Unit IV Laplace Transforms

  • View
    224

  • Download
    2

Embed Size (px)

Text of Unit IV Laplace Transforms

  • 7/29/2019 Unit IV Laplace Transforms

    1/26

    Unit I Laplace Transform

    Laplace Transform

    The Laplace transform is an integral transform that transforms a real valued

    function f of some non-negative variable, say t into a function F(s) for all s for

    which the improper integral

    0

    st dt)t(fe

    converges.

    Definition 1.1 Let f (t) be a given real valued function that is

    defined for

    all t 0. The function defined by

    0 )( dttfe

    st

    for all s for which this improper integral converges is

    called the

    Laplace transform of f (t) and will be denoted by

    F (s) = L (f) =

    0

    )( dttfest

    The operation which yields F (s) from a given real valued f (t) is also called the

    Laplace transformof f (t) while f (t) is called

    the inverse transformof F (s)

    and will be denoted by

    f (t) = )F(L 1

    Example 1 Let f (t) = 1 for t 0. Then find F (s).

    Solution Applying definition 1.1 we get:

    L (f) = L (1) = b

    0

    st dteb

    im =

    0

    bst

    s

    e

    bim

    =

    +

    s

    e

    bim

    s

    1 tb =s

    1for s

    > 0.

    Therefore, L (1) =s1 for s > 0.

    Remark: Let f (t) = k for any scalar k. Then L (f) =s

    kfor s > 0.

    Example 2 Let

    >

    =

    1,0

    10,4)(

    tfor

    tfortf . Then find F (s).

    Prepared by Tekleyohannes Negussie 83

  • 7/29/2019 Unit IV Laplace Transforms

    2/26

    Unit I Laplace Transform

    Solution L (f) = F (s) = dteb

    imb

    tfst 0

    )( = dtest

    1

    0

    4 = ( )ses

    14

    .

    Therefore, L (f) = ( )ses

    14

    for s > 0.

    Example 3 Let f (t) = t for t 0. Then find F (s).

    Solution Applying definition 1.1 we get:

    L (f) = F (s) = b

    0

    st dtetb

    im

    Now applying integration by parts we get:

    b

    0

    st dtetb

    im =0

    2

    bstst

    s

    ee

    s

    t

    bim

    =2s

    1for s > 0.

    Therefore, L (f) = 21

    s

    for s > 0.

    Example 4 Prove that for any natural number n, L ( nt ) = 1!+n

    s

    nfor t 0.

    Solution We need to proceed by applying the principle of mathematical

    induction on n

    i) For n = 1 it follows from example 3.

    ii) Assume that it holds true for n = k, i.e.

    L ( kt ) = 1!+k

    s

    kwhere t 0

    iii) We need to show that it holds true for n = k + 1. i.e. L ( 1+k

    t ) = 2!)1(

    ++

    ks

    k

    for t 0.Now by the definition of the Laplace transform

    L ( 1+k

    t ) = +b

    tskdtet

    bim

    0

    1

    Applying integration by parts we get:

    u = 1+k

    t and 'v = tse dt

    then 'u = ktk )1( + dt and v =ts

    es

    1 .

    Hence, +b

    tskdtet

    bim

    0

    1 =

    ++

    + btskts

    k

    dtets

    kb

    es

    t

    bim

    0

    11

    0

    Prepared by Tekleyohannes Negussie 84

  • 7/29/2019 Unit IV Laplace Transforms

    3/26

    Unit I Laplace Transform

    =

    + b tsk dtet

    bim

    s

    k

    0

    1 =

    +

    +1!1

    ks

    k

    s

    k, by our

    assumption

    = 2!)1(

    +

    +k

    s

    kfor s > 0.

    Thus, by the principle of mathematical induction it holds true for any naturalnumber n.

    Therefore, L ( nt ) = 1!+n

    s

    nfor t 0.

    Example 5 Let f (t) =t

    e

    for t 0. Then find F (s).

    Solution Applying definition 1.1 we get:

    L (f) = L (t

    e

    ) = b

    0

    t)s( dteb

    im =

    0

    bt)s(

    s

    e

    bim

    =

    +

    s

    1

    s

    e

    bim

    b)s( =

    s1

    for s > .

    Therefore, L (t

    e

    ) =s

    1for s > .

    Theorem 1.1 (linearity of the Laplace

    transform)

    The Laplace transform is a linear operation, that

    is forany real valued functions f (t) and g (t) whose

    Laplace

    transform exists and any scalars and

    L ( f + g) = L (f ) + L ( g)

    ProofBy definition 1.1

    L ( f + g) = { } +b

    tsdte

    bim tgtf

    0

    )()(

    = b

    tsdttfe

    bim

    0

    )( + b

    tsdttge

    bim

    0

    )(

    = h L (f) + k L ( g).

    Therefore, Laplace transform is a linear operation.

    Example 6 Find the Laplace transform of the hyperbolic functions f (t) = cosh

    t and

    Prepared by Tekleyohannes Negussie 85

  • 7/29/2019 Unit IV Laplace Transforms

    4/26

    Unit I Laplace Transform

    g (t) = sinh t for t 0.

    Solution From theorem 1.1 and the result of example 2 we get:

    L (f ) = L (cosh t) = b

    tsdte

    bim

    0

    )(

    2

    1 + +b

    tsdte

    bim

    0

    )(

    2

    1

    =

    02

    1)()( b

    s

    e

    s

    e

    bim

    tsts

    +

    +

    =

    +

    +

    s

    e

    s

    e

    bim

    bsbs

    )()(

    2

    1

    +

    ss 11

    2

    1= 22 s

    sfor s

    > .

    Similarly, L (g) = L (sinh t) = b

    tsdte

    bim

    0

    )(

    2

    1 +b

    tsdte

    bim

    0

    )(

    2

    1

    =

    021

    )()( b

    se

    se

    bim

    tsts

    +

    +

    +

    =

    ++

    +

    s

    e

    s

    e

    bim

    bsbs

    )()(

    2

    1

    ++

    ss 11

    2

    1= 22

    sfor s

    > .

    Therefore, L (cosh t) = 22 ss

    and L (sinh t) = 22

    sfor s > .

    Example 7 Find the Laplace transform of functions f (t) = cos t and g (t) = sint for t 0.

    Solution From theorem 1.1 and Eulers formula, where i = 1 we get:

    tie = cos t + i sin t

    L (f ) = L ( tie ) =is

    1=

    )()(

    isis

    is

    ++

    = 2222

    ++

    + si

    s

    s

    On the other hand, L (cos t + i sin t) = L (cos t) + i L (sin t).

    Now equating the real and the imaginary parts we get:

    L (cos t ) = 22 +ss

    and L (sin t) = 22

    +s.

    Therefore, L (cos t ) = 22 +ss and L (sin t) = 22

    +s

    for s > .

    We can also derive these by using definition 1. 1 and integration by parts with

    out going into the operations in complex numbers.

    Example 8 Find L (f), where

    Prepared by Tekleyohannes Negussie 86

  • 7/29/2019 Unit IV Laplace Transforms

    5/26

    Unit I Laplace Transform

    i) f (t) = 3 cos 2t + 2 sin t 34 t ii) f (t) = 4 cosh 5t 3 sinh 2t +

    te

    5

    iii) f (t) = sin 2t cos 4t

    Solutions By the Linearity of the Laplace transform we have

    i) L (f) = 3 L (cos 2t) + 2 L (sin t) )(4 3tL =4

    32 +s

    s+

    1

    22 +s

    4)!3(4

    s.

    Therefore, L [3 cos 2t + 2 sin t 34 t ] =4

    32 +s

    s+

    1

    22 +s

    424

    sfor s > 0.

    ii) L (f) = 4 L (cosh 5t) 3 L (sinh 2t) + )(5 teL =25

    42 s

    s

    4

    62 s

    +s

    5.

    Therefore, L [4 cosh 5t 3 sinh 2t + te

    5 ] =25

    42

    s

    s

    4

    62

    s

    +s

    5for

    s > 0.iii) First observe that:

    sin 2t cos 4t = tt )42(sin2

    1)42(sin

    2

    1++ = tt 6sin

    2

    12sin

    2

    1+

    Then by the Linearity of the Laplace transform we get:

    L [sin 2t cos 4t] = ]6[sin2

    1]2[sin

    2

    1tLtL + =

    ++

    +

    36

    6

    2

    1

    4

    2

    2

    122 ss

    .

    Therefore, L [sin 2t cos 4t] =

    ++

    +

    36

    3

    4

    122 ss

    for s > 0.

    Example 9 For any real valued functions f (t) and g (t) whose Laplace transform

    exists and

    any real numbers and , show that the inverse Laplace transform

    is linear.

    Solution From the linearity of the Laplace transform we get:

    L ( f (t) + g (t)) = L (f ) + L ( g ) = F (s ) + F ( s )

    Hence, ))()((1

    sFsFL + = f (t) + g (t)) = )(1)(1 sGsF LL + .

    Therefore, the inverse Laplace transform is linear.

    Example 10 Find the inverse Laplace transforms of

    i)s2

    3ii) 3

    40

    siii)

    1

    22 ++

    s

    siv)

    9

    42 +

    s

    s

    Solutions i) From the linearity of the inverse Laplace transform we get:

    Prepared by Tekleyohannes Negussie 87

  • 7/29/2019 Unit IV Laplace Transforms

    6/26

    Unit I Laplace Transform

    sL

    2

    31=

    s

    L11

    2

    3=

    2

    3.

    Therefore,

    s

    L2

    31=

    2

    3.

    ii) Now 340

    s= 3

    !220

    sand 23

    !21t

    sL = .

    Hence,

    3

    401

    s

    L =

    3

    !2120

    s

    L = 220 t .

    Therefore,

    3

    401

    s

    L = 220 t .

    iii)

    ++

    1

    22

    1

    s

    sL =

    +

    121

    s

    sL +

    +

    1

    12

    2

    1

    sL = cos t + 2 sin t.

    Therefore,

    +

    +1

    2

    2

    1

    s

    s

    L = cos t + 2 sin t.

    iv)

    +

    9

    42

    1

    s

    sL =

    921

    s

    sL +

    9

    3

    3

    42

    1

    sL = cosh 3t +

    3

    4sinh

    3t.

    Therefore,

    +

    9

    42

    1

    s

    sL = cosh 3t +

    3

    4sinh 3t.

    Example 11 Find f (t) if

    i) F (s) =ss 3

    92 +

    ii) F (s) =9

    )1(42 +

    s

    s

    Solution By partial fraction reduction we get:

    i)ss 3

    92 +

    =s

    A+

    3+sB

    =ss

    sBsA

    3

    )3(2 +

    ++A + B = 0 and 3 A = 9 A = 3

    and B = 3

    Thus, F (s) =s

    3

    3

    3

    +s=

    +

    33

    11113

    ssLL = 3 3 te 3 .

    Therefore, f (t) = 3 3 te 3 .

    ii)9

    )1(4

    2

    +

    s

    s=

    3s

    A+

    3+s

    B=

    9

    )3()3(

    2

    ++

    s

    sBsA

    A + B = 4 and 3 (A B) = 4 A =3

    8and B =

    3

    4

    Thus, F (s) =3

    8

    31

    s+

    3

    4

    + 31

    s=

    +

    +

    33

    4

    3

    1111

    3

    8

    ssLL =

    te 3

    3

    8+

    3

    4 te

    3.

    Prepared by Tekleyohannes Negussie 88

  • 7/29/2019 Unit IV Laplace Transforms

    7/26

    Unit I Laplace Transform

    Therefore, f (t) =te 3

    3

    8+

    3

    4 te

    3.

    Example 12 For any 0, the Gamma function () is defined by:

    () =

    0

    1dtte

    t

    Show that )( atL = 1)1(

    ++

    as

    afor a > 0 and in particular (n + 1) = n!,

    for any non-negative integer n.

    Solution Put x = st, so that dt =s

    dx.

    Thus, )( atL =s

    dx

    s

    xe

    bim

    b ax

    0

    = dxxeb

    im

    s

    bax

    a

    + 01

    1 =

    1

    )1(

    +

    +a

    s

    a

    .

    Therefore, )( atL = 1)1(

    ++

    as

    afor a > 0.

    Further more, for any natural number n

    (n + 1) = dteb

    imb

    nt

    t 0 =

    0

    bnt

    te

    bim

    + n

    dteb

    imb

    nt

    t 01

    = n dteb

    imb

    nt

    t 01 = n (n)

    Thus, (n + 1) = n (n) = n (n 1) (n 1) = n (n 1) (n 2) (n2) = . . . =

    n! (1) = n!

    We summarize the Laplace Transforms of some functions for future reference in

    the table below.

    N

    o

    Function f

    (t)

    Laplace

    transform

    No Function f

    (t)

    Laplace

    transform

    1 1s

    16

    te

    s1

    2 k for k s

    k7 cos t

    22 +ss

    3 t2

    1

    s8 sin t

    22

    +s

    Prepared by Tekleyohannes Negussie 89

  • 7/29/2019 Unit IV Laplace Transforms

    8/26

    Unit I Laplace Transform

    4nt for

    n = 0, 1,2, . . .

    1

    !+n

    s

    n9 cosh t

    22 ss

    5 at for a

    01

    )1(++

    as

    a10

    sinh t22

    s

    Existence of Laplace Transforms

    Before we define the theorem that guarantees the existence of Laplace

    transform, let us see the definition of piecewise continuity.

    Definition 1.2 A function f (t) ispiecewise continuous on a

    finite

    interval a t b if f (t) is defined on the interval and is

    such that

    the interval can be divided into finitely many subintervals ,

    in

    each of which f (t) is continuous and has finite limits as t

    approaches either endpoint of the subintervals from the

    interior.

    By definition 1.2, it follows that finite jumps are the only discontinuities that a

    piecewise

    continuous function may have.

    Theorem 1.2 (Existence of Laplace Transforms)

    Let f (t) be a real valued function that is piecewise

    continuous

    on every finite interval in [0, ) and satisfies the

    Prepared by Tekleyohannes Negussie

    x

    y

    Figure 1.1 a piecewise continuous function

    90

  • 7/29/2019 Unit IV Laplace Transforms

    9/26

    Unit I Laplace Transform

    inequality

    )t(f teM

    for all t 0 and for some constants and M.

    Then there

    is a unique Laplace transform of f (t) for all s >

    .

    Proof Since f (t) is piecewise continuous, )t(fe st is integrable over any finite

    interval on

    the t-axis. Thus, for s > we have

    )f(L = b

    0

    )t(fst dte

    bim

    b

    0

    )t(fst dte

    bim

    b

    sttdteeM

    bim

    0

    =s

    Mfor s > .

    Therefore, )( fL is finite, and hence it exists for s > .

    Example 13 Let f (t) =t

    1for t > 0. Then show that L [f] doesnt exist.

    Solution We prove this by showing that dtet

    ts

    0

    1diverges. For some c > 0,

    dte

    t

    ts

    0

    1= dte

    tb

    imc

    b

    ts

    +

    1

    0

    + dte

    td

    imd

    c

    ts

    1

    To this end, let

    u =ts

    e

    and 'v = dtt

    1. Then du = dt

    tses and v = tn .

    Hence, dtetb

    imc

    b

    ts +1

    0

    = { }

    b

    ctstne

    b

    im

    +0

    + dttnesb

    imc

    b

    ts

    +0 .

    But, { }b

    ctstne

    b

    im

    +0

    = { }bnecneb

    im bscs

    +0

    =

    cse

    cn { }bnbse

    b

    im

    +

    0= for any c > 0.

    Consequently, dtet

    ts

    0

    1diverges.

    Therefore, the Laplace transform of f (t) =t

    1for t > 0 doesnt exist.

    Remark: If the Laplace transform of a given real valued function exists, then it

    is unique.

    Prepared by Tekleyohannes Negussie 91

  • 7/29/2019 Unit IV Laplace Transforms

    10/26

    Unit I Laplace Transform

    Conversely, if two functions have the same Laplace transform, then these

    functions

    are equal over any interval of positive length (i.e. they differ at various

    isolated points).

    Exercises 1.1

    In exercises 1-8, find the Laplace transforms of the following functions by

    showing the details of your steps. (a, b, c, and are constants) .

    1. f (t) = 2tctba ++ 2. f (t) = sin t cos t 3. f (t) = tcos 2 4. f (t) =

    t3coshe t

    5. f (t) = )t(sin + 6.

    =1x0for1

    1x0for0)t(f

    7. 2x0for1x1)t(f = 8.

    =1x0for11x0fort)t(f

    In exercises 9 13, given F (s) = L (f), find f (t) by showing the details of your

    steps. (, , and are constants) .

    9.4s

    4s

    2

    10.25s

    s5

    2 11.

    4s

    112.

    14

    3

    s

    s2

    13.

    2ss

    10s

    2

    In exercises 14 17, find the Laplace transforms of the following functions by

    showing the details of your steps. (a, b, c, and are constants) .14. sinh t cos t 15. cosh t sin t 16. t2sinhe5 t2 17. tet 2)1( +

    In exercises 18 20, find the inverse Laplace transforms of the following by

    showing the details of your steps.

    18.2)1s(

    1

    +19.

    4)3s(

    12

    20.

    18s6s

    3

    2 ++

    1.2 More on Transforms of Functions

    1.2.1 Laplace Transforms of Derivatives

    Theorem 1.3 ( Laplace Transform of the Derivative of f

    (t))

    Suppose that f (t) is a piecewise continuous real valued

    Prepared by Tekleyohannes Negussie 92

  • 7/29/2019 Unit IV Laplace Transforms

    11/26

    Unit I Laplace Transform

    function

    for all t 0 and )t(f teM

    for some constants

    and M,

    and has a derivative )t(f' that is piecewise continuouson every

    finite interval on [0, ). Then the Laplace transform of

    )t(f'

    exists when s >, and

    )f(L ' = sL (f) f (0), for all s >.

    ProofApplying integration by parts we get:

    )]([ tfL =

    0

    )( dttfe st =

    +

    00)()( dtetfsetf

    b

    im stbst

    = [ ] ].[)0()( fLsfebfb

    im sb +

    Sinceb

    Mebf)( , we have sbebf )( bsbeMebf )( = bsMe )(

    But,bs

    Meb

    im )(

    = bse

    M

    b

    im)(

    = 0, for >s .

    This implies thatsb

    ebfb

    im

    )(

    = 0, and hencesb

    ebfb

    im

    )(

    = 0.

    Consequently,

    )0()]([)]([ ftfsLtfL =

    Therefore, )f(L ' = sL (f) f (0), for all s >.

    Theorem 1.4 (Laplace Transforms of Higher order

    Derivatives)

    Let )(tf and its derivatives )(tf , )(tf,

    ,

    )()1(

    tfn

    be real

    valued

    continuousfunctionsfor all 0t and satisfying )()( tf k

    teM

    for some and M, and k 0 and let the derivative )()( tf n be

    Prepared by Tekleyohannes Negussie 93

  • 7/29/2019 Unit IV Laplace Transforms

    12/26

    Unit I Laplace Transform

    piecewise

    continuous on every finite interval contained in ),0[ . Then the

    Laplace

    transform of )(

    )(

    tf

    n

    exists when>s

    , and is given by

    ))(( )( tfL n = )0()0()0())(( )1(21 nnnn ffsfstfLs

    Proof( by the principle of mathematical induction on n)

    For n = 1, it holds by theorem 1.4 i.e

    )0()]([)]([ ftfsLtfL = .

    Assume that it holds true for n = k

    ][)(k

    fL = )0(...)0(')0(][)1(21 kkkk ffsfsfLs

    Now we need to show that it holds true for n = k + 1

    ][)1( +k

    fL = )0(][)()( kk

    ffLs

    = )0(...)0(')0(][)1(21 kkkk ffsfsfLss

    = )0(...)0(')0(][)(11 kkkk ffsfsfLs +

    Therefore, by the principle of mathematical induction

    ))(()(

    tfLn = )0()0()0())((

    )1(21 nnnn ffsfstfLs for any natural number n.

    Example 14 Let .)(3

    ttf = Then find ].[ fLSolution )(tf = 23t , 0)0( =f , )(tf = t6 , 0)0( =f , 6)( = tf and .6)0( =f

    Thus, )]([ tfL = )0()0()0()]([ 23 ffsfstfLs , and also, )]([ tfL =s

    L6

    )6( = .

    Equating these values we get:

    ][3 fLs =s

    6 4

    6][

    sfL =

    Therefore, ][ 3tL = 46

    s.

    Remark: It can be shown by induction on n that

    )( ntL = 1!+n

    s

    nfor n = 0, 1, 2, and s > 0.

    Example 15Find ).(cos tL

    SolutionLet )(tf = .cos t Then )0(f = 1, )(tf = tsin , )0(f = 0,

    )(tf = t cos2 = ).(2 tf

    Prepared by Tekleyohannes Negussie 94

  • 7/29/2019 Unit IV Laplace Transforms

    13/26

    Unit I Laplace Transform

    Now we have ))(( tfL = )0()0())((2 fsftfLs and )]([ tfL = ][2 fL

    )]([ tfL = 22 +ss

    .

    Therefore, )(cos tL = .22

    +s

    s

    Example 16 Find L [ t2

    cos ].

    Solution Let )(tf = t2

    cos . Then f (0) = 1, )(tf = ttsincos2 , )0(f = 0 and

    )(tf = tt 22 cos2sin2 = ]cos[sin2 22 tt

    = ]coscos1[2 22 tt = t2cos42 = ).(42 tf

    Now, since ))(( tfL = ),0()0())((2 fsftfLs

    we have

    ]2)(4[ + tfL = )0()0())((2 fsftfLs and ]2)(4[ + tfL = )2()]([4 LtfL +

    sfLs )(2 = )2(][4 LfL + L[f] =)4(

    22

    2

    ++

    ss

    s.

    Therefore, )(cos2 tL =)4(

    22

    2

    ++

    ss

    sfor s > 0.

    Example 17 Show that )(sin2

    tL =)4(

    22 +ss

    .

    Solution Let f (t) = )(sin 2 tL . Then f (0) = 0, )(tf = ttsincos2 , )0(f = 0 and

    )(tf = tt 22 sin2cos2 = ]sin21[2 2 t = )(42sin42 2 tft = .

    Now, since ))(( tfL = ),0()0())((2 fsftfLs

    we have: ]2)(4[ + tfL = )0()0())((2 fsftfLs and ]2)(4[ + tfL = )2()]([4 LtfL +

    )(2 fLs = )2(][4 LfL + L[f] =)4(

    22 +ss .

    Therefore, )(sin 2 tL =)4(

    22 +ss

    for s > 0.

    Example 18 Find )cos( ttL .

    Solution Let )(tf = tt cos . Then f (0) = 0, )(tf = ttt sincos , )0(f = 1and )(tf = tttt cossinsin 2 = ).(sin2 2 tft

    Thus, )]([ tfL = )](sin2[ 2 tftL

    )0()0())((2 fsftfLs = )].([)(sin2 2 tfLtL

    1)(2 fLs = ].[2 222 fLs

    +

    Prepared by Tekleyohannes Negussie 95

  • 7/29/2019 Unit IV Laplace Transforms

    14/26

    Unit I Laplace Transform

    )()( 22 fLs + = 12

    22

    2

    ++

    s

    = .

    22

    22

    +

    s

    s

    Therefore, )cos( ttL = )]([ tfL =22

    22

    )( 2

    +

    s

    s

    Similarly, it can be shown that

    )cos( ttL =222

    22

    )(

    +

    s

    sand )sin( ttL = 222 )(

    2

    +ss

    .

    1.2.2 Laplace Transforms of the Integral

    Theorem 1.5 (Integration of f (t))

    Let F (s) be the Laplace transform of the real valued

    function

    f (t). If f (t) is piecewise continuous and satisfies the

    inequality

    teMtf)(

    for some constants and M, then

    )(1

    )(0

    sFs

    dfLt

    =

    for s > 0 and s > .

    Proof Suppose that f (t) is piecewise continuous andt

    eMtf

    )( for some

    constants

    and M . Then the integral

    g (x) = t

    df0

    )(

    is continuous and for any positive number t

    =t

    dtftg0

    )()( t

    deM0

    = )1( ek

    M

    e

    k

    Mfor s > .

    This shows that g (t) also satisfies an inequality of the form

    )(tg

    ek

    Mfor s > .

    Also, )(' tg = f (t), except for points at which f (t) is discontinuous. Hence,

    )(' tg is

    piecewise continuous on each finite interval, and by theorem 1.4,

    Prepared by Tekleyohannes Negussie 96

  • 7/29/2019 Unit IV Laplace Transforms

    15/26

    Unit I Laplace Transform

    L[f (t)] = )]([ ' tgL = s L [g (t)] g (0) for s > .

    Hence, g (0) = 0, so that L [f (t)] = s L [g (t)] for s > 0 and s > .

    Therefore, )(1

    )(0

    sFs

    dfLt

    =

    for s > 0 and s > .

    Example 19 Let F (s) =)(

    122 +ss

    . Find f (t).

    Solution From the result of example 7 we get

    +

    )(

    122

    1

    ssL = t

    sin

    1

    Again, using theorem 1.7 we have

    +

    )(

    122

    1

    ssL =

    d

    t

    0

    sin1

    = )cos1(12

    t

    Therefore, f (t) = )cos1(1

    2t

    .

    Example 20 Let F (s) =)(

    1222 +ss

    . Find f (t).

    Solution From the result of example 19 we get

    +

    )(

    1222

    1

    ssL =

    d

    t

    0

    2)cos1(

    1= )

    sin(

    12

    tt

    Therefore, f (t) = )sin

    (12

    tt .

    1.2.3 S-shifting, Unit step functions and t-shiftingThe first shifting theorem

    If we replace s by s a, in the definition of the Laplace transform we get the

    following important result.

    Theorem 1.6 (First shifting theorem)

    If f (t) has the transform F (s) where s > k, then )t(feat

    Prepared by Tekleyohannes Negussie 97

  • 7/29/2019 Unit IV Laplace Transforms

    16/26

    Unit I Laplace Transform

    has the

    transform F (s a) where s a > k. Symbolically,

    L { )t(feat } = F (s a) or )t(feat = )}as(F{L 1

    ProofF (s a) can be obtained by replacing s by s a in definition 1.1 as

    follows:

    F (s a) = b

    0

    )t(ft)as( dte

    bim = [ ]

    b

    0

    )t(ftaets dte

    bim = L { )t(feat }.

    Now if F (s) exists for s greater than some k, then our integral exists for

    s a > k.

    Example 21 Find the Laplace transform of f (t) = ate cos t and g (t) = ate sin t

    for t 0.

    Solution Applying theorem 1.6 on the result of example 5 we get:

    L (f ) = L ( ate cos t) =22)as(

    as

    +

    and L (g ) = L ( ate sin t) =

    22)as( +

    Unit step functions and t-shifting Theorem

    The unit step function defined below is a typical engineering function made to

    measure for engineering applications, which often involve function that are off

    and on.

    Definition 1.3The function u

    defined by

    > 0.

    Solution Applying definition 1.1 we get:

    L [u (t a)] =

    b

    tsdtatue

    b

    im

    0

    )(

    =

    b

    a

    tsdte

    b

    im=

    b

    im

    a

    b

    s

    esa

    =

    s

    esa

    .

    Therefore, L [u (t a)] =s

    esa

    for s > 0.

    Theorem 1.7 (The second shifting theorem, t-shifting

    theorem)

    If f (t) has the Laplace transform F (s), then the shiftedfunction

    )(~

    tf = f (t a) u (t a) =

    atfor

    atfor

    1

    0

    has the Laplace transform )(sFeas . That is

    L[f (t a) u (t a)] = )(sFe as .

    ProofFrom the definition of the Laplace transform we have

    )(sFe as = b

    dfeb

    ime sas

    0

    )( = +b

    dfeb

    im as

    0

    )( )(

    Substituting + a = t in the integral we get:

    )(sFeas =

    +

    ba

    dtatfeb

    im

    a

    ts)(

    =

    b

    dtatuatfeb

    im ts

    0

    )()(

    = L[f (t a) u (t a)].

    Prepared by Tekleyohannes Negussie

    t

    f(t)

    t

    f (t)

    1 1

    1

    f (t) = u (t 1)

    1

    f (t) = t [u (t) u (t 1)] + u (t 1)

    99

  • 7/29/2019 Unit IV Laplace Transforms

    18/26

    Unit I Laplace Transform

    Therefore, )(sFeas = L[f (t a) u (t a)].

    Example 24 Find L [f], where

    =

    2sin

    20

    02

    )(

    tift

    tif

    tif

    tf

    Solution We write f (t) in terms of unit step functions. For 0 < t < , we take 2 u

    (t).

    For < t < 2 we want 0, so we must subtract the step function 2 u (t

    ) and for

    t > 2 we need to add 2 u (t ) sin t.

    Hence, f (t) = 2 u (t) 2 u (t ) 2 u (t 2) sin t. f (t) = 2 u (t) 2 u (t ) 2 u

    (t 2) sin (t 2). (since sine is a periodic function with period 2)

    Therefore, L [f] =s

    e

    s

    s

    22

    +12

    2

    +

    s

    es

    .

    Example 25Find the inverse Laplace transform of

    F (s) =2

    2

    2

    22

    s

    e

    s

    s

    s

    es2

    4

    +1

    2

    +

    s

    ess

    .

    Solution Without the exponential functions the four terms of F (s) would have

    the inverses

    2t, 2t, 4 and cos t. Hence, by theorem 1.7

    f (t) = 2t 2(t 2) u(t 2) 4 u(t 2) + u (t ) cos (t )= 2t 2t u (t 2) u (t

    )cos t.

    Therefore,

    =

    tift

    tif

    tift

    tf

    cos

    20

    202

    )( .

    1.2.3 LaplaceTransforms ofPeriodic Functions

    ProofBy definition 1.1

    Prepared by Tekleyohannes Negussie

    Theorem 1.8 (Laplace Transforms of Periodic

    Functions)

    The Laplace transform of a piecewise

    continuous

    function f (t) with period p is

    L [f] =

    p

    tsps dttfee 0

    )(1

    1

    100

  • 7/29/2019 Unit IV Laplace Transforms

    19/26

    Unit I Laplace Transform

    L [f] = =

    +

    m

    n

    pn

    np

    tsdttfe

    m

    im

    0

    )1(

    )(

    = ( )

    =

    0 0)(

    n

    ptsnps

    dttfeem

    im(put t = t + np since f (t) is periodic with

    period p)

    =

    pts

    ts dttfee 0

    )(1

    1, because it is a geometric series.

    Therefore, L [f] =

    pts

    ts dttfee 0

    )(1

    1.

    Example 26 Find the Laplace transform of the saw-tooth wave given by

    p

    ktf =)(

    where 0 t p and f (t + p) = f (t) for t 0.

    Solution From theorem 1.8 we have

    L [f] =

    p tsts

    dteep

    k

    01

    1=

    01

    1p

    s

    e

    ep

    kts

    ts = ps

    k

    Therefore,

    t

    p

    kL =

    ps

    k.

    1.3 Differential Equations

    1.3.1 Ordinary Linear Differential Equations

    We shall now discuss how the Laplace transform method solves differential

    equations. We began with an initial value problem

    )(trbyyay =++ , y (0) = 0k and 1)0( ky =

    (1)

    with constants a and b. Here r (t) is the input (driving force) applied to the

    mechanical system and y (t) is the output (response of the system). In Laplace

    method we do three steps:

    Step 1 We transfer (1) by means of theorem 1.3 and 1.4, writing Y = L (y) and R

    = L (r). This gives

    [ ] )()0()0()0(2 sRbYysYayysYs =++

    (2)

    This is called the subsidiary equation.

    Collecting Y terms we have

    Prepared by Tekleyohannes Negussie 101

  • 7/29/2019 Unit IV Laplace Transforms

    20/26

    Unit I Laplace Transform

    ( ) ( ) )()0()0(2 sRyyasYbass +++=++

    (3)

    Step II We solve the subsidiary equation algebraically for Y. Division by

    bass ++2 and use

    of the so- called the transfer function

    basssQ

    ++=

    2

    1)(

    (4)gives the solution

    ( )[ ] )()()0()0( )( sQsQyyasY sR+++=

    (5)

    If y (0) = )0(y = 0, then this implies Y = QR; thus Q is the quotient

    R

    YsQ =)( =

    )(

    (

    inputL

    outputL

    (6)and this explains the name of Q.

    Note that: Q depends only on a and b, but neither on r (t) nor on the

    initial conditions.

    Step III We reduce (5) to a sum of terms whose inverse can be found from the

    table, so that

    the solution

    y (t) = ][1

    YL

    .

    Example 27 (initial value problem)

    Solve tyy = , y (0) = 1 and )0(y = 1

    Solution The subsidiary equation becomes

    21)0()0(2

    sYyysYs =

    Thus,2

    11)1( 2

    ssYs += ( )1

    1

    1

    122 2

    +

    =sss

    sY

    Prepared by Tekleyohannes Negussie 102

  • 7/29/2019 Unit IV Laplace Transforms

    21/26

    Unit I Laplace Transform

    ( )11

    1

    122

    ++

    =sss

    Y =

    +

    +

    2

    1

    1

    1

    1

    12

    sss

    Nowt

    es

    L =

    +111

    , ts

    L sinh1

    12

    1 =

    and ts

    L =

    2

    1 1.

    Therefore, y (t) = t te

    + sinh t.

    Example 28 (initial value problem)

    Solve teyyy=++ 2 , y (0) = 1 and )0(y = 1

    Solution The subsidiary equation becomes

    ( ) ( ) 11

    121

    2

    +=++++ sYsYsYs

    Thus,1

    11)12( 2

    ++=++

    ssYss 3)1(

    1

    )1(

    12 ++

    ++

    =ss

    sY

    3)1(

    1

    1

    1

    ++

    +=

    ssY

    Nowt

    es

    L =

    +111

    ,2

    3

    1

    2

    1

    )1(

    1t

    sL =

    + t

    e

    .

    Therefore, y (t) =t

    et

    12

    1 2.

    Example 29 (initial value problem)

    Solve tetyyy 3423 +=+ , y (0) = 1 and )0(y = 1

    Solution The subsidiary equation becomes

    ( ) ( )3

    142131

    2

    2

    +=++

    ssYsYsYs

    Thus,3

    144)23(

    2

    2

    ++=+

    sssYss

    ( ) ( )23)3(1

    23

    4

    23

    42222 +++++

    = ssssssss

    s

    Y

    Now2123

    42

    +

    =+

    =

    s

    B

    s

    A

    ss

    sY A + B = 1 and 2A + B = 4 A = 3 and

    B = 2

    ( )234

    22 + sss=

    s

    A+ 2s

    B+

    1sC

    +2s

    DA = 3, B = 2, C = 4 and D =

    1

    Prepared by Tekleyohannes Negussie 103

  • 7/29/2019 Unit IV Laplace Transforms

    22/26

    Unit I Laplace Transform

    and ( )23)3(12 + sss

    =3s

    A+

    1sB

    +2s

    CA = 0.5, B = 0.5 and C = 1.

    Hence, ( ) ( )23)3(1

    23

    4

    23

    42222 +

    ++

    ++

    ssssssss

    s

    =s3 + 2

    2s

    11

    21

    s

    22s

    +

    31

    21

    s

    Now 331 =

    s

    L , ts

    L 222

    1 =

    ,

    1

    1

    2

    11

    sL =

    te2

    1and

    3

    1

    2

    11

    sL =

    te 3

    2

    1.

    Therefore, y (t) = 3 + 2t te2

    1 te 22 +

    te 3

    2

    1.

    Example 30 (initial value problem)

    Solve tt

    eyyy sin52=++ , y (0) = 0 and )0(y = 1.

    Solution The subsidiary equation becomes

    ( ) ( )1)1(

    1521

    2

    2

    ++=++

    sYsYYs

    Thus, ( )1)1(

    114)1(

    2

    2

    +++=++

    sYs ( )( )4)1(1)1(

    1

    4)1(

    1222 ++++

    +++

    =sss

    Y

    Now ( )( )4)1(1)1(1

    22 ++++ ss=

    4)1(2 +++

    s

    BAs+

    1)1(2 ++

    +s

    DCs

    1)52()2522()22()( 23 =+++++++++++ DBsDCBAsDCBAsCA

    A = C = 0, B =3

    1 and D =

    3

    1.

    Hence, Y =

    ++

    ++

    +

    ++ 4)1(1

    1)1(

    1

    3

    1

    4)1(

    1222 sss

    =

    ++

    +++

    4)1(

    2

    1)1(

    1

    3

    122 ss

    Now

    ++

    42)1(

    21

    sL = tte 2sin and

    ++

    12)1(

    11

    sL = tte sin

    Therefore, y (t) =t

    ett+ )2sinsin(

    3

    1.

    Example 31 (initial value problem)

    Solve tyyy sin1022 =++ , y (0) = 0 and )0(y = 1.

    Solution The subsidiary equation becomes

    Prepared by Tekleyohannes Negussie 104

  • 7/29/2019 Unit IV Laplace Transforms

    23/26

    Unit I Laplace Transform

    ( )1

    10221

    22

    +=++

    sYsYYs

    Thus, ( )1

    1011)1(

    2

    2

    ++=++

    sYs ( )( )11)1(

    10

    1)1(

    1222 +++

    +++

    =sss

    Y

    Now ( )( )11)1( 10 22 +++ ss = 1)1( 2 ++ +s BAs + 12 ++s DCs

    10)2()22()2()( 23 =++++++++ + DBsDCAsDCBsCA

    A = 4, B = 6, C = 4 and D = 2.

    Hence, Y =

    +

    +

    +

    ++

    ++

    ++ 11

    211)1(

    14

    1)1(

    32222 ss

    s

    s

    s

    s

    Now

    +

    12

    11

    s

    L = tsin and

    ++

    +

    12)1(

    11

    s

    sL = t

    te sin

    Therefore, y (t) = tetttt ++ )cos4sin3(cos4sin2 .

    Example 32 (initial value problem)

    Solve tyy cosh189 =+ , y (0) = 2 and )0(y = 0.

    Solution The subsidiary equation becomes

    ( )1

    1892

    22

    =+

    s

    sYsYs

    Thus, ( )1

    1829

    2

    2

    +=+

    s

    ssYs ( )( )19

    18

    9

    2222 +

    ++

    =ss

    s

    s

    sY

    Now ( )( )1918

    22 + sss

    =92 +

    +s

    BAs+

    12 +

    s

    DCs

    sDBsCAsDBsCA 18)()9()()( 23 =++++++ +

    B = D = 0, A =5

    9 and B =

    5

    9.

    Hence, Y =

    +

    +

    + 9159

    9

    2222 s

    s

    s

    s

    s

    s=

    +

    +

    91

    9

    5

    122 s

    s

    s

    s

    Now

    121

    s

    sL = tshco and

    +

    921

    s

    sL = t3cos

    Therefore, y (t) = tt 3cos5

    1cosh

    5

    9+ .

    Example 33 (shifted data problem)

    Solve tyy 2''1

    =+ ,24

    =

    y and 224

    ' =

    y .

    Prepared by Tekleyohannes Negussie 105

  • 7/29/2019 Unit IV Laplace Transforms

    24/26

    Unit I Laplace Transform

    Solution set t = 0~

    tt + , so that4

    0

    =t .

    Now the subsidiary equation in terms of the new variable t~ becomes

    ssYyysYs

    2

    2~)0('~)0(~

    ~

    2

    2

    +=+

    ss

    YsYs2

    2~22

    2

    ~2

    2 +=++

    ss

    sYs2

    222

    2

    ~)1(

    22 +++=+

    )1(2)1(

    2

    1

    2

    1

    2

    )1(2

    ~222222 +

    ++

    ++

    +

    ++

    =sssssss

    sY

    Now)1(

    222 +ss=

    + 1

    11222 ss

    and)1(2 2 +ss

    =

    +

    11

    2 2ss

    s

    Hence,1

    2

    1

    2

    )1(2

    ~222 +

    +

    ++

    =sss

    sY

    +

    +

    1

    112

    22 ss+

    +

    1

    1

    2 2s

    s

    s

    =1

    22 +

    s

    + 22

    s+

    s2

    Therefore, y~ ( t~ ) = t

    ~sin2 + 2 t~ +

    2

    .

    Now in terms of the original variable t we get:

    y(t) = 2t + cos t sin t.

    Therefore, y(t) = 2t + cos t sin t.

    1.3.2 Systems of Linear Differential Equations

    Example 34 Solve211

    ' yyy += ,212

    ' yyy = , 1)0(1

    =y and 0)0(2

    =y .

    Solution The subsidiary equations become

    2111 )0( YYysY += and 2122 )0( YYysY = 1)1( 21 =+ YYs and0)1(

    12 =++YYs

    Solving for 1Y and 2Y algebraically we get

    )1()1()1( 212 +=++ sYsYs and 0)1( 12 =++ YYs

    )1(1)1( 12 +=++ sYs and 12

    1

    1Y

    sY

    +

    =

    1)1(

    121 ++

    +=

    s

    sY and

    1)1(

    122 ++

    =s

    Y

    Prepared by Tekleyohannes Negussie 106

  • 7/29/2019 Unit IV Laplace Transforms

    25/26

    Unit I Laplace Transform

    Now,

    +++

    1)1(

    12

    1

    s

    sL =

    te

    cos t and

    ++

    1)1(

    12

    1

    sL =

    te

    sin t

    Therefore, tt

    ety cos1 )(= and ttety sin2 )(

    = .

    Example 35 Solve211

    5' yyy += ,212

    5' yyy += , 3)0(1

    =y and 7)0(2

    =y .

    Solution The subsidiary equations become

    2111 5)0( YYysY += and 2122 5)0( YYysY +=

    3)5( 21 = YYs and 7)5( 12 = YYs

    Solving for 1Y and 2Y algebraically we get

    )5(3)5()5( 212 = sYsYs and 7)5( 12 = YYs

    )5(371)5( 12 = sYs and 12

    5

    1

    5

    7Y

    ssY

    +

    =

    = 1)5(5

    31)5(

    7221 s

    s

    sY and

    ( )

    +

    =

    1)5(

    3

    5

    7

    1)5()5(

    7222 ssss

    Y

    Now( ) 5

    7

    1)5(

    57

    1)5()5(

    722

    = ss

    s

    ss.

    Hence,

    =1)5(

    3

    1)5(

    57

    222 ss

    sY .

    Now,

    1)5(52

    1

    s

    sL = te 5 cosh t and

    1)5(1

    2

    1

    sL = te 5 sinh t

    Therefore, )cosh3sinh7(5

    1 )( ttt

    ety = and )sinh3cosh7(52 )( ttt

    ety = .

    Example 36 Solve tyy 2cos5''21

    =+ , tyy 2cos5''12=+ , 1)0(

    1=y , 1)0('

    1=y , 1)0(

    2=y

    and 1)0('2

    =y .

    Solution The subsidiary equations become

    4

    5)0(')0(

    22111

    2

    +=+

    s

    sYyysYs and

    4

    5)0(')0(

    21222

    2

    +=+

    s

    sYyysYs

    ( )45

    1221

    2

    ++=+

    s

    ssYYs and

    4

    51

    2122

    ++=+

    s

    ssYYs

    Solving for 1Y and 2Y algebraically we get

    ( )4

    52

    332

    22

    14

    ++=+

    s

    sssYsYs and

    4

    51

    2122

    ++=

    s

    ssYYs

    Prepared by Tekleyohannes Negussie 107

  • 7/29/2019 Unit IV Laplace Transforms

    26/26

    Unit I Laplace Transform

    4

    5

    4

    51)1(

    22

    332

    14

    +

    ++++=

    s

    s

    s

    ssssYs and ( ) 1

    222 4

    51 Ys

    s

    ssY

    ++=

    +

    +

    +=

    4

    5

    )1(

    1

    11

    122221 s

    s

    ss

    s

    sY

    Now41)4)(1(

    52222 +++

    +=

    + sDCs

    s

    BAs

    ss

    s 41)4)(1(

    52222 +

    =+ s

    s

    s

    s

    ss

    s

    Hence,41

    1221 +

    ++

    =s

    s

    sY and

    41

    1222 +

    +

    =s

    s

    sY

    Therefore, ttty 2cossin)(1 += and ttty 2cossin)(2 = .

    Example 37 Solve211

    3'' yyy += , teyy 44''12= , 2)0(

    1=y , 3)0('

    1=y , 1)0(

    2=y and

    2)0('2

    =y . Solution The subsidiary equations become

    211112 3)0(')0( YYyysYs += and

    144)0(')0( 1222

    2

    =

    sYyysYs

    sYYs 233)1( 212 += and

    1

    424 12

    2

    +=

    ssYYs

    Solving for 1Y and 2Y algebraically we get

    sYYs 81212)1(4 212 += and

    )1(4)1()2()1(4)1( 212

    222 ++= sssYsYss

    2

    12 =

    sY and

    2

    1

    1

    11

    +

    =ss

    Y

    Therefore, tt eety 21 )( += andtety 22 )( = .