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Diff eq problem need help

  1. Oct 11, 2006 #1
    (s+1)/s*(s^2+s+1)
    find laplace transform..
    i am confused how to solve this problem
    i did numerator seperation and i did A/s + (Bs+c)/(s^2+s+1) but still doenst work..so experts..i do really need ur help..thnx in advance
     
  2. jcsd
  3. Oct 11, 2006 #2

    benorin

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    Homework Helper

    [tex]\frac{s+1}{s(s^2+s+1)}=\frac{1}{s}-\frac{s}{s^2+s+1}=\frac{1}{s}-\frac{s\frac{1}{2}}{\left(s+\frac{1}{2}\right)^2+\frac{3}{4}}+\frac{\frac{1}{2}}{\left(s+\frac{1}{2}\right)^2+\frac{3}{4}}=\frac{1}{s}-\frac{s+\frac{1}{2}}{\left(s+\frac{1}{2}\right)^2+\frac{3}{4}}+\frac{1}{\sqrt{3}}\cdot\frac{\frac{\sqrt{3}}{2}}{\left(s+\frac{1}{2}\right)^2+\frac{3}{4}}[/tex]​

    so the inverse transformation is

    [tex]\mathcal{L}^{-1}\left\{\frac{s+1}{s(s^2+s+1)}\right\}=\mathcal{L}^{-1}\left\{\frac{1}{s}-\frac{s+\frac{1}{2}}{\left(s+\frac{1}{2}\right)^2+\frac{3}{4}}+\frac{1}{\sqrt{3}}\cdot\frac{\frac{\sqrt{3}}{2}}{\left(s+\frac{1}{2}\right)^2+\frac{3}{4}}\right\}=u(t)-u(t)e^{-\frac{1}{2}t}\cos\left(\frac{\sqrt{3}}{2}t}\right)+u(t)e^{-\frac{1}{2}t}\sin\left(\frac{\sqrt{3}}{2}t}\right)[/tex]​

    where [tex]u(t)[/tex] is the unit step function and I got the inverse transforms from here.
     
    Last edited: Oct 11, 2006
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