Partial Fraction with Unknown Constants

Question

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.

Answer 1

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

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

Answer 2

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.

Answer 3

I'd like to expand on Mark Viola's answer. Let's start with this variant of his hint: $$ x(s) = \frac{F_0 \omega}{(s+i\omega)(s-i\omega)(s+\omega)(s-\omega)} = \frac{A}{s+i\omega} + \frac{B}{s-i\omega} + \frac{C}{s+\omega} + \frac{D}{s-\omega}.$$ Whenever all the factors in the denominator of a rational function are simple (i.e., not raised to a power $> 1$) and the numerator has degree lower than the denominator, the partial fraction expansion has the simple form above. In this case one may solve for the coefficients using the cover-up method: $$\frac{F_0 \omega}{1\cdot(s-i\omega)(s+\omega)(s-\omega)}\bigg|_{s = -i\omega} = \frac{A}{1}. $$ That is, to use the cover-up method to find the coefficient $A$, you first find the zero of its corresponding denominator $s+i\omega$, which is just $s=-i\omega$. You then

• Delete the other ($B$, $C$, and $D$) terms on the RHS,
• "Cover-up" the $s+i\omega$ on both sides of the equation (replacing it with $1$), and
• Substitute $s = -i\omega$ into the resulting equation.

This yields a simple calculation for $A$ in lieu of setting up and solving a system of equations (i.e., the method of undetermined coefficients): $$ A = \frac{F_0 \omega}{1\cdot(-i\omega-i\omega)(-i\omega+\omega)(-i\omega-\omega)} = -\frac{iF_0}{4\omega^2}. $$ The other coefficients may be determined in the same way, of course.

If you're going to use the cover-up method, you owe it to yourself and every math teacher you've had to understand why it works. It's a simple matter of multiplying the original equation by $s+i\omega$. One you see it, you should have an "aha!" moment and think, "oh, that's trivial." It will then be easy to recall, with no need to memorize it as the sequence of steps above.

As a test of your understanding, consider what would happen if some of the factors in the denominator of $x(s)$ were raised to powers $> 1$. Which coefficients of the partial fraction expansion could you still determine using the cover-up method?