Inverse Laplace Transform(s/((s^2)+1)^2

  1. 1. The problem statement, all variables and given/known data
    Find inverse Laplace transform of:
    s/((s^2)+1)^2

    2. Relevant equations



    3. The attempt at a solution
    The answer is:
    (t/2)sin t
    First shifting theorem or partial fractions don't work, I think. From a Laplace transforms table, sin at = a/((s^2)+(a^2)), which is almost the form that the problem is in.
    Thanks
     
  2. jcsd
  3. hunt_mat

    hunt_mat 1,583
    Homework Helper

    Here is something that may help, take the sin transform and differentiate it w.r.t s and what do you get?

    Mat
     
  4. Thanks for replying.
    d/ds of sin transform = -2sa/(s^2+a^2)^2
    This would give an inverse of the problem of:
    (d/ds)-sin(t)/2
    I'm not sure I did that right.
     
  5. hunt_mat

    hunt_mat 1,583
    Homework Helper

    Set a=-1 and what do you get?
     
  6. hunt_mat

    hunt_mat 1,583
    Homework Helper

    Not quite, we have:
    [tex]
    \frac{1}{s^{2}+a^{2}}=\int_{0}^{\infty}e^{-st}\sin atdt
    [/tex]
    Differentiate this w.r.t and tell me what you get?
     
  7. I'm getting:
    1/(a-s)
    with integration-by-parts twice.
     
  8. Mark44

    Staff: Mentor

    No, hunt_mat said to differentiate, not integrate.
     
  9. Oh, right, sorry.
    (d/ds)1/(s^2+a^2)=-2s/(s^2+a^2)^2
     
  10. hunt_mat

    hunt_mat 1,583
    Homework Helper

    and on the RHS?
     
  11. RHS:
    integral from 0 to infinity(-s(e^-st)sin atdt
     
  12. hunt_mat

    hunt_mat 1,583
    Homework Helper

    set a=-1 and what have you got?
     
  13. If a=-1:
    -2s/(s^2+a^2)^2 = -s/(-s-1)
     
  14. Sorry:
    -2s/(s^2+1)^2=-s/(-s-1)
     
  15. hunt_mat

    hunt_mat 1,583
    Homework Helper

    No, if a=-1:
    [tex]
    \frac{s}{(s^{2}+1)^{2}}=\int_{0}^{\infty}\frac{t}{2}e^{-st}\sin tdt
    [/tex]
    What does that tell you?
     
  16. I know then that:
    L{tsin(at)/2}=s/(s^2+a^2)?
    I'm not sure.
     
  17. hunt_mat

    hunt_mat 1,583
    Homework Helper

    Let me re-arrange the RHS:
    [tex]
    \int_{0}^{\infty}\frac{t\sin t}{2}e^{-st}dt
    [/tex]
    Does it make sense now?
     
  18. No, I'm sorry, I'm just not seeing it. Thanks for all your help, though.
     
  19. hunt_mat

    hunt_mat 1,583
    Homework Helper

    Think of it like this, you know that:
    [tex]
    \frac{1}{1-x}=1+x+x^{2}+x^{3}+x^{4}+\cdots
    [/tex]
    Now differentiate both sides to show that:
    [tex]
    \frac{1}{(1-x)^{2}}=1+2x+3x^{2}+4x^{3}+\cdots
    [/tex]
    and now you know the power series for [tex](1-x)^{2}[/tex] is just by differentiating the series.
     
  20. So I know that differentiating both sides of the sin(at) LT will give me the square that I need in the denominator, but then I'm stuck with the d/ds on the sin side. I'm not sure what to do after this.
    Thanks
     
  21. hunt_mat

    hunt_mat 1,583
    Homework Helper

    when you differentiate the integral you take in differentiation under the integral sign:
    [tex]
    \frac{d}{ds}\int_{0}^{\infty}e^{-st}\sin tdt=\int_{0}^{\infty}\frac{d}{ds}\left( e^{-st}\sin t\right) dt
    [/tex]
    Differentiating the integrand:
    [tex]
    \frac{d}{ds}e^{-st}\sin t=-te^{-st}\sin t
    [/tex]
    As now we regard s as variable and t is fixed. So the above can be written as:
    [tex]
    (-t\sin t)e^{-st}
    [/tex]
    and so you're taking the Laplace transform of something different now. That something is the answer to your question. Is this making sense now?
     
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