k 25k Req = 5 kΩ, 6


Option (D) is correct.

vC (t) = 25e−2000t V, vC (1 ms) = 3.383 V

for 0 < t < 1 ms for t > 1 ms,

vC (t) = 3.38e− 2.5k # 0.6μ

(t − 1m)

Page 11

EC_Practice Paper C

Chapter 1

vC (4 ms) = 2.77 V
SOL 1.46

Option (C) is correct. For the line to be matched, it is required that the sum of normalized input admittance of the shorted stub and main line at the point where the stub is connected by unity. For λ shorted stub, zL = 0 and 10 z + j tan βl zins = L = j tan βl 1 + zL j tan βl yins =− j cot βl =− j cot b 2π λ l =− j1.3764 λ 10 For line to be matched at junction normalized input admittance of line must be 1 + j1.3764 tan βl = tan b 2π λ l = 3 λ 4 yin = zL , zL = 1 + j1.3764 , ZL = Z 0 zL = 300 + j 413

SOL 1.47

Option (D) is correct. We know that IC = Is exp bVBE lb1 + VCE l VT VT gmo = 1 = dIC = Is exp bVBE l 1 ro dVCE VT VA VBE ;Is exp b VT lE 1 = . IC ro VA VA ro = bVA l, ro > 10 kΩ IC VA > 10 kΩ, V > 10 103 I # # C A IC VA > 10 # 103 # 2 # 10−3 & VA > 20 V

SOL 1.48

Option (B) is correct.
− 80 # 3 + vC (0 ) 200 vC (0−) = 400 3 + 1 + 1 400 200 100

SOL 1.49

vC (0−) = 34.29 V = vC (0+) i1 (0+) = 34.29 = 0.17 A 200 Option (D) is correct. The circuit is as shown below

Page 12

EC_Practice Paper C

Chapter 1

vtest + vtest − 0.25vtest = 1 100 133.3 200 vtest = 50 V, RTH = 50 Ω , τ = 50 # 0.5 m t vC (t) = 34.29e− 50 # 0.5m
− 25m i1 (80m) = 34.29e = 6.98 mA 200 Option (D) is correct. 80m

SOL 1.50

PXX = 1 2π


−3

3

XX

(ω) dω

SOL 1.51

dω = 3 = 3 2π 49 + ω2 14 −3 Option (B) is correct.

#

3

SOL 1.52

2 = H2 (ω) (7 + jω) 3 2 sYY (ω) = sXX (ω) = H1 (ω) H2 (ω) 2 = 12ω 2 4 (49 + ω ) Option (A) is correct. o(t) = Ax (t) We have x p q Let A => H r s h2 = 49 (t) t2 e−7t
F

1 e−2t For initial state error x (0) = > H the system response is x (t) = > H −2 − 2e−2t V R d S e−2t W p q 1 dt W = > H> H Thus S d −2t r s −2 S (− 2e )W W Sdt Xt = 0 T −2 (0) p q 1 − 2c or > 4e−2 (0) H = >r sH>− 2H −2 p 2q > 4 H = >r − 2sH We get p − 2q =− 2 and r − 2s = 4 1 From initial state vector x (0) = > H −1 ....(1)

Page 13

EC_Practice Paper C

Chapter 1

e−t the system response is x (t) = > −tH −e V R d S e−t W p q 1 dt W = > H> H Thus S d −t r s −1 S (− e )W W Sdt t=0 X T − (0) p q 1 −c or > e− (0) H = >r sH>− 1H 1 p −q >− 1H = >r − s H We get p − q =− 1 and r − s = 1 Solving (1) and (2) set of equations we get 0 1 q q >r sH = >− 2 − 3H The characteristic equation λI − A = 0 λ −1 =0 1 λ+3 or λ (λ + 3) + 2 = 0 or λ =− 1, − 2 Thus Eigen values are − 1 and − 2 Eigen vectors for λ1 =− 1 (λ1 1 − A) X1 = 0 or λ1 − 1 x11 > 2 λ + 3H>x H = 0 1 21 − 1 − 1 x11 > 2 2 H>x H = 0 21 or − 2x12 − x22 = 0 We have only one independent equation x22 =− 2x21 Let x12 = K, then x22 =− 2K . Thus Eigen vector will be x12 K 1 >x H = >− 2K H = K >− 2H 22 Option (D) is correct. As shown in previous solution the system matrix is 0 1 A => − 2 − 3H Option (B) is correct. x (t) = sin c (4000t) .....(2)

SOL 1.53

SOL 1.54

Page 14

EC_Practice Paper C

Chapter 1

Fourier transform of x (t) is given as f X (f ) = 1 rect b 4000 4000 l Spectrum of X (f ) is

Bandwidth B = 2 kHz For ideal sampling we consider a impulse train function as 3 H (f ) = 1 δ (f − k/Ts) Ts k =− 3

/

H (f ) = 4000

k =− 3

/ δ (f − 4000k)

3

Sampling frequency fs = 4 kHz = 2 B (critical sampling) Sampled signal is s (f ) = X (f ) * H (f ) = =
3 1 rect f * 4000δ (f − 4000k) b l 4000 4000 k =− 3

/

/
k =− 3

3

rect b

f − 4000k l 4000

So sampled signal will contain replicas of X (f ) at f = 0, ! 4 kHz, ! 8 kHz, ! 6 kHz ...

SOL 1.55

Option (C) is correct. For Ts = 0.4 ms 1 = 2.5 kHz 0.4 # 10−3 fs < 2B (Under sampling) So the output spectrum is Sampling frequency fs =

Page 15

EC_Practice Paper C

Chapter 1

s (f ) =

3 1 rect f 2500δ (f − 2500k) b l * 4000 4000 k =− 3 f = 0.625rect b − 2500k l 4000

/

So s (f ) is

SOLUTIONS EC_Practice Paper D

SOL 1.1

Option (A) is correct. 0.6 T (s) = 1 6 (s + 0.8) 2 + (0.6) 2 ωn 1 − ξ 2 = 0.6 , ξωn = 0.8 Hence ωn = 1, ξ = 0.8

SOL 1.2

Option (D) is correct. (2) Not symmetric; root locus does not exist on real axis to the left of an odd number of poles and zero. (3) Not symmetric; root locus exist on real axis the left of even number of pole and zero.

SOL 1.3 SOL 1.4

Option ( ) is correct. Option (C) is correct. Area = ;2

#

135c

#
2

4

45c

ρdρdφ +

# #
3

4

135c

45c

2dφdz

+

# #
3

4

135c

45c

4dφdz + 2

# #
3 2

4

4

dρdz E

SOL 1.5

ρ2 4 = 2 ; E a π k + (2) (1) a π k = 32.27 2 2 2 2 Option (B) is correct. emf =− dΦ =− d # B : dS dt dt = 2π (0.2) 2 (20) (377) sin 377t mV = 0.95 sin 377t V Option (C) is correct. The beam-width of Hertzian dipole is 180c and its half power beam-width is 90c. Option (B) is correct. Put z = 0 + t, f (z) = z cos 1 z = t cos 1 = t b1 − 1 1 + 1 1 − ...l t 2! t2 4! t 4

SOL 1.6

SOL 1.7

Page 2

EC_Practice Paper D

Chapter 1

= t − 1 + 1 3 − ... 2t 24t Residue of f (z) at z = 0 is the coefficient of 1 i.e. − 1 t 2 Option (B) is correct. Here p = 3 = 1 , q = b1 − 1 l = 1 and n = 100 5 2 2 2 Thus Variance = npq = b100 # 1 # 1 l = 25 2 2 Option (C) is correct. This differential equation is linear, Hence I.F. = e2 # tan x dx = e2 log sec x = sec2 x Option (B) is correct. From linear algebra for A n # n triangular matrix det A = aii, . The product of the i=1 diagonal entries of A
SOL 1.11

SOL 1.8

SOL 1.9

SOL 1.10

%

n

Option (C) is correct. 48 10 = 00110000 2 11001111 − 48 10 = + 1 11010000 Option (B) is correct. y 3 = x 3 (MSB) y2 = x 3 5 x2, y1 = y2 5 x1, y 0 = y1 5 x 0 So, the conversion is gray code to Binary code Option (C) is correct. Step size = RF # 5 = 0.5, RF = 800 Ω 8k Option (A) is correct. The minimum number of isolation region is two. One for transistor and one for resistor. Option (B) is correct. Other three circuits can be drawn on plane without crossing as shown below

SOL 1.12

SOL 1.13

SOL 1.14

SOL 1.15

Page 3 SOL 1.16

EC_Practice Paper D

Chapter 1

Option (A) is correct. 1017 # 1014 = 0.63 V d Vbi = Vt ln c Na N 0.0259 ln n i2 m (1.5 # 1010) 2

SOL 1.17

Option (B) is correct. Using same potential technique, there are two node only

Req = 4 + 25 50 50 + 6 = 22.5 Ω
SOL 1.18

Option (B) is correct. For series critically damped circuit R 0 = 4L = 4 # 4 = 40 Ω 10m C R 120 = 40 Ω , R = 60 Ω

SOL 1.19

Option (D) is correct. For t < 0 , s (t) = 0 For t > 0 , s (t) = δ(2) (t) dt = δ (t) # −3 P = lim
n=N 1 / (u [n]) 2 N " 32N + 1 n =− N = N+1 = 1 2N + 1 2 t

SOL 1.20

Option (C) is correct.

SOL 1.21 SOL 1.22

Option ( ) is correct. Option (C) is correct. D2 and D 3 are ON. If D 3 is ON, then D1 is OFF. vo = 5 − 0.6 = 4.4 V, iD = 4.4 − 0.6 = 7.6 mA 0.5k Option (D) is correct. Vo = VZ + VBE = 8.2 + 0.7 = 8.9 V
2

SOL 1.23

SOL 1.24

Option (B) is correct. The figure as shown below

Page 4

EC_Practice Paper D

Chapter 1

SOL 1.25 SOL 1.26

Option ( ) is correct. Option (C) is correct. E [Y] = E

#
0 0

2

X (t) dt =
2 2

#
0

2

E [X (t)] dt = 3

# dt = 6
0

2

E [Y] 2 = E [ X (t) dt
2 2

#

# X (u) du]
0

= = =

# # E [X (t) X (u) dudt]
0 2 0 2

# #R
0 2 0 2 0 0

XX

(t − u) dtdu
− t−u

# # [9 + 2e
2 2 0 0

] dtdu dtdu = 4 (10 + e−2)

= 36 + 2
2 Y 2

# #e

− t−u

σ = E [Y ] − (E [Y]) 2 = 4 (1 + e−2) = 4.541
SOL 1.27

Option (B) is correct. R = 1000 Ω , L = 100 μH ^Rh^ jωL h , Z = R + jω L
2 2 2 ωL Re [Z] = R 2 R + ω2 L2

Page 5

EC_Practice Paper D
2 2 2 ω L df 2 = 4kT # Re [Z] df = 4kT # R vn 2 R + ω2 L2 0 0 By putting ωL = Ry y Ry2 4 kTR 2 dy where y 0 = b 2πL l f0 vn = # 2πL 0 1 + y2 R
0

Chapter 1
f

f0

2 2 = 4kTR ^y 0 − tan−1 y 0h vn 2πL −6 y 0 = 2π # 1003# 10 # 159 # 103 = 0.1 10 2 = vn

Vnrms
SOL 1.28

4 # 1.38 # 10−23 # 290 # ^103h2 −1 ^0.1 − tan 0.1h 2π # 100 # 10−6 = 8.5 # 10−15 V2 = 9.22 # 10−8 V

Option (D) is correct. Instantaneous frequency ω1 = d θ (t) dt = d 2π 62 # 106 t + 30 sin 150t + 40 cos 150t@ dt = 2π (2 # 106 + 4500 cos 150t − 6000 sin 150t) 3 ω = ωi − ωc = 3000π (3 cos 150t − 4 sin 150t) = 15000π cos (150t + α) 3 ω max Maximum frequency deviation 3 f = 2ω = 15000π = 7.5 kHz 2π and φ (t) = 2π (30 sin 150t + 40 cos 150t) = 100π sin (150t + α') Thus maximum phase deviation φ (t) max = 100π

SOL 1.29

Option (B) is correct. The processing gain is given as W = 500 or 27 dB R The εb /J 0 required to obtain an error probability of 10−5 for binary PSK is 9.5 dB. Hence, the jamming margin is εb Jav W b Pav l = b R l − a J 0 k = 27.95 or 17.5 dB dB dB dB

SOL 1.30

Option (C) is correct. For case 1, μ = B1 = 2 1200 H μ 1 = 1 # = 1326.3 μr1 = μ0 600 4π # 10−7

Page 6

EC_Practice Paper D

Chapter 1

χm = μr − 1 = 1325.3 M1 = χm H1 = 1.590 # 106 A/m For case 2, μ = B2 = 1.4 H2 400 1.4 = 2785.2 400 # 4π # 10−7 χm = 2784.2 M2 = χm H2 = 1.114 # 106 A/m 3 M = (1.590 − 1.114) # 106 = 476 kA/m μr1 =
SOL 1.31

Option (B) is correct. Using the method of images, the configuration is as shown below

Here Array factor is

d = λ, α = π, thus βd = 2π βd cos ψ + α D 2 2π cos ψ + π = cos : D 2 = cos : = sin (π cos ψ)

SOL 1.32

Option (A) is correct. y y Let v = φ a k and ω = xΨ a k x x Then u = v + w Now v is homogeneous of degree zero and ω is homogeneous of degree one 2 2 2 v + y 22 2 v = 0 & x 22 v 2 + 2xy 2x2y 2x 2y 2 and x22 ω + 2xy 2 ω (v + ω) + y2 2 2 (v + w) = 0 2x2y 2x 2 2y 2 2 2 2 u + y 22 u = 0 & x 22 u 2 + 2xy 2x2y 2x 2y2
2 2 2

...(i)

SOL 1.33

Option (C) is correct.

Page 7

EC_Practice Paper D

Chapter 1

# 1 +dx sin x

=

# #

=

dx x x x 2x a sin 2 + cos 2 k + 2 sin 2 cos 2 sec2 x dx 2 = dx # 2 x x x 2 + + cos sin 1 tan a a 2 2k 2k
2

Put 1 + tan x = t 2 & sec2 x dx = 2dt & 2

2 # 2tdt 2 dt =− t + K

− 2 cos x − 2 2 +K = +K = x x 1 + tan cos + sin x 2 2 2 − 2 cos x cos x − sin x 2 # 2 2 +K = x x x x cos + sin cos − sin 2 2 2 2 − 2 cos2 x + 2 sin x cos x 2 2 2 +K = 2x 2x cos − sin 2 2 − (1 + cos x) + sin x = + K = tan x − sec x − 1 + K cos x = tan x − sec x + c
SOL 1.34

Option (C) is correct. The point (1 − i) lies within circle z = 2 (... the distance of 1 − i i.e., (1, 1) from the origin is 2 which is less than 2, the radius of the circle). Let φ (z) = 3z2 + 7z + 1 then by Cauchy’s integral formula z + 1 dz # 3z z+−7z 0
2

= 2πiφ (z 0) = 2πiφ (z 0) & f l (z 0) = 2πiφl (z 0) = 2πiφm (z 0) = 3z2 + 7z + 1 = 6z + 7 and φm (z) = 6 = 2πi [6 (1 − i) + 7] = 2π (5 + 13i)

c

& and since, &
SOL 1.35

f (z 0) f m (z 0) φ (z) φl (z) f l (1 − i)

Option (C) is correct. Here f (x, y) = x2 + y2, x− = 0, y 0 = 0 We have, by Picard’s method y = y0 + y(1) = y 0 +

#
x0

x

f (x, y) dx

...(i)

The first approximation to y is given by

#
x0

x

f (x, y 0) dx

Page 8

EC_Practice Paper D

Chapter 1
x

where

y0 = 0 +

#
0

x

f (x, 0) dx =
x

#
0

x2 dx

...(ii) f bx, x l dx 3
3

The second approximation to y is given by y(2) = y 0 + = 0+

#
x0

f (x, y(1)) dx = 0 +
6 3

#
0

x

#
0

x

x x x7 2 bx + 9 l dx = 3 + 63

SOL 1.36

(0.4) 3 (0.4) 7 Now, + = 0.02135 y (0.4) = 3 63 Option (A) is correct. vo will be negative for vs $ 0; D2 off D1 on output vo is vo =− vs . R2 =− vs . 10 =− 5vs 2 R1 For vs < 0, vo will be positive vs < 0; D2 on, D1 off, vo = 0

SOL 1.37

Option (D) is correct. By equivalent circuit of an inverting amplifier

Ro " output resistance of op-amp KCL at output node vo + vo − (− AOL v1) + vo − v1 = 0 RL Ro R2 − v1 b AOL − 1 l Ro R2 vo = 1 + 1 + 1 RL Ro R2 KCL at input node

...(i)

Page 9

EC_Practice Paper D

Chapter 1

i1 = v1 + v1 − vo Roi R2 From eq. (i) and (ii) Resistance at inverting terminal of op-amp is J1 + A + Ro N OL RL O 1 = i1 = 1 + 1 K O v1 Roi R2 K Rif R R K1+ o + o O R R L 2 L P In the given problem R2 = 10 kΩ , R1 = 2 kΩ , Ro = 0 , Roi = 10 kΩ , AOL = 105 So, 1 = 1 + 1 1 + 105 + 0 = 10 Ω Rif 10 kΩ 10 kΩ ; 1 + 0 + 0 E Rif = 0.1 Ω
SOL 1.38

...(ii)

Option (C) is correct. Since transistor are matched VBE 1 = VBE 2, IB1 = IB 2 Thus the circuit is as follows

Iref = IC 1 + IB1 + IB 2 = IC 1 + 2IB 2 = IC 2 + 2 IC 2 β Io = IC 2 , Iref = Io c1 + 2 m β I Io = ref = 1m = 962 μA 1+2 1+ 2 50 β Option (B) is correct. IS = 50 μA = ID, ID = Kn (VGS − VTN ) 2 & 50 # 10−6 = 0.5 # 10−3 (VGS − 1.2) 2 & VGS = 1.516 V, VG = 0 , VS = VG − VGS =− 1.516 V VDS = VD − VS = 5 − (− 1.516) = 6.516 V

SOL 1.39

Page 10 SOL 1.40

EC_Practice Paper D

Chapter 1

Option (D) is correct. σ1 = eni (μn + μp) 10−6 = (1.6 # 10−19) (1000 + 600) ni At T = 300 K, ni = 3.91 # 109 cm−3 E v n i2 = Nc Nv e−c kT m & Eg = kT ln c Nc N n i2 m 1019 & Eg = 2 (0.0259) ln c = 1.122 eV 3.91 # 109 m At T = 500 K, kT = 0.0259 b 500 l = 0.0432 eV, 300
g

&

n i2 = (1019) e−b 0.0432 l cm−3 . ni = 2.29 # 1013 cm−3 = (1.6 # 10−19) (2.29 # 1013) (1000 + 600) = 5.86 # 10−3 (Ω − cm) −1
1.122

SOL 1.41

Option (C) is correct. VDS (sat) = VGS − VTN = 0 − (− 2) = 2 V VDS > VDS (sat) Therefore biased in saturation ' ID = k n bW l [VGS − VTN ] 2 2 L &
−3 1.5 = 80 # 10 bW l [0 − (− 2)] 2 , & W = 9.375 mA 2 L L Option (B) is correct. C Vbi = Vt ln c NB N n i2 m 16 17 # 10 = 0.824 V = 0.02591n e 3 # 10 # 5 10 o (1.5 # 10 ) 2 At punch-through

SOL 1.42

xB = 0.7 # 10−4 = x p (VBC = Vpt) − x p (VBC = 0) = >) =)
1 2 2ε (Vbi + Vpt) NC 2εVbi NC 2 1 1 − ' e NB (NC + NB) 1 H e NB (NC + NB) 3 1

2 (11.7 # 8.85 # 10 -14) (Vi + Vpt) 1.6 # 10 -19
1

2 2 (11.7 # 8.85) (0.824) 5 # 1017 1 c 3 1016 m (5 1017 + 3 1016) 3 − ' # # # 1.6 # 10 -19 2 5 # 1017 1 c 3 1016 m (5 1017 + 3 1016) 3 # # # 1

&

0.7 # 10−4 = 2.02 # 10−5 (Vbi + Vpt) − 1.83 # 10−5 Vbi + Vpt = 19.11 & Vpt = 19.11 − 0.824 = 18.3 V

SOL 1.43

Option (D) is correct.

Page 11

EC_Practice Paper D

Chapter 1

In the circuit D0 = Q 0 , D 1 = Q 0 5 Q1 Initial state " 00 Q0 Q1 D0 D1 0 0 1 1 1 1 0 1 0 1 1 0 1 0 0 0 0 0 1 1 (repeat) So, the state transition sequence Q 0 Q1 is

SOL 1.44

Option (B) is correct. The operation of this circuit is given below : ABC ## 0 001 #1 1 1#1 PA PB PC ## ON ON ON OFF # OFF OFF OFF # OFF Y = (A + B ) C NA NB NC # # OFF OFF OFF ON # ON ON ON # ON Y HIGH HIGH LOW LOW

SOL 1.45

Option (A) is correct. GH (jω) = 1+ω
2

K , 1 + 4ω2 1 + 9ω2

+GH (j ω) =− 90c − tan−1 ω − tan−1 2ω − tan−1 3ω , For ω = 0 , GH (jω) = 3+ − 90c, For ω = 3, GH (jω) = 0+ − 360c,
SOL 1.46

Option (C) is correct. If R (s) = 0 K2 s (s + 4) TD (s) = K1 K2 (s + 2) 1+ s (s + 4) (s + 3) = K2 (s + 3) s (s + 3) (s + 4) + K1 K2 (s + 2)

Error in output due to disturbance E (s) = TD (s) D (s),

Page 12

EC_Practice Paper D

Chapter 1

If D (s) = 1 , s

essD = lim sE (s) = lim s # 1 # TD (s) = lim TD (s) = 3 s s"0 s"0 s"0 2K1 3 = 0.000012 & K = 125 103 # 1 2K1 sR (s) , 1 + G (s) K1 K2 (s + 2) R (s) = 1 2 , G (s) = s ( s + 3) (s + 4) s 1 ess = lim = 6 K1 K 2 s"0 K1 K2 (s + 2) s+ (s + 3) (s + 4) ess = lim sE (s) = lim
s"0 s"0

Error due to ramp input

6 = 0.003 & K2 = 0.016 125 # 103 K2
SOL 1.47

Option (B) is correct. LXI SP, EFFF H ; Load SP with data EFFH CALL 3000 H ; Jump to location 3000H : : : 3000H LXI H, 3CF4 ; Load HL with data 3CF4H PUSH PSW ; Store contnets of PSW to Stack SPHL ; Copy contents of HL to SP (3CF4H) POP PSW ; Restore contents of RSW from stack RET ; stop Before instruction SPHL the contents of SP is 3CF4H. After execution of POP PSW, SP + 2 " SP is 3CF4H. After execution of RET, SP + 2 " SP Thus the contents of SP will be 3CF4H + 4 = 3CF8H Option (B) is correct. DFT of sequence g [n] is GDFT [k] = XDFT [k] e 4 (time-shift property) = XDFT [k] e−jkπ k = 0, 1, 2, 3 GDFT [0] = XDFT [0] e0 = 1 GDFT [1] = XDFT [2] e−jπ = 2 (− 1) =− 2 GDFT [2] = XDFT [2] e−j 2π = 3 (1) = 3 GDFT [3] = XDFT [3] e−j 3π = 4 (− 1) =− 4 GDFT [k] = {1, − 2, 3, − 4}
−jk 2π (2)

SOL 1.48

So
SOL 1.49

Option (D) is correct.

Page 13

EC_Practice Paper D

Chapter 1

Let

So
SOL 1.50

x [− n] + XDFT [− k] = XDFT [N − k] (periodic extension of folded version) g [n] = x [− n] then GDFT [k ] = XDFT [N − k], N=4 GDFT [0] = XDFT [4] = 1 GDFT [1] = XDFT [4 − 1] = XDFT [3] & GDFT [1] = 4 GDFT [2] = XDFT [4 − 2] = XDFT [2] & GDFT [2] = 3 GDFT [3] = XDFT [4 − 3] = XDFT [1] & GDFT [3] = 2 GDFT [k ] = {1, 4, 3, 2}

Option (C) is correct. The circuit is as shown below

j j 5+0c = i1 b j4 + 1 + 1 − l − i2 b1 − l 4 4 & (8 + j15) i1 − (4 − j) i2 = 20+0 j j − 10+ − 30c = i2 b1 + j4 + 1 − l − i1 b1 − l 4 4 & (4 − j) i1 − (8 + j15) i2 i1 [(8 + j15) 2 − (4 − j) 2] i1 (− 176 + j 248) i1 i2 = &
SOL 1.52

...(i)

&
SOL 1.51

= 40+ − 30c = (20+0) (8 + j15) − (40+ − 30c) (4 − j) = 41.43 + j 414.64 = 1.03 − j 0.9 = 1.37+ − 41.07

...(ii)

Option (A) is correct.

(8 + j15) (1.03 − j 0.9) − 20+0c 4−j

=− 0.076 + j 2.04 i2 = 2.04+92.13c

Option (A) is correct. C (s) b0 24 = 3 = 2 2 2 R (s) s + a2 s + a1 s + a 0 (s + 9s + 26s + 24) 3 2 (s + a2 s + a1 s + a 0) C (s) = b 0 R (s) Taking the inverse Laplace transform assuming zero initial conditions q + a2 c p + a0 c = b0 r c o, x 3 = c p x1 = c = y , x 2 = c o = x2, x o2 = c p = x3 o1 = c x o o p = x3 o x 3 = c = x2, x2 = c

Page 14

EC_Practice Paper D

Chapter 1

q = b 0 r − a2 c p − a1 c o − a0 c o3 = c x =− a 0 x1 − a1 x2 − a2 x 3 + b 0 r , Ro V R R V R V x 1 0V S 1W S 0 WSx1W S 0 W o2W = S 0 Sx 0 1 WSx2W + S 0 W r S W S o3W S− a 0 − a1 − a1W Sx W W S WS Sb 0W Sx 3W T X T XT X T X a 0 = 24 , a1 = 26 , a2 = 9 , b 0 = 24 Ro V R R V R V 1 0V Sx1W S 0 WSx1W S 0 W o2W = S 0 Sx 0 1WSx2W + S 0 W r S W S o3W S− 24 − 26 − 9W Sx W W S WS S24W Sx 3W T X T XT X T X Rx V S 1W y = 81 0 0BSx2W S W Sx 3W T X

SOL 1.53

Option (B) is correct. Fourth order hence four state variable R R V V 0 0 W 1 S 0 S0W S 0 S0W 1 0 W 0 o =S x x = S W r , y = 81 0 0 0B x W 0 0 1 W S 0 S0W S− a 0 − a1 − a2 − a 3W Sb 0W T T X X a 0 = 100, a1 = 7 , a2 = 10 , a 3 = 20 , b 0 = 100 Option (B) is correct. fc = fc10 fc01 2 GHz < f < 3 GHz c 2 εr 3 # 108 = = 2 GHz = c 2 # 1.25 # 0.06 a2 εr 3 # 108 = c = = 3 GHz 2 # 1.25 # 0.04 b2 εr m 2 n 2 a a k +ab k

SOL 1.54

SOL 1.55

Option (C) is correct. fc11 = 3 GHz < f < 3.6 GHz 3 # 108 2 # 1.25 # 10−2 1 2 1 2 b 6 l + b 4 l = 3.6 GHz

SOLUTIONS EC_Practice Paper E

SOL 1.1

Option (D) is correct. Let I = = # e dx # exdx 1 − e−x −1
−x

SOL 1.2

Put 1 − e−x = t & e−x dx = dt I = # dt = log t = log (1 − e−x ) t Option (B) is correct. −1 z < 1, 1 − 1 =− 1 a1 − z k + (1 − z) −1 z−2 z−1 2 2
2 3 =− 1 :1 + z + z + z + ...D + (1 + z + z2 + z3 + ...) 2 2 4 8 f (z) = 1 + 3 z + 7 z2 + 15 z3 + ... 2 4 8 16 Option (A) is correct. Since A (adj A ) = A I 3

SOL 1.3

SOL 1.4

R1 0 0V R2 0 0V W W S S & A (adj A ) = 2 S0 1 0W = S0 2 0W S W W S S0 0 2W S0 0 1W X X T T Option (D) is correct. K (s + 3) T (s) = 5 s + 2s2 + Ks + 1 Always unstable since s 4 and s3 term are missing. Option (A) is correct. X (s) = 2 − 1 (s + 2) (s + 3) = 2− 1 + 1 (s + 2) (s + 3)

SOL 1.5

x (t) = 2δ (t) + (e−3t − e−2t) u (t)
SOL 1.6

Option (B) is correct.

αc = 0 − 1 − 2 =− 1 3−0

Page 2 SOL 1.7

EC_Practice Paper E

Chapter 1

Option (A) is correct. The function has poles at z = 1, 3 . Thus final value theorem applies. 4
n"3

lim x (n) = lim (z − 1) X (z)
z"1

= (z − 1)
SOL 1.8

z (2z − 7 4) =1 (z − 1) (z − 3 4)

Option (B) is correct. There is one forward path G1 G2 . Four loops − G1 G 4, − G1 G2 G 3, − G1 G2 G5 G7 and − G1 G2 G 3 G6 G7 .

There is no non-touching loop.
SOL 1.9

Option (B) is correct. At t = 1, the slope of function changes from 0 to 2 At t = 2 , the slope of function changes for a changes in magnitude of − 2 from 2 to 0; At t = 3 , the slope of function changes for a changes in magnitude of 2 from 0 to 2; Alternatively : For 1 < t < 2 , x (t) = 2 (t − 1) For 2 < t < 3 , x (t) = 2 For 3 < t , x (t) = 2t − 2

SOL 1.10 SOL 1.11

Option ( ) is correct. Option (C) is correct. 2x2 + x + 1 = 64 + 5 # 8 + 2 & x =7 Option (B) is correct. In negative cycle diode will be ON and capacitor will be charged. In positive cycle + 10 V of vi will be added to capacitor voltage.

SOL 1.12

Page 3

EC_Practice Paper E

Chapter 1

Thus vo = 20 V.
SOL 1.13

Option (B) is correct. Id = Is eb−V l ,
V
t

SOL 1.14

(V − V ) Id1 = e = e− V Id2 e V1 − V2 = Vt ln c Id2 m = 0.02591n10 = 59.6 mV Id1 Option (B) is correct. We have A qD n 2 IC = E n i (eV /V − 1) NE WB IC α 1 WB
1 2 t BE T

V − 1 Vt V − 2 Vt

So, If WB increases by a factor of three, then IC is decreased by a factor of three
SOL 1.15

Option (B) is correct.
' ID = k n bW l (VGS − VTN ) 2 2 L −3 0.5 = 80 # 10 (5) (VGS − 0.8) 2 2

&
SOL 1.16

VGS =

0.5 + 0.8 = 2.38 V 0.2

Option (C) is correct. 4i2 + 6i 3 − 2i1 = 0 , i1 + i 2 = 2 i2 = 5 + i 3 & i1 =− 5 A 6

Page 4 SOL 1.17

EC_Practice Paper E

Chapter 1

Option (C) is correct. The circuit is as shown below

RTH = 7 5 + 6 9 = 6.52 Ω For maximum power transfer RL = RTH = 6.52 Ω
SOL 1.18

Option (B) is correct. Initially i (0−) = 0 . Due to inductor i (0+) = 0 Thus all current Is will flow in resistor Rs and voltage across inductor will be di (0+) vL (0+) = Is Rs but vL (0+) = L , dt di (0+) = Is Rs L dt Option (D) is correct. 2ρv =−4: J = 1 2 (ρJ ρ) + 2Jz ρ 2ρ 2z 2t = 1 2 (25) + 2 c 2 − 20 m = 0 ρ 2ρ 2z ρ + 0.01 Thus Option (D) is correct. For Ey = 0 , 2y sin 2x = 0 & y = 0

SOL 1.19

SOL 1.20

SOL 1.21

sin 2x = 0 , & 2x = 0 , π, 3π, & x = 0, 3π 2 Option (C) is correct. For nonmagnetic medium μr = 1 β = ω = ω εr , v c ω = 109, β = 8, 9 8 = 10 # 108 εr & εr = 5.76 3 Option (A) is correct. It = Ic b1 + α l or 20 = 18 b1 + α l or α = 0.68 2 2
1 2 2 1 2 2

SOL 1.22

SOL 1.23

Option (D) is correct. Initial BW = 4 # 10 = 40 kHz Now if deviation is d , then 3 BW = 2d = 2 # 2 = 4 kHz

Page 5

EC_Practice Paper E

Chapter 1

New Bandwidth BW = 40 + 4 = 44 kHz
SOL 1.24

SOL 1.25

Option (C) is correct. p = 2 = 1 , q = b1 − 1 l = 2 and n = 3 3 3 6 3 Thus Mean np = 1 # 3 = 1 3 Option (C) is correct. f (1) − f (0) f l (c) = & ec = e − 1 (1 − 0) & c = log (e − 1) Option (A) is correct. 0.1 0.2 0.3 0.4 x : 0 Euler’s method gives yn + 1 = yn + h (xn, yn) n = 0 in (1) gives y1 = y0 + hf (x0, y0) Here x0 = 0 , y0 = 1, h = 0.1 y1 = 1 + 0.1f (0, 1) = 1 + 0 = 1 n = 0 in (1) gives y2 = y1 + hf (x1, y1) = 1 + 0.1f (0.1, 1) = 1 + 0.1 (0.1) = 1 + 0.01 Thus y2 = y(0.2) = 1.01 n = 2 in (1) gives y 3 = y2 + hf (x2, y2) = 1.01 + 0.1f (0.2, 1.01) y 3 = y(0.3) = 1.01 + 0.0202 = 1.0302 n = 3 in (1) gives y 3 = y2 + hf (x2, y2) = 1.01 + 0.1f (0.2, 1.01) y 3 = y(0.3) = 1.01 + 0.0202 = 1.0302 n = 3 in (1) gives y 4 = y 3 + hf (x 3, y 3) = 1.0302 + 0.1f (0.3, 1.0302) = 1.0302 + 0.03090 y 4 = y(0.4) = 1.0611 Hence y(0.4) = 1.0611 Option (B) is correct. dy − y tan x = y 2 sec x dx dy or, y−2 − y−1 tan x = sec x dx Put y−1 = v & − y−2 dy dv = dx dx

SOL 1.26

...(i)

SOL 1.27

Substituting in the given equation, we get − dv − v tan x = sec x or, dv + (tan x) : v =− sec x dx dx

Page 6

EC_Practice Paper E

Chapter 1

I.F. = e # tan x dx = e logsec x = sec x v : sec x =− sec2 x dx + c =− tan x + c 1 =− sin x + c cos x y or
SOL 1.28

#

y−1 =− sin x + c cos x
1 1

Option (A) is correct.

# # # 0 0 0

1

ex + y + z dxdydz = = = =

# # 0 0 # 0 # 0 # 0
1 1

1

1

[e x + y + z ] 1 0 dydz =

# # 0 0

1

1

[e1 + y + z − ey + z ] dydz

[e1 + y + z − ey + z ] 1 0 dz [(e2 + z − e1 + z ) − (e1 + z − ez )] dz (e2 + z − 2e1 + z + ez ) dz = [e2 + z − 2e1 + z + ez ] 1 0

1

= (e3 − 2e2 + e) − (e2 − 2e + 1) = e3 − 3e2 + 3e − 1 = (e − 1) 3
SOL 1.29

Option (B) is correct.

βl = 2π (1.25λ) = π + 360c, tan π = 3 2 2 λ Z + jZ 0 tan βl Zin = Z 0 c L Z 0 + jZL tan βl m

If

tan βl = 3, 2 2 Z0 = 80 = 51.2 Ω 125 ZL 4# H =−2H z u z = ε2E 2y 2t = 6β sin (βx) cos (108 t) u y E = 1 # 6β sin (βx) sin (108 t) u y ε

SOL 1.30

Option (A) is correct.

8 ε = 6.25ε0 , β = ω = ω εr = 10 8 6.25 = 0.833 v c 3 # 10 6 (0.833) E = sin (βx) sin (108 t) u y V/m 6.25ε0 # 108 = 903 sin (0.83x) sin (108 t) u y V/m

SOL 1.31

Option (B) is correct. At the boundary normal unit vector 4 (2x + 3y − 4z) 2u + 3u y − 4u z = x un = 4 (2x + 3y − 4z) 29 = 0.37u x + 0.56u y − 0.74u z Since this vector is found through the gradient, it will point in the direction of

Page 7

EC_Practice Paper E

Chapter 1

increasing values of 2x + 3y − 4z , and so will be directed into region 1. Thus u n = u n21 The normal component of H 1 is H N1 = (H 1 : u N21) u N21 H 1 : u N21 = (50u x − 30u y + 20u z ) : (0.37u x + 0.56u y − 0.74u z ) = 18.5 − 16.8 − 14.8 =− 13.1 (H 1 : u N21) u N21 = (− 13.1) (0.37u x + 0.56u y − 0.74u z ) =− 4.83u x − 7.24u y + 9.66u z A/m Tangential component of H 1 at the boundary HT1 = H 1 − H N1 = (50u x − 30u y + 20u z ) − (− 4.83u x − 7.24u y + 9.66u z ) = 54.83u x − 22.76u y + 10.34u z A/m HT2 = HT1 μ H N2 = r1 H N1 = 2 (− 4.83u x − 7.24u y + 9.66u z ) μr2 5 =− 193u x − 2.90u y + 3.86u z A/m H 2 = 6HT2 + H N2 = (54.83u x − 22.76u y + 10.34u z ) + (− 193u x − 2.9u y + 3.86u z )@ = 52.9u x − 25.66u y + 14.2u z
SOL 1.32

Option (C) is correct. We assume that the interference is characterized density a zero-mean AWGN process with power spectral density J 0 . To achieve an error probability of 10−5 , the required εb /J 0 = 10 we have W/R W/R = = εb Nu − 1 J 0 Jav /Pav W = εb (N − 1) a J0 k u R W = R a εb k (Nu − 1) J0 where R = 10 4 bps, Nu = 30 and εb = 10 J0 Therefore, W = 2.9 # 106 Hz The minimum chip rate is 1 = W = 2.9 # 106 chips/sec Tc Option (D) is correct. y I (x j y) = H (y) − H a k x where
2 2 y y H a k =− p (xi, y j ) log 2 p b j l x xi i=1 j=1 y H a k =− αp log 2 p − (1 − α) q log 2 p − αq log 2 q − (1 − α) q log 2 q x y H a k =− p log 2 p − q log 2 q x

SOL 1.33

//

or

Page 8

EC_Practice Paper E

Chapter 1

Thus I (x j y) = H (y) + p log 2 p + q log 2 q Which is maximum when H (y) is maximum. Since the system output is binary, H (y) is maximum when each output has a probability of 1/2 and, for a symmetrical channel, is achieved for equally likely inputs. For this case H (y) is unity and the channel capacity is C = 1 + p log 2 p + q log 2 q
SOL 1.34

Option (C) is correct. Nyquist Rate = 2 MHz 50% higher rate = 3 MHz, L = 256 = 28 Thus transmission bandwidth is 3 MHz # 8 = 24 Mbits/s. New sampling rate is at 20% above the Nyquist rate. Sampling rate = 1.2 # 2 = 2.4 MHz. bits Bits per second = 24 M sec = 10 bits 2.4 MHz Level = 210 = 1024, 2 S 0 = 3 (1024) = 102300 = 50.1 dB N0 (ln 256) 2

SOL 1.35

Option (C) is correct. Since 88 MHz < fc < 108 MHz and fc − f c' = 2fIF if fIF < fLO We conclude that in order for the image frequency f c' to fall outside the interval [88, 108] MHz, the minimum frequency fIF is such that 2fIF = 108 − 88 or fIf = 10 MHz If fIF = 10 MHz, then the range of fLO is [88 + 10, 108 + 10] = [98, 118] MHz Option (B) is correct. 0 1 s , (sI − A) −1 = 2 1 > A => − 2 0H − s +2 2 R S cos 2t + sin 2t Φ (t) = L−1 x (0) = S 2 S S− 2 sin 2t + cos T y = x1 − x2 = 3 sin 2t 2 Option (C) is correct. K GH (j ω) = j ω (− ω2 + 2j ω + 2) +GH (j ω) =− tan−1 2ω 2 − 90c 2−ω K GH (j ω) = ω (2 − ω2) 2 + 4ω2 At ω = 0 , GH (jω) = 3+ − 90c, 1 sH V W W W 2t W X

SOL 1.36

SOL 1.37

Page 9

EC_Practice Paper E

Chapter 1

At ω = 3, GH (jω) = 0+ − 270c, At ω = 1, GH (j ω) = K + − 153.43c, 5 At ω = 2 , GH (j ω) = K + − 206.6c, 2 18 Due to s there will be a infinite semicircle.
SOL 1.38

Option (B) is correct. The equivalent forward transfer function Ge (s) = = 1+
K s (s + 1)(s + 4) K (s + α − 1) s (s + 1)(s + 4)

SOL 1.39

K s3 + 5s2 + (K − 4) s + K (α − 1) e (3) = 1 , 1 + Kp K p = lim G (s) = 1 α−1 s"0 1 e (3) = = α−1 1 α 1+ α−1 α − (α − 1) 1 Se: α = α 2e = α c = α− m 2 2 e 2α α α α α−1 Option (B) is correct. Since x (t) is real and odd, X [k] is purely imaginary and odd. Therefore X [k] =− X [− k] and X [0] = 0 . Since X {k] = 0 for k > 1, they only unknown coefficient are X [1] and X [− 1] 3 1 x (t) 2 dt = / X [k] 2 # T k =− 3
T

For x (t), 1 2
2

# 0

2

x (t) 2 dt =
2

/
k =− 1

1

X [k ] 2

& X [ 1] + X [ − 1] = 1 , 2 X [ 1] 2 = 1 X [1] =− X [− 1] = −j j or X [1] =− X [− 1] = 2 2

SOL 1.40 SOL 1.41

Thus there are two solutions j jb 22π lt j −jb 22π lt x1 (t) = − =− 2 sin πt e e 2j j 2π t 2j −j 2π t x2 (t) =− eb2l + e b 2 l = 2 sin πt 2 2 Option ( ) is correct. Option (B) is correct. Length of x [n] is 8 − 4 + 1 = 5 we know that

Page 10

EC_Practice Paper E

Chapter 1

y [ n ] = h [ n ] * x [ n] = y [n] =
3

k =− 3

/ h [ k ] x [n − k ]

3

k=0

/ h [ k] x [n − k ]
k=1 3

y [n] = h [0] x [n] + So

/ h [k ] x [n − k ] / /
3 3

3

x [n] = 1 =y [n] − h [k] x [n − k ]G h [0] k=1 x [0] = 1 =y [n] − h [k ] x [− k ]G h [0] k=1 = 1 {y [0] − h [1] x [− 1] − h [− 2] ...} = 1 # 6 = 3 2 h [0]

Similarly

x [1] = 1 =y [1] − h [k ] x [1 − k ]G h [0] k=1 1 = {y [1] − h [1] x [0] − h [2] x [− 1] ...} h [0] = 1 {11 − 5 # 3} 2 = − 4 =− 2 2

/

Similarly we can find the sequence.
SOL 1.42

Option (A) is correct. MVI A, DATA1 ; DATA1 " A MOV B, A ; A"B SUI 51H ; A - 51H " A JC DLT ; If CY=1, Jump in DLT MOV A, B ; B"A SUI 82H ; A - 82H " A JC DSPLY ; If CY = 1, Jump on DSPLY DLT : XRA A ; Clar A OUT PORT1 ; A " PORT1 DSPLY:MOVA, B ; B"A OUT PORT2 ; A " PORT2 HLT If DATA1 is less than 51H, SUI 51H will set the CY flag and execution will jump on DLT. After this A will be cleared and output at PORT1 will be 00. If DATA1 is greater than 51H and less than or equal to 82H, execution will jump on DSPLY and DATA1 will be displayed at PORT2. Option (B) is correct. 12 + Vo 6 .8k 24k V+ = 1 + 1 + 1 6.8k 6.8k 24k

SOL 1.43

Page 11

EC_Practice Paper E

Chapter 1

For Vo =+ 12 V, For Vo = 0 V V+ = 6.74 V V+ = 5.76 V vc (t) = Vf − (Vf − Vi) e− RC & 6.74 = 12 − (12 − 5.76) e− RC t1 = RC ln 6.74 5.26 RC = 6.2k # 0.033 μ = 2.046 # 10−4 t1 = 2.046 # 10 4 ln b 6.74 l = 50.7 μs 5.26 5.26 = 0 − (0 − 6.74) e− RC t2 = 2.046 # 10−4 ln b 6.74 l = 50.7 μs 5.26 T = t1 + t2 = 2 # 50.7μ = 101.4 μs f = 1 = 9.86 kHz T
SOL 1.44
t2 t1 t

&

Option (B) is correct. It is a down counter because 0 state of previous FFs change the state of next FF. You may trace the following sequence, let initial state be 0 0 0 FF C J K C 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 0 0 1 0 0 1 0 0 0 FF B J K B 1 1 1 0 0 0 1 1 0 0 0 1 1 1 1 0 0 0 1 1 0 0 0 1 FF A J K A 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 0 C+ B+ A+ 1 1 1 1 1 0 1 0 1 1 0 0 0 1 1 0 1 0 0 0 1 0 0 0

SOL 1.45

Option (B) is correct. Let P = 1001 and Q = 1010 then Yn = Pn 5 An 5 Rn , Zn = Rn Qn + Pn Rn + Qn Pn output is 1111 which is 2’s complement of − 1. So it gives P − Q . Let another example P = 1101 and Q = 0110 then output is 00111. It gives P − Q .

Page 12

EC_Practice Paper E

Chapter 1

SOL 1.46

Option (C) is correct. By solving the circuit, output voltage vo is given as J R4 N O R 2 K vo = b1 + lK R 3 O v2 − b R2 l v1 R1 K 1 + R 4 O R1 R 3 L P 11 vo = (1 + 10) b v − 10v1 1 + 11 l 2 vo = b11 # 11 l v2 − v1 = 10.0833v2 − 10v1 12

So,

...(i)

Difference mode input voltage is vo = v1 − v2 Common mode input voltage is vcm = v1 + v2 2 Combining equation (ii) and (iii) v1 = vcm − vd and v2 = vm + vd 2 2 so, vo = (10.0833) avcm + vd k − (10) avcm − vd k 2 2 = 10.042vd + 0.0833vcm

...(ii) ...(iii)

Page 13

EC_Practice Paper E

Chapter 1

SOL 1.47

vo = Ad vd = Acm vm Difference mode gain Ad = 10.042 Common-mode gain Acm = 0.0833 CMRR(dB) = 20 log 10 c Ad m Acm = 20 log 10 b 10.042 l = 41.6 dB 0.0833 Option (C) is correct. DC Analysis : ICQ = IEQ VCEQ = 5 = 10 − ICQ (RC + RE ) & 5 = 10 − ICQ (1.2k + 0.2k) & ICQ = 3.57 mA IBQ = 3.57 = 23.8 μA 150 AC Analysis :

(0.0259) βVt = (150) = 1.09 kΩ, ro = 3 3.57m ICQ − (βIb) RC , Av = vo = vs vs rπ = vs = Ib rπ + (β + 1) RE Ib − βRC Av = rπ + (1 + β) RE − (150) (1.2) k =− 5.75 = 1.09k + (151) (0.2k) Option (B) is correct. For TE10 mode π fμ π # 9 # 109 # 4π # 10−7 Rs = 1 = = σc σc δ 1.1 # 107 = 0.0568 f 2 2 # 1.5 3.876 2 Rs e 1 + 2b c c m o 0 . 0568 1 + c a f 2. 4 b 9 l m = αc = 2 f 2 15 # 10−3 # 233.8 1 − b 3.876 l bη 1 − c c m f 9 = 0.022

SOL 1.48

Page 14 SOL 1.49

EC_Practice Paper E

Chapter 1

Option (B) is correct. σd = 10−15 10−15 = 1.3 ωε 2π # 9 # 109 # 2.6 # 8.85 # 10−12 σd << 1, hence v . 1 = c , η . 377 = 233.8 ωε με 2.6 2. 6 8 3 # 10 fc = = 3.876 GHz 2 # 2.4 # 10−2 2.6 σd η 10−15 # 233.8 = 1.3 10−13 Np/m αd = = # 2 fc 2 3 . 876 2 1 −c m 1 −b 9 l f Option (D) is correct. y (t) = 4 [m (t) + cos ωc t] + 10 [m (t) + cos ωc t] 2 = 4m (t) + 4 cos ωc t + 10m2 (t) + 20m (t) cos ωc t + 5 + 5 cos 2ωc t = 5 + 4m (t) + 10m2 (t) + 4 [1 + 5m (t)] cos ωc t + 5 cos 2ωc t The AM signal is, xc (t) = 4 [1 + 5m (t)] cos ωc t m (t) = Mmn (t) xc (t) = 4 [1 + 5Mmn (t)] cos ωc t 5M = 0.8 or M = 0.16 Option (C) is correct. The filter characteristic is shown below

SOL 1.50

SOL 1.51

fc − W > 2W or fc > 3W , fc + W < 2f or fc > W Therefore fc > 3W
SOL 1.52

Option (B) is correct. For n+ polysilicon gate E Φms =−b g − Φfn l 2e − 0.35 =−b 1.11 − Φfn l & Φfn = 0.205 2 Φfn = Vt ln b Nd l ni & Nd = 1.5 # 1010 eb 0.0259 l = 4 # 1013 cm−3
0.205

Page 15 SOL 1.53

EC_Practice Paper E

Chapter 1

Option (D) is correct. For p+ polysilicion gate Eg + Φfs l 2e − 0.35 = b 1.11 + Φfn l & Φfn =− 0.905 V 2 Φms = b

&

The is impossible, cannot use a p+ polysilicon gate.
SOL 1.54

Option (D) is correct. Assume transistor is in saturation VS =− VGS 2 ID = 0 − VS = VGS = IDSS b−VGS l , RS RS VP R S = 1 kΩ VGS = 6m 1 − VGS 2 b 4 l 1k & VGS = 8.86, 181 V VGS = 8.86 V is impossible. ID = VGS = 181 = 1.81 mA RS 1k Option (B) is correct. VD = ID RD − 5 = (1.18 m) (0.4 k) − 5 =− 4.276 V VSD = VS − VD = 1.81 − (− 4.276) = 2.47 V VSD (sat) = VP − VGS = 4 − 1.81 = 2.19 V VSD > VSD (sat) , Assumption is correct.

SOL 1.55

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