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:

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

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

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

$$B = F(-70+j\cdot10\sqrt{51}) \approx 6.31514\cdot10^9 + j\cdot 6.44274\cdot10^9$$

Where the last step was done in Mathematica.

Answer, according to several other people, is supposed to be $B = -70 + j\cdot71.41$, which looks a lot nicer, but I'm not sure HOW they got to that answer.

• This is an error of your calculator: if you tried putting the values into $F$ you would have noticed that the formula you are using is missing something as you are trying to find the value where the function is not defined. – N74 May 16 '16 at 15:04

Note that the partial fraction expansion of $F(s)$ is

$$F(s)=\frac{A}{s}+\frac{B}{s-s_0}+\frac{B^*}{s-s_0^*}$$

where $s_0= -70+i10\sqrt{51}$ and $s_0^*=s_0= -70-i10\sqrt{51}$ are the complex-conjugate roots of $s^2+140s+10^4$.

Note that we have

\begin{align} A&=\lim_{s\to 0}sF(s)\\\\ &=10^{-4}\\\\ \end{align}

and

\begin{align} B&=\lim_{s\to s_0}(s-s_0)F(s)\\\\ &=\frac{1}{s_0(s_0-s_0^*)}\\\\ &=\frac{-70-i10\sqrt{51}}{i20\sqrt{51}10^4 }\\\\ &=\frac{-51+i7\sqrt{51}}{1020000} \end{align}

• Excellent! One question if you don't mind. I see that $(s_0 - s^*_0) = i20\sqrt{51}$ but how did you convert the factor $1/s_0$ into $(-70-i10\sqrt{51}) / 10^4$? – B. Lee May 16 '16 at 15:45
• We have $$\frac{1}{s_0}=\frac{s_0^*}{|s_0|^2}$$ – Mark Viola May 16 '16 at 16:01