How can we take the LaPlace transform of a piecewise function?
How can we take the LaPlace transform of a function, given piece-wise function notation?
For example, $f(t)=\begin{cases} 0 &\mbox{for } 0<t<2\\ t&\mbox{ for } 2<t \end{cases}$
Frankly, I've read about step-functions but I can't find anything that really breaks down how these should be solved.
This question gives hints: Laplace Transformations of a piecewise function
But how do we REALLY come up with these integrals?
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$\begingroup$From the definition of the Laplace transform, and letting $f$ be the example function
$$f(t) = \left\{\begin{array}{cc} 0 & 0 < t < 2 \\ t & t \ge 2\end{array}\right.$$ we have
\begin{align*} L\{f(t)\} &= \int_0^{\infty} e^{-st} f(t) dt \\ &= \int_0^2 e^{-st} f(t) dt + \int_2^{\infty} e^{-st} f(t) dt \\ &= \int_0^2 e^{-st} \cdot 0 dt + \int_2^{\infty} e^{-st} t dt \\ &= \int_2^{\infty} t e^{-st} dt \end{align*}
and the Laplace transform follows from just computing the integral. For any general piecewise function for which the integrals make sense, one just integrates the function on each separate interval of definition.
$\endgroup$ 4 $\begingroup$A more complex example, in the step-function form: $$ f(t) = \left\{\begin{array}{cc} \sin t & 0 < t < 2 \\ e^{-t} & t \ge 2\end{array}\right. $$ Then $f(t) = \sin t + (e^{-t}-\sin t)u(t-2)$
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