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Laplace transform integral problem

  1. Mar 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Hey guys, i've attached the question that's troubling me. I've also attached the table and formulas of Laplace transform for you convenience.




    2. Relevant equations

    attached

    3. The attempt at a solution

    Right now, im thinking the best way to do this problem is by working backwards. I'm using the convolution integral formula.
    I've already established that G(s) is 1/(s+4) using inverse Laplace transforms but im not sure if that will get me an answer for f(t) rather F(s).

    thats my inverse laplace------->invL
    and laplace is ------> L
     

    Attached Files:

  2. jcsd
  3. Mar 23, 2012 #2
    When you take the Laplace transform of both sides of the equation, what are you getting? Where are you getting stuck?
     
  4. Mar 23, 2012 #3
    I think I'm stuck because I'm not sure how to approach a problem like this. anyway, Laplace transform of both sides just makes it L{f(t)}= F(s)G(s) right???? no thats not right coz there the 6t-5
     
    Last edited: Mar 23, 2012
  5. Mar 23, 2012 #4
    edit:
    I am incorrect. Sorry.
     
    Last edited: Mar 23, 2012
  6. Mar 23, 2012 #5
    so that means h(t)= 6t so H(t) = 6/s^2

    therefore
    F(s) = 6/s^2 + 5/(s+4) F(s),
    rearrange and F(s) = 6(s+4) / s^2(s-1)
    and inverse Laplace it and f(t) = 6*(-4t + 5 e^t - 5)

    is that right, im not sure if my calculations are correct.
     
    Last edited: Mar 23, 2012
  7. Mar 23, 2012 #6
    i just saw your last post....r u sure ur wrong coz it actually made sense to me...
     
  8. Mar 24, 2012 #7
    The limits of integration in the problem are from 0 to t, which makes me feel uncomfortable about my answer. If it were convolution, the limits would be -infinity to infinity. If f(u) is nonzero only for u > 0, then the limits could be from 0 to infinity. I'm just unsure why the top limit is t.
     
  9. Mar 24, 2012 #8
    in the formula table i've provided, for convolution the limits says 0 to t.
     
  10. Mar 24, 2012 #9
    If that's the case, then what I said made sense. I suppose G(i) is also zero for i < 0. So G(t-u) is zero outside u > t.
     
  11. Mar 24, 2012 #10
    [tex]F(s) = \frac{6}{s^2} - 5 \frac{1}{s+4} F(s)[/tex]
    Note the minus sign. Your next step would have been right if it had been addition (assuming a/bc is atually a/b/c). It is actually:
    [tex]F(s) = 6\frac{s+4}{s^2(s+9)}[/tex]

    Your final answer also WOULD have been right without the sign error. Give it a try again with the proper minus sign this time.
     
  12. Mar 24, 2012 #11
    ah i see so then the value of f(t) is 6 ((4t)/9 - 5e^(-9 t))/81 + 5/81 correct?
     
  13. Mar 24, 2012 #12
    or (8*t)/3 - (10*exp(-9*t)/27) + 10/27
     
  14. Mar 24, 2012 #13
    This has a lot of right terms in it, but it looks like you messed up on your parentheses.

    [tex]6(\frac{4t}{9} - \frac{5e^{-9t}}{81}+\frac{5}{81})[/tex]
     
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