Unit IV Laplace Transforms

  • View
    215

  • Download
    0

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