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Homework Help: Inverse Laplace Tranform

  1. Dec 6, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]Y(s) = \frac{1}{(s-2)^{2}(s+1)^{3}}[/tex]


    2. Relevant equations
    ...


    3. The attempt at a solution

    Ok, so my goal is to find the Laplace transform of the above. I need to use partial fraction decomposition to break up the denominator. With that in mind, I have the following:


    [tex]Y(s) = \frac{1}{(s-2)^{2}(s+1)^{3}} = \frac{A}{s-2}+ \frac{B}{(s-2)^{2}} + \frac{C}{s+1} + \frac{D}{(s+1)^{2}}+ \frac{E}{(s+1)^{3}}[/tex]

    After this I multiplied by the common denominator and came up with 5 equations. Those equations I came up with are:

    A + C = 0
    A + B - 2C + D = 0
    -3A + 3B - 3C -3D + E = 0
    -5A + 3B + 4C -4E = 0
    -2A + B + 4C + 4D + 4E = 1

    Now, once solving those I get:

    A = 4/27
    B = 8/27
    C = -4/27
    D = 7/27
    E = 3/27

    However, that's wrong, I am wondering if my setup is correct. I can post all my algebraic work if necessary.
     
  2. jcsd
  3. Dec 6, 2008 #2

    gabbagabbahey

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    so far so good:smile:...

    This can't be right; using these values, [itex]A+B-2C+D==27/27=1\neq0[/itex] which clearly contradicts your second equation....
     
  4. Dec 6, 2008 #3
    This post...unnecessary.
     
  5. Dec 6, 2008 #4
    Hmm...well, let me double check my matrix in my calculator. Thanks for pointing that out to me.

    I found my problem, I had swapped a 1 with a 0 in my matrix on my calculator. My new values are:

    A = -1/27
    B = 1/27
    C = 1/27
    D = 2/27
    E = 1/9

    Thank you, after looking at my work for awhile, and double checking the work tends to blur together. I needed an extra pair of eyes. Thanks again!
     
    Last edited: Dec 6, 2008
  6. Dec 6, 2008 #5

    gabbagabbahey

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    That's better :smile:
     
  7. Dec 6, 2008 #6
    Yeah, I have yet to do the inverse transform, but those values are definitely better and look like they will work.
     
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