 Janet Forbes

2022-07-03

Partial Fraction Decomposition with complex poles
I have a function which I'd like to perform partial fraction decomposition on, to allow easier inverse laplace transform.
$F\left(s\right)=\frac{1}{s\left({s}^{2}+140s+{10}^{4}\right)}$
I begin with finding the poles
$s=0,s=-70±j\cdot 10\sqrt{51}$
To which I then try putting $F\left(s\right)$ in this form:
$F\left(s\right)=\frac{A}{s}+\frac{B}{s-70+j\cdot 10\sqrt{51}}+\frac{{B}^{\ast }}{s-70-j\cdot 10\sqrt{51}}$
Because one unknown (${B}^{\ast }$) is just the complex conjugate of $B$, I only need to find out what $A$, and $B$ is.
$A=F\left(0\right)={10}^{-4}$
$B=F\left(-70+j\cdot 10\sqrt{51}\right)\approx 6.31514\cdot {10}^{9}+j\cdot 6.44274\cdot {10}^{9}$
Where the last step was done in Mathematica.
Answer, according to several other people, is supposed to be $B=-70+j\cdot 71.41$, which looks a lot nicer, but I'm not sure HOW they got to that answer. Perman7z

Expert

Note that the partial fraction expansion of $F\left(s\right)$ is:
$F\left(s\right)=\frac{A}{s}+\frac{B}{s-{s}_{0}}+\frac{{B}^{\ast }}{s-{s}_{0}^{\ast }}$
where ${s}_{0}=-70+i10\sqrt{51}$ and ${s}_{0}^{\ast }={s}_{0}=-70-i10\sqrt{51}$ are the complex-conjugate roots of ${s}^{2}+140s+{10}^{4}$
Note that we have
$\begin{array}{rl}A& =\underset{s\to 0}{lim}sF\left(s\right)\\ \\ & ={10}^{-4}\\ \end{array}$
and
$\begin{array}{rl}B& =\underset{s\to {s}_{0}}{lim}\left(s-{s}_{0}\right)F\left(s\right)\\ \\ & =\frac{1}{{s}_{0}\left({s}_{0}-{s}_{0}^{\ast }\right)}\\ \\ & =\frac{-70-i10\sqrt{51}}{i20\sqrt{51}{10}^{4}}\\ \\ & =\frac{-51+i7\sqrt{51}}{1020000}\end{array}$

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