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Homework Help: 2 Questions on inverse Laplace Transformation

  1. Nov 26, 2006 #1
    1] Find the inverse Laplace transform of the given function, i. e.
    [tex]L^{-1}\left[1\over {s^3 + 1}}\right][/tex]

    And this is how I proceeded about:
    [tex]L^{-1}\left[1\over {s^3 + 1}}\right][/tex]

    [tex]= L^{-1}\left[\frac{1}{(s+1)(s^2 - s +1)}\right][/tex] (By factorising the denominator)

    I am stuck at this step and can't proceed furthur. Someone help.

    2]Find the inverse Laplace Transform of the given function, i. e.
    [tex]L^{-1}\left[9\over {s(s^2 + 9)}}\right][/tex]

    I did solve this one, just want to know if its correct.
    [tex]L^{-1}\left[9\over {s(s^2 + 9)}}\right][/tex]

    [tex]= 3L^{-1}\left[3\over {s(s^2 + 9)}}\right][/tex]

    We know:
    [tex]L^{-1}\left[3\over {s^2 + 9}}\right] = \sin 3t = F(t)[/tex]
    [tex]L^{-1}\left[1\over s}\right] = 1 = G(t)[/tex]

    By the convolution theorem;
    [tex]L^{-1}{f(s)} = F(t) * G(t) = \int_0^t G(\tau)F(t - \tau)d\tau[/tex]

    [tex]L^{-1}\left[9\over {s(s^2 + 9)}}\right][/tex]
    [tex]=3\int_0^t sin3(t - \tau)d\tau = (\cos 3\tau - 1)[/tex]

    Please care to see if this is right. Thanks in advance.
    Last edited: Nov 26, 2006
  2. jcsd
  3. Nov 26, 2006 #2


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    Partial fractions, of course!
    [tex]\frac{1}{(s+1)(s^2-s+1)}= \frac{A}{s+1}+ \frac{Bs+ C}{s^2- s+ 1}[/tex]
    s2- s+ 1 cannot be factored using real coefficients but you can complete the square: s2- s+ 1= (s- 1/2)2+ 3/4.

    Well, it is [itex] cos (3t)- 1[/itex], not [itex]cos(3\tau )-1[/itex]!

    Again, I would have done that by partial fractions:
    [tex]\frac{9}{s(s^2+ 9)}= \frac{1}{s}- \frac{s}{s^2+ 9}[/itex]
    which then gives what you have.
  4. Nov 26, 2006 #3
    Cool, that makes sense. I will follow the technique you've suggested. Thank you so very much for your time.
  5. Dec 1, 2006 #4

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    prakash chennai
  6. Dec 2, 2006 #5
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