Unit step function from piecewise

  • Thread starter cdotter
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  • #1
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Homework Statement


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

Homework Equations


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

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?
 

Answers and Replies

  • #2
305
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Is [itex]u_{\pi}(t)(t-\pi) - u_{2 \pi}(t)(t- \pi)[/itex] allowed? That does work but the "c" doesn't match.
 
  • #3
305
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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]
 
  • #4
SammyS
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What the exact wording of the question?
 
  • #5
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What the exact wording of the question?
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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