# Inverse Laplace Transform

1. Jan 26, 2012

### sandy.bridge

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:
$$F(s)=\frac{s+3}{s^3+3s^2+6s+4}$$
I tried two different ways to tackle it. First method I divided it right away:
$$F(s)=\frac{s+3}{s^3+3s^2+6s+4}\rightarrow{s^2+6-\frac{14}{s+3}}$$
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:
$$F(s)=\frac{s+3}{s^3+3s^2+6s+4}=\frac{s+3}{(s+1)(s+1-\sqrt{3}j)(s+1+\sqrt{3}j)}=\frac{A}{s+1}+\frac{B}{s+1-\sqrt{3}j}+\frac{C}{s+1+\sqrt{3}j}$$
where j is a complex number.
Thus,
$$A(s^2+2s+4)+B(s+1)(s+1+\sqrt{3}j)+C(s+1)(s+1-\sqrt{3}j)=s+3$$
$$As^2+Bs^2+Cs^2=A+B+C=0$$
The part that I do not know how to do is from here.
Would it be:
$$2As+2Bs+2Cs+j\sqrt{3}Bs+j\sqrt{3}Cs=s$$
or are the complex numbers treated separetely?

Or is there an easier way, altogether?

Last edited: Jan 26, 2012
2. Jan 26, 2012

### sandy.bridge

got it. Solved via matrix for A, B, C

3. Jan 26, 2012

### vela

Staff Emeritus
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
$$s+3 = A(s^2+2s+4) + (Bs+C)(s+1)$$

4. Jan 26, 2012

### 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
$$\frac{s + 3}{s^3 + 3s^2 + 6s + 4} = \frac{A}{s+1} + \frac{Bs + C}{s^2 + 2s + 4}$$

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.