PDA

View Full Version : Partial-fractions problem


tandoorichicken
May29-05, 12:14 AM
The whole problem reads:
Find the inverse Laplace transform of
\frac{s-2}{s^3+2s^2+2s}

I think once I get the expression simplified I can do the rest by myself. I started to separate this expression out by partial fractions and got as far as this:
\frac{s-2}{s^3+2s^2+2s}=\frac{s-2}{(s^2+2s+2)s}=\frac{As+B}{s^2+2s+2} +\frac{C}{s}

(As+B)s+C(s^2+2s+2)=s-2
From this expression I got C=-1, which I checked was correct by using my calculator, but I still don't know how to find A, or B. Can anyone help with this please?

Any ideas or hints on how to do the inverse would be also be appreciated.

Corneo
May29-05, 12:20 AM
Note that

s^2+2s+2 = (s+1)^2+1

Opps major brain laspe. That should be fixed

whozum
May29-05, 12:22 AM
Note that

s^2+2s+2 = (s+1)^2

Nope, it equals [itex](s+1)^2 + 1.[/tex]

edit: tex

Corneo
May29-05, 12:25 AM
Or perhaps that wasn't too helpful. Here is an trick I learned when doing these partial fractions problems.

Consider what you have
(As+B)s+C(s^2+2s+2)=s-2

Multiply out and gather the terms.
(A+C)s^2 + (2C+B)s + 2C = s-2

Is there a s^2 term on the right hand side of the equation? What does this tell you about A+C? What about 2C, what should that be equal to? And 2C+B?

tandoorichicken
May29-05, 12:37 AM
Thanks, Corneo.