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Laplace Transforms

  1. Jul 23, 2011 #1
    Ok, so I have uploaded/attached the question and the solution. I just need help understanding the solution please. I understand how to calculate the initial inverse transform, but I included it as the reference to the second part of the question regarding the y'' + 4y' = H(t-3)

    Can someone please explain the full steps to invert the laplace transform Y to f, like the solution shows in the last step?
    I have got to the following point, but I think I may be forgetting some Laplace Transform identities needed to make my life easier?

    I can split the equation into parts where I recognise a few Laplace Transforms but not sure about the rest, cheers...

    Y(s) = (1/s).e-3s.[1/(s2+22)] + [s/(s2+22)] - [2/s2+22]


    I recognise a few inverse laplace transforms there but without adding my confusion to the mess can someone please clarify how they got the answer? many thanks in advance.
    hopefully I didnt make this post toooo convoluted with my thoughts!
     

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    Last edited: Jul 23, 2011
  2. jcsd
  3. Jul 23, 2011 #2
    U are confused because they have skipped a step.

    I bet u must have recognised this part: [s/(s2+22)] - [2/s2+22]

    For this part: (1/s).e-3s.[1/(s2+22)] :
    (1/(s(s²+2²)))
    =(1/4)((1/s)-(s/(s²+2²)))
    =(1/4)(H(t-3)-cos(2(t-3)))
     
  4. Jul 23, 2011 #3

    How did you get rid of the e^(-3s) ?

    thanks for helping.
     
  5. Jul 23, 2011 #4
    You are most welcome adam :)

    [tex]L^{-1}(e^{-cs}F(s))=f(t-c)[/tex]
    Given :
    [tex] L^{-1}(F(s))=f(t) [/tex]

    :smile:
     
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