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Partical Fractions

  1. Mar 10, 2008 #1
    1. The problem statement, all variables and given/known data

    Turn this into partial fraction.
    k1b1/[((k1+b1*s)(k2+b2*s))-b1[tex]^{2}[/tex]s[tex]^{2}[/tex]]

    2. Relevant equations

    n/a

    3. The attempt at a solution

    original question was to find the transfer function with springs and a damper and I reduced it to this far but I cant get the partial fraction.
    once i get that partical fractions, i take the inverse laplace transform and get the answer.
     
    Last edited: Mar 10, 2008
  2. jcsd
  3. Mar 10, 2008 #2
    I'm getting dizzy reading it ...

    [tex]\frac{k_1b_1}{(k_1+b_1s)(k_2+b_2s)-b_1^2s^2}[/tex]

    Yes?
     
  4. Mar 10, 2008 #3
    yeap :)
     
  5. Mar 10, 2008 #4
    well is this impossible to separate?
    i did other problems but i am just stuck on this one.
    let me know if you need the actual problem statement...
     
  6. Mar 10, 2008 #5

    HallsofIvy

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    you want, of course, to factor the denominator. I think I would be inclined to multiply out that first part and combine coefficients of like powers. It will be, of course, a quadratic. At worst, you could set the denominator equal to 0 and solve the equation by the quadratic formula.
     
  7. Mar 10, 2008 #6

    tiny-tim

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    [tex]\frac{k_1b_1}{(k_1+b_1s)(k_2+b_2s)-b_1^2s^2}[/tex]

    krnhseya, just expand the bottom line into the form [tex]as^2\,+\,bs\,+\,c[/tex], and then factor it using the good ol' (-b ±√b^2 - 4ac)/2a. :smile:
     
  8. Mar 10, 2008 #7
    well that b1 squared and s squared at the end...it cancells the expansion of the squared part...
     
  9. Mar 11, 2008 #8

    tiny-tim

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    No, it doesn't …

    It's [tex]k_1k_2\,+\,(b_1k_2\,+\,b_2k_1)s\,+\,b_1(b_2\,-\,b_1)s^2[/tex]
     
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