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Math Help - Solving Laplace transforms from a given, linear graph

  1. #1
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    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?
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    Quote Originally Posted by zepto- View Post
    Given a discontinuous linear graph with a range of (0,1), how would go about finding the Laplace transform?
    How discontinous? if it is a finite set t_0,t_1,...t_n
    then just break up the integral to get

    \int_{0}^{t_0}e^{-st}f(t)dt+\int_{t_0}^{t_1}e^{-st}f(t)dt+...+\int_{t_{n-1}}^{t_n}e^{-st}f(t)dt
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    I think my probelm might be writing the Heaviside / Unit step functions in U(t) format.
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    Quote Originally Posted by zepto- View Post
    I think my probelm might be writing the Heaviside functions in U(t) format.
    Okay well Here is an example I guess

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

    The Lapalce transform is

    \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

    \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

    \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}=
    -\frac{e^{-(s-1)}}{s-1}+\frac{1}{s-1}+\frac{1}{s}e^{-s}
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by zepto- View Post
    Given a discontinuous linear graph with a range of (0,1), how would go about finding the Laplace transform?
    Can you post it?

    CB
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