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  • MA6351-TRANSFORMS AND PARTIAL

    DIFFERENTIAL EQUATIONS

    SUBJECT NOTES

    Department of Mathematics

    FATIMA MICHAEL

    COLLEGE OF ENGINEERING &

    TECHNOLOGY

    MADURAI 625 020, Tamilnadu, India

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  • Department of Mathematics FMCET MADURAI

    Pag

    e2

    Basic Formulae

    DIFFERENTIATION &INTEGRATION FORMULAE

    0 Function

    ( )y f x

    Differentiation

    dy

    dx

    1 nx 1nnx

    2 log x 1

    x

    3 sin x cos x

    4 cos x sin x

    5 axe a xe

    6 C (constant) 0

    7 tan x 2sec x

    8 sec x sec tanx x

    9 cot x 2cos ec x

    10 cos ecx cos cotecx x

    11 x 1

    2 x

    12 1sin x 2

    1

    1 x

    13 1cos x 2

    1

    1 x

    14 1tan x 2

    1

    1 x

    15 1sec x 2

    1

    1x x

    16 1cosec x 2

    1

    1x x

    17 1cot x 2

    1

    1 x

    18 xa logxa a

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  • Department of Mathematics FMCET MADURAI

    Pag

    e3

    19. If y uv , then dy du dv

    v udx dx dx

    20. If u

    yv

    , then2

    du dvv u

    dy dx dx

    dx v

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  • Department of Mathematics FMCET MADURAI

    Pag

    e4

    1

    2

    1

    2 2

    2 2

    1.1

    2. , &

    cos3. sin cos & sin

    sin4. cos sin & cos

    5. tan log sec log cos

    6. sec tan

    17. tan

    18. log

    2

    nn

    ax axx x ax ax

    xx dx

    n

    e ee dx e e dx e dx

    a a

    axxdx x axdx

    a

    axxdx x axdx

    a

    xdx x x

    xdx x

    dx xdx

    x a a a

    dx x adx

    x a a x

    1

    2 2

    1

    2 2

    1

    2 2

    22 2 2 2 1

    22 2 2 2 1

    22 2 2 2 1

    2 2

    2 2

    9. sin

    10. sinh

    11. cosh

    12. sin2 2

    13. sinh2 2

    14. cosh2 2

    15. log

    216. log

    a

    dx xdx

    aa x

    dx xdx

    aa x

    dx xdx

    ax a

    x a xa x dx a x

    a

    x a xa x dx a x

    a

    x a xx a dx x a

    a

    dxx

    x

    xdxx a

    x a

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  • Department of Mathematics FMCET MADURAI

    Pag

    e5

    3

    2

    3

    2

    17. log log

    18.3

    19.3

    120. 2

    xdx x x x

    a xa x dx

    a xa x dx

    dx xx

    21. 2 2

    cos cos sinax

    ax ee bxdx a bx b bxa b

    22.2 2

    sin sin cosax

    ax ee bxdx a bx b bxa b

    23.1 2 3........udv uv u v u v u v

    24.0

    ( ) 2 ( )

    a a

    a

    f x dx f x dx when f(x) is even

    25. ( ) 0

    a

    a

    f x dx when f(x) is odd

    26. 2 2

    0

    cosaxa

    e bxdxa b

    27. 2 2

    0

    sinaxb

    e bxdxa b

    TRIGNOMETRY FORMULA

    1. sin2 2sin cosA A A

    2 2

    2

    2

    2.cos 2 cos sin

    1 2sin

    2cos 1

    A A A

    A

    A

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  • Department of Mathematics FMCET MADURAI

    Pag

    e6

    3. 21 cos 2

    cos2

    xx & 2

    1 cos 2sin

    2

    xx

    4. sin( ) sin cos cos sin

    sin( ) sin cos cos sin

    cos( ) cos cos sin sin

    cos( ) cos cos sin sin

    A B A B A B

    A B A B A B

    A B A B A B

    A B A B A B

    15.sin cos sin( ) sin( )

    2

    1cos sin sin( ) sin( )

    2

    1cos cos cos( ) cos( )

    2

    1sin sin cos( ) cos( )

    2

    A B A B A B

    A B A B A B

    A B A B A B

    A B A B A B

    3

    3

    16. sin 3sin sin 3

    4

    1cos 3cos cos3

    4

    A A A

    A A A

    2 2

    2 2

    7.sin 2sin cos2 2

    cos cos sin2 2

    1 2sin 1 cos 2sin2 2

    A AA

    A AA

    A AA

    LOGRATHEMIC FORMULA

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  • Department of Mathematics FMCET MADURAI

    Pag

    e7

    log

    log log log

    log log log

    log log

    log 1 0

    log 0

    log 1

    n

    a

    a

    a

    x

    mn m n

    mm n

    n

    m n m

    a

    e x

    UNIT - 1

    PARTIAL DIFFRENTIAL EQUATIONS

    PARTIAL DIFFERENTIAL EQUTIONS

    Notations

    zp

    x

    zq

    y

    2

    2

    zr

    x

    2zs

    x y

    2

    2

    zt

    y

    Formation PDE by Eliminating arbitrary functions

    Suppose we are given f(u,v) = 0

    Then it can be written as u = g(v) or v = g(u)

    LAGRANGES LINEAR EQUATION

    (Method of Multipliers)

    General form

    Pp + Qq = R

    Subsidiary Equation

    dx dy dz

    P Q R

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  • Department of Mathematics FMCET MADURAI

    Pag

    e8

    dx dy dz x my nz

    P Q R P mQ nR

    Where ( , m ,n) are the Lagrangian Multipliers

    Choose , m, n such that P + mQ + nR = 0

    Then dx + m dy + n dz = 0

    On Integration we get a solution u = a

    Similarly, We can find another solution v = a for another multiplier

    The solution is (u, v) = 0

    TYPE 2 (Clairuts form)

    General form

    Z = px + qy + f(p,q) (1)

    Complete integral

    Put p = a & q = b in (1), We get (2) Which is the Complete integral

    Singular Integral

    Diff (2) Partially w.r.t a We get (3)

    Diff (2) Partially w.r.t b We get (4)

    Using (3) & (4) Find a & b and sub in (2) we get Singular Integral

    REDUCIBLE FORM

    F(xm

    p ,ynq) = 0 (1) (or)

    F( xm

    p, ynq, z)=0 (1)

    F( zkp, z

    kq)=0

    If 1& 1m n then X = x

    1-m & Y = y

    1-n

    xm

    p = P(1-m) & yn q = Q(1-n)

    Using the above in (1)we get

    F(P,Q) = 0 (or) F(P,Q,z) = 0

    If 1k then Z = zk+1

    1

    k Qz qk

    Using the above in (1) We get

    F(P,Q) = 0

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  • Department of Mathematics FMCET MADURAI

    Pag

    e9

    If m=1 & n=1 then

    X= logx & Y= logy

    xp = P & yq = Q

    Using the above in (1) we get

    F(P,Q) (or) F(P, Q, z) = 0

    If k =-1 then Z = log z

    p

    Pz

    & q

    Qz

    Using the above in (1)

    we get

    F(P,Q) = 0

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  • Department of Mathematics FMCET MADURAI

    Pag

    e10

    STANDARD TYPES

    TYPE 1

    TYPE 3(a)

    TYPE 3(b)

    TYPE 3(c)

    TYPE 4

    General form

    F(p,q) = 0 (1)

    General form

    F(x,p,q) = 0 (1)

    General form

    F(y,p,q) = 0

    (1)

    General form

    F(z,p,q) = 0

    (1)

    General form

    F(x,y,p,q) = 0

    (1)

    Complete

    Integral Put p = a and q =

    b in (1)

    Find b in terms

    of a

    Then sub b in

    z = ax + by + c

    we get (2)

    which is the

    Complete

    Integral

    Complete

    Integral

    Put q = a in (1)

    Then, find p and

    sub in

    dz = p dx + q dy

    Integrating ,

    We get (2)

    which is the

    Complete

    integral

    Complete

    Integral

    Put p = a in (1)

    Then, find q and

    sub in dz = p dx

    + q dy

    Integrating ,

    We get (2)

    which is the

    Complete

    integral

    Complete

    Integral

    Put q = ap in (1)

    Then, find p and

    sub in

    dz = p dx + q dy

    Integrating,

    We get (2) which

    is the

    Complete

    integral

    Complete

    Integral

    (1) Can be written

    as,

    f(x,p) =f(y,q) = a

    Then, find p and q

    sub in

    dz = p dx + qD y

    Integrating,

    We get (2) which

    is the

    Complete integral

    Singular

    Integral

    Diff (2) partially

    w.r.t cWe get,0

    =1 (absurdThere

    is no Singular

    Integral

    Singular

    Integral

    Diff (2) partially

    w.r.t cWe get,0

    =1 (absurdThere

    is no Singular

    Integral

    Singular

    Integral

    Diff (2) partially

    w.r.t cWe get,0

    =1 (absurdThere

    is no Singular

    Integral

    Singular

    Integral

    Diff (2) partially

    w.r.t cWe get,0

    =1 (absurdThere

    is no Singular

    Integral

    Singular Integral

    Diff (2) partially

    w.r.t cWe get,0

    =1 (absurd

    There is no

    Singular Integral

    General

    Integral

    Put c = (a) in

    (2)We get

    (3)Diff (3)

    partially w.r.t

    aWe get

    (4)Eliminating a

    from (3) and (4)

    we get General

    Integral

    General

    Integral

    Put c = (a) in

    (2)We get

    (3)Diff (3)

    partially w.r.t

    aWe get

    (4)Eliminating a

    rom (3) and (4)

    we get General

    Integral

    General

    Integral

    Put c = (a) in

    (2)We get (3)

    Diff (3) partially

    w.r.t aWe get (4)

    Eliminating a

    from (3) and (4)

    we get General

    Integral

    General

    Integral

    Put c = (a) in

    (2)We get

    (3)Diff (3)

    partially w.r.t

    aWe get

    (4)Eliminating a

    from (3) and (4)

    we get General

    Integral

    General Integral

    Put c = (a) in

    (2)We get (3)

    Diff (3) partially

    w.r.t aWe get

    (4)Eliminating a

    from (3) and (4)

    we get General

    Integral

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  • Department of Mathematics FMCET MADURAI

    Pag

    e11

    HOMOGENEOUS LINEAR EQUATION

    General form

    33 2 2( ) ( , )aD bD D cDD dD z f x y (1)

    To Find Complementary Function

    Auxiliary Equation

    Put D = m & D = 1 in (1)

    Solving we get the roots m1

    , m2

    , m3

    Case (1)

    If the roots are distinct then

    C.F. = 1 1 2 2 3 3( ) ( ) ( )y m x y m x y m x

    Case (2)

    If the roots are same then

    C.F. = 2

    1 2 3( ) ( ) ( )y mx x y mx x y mx

    Case (3)

    If the two roots are same and one is distinct, then

    C.F = 1 2 3 3( ) ( ) ( )y mx x y mx y m x

    Function PI =

    1

    1( , )

    ( , )F x y

    F D D

    F(x,y) = eax+by

    Put D = a & D1

    = b

    F(x,y)= sin(ax+by)(or)

    Cos (ax+by)

    Put

    22 2 2( ), ( )& ( )D a DD ab D b

    F(x,y) = xr

    ys

    PI=

    1

    ( , ) r sF D D x y

    Expand and operating D & D

    on xr

    ys

    F(x,y) = eax+by

    f(x,y) Put D = D+a & D = D +b

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  • Department of Mathematics FMCET MADURAI

    Pag

    e12

    Particular Integral

    F(x,y)=ex+y

    cosh(x+y) F(x ,y)=2 21

    2

    x ye e

    F(x,y)=ex+y

    sinh(x+y) F(x, y) =2 21

    2

    x ye e

    F(x,y)=sin x cos y

    1( , ) sin( ) sin( )

    2F x y x y x y

    F(x,y)= cos x sin y

    1( , ) sin( ) sin( )

    2F x y x y x y

    F(x,y)= cos x cos y

    1( , ) s( ) s( )

    2F x y co x y co x y

    F(x,y)= sin x sin y

    1( , ) cos( ) cos( )

    2F x y x y x y

    Note:

    D represents differentiation with respect to x

    D represents differentiation with respect to y

    1

    D

    represents integration with respect to x

    1

    D represents integration with respect to y

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  • Department of Mathematics FMCET MADURAI

    Pag

    e13

    PARTIAL DIFFRENTIAL EQUATIONS

    1. Eliminate the arbitrary constants a & b from

    z = (x2

    + a)(y2

    + b)

    Answer:

    z = (x2

    + a)(y2

    + b)

    Diff partially w.r.to x & y here &z z

    p qx y

    p = 2x(y2

    + b) ; q = (x2

    + a) 2y

    (y2

    + b) = p/2x ; (x2

    + a) = q/2y

    z = (p/2x)(q/2y)

    4xyz = pq

    2. Form the PDE by eliminating the arbitrary function from z = f(xy)

    Answer:

    z = f(xy)

    Diff partially w.r.to x & y here &z z

    p qx y

    p = ( ).f xy y q = ( ).f xy x

    p/q = y/x px qy = 0

    3. Form the PDE by eliminating the constants a and b from z = axn

    + byn

    Answer:

    z = axn

    + byn

    Diff. w .r. t. x and y here &z z

    p qx y

    p = naxn-1

    ; q = nbyn-1

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  • Department of Mathematics FMCET MADURAI

    Pag

    e14

    1 1

    1 1

    ;n n

    n n

    n n

    p qa b

    nx ny

    p qz x y

    nx ny

    nz px qy

    4. Eliminate the arbitrary function f from xy

    z fz

    and form the PDE

    Answer:

    xy

    z fz

    Diff. w .r. t. and y here &z z

    p qx y

    2

    2

    .

    .

    .

    xy z xpp f y

    z z

    xy z yqq f x

    z z

    p y z xp

    q x z yq

    0

    pxz pqxy qyz pqxy

    px qy

    5. Find the complete integral of p + q =pq

    Answer:

    Put p = a, q = b

    p + q =pq a+b=ab

    b ab = -a 1 1

    a ab

    a a

    The complete integral is z= ax+

    1

    a

    ay +c

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  • Department of Mathematics FMCET MADURAI

    Pag

    e15

    6. Find the solution of 1p q

    Answer:

    z = ax+by+c ----(1) is the required solution

    given 1p q -----(2)

    put p=a, q = b in (2)

    2

    2

    1 1 (1 )

    (1 )

    a b b a b a

    z a x a y c

    7. Find the General solution of p tanx + q tany = tanz.

    Answer:

    1 2

    1 2

    tan tan tan

    cot cot cot

    cot cot cot cot

    logsin logsin log logsin logsin log

    sin sin

    sin sin

    sin sin, 0

    sin sin

    dx dy dz

    x y z

    x dx y dy z dz

    take x dx y dy y dy zdz

    x y c y z c

    x yc c

    y z

    x y

    y z

    8. Eliminate the arbitrary function f from 2 2z f x y and form the PDE.

    Answer:

    2 2z f x y

    2 2 2 22 ; ( 2 )

    20

    2

    p f x y x q f x y y

    p xpy qx

    q y

    9. Find the equation of the plane whose centre lie on the z-axis

    Answer:

    General form of the sphere equation is

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  • Department of Mathematics FMCET MADURAI

    Pag

    e16

    22 2 2x y z c r (1)

    Where r is a constant. From (1)

    2x+2(z-c) p=0 (2)

    2y +2(z-c) q = 0 (3)

    From (2) and (3)

    x y

    p q

    That is py -qx =0 which is a required PDE.

    10. Eliminate the arbitrary constants 2 2z ax by a b and form the PDE.

    Answer:

    2 2z ax by a b

    2 2

    ;p a q b

    z px qy p q

    11. Find the singular integral of z px qy pq

    Answer:

    The complete solution is z ax by ab

    0 ; 0

    ;

    ( ) ( ) ( . )

    0

    z zx b y a

    a b

    b x a y

    z y x x y y x

    xy xy xy

    xy

    xy z

    12. Find the general solution of px+qy=z

    Answer:

    The auxiliary equation is

    dx dy dz

    x y z

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  • Department of Mathematics FMCET MADURAI

    Pag

    e17

    From

    dx dy

    x y Integrating we get log x = log y + log c

    on simplifying 1

    xc

    y.

    2

    dy dz yc

    y z z

    Therefore , 0

    x y

    y z is general solution.

    13. Find the general solution of px2

    +qy2

    =z2

    Answer:

    The auxiliary equation is 2 2 2

    dx dy dz

    x y z

    From 2 2

    dx dy

    x y Integrating we get 1

    1 1c

    y x

    Also 2 2

    dy dz

    y z Integrating we get 2

    1 1c

    z y

    Therefore 1 1 1 1

    , 0y x z y

    is general solution.

    14. Solve 2 22 3 0D DD D z

    Answer:

    Auxiliary equation is

    2 2 3 0m m 3 1 0m m

    1, 3m m

    The solution is 1 2 3z f y x f y x

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  • Department of Mathematics FMCET MADURAI

    Pag

    e18

    15. Solve 2 24 3 x yD DD D z e

    Answer:

    Auxiliary equation is 2 4 3 0m m

    3 1 0m m

    1, 3m m

    The CF is 1 2 3CF f y x f y x

    2 2

    1

    4 3

    x yPI eD DD D

    Put 1, 1D D Denominator =0.

    2 4

    x yxPI eD D

    2

    x yxe

    Z=CF + PI

    1 2 3z f y x f y x

    2

    x yxe

    16. Solve. 2 23 4 x yD DD D z e

    Answer:

    Auxiliary equation is 2 3 4 0m m

    4 1 0m m

    C.F is = f1

    (y + 4x) + f2

    (y - x)

    2 2

    1 1 1

    3 4 1 3 4 6

    x y x y x yPI e e eD DD D

    17. Find P.I 2 2 24 4 x yD DD D z e

    Answer:

    2

    2 2

    1

    4 4

    x yPI eD DD D

    Put 2, 1D D

    2

    2

    1

    2

    x yPI eD D

    2

    2

    1

    2 2

    x ye

    2

    16

    x ye

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  • Department of Mathematics FMCET MADURAI

    Pag

    e19

    18. Find P.I 2 2 26D DD D z x y

    Answer:

    2

    22

    2

    1

    61

    PI x yD D

    DD D

    2

    2

    11

    Dx y

    D D

    32

    2

    1

    3

    xx y

    D

    4 5

    12 60

    x y x

    19. Find P.I

    2 2

    2sin

    z zx y

    x x y

    Answer:

    2

    1PI Sin x y

    D DD Put

    2 1, (1)( 1) 1D DD

    1

    1 1Sin x y

    2

    Sin x y

    20. Solve 3 33 2 0D DD D Z

    Answer:

    Auxiliary equation is 3 3 2 0m m

    21 2 0m m m

    1 2 1 0m m m

    1,1 2m m

    The Solution is 1 2 3 2CF f y x x f y x f y x

    FOR PRACTICE:

    1. Eliminating arbitrary constants

    2 2 2

    2 2 21

    x y z

    a b c

    2. Solve

    2

    2sin

    zy

    x

    3. Find the complete the solution of p. d .e 2 2 4 0p q pq

    4. Form p.d.e eliminating arbitrary function from 2 ,

    2

    xz xy

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  • Department of Mathematics FMCET MADURAI

    Pag

    e20

    5. Find the singular soln of 2 2 1z px qy p q

    1. (i) Solve 2 2 2x y z p y z x q z x y

    (ii) Solve 2 2 2 2 2 2x z y p y x z q z y x

    2. (i) Solve z z

    mz ny nx lz ly mxx y

    (ii) Solve 3 4 4 2 2 3z y p x z q y x

    3. (i) Solve 2 2 2 2 2x y z p xyq xz

    (ii) Solve 2 2 2 2 2 0y z x p xyq zx

    4. (i) Solve y z p z x q x y

    (ii) Solve y z p z x q x y

    5. Solve 2 2 3 23 2 sin(3 2 )x yD DD D e x y

    6. Solve 2 2

    2cos cos 2

    z zx y

    x x y

    7. Solve 2 26 cosD DD D z y x

    8. Solve 2 2 630 x yD DD D z xy e

    9. Solve 2 26 5 sinhxD DD D z e y xy

    10. Solve 2 2 24 4 x yD DD D z e

    11. Solve 3 2 2 3 2 cos( )x yD D D DD D z e x y

    12. Solve (i) 2 21z px qy p q

    (ii) 2 2z px qy p q

    13. Solve 2 2 21z p q

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  • Department of Mathematics FMCET MADURAI

    Pag

    e21

    14. Solve 2 2 2 2( ) 1z p x q

    15. Solve (i) 2 2 2 2( )z p q x y

    (ii) 2 2 2 2 2( )z p q x y

    UNIT - 2

    FOURIER SERIES

    0

    1

    ( ) cos sin2

    n n

    n

    af x a nx b nx

    (0,2 ) ( - , )

    Even (or) Half range

    Fourier co sine series

    Odd (or) Half range

    Fourier sine series

    Neither even nor odd

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  • Department of Mathematics FMCET MADURAI

    Pag

    e22

    2

    0

    0

    1( )a f x dx 0

    0

    2( )a f x dx

    0 0a 0

    1( )a f x dx

    2

    0

    1( )cosna f x nxdx

    0

    2( )cosna f x nxdx

    0na 1( )cosna f x nxdx

    2

    0

    1( )snb f x innxdx

    bn=0

    0

    2( )snb f x innxdx

    1( )snb f x innxdx

    0

    1

    ( ) cos sin2

    n n

    n

    a n x n xf x a b

    (0,2 ) ( - , )

    Even (or) Half range

    Fourier cosine series

    Odd (or) Half range

    Fourier sine series

    Neither even nor odd

    2

    0

    0

    1( )a f x dx

    0

    0

    2( )a f x dx

    0 0a 0

    1( )a f x dx

    2

    0

    1( )cosn

    n xa f x dx

    0

    2( )cosn

    n xa f x dx

    0na 1( )cosn

    n xa f x dx

    2

    0

    1( )sn

    n xb f x in dx

    bn=0

    0

    2( )sn

    n xb f x in dx

    1( )sn

    n xb f x in dx

    Even and odd function:

    Even function:

    f(-x)=f(x)

    eg : cosx,x2

    , , , sin , cosx x x are even functions

    Odd function:

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  • Department of Mathematics FMCET MADURAI

    Pag

    e23

    f(-x)=-f(x)

    eg : sinx,x3 ,sinhx, tanx are odd functions

    For deduction

    In the interval (0,2 ) if x = 0 or x = 2 then

    f(0) = f(2 ) = (0) (2 )

    2

    f f

    In the interval (0,2 ) if x = 0 or x = 2 then

    f(0) = f(2 ) = (0) (2 )

    2

    f f

    In the interval (- , ) if x = - or x = then

    f(- ) = f( ) = ( ) ( )

    2

    f f

    In the interval (- , ) if x = - or x = then

    f(- ) = f( ) = ( ) ( )

    2

    f f

    HARMONIC ANALYSIS

    f(x)= 0

    2

    a + a1 cosx +b1sinx + a2cos2x + b2sin2x for form

    0 2y

    an

    1cos

    2y x

    an

    , 2cos 2

    2y x

    an

    1sin

    2y x

    bn

    , 2sin 2

    2y x

    bn

    f(x)= 0

    2

    a + a1 cos

    x

    +b1

    xsin

    + a2

    2cos

    x

    + b2

    2 xsin

    ( form)

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  • Department of Mathematics FMCET MADURAI

    Pag

    e24

    0 2y

    an

    1

    cos

    2

    xy

    an

    2

    2cos

    2

    xy

    an

    1

    sin

    2

    xy

    bn

    ,2

    2sin

    2

    xy

    bn

    1. Define R.M.S value.

    If let f(x) be a function defined in the interval (a, b), then the R.M.S value of

    f(x) is defined by

    21( )

    b

    a

    y f x dxb a

    2. State Parsevals Theorem.

    Let f(x) be periodic function with period 2l defined in the interval (c, c+2l).

    2 22 2 2

    1

    1 1( )

    2 4 2

    c l

    on n

    nc

    af x dx a b

    lWhere , &o n na a b are Fourier constants

    3. Define periodic function with example.

    If a function f(x) satisfies the condition that f(x + T) = f(x), then we say f(x) is a periodic

    function with the period T.

    Example:-

    i) Sinx, cosx are periodic function with period 2

    ii) tanx is are periodic function with period

    4. State Dirichlets condition.

    (i) f(x) is single valued periodic and well defined except possibly at a

    Finite number of points.

    (ii) f (x) has at most a finite number of finite discontinuous and no infinite

    Discontinuous.

    (iii) f (x) has at most a finite number of maxima and minima.

    5. State Eulers formula.

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  • Department of Mathematics FMCET MADURAI

    Pag

    e25

    Answer:

    ( , 2 )

    cos sin2

    on n

    In c c l

    af x a nx b nx

    2

    2

    2

    1( )

    1( )cos

    1( )sin

    c l

    o

    c

    c l

    n

    c

    c l

    n

    c

    where a f x dx

    a f x nxdx

    b f x nxdx

    6. Write Fourier constant formula for f(x) in the interval (0,2 )

    Answer:

    2

    0

    2

    0

    2

    0

    1( )

    1( )cos

    1( )sin

    o

    n

    n

    a f x dx

    a f x nxdx

    b f x nxdx

    7. In the Fourier expansion of

    f(x) =

    21 , 0

    21 ,0

    xx

    xx

    in (- , ), find the value of nb

    Since f(-x)=f(x) then f(x) is an even function. Hence nb = 0

    8. If f(x) = x3

    in < x < , find the constant term of its Fourier series.

    Answer:

    Given f(x) = x3

    f(-x) = (- x)3

    = - x3

    = - f(x)

    Hence f(x) is an odd function

    The required constant term of the Fourier series = ao

    = 0

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  • Department of Mathematics FMCET MADURAI

    Pag

    e26

    9. What are the constant term a0

    and the coefficient of cosnx in the Fourier

    Expansion f(x) = x x3 in < x <

    Answer:

    Given f(x) = x x3 f(-x) = -x - (- x)3= - [x - x3] = - f(x)

    Hence f(x) is an odd function

    The required constant term of the Fourier series = 0a = 0

    10. Find the value of a0

    for f(x) = 1+x+x2

    in ( 0 ,2 )

    Answer:

    2

    0

    1( )oa f x dx

    22 2 32

    0 0

    2 3 2

    1 1(1 )

    2 3

    1 4 8 82 2 2

    2 3 3

    x xx x dx x

    11. (i)Find bn

    in the expansion of x2

    as a Fourier series in ( , )

    (ii)Find bn

    in the expansion of xsinx a Fourier series in ( , )

    Answer:

    (i) Given f(x) = x2

    f(-x) = x2

    = f(x)

    Hence f(x) is an even function

    In the Fourier series nb = 0

    (ii) Given f(x) = xsinx f(-x) = (-x)sin(-x)

    = xsinx = f(x)

    Hence f(x) is an even function

    In the Fourier series nb = 0

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  • Department of Mathematics FMCET MADURAI

    Pag

    e27

    12. Obtain the sine series for

    0 / 2

    / 2

    x x lf x

    l x l x l

    Given

    0 / 2

    / 2

    x x lf x

    l x l x l

    Answer:

    Given

    0 / 2

    / 2

    x x lf x

    l x l x l

    Fourier sine series is sinnnx

    f x bl

    0

    2

    0 2

    2

    2 2

    2 2

    0 2

    2( )sin

    2sin ( )sin

    cos sin cos sin2

    (1) ( ) ( 1)

    l

    n

    l l

    l

    l l

    l

    nxb f x dx

    l l

    nx nxx dx l x dx

    l l l

    nx nx nx nx

    l l l llx l l l x ll n n n n

    2 2 2 2

    2 2 2 2

    2

    2 2 2 2

    2 cos 2 sin 2 cos 2 sin 2

    2 2

    2 2 sin 2 4 sin 2

    l n l n l nl l n

    l n n n n

    l n l n

    l n n

    Fourier series is 2 21

    4 sin 2sin

    n

    l n n xf x

    n l

    13. If f(x) is odd function in ,l l . What are the value of a0

    &an

    Answer:

    If f(x) is an odd function, ao

    = 0, an

    = 0

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  • Department of Mathematics FMCET MADURAI

    Pag

    e28

    14. In the Expansion f(x) = |x| as a Fourier series in (- . ) find the value of a0

    Answer:

    Given f(x) = |x|

    f(-x) = |-x|

    = |x| = f(x)

    Hence f(x) is an even function

    2 2

    0 0

    2 2 2

    2 2o

    xa xdx

    15. Find half range cosine series of f(x) = x, in 0 x

    Answer:

    2 2

    0 0

    2 2 2

    2 2o

    xa xdx

    2

    0 0

    1

    2 1 cos sinsin (1)

    1 11 cos0 0

    n

    n n

    nx nxa x nxdx x

    n n

    n

    n n n

    Fourier series is

    0

    1

    0

    cos2

    1cos

    2

    on

    n

    n

    n

    af x a nx

    nxn

    16. Find the RMS value of f(x) = x2

    , 0 1x

    Answer:

    Given f(x) = x2

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  • Department of Mathematics FMCET MADURAI

    Pag

    e29

    R.M.S value

    2 122 2

    0 0

    15

    0

    1 1( )

    1 2

    22

    5 5

    l

    y f x dx x dxl

    x

    17. Find the half range sine series of ( )f x x in (0, )

    Answer:

    0

    2( )sinnb f x nxdx

    2

    0 0

    1

    2 2 cos sinsin (1)

    2 ( 1) 2( 1)n n

    nx nxx nxdx x

    n n

    n n

    Half range Fourier sine series is

    1

    0

    2( 1)sin

    n

    n

    f x nxn

    18. Find the value of a0

    in the cosine series of ( )f x x in (0, 5)

    Answer:

    55 2 2

    0 0

    2 2 2 55

    5 5 2 5 2o

    xa xdx

    19. Define odd and even function with example.

    Answer:

    (i) If ( ) ( )f x f x then the function is an even function.

    eg : cosx ,x2 , , sin , cosx x x are even functions

    (ii) If ( ) ( )f x f x then the function is an odd function.

    eg : sinx,x3 ,sinhx, tanx are odd functions

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  • Department of Mathematics FMCET MADURAI

    Pag

    e30

    20. Write the first two harmonic.

    Answer:

    1 1 2 2cos sin cos 2 sin 22

    o

    The first twoharmonics are

    af x a x b x a x b x

    FOURIER SERIES

    1. Expand (0, )

    ( )2 ( ,2 )

    xf x

    x as Fourier series

    and hence deduce that

    2

    2 2 2

    1 1 1.........

    1 3 5 8

    2. Find the Fourier series for f(x) = x2 in (- . ) and also prove that

    (i)

    2

    2 2 2

    1 1 1.........

    1 2 3 6 (ii)

    2

    2 2 2

    1 1 1.........

    1 2 3 12

    3. (i) Expand f(x) = | cosx | as Fourier series in (- . ).

    (ii) Find cosine series for f(x) = x in (0, ) use Parsevals identity to

    Show that

    4

    4 4 4

    1 1 1.........

    1 2 3 90

    4. (i) Expand f(x) = xsinx as a Fourier series in (0,2 )

    (ii) Expand f(x) = |x| as a Fourier series in (- . ) and deduce to 2

    2 2 2

    1 1 1.........

    1 3 5 8

    5. If 0 , ( ,0)

    ( )sin , (0, )

    f xx

    Find the Fourier series and hence deduce that

    1 1 1 2

    .........1.3 3.5 5.7 4

    6. (i) Find the Fourier series up to second harmonic

    X 0 1 2 3 4 5

    Y 9 18 24 28 26 20

    (ii)Find the Fourier series up to third harmonic

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  • Department of Mathematics FMCET MADURAI

    Pag

    e31

    X 0 /3 2/3 4/3 5/3 2

    F(x) 10 14 19 17 15 12 10

    7. (i) Find the Fourier expansion of 2( ) ( )f x x in (0,2 ) and

    Hence deduce that

    2

    2 2 2

    1 1 1.........

    1 2 3 6

    (ii). Find a Fourier series to represent

    2( ) 2f x x x with period 3

    in the range (0,3)

    (ii) Find the Fourier series of xf x e in ( , ) .

    (ii) Find the Fourier series for

    1 (0, )

    2 ( ,2 )

    inf x

    in

    and hence show that

    2

    2 2 2

    1 1 1.........

    1 3 5 8

    8. (i) Find the the half range sine series for f x x x in the interval (0, ) and deduce

    that 3 3 3

    1 1 1....

    1 3 5

    (ii) Obtain the half range cosine series for 2

    1f x x in (0,1)

    and also deduce that

    2

    2 2 2

    1 1 1.........

    1 2 3 6

    9. (i) Find the Fourier series for f(x) = x2 in (- . ) and also prove that

    4

    4 4

    1 11 .........

    1 2 90 (use P.I)

    (ii) Find the Fourier series for f(x) = x in (- . ) and also prove that

    4

    4 4

    1 11 .........

    1 3 96 (use P.I)

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  • Department of Mathematics FMCET MADURAI

    Pag

    e32

    10(i)Obtain the sine series for

    ,02

    ,2

    lcx x

    f xl

    c l x x l

    (ii). Find the Fourier series for the function

    ,02

    2 ,2

    lkx x

    f xl

    k l x x l

    11.(i).Find the Fourier series for the function 21 ( , )f x x x in and also

    deduce that

    2

    2 2 2

    1 1 1.........

    1 2 3 6

    (ii) Find the Fourier expansion of

    f(x) =

    21 , 0

    21 ,0

    xx

    xx

    in (- , ), and also deduce that 2

    2 2 2

    1 1 1.........

    1 3 5 8

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  • Department of Mathematics FMCET MADURAI

    Pag

    e33

    UNIT - 3

    APPLICATIONS OF P.D.E

    S.

    N

    O

    ONE DIMENSIONAL WAVE EQUATION

    VELOCITY MODEL INITIAL POSITION MODEL

    1 STEP-1

    One Dimensional wave equation

    is

    2 22

    2 21

    y ya

    t x

    STEP-1

    One Dimensional wave equation

    is

    2 22

    2 21

    y ya

    t x

    2 STEP-2

    Boundary conditions

    1. y(0,t) = 0 for 0t

    2. y( , t) = 0 for 0t 3. y(x,0) = 0 for 0 < x <

    4. 0t

    y

    t= f(x) for 0 < x <

    STEP-2

    Boundary conditions

    1. y(0,t) = 0 for 0t

    2. y( , t) = 0 for 0t

    3. 0t

    y

    t = 0 for 0 < x <

    4. y(x,0) = f(x) for 0 < x <

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  • Department of Mathematics FMCET MADURAI

    Pag

    e34

    3 STEP-3

    The possible solutions are

    y(x,t) = (A ex + B e

    - x) (C e

    at + D e

    - at)

    y(x,t) = (A cos x + B sin x )( C cos at + D sin

    at)

    y(x,t) = (Ax + B) ( Ct + D)

    STEP-3

    The possible solutions are

    y(x,t) = (A ex + B e

    - x) (C e

    at + D e

    - at)

    y(x,t) = (A cos x + B sin x )( C cos at + D

    sin at)

    y(x,t) = (Ax + B) ( Ct + D)

    4 STEP-4

    The suitable solution for the given

    boundary condition is

    y(x,t) = (Acos x+B sin x )(Ccos at+D sin at)

    (2)

    STEP-4

    The suitable solution for the given

    boundary condition is

    y(x,t) = (Acos x+B sin x )(Ccos at+D sin at)

    (2)

    5 STEP-5

    Using Boundary condition 1

    y(0,t) = 0

    Then (2) becomes,

    y(0,t) = (A cos 0 +B sin 0 ) ( C cos at + Dsin at) =0

    (A) ( C cos at + D sin at)=0

    A = 0

    Using A = 0 in (2)

    y(x,t) = ( B sin x) ( C cos at + D sin at) (3)

    STEP-5

    Using Boundary condition 1

    y(0,t) = 0

    Then (2) becomes,

    y(0,t) = (A cos 0 +B sin 0 ) ( C cos at + D sin at)

    =0

    (A) ( C cos at + D sin at)=0

    A = 0

    Using A = 0 in (2)

    y(x,t) = ( B sin x) ( C cos at + D sin at) (3)

    6

    STEP-6

    Using Boundary condition 2

    y( ,t) = 0 Then (3) becomes,

    y( ,t) = (B sin ) ( C cos at + D sin at)=0

    (B sin ) ( C cos at + D sin at)=0

    = n

    Then (3) becomes,

    ( , ) sin( ) cos( ) sin( )n x n at n at

    y x t B C D

    (4)

    STEP-6

    Using Boundary condition 2

    y( ,t) = 0 Then (3) becomes,

    y( ,t) = (B sin ) ( C cos at + D sin at)=0

    (B sin ) ( C cos at + D sin at)=0

    = n

    Then (3) becomes,

    ( , ) sin( ) cos( ) sin( )n x n at n at

    y x t B C D

    (4)

    7 STEP-7

    Using Boundary condition 3

    y(x,0) = 0

    Then (4) becomes,

    STEP-7

    Using Boundary condition 3

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  • Department of Mathematics FMCET MADURAI

    Pag

    e35

    ( , ) sin( ) cos0 sin 0n x

    y x t B C D

    =0

    sin( ) 0n x

    B C

    C = 0 Then (4) becomes,

    ( , ) sin( ) sin( )n x n at

    y x t B D

    The most general solution is

    1

    ( , ) sin( )sin( )nn

    n x n aty x t B

    (5)

    0t

    y

    t= 0Then (4) becomes,

    Differentiating (5) partially w.r.to t and put t =0

    0

    sin( ) sin( ) cos( )t

    y n x n at n at n aB C D

    t

    sin( ) 0n x n a

    B D

    D = 0 Then (4) becomes,

    ( , ) sin( ) cos( )n x n at

    y x t B C

    The most general solution is

    1

    ( , ) sin( ) cos( )nn

    n x n aty x t B

    (5)

    8 STEP-8

    Differentiating (5) partially w.r.to t

    1

    sin( ) cos( )nn

    y n x n at n aB

    t

    Using Boundary condition (4),

    0t

    y

    t= f(x)

    1

    ( ) sin( )nn

    n x n af x B

    This is the Half Range Fourier Sine Series.

    0

    2( )sin( )n

    n a n xB f x

    0

    2( )sin( )n

    n xB f x dx

    n a

    STEP-8

    Using Boundary condition (4),

    y(x,0) = f(x)

    1

    ( ,0) sin( )cos(0)nn

    n xy x B

    1

    ( ) sin( )nn

    n xf x B

    This is the Half Range Fourier Sine Series.

    0

    2( )sin( )n

    n xB f x

    9 STEP-9

    The required solution is

    1

    ( , ) sin( )sin( )nn

    n x n aty x t B

    Where

    0

    2( )sin( )n

    n xB f x dx

    n a

    STEP-9

    The required solution is

    1

    ( , ) sin( )sin( )nn

    n x n aty x t B

    Where

    0

    2( )sin( )n

    n xB f x dx

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  • Department of Mathematics FMCET MADURAI

    Pag

    e36

    ONE DIMENSIONAL HEAT EQUATION

    TWO DIMENSIONAL HEAT FLOW EQUATION

    1 The one dimensional heat equation is

    22

    2

    u u

    t x

    The two Dimensional equation is

    2 2

    2 20

    u u

    x y

    2 Boundary conditions

    1.u(0,t) = 0 for 0t

    2.u( ,t) = 0 for 0t 3.u(x,t) = f(x) for 0

  • Department of Mathematics FMCET MADURAI

    Pag

    e37

    7 Using Boundary condition 3

    u(x,0) = f(x)

    1

    ( ,0) sin( )nn

    n xu x B

    1

    ( ) sin( )nn

    n xf x B

    This the Half range Fourier sine series

    0

    2( )sin( )n

    n xB f x dx

    Using Boundary condition 3

    u(x, ) =0

    ( , ) ( sin )( )n x

    u x B Ce De

    0 ( sin )( 0)n x

    B C D

    C=0

    then (3) becomes

    ( , ) ( sin )( )n y

    n xu x y B De

    The most general solution is

    1

    ( , ) sin( )n y

    n

    n

    n xu x y B e

    (5

    8 The required solution is 2 2 2

    2

    1

    ( , ) sin( )

    n t

    n

    n

    n xu x t B e

    Where

    0

    2( )sin( )n

    n xB f x dx

    Using Boundary condition 4 y(x,0) = f(x)

    0

    1

    ( ,0) sin( )nn

    n xu x B e

    1

    ( ) sin( )nn

    n xf x B

    This the Half range Fourier sine series

    0

    2( )sin( )n

    n xB f x dx

    The required solution is

    1

    ( , ) sin( )n y

    n

    n

    n xu x y B e

    Where

    0

    2( )sin( )n

    n xB f x dx

    QUESTION WITH ANSWER

    1. Classify the Partial Differential Equation i)

    2 2

    2 2

    u u

    x y

    Answer:

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  • Department of Mathematics FMCET MADURAI

    Pag

    e38

    2 2

    2 2

    u u

    x y here A=1,B=0,&C=-1

    B2

    - 4AC=0-4(1)(-1)=4>0

    The Partial Differential Equation is hyperbolic

    2. Classify the Partial Differential Equation

    2u u uxy

    x y y x

    Answer:

    2u u uxy

    x y y x here A=0,B=1,&C=0

    B2

    -4AC=1-4(0)(0)=1>0

    The Partial Differential Equation is hyperbolic

    3. Classify the following second order Partial Differential equation

    2 22 2

    2 2

    u u u u

    x y y x

    Answer:

    2 22 2

    2 2

    u u u u

    x y y x here A=1,B=0,&C=1

    B2

    -4AC=0-4(1)(1)=-4

  • Department of Mathematics FMCET MADURAI

    Pag

    e39

    ii) 2 2 2 7 0xx yy x yy u u u u

    Answer:

    i) Parabolic ii) Hyperbolic (If y = 0)

    iii)Elliptic (If y may be +ve or ve)

    6. In the wave equation

    2 22

    2 2

    y yc

    t x what does c

    2

    stands for?

    Answer:

    2 22

    2 2

    y yc

    t x

    here 2 Ta

    m T-Tension and m- Mass

    7. In one dimensional heat equation ut

    = 2 uxx

    what does 2 stands for?

    Answer:-

    22

    2

    u u

    t x

    2

    =k

    c is called diffusivity of the substance

    Where k Thermal conductivity

    - Density c Specific heat

    8. State any two laws which are assumed to derive one dimensional heat equation

    Answer:

    i) Heat flows from higher to lower temp

    ii) The rate at which heat flows across any area is proportional to the area

    and to the temperature gradient normal to the curve. This constant of

    proportionality is known as the conductivity of the material. It is known as

    Fourier law of heat conduction

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  • Department of Mathematics FMCET MADURAI

    Pag

    e40

    9. A tightly stretched string of length 2 is fastened at both ends. The midpoint of the string is displaced to a distance b and released from rest in this position. Write the initial conditions.

    Answer:

    (i) y(0 , t) = 0

    (ii) y(2 ,t) = 0

    (iii)

    0

    0t

    y

    t

    (iv) y(x , 0 ) =

    0

    (2 ) 2

    bx x

    bx x

    10. What are the possible solutions of one dimensional Wave equation?

    The possible solutions are

    Answer:

    y(x,t) = (A ex

    + B e- x

    ) (C eat

    + D e- at

    )

    y(x,t) = (A cos x + B sin x )( C cos at + D sin at)

    y(x,t) = (Ax + B) ( Ct + D)

    11. What are the possible solutions of one dimensional head flow equation?

    Answer:

    The possible solutions are

    2 2

    2 2

    ( , ) ( )

    ( , ) ( cos sin )

    ( , ) ( )

    x x t

    t

    u x t Ae Be Ce

    u x t A x B x Ce

    u x t Ax B C

    12. State Fourier law of heat conduction

    Answer:

    uQ kA

    x

    (the rate at which heat flows across an area A at a distance from one end of a bar is

    proportional to temperature gradient)

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  • Department of Mathematics FMCET MADURAI

    Pag

    e41

    Q=Quantity of heat flowing

    k Thermal conductivity

    A=area of cross section ;u

    x=Temperature gradient

    13. What are the possible solutions of two dimensional head flow equation?

    Answer:

    The possible solutions are

    ( , ) ( )( cos sin )

    ( , ) ( cos sin )( )

    ( , ) ( )( )

    x x

    y y

    u x y Ae Be C y D y

    u x y A x B x Ce De

    u x y Ax B Cy D

    14. The steady state temperature distribution is considered in a square plate with sides x

    = 0 , y = 0 , x = a and y = a. The edge y = 0 is kept at a constant temperature T and the

    three edges are insulated. The same state is continued subsequently. Express the

    problem mathematically.

    Answer:

    U(0,y) = 0 , U(a,y) = 0 ,U(x,a) = 0, U(x,0) = T

    15. An insulated rod of length 60cm has its ends A and B maintained 20C and

    80C respectively. Find the steady state solution of the rod

    Answer:

    Here a=20C & b=80C

    In Steady state condition The Temperature ( , )b a x

    u x t al

    80 2020

    60

    x

    ( , ) 20u x t x

    16. Write the DAlemberts solution of the one dimensional wave equation?

    Answer:

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  • Department of Mathematics FMCET MADURAI

    Pag

    e42

    1 1( )

    2 2

    x at

    x at

    y x at x at v da

    here x f x g x

    v x ax f ag

    17. What are the boundary conditions of one dimensional Wave equation?

    Answer:

    Boundary conditions

    1. y(0,t) = 0 for 0t

    2. y( , t) = 0 for 0t 3. y(x,0) = 0 for 0 < x <

    4.

    0t

    y

    t= f(x) for 0 < x <

    18. What are the boundary conditions of one dimensional heat equation?

    Answer:

    Boundary conditions

    1.u(0,t) = 0 for 0t

    2.u( ,t) = 0 for 0t

    3.u(x,t) = f(x) for 0

  • Department of Mathematics FMCET MADURAI

    Pag

    e43

    20.T he ends A and B has 30cm long have their temperatures 30c and 80c until steady

    state prevails. If the temperature A is raised to40c and Reduced to 60C, find the

    transient state temperature

    Answer:

    Here a=30C & b=80C

    In Steady state condition The Temperature ( , )b a x

    u x t al

    Here a=40C & b=60C

    60 40 240 40

    30 3t

    xu x

    PART-B QUESTION BANK APPLICATIONS OF PDE

    1. A tightly stretched string with fixed end points x = 0 and x = l is initially at rest in its

    equilibrium position. If it is set vibrating giving each point a velocity 3x (l-x). Find the

    displacement.

    2. A string is stretched and fastened to two points and apart. Motion is started by displacing the string into the form y = K(lx-x

    2

    ) from which it is released at time t = 0. Find the

    displacement at any point of the string.

    3. A taut string of length 2l is fastened at both ends. The midpoint of string is taken to a height b and then released from rest in that position. Find the displacement of the string.

    4. A tightly stretched string with fixed end points x = 0 and x = l is initially at rest in its

    position given by y(x, 0) =3

    0 sinx

    yl

    . If it is released from rest find the displacement.

    5. A string is stretched between two fixed points at a distance 2l apart and points of the

    string are given initial velocities where

    0 < x < 1

    (2 ) 0 < x < 1

    cx

    lV

    cl x

    l

    Find the

    displacement.

    6. Derive all possible solution of one dimensional wave equation. Derive all possible solution of one dimensional heat equation. Derive all possible solution of two dimensional heat

    equations.

    7. A rod 30 cm long has its end A and B kept at 20oC and 80oC, respectively until steady state condition prevails. The temperature at each end is then reduced to 0

    o

    C and kept so. Find

    the resulting temperature u(x, t) taking x = 0.

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  • Department of Mathematics FMCET MADURAI

    Pag

    e44

    8. A bar 10 cm long , with insulated sides has its end A & B kept at 20oC and 40oC respectively until the steady state condition prevails. The temperature at A is suddenly raised to 50

    o

    C

    and B is lowered to 10o

    C. Find the subsequent temperature function u(x , t).

    9. A rectangular plate with insulated surface is 8 cm wide so long compared to its width that it may be considered as an infinite plate. If the temperature along short edge y = 0 is u (

    x ,0) = 100sin8

    x 0 < x < 1. While two edges x = 0 and x = 8 as well as the other short

    edges are kept at 0o

    C. Find the steady state temperature.

    10. A rectangular plate with insulated surface is10 cm wide so long compared to its width that it may be considered as an infinite plate. If the temperature along short edge y = 0 is given

    by

    20 0 x 5

    20(10 ) 5 x 10

    xu

    x and all other three edges are kept at 0

    o

    C. Find the steady

    state temperature at any point of the plate.

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  • Department of Mathematics FMCET MADURAI

    Pag

    e45

    Unit - 4

    FOURIER TRANSFORMS

    FORMULAE

    1. Fourier Transform of f(x) is isx

    -

    1[ ( )] f(x)e

    2F f x dx

    2. The inversion formula -isx

    -

    1( ) ( )e

    2f x F s ds

    3. Fourier cosine Transform Fc

    [f(x)] = Fc

    (s) =

    0

    2( )cosf x sxdx

    4. Inversion formula f(x) =

    0

    2( )coscF s sxds

    5. Fourier sine Transform (FST) Fs

    [f(x)] = Fs

    (s) =

    0

    2( )sinf x sxdx

    6. Inversion formula f(x) =

    0

    2( )sinsF s sxds

    7. Parsevals Identity

    2

    2( ) ( )f x dx F s ds

    8. Gamma function 1

    0

    1, 1 &

    2

    n xn x e dx n n n

    9. 2 2

    0

    cosaxa

    e bxdxa b

    10 2 2

    0

    sinaxb

    e bxdxa b

    11. 0

    sin

    2

    axdx

    x

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  • Department of Mathematics FMCET MADURAI

    Pag

    e46

    12. 2 2

    0

    &2

    x xe dx e dx

    13. cos & sin2 2

    iax iax iax iaxe e e eax ax

    ORKING RULE TO FIND THE FOURIER TRANSFORM

    Step1: Write the FT formula.

    Step2: Substitute given f(x) with their limits.

    Step3: Expand isxe as cos sx + isin sx and use Even & odd property

    Step4: Integrate by using Bernoullis formula then we get F(s)

    WORKING RULE TO FIND THE INVERSE FOURIER TRANSFORM

    Step1: Write the Inverse FT formula

    Step2: Sub f(x) & F(s) with limit , in the formula

    Step3: Expand isxe as cos sx -isin sx and equate real part

    Step4: Simplify we get result

    WORKING RULE FOR PARSEVALS IDENTITY

    If F(s) is the Fourier transform of f(x) then

    2

    2( ) ( )f x dx F s ds is known as Parsevals identity.

    Step1: Sub f(x) & F(s) With their limits in the above formula

    Step2: Simplify we get result

    WORKING RULE TO FIND FCT

    Step1: Write the FCT formula & Sub f(x) with its limit in the formula

    Step2: Simplify, we get ( )C

    F S

    WORKING RULE TO FIND INVERSE FCT

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  • Department of Mathematics FMCET MADURAI

    Pag

    e47

    Step1: Write the inverse FCT formula & Sub ( )C

    F S with its limit in the formula

    Step2: Simplify, we get f(x)

    WORKING RULE TO FIND FST

    Step1: Write the FST formula & Sub f(x) with its limit in the formula

    Step2: Simplify, we get ( )SF S

    WORKING RULE TO FIND INVERSE FCT

    Step1: Write the inverse FST formula & Sub ( )sF S with limit in the formula

    Step2: Simplify, we get f(x)

    WORKING RULE FOR f(x) = axe

    Step:1 First we follow the above FCT & FST working rule and then we get this

    result

    Fc

    (e-ax

    ) = 2 2

    2 a

    a s F

    s

    (e-ax

    ) = 2 2

    2 s

    a s

    By Inversion formula, By Inversion formula,

    2 2

    0

    cos

    2

    axsx ds ea s a

    2 2

    0

    sin2

    axs sxds ea s

    TYPE-I : If problems of the form i)2 2

    x

    x a ii)

    2 2

    1

    x a , then use Inversion formula

    TYPE-II: If problems of the form i)

    2

    22 2

    0

    xdx

    x a ii)

    22 2

    0

    dx

    x a, then use Parsevals Identity

    TYPE-III

    2 2 2 2

    0

    dx

    x a x b , then use

    0 0

    ( ) ( ) ( ) ( )C Cf x g x dx F f x F g x dx

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  • Department of Mathematics FMCET MADURAI

    Pag

    e48

    UNIT - 4

    FOURIER TRANSFORM

    1. State Fourier Integral Theorem.

    Answer:

    If ( )f x is piece wise continuously differentiable and absolutely on , then,

    ( )1( )2

    i x t sf x f t e dt ds.

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  • Department of Mathematics FMCET MADURAI

    Pag

    e49

    2. StateandproveModulation

    theorem.

    1cos

    2F f x ax F s a F s a

    Proof:

    ( ) ( )

    1cos cos

    2

    1

    22

    1 1 1 1

    2 22 2

    1 1

    2 2

    1cos

    2

    isx

    iax iaxisx

    i s a x i s a x

    F f x ax f x ax e dx

    e ef x e dx

    f x e dx f x e dx

    F s a F s a

    F f x ax F s a F s a

    3. State Parsevals Identity. Answer:

    If F s is a Fourier transform of f x , then

    2 2

    F s ds f x dx

    4. State Convolution theorem.

    Answer:

    The Fourier transform of Convolution of f x and g x is the product of their Fourier

    transforms.

    F f g F s G s

    5. State and prove Change of scale of property.

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  • Department of Mathematics FMCET MADURAI

    Pag

    e50

    Answer:

    If ,F s F f x then 1

    saF f ax F

    a

    1

    2

    1; where

    2

    sa

    isx

    i t

    F f ax f ax e dx

    dtf t e t ax

    a

    1 sF f ax Faa

    6. Prove that if F[f(x)] = F(s) then ( ) ( ) ( )n

    n n

    n

    dF x f x i F s

    ds

    Answer:

    1

    2

    isxF s f x e dx

    Diff w.r.t s n times

    1

    2

    1( ) ( )

    2

    nn isx

    n

    n n isx

    dF s f x ix e dx

    ds

    f x i x e dx

    1 1( )

    ( ) 2

    1( ) ( )

    2

    nn isx

    n n

    nn n isx

    n

    nnn

    n

    dF s x f x e dx

    i ds

    di F s x f x e dx

    ds

    dF x f x i F s

    ds

    7. Solve for f(x) from the integral equation 0

    ( )cos sf x sxdx e

    Answer:

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  • Department of Mathematics FMCET MADURAI

    Pag

    e51

    0

    ( )cos sf x sxdx e

    0

    2cos

    2

    c

    s

    c

    F f x f x sx dx

    F f x e

    0

    0

    2

    0

    2( ) cos

    2 2cos

    2 2 1cos

    1

    c

    s

    s

    f x F f x sx ds

    e sx ds

    e sx dsx

    2 2

    0

    cos

    1,

    ax ae bx dxa b

    a b x

    8. Find the complex Fourier Transform of 1

    ( )0 0

    x af x

    x a

    Answer:

    1

    2

    isxF f x f x e dx

    ;x a a x a

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  • Department of Mathematics FMCET MADURAI

    Pag

    e52

    11

    2

    1(cos sin )

    2

    a

    isx

    a

    a

    a

    F f x e dx

    sx i sx dx

    00

    2 2 sin(cos )

    2 2

    2 sin

    aasx

    sx dxs

    as

    s

    [Use even and odd property second term become zero]

    9. Find the complex Fourier Transform of ( )0 0

    x x af x

    x a

    Answer:

    1

    2

    1

    2

    1(cos sin )

    2

    isx

    a

    isx

    a

    a

    a

    F f x f x e dx

    xe dx

    x sx i sx dx

    ;x a a x a

    2

    0 0

    2

    2 2 cos sin( ( sin ) (1)

    2 2

    2 cos sin

    aai sx sx

    x i sx dx xs s

    as as asi

    s

    [Use even and odd property first term become zero]

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  • Department of Mathematics FMCET MADURAI

    Pag

    e53

    10. Write Fourier Transform pair.

    Answer:

    If ( )f x is defined in , , then its Fourier transform is defined as

    1

    2

    isxF s f x e dx

    If F s is an Fourier transform of f x , then at every point of Continuity of f x , we

    have

    1

    2

    isxf x F s e ds .

    11. Find the Fourier cosine Transform of f(x) = e-x

    Answer:

    0

    0

    2

    2cos

    2cos

    2 1

    1

    c

    x x

    c

    x

    c

    F f x f x sx dx

    F e e sx dx

    F es

    2 2

    0

    cosaxa

    e bx dxa b

    12. Find the Fourier Transform of ,

    ( )0,

    imxe a x bf x

    otherwise

    Answer:

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  • Department of Mathematics FMCET MADURAI

    Pag

    e54

    1

    2

    1 1

    2 2

    isx

    b bi m s ximx isx

    a a

    F f x f x e dx

    e e dx e dx

    1 1 1

    2 2

    bi m s x

    i m s b i m s a

    a

    ee e

    i m s i m s

    13. Find the Fourier sine Transform of 1

    x.

    Answer:

    0

    0

    2sin

    2 sin 2

    2

    1

    2

    s

    s

    F f x f x sx dx

    sxdx

    x

    Fx

    14. Find the Fourier sine transform of xe

    Answer:

    0

    0

    2

    2sin

    2sin

    2

    1

    s

    x x

    s

    x

    s

    F f x f x sx dx

    F e e sx dx

    sF e

    s

    2 2

    0

    sinaxb

    e bx dxa b

    15. Find the Fourier cosine transform of 2 2x xe e

    Answer:

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  • Department of Mathematics FMCET MADURAI

    Pag

    e55

    0

    2coscF f x f x sx dx

    2 2

    0

    2

    0 0

    22 2 cos

    2cos 2 cos

    x x x x

    c

    x x

    F e e e e sx dx

    e sx dx e sx dx

    2 2 2 2

    2 2 1 2 1 12 2

    4 1 4 1s s s s

    16. Find the Fourier sine transform of 1 , 0 1

    ( )0 1

    xf x

    x

    Answer:

    0

    1

    0 1

    11

    00

    2sin

    2sin sin

    2 2 cos1sin 0

    2 1 cos

    s

    s

    F f x f x sx dx

    F f x f x sx dx f x sx dx

    sxsx dx

    s

    s

    s

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  • Department of Mathematics FMCET MADURAI

    Pag

    e56

    17. Obtain the Fourier sine transform of

    , 1

    ( ) 2 , 1 2

    0, 2

    x o x

    f x x x

    x.

    Answer:

    0

    1 2

    0 1

    1 2

    2 2

    0 1

    2sin

    2sin 2 sin

    2 cos sin cos sin2

    sF f x f x sx dx

    x sx dx x sx dx

    sx sx sx sxx x

    s s s s

    2 2 2

    2

    2 cos sin sin 2 cos sin

    2 2sin sin 2

    s

    s s s s sF f x

    s s s s s

    s s

    s

    18. Define self reciprocal and give example.

    If the transform of f x is equal to f s , then the function f x is called self

    reciprocal.

    2

    2x

    e is self reciprocal under Fourier transform.

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  • Department of Mathematics FMCET MADURAI

    Pag

    e57

    19. Find the Fourier cosine Transform of 0

    ( )0

    x xf x

    x

    Answer:

    0 0

    2 2 2

    0

    2

    2 2cos cos

    2 sin cos 2 cos 1sin

    2 sin cos 1

    cF f x f x sx dx x sx dx

    sx sx sx s

    s s s s s

    s s s

    s

    20. Find the Fourier sine transform of 2 2

    x

    x a.

    Answer:

    L et

    axf x e

    2 2

    2axs

    sF e

    s a

    Using Inverse formula for Fourier sine transforms

    2 2

    0

    2 2

    0

    2 2sin

    ( ) sin , 02

    ax

    ax

    se sx ds

    s a

    sie sx ds e a

    s a

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  • Department of Mathematics FMCET MADURAI

    Pag

    e58

    Change x and s, we get 2 2

    0

    sin2

    asx sx dx ex a

    2 2 2 2

    0

    2sin

    2

    2 2

    s

    as as

    x xF sx dx

    x a x a

    e e

    FOURIER TRANSFORM

    PART-B

    1. (i)Find the Fourier Transform of

    21 1( )

    0 1

    x if xf x

    if x and hence

    deduce that (i)3

    0

    cos sin 3cos

    2 16

    x x x xdx

    x (ii)

    2

    3

    0

    sin cos

    15

    x x xdx

    x

    (ii). Find the Fourier Transform of

    2 2

    ( )0 0

    a x x af x

    x a . hence

    deduce that 3

    0

    sin cos

    4

    x x xdx

    x

    2. Find the Fourier Transform of

    1( )

    0

    if x af x

    if x a and hence evaluate

    0

    sin)

    xi dx

    x ii)

    2

    0

    sin xdx

    x

    4. Find Fourier Transform of

    1 1( )

    0 1

    x if xf x

    if x and hence evaluate

    i)

    2

    0

    sin xdx

    x ii)

    4

    0

    sin xdx

    x

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  • Department of Mathematics FMCET MADURAI

    Pag

    e59

    5. Evaluate i)

    2

    22 2

    0

    xdx

    x a ii) 2

    2 20

    dx

    x a

    6 i). Evaluate (a) 2 2 2 2

    0

    dx

    x a x b(b)

    2

    2 2 2 2

    0

    x dx

    x a x b

    ii). Evaluate (a) 2 2

    0 1 4

    dx

    x x(b)

    2

    2 2

    0 4 9

    t dt

    t t

    7. (i)Find the Fourier sine transform of

    sin ;( )

    0 ;

    x when o xf x

    whenx

    (ii) Find the Fourier cosine transform of

    cos ;( )

    0 ;

    x when o x af x

    whenx a

    8. (i) Show that Fourier transform

    2

    2

    x

    e is

    2

    2

    s

    e

    (ii)Obtain Fourier cosine Transform of

    2 2a xe and hence find Fourier sine Transform x2 2a xe

    9. (i) Solve for f(x) from the integral equation

    0

    ( )cosf x xdx e

    (ii) Solve for f(x) from the integral equation

    0

    1 ,0 1

    ( )sin 2 ,1 2

    0 , 2

    t

    f x tx dx t

    t

    10. (i) Find Fourier sine Transform of xe , x>0 and hence deduce that

    2

    0

    sin

    1

    x xdx

    x

    (ii) Find Fourier cosine and sine Transform of4xe , x>0 and hence deduce

    that8 8

    2 2

    0 0

    cos2 sin 2( ) ( )

    16 8 16 8

    x x xi dx e ii dx e

    x x

    11.(i)Find &ax axS cF xe F xe

    (ii) Find &ax ax

    S c

    e eF F

    x x

    (iii) Find the Fourier cosine transform of ( ) cosaxf x e ax

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  • Department of Mathematics FMCET MADURAI

    Pag

    e60

    Z - TRANSFORMS

    Definition of Z Transform

    Let {f(n)} be a sequence defined for n = 0, 1,2 and f(n) = 0 for n< 0 then its Z Transform is defined as

    ( ) ( )n

    n

    Z f n F z f n z (Two sided z transform)

    0

    ( ) ( ) n

    n

    Z f n F z f n z (One sided z transform)

    Unit sample and Unit step sequence

    The unit sample sequence is defined as follows

    1 0( )

    0 0

    for nn

    for n

    The unit step sequence is defined as follows

    1 0( )

    0 0

    for nu n

    for n

    Properties

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    1. Z Transform is linear

    (i) Z {a f(n) + b g(n)} = a Z{f(n)} + b Z{g(n)}

    2. First Shifting Theorem

    (i) If Z {f(t)} = F(z),

    then aT

    at

    z zeZ e f t F z

    (ii) If Z {f(n)} = F(z),

    then n zZ a f n F

    a

    3. Second Shifting Theorem

    If Z[f(n)]= F(z) then

    (i)Z[f(n +1)] = z[ F(z) f(0)]

    (ii)Z[f(n +2)] = 2z [ F(z) f(0)-f(1) 1z ]

    (iii)Z[f(n +k)] = kz [ F(z) f(0)-f(1) 1z - f(2) 2z - f(k-1) ( 1)kz ]

    (iv) Z[f(n -k)] = kz F(z)

    4. Initial Value Theorem

    If Z[f(n)] = F(z) then f(0) = lim ( )z

    F z

    5. Final Value Theorem

    If Z[f(n)] = F(z) then 1

    lim ( ) lim( 1) ( )n z

    f n z F z

    PARTIAL FRACTION METHODS

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    Model:I

    1 A B

    z a z b z a z b

    Model:II

    2 2

    1

    ( )

    A B C

    z a z b z bz a z b

    Model:III

    22

    1 A Bz C

    z a z bz a z b

    Convolution of Two Sequences

    Convolution of Two Sequences {f(n)} and {g(n)} is defined as

    0

    { ( )* ( )} ( ) ( )n

    K

    f n g n f K g n K

    Convolution Theorem

    If Z[f(n)] = F(z) and Z[g(n)] = G(z) then Z{f(n)*g(n)} = F(z).G(z)

    WORKING RULE TO FIND INVERSE Z-TRANSFORM USING CONVOLUTION THEOREM

    Step: 1 Split given function as two terms

    Step: 2 Take 1z both terms

    Step: 3 Apply 1z formula

    Step: 4 Simplifying we get answer

    Note:

    12 11 .......

    1

    nn aa a a

    a

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    1

    2 11 .......

    1 ( )

    n

    n aa a a

    a

    Solution of difference equations

    Formula

    i) Z[y(n)] = F(z)

    ii) Z[y(n +1)] = z[ F(z) y(0)]

    iii) Z[y(n +2)] = 2z [ F(z) y(0)- y(1) 1z ]

    iv) Z[y(n +3)] = 3z [ F(z) y(0)- y(1) 1z + y(2) 2z ]

    WORKING RULE TO SOLVE DIFFERENCE EQUATION:

    Step: 1 Take z transform on both sides

    Step: 2 Apply formula and values of y(0) and y(1).

    Step: 3 Simplify and we get F(Z)

    Step:4 Find y(n) by using inverse method

    Z - Transform Table

    No.

    f(n)

    Z[f(n)]

    1. 1

    1

    z

    z

    2. an

    z

    z a

    3. n

    2( 1)

    z

    z

    4. n2

    2

    3( 1)

    z z

    z

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    6. 1

    n log

    ( 1)

    z

    z

    7. 1

    1n log

    ( 1)

    zz

    z

    8. 1

    1n

    1log

    ( 1)

    z

    z z

    9. ean

    ( )az

    z e

    10. 1

    !n

    1

    ze

    11. Cos n

    2

    ( cos )

    2 cos 1

    z z

    z z

    12. sin n

    2

    sin

    2 cos 1

    z

    z z

    13.

    cos2

    n

    2

    2 1

    z

    z

    sin2

    n

    2 1

    z

    z

    14. nna 2( )

    az

    z a

    f(t) Z(f(t)

    1 t

    2( 1)

    Tz

    z

    2. t2

    2

    3

    ( 1)

    ( 1)

    T z z

    z

    3 eat

    ( )aTz

    z e

    4. Sin t

    2

    sin

    2 cos 1

    z T

    z z T

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    5. cos t

    2

    ( cos )

    2 cos 1

    z z T

    z z T

    TWO MARKS QUESTIONS WITH ANSWER

    1. Define Z transform

    Answer:

    Let {f(n)} be a sequence defined for n = 0, 1,2 and f(n) = 0 for n< 0 then its Z Transform is defined as

    ( ) ( )n

    n

    Z f n F z f n z (Two sided z transform)

    0

    ( ) ( ) n

    n

    Z f n F z f n z (One sided z transform)

    Find the Z Transform of 1

    Answer:

    0

    n

    n

    Z f n f n z

    0

    1 (1) n

    n

    Z z

    1 21 ....z z

    111 z

    1 11 1

    11

    z z

    z z z

    11

    zZ

    z

    2. Find the Z Transform of n

    Answer:

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    0

    0

    1 2 3

    0

    2 22

    1 1

    2

    0 2 3 ...

    1 1 11 1

    1

    1

    n

    n

    n

    n

    n

    n

    Z f n f n z

    Z n nz

    nz z z z

    zz z

    z z z z

    z

    z

    3. Find the Z Transform of n2.

    Answer:

    2 dZ n Z nn z Z ndz

    , by the property,

    2 2

    2 4 3

    1 2 1( )

    ( 1)1 1

    z z zd z z zz zdz zz z

    4. State Initial & Final value theorem on Z Transform

    Initial Value Theorem

    If Z [f (n)] = F (z) then f (0) = lim ( )z

    F z

    Final Value Theorem

    If Z [f (n)] = F (z) then 1

    lim ( ) lim( 1) ( )n z

    f n z F z

    6. State convolution theorem of Z- Transform.

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    Answer:

    Z[f(n)] = F(z) and Z[g(n)] = G(z) then Z{f(n)*g(n)} = F(z) G(z)

    7. Find Z Transform of nna

    Answer:

    0

    0

    1 2 3

    0

    2

    2

    0 2 3 ...

    1

    n

    n

    n n n

    n

    n

    n

    Z f n f n z

    Z na na z

    a a a an

    z z z z

    a a az

    z z z a

    8. Find Z Transform of cos2

    n and

    sin2

    n

    Answer:

    We know that

    0

    n

    n

    Z f n f n z

    2

    coscos

    2 cos 1

    z zZ n

    z z

    2

    22

    cos2

    cos2 1

    2 cos 12

    z zz

    Z nz

    z z

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    Similarly 2

    sinsin

    2 cos 1

    zZ n

    z z

    22

    sin2sin

    2 12 cos 1

    2

    zz

    Z nz

    z z

    9. Find Z Transform of 1

    n

    Answer:

    0

    n

    n

    Z f n f n z

    0

    1 2 3

    1

    1

    1 1

    1....

    1 2 3

    1 1log 1 log

    log1

    n

    n

    n

    n

    Z zn n

    z z zz

    n

    z

    z z

    z

    z

    10. Find Z Transform of 1

    !n

    Answer:

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    1

    0

    0

    1 2 3

    0

    1

    1 1

    ! !

    11 ....

    ! 1! 2! 3!

    n

    n

    n

    n

    n

    n

    z z

    Z f n f n z

    Z zn n

    z z zz

    n

    e e

    11. Find Z Transform of 1

    1n

    Answer:

    0

    0

    ( 1)

    0

    2 31

    1 1

    1 1

    1

    1

    ....2 3

    1log 1

    log1

    n

    n

    n

    n

    n

    n

    Z f n f n z

    Z zn n

    z zn

    z zz z

    zz

    zz

    z

    12. Find Z Transform of an

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    Answer:

    0

    0 0

    1 2 3

    1

    1

    1 ...

    1

    n

    n

    nnn

    nn n

    Z f n f n z

    a aZ a

    z z

    a a a

    z z z

    a

    z

    z a z

    z z a

    13. State and prove First shifting theorem

    Statement: IfZ f t F z , then ( )at aTZ e f t F ze

    Proof:

    0

    ( ) ( )at anT n

    n

    Z e f t e f nT z

    As f(t) is a function defined for discrete values of t, where t = nT,

    then the Z-transform is

    0

    ( ) ( ) ( )n

    n

    Z f t f nT z F z ).

    0

    ( ) ( ) ( )n

    at aT aT

    n

    Z e f t f nT ze F ze

    14. Define unit impulse function and unit step function.

    The unit sample sequence is defined as follows:

    1 0( )

    0 0

    for nn

    for n

    The unit step sequence is defined as follows:

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    1 0( )

    0 0

    for nu n

    for n

    15. Find Z Transform of atZ e

    Answer:

    0 0

    n nat anT n aT n aT

    n n

    n

    aT

    Z e e z e z z e

    z zz a

    z e z a

    [Using First shifting theorem]

    16. Find Z Transform of 2tZ te

    Answer:

    2

    2

    2

    2

    2

    22

    1

    1

    T

    T

    t

    z ze

    z ze

    T

    T

    TzZ te Z t

    z

    Tze

    ze

    [Using First shifting theorem]

    17. Find Z Transform of cos2tZ e t

    Answer:

    2

    coscos 2 cos 2

    2cos 1T

    T

    t

    z ze

    z ze

    z zZ e t Z t

    z z

    2

    cos

    2cos 1

    T T

    T T

    ze ze T

    ze T ze

    [Using First shifting theorem]

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    18. Find Z Transform of 2 t T

    Z e

    Answer:

    Let f (t) = e2t

    , by second sifting theorem

    2( )

    2

    2 2

    ( ) ( ) (0)

    11

    1 1

    t T

    T

    T T

    Z e Z f t T z F z f

    zez z

    ze ze

    19. Find Z Transform of sinZ t T

    Answer:

    Let f (t) = sint , by second sifting theorem

    2

    2 2

    sin( ) ( ) ( ) (0)

    sin sin0

    2cos 1 2cos 1

    Z t T Z f t T z F z f

    z t z tz

    z t z z t z

    20. Find Z transform of 1 2n n

    Answer:

    0

    n

    n

    Z f n f n z

    2

    2 2

    2

    3 2

    1 2 2 2

    3 2 3 2 1

    3 211 1

    Z n n Z n n n

    Z n n z n z n z

    z z z z

    zz z

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    QUESTION BANK

    Z-TRANSFORMS

    1. (i)Find 2

    1 8

    (2 1)(4 1)

    zZ

    z z &

    21 8

    (2 1)(4 1)

    zZ

    z zby convolution theorem.

    (ii) Find

    21

    ( )( )

    zZ

    z a z b &

    21

    ( 1)( 3)

    zZ

    z zby convolution theorem

    2. (i) Find 2

    1

    2( )

    zZ

    z a &

    21

    2( )

    zZ

    z aby convolution theorem

    3. (i ) State and prove Initial & Final value theorem.

    (ii) State and prove Second shifting theorem

    (i) Find the Z transform of 1

    ( 1)( 2)n n&

    2 3

    ( 1)( 2)

    n

    n n

    4. (i) Find 2

    1

    2( 4)

    zZ

    z by residues.

    (ii) Find the inverse Z transform of

    2

    21 ( 1)

    z z

    z zby partial fractions.

    5. (i) Find 12 2 2

    zZ

    z z &

    21

    2 7 10

    zZ

    z z

    6. (i)Find the Z transform of 1

    ( )!

    f nn

    Hence find 1

    ( 1)!Z

    n and

    1

    ( 2)!Z

    n.

    (ii) Find 1

    !Z

    n and also find the value of sin( 1)n and cos( 1)n .

    7. (i)Solve 2 6 1 9 2 0 0& 1 0ny n y n y n with y y

    (ii) Solve 2 4 1 4 0y n y n y n y(0) = 1 ,y(1) =0

    8. (i )Solve 3 1 4 2 0, 2 (0) 3& (1) 2y n y n y n n given y y

    (ii) Solve 3 3 1 2 0, 0 4, 1 0& 2 8y n y n y n y y y ,

    9. (i)Find cos & sinZ n Z n and also find cos & sinn nZ a n Z a n

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