Unit step function from piecewise

[cdotter]

Homework Statement

<br /> f(x) = \begin{cases}<br /> 0, &amp; t &lt; \pi \\<br /> t - \pi , &amp; \pi \leq t &lt; 2 \pi \\<br /> 0, &amp; t \geq 2 \pi<br /> \end{cases}<br />

Homework Equations

Unit step function:
u_c(t) = \begin{cases}<br /> 0, &amp; t &lt; c \\<br /> 1 , &amp; t \geq c \\<br /> \end{cases}<br />

The Attempt at a Solution

u_{\pi}(t)(t-\pi) - u_{2 \pi}(t)(t-2 \pi) gives me:

<br /> f(x) = \begin{cases}<br /> 0, &amp; t &lt; \pi \\<br /> t - \pi , &amp; \pi \leq t &lt; 2 \pi \\<br /> \pi, &amp; t \geq 2 \pi<br /> \end{cases}<br />

I'm not sure what to do to get the last inequality correct?

 

[cdotter]

Is u_{\pi}(t)(t-\pi) - u_{2 \pi}(t)(t- \pi) allowed? That does work but the "c" doesn't match.

 

[cdotter]

I figured it out. It is allowed, it just needs forcing.

u_{\pi}(t)(t-\pi) - u_{2 \pi}(t)(t- \pi) \rightarrow u_{\pi}(t)(t-\pi) - u_{2 \pi}(t)((t- \pi) - \pi)

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

<br /> \begin{aligned}<br /> F(s) &amp;= L\{u_{\pi}t(t- \pi)-u_{2 \pi}(t)((t- \pi)- \pi)\}\\<br /> F(s) &amp;= e^{- \pi s}L\{t\} - e^{-2 \pi s}L\{t- \pi\}\\<br /> F(s) &amp;= \frac{e^{- \pi s}}{s^2} - e^{- 2 \pi s} \cdot (\frac{1}{s^2} - \frac{\pi}{s})<br /> \end{aligned}<br />

 

[SammyS]

What the exact wording of the question?

 

[cdotter]

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 L\{u_c(t)f(t-c)\} = e^{-cs}L\{f(t)\}, which I figured out.