# Partical Fractions

1. Mar 10, 2008

### krnhseya

1. The problem statement, all variables and given/known data

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

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. Mar 10, 2008

### rocomath

I'm getting dizzy reading it ...

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

Yes?

3. Mar 10, 2008

### krnhseya

yeap :)

4. Mar 10, 2008

### 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...

5. Mar 10, 2008

### HallsofIvy

Staff Emeritus
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.

6. Mar 10, 2008

### 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.

7. Mar 10, 2008

### krnhseya

well that b1 squared and s squared at the end...it cancells the expansion of the squared part...

8. Mar 11, 2008

### tiny-tim

No, it doesn't …

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