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Math Help - Simple Laplace Transform w/ Final Value Theorem

  1. #1
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    Unhappy Simple Laplace Transform w/ Final Value Theorem

    Help! This is the last problem on my math homework and we are starting a section on Fourier transforms which I understand, but we never learned Laplace transforms. My book doesn't give any examples of how do a problem similar to this one and I can't find any close solved examples on the internet. I only have to do three of these so I'm posting the one I could learn the most from:

    about the only think I found on the internet is that the final value theorem is lim y(t)=lim Sy(s) which at this point doesn't help because I don' know how to do the rest of the problem.

    Any steps to this problem are greatly appreciated,
    Evan
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  2. #2
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    Quote Originally Posted by 360modina View Post
    Help! This is the last problem on my math homework and we are starting a section on Fourier transforms which I understand, but we never learned Laplace transforms. My book doesn't give any examples of how do a problem similar to this one and I can't find any close solved examples on the internet. I only have to do three of these so I'm posting the one I could learn the most from:

    about the only think I found on the internet is that the final value theorem is lim y(t)=lim Sy(s) which at this point doesn't help because I don' know how to do the rest of the problem.

    Any steps to this problem are greatly appreciated,
    Evan
    Why not use the transfer function?

    Remember that:

     \mathcal L({f''}) = s^{2}F(s) - sf(0) - f'(0)
     \mathcal L({f'}) = sF(s) - f(0)

    The transfer function,  Q(S) , is a solution of the equation and it is given by:

     \frac{Y(s)}{R(s)}

    EDIT: In fact, with your initial conditions, you don't need to consider the transfer function, you just need the above! ^^^
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  3. #3
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    L[y''] = L [(y')'] = p L [y'] - y'(o) = p^2 L[y] - p y(o) - y'(0)
    Then you're plugging it in and solve for y using L^-1
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  4. #4
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    I was trying to solve this problem, as I've studied this before and need to reivew for the final. I also got stuck.

    on the right hand side:
    L [2r(t)] = 2 L [ r(t) ] = 2 L [ y/ sint ] can some body help? L [ y/sint ] = d/dp L [ (sint)^-1] = ???
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  5. #5
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    Quote Originally Posted by tennis14321 View Post
    I was trying to solve this problem, as I've studied this before and need to reivew for the final. I also got stuck.

    on the right hand side:
    L [2r(t)] = 2 L [ r(t) ] = 2 L [ y/ sint ] can some body help? L [ y/sint ] = d/dp L [ (sint)^-1] = ???
    Where are you getting sine from?!

    \mathcal L (\frac{d^2y}{dt^2}+4\frac{dy}{dt}+3y)=\mathcal L (2r(t))


     s^{2}Y(s)-sy(0)-y'(0) +4sY(s)-y(0)+3Y(s)=2R(s)

    We know that  y(0) = 1, y'(0) = 0, r(t) = 1.

    If  r(t) = 1, then R(s) = \mathcal L (1) = \frac{1}{s}. Plug these all in:

     s^{2}Y(s)-s+4sY(s)-1+3Y(s)=2(\frac{1}{s})


    Y(s)(s^{2}+4s+3)-s-1=\frac{2}{s}


    Y(s)(s^{2}+4s+3)=\frac{2}{s} + s +1



    Y(s)(s^{2}+4s+3)=\frac{2+s^2+s}{s}


    Y(s)=\frac{s^2 +2+s}{s}\times \frac{1}{s^{2}+4s+3}

    Multiply it out and use partial fractions to get it into a recognisable form...
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  6. #6
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    Quote Originally Posted by Mush View Post

    Y(s)=\frac{s^2 +2+s}{s}\times \frac{1}{s^{2}+4s+3}

    Multiply it out and use partial fractions to get it into a recognisable form...
    Y(s)=\frac{1}{s^2+4s+3} + \frac{2}{s(s^2+4s+3)} + \frac{s}{s^2+4s+3}


    Y(s)=\frac{1}{(s+1)(s+3)} + \frac{2}{s(s+1)(s+3)} + \frac{s}{(s+1)(s+3)}

    That was extremely helpful, do you mind taking all the way through the final value theorem, Mush you are good at this.
    Last edited by 360modina; December 11th 2008 at 05:49 PM.
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  7. #7
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    Quote Originally Posted by 360modina View Post
    Y(s)=\frac{1}{s^2+4s+3} + \frac{2}{s(s^2+4s+3)} + \frac{s}{s^2+4s+3}

    That was extremely helpful, do you mind taking all the way through the final value theorem, Mush you are good at this.
    Final value theorem states:

     \lim_{s \to 0} sF(s) = \lim_{t \to \infty} f(t)

    Or, in our case:

     \lim_{s \to 0} sY(s) = \lim_{t \to \infty} y(t)

    Y(s)=\frac{1}{s^2+4s+3} + \frac{2}{s(s^2+4s+3)} + \frac{s}{s^2+4s+3}

    Multiply this by s to get sY(s):


    sY(s)=\frac{s}{s^2+4s+3} + \frac{2}{(s^2+4s+3)} + \frac{s^2}{s^2+4s+3}

    As s tends towards zero, the first and third terms of this equation disappear, because there is s and s^2 on the numerator. However, the 2nd term does not disappear! The s^2 and 4s in the denominator disappear leaving only 3 in the denominator, and the 2 in the numerator remains.

    Hence:

    \lim_{s \to 0} sY(s) = \lim_{t \to \infty} y(t) = \frac{2}{3}
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  8. #8
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    ohh,
    r(t) = 1
    I was thinking y = r sin (t)

    Ok. it's easier than i thought
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