I am trying to work through the steps deriving the Bode Integral in the following paper:
https://www.sciencedirect.com/science/article/pii/S2405896315025045
I am stuck on page 261, left column, on the section for the Gamma_pk contour integral, specifically the following unjustified statement:
$$ S(s)=S'(s)(p_{k}-s)$$
S(s) is the sensitivity transfer function. Pk is a pole of the open-loop transfer function, and therefore a zero of S(s), which the contour was chosen to go around. Please see figure 3 in the paper for a visual depiction.
Anyhow, I think it's just the complex plane throwing me off here, but I am unable to arrive at, or justify, that statement. I come close, but can't quite get all the way there.
Thanks for any insights!
It's helpful to consider one of the examples provided in the paper, namely that of an open-loop unstable system with sensitivity function $S(s)=\dfrac{s^2-1}{s^2+s+9}$. Then $$\frac{S'(s)}{S(s)}=\frac{d}{ds}\ln S(s)=\frac{2s}{s^2-1}-\frac{2s+1}{s^2+s+9}=\frac{s^2+20s+1}{(s^2-1)(s^2+s+9)}$$
or $$S(s)=S'(s)\cdot \frac{(s^2-1)(s^2+s+9)}{s^2+20s+1}.$$
It should be evident that this doesn't agree with their stated formula. That said, the reason $S(s)$ is open-loop unstable is due to the zero at $s=1$, and with this in mind we write $S(s)=S'(s)(s-1)\cdot f(s)$ where $$f(s)=\frac{(s+1)(s^2+s+9)}{s^2+20s+1}$$ is analytic in the right half-plane. This form seems entirely sufficient for the purposes of their argument.
Moreover, if we look to Lewis's notes on control theory (which your paper cites!) then the above is a case of their first equation on page 366: Given a zero of $S_L(s)$ at $s=p_j$ with positive real part, one may write
$$\ln\frac{S_L(s)}{S_L(\infty)}=\ln(s-p_j)+\ln f_j(s)$$ with $f_j(s)$ being analytic on and within the contour which runs around $p_j$. So I would discount the paper's arguments for the Bode integral formula in favor of those in Lewis's notes.