Using Laplace, getting stuck on decomposition

**liquidFuzz** · December 12th 2011, 12:50 PM

I think I have a fair idea of how to solve my equations, using Laplace transform. But the partial fraction decomposition seems to elude me...

I get X and Y as such.

$\displaystyle Y(s) = \frac{s^3+3s}{s^4+5s^2+4}$
$\displaystyle X(s) = \frac{2}{s^4+5s^2+4}$

I get this:

$\displaystyle Y(s) = \frac{1}{s^2+1} - \frac{1}{s^2+4}$

I pulled the handbrake here because wolfram alpha got something else...

Wolfram Alpha, see: Partition fraction expansion

If anyone could find it in their heart to ease me throw this one, so that I can solve my system of differential equations...

**Siron** · December 12th 2011, 01:07 PM

Your answer can't be right because:
$Y(s)=\frac{1}{s^2+1}-\frac{1}{s^2+4}=\frac{(s^2+4)-(s^2+1)}{(s^2+1)(s^2+4)}=\frac{3}{(s^2+1)\cdot(s^2 +4)}$
which is different from the original given $Y(s)$ .

We have $Y(s)=\frac{s^3+3s}{(s^2+1)(s^2+4)}=\frac{As+B}{s^2 +1}+\frac{Cs+D}{s^2+4}$

If we proceed:
$\frac{(As+B)(s^2+4)+(Cs+D)(s^2+1)}{(s^2+1)(s^2+4)}$
$=\frac{As^3+Bs^2+4As+4B+Cs^3+Ds^2+Cs+D}{(s^2+1)(s^ 2+4)}$
$=\frac{(A+C)s^3+(B+D)s^2+(4A+C)s+(4B+D)}{(s^2+1)(s ^2+4)}$

This means: $A+C=1$ , $B+D=0$ , $4A+C=3$ and $4B+D=0$

Can you proceed?

**liquidFuzz** · December 13th 2011, 01:44 AM

Thank you very much! I got it from here. I didn't write the proposed denominators correctly.

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