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Homework Help: Finding the inverse laplace transform

  1. Mar 29, 2012 #1
    I'm attempting to find the inverse laplace transform of

    [tex] \frac{25}{(1-s)^2*(4+s^2)} [/tex]

    I get to this point but can't get the values of A B and C when equating coefficients.


    Also for a separate question: (s+1)/(5-4s+s^2)

    How do you find the inverse laplace transform when you cant perform partial fraction expansion?
  2. jcsd
  3. Mar 29, 2012 #2
    You are not expanding the first one right, because you are ignoring the effect of repeated roots.

    As for the other, my hint is to complete the square in the denominator. These, if I remember correctly, turn out to be a sine plus a cosine (or sometimes just a cosine or a sine).
  4. Mar 29, 2012 #3
    when you complete the square for the 2nd one you get 1+(s-2)^2
    so would you just have one residual and that term in the denominator and solve the inverse laplace from there?
  5. Mar 29, 2012 #4
    Once you complete the square, you should already be very close to a form ready for an inverse Laplace.

    [tex]e^{-at}cos(bt)u(t) \leftrightarrow \frac{s+a}{(s+a)^2+b^2} [/tex]
    [tex]e^{-at}sin(bt)u(t) \leftrightarrow \frac{b}{(s+a)^2+b^2} [/tex]
    So in general, if you have
    You can do the following manipulations:
    [tex]\frac{s+a - a+ U}{(s+a)^2+b^2}=\frac{s+a}{(s+a)^2+b^2}+\frac{U-a}{(s+a)^2+b^2}=\frac{s+a}{(s+a)^2+b^2}+\frac{U-a}{b}\frac{b}{(s+a)^2+b^2} \leftrightarrow e^{-at}cos(bt)u(t)+\frac{U-a}{b}e^{-at}sin(bt)u(t)[/tex]
    where U is a constant. But I see an issue, your a is -2, resulting in e^(2t). Are you doing right-handed Laplace transforms only?
    Last edited: Mar 29, 2012
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