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Mo, Yilin edited algorithm.tex
over 8 years ago
Commit id: 36d1848893258b8bb87051d8c31ef2ec3d7a0d08
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%% in which $\mZ(z)$ is the one-sided z-transform of $\mathbb{E}\{y(k) u(k)\}$
%in which $D=\begin{bmatrix} Q & 0 \\ 0 & R \end{bmatrix}$.
Since the feedback system is asymptotic stable, $\Phi_{y,u}(z)$ has no poles on the unit circle. Consider a Mobius transform $z=\frac{1+s}{1-s}$ and let $\Psi_{y,u}(s)=\Phi_{y,u}\left(\frac{1+s}{1-s}\right)$, then for $\Psi_{y,u}(s)$ there exists a positive real matrix $\mS(s)$
\cite{keith}, \cite{Keith}, such that
\begin{equation} \label{Zdef}
\mS(s) + \mS^T(-s) = \Psi(s)=\mW(s)\mW^T(-s).
\end{equation}