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Thread: Laplace transform - stuck on the partial fraction

  1. #1
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    Laplace transform - stuck on the partial fraction

    In solving a problem, I ended up having to take the inverse laplace of (e^(-2s))/((x^2)(x-1))
    I know what formula I need to use, but for some reason I'm stuck and can't remember how to do partial fractions with a denominator like that (I separated them so the numerator = 1), can anyone help?
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  2. #2
    Senior Member Peritus's Avatar
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    $\displaystyle
    \frac{1}
    {{x^2 \left( {x - 1} \right)}} = \frac{{ax + b}}
    {{x^2 }} + \frac{c}
    {{x - 1}}\quad \quad a,b,c \in Z

    $

    now all you have to do is find the constant coefficients, good luck.
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  3. #3
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    Quote Originally Posted by mistykz View Post
    In solving a problem, I ended up having to take the inverse laplace of (e^(-2s))/((x^2)(x-1))
    I know what formula I need to use, but for some reason I'm stuck and can't remember how to do partial fractions with a denominator like that (I separated them so the numerator = 1), can anyone help?
    That looks pretty standard to me. The partial fractions form is $\displaystyle \frac{1}{x^2(x-1)}= \frac{A}{x}+ \frac{B}{x^2}+ \frac{C}{x-1}$. Multiplying through by $\displaystyle x^2(x- 1)$ gives $\displaystyle 1= Ax(x-1)+ B(x-1)+ Cx^2$. Taking x= 0, $\displaystyle 1= -B$. Taking x= 1, $\displaystyle 1= C$. Those are the only values of x that will reduce so simply but any other value for x will give a third equation. For example, taking x= 2, $\displaystyle 1= A(2)(2-1)+ B(2-1)+ C(4)$. Since we know that B= -1 and C= 1, that becomes $\displaystyle 2A- 1+ 4= 1$ or A= -1.

    Another way to do any partial fractions, though harder since we are not taking advantage of those special values, x= 0 and x= 1, is to combine the fractions: $\displaystyle \frac{1}{x^2(x-1)}= \frac{A}{x}+ \frac{B}{x^2}+ \frac{C}{x-1}= \frac{A(x)(x-1)}{x^2(x-1)}+ \frac{B(x-1)}{x^2(x-1)}+ \frac{Cx^2}{x^2(x-1)}$$\displaystyle \frac{0x^2+ 0x+ 1}{x^2(x-1)}= \frac{(A+ C)x^2+ (-A+ B)x- B}{x^2(x-1)}$.

    Since this must be true for all x, we must have A+ C= 0, -A+ B= 0 and -B= 1. Again, we get B= -1, A= -1, and C= 1.

    (Note that $\displaystyle \frac{ax+ b}{x^2}= \frac{ax}{x^2}+ \frac{b}{x^2}= \frac{a}{x}+ \frac{b}{x^2}$ so Peritus' form is the same as mine.)
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  4. #4
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    ah, that's what i did wrong, i multiplied through by x, x^2, and (x-1)...didnt need that single x. oops! thanks!
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  5. #5
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    Quote Originally Posted by mistykz View Post
    In solving a problem, I ended up having to take the inverse laplace of (e^(-2s))/((x^2)(x-1))
    I know what formula I need to use, but for some reason I'm stuck and can't remember how to do partial fractions with a denominator like that (I separated them so the numerator = 1), can anyone help?
    Just FYI you can invert this with out partial fractions.

    Remember that $\displaystyle \mathcal{L}^{-1}=\frac{F(s)}{s}=\int_{0}^{t}f(\tau)d\tau$

    Where $\displaystyle \mathcal{L}(f(t))=F(s)$

    Note that $\displaystyle \mathcal{L}^{-1}\left( \frac{e^{-2s}}{(s-1)}\right)=H(t-2)e^{t-2}$

    Where $\displaystyle H(t)$ is the heaviside function

    Now if we integrate this twice we will get the inverse transform

    $\displaystyle \int_{0}^{t} H(\tau-2)e^{\tau -2}d \tau=\int_{2}^{t}e^{\tau-2}d\tau=e^{\tau-2}\bigg|_{2}^{t}=e^{t-2}-1$

    Integrating one more time we get

    $\displaystyle \int_{2}^{t} e^{\tau-2}-1 d\tau=e^{\tau - 2} -\tau \bigg|_{2}^{t}=e^{t-2}-t-(1-2)=1-t+e^{t-2}$

    so

    $\displaystyle f(t)=H(t-2)(1-t+e^{t-2})$
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