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Math Help - I can't understand Heaviside

  1. #1
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    I can't understand Heaviside

    Find the Laplace transformation:



    Simply telling me what to do won't help me. I went through the definitions and more than 5 examples and I can't understand how they undergo the Heaviside calculation steps. If someone could be kind enough to explain each step of what he's doing would be really nice!

    Thanx
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  2. #2
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    the heaviside function is defined by
    u_{c}(t)=\left\{<br />
\begin{array}<br />
[c]{cc}%<br />
0, & t<c\\<br />
1, & t\geq c<br />
\end{array}<br />
\right.

    using this, we get
    f(t)=t\left[  u_{0}(t)-u_{1}(t)\right]  +u_{1}(t)

    Calculate the Laplace transform of f(t) using the identity for Laplace transform of heaviside function.
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  3. #3
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    it's easy bro,

    i got

    F(s)=\frac{1-e^{-s}}{s^{2}}
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  4. #4
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    Quote Originally Posted by Provoke View Post
    Find the Laplace transformation:



    Simply telling me what to do won't help me. I went through the definitions and more than 5 examples and I can't understand how they undergo the Heaviside calculation steps. If someone could be kind enough to explain each step of what he's doing would be really nice!

    Thanx
    The Laplace transform of a function, f(x) is
    \int_0^\infty f(x)e^{-sx}dx

    f(t)=t\left[ u_{0}(t)-u_{1}(t)\right] +u_{1}(t)
    For t< 0 both u_0 and u_1 are 0 so f(x)= t(0- 0)+ 0= 0.

    For 0\le t< 1 u_0 is 1 and u_1 is 0 so f(x)= t(1- 0)+ 0= t.

    for 1\le t both u_0 and u_1 are 1 so f(x)= t(1-1)+ 1= 1.

    The Laplace transform of f is \int_0^1 t e^{-st}dt+ \int_1^\infty e^{-st}dt

    Integrate the first by parts with u= t, dv= e^{-st}dt so that du= dt and v= -\frac{1}{s}e^{-st}.

    \int_0^1 te^{-st}dt= \left[-\frac{1}{s}te^{-st}\right]_0^1+ \frac{1}{s}\int_0^1 e^{-st}dt
    = -\frac{1}{s}e^{-s}- \left[\frac{1}{s^2}(e^{-st}-1)\right]_0^1
    = -\frac{1}{s}e^{-s}- \frac{1}{s^2}(e^{-s}-1)

    The second integrates easily
    \int_1^\infty e^{-st}dt= \left[-\frac{1}{s}e^{-st}\right]_1^\infty
    = \frac{1}{s}e^{-s}

    The \frac{1}{s}e^{-s} terms cancel when we add those integrals, leaving -\frac{e^{-s}-1}{s^2} which is the same as dedust's result.
    Last edited by HallsofIvy; November 26th 2009 at 03:09 AM.
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