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Laplace transform Heaviside step function

  1. Feb 28, 2009 #1
    1. The problem statement, all variables and given/known data
    What is the Laplace transform of f(t) = 0 for 0 < t < 2 and f(t) = (4-t) for 2 < t < 3 and f(t) = 1 for 3 < t < 4 and f(t) = (5-t) for 4 < t < 5 and f(t) = 0 for t > 5?

    2. Relevant equations
    7410747ec7563eab51f608f2c80a9497.png

    3. The attempt at a solution
    f(t) = H(t-2)(4-t) - H(t-3)(4-t) + H(t-3) - H(t-4) + H(t-4)(5-t) - H(t-5)(5-t)
    f(t) = H(t-2)(4-t) + H(t-3)(2t-7) - H(t-4)(t-4) + H(t-5)(t-5)

    How can I rewrite H(t-2)(4-t) + H(t-3)(2t-7) to a form suitable for Laplace transform?
     
  2. jcsd
  3. Feb 28, 2009 #2

    Tom Mattson

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    For [itex]H(t-2)(4-t)[/itex], you'd like to see a [itex]t-2[/itex] in the second factor right? So just put one there, and balance it out elsewhere in the expression. Like this:

    [itex]H(t-2)(4-(t-2)-2)[/itex]

    See how I added and subtracted 2? This simplifies to:

    [itex]H(t-2)(2-(t-2))[/itex]

    Do the same for [itex]H(t-3)(2t-7)[/itex].
     
  4. Feb 28, 2009 #3
    Thank you for your reply!

    Now I am getting:
    [itex]
    H(t-2)(2-(t-2)) + H(t-3)(2(t-3)-1) - H(t-4)(t-4) + H(t-5)(t-5)
    [/itex]

    And the Laplace transform:
    [itex]
    ((2/s)-(1/s^2))e^(-2s) + ((2/s^2)-(1/s))e^(-3s) + ((-1/s^2)e^(-4s)) + ((1/s^2)e^(-5s))
    [/itex]

    Is this correct?
     
  5. Feb 28, 2009 #4

    Tom Mattson

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    Looks good to me.
     
  6. Feb 28, 2009 #5
    Thanks a lot!
     
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