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Laplace transform of Heaviside function

  1. Nov 13, 2013 #1
    1. The problem statement, all variables and given/known data

    What is the laplace transform of H(-t-17)

    2. Relevant equations

    Shifting theorem:

    L(H(t-a)) = (e^-as)/s

    3. The attempt at a solution

    This is the only part of the problem that I can not get (this part is from a larger differential equation I'm trying to solve). I'm can't seem to figure out how the Laplace transform changes if there is a negative sign in front of the t. I'm not sure if there are some properties of the Heaviside function that will allow me to solve this that I simply don't know and that we didn't go over in class.
     
    Last edited: Nov 13, 2013
  2. jcsd
  3. Nov 13, 2013 #2

    Mark44

    Staff: Mentor

    It's "Heaviside" which is now fixed.

    Your relevant equation is incorrect and has two errors. It should be L(H(t - a)) = (e-as)/s.

    H(t) is the unit step function. H(t - a) is the translation of the unit step function by a units. H(-t - 17) = H(-(t + 17)). This involves a reflection across the vertical axis, followed by a translation to the left by 17 units.
     
  4. Nov 13, 2013 #3

    pasmith

    User Avatar
    Homework Helper

    If [itex]t > 0[/itex] then [itex]-t -17 < 0[/itex] so [itex]H(-t-17) = 0[/itex] for all [itex]t > 0[/itex]. Since the Laplace transform only cares about the values of a function in [itex]t > 0[/itex] the Laplace transform of [itex]H(-t-17)[/itex] is equal to the Laplace transform of the zero function.
     
  5. Nov 13, 2013 #4
    Thank you both.


    I was actually able to pull up a nice identity online for H(x), where H(x) = (x + |x|)/2x. It helped a lot as well.
     
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