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Homework Help: Laplace transform with 2s(s^2+9) in denominator

  1. Feb 13, 2014 #1
    1. The problem statement, all variables and given/known data

    Hi I would like to know how to expand 20/2s(s^2+9) in order to find the inverse Laplace Transform of the function K(s) to gt k(t). The (s^2+9) term in the denominator is throwing my calculations off for the constants because of the s^2 term.

    2. Relevant equations

    3. The attempt at a solution

    I attempted to use partial fraction decomposition and expanded like so

    c1/2 + c2/s + c3/(s^2+9)

    from there I multiplied both sides by the common denominator, attempted to distribute and collect like terms to equate coefficients in order to find the proper system of equations to find the constants in order to take the inverse laplace of each term to find k(t) however I don't think I am expanding this problem right in the partial decomp and s^2+9 is not factorable any further. Please help!
  2. jcsd
  3. Feb 13, 2014 #2
    Replace c3 with c3s, and lose the first term.

  4. Feb 13, 2014 #3
    Ok, so I expanded it like this

    c1/2 + c2/s + (c3(s)+c4)/(s^2+9)

    got my system of eqs got my constants



    -20/7*(1/s) --> L^-1[ -20/7*(1/s) ] = -20/7*Us(t) from table

    but I am a little unsure how to find inverse laplace for the next term

    L-1[ 20/7*(s) / (s^2+9) ] where 20/7 is c3 the const found c4 was 0 so its gone.
  5. Feb 13, 2014 #4
    Nevermind we got it. It's simply 20/7(s/s^2+3^2) which is 20/7cos(3t).....do you agree?
  6. Feb 13, 2014 #5

    Ray Vickson

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    You could also do it by integrating, using the well-known fact that if ##g(s)## is the LT of ##f(t)##, then ##g(s)/s## is the LT of ##\int_0^t f(u) \, du##. So, you can get the inverse transform of ##1/(s^2+9)##, then integrate it.
  7. Feb 13, 2014 #6
    Thanks for all the help, let me ask you guys this,

    If you had F(s)=1/((s)^2+16)^2 could you expand it like so to find your constants

    [ (c1(s)+c2) / ((s)^2+16)] + [(c3(s)+c4) / ((s)^2+16)^2]
  8. Feb 13, 2014 #7

    Ray Vickson

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    If you restrict yourself to real constants, the function 1/(s^2 + 16)^2 is already in 'partial fraction' form. The only way you can break it down further is to use complex numbers, writing s^2 + 16 = (s+4*i)*(s-4*i), where i = √(-1). You would then get:
    [tex] \frac{1}{(s^2+16)^2} = \frac{i}{256}\frac{1}{s+4i} -\frac{i}{256}\frac{1}{s-4i}
    -\frac{1}{64(s-4i)^2}-\frac{1}{64(s+4i)^2} [/tex]
  9. Feb 13, 2014 #8
    Thanks for your reply Vickson, but to be honest I have no idea how to understand that as of right now.....how would you recommend I tackle this problem to find the inverse laplace to get the function back to the time domain?
  10. Feb 13, 2014 #9
    Does anyone know how to find the constants If you had

    F(s)= 1 / ((s)^2+16)^2 ? I need to find the inverse laplace transform of F(s)
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