# Partial Fraction with Unknown Constants

I'm having trouble doing the partial fraction decomposition here due to the unknown constants. I need to break down $$x(s)=\frac{F_0\omega}{(s^2+\omega^2)(s^2-\omega_0^2)}$$ where $F_0, \omega,$ and $\omega_0$ are all constants.

HINT:

$$\frac{1}{(s^2+\omega^2)(s^2-\omega_0^2)}=\frac{A}{s+i\omega}+\frac{B}{s-i\omega}+\frac{C}{s+\omega_0}+\frac{D}{s-\omega_0}$$

Then, as an example, we can find $A$ as the limit

$$A=\lim_{s\to -i\omega}\frac{(s+i\omega)}{(s^2+\omega^2)(s^2-\omega_0^2)}=\lim_{s\to -i\omega}\frac{1}{(s-i\omega)(s^2-\omega_0^2)}=\frac{1}{i2\omega(\omega^2+\omega_0^2)}$$

• Please let me know how I can improve my answer. I really want to given you the best answer I can. -Mark – Mark Viola Apr 12 '17 at 17:09

You do it the same way as if they were numbers. First $s^2-\omega_0^2=(s-\omega)(s+\omega)$ so write $$\frac{F_0\omega}{(s^2+\omega^2)(s^2-\omega_0^2)}=\frac{a+bs}{s^2+\omega^2}+\frac c{s-\omega}+\frac d{s+\omega}$$ and use the techniques you are used to. You can clear fractions and equate powers of $s$, you can take limits as $s \to \pm \omega$, and so on. Clearly $a,b,c,d$ are proportional to $F_0$ so you can start by ignoring it and multiply by it at the end if you want.

• So I could then multiply by the denominator to eliminate the fractions and simplify from there? Will that be enough? – H. Rumo Mar 13 '17 at 4:04
• Yes, that will work. You will get four equations in four unknowns, $a,b,c,d$, to solve, one for each power of $s$. – Ross Millikan Mar 13 '17 at 4:07
• Should I disregard the constants somehow? When I start to simplify, I get a very complicated equation that I'm not sure how to solve. How would I proceed from here? $$(b+c+d){s^3}+(a+c\omega_0-d\omega_0){s^2}+(-b\omega_0+c{\omega^2}+d{\omega^2})s+(c-d)({\omega^2}\omega_0)-a{\omega_0^2}$$ – H. Rumo Mar 13 '17 at 4:22
• You did not show the other side of the equation. Presumably it is $F_0\omega$. You can now split it into four equations based on the powers of $s$ and solve them simultaneously for $a,b,c,d$. This gives $b+c+d=0, a+c\omega_0-d\omega_0=0$ and so on. – Ross Millikan Mar 13 '17 at 4:36
• Okay, that makes sense. However, are they all equal to $0$? If so, why is that? Because the other side is just a constant? – H. Rumo Mar 13 '17 at 4:39