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Partial Fraction Decomposition

  1. Jul 25, 2009 #1
    1. The problem statement, all variables and given/known data
    I just cant understand it ive read plenty of guides online, I just cant figure it out. How do you do partial fraction decomposition the farthest i can get is below
    1/[(s^2 + 1)(s+1)]


    2. Relevant equations



    3. The attempt at a solution
    1/[(s^2 + 1)(s+1)] = A/(s+1) + (Bs + C)/(s^2 + 1)
    1 = A(s^2 + 1) + (Bs + C)(s+1)

    set s = -1

    1 = A(2)

    A = 1/2

    I cant figure anything else out.
     
  2. jcsd
  3. Jul 25, 2009 #2
    Yes, you're right. A = 1/2. So that's one less "variable" you have to worry about. Substitute it back into the equation to get:

    (1/2)(s^2 + 1) + (Bs + C)(s+1) = 1

    You can easily get rid of B, and the rest should be pretty simple.
     
  4. Jul 25, 2009 #3
    EDIT : sorry cse63146 has given a nice clue

    [tex]\frac{1}{(s^{2}+1)(s+1)} = \frac{A}{s+1} + \frac{Bs+C}{s^{2}+1}[/tex]

    [tex] 1 = A (s^{2}+1) + (Bs+C) (s+1)[/tex]

    [tex] 1 = A s^{2} + A + B s^{2} + B s + Cs + C [/tex]

    [tex] 1 = (A + B) s^{2} + (B+C)s + (A+C) [/tex]

    then, comparing coefficients between of the RHS and LHS :

    A+B = 0, because the LHS doesn't have s^{2}

    B+C = 0, because the LHS doesn't have s

    A+C = 1
     
  5. Jul 25, 2009 #4
    Ah im sorry im still confused : songoku, as for cse how do you get rid of b? Im past calculus III but i just cant remember how to do it.
     
  6. Jul 25, 2009 #5
    You choose one value of s that will make B = 0
     
  7. Jul 25, 2009 #6
    Im still confused do you mean I move all of the equation to one side with B on the other?
     
  8. Jul 25, 2009 #7
    just worry about this for now: (Bs + C)(s+1)

    what value of s would eliminate B?
     
  9. Jul 25, 2009 #8
  10. Jul 25, 2009 #9
    and out of 0 and -1, which what would ONLY eliminate B, and leave C?
     
  11. Jul 25, 2009 #10
    Yes, that's right

    Because you have used -1, try 0
     
  12. Jul 25, 2009 #11
    I mean if i made s=0 then C = -A(1) C=-1/2
     
  13. Jul 25, 2009 #12
    Yes, now that you have the value of C, substitute it back to get:

    (1/2)(s^2 + 1) + (Bs + -1/2)(s+1) = 1

    now just let s = 1, and you can get the value of B.

    Out of curiousity, is this a laplace transformation
     
  14. Jul 25, 2009 #13
    (1/2)(s^2 + 1) + (Bs + -1/2)(s+1) = 1 s = 1
    (1/2)(2) + (B - 1/2) = 1
    B - 1/2 = 0
    B = +1/2

    Yes it is and its a summer class to make things work, we dont get examples, he just does proofs >_>
     
  15. Jul 25, 2009 #14
    So does 1/[(s^2 + 1)(s + 1)] = (1/2)/(s + 1) + [(1/2)s + -1/2]/(s^2 + 1)

    book says it should be 1/[(s^2 + 1)(s + 1)] = (1/2)/(s + 1) + [-(1/2)s + 1/2]/(s^2 + 1)
     
  16. Jul 25, 2009 #15
    I would double check your value for C, I keep getting positive 1/2.

    (1/2) (0^2 + 1) + (B(0)+C)(0 + 1) = 1
    (1/2)(1) + C(1) = 1
    C = 1/2

    which would also affect your value for B.
     
  17. Jul 25, 2009 #16
    Ah yes! Thats what I did wrong! Now it makes since. Do i always use 1 to find the Bs in any case?
     
  18. Jul 25, 2009 #17
    Not always. The idea behind this method is to eliminate all but 1 variable by assigning values for s.

    To determine the value of A, you made s = -1, which eliminated B and C, which gave you the value of A.

    Now that you have the value of A, it's no longer a variable; it's now a constant, which you can substitute back in your equation.

    By assigning s = 0, you eliminated B, leaving you only C to worry about. The idea is to keep assigning values to s in order to eliminate variables until you have one variable left (in this case you found A and C, leaving you only B), and now you have to assign a value which would "get rid" of s, and leave only B and constants.

    You could also try songoku's method, which would give you the same answer.

    http://www.exampleproblems.com/wiki/index.php/Algebra-Partial_Fraction_Decomposition has a lot of examples, and step by step solutions (with the method you used).

    Also, there are some PFD's that you might need to know some matrix row reduction to solve, so you might want to get a refresher on that if you need it.

    Good luck with Differential Equations. Had to do that last semester; it was a pain.
     
  19. Jul 25, 2009 #18
    Cheers Mate!
     
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