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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!

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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.

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  • $\begingroup$ Thank you for the response...I actually didn't ask a well-formed question...But given the question, the answer is spot on indeed... $\endgroup$ – Zero Jul 8 '16 at 12:38
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I would like to thank WG- for the response. To answer the question in a more general setting, I hereby present Theorem 3.1.5 from the book Fundamental Limitations in Filtering and Control by Maria M. Seron, Julio H. Braslavsky and Graham C. Goodwin.

The theroem states that: Given the system $L(s)$ has set of zeros $\{q_i : i = 1,\ldots,n_q\}$ in the ORHP and $L(0)\neq 0$, then, assuming closed loop stability

  1. if $L$ is a proper rational function

$$\int_0^{\infty} \log \left \vert \dfrac{\Gamma(j\omega)}{\Gamma(0)} \right \vert \dfrac{d\omega}{\omega^2} = \dfrac{\pi}{2}\dfrac{1}{\Gamma(0)}\lim_{s \to 0}\dfrac{d\Gamma(s)}{ds} + \pi \sum_{i=1}^{n_q} \dfrac{1}{q_i}$$

  1. if $L(s) = L_0(s)e^{-s\tau}$, where $L_0(s)$ is a strictly proper rational function and $\tau>0$,

$$\int_0^{\infty} \log \left \vert \dfrac{\Gamma(j\omega)}{\Gamma(0)} \right \vert \dfrac{d\omega}{\omega^2} = \dfrac{\pi}{2}\dfrac{1}{\Gamma(0)}\lim_{s \to 0}\dfrac{d\Gamma(s)}{ds} + \pi \sum_{i=1}^{n_q} \dfrac{1}{q_i} + \dfrac{\pi}{2}\tau$$

It can also be shown that if the closed loop function is type II, i.e. it has two poles at the origin, then similar to Bode Sensitivity Integral, area for suppression and amplification for $\Gamma(s)$ must be same. Presence of delay worsens the trade-off.

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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.

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  • $\begingroup$ Thank you for the response but I dont't agree...the sensitivity and complementary sensitivity are, as the name suggests, complementary. They do not hold an inverse relationship (implied by rise and fall comparion). At higher frequencies i.e. above bandwidth, the CS function is close to zero for most motion systems. I, sincerely, do not see an answer to my question. $\endgroup$ – Zero Jul 1 '16 at 14:37

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