Anonymous asked in Science & MathematicsEngineering · 6 years ago

How to calculate f (t) = cos at, t >= 0 by the definition of Laplace?

3 Answers

  • 6 years ago
    Favorite Answer

    F(s) = integral from t = 0 to infinity of

    [cos(at)] * e^(-st) dt

    Integrate by parts, letting

    u = cos(at) and dv = e^(-st) dt


    du = -a sin(at) dt and v = (-1/s) e^(-st)

    The integral is

    uv - integral of v du

    = [cos(at) * (-1/s)e^(-st)] +

    integral of (-a/s) [sin(at)*e^(-st)] dt

    The first term, evaluated at t = infinity and zero,

    comes out to just 1/s.

    The second term must be integrated by parts a second time,

    using u2 = a sin(at) and dv2 = (-1/s)e^(-st) dt, so

    du2 = a^2 cos(at) and v2 = (-1/s^2)e^(-st)

    Then the original integral must be

    1/s - (a/s^2)*sin(at)e^(-st) -

    (a^2/s^2) integral of [ (cos at) e^(-st) ] dt

    ADD the integral from the RHS to the original integral on the LHS, obtaining:

    [1 + (a^2/s^2)] * F(s) =

    1/s + [(a/s^2)*(sin at)*e^(-st)]

    but note that the last term will be zero BOTH at t = infinity and t = 0.

    So you have

    F(s) = (1/s) / (1 + a^2/s^2) ]

    = s / (s^2 + a^2)

    and luckily this coincides with the expected Laplace transform of cos(at).

  • Rick
    Lv 7
    6 years ago

    1. Compute the Laplace transform of f(t).

    2. Now use the initial value theorem. See:

  • Anonymous
    6 years ago

    Step by step please.


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