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Unit step function from piecewise

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data
    [tex]
    f(x) = \begin{cases}
    0, & t < \pi \\
    t - \pi , & \pi \leq t < 2 \pi \\
    0, & t \geq 2 \pi
    \end{cases}
    [/tex]

    2. Relevant equations
    Unit step function:
    [tex]u_c(t) = \begin{cases}
    0, & t < c \\
    1 , & t \geq c \\
    \end{cases}
    [/tex]

    3. The attempt at a solution

    [itex]u_{\pi}(t)(t-\pi) - u_{2 \pi}(t)(t-2 \pi)[/itex] gives me:

    [tex]
    f(x) = \begin{cases}
    0, & t < \pi \\
    t - \pi , & \pi \leq t < 2 \pi \\
    \pi, & t \geq 2 \pi
    \end{cases}
    [/tex]

    I'm not sure what to do to get the last inequality correct?
     
  2. jcsd
  3. Oct 24, 2011 #2
    Is [itex]u_{\pi}(t)(t-\pi) - u_{2 \pi}(t)(t- \pi)[/itex] allowed? That does work but the "c" doesn't match.
     
  4. Oct 24, 2011 #3
    I figured it out. :smile: It is allowed, it just needs forcing.

    [itex]u_{\pi}(t)(t-\pi) - u_{2 \pi}(t)(t- \pi) \rightarrow u_{\pi}(t)(t-\pi) - u_{2 \pi}(t)((t- \pi) - \pi)[/itex]

    Now that it's in the form [itex]u_c(t)(t-c)[/itex], the Laplace transform can be taken:

    [tex]
    \begin{aligned}
    F(s) &= L\{u_{\pi}t(t- \pi)-u_{2 \pi}(t)((t- \pi)- \pi)\}\\
    F(s) &= e^{- \pi s}L\{t\} - e^{-2 \pi s}L\{t- \pi\}\\
    F(s) &= \frac{e^{- \pi s}}{s^2} - e^{- 2 \pi s} \cdot (\frac{1}{s^2} - \frac{\pi}{s})
    \end{aligned}
    [/tex]
     
  5. Oct 24, 2011 #4

    SammyS

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    What the exact wording of the question?
     
  6. Oct 25, 2011 #5
    The question asked to take the Laplace transform of the piecewise function. I had to create a step function and use [itex]L\{u_c(t)f(t-c)\} = e^{-cs}L\{f(t)\}[/itex], which I figured out.
     
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