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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.

    25/(s-1)=A(4+s^2)+(Bs+C)(s-1)


    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.
    [tex]\frac{25}{(1-s)^2(s^2+4)}=\frac{A}{1-s}+\frac{B}{(1-s)^2}+\frac{Cs+D}{s^2+4}[/tex]

    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
    [tex]\frac{s+U}{(s+a)^2+b^2}[/tex]
    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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