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Math Help - Laplace Transforms

  1. #1
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    Laplace Transforms

    find the laplace transform F(s) of f(t) = sin wt

    the w is actually supposed to be omega

    what i got was

    integeral from 0 to infinity of: (sin wt)*(e^-st) dt

    then i integrated by parts and obtained a large expression
    which involves the limit to infinity of sin and cos ... how do i solve this limit
    can someone post a step by step explanation starting from the limit part
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  2. #2
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    L{sin(wt)} = w/ (s^2 + w^2)

    It's a formula.. I didn't do the integration.
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    really its a formula
    but i think i have to show it because i haven't yet learned that formula
    is it hard to show or is there some easy way to show it???
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by razorfever View Post
    really its a formula
    but i think i have to show it because i haven't yet learned that formula
    is it hard to show or is there some easy way to show it???
    \int_0^{\infty}e^{-st}\sin\!\left(\omega t\right)\,dt

    By parts, let u=\sin\!\left(\omega t\right) and \,dv=e^{-st}\,dt. Thus, \,du=\omega\cos\left(\omega t\right)\,dt and v=-\frac{1}{s}e^{-st}

    Thus, we have \int_0^{\infty}e^{-st}\sin\!\left(\omega t\right)\,dt=\left.\left[-\frac{1}{s}e^{-st}\sin\!\left(\omega t\right)\right]\right|_0^{\infty}+\frac{\omega}{s}\int_0^{\infty}  e^{-st}\cos\!\left(\omega t\right)\,dt =\frac{\omega}{s}\int_0^{\infty}e^{-st}\cos\!\left(\omega t\right)\,dt, since \lim_{t\to\infty}e^{-st}\sin\!\left(\omega t\right)=0 (You can take note that sine oscilates bewteen -1 and 1, but e^{-st} approaches zero very quicly. Thus its the term we are most concerned with when evaluating the limit.)

    Now, let u=\cos\!\left(\omega t\right) and \,dv=e^{-st}\,dt. Thus, \,du=-\omega\sin\!\left(\omega t\right)\,dt and v=-\frac{1}{s}e^{-st}

    Thus, we now see that \int_0^{\infty}e^{-st}\sin\!\left(\omega t\right)\,dt=\frac{\omega}{s}\left[\left.\left[-\frac{1}{s}e^{-st}\cos\!\left(\omega t\right)\right]\right|_0^{\infty}-\frac{\omega}{s}\int_0^{\infty}e^{-st}\sin\!\left(\omega t\right)\,dt\right] \implies \left(1+\frac{\omega^2}{s^2}\right)\int_0^{\infty}  e^{-st}\sin\!\left(\omega t\right)\,dt = \frac{\omega}{s^2} since \lim_{t\to{\infty}}e^{-sy}\cos\!\left(\omega t\right)=0 by similar explanation above. Also note that \lim_{t\to0}e^{-st}\cos\!\left(\omega t\right)=1.

    Thus, it is now evident that

    \left(1+\frac{\omega^2}{s^2}\right)\int_0^{\infty}  e^{-st}\sin\!\left(\omega t\right)\,dt = \frac{\omega}{s^2}\implies\frac{s^2+\omega^2}{s^2}  \int_0^{\infty}e^{-st}\sin\!\left(\omega t\right)\,dt = \frac{\omega}{s^2} \implies \int_0^{\infty}e^{-st}\sin\!\left(\omega t\right)\,dt = \frac{s^2}{\omega^2+s^2}\cdot\frac{\omega}{s^2} \implies \color{red}\boxed{\mathcal{L}\left\{\sin\!\left(\o  mega t\right)\right\}=\int_0^{\infty}e^{-st}\sin\!\left(\omega t\right)\,dt=\frac{\omega}{s^2+\omega^2}}

    Keep in mind this is only true when s>0.

    Does this make sense?
    Last edited by Chris L T521; January 28th 2009 at 09:20 PM.
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  5. #5
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    thanks for that explanation i almost made it to the end but i just had a hard time evaluating the limit for sin

    the second part is:

    show that

    L[e^at * sin wt] = F(s-a)

    how do i get started on this

    and how do i get the answer in the form of F()???
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by razorfever View Post
    thanks for that explanation i almost made it to the end but i just had a hard time evaluating the limit for sin

    the second part is:

    show that

    L[e^at * sin wt] = F(s-a)

    how do i get started on this

    and how do i get the answer in the form of F()???
    Recall that \mathcal{L}\left\{f\!\left(t\right)\right\}=\int_0  ^{\infty}e^{-st}f\!\left(t\right)\,dt=F\!\left(s\right)

    In general, if you are asked to find \mathcal{L}\left\{e^{at}f\!\left(t\right)\right\}, you should see that \mathcal{L}\left\{e^{at}f\!\left(t\right)\right\}=  \int_0^{\infty}e^{-st}e^{at}f\!\left(t\right)\,dt=\int_0^{\infty}e^{\  left(a-s\right)t} f\!\left(t\right)\,dt =\int_0^{\infty}e^{-\left(s-a\right)t}f\left(t\right)\,dt=F\!\left(s-a\right).

    All you're really doing is manipilating the definition of the Laplace Transform.

    Does this make sense?
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  7. #7
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    i don't undertand the last step ... can you go into more detail ...
    i really want to undertand this and my prof didn't go into much detail over this last week
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  8. #8
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    Quote Originally Posted by razorfever View Post
    i don't undertand the last step ... can you go into more detail ...
    i really want to undertand this and my prof didn't go into much detail over this last week
    The last step is by definition of the Laplace Transform.

    If what I'm answering is not what you're asking then it would help if you quoted what it is that you're refering to.
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