Janet Forbes

Answered

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.

Answer & Explanation

Perman7z

Expert

2022-07-04Added 13 answers

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
$\begin{array}{rl}B& =\underset{s\to {s}_{0}}{lim}\left(s-{s}_{0}\right)F\left(s\right)\\ \\ & =\frac{1}{{s}_{0}\left({s}_{0}-{s}_{0}^{\ast }\right)}\\ \\ & =\frac{-70-i10\sqrt{51}}{i20\sqrt{51}{10}^{4}}\\ \\ & =\frac{-51+i7\sqrt{51}}{1020000}\end{array}$

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?