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Thread: Partial Fraction

  1. #1
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    Partial Fraction

    havin trouble solving this partial fraction:

    $\displaystyle 1/s^2-2s+2$

    i think i can break it up into:

    $\displaystyle 1/(s-1)^2+1$
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  2. #2
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by Chadly724 View Post
    havin trouble solving this partial fraction:

    $\displaystyle 1/s^2-2s+2$

    i think i can break it up into:

    $\displaystyle 1/(s-1)^2+1$
    is it: $\displaystyle \frac{1}{s^2}-2s+2$ or $\displaystyle \frac{1}{s^2-2s+2}$?
    Last edited by Quick; Nov 16th 2006 at 06:12 PM.
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    correction

    sorry there should be partheses around the entire bottom function
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    Quote Originally Posted by Chadly724 View Post
    havin trouble solving this partial fraction:

    $\displaystyle 1/s^2-2s+2$

    i think i can break it up into:

    $\displaystyle 1/(s-1)^2+1$
    You have,
    $\displaystyle \frac{1}{s^2-2s+2}$

    Express as,
    $\displaystyle \frac{1}{s^2-2s+1+1}$

    Thus,
    $\displaystyle \frac{1}{(s-1)^2+1}$

    Now, I presume you have an inverse Laplace transform for $\displaystyle s>1$.

    In that case if you have,
    $\displaystyle \frac{1}{s^2+1}$ $\displaystyle s>0$
    Then, that is the function,
    $\displaystyle f(x)=\sin x$
    Since you have a substitution you use the shift rule.
    Thus,
    $\displaystyle f(x)=e^x\sin x$
    If the unique inverse LaPlace Transform.
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  5. #5
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    Quote Originally Posted by Chadly724 View Post
    havin trouble solving this partial fraction:

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

    i think i can break it up into:

    $\displaystyle \frac{1}{(s-1)^2+1}$
    Do you want to break this up into partial fractions or just transform into a
    standard form of LT that can be inverted by table look-up?

    My table of LT's has:

    $\displaystyle
    \frac{1}{(s-b)^2+a^2}=\mathcal{L}\ \frac{e^{bt}\sin(at)}{a}
    $

    which should take case of your last form.

    However we can do the same thing with partial fractions if you will:

    $\displaystyle
    \frac{1}{s^2-2s+2}=\frac{1}{ (s-(1+\bold{i})) (s-(1-\bold{i}))}=
    \frac{\bold{i}}{2(s-(1-\bold{i})) } - \frac{\bold{i}}{2(s-(1+\bold{i})) }
    $

    (please check the arithmetic)

    RonL
    Last edited by CaptainBlack; Nov 17th 2006 at 08:25 AM.
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  6. #6
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    Hello, Chadly724!

    Havin trouble solving this partial fraction: $\displaystyle \frac{1}{s^2-2s+2}$

    i think i can break it up into: $\displaystyle \frac{1}{(s-1)^2+1}$. . Yes, you can . . . but why?

    Partial Fractions are used when the denominator can be factored.
    . . Like: .$\displaystyle \frac{1}{x^2+2x-15}\:=\:\frac{1}{(x-3)(x+5)}$ or: .$\displaystyle \frac{1}{x^3+4x} \:=\:\frac{1}{x(x^2 +4)}$


    As Captain Black pointed out, it can be factored: .$\displaystyle (s - 1 - i)(s - 1 + i)$
    . . but I doubt that you want to introduce complex numbers.


    If this is an integration problem:

    . . $\displaystyle \int\frac{ds}{s^2 - 2x + 2} \;=\;\int\frac{ds}{(s-1)^2+1} \;=\;\tan^{-1}(s-1) + C $

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  7. #7
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    Quote Originally Posted by Soroban View Post
    Hello, Chadly724!


    Partial Fractions are used when the denominator can be factored.
    . . Like: .$\displaystyle \frac{1}{x^2+2x-15}\:=\:\frac{1}{(x-3)(x+5)}$ or: .$\displaystyle \frac{1}{x^3+4x} \:=\:\frac{1}{x(x^2 +4)}$


    As Captain Black pointed out, it can be factored: .$\displaystyle (s - 1 - i)(s - 1 + i)$
    . . but I doubt that you want to introduce complex numbers.


    If this is an integration problem:

    . . $\displaystyle \int\frac{ds}{s^2 - 2x + 2} \;=\;\int\frac{ds}{(s-1)^2+1} \;=\;\tan^{-1}(s-1) + C $
    The variable $\displaystyle s$ points in the direction of a Laplace transform problem, not an intrgration.

    RonL
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