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Math Help - Inverse Laplace transform problem 1

  1. #1
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    Inverse Laplace transform problem 1

    Hello,

    I got the following differential equation :
    E_2'(t)+2{}E_2(t) = E_1(t) + E_1'(t)

    Transforming this into a Laplace equation and computing the transfer function, I get :
    H(s) = {{s+1} \over {s+2}}

    Where I get into trouble is to obtain the inverse transformation. It seems probably silly. I do not have any problem solving more complex equations using partial fractions, but trying, this does not seem to apply here :
    {{s+1} \over {s+2}} = {A \over {s+2}}



    What am I missing here?

    Thanks,

    Jurgen
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Take into account that:


    \displaystyle\lim_{s \to{+}\infty}{\dfrac{s+1}{s+2}}\neq 0



    Fernando Revilla
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  3. #3
    MHF Contributor chisigma's Avatar
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    ... and also that is...

    \displaystyle \frac{s+1}{s+2}= 1-\frac{1}{s+2}

    Kind regards

    \chi \sigma
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    ... and also we deduce that


    \mathcal{L}^{-1}\left\{{ \dfrac{s+1}{s+2} }\right\}

    does not exist.


    Fernando Revilla
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  5. #5
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    Reply : Inverse Laplace transform problem 1

    Why not? I have controlled the answer above yours, which says
        \displaystyle \frac{s+1}{s+2}= 1-\frac{1}{s+2}

    which then translates to

    \delta{(t)}-e^{-2t}
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  6. #6
    MHF Contributor chisigma's Avatar
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    It is important to observe that H(s) is a Transfer function, i.e. the ratio of the L-transforms of the 'output signal' e_{2} (t) and the 'input signal' e_{1} (t) . In this case the H(s) is a [simple] first order 'high pass filter'...

    Kind regards

    \chi \sigma
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  7. #7
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by chthon View Post
    \delta{(t)}-e^{-2t}
    Of course. The misunderstanding on my part: I read your problem only about the decomposition of:

    \dfrac{s+1}{s+2}

    into simple fractions to find its inverse as a piecewise continuous and exponential order function not as a distribution. For that reason I talked about the non zero limit.


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