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Math Help - Tricky Inverse Laplace! Convolution perhaps?

  1. #1
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    Tricky Inverse Laplace! Convolution perhaps?

    Hello All,

    I have been working at the following for a while, trying to find the inverse of it, my working to date is shown below, although I suspect I may be approaching it wrong.

    <br /> <br />
F(s) = \frac{3e^{-3s}}{s(s+3)}<br /> <br />

    My train of thought so far is as follows;

    By using the convolution theorem and treating
    <br /> <br />
\mathcal{L}^{-1}(F(s)G(s)) = \int^{t}_{0} f(t)g(\tau-t)dt<br /> <br />

    <br />
G(s) = \frac{3}{s}<br /> <br />
    <br />
F(s) = \frac{e^{-3s}}{s+3}<br /> <br />
    from identity

    <br /> <br />
\mathcal {L}[H(t-a)y(t-a)]=e^{-as}Y(s)<br /> <br />

    let  Y(s)=\frac{1}{s+3}
    therefore
     {L}[H(t-3)(e^{-3t}-3)]=\frac{e^{-3s}}{s+3}<br />

    <br /> <br />
{L}^{-1}\left\{\frac{3e^{-3s}}{s\left(s+3\right)}\right\}= 3\int_0^t H(t-3)(e^{-3t}-3)dt<br /> <br />

    Any solutions or help would be appreciated, or pointers as to any possible errors.

    Thanks, Alex.
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  2. #2
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    Quote Originally Posted by arickard1988 View Post
    Hello All,

    I have been working at the following for a while, trying to find the inverse of it, my working to date is shown below, although I suspect I may be approaching it wrong.

    <br /> <br />
F(s) = \frac{3e^{-3s}}{s(s+3)}<br /> <br />

    My train of thought so far is as follows;

    By using the convolution theorem and treating
    <br /> <br />
\mathcal{L}^{-1}(F(s)G(s)) = \int^{t}_{0} f(t)g(\tau-t)dt<br /> <br />

    <br />
G(s) = \frac{3}{s}<br /> <br />
    <br />
F(s) = \frac{e^{-3s}}{s+3}<br /> <br />
    from identity

    <br /> <br />
\mathcal {L}[H(t-a)y(t-a)]=e^{-as}Y(s)<br /> <br />

    let  Y(s)=\frac{1}{s+3}
    therefore
     {L}[H(t-3)(e^{-3t}-3)]=\frac{e^{-3s}}{s+3}<br />

    <br /> <br />
{L}^{-1}\left\{\frac{3e^{-3s}}{s\left(s+3\right)}\right\}= 3\int_0^t H(t-3)(e^{-3t}-3)dt<br /> <br />

    Any solutions or help would be appreciated, or pointers as to any possible errors.

    Thanks, Alex.
    Can't you just use LT^{-1}\left[ e^{-as} G(s) \right] = H(t - a) g(t - a) where g(t) = LT^{-1} [G(s)].
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  3. #3
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    fantastic,

    cool, using that;

    <br /> <br />
\mathcal{L}^{-1}(F(s))=f(t)=3H(t-3)[1-e^{-(t-3)}]<br /> <br />

    However, using Matlab to give an answer, it gives something different, but similar; the following:

    <br /> <br />
f(t)=-3H(t-1)(\frac{e^{3-3t}}{\frac{8}{3}})<br /> <br />

    I am not quite sure where I am making an error, please give a pointer.

    Thanks, Alex.
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  4. #4
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    Quote Originally Posted by arickard1988 View Post
    fantastic,

    cool, using that;

    <br /> <br />
\mathcal{L}^{-1}(F(s))=f(t)=3H(t-3)[1-e^{-(t-3)}]<br /> <br />

    However, using Matlab to give an answer, it gives something different, but similar; the following:

    <br /> <br />
f(t)=-3H(t-1)(\frac{e^{3-3t}}{\frac{8}{3}})<br /> <br />

    I am not quite sure where I am making an error, please give a pointer.

    Thanks, Alex.
    You will need to show your working because I get H(t - 3) \left(1 - e^{-3(t-3)}\right) (and this is the correct answer ).

    Your Matlab result is wrong (probbaly because of a data entry error).
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  5. #5
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    Fantastic,

    Thanku, I have looked over the syntax and corrected it. I simply had the factor of 3 in the wrong place, al sorted now.

    Alex
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