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Partial Fractions

  1. Apr 21, 2007 #1
    if you have

    [tex]\frac{3s + 1}{(s+2)^2 + 4^2}[/tex]

    does it become...

    [tex]3s + 1 = \frac{A}{(s+2)} + \frac{B}{(s+2)^2} + \frac{C}{4} + \frac{D}{4^2}[/tex]

    or...

    [tex]3s + 1 = \frac{A}{(s+2)} + \frac{B}{(s+2)^2} + \frac{C}{4^2}[/tex]
     
  2. jcsd
  3. Apr 21, 2007 #2

    arildno

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    What is the difference between a FACTOR, and a TERM?
     
  4. Apr 21, 2007 #3
    no idea? I take it your saying its just [tex]\frac{C}{4^2}[/tex]

    Edit: oh nvm, you just ignore the part after B.. then add it in later...
     
    Last edited: Apr 21, 2007
  5. Apr 21, 2007 #4
    [tex]3s + 1 = \frac{A}{(s+2)} + \frac{B}{(s+2)^2} + \frac{C}{4} + \frac{D}{4^2}[/tex]

    seems the more logic answer to me ; but I personally have not yet studies partial fractions . If , underneath you have :
    (s+2)² + 4²
    , then why would it not be A divided by (s+2) + B divided by (s+2) instead of B being divided by (s+2)² ?

    thought Like I said, I have not yet learnt these. If anybody can correct me , as I'm probably wrong , it would be appreciated of course.
     
  6. Apr 21, 2007 #5

    Gib Z

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    Partial Fractions Only work like that on FACTORS of the denominator. Is 4 a factor or just a term?
     
  7. Apr 21, 2007 #6

    Looks to me like a Lapace Transform. No partial fractions required, split them and do some algebra work and you shell get your transforms. Yeah, definately a Lapace Transform, hey If you need any help, you are welcome to pm me, I can help you out. I saw your recent threads, that give me alittle clue. Here is a big hint, there will be a cosine and a sine answer.
     
    Last edited: Apr 22, 2007
  8. Apr 22, 2007 #7
    This has an interesting solution which comes from splitting (s+2)^2+4^2 into imaginary parts. However we then have a new term in i, so that we need three equations to solve (for the numerical value, for the value in s and for the value in i):

    [tex]\frac{3s+1}{(s+2)^2+4^2}=\frac{A}{s+2+4i}+\frac{B}{s+2-4i}+\frac{C}{(s+2)^2+4^2}[/tex]

    Then cross multiplying on the first two terms and adding the third gives:
    A(s+2-4i)+B(s+2+4i)+C=3s+1. Giving A=B=3/2, C=-5.

    Now should this be done in order to facilate integration, since A and B are the same, by cross multiplying the first two terms, we can eliminate the i term and are left with:

    [tex]\frac{3(s+2)}{(s+2)^2+4^2}-\frac{5}{(s+2)^2+4^2}[/tex] So that integration is easy to perform, the first resulting in a log form and the second ,with the substitution 4u=s+2, in an arctangent form:

    (3/2)In(s+2)^2+4^2) -(5/4)arctangent((s+2)/4)

    Of course for integration it is now clear the only form needed is: [tex]\frac{3(s+2)}{(s+2)^2+4^2}-\frac{5}{(s+2)^2+4^2}[/tex]
     
    Last edited: Apr 22, 2007
  9. Apr 22, 2007 #8

    Gib Z

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    Umm Tiny error robert, (s+2)^2 + 4^2 factors into (s+2+2i)(s+2-2i). And Why did you put the 3rd fraction?
     
  10. Apr 22, 2007 #9
    My bad, I didn't mean separate them into two equations. I mean split them and use the Lapace Transform identity of sine and cosine functions.
     
  11. Apr 22, 2007 #10

    Gib Z

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    Ahh actually I see Why We need the C, because I tried it without that and It turned out 3 had to equal 1/2, So i needed a constant term..Let me do it one second
     
  12. Apr 22, 2007 #11
    why go through all that robert? you can do it so much quicker by separation?
     
  13. Apr 22, 2007 #12

    Gib Z

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    [tex]\frac{3s+1}{(s+2)^2+4^2}=\frac{A}{s+2+4i}+\frac{B} {s+2-4i}+\frac{C}{(s+2)^2+4^2}[/tex]

    Cross multiply, equate co efficents.

    [tex]3s+1 = (A+B)s + (2A+2B + C) + (4B-4A)i[/tex]

    A+B=3
    2A+2B + C =1
    Since there are no imaginary terms in 3s+1, 4B-4A = 0.

    Since 4B-4A=0, B=A. But A+B=3, A=B=1.5
    Putting those values is to 2A+2B+C=1 gives C= -5.

    [tex]\frac{3s+1}{(s+2)^2+4^2}=\frac{3}{2(s+2+4i)}+\frac{3} {2(s+2-4i)}-\frac{5}{(s+2)^2+4^2}[/tex]
     
  14. Apr 22, 2007 #13

    Gib Z

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    Ok thanks for that, I just thought i could be treated like any other constant therefore didnt need any special treatment, but I see why im wrong thanks.
     
  15. Apr 22, 2007 #14
    mathPimpDaddy:why go through all that robert? you can do it so much quicker by separation?

    I don't know, except the original question ask to find a partial fraction form. I am adding the assumption that this is for integration, since that is frequently the case.
     
  16. Apr 22, 2007 #15
    I've checked his previous threads, they are all forms of laplace transforms. Look at his original form, doesn't it look familiar?
     
    Last edited: Apr 22, 2007
  17. Apr 26, 2007 #16
    Personally, I avoid imaginary terms whenever possible, so factor the 3 out of the numerator and proceed with the common denominator.

    [tex]3\(\frac{s+\frac{1}{3}} {(s+2)^2+4^2}\)[/tex]
     
    Last edited: Apr 26, 2007
  18. Apr 27, 2007 #17
    simple partial fractions help (warning complex analysis :P )

    1. The problem statement, all variables and given/known data
    the question can be ignored - it involves laplace and Z transforms of RLC ckts.

    Code (Text):

    Vc(s) =          0.2
               -----------------
                 s^2 + 0.2s + 1
     
    find the partial fraction equivalent such that it is :
    Code (Text):

      -j(0.1005)     +    j (0.1005)
    --------------    ------------------
    s + 0.1-(0.995)    s + 0.1 + j(0.995)
     
    2. Relevant equations

    none

    3. The attempt at a solution
    Code (Text):

          0.2                      A                     B
    ---------------  =  ---------------------  +  -------------------
    s^2 + 0.25 + 1      s + (0.1 - j(0.995)))     s + (0.1 + j(0.995))
     
    0.2 = A(s + 0.1 + j(0.995)) + B(s + (0.1 - j0.995))

    0.2 = As + A(0.1 + j(0.995)) + Bs + B(0.1 - j0.995)

    so As + Bs = 0
    or (A + B) = 0
    or A = -B
    so
    0.2 = j(0.995A) - j(0.995B)

    somethings not right - if i evaluate this I don't get anywhere near the answer
     
  19. Jul 22, 2007 #18
    The first one seems right, but I think you should simplify the denominator:

    [tex]\frac{3s + 1}{s^2 + 2s + 4 + 16} = \frac{3s + 1}{s^2 + 2s + 20} = \frac{3s + 1}{(s + 1 + \sqrt{19}j)(s + 1 - \sqrt{19}j)} [/tex]

    So now you can do partial fractions:

    [tex]\frac{3s + 1}{(s + 1 + \sqrt{19}j)(s + 1 - \sqrt{19}j)} = \frac{A}{(s + 1 + \sqrt{19}j)} + \frac{A^*}{(s + 1 - \sqrt{19}j)}[/tex]

    I'm going to assume you know how to solve for A and A* so i'll just post the answer that I got:

    [tex]3.078e^{-t}cos(\sqrt{19}t - 0.225)[/tex]
     
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