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Math Help - How to verify a Laplace transform?

  1. #1
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    How to verify a Laplace transform?

    <br /> <br />
\mathcal{L}\left\{e^{at}\cos\!\left(kt\right)\righ  t\}=\frac{s-a}{\left(s-a\right)^2+k^2}<br />

    What is the proof of this? Or rather, could somebody get me started and point me in the right direction for this?

    Thanks! Any help is appreciated.
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  2. #2
    MHF Contributor chisigma's Avatar
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    One possible start is the computation of the integral...

    \mathcal{L} \{\cos kt\} = \int_{0}^{\infty} \cos kt\cdot e^{-st}\cdot dt

    ... that can be done by parts...

    Kind regards

    \chi \sigma
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  3. #3
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    But what do you do with the e^at? Does that fit somewhere in the integral or no?
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by tibetan-knight View Post

    But what do you do with the e^at? Does that fit somewhere in the integral or no?
    yes...

    \int_{0}^{\infty}e^{at}e^{-st}cos(kt)dt=\int_{0}^{\infty}e^{-(s-a)t}cos(kt)dt


    Now just use integration by parts on this

    Or note in general that

    \mathcal{L}(e^{at}f(t))=\int_{0}^{\infty}e^{at}e^{-st}f(t)dt

    \mathcal{L}(e^{at}f(t))=\int_{0}^{\infty}e^{-(s-a)t}f(t)dt

    Let s-a=m then we get

    \mathcal{L}(e^{at}f(t))=\int_{0}^{\infty}e^{-mt}f(t)dt=F(m)
    The above integral is the form of the laplace transform so we get
    Where F(m)=F(s-a)

    This is commonly know as the s-axis shift theorem.

    So the above would give

    \mathcal{L}(e^{at}cos(kt))=\mathcal{L}(\cos(kt))\b  igg|_{s\to s-a}=\frac{s}{s^2+k^2}\bigg|_{s \to s-a}=\frac{s-a}{(s-a)^2+k^2}
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  5. #5
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    Ooh. Shift theorem is helpful. Thank you.

    But, to integrate the former by parts, which would be the 'u' part and which would be the 'dv' part?
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  6. #6
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    Wat the "TheEmptySet" has showed is perfectly right but i like to show the same proof in another way ...We know by theorem Laplace {{e^at} f(t)}=F(s-a)
    Proof of the theorem:
    We know L{f(t)}=Intergral 0-infinity ({e^-st} f(t) dt)=F(s)
    Therefore Intergral {e^at f(t)} = Intergral 0-infinity ({e^-st}{e^at} f(t) dt)
    =Intergral 0-infinity ({e^-(s-a)t} f(t) dt)
    =F(s-a) {Reason:Since Intergral 0-infinity ({e^-st} f(t) dt)=F(s)}
    Hence Laplace {{e^at} f(t)}=F(s-a)..
    In the qn here in the place of f(t) we have coskt..
    Therefore we know laplace {coskt}=s/(s^2-k^2) = F(s)
    Therefore F(s-a)=(s-a)/{(s-a)^2-k^2}
    I think ive done my best to make the script understandable..
    If not im really sorry about the inconvenience..
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