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Homework Help: Laplace transforms, heaviside step functions

  1. Mar 8, 2010 #1
    Hey guys! I just have a question regarding some notes I've been looking at and trying to understand...

    1. The problem statement, all variables and given/known data
    eq0018MP.gif
    Taking the transform:

    eq0019MP.gif

    Just wondering, can someone explain to me why the e-3s/s is positive, and not negative because it is originally -u3(t)...

    Another quick one:

    Write this in terms of Heaviside function:

    f(t) =
    { 0, 0 <= t <= pi
    { -sin3t, t >= pi

    1) So you would get f(t) = -(upit)sin3t

    And somehow it ends up as f(t) = (upit)sin(3(t-pi))
    Can someone explain this step as well? Just what happened to the negative in front of it, and why it became 3(t-pi)

    Thanks!
     
  2. jcsd
  3. Mar 9, 2010 #2

    vela

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    It's probably just a typo in the original problem.

    Try expanding sin[3(t-pi)] using the angle addition formula for sine. The reason it was written this way is probably because it's now in the form where the time-shift property of the Laplace transform can be applied.
     

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  4. Mar 9, 2010 #3
    Thanks for the first one that makes sense.

    For the second one, I attempted using the identity and get:

    f(t) = -upi(t)sin(3t)
    = -L{upi(t)sin3(t-pi+pi)}
    = -L{upi(t)[sin3(t-pi)cos(3pi) + sin(3pi)cos3(t-pi)]}
    taking the transform:
    = -e-pi*s(3cos(3pi) + 3s*sin(3pi) / s2 + 9

    Hmm I don't understand how it is suppose to be just upi(t)sin(3(t-pi)) = e-pi*s(3) / (s2 + 9)(s2 + 1)

    What happened when you add the pi and subtract it, and what happened to the negative in front of it all?
     
  5. Mar 9, 2010 #4

    vela

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    What are cos(3π) and sin(3π) equal to? Also, I didn't mean that you needed to use the time-shift property to get that result. It's just basic trig.
     
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