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Homework Help: Heavyside function for Sine

  1. Dec 10, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex] f(t) = \left\{ \begin{array}{rcl}
    5sin(t) & \mbox{for}
    & 0 < t < 2\pi \\
    0 & \mbox{for} & t > 2\pi
    \end{array}\right. [/tex]

    Now, the problem is about rewriting f(t). My friend and I decided that it had to be

    [tex] \dfrac{10 - 5e^{-2\pi s}}{s^2 + 1} [/tex]

    However, the answer turned out to be


    [tex] \dfrac{5 - 5e^{-2\pi s}}{s^2 + 1} [/tex]

    Any help towards understanding this would be greatly appreciated! (We assumed they divided the first part by [tex]2\pi[/tex] when we extended [tex]5sin(t)[/tex] to [tex]5sin(t - 2\pi)[/tex], but we don't understand why!)
     
    Last edited: Dec 10, 2009
  2. jcsd
  3. Dec 10, 2009 #2

    LCKurtz

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    Did you begin by writing

    [tex]f(t) = 5\sin t(u(t) - u(t-2\pi) )[/tex]?
     
  4. Dec 10, 2009 #3
    Yeah, but we thought in this case that u(t) was [tex]2\pi[/tex], I guess that wasn't the case? :P
     
  5. Dec 10, 2009 #4

    LCKurtz

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    Nope, I guess not. u(t) is either 0 or 1.
     
  6. Dec 10, 2009 #5
    Thank you very much! :)
     
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