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Inverse Laplace Transform

  1. Jan 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Hello all,
    Having difficulty with this one question that involves complex roots. Here it is:
    [tex]F(s)=\frac{s+3}{s^3+3s^2+6s+4}[/tex]
    I tried two different ways to tackle it. First method I divided it right away:
    [tex]F(s)=\frac{s+3}{s^3+3s^2+6s+4}\rightarrow{s^2+6-\frac{14}{s+3}}[/tex]
    Is there some sort of approach to something like s^2? I have not taken a differential equations course, and this is in one of my classes for modelling circuits.
    If I decide not to do it this way, I can break it up via partial fractions:
    [tex]F(s)=\frac{s+3}{s^3+3s^2+6s+4}=\frac{s+3}{(s+1)(s+1-\sqrt{3}j)(s+1+\sqrt{3}j)}=\frac{A}{s+1}+\frac{B}{s+1-\sqrt{3}j}+\frac{C}{s+1+\sqrt{3}j}[/tex]
    where j is a complex number.
    Thus,
    [tex]A(s^2+2s+4)+B(s+1)(s+1+\sqrt{3}j)+C(s+1)(s+1-\sqrt{3}j)=s+3[/tex]
    [tex]As^2+Bs^2+Cs^2=A+B+C=0[/tex]
    The part that I do not know how to do is from here.
    Would it be:
    [tex]2As+2Bs+2Cs+j\sqrt{3}Bs+j\sqrt{3}Cs=s[/tex]
    or are the complex numbers treated separetely?

    Or is there an easier way, altogether?
     
    Last edited: Jan 26, 2012
  2. jcsd
  3. Jan 26, 2012 #2
    got it. Solved via matrix for A, B, C
     
  4. Jan 26, 2012 #3

    vela

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    Generally, I try to avoid imaginary numbers if I can. Factors of j are just like mistakes waiting to happen. I'd instead expand it as
    $$F(s) = \frac{s+3}{s^3+3s+6s+4} = \frac{A}{s+1}+\frac{Bs + C}{s^2+2s+4}$$which leads to
    $$s+3 = A(s^2+2s+4) + (Bs+C)(s+1)$$
     
  5. Jan 26, 2012 #4

    Mark44

    Staff: Mentor

    Edit: vela beat me to the punch!

    Breaking the denominator into all linear factors seems like it might be the harder way to go. You can factor s3 + 3s2 + 6s + 4 into (s + 1)(s2 + 2s + 4).

    When you use partial fractions, you'll be solving for constants A, B, and C so that
    [tex]\frac{s + 3}{s^3 + 3s^2 + 6s + 4} = \frac{A}{s+1} + \frac{Bs + C}{s^2 + 2s + 4}[/tex]

    That last denominator can be rewritten as s2 + 2s + 1 + 3 = (s + 1)2 + (√3)2, if need be, which lends itself to a trig substitution.
     
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