Let \(\Sigma_{1}\) and \(\Sigma_{2}\) be two systems, and let $$ k_{i}: y_{1} \times y_{2} \rightarrow \mathcal{U}_{i}, \quad i=1,2 $$ be two maps. We say that the interconnection of \(\Sigma_{1}\) and \(\Sigma_{2}\) through \(k_{1}\) and \(k_{2}\) is well-posed if for each $\left(x_{1}, x_{2}\right) \in X_{1} \times \mathcal{X}_{2}$ there exist unique functions $$ \xi_{i}:[0, \infty) \rightarrow X_{i}, \quad i=1,2 $$ such that, with the notations $$ \begin{aligned} \eta_{i}(t) &:=h_{i}\left(\xi_{i}(t)\right) \\ \omega_{i}(t) &:=k_{i}\left(\eta_{1}(t), \eta_{2}(t)\right) \end{aligned} $$ it holds that \(\xi_{i}=\psi_{i}\left(x_{i}, \omega_{i}\right)\) for \(i=1,2\).

If \(w\) is stable, then for each \(k \neq 0\) such that $\frac{1}{k} \notin \Gamma\(, the closed-loop system with \)v=k$ is stable if and only if \(\mathrm{cw}\left(\Gamma, \frac{1}{k}\right)=0\).

As a trivial illustration, take the system \(\dot{x}=u, y=x\). Its transfer function is \(w=1 / s\), which admits the coprime fractional representation $$ w=\frac{1 /(s+1)}{s /(s+1)}, $$ and we may pick \(\alpha=\beta=1\). According to the Theorem, every possible stabilizer is of the form $$ v=\frac{\mu s /(s+1)-1}{\mu /(s+1)+1}=\frac{\mu s-s-1}{\mu+s+1} $$ with \(\mu\) ranging over all possible \(\mu \in \mathbf{R H}_{\infty}\). Such \(v\) 's are guaranteed to be proper and to stabilize. For instance, for \(\mu=0\) one obtains \(v=-1\), which corresponds to the obvious stabilizing law \(u=-y\).

The time-invariant complete system \(\Sigma\) is asymptotically observable (with respect to \(x^{0}, u^{0}, y^{0}\) ) if \(\Sigma^{0}\) is globally asymptotically stable (with respect to \(x^{0}\) ).

Assume that \(w\) is rational and that \(v=k \neq 0\) is a constant. Then the poles of the entries of \(W_{\mathrm{cl}}\) (that is, the elements in the list \((7.14),\), are precisely the zeros of \(w-\frac{1}{k}\).

For linear time-invariant continuous-time or discrete-time systems, \((A, C)\) is asymptotically observable if and only if $\left(A^{\prime}, C^{\prime}\right)$ is asymptotically controllable.