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Math Help - 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;

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

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

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

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

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

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

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

    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; November 27th 2010 at 02:03 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.

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

    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!

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

    CB
    Err, I feel foolish. Thanks!
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