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Homework Help: Inverse Laplace Transformation

  1. Mar 28, 2006 #1
    Okay I really really need somebody to help me :cry:

    Find the Inverse Laplace Transformation of [itex]\{ 1/(s^3 + 1)\}(t)[/itex]

    (for those of you who don't know, you look it up in a table. the closest thing that I can find is [itex]\{ 1/(s^2 + 1)\}(t)[/itex] which is sin(t) )

    Well, I started of by breaking up [itex]s^3 + 1[/itex] into [itex](s+1)(s^2 - s + 1)[/itex]

    After this, im so lost because [itex]s^2 - s + 1[/itex] cannot be broken up any further....

    anybody have any ideas?
    Last edited by a moderator: Mar 28, 2006
  2. jcsd
  3. Mar 28, 2006 #2


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    s2-s+1 can be written as a perfect square plus something:
    s2-s+ 1/4+ 1-1/4= s2-s+ 1/4+ 3/4
    = (s- 1/2)2 + 3/4
  4. Mar 28, 2006 #3
    HallsofIvy, thank you for that.
    I was thinking of completeing the square, but it made things even more messy.

    The question no becomes:

    Take the inverse leplace transformation of

    [itex]\{ 1/(s^3 + 1)\}(t)[/itex]
    =[itex]\{ 1/(s+1)((s- 1/2)^2 + 3/4)\}(t)[/itex]
    =[itex]\{ 1/(s+1)(s- 1/2)^2\}[/tex] + [itex]\{ 1/(s+1)(3/4))\}(t)[/itex]

    and this inverse leplace transformation is too messy: [itex]\{ 1/(s+1)(s- 1/2)^2\}[/itex]
    Last edited by a moderator: Mar 28, 2006
  5. Mar 28, 2006 #4
    no that's not too messy man! You can do it! You mean:

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

    In fact you don't don't even have to bother finding any constants, these should be right in your table.
    Last edited: Mar 28, 2006
  6. Mar 28, 2006 #5
    you made an error, im looking for the Inverse Laplace Transformation of:
    [tex] \frac{1}{(s+1)(s-(1/2))^2} [/tex]

    I looked that up in my text book, and no, its not in the tabel :(

    Then i tried partial fractions to break it up and boy, that too isn't any easier...

    So I have absolutly no idea what to do with this?
    Last edited by a moderator: Mar 28, 2006
  7. Mar 28, 2006 #6
    [tex]\frac{t^{n}}{n!}e^{-\alpha t} \cdot u(t) =\frac{1}{(s+\alpha)^{n+1}}[/tex]

    Don't be upset if that one's not in your book, I had to fish around to find it my first time too. From there I think you can do a partial fraction expantion right? maybe?
  8. Mar 28, 2006 #7
    Im so lost right now..... :(
  9. Mar 28, 2006 #8
    I want to break [tex] \frac{1}{(s+1)(s-(1/2))^2} [/tex] = [tex] \frac{A}{(s+1)} + \frac{B}{(s-(1/2)} + \frac{C}{(s-(1/2))^2} [/tex] = [tex] \frac{A}{(s+1)} + \frac{B((s-(1/2)) + C}{(s-(1/2))^2} [/tex]

    I tried solving for A, B and C like 3 times and got different answers :(
    Last edited by a moderator: Mar 28, 2006
  10. Mar 29, 2006 #9


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    Obviously you aren't going to find an inverse transform for
    in a table- reduce it by partial fractions.

    As you say, you want
    [tex]\frac{1}{(s+1)(s-\frac{1}{2})^2}= \frac{A}{s+1}+ \frac{B}{(s-\frac{1}{2})}+ \frac{C}{(s-\frac{1}{2})^2}[/tex]

    Multiply on both sides by the common denominator and you get
    [tex]1= A(s-\frac{1}{2})^2+ B(s+1)(s-\frac{1}{2})+ C(s+1)[/tex]
    Taking x= -1 gives 1= A(-3/2) so A= -2/3.
    Taking x= 1/2 gives 1= c(3/2) so C= 2/3.
    Taking x= 0 gives 1= -A/2- B/2+ C or 1= -1/3- B/2+ 2/3 so B/2= 2/3- 1/3- 1= -2/3 and B= -1/3.
  11. Mar 29, 2006 #10
    You forgot to square the factor in your A equation, and you solved for B incorrectly.
    s=-1 gives 1 = A(-3/2)^2 so A = 4/9.
    s=1/2 gives 1 = C(3/2) so C = 2/3
    s=0 gives 1 = A(-1/2)^2 + B(-1/2) + C or 1 = A/4 - B/2 + C so B/2 = -1 + 1/9 + 2/3 = -2/9. So B = -4/9.

  12. Mar 29, 2006 #11
    omg, i was actually expanding this:

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

    and trying to equate cooeffiecnts... which got soooo messy....
  13. Mar 29, 2006 #12
    okay, this is getting ridiculus.

    After solving for A, B, C, and substituting the values back into the original equation, we get...

    [tex]\frac{1}{(s+1)(s-\frac{1}{2})^2}= \frac{(4/9)}{s+1}+ \frac{(-4/9)}{(s-\frac{1}{2})}+ \frac{(2/3)}{(s-\frac{1}{2})^2}[/tex]

    The first two terms for an Inverse Leplace Transform are easy, however, the thrid term is messy again.

    How on earth do you take the inverse leplace transform of this:


    do you complete the square again?
  14. Mar 29, 2006 #13
    Never mind, I figured this one out.

    You have to play around with one of the transforms in the tabel in the book, then it works.
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