A (strong) observer for \(\Sigma\) consists of a (time-invariant and complete)
system \(\hat{\Sigma}\) having input value space \(u \times y\)
together with a map
$$
\theta: \mathcal{Z} \times y \rightarrow x
$$
(where \(\mathcal{Z}\) is the state space of \(\hat{\Sigma}\) ) so that the
following property holds. For each \(x \in X\), each \(z \in \mathcal{Z}\), and
each \(\omega \in U[0, \infty)\), we let \(\xi\) be the infinite path resulting
from initial state \(x\) and control \(\omega\), that is,
$$
\xi(t)=\phi\left(t, 0, x,\left.\omega\right|_{[0, t)}\right)
$$
for all \(t \in \mathcal{T}_{+}\), and write \(\eta(t)=h(\xi(t))\); we also let
\(\zeta\) be the infinite path resulting from initial state \(z\) and control
\((\omega, \eta)\) for the system \(\hat{\Sigma}\),
$$
\zeta(t)=\hat{\phi}\left(t, 0, z,\left(\left.\omega\right|_{(0,
t)},\left.\eta\right|_{[0, t)}\right)\right),
$$
and finally we write
$$
\hat{\xi}(t):=\theta(\zeta(t), \eta(t)) .
$$
Then it is required that, for every \(x, z\) and every \(\omega\),
$$
\mathrm{d}(\xi(t), \hat{\xi}(t)) \rightarrow 0 \text { as } t \rightarrow
\infty
$$
(global convergence of estimate) as well as that for each \(\varepsilon>0\)
there be some \(\delta>0\) so that for every \(x, z, \omega\)
$$
\mathrm{d}(x, \theta(z, h(x)))<\delta \Rightarrow \mathrm{d}(\xi(t),
\hat{\xi}(t))<\varepsilon \text { for all } t \geq 0
$$
Explanations 
Textbooks 
Exams 
Magazine 
