Cases where Heaviside cover up method fails for partial fraction expansion.
Why can't I use Heaviside coverup method to find partial fraction expansion of:
$$F(s) = \frac{s^3}{(s^2+1)^2}$$
For example:
$$F(s) = \frac{s^3}{(s^2+1)^2} = \frac{r_1}{(s^2+1)^2} + \frac{r_2}{(s^2+1)}$$
$$r_1 = (s-\alpha)^r~ F(s) ~\bigg|_{s=\alpha}$$
$$r_1= s^3 \bigg|_{s=-1} = -1$$
$$r_2= \frac{d}{ds}[s^3] \bigg|_{s=-1} = \bigg[3s^2\bigg]_{s=-1} = 3$$
$$r_2 = \bigg( \frac{d}{ds} \Big[ (s-\alpha)^r ~ F(s)\Big] \bigg)~\bigg|_{s=\alpha}$$
$$F(s) = \frac{s^3}{(s^2+1)^2} = \frac{-1}{(s^2+1)^2} + \frac{3}{(s^2+1)}$$
however, if I check the result:
$$F(s) = \frac{-1}{(s^2+1)^2} + \frac{3(s^2+1)}{(s^2+1)(s^2+1)}$$
$$F(s) = \frac{-1 +3s^2+3}{(s^2+1)^2}$$
$$F(s) = \frac{3s^2+2}{(s^2+1)^2} \ne \frac{s^3}{(s^2+1)^2}$$
Why does heaviside coverup fail for this case?
I look at order of polynomial in numerator, its 3, and the polynomial order for the denominator is 4, so it should work, but it doesn't. Does heaviside coverup method only work for first order polynomial factors?
$\endgroup$ 61 Answer
$\begingroup$$$F(s) = \frac{s^3}{(s^2+1)^2}$$Is easy to decompose:$$F(s) = \frac{s^3+s-s}{(s^2+1)^2}$$$$F(s) = \frac{s(s^2+1)-s}{(s^2+1)^2}$$$$F(s) = \frac{s}{(s^2+1)}-\frac{s}{(s^2+1)^2}$$
$\endgroup$
