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Math Help - [SOLVED] inverse of a laplace transform

  1. #1
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    [SOLVED] inverse of a laplace transform

    Hey, I cannot figure out how to get these two probelms figured out. I need to find the inverse of the laplace transform.

    Here is what ive tried: given H(s) find h(t). (take inverse laplace transform)
    1) H(s)=\dfrac{s^{2}-3s+9}{(s^{2}+16)^{2}}

    using partial fractions I split this up into:
    \dfrac{1}{s^{2}+16}+\dfrac{-3s-7}{(s^{2}+16)^{2}}

    I know:
    \dfrac{1}{s^{2}+16}=\dfrac{1}{4}sin(4t)

    which leaves me with:
    \dfrac{-3s-7}{(s^{2}+16)^{2}}

    From here i cant seem to split this up any further into a sin/cos/e^(x) or other simple function. Any help on the next step, or corrections to what I have already done would be great.


    2) given K(s) find k(t) (take inverse of laplace transform).
    K(s)=\dfrac{2s}{s^2-6s+25}

    I cant factor the denominator so i tried writing the demoninator in various ways which didnt lead me anywhere. Without being able to split the denominator up I also cannot use partial fractions to help simplify this. Any ideas?
    K(s)=\dfrac{2s}{(s^2+16)+(-6s+9)}, didn't seem to help.
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  2. #2
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    \frac{3s}{(s^{2}+16)^{2}}

    Gives

    \frac{3}{8}\cdot t \cdot \sin(4*t)

    and

    \frac{7}{(s^{2}+16)^{2}}

    Gives

    \frac{7}{128}\cdot (\sin(4*t) - 4 \cdot t \cdot \cos(4*t))

    That one might not be in your book. You may have to do it the hard way.
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  3. #3
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    Nevermind, Im gonig to ask my professor at his office hours later today...Thanks again.
    Last edited by snaes; April 22nd 2010 at 09:53 AM.
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