Given a discontinuous linear graph with a range of (0,1), how would go about finding the Laplace transform?

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- Feb 26th 2010, 02:04 PMzepto-Solving Laplace transforms from a given, linear graph
Given a discontinuous linear graph with a range of (0,1), how would go about finding the Laplace transform?

- Feb 26th 2010, 02:10 PMTheEmptySet
- Feb 26th 2010, 02:18 PMzepto-
I think my probelm might be writing the Heaviside / Unit step functions in U(t) format.

- Feb 26th 2010, 02:31 PMTheEmptySet
Okay well Here is an example I guess

$\displaystyle f(t)=\begin{cases} e^t, 0 \le t\le 1 \\ e, t > 1\end{cases}$

The Lapalce transform is

$\displaystyle \mathcal{L}(f)=\int_{0}^{\infty}e^{-st}dt=\int_{0}^{1}e^{-st}f(t)dt+\int_{1}^{\infty}e^{-st}f(t)dt$

Now using the definition of f gives

$\displaystyle \mathcal{L}(f)=\int_{0}^{1}e^{-st}e^{t}dt+\int_{1}^{\infty}e^{-st}\cdot 1dt$

Now just evaluate each of these integrals to get

$\displaystyle \mathcal{L}(f)=\int_{0}^{1}e^{-(s-1)t}dt+\int_{1}^{\infty}e^{-st}dt=-\frac{1}{s-1}e^{-(s-1)t}\bigg|_{0}^{1}-\frac{1}{s}e^{-st}\bigg|_{1}^{\infty}=$

$\displaystyle -\frac{e^{-(s-1)}}{s-1}+\frac{1}{s-1}+\frac{1}{s}e^{-s}$ - Feb 26th 2010, 11:28 PMCaptainBlack