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.

$F(s)=\frac{1}{s({s}^{2}+140s+{10}^{4})}$

I begin with finding the poles

$s=0,s=-70\pm j\cdot 10\sqrt{51}$

To which I then try putting $F(s)$ in this form:

$F(s)=\frac{A}{s}+\frac{B}{s-70+j\cdot 10\sqrt{51}}+\frac{{B}^{\ast}}{s-70-j\cdot 10\sqrt{51}}$

Because one unknown (${B}^{\ast}$) is just the complex conjugate of $B$, I only need to find out what $A$, and $B$ is.

$A=F(0)={10}^{-4}$

$B=F(-70+j\cdot 10\sqrt{51})\approx 6.31514\cdot {10}^{9}+j\cdot 6.44274\cdot {10}^{9}$

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.

I have a function which I'd like to perform partial fraction decomposition on, to allow easier inverse laplace transform.

$F(s)=\frac{1}{s({s}^{2}+140s+{10}^{4})}$

I begin with finding the poles

$s=0,s=-70\pm j\cdot 10\sqrt{51}$

To which I then try putting $F(s)$ in this form:

$F(s)=\frac{A}{s}+\frac{B}{s-70+j\cdot 10\sqrt{51}}+\frac{{B}^{\ast}}{s-70-j\cdot 10\sqrt{51}}$

Because one unknown (${B}^{\ast}$) is just the complex conjugate of $B$, I only need to find out what $A$, and $B$ is.

$A=F(0)={10}^{-4}$

$B=F(-70+j\cdot 10\sqrt{51})\approx 6.31514\cdot {10}^{9}+j\cdot 6.44274\cdot {10}^{9}$

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 $F(s)$ is:

$F(s)=\frac{A}{s}+\frac{B}{s-{s}_{0}}+\frac{{B}^{\ast}}{s-{s}_{0}^{\ast}}$

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

$\begin{array}{rl}A& =\underset{s\to 0}{lim}sF(s)\\ \\ & ={10}^{-4}\\ \end{array}$

and

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

$F(s)=\frac{A}{s}+\frac{B}{s-{s}_{0}}+\frac{{B}^{\ast}}{s-{s}_{0}^{\ast}}$

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

$\begin{array}{rl}A& =\underset{s\to 0}{lim}sF(s)\\ \\ & ={10}^{-4}\\ \end{array}$

and

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

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