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Thread: Inverse Laplace Transform

  1. #1
    Junior Member hercules's Avatar
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    Inverse Laplace Transform

    I need help finding the inverse laplace transform of

    $\displaystyle \frac {8s^2-4s+12}{s(s^2+4)}
    $


    and

    $\displaystyle \frac {1-2s}{s^2+4s+5}$

    Thank You
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by hercules View Post
    I need help finding the inverse laplace transform of

    $\displaystyle \frac {8s^2-4s+12}{s(s^2+4)}
    $

    By partial fractions, we can write this as:

    $\displaystyle \frac 3s + \frac {5s - 4}{s^2 + 4}$

    or, in other words:

    $\displaystyle 3 \cdot {\color{red} \frac 1s} + 5 \cdot {\color{red} \frac s{s^2 + 2^2}} - 2 \cdot { \color{red}\frac 2{s^2 + 2^2}}$

    now just look up the rules in the table of your text. what are the inverse Laplace transforms for the guys in red?

    and

    $\displaystyle \frac {1-2s}{s^2+4s+5}$

    Thank You
    with some algebraic manipulation (completing the square of the denominator, factoring out a minus 1 from the numerator and writing -1 as 4 - 5) we find that this expression is:

    $\displaystyle -2 \cdot {\color{red}\frac {s + 2}{(s + 2)^2 + 1}} + 5 \cdot {\color{red} \frac 1{(s + 2)^2 + 1}}$

    now this is the same story as the last one. look up the transforms in the table in your text
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  3. #3
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    Quote Originally Posted by hercules View Post
    I need help finding the inverse laplace transform of

    $\displaystyle \frac {8s^2-4s+12}{s(s^2+4)}
    $
    You can write, $\displaystyle \tfrac{3}{s} + \tfrac{5s}{s^2+4} - 2\cdot \tfrac{2}{s^2+4}$.
    Now just recognize the forms, $\displaystyle 3+5\cos (2x) - 2 \sin (2x)$.
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by hercules View Post
    I need help finding the inverse laplace transform of

    $\displaystyle \frac {8s^2-4s+12}{s(s^2+4)}$

    and

    $\displaystyle \frac {1-2s}{s^2+4s+5}$

    Thank You
    by partial fractions (work not shown) we get

    $\displaystyle \frac{8s^2-4s+12}{s(s^2+4)}=\frac{3}{s}+\frac{5s-4}{(s^2+4)}=3\frac{1}{s}+5\frac{s}{s^2+4}+2\frac{2 }{s^2+4}$

    now taking the inverse transform we get

    $\displaystyle \mathcal{L}^{-1}\{ 3\frac{1}{s}+5\frac{s}{s^2+4}+2\frac{2}{s^2+4}\}=$

    by the linearity property we get

    $\displaystyle 3\mathcal{L}^{-1}\{ \frac{1}{s}\} +5\mathcal{L}^{-1}\{ \frac{s}{s^2+4}\}+2\mathcal{L}^{-1}\{ \frac{2}{s^2+4}\} =3+5\cos(2t)+2\sin(2t)$

    Edit: late again
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  5. #5
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    Quote Originally Posted by hercules View Post
    I need help finding the inverse laplace transform of

    $\displaystyle \frac {8s^2-4s+12}{s(s^2+4)}
    $

    and

    $\displaystyle \frac {1-2s}{s^2+4s+5}$

    Thank You
    Hi! I'll be omitting $\displaystyle \theta(t)$, the Heaviside step function, in this post.

    Assuming you know how to do partial fraction expansion we have:

    $\displaystyle
    \frac{8s^2-4s+12}{s(s^2+4)}
    = 3\frac1s + \frac{5s-4}{s^2-2^2}
    = \frac3s + 5\frac{s}{s^2-2^2} - 2\frac2{s^2-2^2}
    $

    and knowing these transformation pairs:

    $\displaystyle \mathcal{L}^{-1}\left\{\frac1s\right\}=1$
    $\displaystyle \mathcal{L}^{-1}\left\{\frac{\omega}{s^2+\omega^2}\right\}=\sin( \omega t)$
    $\displaystyle \mathcal{L}^{-1}\left\{\frac{s}{s^2+\omega^2}\right\}=\cos(\omeg a t)$

    we have

    $\displaystyle \mathcal{L}^{-1}\left\{3\frac1s + 5\frac{s}{s^2-2^2} - 2\frac2{s^2-2^2}\right\}
    = 3\mathcal{L}^{-1}\left\{\frac1s\right\}
    +5\mathcal{L}^{-1}\left\{\frac{s}{s^2-2^2}\right\}
    -2\mathcal{L}^{-1}\left\{\frac2{s^2-2^2}\right\}
    $$\displaystyle =3+5\cos(2t)-2\sin(2t)$

    Edit: Wow, some guys are too fast for me!
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