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Thread: Inverse Laplace transform

  1. #1
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    Inverse Laplace transform

    Hi, i am currently studying laplace transforms and i am arriving at the wrong answer.
    I have attached my working out aswell as the question.
    Inverse Laplace transform-img_20170913_164839.jpg

    I have a feeling that i am getting the partial fractions wrong, although i'm not sure how.
    The correct answer in the textbook is:
    (t/2) - (1/4) + (1/4)(e^(-2t))
    please let me know if you need anything clarified as i know i do not write that neatly.
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  2. #2
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    Re: Inverse Laplace transform

    you have to break up the denominator into all it's factors

    $(s^2 - 4)Y(s) = \dfrac{s-2}{s^2}$

    $s^2-4=(s+2)(s-2)$

    $Y(s) = \dfrac{s-2}{s^2(s+2)(s-2)} = \dfrac{1}{s^2 (s+2)}$

    $\dfrac{1}{s^2 (s+2)} = \dfrac{A}{s} + \dfrac{B}{s^2} + \dfrac{C}{s+2}$

    $1 = As(s+2) + B(s+2) + Cs^2 = s^2(A + C) + s(2A + B) + 2B$

    $0=A+C$

    $0 = 2A+B$

    $1 = 2B$

    $B=\dfrac 1 2$

    $0=2A + \dfrac 1 2$

    $A = -\dfrac 1 4$

    $0=-\dfrac 1 4 + C$

    $C=\dfrac 1 4$

    $Y(s) = \dfrac{1}{2s^2} -\dfrac{1}{4s}+\dfrac{1}{4(2+s)}$

    $y(t) = \dfrac{t}{2}-\dfrac{1}{4}+\dfrac{e^{-2 t}}{4}$
    Thanks from CrossXHair
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  3. #3
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    Re: Inverse Laplace transform

    Thanks heaps, i did not think of adding 1/s - 2/s^2 into one fraction. as well as breaking up the denominator further.
    will definitely keep this in mind for future problems.
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