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Laplace Transforms (Partial Fractions - complex roots)

  1. Dec 9, 2008 #1
    Just when I thought I had Laplace transforms figured out with partial fractions, I once again fall flat on my face.

    The following question from my book labeled " Inverse Laplace Transforms - Partial Fractions (Complex Roots)

    NOTE: Since I can't draw S(s) and T(s) very well, I have replaced them with X(s) and Y(s) respectively; f(s)=f(x),{s=x} and f(t)=f(y),{t=y}

    [​IMG]



    So I first thought this should be easy... Until I noticed normal factoring wouldn't work. I attempted to use the quadratic equation...

    [​IMG]



    Once I got the factors I figured it should be smooth sailing. I should simply be able to solve for A and B. From there it's simply reversing the Laplace transforms...

    [​IMG]



    This is where I hit a dead end road block. I can't find ANYTHING in my book of how to deal with imaginary numbers in formula(equations) you wish to transform(or un-transform). I certainly didn't notice anything that could cancel the imaginary number (i) out.

    [​IMG]



    This is the answer my book gives, but as with the last example, the book proved to be wrong. I wouldn't put it past them for a second error.

    [​IMG]



    Any help with this would be greatly appreciated!

    Thank you for your time!~
     
  2. jcsd
  3. Dec 9, 2008 #2

    rock.freak667

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    Try completing the square of the denominator, then rewrite the numerator as (4x-4)+4
     
  4. Dec 9, 2008 #3
    I don't know why, but for some reason your idea gave me a idea to wrap my attempt up with an answer. By treating the complex number as a mere constant I came up with this;

    [​IMG]


    I can't really do anything with it, but I am pretty sure its a legit answer...


    Any input on that?
     
  5. Dec 9, 2008 #4

    rock.freak667

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    Not sure how beneficial imaginary numbers are in laplace transforms, but if to solve an ordinary differential equation, it would be better to have real numbers.
     
  6. Dec 9, 2008 #5

    gabbagabbahey

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    hmmm...there are a couple of small errors....

    (1)This first one doesn't have any outcome on your final result, but the 'b' in your quadratic equation is -2 not +2, so your roots are [itex]x_{\pm}=+1 \pm i[/itex]....Luckily your next line is correct, since [itex]x^2-2x+2=(x-x_{+})(x-x_{-})=(x-(1+i))(x-(1-i))[/itex]. it seems as though you made two errors which canceled eachother out.

    (2)Your solution for A and B is a little off; go over that part again. This error will propogate through to your final result.
     
  7. Dec 9, 2008 #6

    gabbagabbahey

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    There is nothing wrong with expressing the solution in terms of complex nubers, although one can also express it in terms of sines and cosines.
     
  8. Dec 9, 2008 #7
    Hmm... Ah, I see my errors now! Despite the fact they are small, they would no doubt cost me points on a test...

    Anyway, thanks a lot abbagabbahey on the verification regarding my final answer represented with complex numbers. Makes me feel a a little more comfortable doing these problems
     
  9. Dec 9, 2008 #8

    gabbagabbahey

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    Well your final answer is a little off; you should end up with [tex](2i+2)e^{y(1-i)}-(2i+2)e^{y(1+i)}[/tex] which is equivalent to the answer in your text (you can check this using Euler's formula)
     
  10. Dec 1, 2009 #9
    How did you end up with 2i+2 instead of 2/i+2 like he had?
     
  11. Dec 2, 2009 #10

    gabbagabbahey

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    [tex]\frac{2}{-i}=2i[/tex] :wink:
     
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