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Thread: Laplace Transform of a Continuous Piecewise Function

  1. #1
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    Laplace Transform of a Continuous Piecewise Function

    Good afternoon, the question in question is;

    Consider the piecewise function;

    $\displaystyle f(t) = \left\{
    \begin{array}{c l}
    -4t+8 & t \leq 9\\
    5 & t>9\\
    \end{array}
    \right.
    $

    This can be written using step functions as :
    $\displaystyle f(t)=$ ______ + $\displaystyle u_9(t)$_______ .

    The Laplace Transform is then: $\displaystyle L(f(t))=F(s)$where
    $\displaystyle F(s)=________$
    Now for my work, I started by writing f(t) as a sum of functions.

    $\displaystyle f(t) = -4t+8+H(t-9)(5)$
    Where $\displaystyle H$ is the Heaviside Step function shifted by $\displaystyle +9$.

    Now to take the Laplace transform of this, I am trying to use the Second Shift theorem on the second term,

    $\displaystyle L(f(t-9)H(t-9)) = e^{-9s}F(s)$

    But in my case, $\displaystyle f(t-9)=5$, so shifting it to the right by $\displaystyle +9$ won't do anything to a horizontal line, so f(t)=5. So for my answer, I obtained:

    $\displaystyle L(f(t))=(-\frac{4}{s^2} + \frac{8}{s}) + e^{-9s}\frac{5}{s}$

    Which is not showing as correct, I am not entirely sure where I am going wrong! Any help would be appreciated!
    Last edited by Kasper; Nov 27th 2010 at 01:03 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Kasper View Post
    Good afternoon, the question in question is;



    Now for my work, I started by writing f(t) as a sum of functions.

    $\displaystyle f(t) = -4t+8+H(t-9)(5)$
    Where $\displaystyle H$ is the Heaviside Step function shifted by $\displaystyle +9$.
    No!

    $\displaystyle f(t) = -4t+8+H(t-9)(5-(-4t+8))$

    CB
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    No!

    $\displaystyle f(t) = -4t+8+H(t-9)(5-(-4t+8))$

    CB
    Err, I feel foolish. Thanks!
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