Bode Sensitivity Integral

Question

The bode sensitivity integral is well known for linear control systems. To state it in simplistic terms,

Any system $L$ with relative degree $2$ or more satisfies the following equation for the closed loop sensitivity function $G(s) = (1 + L(s))^{-1}$,

$\int_0^{\infty}log\vert G(j\omega)\vert d\omega = 0$.

Basically, it implies that there is always a frequency range wherein $G(j\omega)>0 \hspace{0.1cm} dB$.

My question is: Does it have any bearing on the complementary sensitivity function, $\Gamma(s) \left(= \dfrac{L(s)}{1 + L(s)}\right)$? i.e. Does there, necessarily, exist a frequency range where $\Gamma(s)>0 \hspace{0.1cm}dB$?

Any theoretical reference, if possible, would be appreciated!

Answer 1

Well first, note that $G(s) + \Gamma(s) = 1$ for all values of $s \in \mathbb{C}$.

Ideally, we would want $|\Gamma(j\omega)| \simeq 1$ and thus $G(j\omega) \simeq 0$. Due to the Bode integral formula, we can only achieve these objectives in some frequency range (low frequencies); for high frequencies $G(s)$ will rise and $\Gamma(s)$ will fall. So the integral implies that $\Gamma(s) < 1$ for some frequency range.

Answer 2

Lets just take this from the other way... Why should the statement hold?

Let just assume that $L(s) = \frac{1}{(s + 1)(s + 2)}$ for whatever particular reason. $L(s)$ has a relative degree of $2$.

Then you will see that $\Gamma(s)$ will never be equal or greater then $0$ [dB]. So no, there does not necessarily exist a frequency range where $\Gamma(s) > 0$ [dB].

However, I have the feeling that this reference is more what you are looking for.