Partial fraction

  • Thread starter kdinser
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  • #1
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This is actually part of a larger problem of solving a diff EQ by use of a laplce transform, but it's the partial fraction part that I'm stuck on.

[tex]\frac{1}{s^2(s^2+w^2)} = \frac{A}{s}+\frac{B}{s^2}+\frac{C*s+D}{s^2+w^2}[/tex]

Good so far?

If I then multiply through by the common denominator I get

EQ.2[tex]1 = A*s(s^2+w^2) + B(s^2+w^2) + s^2(C*s+D)[/tex]

So then I set s=0

[tex]1=B*w^2 \longrightarrow B=1/w^2 [/tex]

Then using the quadratic equation, I find that the roots of s^2+w^2 are +/- jw

letting s=jw I end up with

[tex]1 = -jC*w^3 - Dw^2[/tex]

this is where I get stuck, I'm not sure how to find C and D at this point.

Note: I just noticed a mistake in EQ.2,
 
Last edited:

Answers and Replies

  • #2
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you can equate the various bits, ie the coefficients of the quadratic linear and constant terms. you cut out a bit of that with s=0 which is a good trick but u need the other way to get all the constants (if only in terms of w)

ie [tex]{s^2}(C*s + D) = 0[/tex]

i think
 
  • #3
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Thanks for the reply, but I'm not really sure what you are saying.
 
  • #4
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You can get two equations for [itex]C[/itex] and [itex]D[/itex] by letting [itex]s=\pm i\omega[/itex]. You can then solve for [itex]C[/itex] and [itex]D[/itex].

By the way if you want to know a quick way to do this you can say the following:

[tex]
\frac{1}{xy} = \frac{1}{x-y}\left\{\frac{1}{y}-\frac{1}{x}\right\}
[/itex]

So if I choose [itex]x=s^2+\omega^2[/itex], and [itex]y=s^2[/itex] then your initial fraction can be cast directly into the form on the RHS of your equation.
 
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