Laplace transformation of f(t)=t^2cos(bt) and f(t)=|sint|?

Laplace transformation of f(t)=t^2cos(bt) and f(t)=|sint|

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  • 8 years ago
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    #1 f(t) = t²∙cos(b∙t)

    The Laplace transform for cosine is well known and can be found in every transform table:

    ℒ{ cos(b∙t) } = s/(s² + b²)

    The factor t² corresponds to a differentiation of the transform of cos(b∙t) with respect to s, according to the theorem:

    ℒ{ t∙f(t) } = - d/ds[ ℒ{ f(t) } ]

    =>

    ℒ{ tⁿ∙f(t) } = (-1)ⁿ∙dⁿ/dsⁿ[ ℒ{ f(t) } ]

    Therefore

    ℒ{ t²∙cos(b∙t) } = d²/ds²[ ℒ{ cos(b∙t) } ]

    = d²/ds²[ s/(s² + b²) ]

    = d/ds[ (1∙(s² + b²) - s∙(2∙s))/(s² + b²)² ]

    = d/ds[ ((b² - s²)/(s² + b²) ]

    = ( (2∙s)∙(s² + b²) - (b² - s²)∙2∙(2∙s) ) / (s² + b²)

    = (6∙s³ - 2∙b²∙s)/(s² + b²)

    #2 f(t) = |sin(t)|

    f(t) =

    sin(t) for 0 ≤ t ≤ π

    - sin(t) for π ≤ t ≤ 2∙π

    sin(t) for 2∙π ≤ t ≤ 3∙π

    -sin(t) for 3∙π ≤ t ≤ 4∙π

    ...

    Since

    sin(t + π) = -sin(t)

    you can rewrite the function as simple periodic function with period π

    f(t) = sin(t) 0 ≤ t ≤ π

    and

    f(t + π) = f(t) for t > π

    The Laplace transform of period function with period T such that

    f(t + T ) = f(t)

    is given by

    ........................................ T

    ℒ{ f(t) } = ( 1/(1 - e^(-s∙t)) ) ∙ ∫ e^(-s∙t)∙f(t) dt

    ........................................ 0

    Therefore

    .............................................. π

    ℒ{ |sin(t)| } = ( 1/(1 - e^(-s∙π)) ) ∙ ∫ e^(-s∙t)∙sin(t) dt

    .............................................. 0

    with

    ∫ e^(-s∙t)∙sin(t) dt = -e^(-s∙t)∙(s∙sin(t) + cos(t))/(s² + 1)

    follows

    ℒ{ |sin(t)| }

    = ( 1/(1 - e^(-s∙π)) ) ∙ (-e^(-s∙π)∙(s∙sin(π) + cos(π))/(s² + 1) + e^(-s∙0)∙(s∙sin(0) + cos(0))/(s² + 1) )

    = ( 1/(1 - e^(-s∙π)) ) ∙ (- e^(-s∙π)∙(-1) + 1∙1) / (s² + 1)

    = ( (1 + e^(-s∙π))/(1 - e^(-s∙π)) ) ∙ ( 1/(s² + 1) )

  • Anonymous
    4 years ago

    because g(t) = 0 whilst a million <= t , the Laplace essential: int_{0}^{infty} ( exp( -s*t ) * g(t) ) dt = int_{0}^{a million} ( exp( -s*t ) * (t^3) ) dt Use integration via aspects three times to kill the t^3 area of the integrand. on the different hand, you ought to apply the step function U(t) and say that g(t) = ( t^3 ) * [ U(t) - U(t-a million) ] the place U(t) = 0 whilst t<0 and U(t) = a million whilst t>=0. desire that facilitates.

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