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Help with inverse Laplace transforms

  1. Dec 30, 2008 #1
    1. The problem statement, all variables and given/known data

    I have three inverse laplace transforms I can't solve, they are,

    i)
    (s-1)/(s^2 + 8s + 17)

    ii)
    (s+3)/(s^2 + 4s)

    iii)
    2/[(s+1)*(s^2 + 1)]


    2. Relevant equations

    The laplace transform table,

    http://en.wikipedia.org/wiki/Laplace_transform#Table_of_selected_Laplace_transforms

    3. The attempt at a solution


    i) I completed the square and got,

    (s-1)/(s^2 + 8s + 17)

    -->(s-1)/[(s+4)^2 -1]

    then split up into, s/[(s+4)^2 -1] and -1/[(s+4)^2 -1]

    my answer was e^-4t *cos(t) - e^-4t *sin(t)
    but the answer is e^-4t *cos(t) - 5e^-4t *sin(t) , I do not know where the 5 has come from ?


    ii) I'm not sure where to start on this one

    (s+3)/(s^2 + 4s) , I have taken out a factor of 1/s but im not sure were to go from there.


    iii) For this one, I'm not having problem with the actual laplace but rather partial fractions, what I have done,

    2/[(s+1)*(s^2 + 1)] = A/(s+1) + B/(s^2 + 1)

    A(s^2 + 1) + B(s+1) = 2

    for s=-1 I found A= 1

    but I'm stuck at finding B, how do I make (s^2 + 1) = 0

    Thanks a lot in advance !
     
  2. jcsd
  3. Dec 30, 2008 #2
    Part 3:

    It should be A/(s+1) + (Bs + c)/(s2 + 1)

    Hence

    2 = A.(s2 + 1) + (Bs +c).(s + 1)

    You can then find all the values using s = -1, 0 and 1.
     
  4. Dec 30, 2008 #3
    No, (s+4)^2 - 1 = s^2 + 8s + 16 - 1 = s^2 + 8s + 15, you have 17

    Note that s^2 + 4s = (s+2)^2 - 4. Does that help?

    Your setup is wrong, s^2 + 1 has no real roots therefore the proper way to try to split it up is to rewrite it as A/(s+1) + (Bs + C)/(s^2 + 1). Work with that.
     
  5. Dec 30, 2008 #4
    1) (s+4)^2 -1 = s^2+8s+16-1 .. but that's not 17?
    2) Is s/[(s+4)^2 -1] = sin .. ? You should look at your numerator

    NoMoreExams beat me so only look at 2 ..
     
  6. Dec 30, 2008 #5
    First it is just wrong as NotStine mentioned. Second, you might want to review some partial fractions. You not always make things equal to 0 .. Use linear algebra
     
  7. Dec 30, 2008 #6
    Also if you ARE allowed to use a table, you can use this: http://www.vibrationdata.com/Laplace.htm

    i) (s-1)/(s^2 + 8s + 17) = (s-1)/((s + 4)^2 + 1). By equation 2.27, lambda = -1, alpha = 4, beta = 1. So plugging that in we get e^(-4t)[cos(t) - 5sin(t)]

    etc.
     
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