ECE GATE PAPER 25 ANSWERS

Embed Size (px)

Citation preview

  • 7/25/2019 ECE GATE PAPER 25 ANSWERS

    1/19

    ECTest ID: 161275 TarGATE16 www.gateforum.com

    ICPIntensive Classroom ProgrameGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India

    All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

    1

    Answer Keys:

    General Ability

    1 B 2 C 3 D 4 14 5 B 6 B 7 C

    8 2.5 9 B 10 A

    Electronics and Communication Engineering

    1 2 2 A 3 C 4 70 5 2786 6 8.92 7 5

    8 1.59 9 16 10 B 11 C 12 B 13 D 14 A

    15 A 16 C 17 D 18 A 19 1.524 20 55 21 D

    22 D 23 0.2 24 0.52 25 D 26 D 27 0.5 28 16.67

    29 0.139 30 0.047 31 0 32 C 33 D 34 B 35 A

    36 C 37 C 38 7.465 39 C 40 C 41 B 42 B

    43 B 44 A 45 28.125 46 73.6 47 2 48 2 49 0.5

    50 10 51 B 52 821.467 53 2.5 54 B 55 B

    Explanations:-

    General Ability

    1. A renegade is a disloyal person who betrays or deserts his cause or religion or political party

    or friend etc.2. The given analogy is name of study: field of study. Gerontology is the branch of science that

    deals with the problems of aging. Likewise, otology is the science of ear and its diseases.

    4. All numbers are made up of prime factors. All perfect squares are made up of pairs of primes.

    So, in order for 3150y to be a perfect square, it must consist only of pairs of prime factors.

    3150 breaks up into 315 * 10 (no primes)

    315 breaks up into 5 * 63 (5 is prime, put that aside)

    63 breaks up into 7*9 (7 is prime, put that aside)

    9 breaks up into 3*3 (both prime, put them aside).

    10 breaks up into 2*5 (both prime, put them aside).So, we've put aside: 5*7*3*3*2*5

    writing in order:

    2*3*3*5*5*7

    Our 3s and 5s are paired off, so we don't need any more of those to create a perfect square.

    We have a single 2, so we need a 2.

    We have a single 7, so we need a 7.

    So, the minimum possible value for y is 2*7 = 14

    5. Assume c=a3

    c2- b2=127

  • 7/25/2019 ECE GATE PAPER 25 ANSWERS

    2/19

    ECTest ID: 161275 TarGATE16 www.gateforum.com

    ICPIntensive Classroom ProgrameGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India

    All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

    2

    (c+b)(c-b) =127

    since 127 is a prime number the only possible combination is 1, 127.

    since b and c are positive, c+b =127 and c-b =1 solving which we get c=64, b=63

    Then a = 4 (since c = a3)

    6. Consider each choice. (A) is possible; for instance, 6 + 7 + 8 = 21. (B) is impossible, so it is

    correct. An integer with only even prime factors cannot have 7 as a prime factor; if it doesn't

    have 7 as a prime factor, it can't yield an integer when divided by 7.

    (C) is possible: If either of the odd integers is 7 or divisible by 7, the result is also divisible by

    7. so (E) is possible as well. (B) is the correct choice.

    7. The average of the three variables is (a + b + c)/3. However, we need to solve in terms of a,

    which means we must convert b and c into something in terms of a.

    We're told that a = (1/4)b, which is equivalent to b = 4a. We can plug that in and simplify the

    average to: (a + 4a + c)/3

    We also know that c = 7a, which we can plug directly into the average expression:

    (a + 4a + 7a)/3 = 12a/3 = 4a, choice (C).

    8. The formula for compound interest is

    (final balance) = principal * (1 + (interest rate) / N)^(time * N)

    Where N is the number of times the interest is compounded annually

    After one year, Ricardo's balance is

    final balance = $1,000 * (1 + (.10 / 2))^(1 * 2)

    final balance = $1,000 * 1.05^2

    final balance = $1,000 * 1.1025

    final balance = $1,102.50

    After one year, Poonam's balance is

    final balance = $1,000 * (1 + (.10 / 1))^(1 * 1)

    final balance = $1,000 * (1.10)^1

    final balance = $1,000 * 1.10

    final balance = $1,100

    The difference between the accounts

    difference = $1,102.50 - $1,100

    difference = $2.50

    Electronics and Communication Engineering

    1.

    j5 t j3 t

    j5 t j3 t

    j5 t j3 t

    x t e u t e u t

    x t e u t e u t

    x t e u t e u t

    j5 t j3 t

    cs

    o 0

    x t x t 1 1x t e e

    2 2 2

    2; T 2

  • 7/25/2019 ECE GATE PAPER 25 ANSWERS

    3/19

    ECTest ID: 161275 TarGATE16 www.gateforum.com

    ICPIntensive Classroom ProgrameGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India

    All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

    3

    2. The characteristic equation is

    1+G(S)H(S) = 0

    2

    2

    k s 2

    1 0s s 1

    s3-s2+ks2+4ks+4k = 0

    R-H criteria

    3

    2

    1

    0

    s 1 4k

    s k 1 4k

    4k k 2s 0

    k 1

    s 4k 0

    So, for stability k>0, k>1 and k>2

    i.e., k>2

    3. Even no. of complementer will produce the output as input. So,output N

    5. Reflection coefficient at antenna end

    L 0

    L 0

    Z ZZ Z

    Radiation resistance of quarter wave dipole 36.5

    So36.5 25

    0.18736.5 25

    2

    ref incident

    2

    trans incident

    P P

    P P 1

    tra

    2 2

    ref 2n 2

    1 1p 1So 00P 2756W

    6.

    i n p

    n

    p

    i

    1Since,conductivity

    resistivity

    for,puresilicon, en

    where, electronmobility

    Holemobility

    n carrierconcentration

  • 7/25/2019 ECE GATE PAPER 25 ANSWERS

    4/19

    ECTest ID: 161275 TarGATE16 www.gateforum.com

    ICPIntensive Classroom ProgrameGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India

    All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

    4

    19 6

    p p

    p 19

    8 2

    p

    11.6 10 1.5 10 3

    3500

    1

    1.6 10 1.5 10 3500 4

    2.98 10 m V sec

    n p

    8

    n

    8 2

    n

    since =3

    =3 2.98 10

    8.92 10 m V sec

    7.

    Z ABC

    Z ABC A B C A B . C

    Minimum 5 2 input NOR gates are required to Realize Z ABC.

    8.

    2c c1 c2 1 B 2 B

    c 1 B 2 B 1 B

    I I I I I

    I I I I

    c 1 2 1 2 B

    B

    I I

    4I 1.59mA

    30 80 30 80

    9. f 2 44 t

    x t 2 sint 0 otherwise

    By parsavals energy theorem,

    A

    B

    CZ

    Z

    C

    B

    A

    BI

    cI

    C2I

    EI

    c1I

    E1 B2I I

    X

    44

    2

  • 7/25/2019 ECE GATE PAPER 25 ANSWERS

    5/19

    ECTest ID: 161275 TarGATE16 www.gateforum.com

    ICPIntensive Classroom ProgrameGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India

    All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

    5

    2

    4

    4

    1energy of x t X d

    2

    14 d

    2

    4 8 162

    10. T0= 4; 00

    2

    T 2

    Fundamental Fourier term1 0 1 0a cos t b sin t

    the given signal is even symmetrical b1= 0

    1 1 12

    1 0

    1 1 1

    2 1 1 1 cos ta x t cos t dt cos t dt dt4 2 2 2 2

    1

    1 1a FundamentalFourierterm cos t.

    2 2 2

    12. The characteristic equation of A is

    A I 0

    8 6 2

    6 7 4 02 4 3

    3 2 18 45 0

    0,3,15

    The diagonal matrix corresponding to the given matrix consists the eigen values as its

    principal diagonal elements

    0 0 0

    D 0 3 0

    0 0 15

    13. An invertible system must satisfy a one-one mapping between input & output

    For system 1, if x t 2 & x t 3 the output y(t) = 0.

    For system 2, every input gives a unique output

    15. Number of samples in linear convolution = 4+5-1 = 8

    If the result of linear convolution is same as circular convolution x[n] and h[n] must have 8

    samples

  • 7/25/2019 ECE GATE PAPER 25 ANSWERS

    6/19

    ECTest ID: 161275 TarGATE16 www.gateforum.com

    ICPIntensive Classroom ProgrameGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India

    All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

    6

    Number of zeros required for x[n] = 4

    Number of zeros required for h[n] = 3

    16. cot x

    cot x

    lim (x+1) (1 )

    x 0let y = (x+1) log y = cot x log(x+1)

    2

    lim log y = lim cotx .log (x+1) ( 0)x 0 x 0

    log (x+1) 0 = lim

    tan x 0x 0

    1

    x+1 = lim 1 (applying L'Hospital Rule)

    sec xx 0

    elog y= 1 y = e

    limit value of f(x) at x = 0 is "e"

    given that f(x) is continuous at x = 0

    limit value at x = 0 function value at x = 0

    e K K= e

    17. The parameter value at T = 400k are found as3 2

    19

    V

    19 3

    V

    400N 1.04 10 300

    N 1.601 10 cm

    400kT 0.0259 0.03453eV

    300

    The hole concentration is given by

    V F

    F V

    E E kT

    V

    E E kT19

    19 0.25 0.03453

    16 3

    p N e

    p 1.601 10 e

    p 1.601 10 e

    p 1.148 10 cm

    18. The large value of capacitor is used in a full wave rectifier to increase peak current rating of

    the diode and for low conduction period for diode rectifier.

    19. Along ab, d 0 b

    a

    A.dl 0, A.dl 0

  • 7/25/2019 ECE GATE PAPER 25 ANSWERS

    7/19

    ECTest ID: 161275 TarGATE16 www.gateforum.com

    ICPIntensive Classroom ProgrameGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India

    All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

    7

    Along bc, d 0

    2A.dl d

    2c

    2 2

    b 0

    A.dl 2 4.5

    Along CD

    d 0, 2 2A.dl .d

    2d 2 3

    2

    c 3 3

    A.dl d 6.333

    Along da

    d 0 2A.dl d

    0 22

    2

    A.dl d 22

    A.dl 2.5 6.33 1.524 20. Output = Y = ABC + BCD + ACD + ABD

    =4 units of 3 input gates + 1 unit of 4 input gates

    4 10 1 15 55 Minimum cost of implementation is 55/-

    22. Taking Laplace transform of the given equation

    1

    12sL(y) y(0) 2L(y)

    s

    given y(0) 10

    10 12L(y)

    s 2 s(s 2)

    Ta king inverse Laplace transform onbothsides

    10 12y Ls 2 s(s 2)

    1 1

    t

    2t 2t

    0

    2t

    1

    1 s 210L 12Ls 2 s

    10e 12 e dt,Since by inverseLaplaceof division by's'

    4e 6

    a b

    c

    d

  • 7/25/2019 ECE GATE PAPER 25 ANSWERS

    8/19

    ECTest ID: 161275 TarGATE16 www.gateforum.com

    ICPIntensive Classroom ProgrameGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India

    All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

    8

    23. Use superposition principle.

    1stshort the voltage source.

    ThenX1

    3I 2 1.2A

    5

    Now, open circuiting the

    Current source

    We get X25

    I 1A3 2

    X X1 X2I = I I 0.2A

    24. Pe 32

    3 2 , 6

    pe

    k kG 1 at large stability G 1

    s

    3.14k 0.52

    6 6

    25. The ac equivalent of the circuit is

    Writing KCL

    m gsi g ______(1)

    gs 2iR ______(2)

    Substitute (2) in (1)

    m 2i g iR

    2

    m

    1R i

    g

    26.b

    p

    a

    We have

    1dx converges for P < 1

    (x-a)

    diverges for P 1

    Hence P = 2 The given integral diverges

    i

    R

    2 D

    m gsg gs

    S

    2A 5V

    XII X2I

    1

    2

    3

  • 7/25/2019 ECE GATE PAPER 25 ANSWERS

    9/19

    ECTest ID: 161275 TarGATE16 www.gateforum.com

    ICPIntensive Classroom ProgrameGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India

    All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

    9

    27.1 2

    z12

    2

    1 1 2

    2 1 2

    2 0 2

    V 1I 2I

    1V 2I I ..(1)

    2

    Fromoutput sideV z I ..(2)

    Sub(2)in (1)

    0 2 1 2

    0 2 1

    2 2 1

    1

    0 1 1

    0 0

    1z I 2I I

    2

    1z I 2I

    2

    IGiven 2 I 2I

    I

    1z 2I 2I

    2

    1 1z 1 z

    2 2

    28.

    m

    m

    T2T 4

    2 2 Pt

    m m m0 0

    4V1 4m t m t dt V dt

    T T T

    2

    PV

    3

    2 22 2 22C C P

    actual

    L L

    100 0.5A A VP m t 16.67 W

    R R 3 50 3

    29. Given, n

    n p

    J90% 0.90

    J J

    2 nn i

    a n0

    p2p i

    d po

    nn

    a a an pn p

    d d d

    D1J en

    N

    D1J enN

    DJ 25 5so, 0.90 0.90 0.90

    N N NJ JD D 25 16 5 4

    N N N

    a

    d

    a

    d

    N5 4.5 3.6

    N

    N0.139

    N

  • 7/25/2019 ECE GATE PAPER 25 ANSWERS

    10/19

    ECTest ID: 161275 TarGATE16 www.gateforum.com

    ICPIntensive Classroom ProgrameGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India

    All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

    10

    30. Given that

    2

    1.67

    z

    2

    1.67

    p(X 1.67 ) 0.04746

    p(Z 1.67) 0.04746

    p(Z 1.67) 0.04746

    i.e., f (Z)dz 0.05

    1e .dz=0.0474 0.05

    2

    31. The two networks on both sides of the inductor are not connected to each other completely as

    they have no common point between them. So, no current flows through the inductor.

    32.

    0 0 03 3 3

    2 2 22 2 1 3 3 2 3 1 2 3 1 2

    3 1 3 2

    Q t 1 S RQ Q Q Q Q

    Q t 1 JQ KQ Q Q Q Q Q Q Q Q Q Q Q Q

    Q Q Q Q

    1 2

    10 0 1 0

    Q t 1 D Q

    Q t 1 T Q Q Q Q Q

    After eight clock pulses counter goes to 0000 state.

    3 2 1 0 3 2 1 0Q Q Q Q Q Q Q Q

    0 0 0 0 1 0 0 1

    0 0 0 1 0 0 0 0

    0 0 1 0 1 0 0 0

    0 0 1 1 0 0 0 1

    0 1 0 0 1 0 1 1

    0 1 0 1 0 0 1 0

    0 1 1 0 1 0 1 0

    0 1 1 1 0 0 1 1

    1 0 0 0 1 0 0 1

    1 0 0 1 0 0 0 0

    1 0 1 0 1 1 0 0

    1 0 1 1 0 1 0 1

    1 1 0 0 1 1 1 1

    1 1 0 1 0 1 1 0

    1 1 1 0 1 1 1 0

    1 1 1 1 0 1 1 1

    Present state Next state

    1010 1100

    1111

    0111

    0011

    0001

    0000

    1001

    z 1.67 z 0

    0.04746

    z 1.67

  • 7/25/2019 ECE GATE PAPER 25 ANSWERS

    11/19

    ECTest ID: 161275 TarGATE16 www.gateforum.com

    ICPIntensive Classroom ProgrameGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India

    All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

    11

    33. For input voltage, SV 0

    Diode D1is of. i.e., Reverse biased

    So, current through D2 is

    10 0.7i20 20 k

    9.3i mA

    40

    i 0.232 mA

    and output voltage Vo will be

    o

    3 3

    o

    O

    20Ki

    20 10 0.232 10

    4.65V

    O S Sfor 0 V 4.65V

    For negative values of input voltage, output is juet the negative of positive part.

    34.2

    C(S) k =

    R(S) s + (2+ ak)s+k

    Compare with characteristic equation

    n n = k and 2 = 2 + ak

    k 16

    2 0.7 4 2 16a a = 0.225

    35. 2

    2

    2

    2

    2

    dy y yGiven cos

    dx x x

    ycosxdy - ydx dxx

    x

    y xdy-ydx -dxsec

    x x x

    y y -dxsec d

    x x x

    10V

    20k

    2D1D

    SV

    3D 4D20k

    O

    20k

    10V

  • 7/25/2019 ECE GATE PAPER 25 ANSWERS

    12/19

    ECTest ID: 161275 TarGATE16 www.gateforum.com

    ICPIntensive Classroom ProgrameGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India

    All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

    12

    2 y y dxIntegrating sec d cx x x

    y tan ln x c

    x

    put x = 1 , y = tan ln(1) + c

    4 4

    1 = c

    y tan ln x 1

    x

    y1-tan

    x

    y 1 tan ln x

    x

    x= e

    36. L

    Zi min

    L

    R R 1.2K 330 20V V

    R 1200

    i minV 25.5V

    L ZL

    L L

    V V 20VI 16.67mA

    R R 1.2K

    max zm LI I I 60mA 16.67 mA 76.67 mA

    max R max ZV I R V

    376.67 10 330 20 25.3 20

    i maxV 45.3V

    i25.5 V 45.3V

    37. Consider

    j t

    j t

    1x t X e d

    2

    2 x t X e d

    Replace with ' '

    LV

    20V

    25.5V 45.3V I

    V

    i max25.5 V 45.3V

  • 7/25/2019 ECE GATE PAPER 25 ANSWERS

    13/19

    ECTest ID: 161275 TarGATE16 www.gateforum.com

    ICPIntensive Classroom ProgrameGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India

    All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

    13

    j t2 x t X e d

    y t 2 x t ; x t is oddsignal

    y t 2 x t

    proportionalityconstant 2

    38. The oscillator circuit given in the figure is BJT

    Colpits oscillator circuit

    So, the frequency will be

    r

    T

    1f

    2 LC

    1 2T

    1 2

    C Cwhere C

    C C

    129

    T 6

    0.1 0.01 10C 9.09 10 F

    0.11 10

    r3 9

    1F

    2 50 10 9.09 10

    rF 7.465kHz

    39. For given decision rule, probability of correct reception

    c

    P P y p, x 0 P y q, x 0 P y r, x 1

    P y p / x 0 .P x 0} P y q / x 0 . P x 0} P y r / x 1 P x 1}

    0.5 0.5 0.4 0.5 0.5 0.5 0.25 0.2 0.25 0.7

    Thus probability of error = c1 P 1 0.7 0.3

    40. The output voltage V0will be

    1O Z

    2

    RV V 1R

    Where, 1R Resistance connected to emitter.

    i.e.,12k

  • 7/25/2019 ECE GATE PAPER 25 ANSWERS

    14/19

    ECTest ID: 161275 TarGATE16 www.gateforum.com

    ICPIntensive Classroom ProgrameGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India

    All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

    14

    0

    0

    D CE C

    CE i O

    12V 6 1 8V

    36

    V 8V

    P V I

    V V V 15 8 7V

    C E

    8 8I I 0.966mA

    10 48

    DSo, P 7 0.966 6.762W

    Now unregulated I/P voltage is increased by 30%

    i

    30So, V 15 15 19.5V

    100

    0V will remain same

    CEV 19.5 8 11.5V

    DP 11.5 0.966 11.11 mW

    11.1 6.76% increase 64.29%

    6.76

    41. Odd Parity generator is an even function

    1

    1

    1

    1

    1

    1

    1

    1

    f 1 0 x 1 0 1 y

    f 1 x 1 0 1 y

    f x 1 0 1 y

    f x 0 1 y

    f x 1 y

    f x y

    f x y

    f x y

    1 1f x, y,z zf z f z

    f x, y,z x y z

    1 1

    z f

    0 f zf

    1 1 z

  • 7/25/2019 ECE GATE PAPER 25 ANSWERS

    15/19

    ECTest ID: 161275 TarGATE16 www.gateforum.com

    ICPIntensive Classroom ProgrameGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India

    All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

    15

    42. Periodic signal ranges from to . Thus the system must be non-causal.

    Stability of LTI system

    t

    h d

    If 0h t sin t which is a periodic signal then t ,

    0sin t dt

    Thus the system is unstable.

    43.

    A B C D output q

    0 0 0 0 0 1

    0 0 0 1 1 0

    0 0 1 0 2 0

    0 0 1 1 3 0

    0 1 0 0 4 0

    0 1 0 1 5 0

    0 1 1 0 6 0

    0 1 1 1 7 0

    1 0 0 0 8 1

    1 0 0 1 9 0

    44. x t Ax t

    Atx t e x 0

    h t

    0

    0

    0

    30

    2 t

    O

    1 0 0 0

    0 0 0 0

    X X X X

    1 0 X X

    00 01 11 10CD

    AB00

    01

    11

    10

    q BCD

  • 7/25/2019 ECE GATE PAPER 25 ANSWERS

    16/19

    ECTest ID: 161275 TarGATE16 www.gateforum.com

    ICPIntensive Classroom ProgrameGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India

    All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

    16

    t x 0

    1 1

    1cos2t sin 2t cos2t sin 2tx t 2 2

    12sin 2t cos2t 2sin 2t cos2t

    y t cx t

    1

    cos2t sin 2t0 1 2

    2sin 2t cos 2t

    y t 2sin 2t cos 2t

    45. Since, potential function V is known, so work done Q

    W may be determined as

    ABEd V l

    Q

    PQ

    P

    Q P

    6 0 0 0 0 6

    W Q E.d QV

    Q V V

    10 10 10 510 10 Sin90 Cos60 Sin30 Cos120 10 10 28.125 J

    16 1 32 2

    l

    46. The voltage division is the same as it would be for full right circular cylinders. The segment

    shown with angle

    , will have a capacitance/ 2

    times that of the complete coaxialcapacitor.

    10 r

    1

    10

    10

    2

    1 1 2 2 1 2

    21

    1 2

    2 LC

    2 ln(2.25/ 2.0)

    L(1.5 10 )(F)

    C L(4.2 10 )(F)

    Q C V C V and V V V

    CV V

    C C

    4.2(100)

    1.5 4.2

    74V

  • 7/25/2019 ECE GATE PAPER 25 ANSWERS

    17/19

    ECTest ID: 161275 TarGATE16 www.gateforum.com

    ICPIntensive Classroom ProgrameGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India

    All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

    17

    47.

    j j j j

    jj

    j

    j

    j

    j j j

    j j

    1 ae Y e b e X e

    b eH e

    1 ae

    1

    H e 1; b a b 2

    1e

    12Y e X e ; X e1 1

    1 e 1 e2 2

    j

    j

    j j

    1e

    2Y e1 1

    1 e 1 e2 2

    j0n

    3

    2y n Y e 2

    3 1

    2 2

    48. For

    t

    22

    1x t e u t X

    1 j

    2 1 j 1Y

    1 j2 2 2 j j

    1

    2

    f 2t

    2

    2Y

    2 j

    1 d 1 jFor e u t

    X j d 2 j 2 j

    By using frequency differentiation property,

    1

    f2t

    2

    f 2t

    2

    jjte u t

    2 j

    2For 2te u t

    2 j

    k 2

    49. Noise pdf is as shown below

    3n

    3

    1s 0s

    4

    1

    4

    Nf n

    n4

  • 7/25/2019 ECE GATE PAPER 25 ANSWERS

    18/19

    ECTest ID: 161275 TarGATE16 www.gateforum.com

    ICPIntensive Classroom ProgrameGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India

    All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

    18

    Since channel noise is additive, at receiver we receive signal with pdf as shown below:

    Using MAP criteria

    1r 1P y x 3 P x 3 16y x n 3

    1P y x 3 P x 31 r

    16

    For finding optimal threshold value

    1r 1 3 3r r 0.5

    2

    50.PQ

    QC

    V

    PQ

    2 0 0

    1 2 3

    V E.d

    2 x dx y dy z dz 10 Volts

    l

    51.

    C V

    C v

    E E kT2

    i C V

    2E E kT2 10 1.25 0.025

    C V i

    n N N e

    N N n e 1.6 10 e

    42 6

    C VN N 1.327 10 cm ___(1)

    3 2*

    3 2C nC V*

    V p

    N m4 , N 8N

    N m

    From equation (2)

    2 42

    V

    20 3 21 3

    V C

    8N 1.327 10

    N 4.07 10 cm N 3.26 10 cm

    52. p, diffusion Pdp

    J qD .dx

    1when s is

    transmitted

    0when s is

    transmitted

    7 13 1 73 r

    1

    4

    1

    4

  • 7/25/2019 ECE GATE PAPER 25 ANSWERS

    19/19

    ECTest ID: 161275 TarGATE16 www.gateforum.com

    ICPIntensive Classroom ProgrameGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India

    All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

    19

    15 41 8

    P' x 10 exp cmL 12

    15 4

    4

    1 810 exp cm

    12 10 12

    17 44.278 10 cm

    19 17P,diffusionJ 1.6 10 12 4.278 10 2821.467 mA / cm

    53. We need to find the thevenin equivalent resistance (impedance) across RLby removing it.

    54. O 1 1 1G 90 tan tan 2 tan 3

    2 2 2

    KG

    1 4 1 4 1

    0

    0

    0, G , G 90

    , G 0, G 360

    55. We know that,

    Newton-raphson interation formula to find the reciprocal of

    kk nn 1 n

    xN is x (k 1) Nx

    K

    Putting k=3; N=M; then3n

    n 1 n

    xx 4 Mx

    3