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Where am i going wrong? partial fraction question

  1. Aug 12, 2014 #1
    1. The problem statement, all variables and given/known data

    Doing a homework question and I ran it through wolfram and I get a different answer to what I'm working it out as....and I can't see where I'm going wrong. Anyone able to give a pointer?

    my equation is
    [tex]\frac{1}{s(0.641s + 1)}[/tex]

    Wolfram gives the answer as


    2. Relevant equations

    3. The attempt at a solution

    Using partial fraction decomposition I get (using cover up) -

    [tex]\frac{1}{s(0.641s + 1)} = \frac{A}{s} + \frac{B}{0.641s+1}[/tex]

    [tex]A = \frac{1}{s(0.641s + 1)} ~~~~~~ where~s~ = 0[/tex]

    [tex]=\frac{1}{0.641.0 +1}[/tex]



    [tex]B = \frac{1}{s}[/tex]





    [tex] ∴B=\frac{1}{s}[/tex]

    [tex] ∴B=\frac{1}{-1.56}[/tex]

    [tex] ∴B=-0.641[/tex]

    [tex] Inserting~A~and~B~into~the~decomposition[/tex]

    [tex]\frac{1}{s} + \frac{-0.641}{0.641s+1}[/tex]

    which I get to be

    [tex]\frac{1}{s} - \frac{1}{s+1}[/tex]

    Where am I making my schoolboy error? It's my first go at partial fractions and my text book aint exactly crammed with examples. Anyone able to point out if I've made a glaring error or just something stupid...any help would be appreciated.
  2. jcsd
  3. Aug 12, 2014 #2


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    You might want to look again at setting s = 0. Given it's a factor of the denominator.
  4. Aug 12, 2014 #3


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    Let's begin by factoring ##.641## out of the denominator:$$
    \frac{1}{s(0.641s + 1)}=\frac 1 {.641}\cdot \frac 1 {s(s+\frac 1 {.641})}=
    1.56\cdot \frac 1 {s(s+1.56)}=
    \frac a {s(s+a)}$$where ##a=1.56##. You want the fraction in that form to do the "cover up" method, from which you get$$
    \frac 1 s + \frac {-1}{s+a}$$which agrees with Wolfram. I suggest you take it from there.
  5. Aug 12, 2014 #4


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    There. If you wanted to divide the top and bottom by 0.641, you forgot one of the terms in the fractiom.

    Ignore that. Maybe PeroK doesn't understand the "cover up rule".
  6. Aug 12, 2014 #5


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    If the cover up rule allows you to divide by 0 and then tweak the distributive law, then I certainly don't understand it.

    [tex]A = \frac{1}{s(0.641s + 1)} ~~~~~~ where~s~ = 0[/tex]

    [tex]=\frac{1}{0.641.0 +1}[/tex]


    That's simply two algebraic wrongs making a right!
  7. Aug 12, 2014 #6


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    If A/x+ B/(x- a)= C/(x(x- a))
    Then multiplying by x(x- a) we have A(x- a)+ Bx= C.
    There is now no denominator so we can, without difficulty, let x= 0 and x= a getting
    x= 0, -aA= C so A= -C/a and
    x= a, aB= C so B= C/a.

    Notice that
    [tex]\frac{(-C/a)}{x}+ \frac{C/a}{x- a}= \frac{(-(C/a)x+ C}{x(x- a)}+ \frac{(C/a)x}{x(x= a)}= \frac{C}{x(x- a)}[/tex]
    as desired.
  8. Sep 8, 2014 #7
    Thanks for the replies.......I realised where I went wrong suddenly, ie- had to make it 'standard form' and needed a 1 on the numerator instead of the 0.641....so manipulated the equation and it all fell into place.

    So once again, thankyou all for the help.
  9. Sep 8, 2014 #8

    Ray Vickson

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    If you do it properly, it is more-or-less straightforward. You went at it the wrong way, so made it longer and more complicated than necessary.

    [tex] \frac{1}{s(1+as)} = \frac{A}{s} + \frac{B}{as+1}[/tex]
    where ##A,B## are constants. Put the above over a common denominator:
    [tex] \frac{A}{s} + \frac{B}{as+1} = \frac{A(as+1)+Bs}{s(as+1)}[/tex]
    This supposed to equal ##1/[s(as+1)] ## for ALL ##s##, so the numerators much match exactly (because the denominators match). The numerator is ##A(as+1)+ Bs = (Aa+B)s + A##, so we need
    [tex] (Aa+B)s + A = 1[/tex]
    for ALL ##s##. The polynomial ##P(s) = (Aa+B)s + A## must satisfy ##P(s) = 1## for all ##s##; that means its constant term must = 1 and the coefficient of ##s## must vanish, so ##A = 1## and ##Aa+B = 0##. In other words, we have
    [tex] \frac{1}{s(1+as)}= \frac{1}{s} - \frac{a}{as+1}[/tex]
  10. Sep 8, 2014 #9


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    The OP made an algebra error, and the righthand side should have been
    $$\frac{1}{0.641 s + 1}$$ which is defined for ##s=0##.

    (Of course, to get to that point, you multiply the original equation by ##s## and therefore are assuming that ##s\ne 0##, but that's a minor complication.)
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