# partial fraction expansion with As+B and irreducible quadratic

My question is at the bottom.

In an electrical engineering text on Laplace methods for solving electrical circuits a section on discussing p.f.e. the author says that when the denominator has an irreducible quadratic (does not factorize without imaginary terms), the partial fraction expansion should have a term with As+B for numerator as below,

$$\frac{n(s)}{q(s)(s^2+cs+d)} = \frac{As+B}{s^2+cs+d} + ...$$

I solved a circuit problem (see this question) that had an irreducible quadratic in the denominator of the resulting current, i(s), $$i(s) = \frac{264.15}{s^2+37.74s+37735.85}$$

However, I worked it with simple, complex roots as, $$i(s)=\frac{264.15}{s^2+37.74s+37735.85}=\frac{A}{s+α+jβ}+\frac{B}{s+α-jβ}$$

With result of, $$i(s)=\frac{j0.683}{s+18.87+j193.24}+\frac{-j0.683}{s+18.87-j193.24}$$ My solution after inverse Laplace being, $$i(t)=1.366e^{-18.87t} sin(193.24t)$$

After verifying my analytic solution was correct (verified with numerical simulation), I wondered how it would have turned out if i had used the book recommended method with As+B approach.

$$i(s)=\frac{264.15}{s^2+37.74s+37735.85}=\frac{As+B}{s^2+37.74s+37735.85}$$

But, when i eliminate the fraction and equate coefficients i get, $$A = 0$$ $$B = 264.15$$

Which gets me nowhere. I've read several good pages on this method and i think i'm doing it properly.

My question: Can this "As+B" method be used to solve this problem? If so, where am i going off track.

When $$q(s)$$ is an irreducible quadratic, the term $$\frac{As+B}{q(s)}$$ is considered in final partial-fraction form and is meant to be integrated directly. One can do so using a combination of $$\int q'(s)/q(s)\,ds = \log|q(s)|$$ (which handles the linear term in the numerator, perhaps at the cost of changing the constant term) and then using a change of variables to transform $$\int B/q(s)\,ds$$ into $$\int B'/(u^2+1)\,du$$, which can be evaluated in terms of $$\arctan u$$.
• But you have! $A=0$ and $B=264.15$ gives a numerator of $0s+264.15$. The fact that this is the same as $264.15$ does not stop it from being a valid instance of $As+B$. In other words, $A=0$ is definitely allowed (as is $B=0$). Commented Nov 3, 2020 at 5:44