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Math Help - Inverse Laplace

  1. #1
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    Inverse Laplace

    hi,

    Inverse laplace

    ---(s + 9 )
    ------------
    s^2 + 6s+ 9


    -----(s + 3 )
    --------------- + .........
    (s + 3 )^ 2 + 4


    { e^-3t } * Inverse laplace { s / (s+4) } + .......

    I have problem understand this step... how to take away the +3 in the equation ? i know there is a formula in the sheet to give a direct answer given but the example in the book really make my head crack

    = e^-3t cos2t + ......

    Thank you for any help
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  2. #2
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    Quote Originally Posted by Chris0724 View Post
    hi,

    Inverse laplace

    ---(s + 9 )
    ------------
    s^2 + 6s+ 9


    -----(s + 3 )
    --------------- + .........
    (s + 3 )^ 2 + 4


    { e^-3t } * Inverse laplace { s / (s+4) } + .......

    I have problem understand this step... how to take away the +3 in the equation ? i know there is a formula in the sheet to give a direct answer given but the example in the book really make my head crack

    = e^-3t cos2t + ......

    Thank you for any help
    Err, I think you've got your completing the square wrong.

     s^2 + 6s + 9 \neq (s+3)^2 + 4

     (s+3)^2 + 4 = s^2 + 6s + 9 + 4 = s^2 + 6s + 13.

    The way I see it is:

     \frac{s+9}{s^2 + 6s + 9} = \frac{s+9}{(s+3)^2} = \frac{s+3}{(s+3)^2} + \frac{6}{(s+3)^2} = \frac{1}{s+3} + 6 \bigg(\frac{1}{(s+3)^2} \bigg)

    Then for the inverse, remember the following rules:

     \mathcal{L}^{-1} \bigg( \frac{1}{s+a}\bigg) = e^{-at}

     \mathcal{L}^{-1} \bigg(\frac{n!}{(s+a)^{n+1}} \bigg)= t^n e^{-at}
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  3. #3
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    Quote Originally Posted by Mush View Post
    Err, I think you've got your completing the square wrong.

     s^2 + 6s + 9 \neq (s+3)^2 + 4

     (s+3)^2 + 4 = s^2 + 6s + 9 + 4 = s^2 + 6s + 13.

    The way I see it is:

     \frac{s+9}{s^2 + 6s + 9} = \frac{s+9}{(s+3)^2} = \frac{s+3}{(s+3)^2} + \frac{6}{(s+3)^2} = \frac{1}{s+3} + 6 \bigg(\frac{1}{(s+3)^2} \bigg)

    Then for the inverse, remember the following rules:

     \mathcal{L}^{-1} \bigg( \frac{1}{s+a}\bigg) = e^{-at}

     \mathcal{L}^{-1} \bigg(\frac{n!}{(s+a)^{n+1}} \bigg)= t^n e^{-at}

    thanks mush,

    i got the equation wrongly... it should be

     \frac{s+9}{s^2 + 6s + 13}
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  4. #4
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    Quote Originally Posted by Chris0724 View Post
    thanks mush,

    i got the equation wrongly... it should be

     \frac{s+9}{s^2 + 6s + 13}
    Ah, well in that case, your completing the square was correct! Try this then:

     \frac{s+9}{s^2 + 6s + 13} = \frac{s+9}{(s+3)^2 + 2^2}

    = \frac{s+3}{(s+3)^2 + 2^2} + \frac{6}{(s+3)^2 + 2^2}

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

    The inverse rules you must now remember is that:

     \mathcal{L}^{-1} \frac{s+a}{(s+a)^2 + \omega^2} = e^{-at} \cos(\omega t)

     \mathcal{L}^{-1} \frac{\omega}{(s+a)^2 + \omega ^2} = e^{-at} \sin(\omega t)
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