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Partical Fractions

by krnhseya
Tags: fractions, partical
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krnhseya
#1
Mar10-08, 10:45 AM
P: 104
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.
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rocomath
#2
Mar10-08, 10:48 AM
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P: 1,754
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?
krnhseya
#3
Mar10-08, 10:48 AM
P: 104
yeap :)

krnhseya
#4
Mar10-08, 03:29 PM
P: 104
Partical Fractions

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...
HallsofIvy
#5
Mar10-08, 03:54 PM
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PF Gold
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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.
tiny-tim
#6
Mar10-08, 04:14 PM
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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.
krnhseya
#7
Mar10-08, 07:09 PM
P: 104
Quote Quote by tiny-tim View Post
[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.
well that b1 squared and s squared at the end...it cancells the expansion of the squared part...
tiny-tim
#8
Mar11-08, 03:22 AM
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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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