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

  • Thread starter krnhseya
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In summary, the conversation discusses solving a problem involving finding the transfer function with springs and a damper. The problem has been reduced to a partial fraction, but the speaker is having trouble solving it. They suggest factoring the denominator and using the quadratic formula to solve it.
  • #1
krnhseya
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Homework Statement



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

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.
 
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  • #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?
 
  • #3
yeap :)
 
  • #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...
 
  • #5
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
[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:
 
  • #7
tiny-tim said:
[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:

well that b1 squared and s squared at the end...it cancells the expansion of the squared part...
 
  • #8
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]
 

1. What is the purpose of using partial fractions?

Partial fractions are used to simplify and solve complex algebraic expressions involving rational functions. It allows us to break down a single fraction into multiple simpler fractions, making the overall expression easier to manipulate and solve.

2. How do we know when to use partial fractions?

We use partial fractions when we have a rational expression with a denominator that can be factored into linear or quadratic terms. This technique is especially useful in integration problems where we can use partial fractions to break down the integrand into simpler forms.

3. How do we solve for the unknown coefficients in partial fractions?

To solve for the unknown coefficients, we first factor the denominator into linear or quadratic terms. Then, we equate the partial fraction expression to the original expression and solve for the unknown coefficients by comparing the coefficients of each term on both sides of the equation.

4. Can we use partial fractions to solve for any rational function?

No, we can only use partial fractions to solve for rational functions with proper fractions, meaning the degree of the numerator is less than the degree of the denominator. Improper fractions can be simplified using other techniques such as long division.

5. Are there any special cases when using partial fractions?

Yes, there are special cases when the denominator has repeated or complex roots. In these cases, we use slightly different methods to solve for the unknown coefficients. Additionally, we must be careful when canceling out terms in the partial fraction expression to avoid losing solutions.

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