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Math Help - Laplace Transform

  1. #1
    Super Member Aryth's Avatar
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    Laplace Transform

    I have only had a one hour lecture on these, so it is hard to think about these.

    Show that \cal{L} [f'(t)] = sF(s) - f(0)
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    MHF Contributor chisigma's Avatar
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    Integrating by parts we obtain...

    \mathcal {L} \{f^{'}(t)\}= \int_{0}^{\infty} e^{-st}\cdot f^{'}(t)\cdot dt =

    = |e^{-st}\cdot f(t)|_{0}^{\infty} + s\cdot \int_{0}^{\infty} e^{-st}\cdot f(t)\cdot dt =

     s\cdot \int_{0}^{\infty} e^{-st}\cdot f(t)\cdot dt - f(0)= s\cdot F(s) - f(0) (1)

    ... where we use the fact that if \mathcal{L} \{f(t)\} exists, then is...

    \lim_{p \rightarrow \infty} e^{-sp} f(p) =0 (2)

    Kind regards

    \chi \sigma
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  3. #3
    Super Member Aryth's Avatar
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    I appreciate the help. Thanks.
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    Super Member Aryth's Avatar
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    I have one question with inverse transforms.

    \cal{L} \left[\frac{s-1}{(s-3)(s+1)}\right]
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  5. #5
    MHF Contributor chisigma's Avatar
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    Applying the residue theorem we find...

    F(s)=\frac{s-1}{(s-3)\cdot (s+1)} = \frac{r(3)}{s-3} + \frac{r(-1)}{s+1} (1)

    ... where...

    r(3)= \lim_{s \rightarrow 3} F(s)\cdot (s-3)= \frac{1}{2}

    r(-1)= \lim_{s \rightarrow -1} F(s)\cdot (s+1)= \frac{1}{2} (2)

    ... so that is...

    \mathcal {L}^{-1} \{\frac{s-1}{(s-3)\cdot (s+1)} \}= \frac{1}{2}\cdot e^{-t} + \frac{1}{2}\cdot e^{3t} (3)

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    \chi \sigma
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    for the inverse transform use partial fractions to split the expression up. you can check if i did it correctly but i got (1/2)/(s-3) + (1/2)/(s+1). so you should be able to recognize these as laplace transforms of the exponential function. so you should get (1/2)e^(3t) + (1/2)e^(-t).
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  7. #7
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    the laplace transform of te^(-t) should be 1/(s+1)^2, not 1.
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  8. #8
    Super Member Aryth's Avatar
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    Quote Originally Posted by oblixps View Post
    the laplace transform of te^(-t) should be 1/(s+1)^2, not 1.
    I figured out that my solution was riddled with errors. Thanks for the help though. I appreciate it.
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