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Math Help - Laplace transform integral equation

  1. #1
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    Laplace transform integral equation

    Determine i(t)

    {i}'(t) + 4i(t) + 5\int_{0}^{t}i(\tau )d\tau = \delta (t-2)

    t>0, i(0)=0 where \delta (t-2) is a dirac delta function


    i(t) is the current, with laplace transformation i got i(t) = \frac{1}{2}e^{((-2-i)(t-2)))}((1+2i)e^{(2i(t-2))}+(1-2i))\theta (t-2)
    where \theta (t-2) is a heaviside function

    then they ask of \lim_{t \to \infty }\int_{0}^{t}i(\tau )d\tau i think it will be 0 but how do i see that? do i really need to integrate my i(t)?

    regards!
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  2. #2
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Laplace transform integral equation

     \lim_{t \to \infty }\int_{0}^{t}i(\tau )d\tau \to 0

    Attached Thumbnails Attached Thumbnails Laplace transform integral equation-laplace-transform.png  
    Last edited by MaxJasper; October 4th 2012 at 02:13 PM.
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  3. #3
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    Re: Laplace transform integral equation

    Quote Originally Posted by Skruven View Post
    Determine i(t)

    {i}'(t) + 4i(t) + 5\int_{0}^{t}i(\tau )d\tau = \delta (t-2)

    t>0, i(0)=0 where \delta (t-2) is a dirac delta function


    i(t) is the current, with laplace transformation i got i(t) = \frac{1}{2}e^{((-2-i)(t-2)))}((1+2i)e^{(2i(t-2))}+(1-2i))\theta (t-2)
    where \theta (t-2) is a heaviside function

    then they ask of \lim_{t \to \infty }\int_{0}^{t}i(\tau )d\tau i think it will be 0 but how do i see that? do i really need to integrate my i(t)?

    regards!
    Note: It is really bad style or notation to use i as a function name and then use i as the imaginary unit.

    I don't recall the name of the theorem but If all of the poles of the laplace transform are in the left half plane, then

    \lim_{t \to \infty}f(t) =\lim_{s \to 0^{+}}sF(s)

    So in your case if you define g(t) =\int_{0}^{t}i(\tau)d\tau then

    \lim_{t \to \infty}g(t)=\lim_{s \to 0^+}s\mathcal{L} \{\int_{0}^{t}i(\tau)d\tau\}=\lim_{s \to 0^+}I(s)

    Now if you take the laplace transform you get

    If you take the laplace transform you get

    sI+4I+\frac{5}{s}I=e^{2s} \iff (s^2+4s+5)I=se^{2s}

    I = \frac{s}{(s+2)^2+1}e^{2s}

    So

    \lim_{s \to 0^+}\frac{s}{(s+2)^2+1}e^{2s}=0
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