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Math Help - Laplace Transformation

  1. #1
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    Laplace Transformation

    Hi,

    I want to find the Laplace inverse of the following function:

    1/(s^2 +2zws + w^2) , where z and w are constants.

    I know that the result should have an exponintial part and a sin or cos, but I don't know how to factor the function to use the table.

    Thanks,
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  2. #2
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    This is what I get after completing the square:
    <br />
\frac {1}{(s+ \frac{zw}{2})^2 - (\frac{zw}{2})^2 + w^2}

    I still don't know what to do next
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Musab View Post
    This is what I get after completing the square:
    <br />
\frac {1}{(s+ \frac{zw}{2})^2 - (\frac{zw}{2})^2 + w^2}

    I still don't know what to do next
    Um, that's not completing the square...

    if you complete the square, you get

    \displaystyle \frac 1{(s + zw)^2 + w^2(1 - z^2)}

    now what?
    Last edited by Jhevon; October 30th 2010 at 08:43 PM. Reason: z^2 - 1 should be 1 - z^2
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  4. #4
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    Thx

    I think you got the signs wrong, it should be \displaystyle \frac 1{(s + zw)^2 + w^2(-z^2 + 1)}

    Now I should do the following:

    \frac{1}{(w^2(-z^2+1))^{0.5}} .  \frac{(w^2(-z^2+1))^{0.5}}{(s + zw)^2 + w^2(-z^2 + 1)}

    and use the laplace table
    is that right ?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Musab View Post
    Thx

    I think you got the signs wrong, it should be \displaystyle \frac 1{(s + zw)^2 + w^2(-z^2 + 1)}
    yes

    Now I should do the following:

    \frac{1}{(w^2(-z^2+1))^{0.5}} .  \frac{(w^2(-z^2+1))^{0.5}}{(s + zw)^2 + w^2(-z^2 + 1)}

    and use the laplace table
    is that right ?
    and yes
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