Partial Fractions: Solve k1b1/[((k1+b1*s)(k2+b2*s))-b1^{2}s^{2}]

[krnhseya]

Homework Statement

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

Homework Equations

n/a

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 can't get the partial fraction.
once i get that particle fractions, i take the inverse laplace transform and get the answer.

 

[rocomath]

I'm getting dizzy reading it ...

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

Yes?

 

[krnhseya]

yeap :)

 

[krnhseya]

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]

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]

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

krnhseya, just expand the bottom line into the form as^2\,+\,bs\,+\,c, and then factor it using the good ol' (-b ±√b^2 - 4ac)/2a.

 

[krnhseya]

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

krnhseya, just expand the bottom line into the form as^2\,+\,bs\,+\,c, 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]

No, it doesn't …

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