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Inverse Laplace Transforms

  1. Nov 24, 2009 #1
    1. The problem statement, all variables and given/known data
    F(s)=(−2s−2)/((s−1)2+9)

    Find f(x) by computing L-1[F(s)] = f(x).


    2. Relevant equations



    3. The attempt at a solution
    I made put (s-1) in place of s in the numerator, then compensated for that by subtracting 2:
    F(s) = (-2(s-1) - 2 -2 ) / ((s-1)2 + 9)

    Then s-1 becomes s...

    F(s) = (-2s - 4) / (s2 + 32)

    Split this into 2 parts:

    -2s / s2 + 32 and (-4/3)(3 / s2 + 32)

    This gives -2cos(3x) - (4/3)sin(3x) as f(x). But it's wrong... what should I have done differently?
     
  2. jcsd
  3. Nov 24, 2009 #2

    rock.freak667

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    [tex]\frac{-2s-2}{(s-1)^2+9} =-2 \frac{s+1}{(s-1)^2+9}[/tex]


    Now use a similar principle of (a+b)/c = a/c + b/c


    (shift formula will help you here)
     
  4. Nov 24, 2009 #3
    Okay so I split it up:

    -2 [ s / ((s-1)2 + 9) + 1 / (s-1)2 + 9)

    I'll work with s / (s-1)2 + 9 first. Rewrite the denominator as s - 1 + 1. Then split it up into two, so I have:

    s-1 / (s-1)2 + 9 and 1 / (s-1)2 + 9)

    Now working with just the first part of the above. If the shift formula was used then a = 1, and the other part was cos(3t). So this part of f(t) must have been etcos(3t).

    Now for 1 / (s-1)2 + 9. if I multiply the whole thing by 1/3 then I can put a 3 in the denominator, so this part of f(t) is 1/3sin(3t). Since I have two of those (one from the second split) I get 2/3sin(3t).

    Now I multiply everything by -2.

    Final answer: -2etcos(3t) -4/3sin(3t).

    Is that correct?
     
  5. Nov 24, 2009 #4
    I think there's still a problem with the sin(3t). Because I have 1/ (s-1)^2 +9, not just s in the denominator. Did I need to use the shift formula there too? I'm not sure how as I have no s in the numerator to work with!
     
  6. Nov 24, 2009 #5

    rock.freak667

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    yes you have to shift that as well

    [tex]L[sinkt]=\frac{k}{s^2+k^2}[/tex]
     
  7. Nov 24, 2009 #6
    How do you shift it though? I can't put 1-s+s in the numerator, that doesn't help...
     
  8. Nov 24, 2009 #7

    rock.freak667

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    you don't need to, the second term (befor multiplying by -2) is


    [tex]\frac{2}{(s-1)^2+9} = \frac{2}{3}\frac{3}{(s-1)^2+(3^2)}[/tex]


    So you just shift it the same way with a=1
     
  9. Nov 24, 2009 #8
    So -2[etcos(3t) - 0.67etsin(3t)]?
     
  10. Nov 24, 2009 #9

    rock.freak667

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    -2[etcos(3t) + 0.67etsin(3t)]
     
  11. Nov 25, 2009 #10
    Oops yeah, silly mistake. I think I finally understand, yay!! Thanks very much for your help!
     
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