Partial Fractions with Two Repeated Linear Terms

Question

Rewrite the expression below into partial fractions

$$\frac{\omega s}{(s+\omega)^2(s-\omega)^2}$$

I started by taking the general form

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

Then, using the Heaviside "cover-up" method, I was able to find

$$B = -\frac{1}{4}, \quad D = \frac{1}{4}$$

and by letting $s = 0$, I got the following relation

$$A = C$$

I know that $B$ and $D$ are sufficient for this problem, i.e. $A = C = 0$, because I took the sum of $B$ and $D$ terms alone and it gave me the initial expression. However, I would like to know, without checking if $B$ and $C$ are sufficient, how to find $A$ or $C$.

Answer 1

I would recommend you to notice that \begin{align*} 4\omega s & = (s^{2} + 2\omega s + \omega^{2}) - (s^{2} - 2\omega s + \omega^{2})\\\\ & = (s + \omega)^{2} - (s - \omega)^{2} \end{align*}

Consequently, it results that \begin{align*} \frac{\omega s}{(s + \omega)^{2}(s - \omega)^{2}} & = \frac{1}{4}\times\frac{(s + \omega)^{2} - (s - \omega)^{2}}{(s + \omega)^{2}(s - \omega)^{2}}\\\\ & = \frac{1}{4}\times\left[\frac{1}{(s - \omega)^{2}} - \frac{1}{(s + \omega)^{2}}\right] \end{align*}

Hence $A = C = 0$, and we are done.

Hopefully this helps !