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Inverse Laplace Transform

  1. Aug 31, 2009 #1
    My books aren't clear at running through these inverse laplace transforms and this ones got me snookered. I'm trying to perform the laplace transform of:

    1/s + 40/(24s^2 + 40s + 40)

    Factor out 24 becomes

    1/s + 1.667/(s^2 +1.667s +1.667)

    Finding A, B, C

    A/s + Bs+C/(s^2 +1.667s +1.667)

    A=1
    B=-1
    C=-1.667

    Substituting back into Eq becomes

    1/s - (s+1.667)/(s^2 +1.667s +1.667)

    However, I'm stuck here as the Eq doesnt look like its in the correct form to use in the inverse laplace tables? Do I need to manipulate denominator to present as (s+...)(s+...) using roots? Roots are found to be -0.833 +/- j0.986

    Or split numerator into two parts ie 1/s - (s+1)/(s^2 +1.667s +1.667) + 1.667/(s^2 +1.667s +1.667)

    Or both??

    1/s - (s+1)/(s+(-0.833)) + j0.986) + 1.667/(s+(-0.833)) - j0.986)

    Using the transform table at:

    http://www.swarthmore.edu/NatSci/ec...nMethods/LaplaceZTable/LaplaceZFuncTable.html

    Item 16 in the table looks like similiar format but I believe Item 18 is closer after finding wn=1.29 & zeta=0.646 however, using item 18 doesn't account for the additional s in the numerator??

    I feel I'm close but not quite there to finding this solution and its frustrating me!! Any help or direction would be much appriciated.
     
  2. jcsd
  3. Sep 1, 2009 #2

    gabbagabbahey

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    I'm not quite sure what you are trying to do here, but it seems like you are attempting to use partial fraction decomposition, without actually decomposing the denominator. Where did you get those values of 'A', B' and 'C' from? It should be clear that A=1, B=0 and C=5/3 reproduces your original expression, so the values you've found are clearly incorrect!

    Instead of partial fraction decomposition, try completing the square on the denominator of your [itex]\frac{5}{3(s^2+\frac{5}{3}s+\frac{5}{3})}[/itex] term and then applying the frequency shift rule.

    After all, inverse Laplace Transforms are linear, so you should know that

    [tex]\mathcal{L}^{-1}\left[\frac{1}{s}+\frac{5}{3(s^2+\frac{5}{3}s+\frac{5}{3})}\right]=\mathcal{L}^{-1}\left[\frac{1}{s}\right]+\mathcal{L}^{-1}\left[\frac{5}{3(s^2+\frac{5}{3}s+\frac{5}{3})}\right][/tex]

    and so the only difficulty you should have is with the second term.
     
    Last edited: Sep 1, 2009
  4. Sep 2, 2009 #3
    I found A,B,C using partial fraction expansion of second order polynomial where 5/3 = 1.667 (from my earlier post) i.e.

    1/s + (5/3)/(s^2 + 5/3s + 5/3) = A/s + (Bs + C)/(s^2 + 5/3s + 5/3)

    5/3 = A (s^2 + 5/3s + 5/3) + (Bs + C) s

    5/3 = As^2 + 5/3As + 5/3A + Bs^2 + Cs

    5/3 = s^2 (A+B) + s (5/3A + C) + 5/3A

    0 = s^2 (A + B)
    0 = s^1 (5/3A + C)
    5/3 = s^0 (5/3A)

    Therefore:

    5/3 = 5/3A
    A = 1

    0 = A + B
    0 = 1 +B
    B = -1

    0 = 5/3A + C
    0 = 5/3 + C
    C = -5/3

    A/s + (Bs + C)/(s^2 + 5/3s + 5/3) = 1/s - (s+5/3) /(s^2 + 5/3s +5/3)

    Is this method not correct to find A,B,C? I know there are different methods out there but this one seems the quickest to perform.
     
  5. Sep 2, 2009 #4

    gabbagabbahey

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    No, it's not correct.

    For starters, the point of partial fraction decomposition is to break compound fractions into simpler fractions... that means that you want to break factions with denominators of order 2 or higher into fractions with denominators of lower order. What you're trying to do here:

    Takes a fraction with a quadratic denominator and decomposes it into another fraction with a quadratic denominator...what's the use of that?

    Not only that, but the values of A,B and C you come up with are incorrect....just plug in s=1 to your original expression, and your final expression...do you get the same number?

    Instead of using partial fractions, just complete the square on the denominator of your second term
     
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