rsmc_report.tex.bak

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\begin{itemize}
\vspace{-10mm}\item[\begin{footnotesize}\ding{226}\end{footnotesize}]\xiaoerhao\li\sihao\textcolor[rgb]{0.66,0.14,0.44}{Proof of inequality (\ref{eq.uncer}):1}\\
\xiaosihao Note that (\ref{eq.tildeaad}), and let $V=C\tilde P$ \\
$\tilde\Phi_{11}=\ol A_{11}\tilde P-\ol A_{12}V+\tilde P\ol A^T_{11}-V^T\ol A^T_{12}+\tilde M+\tilde M^T+\tilde Q+\ol\tau\tilde X+\quad\textcolor[rgb]{0.66,0.14,0.44}{\tilde P U_2^TH^TD^TG^TU_2+U_2^TGDHU_2^T\tilde P-V^TU_1^TH^TD^TG^TU_2+U_2^TGDHU_1V}$; \\
$\tilde\Phi_{21}=\tilde\Phi_{21}^T=\tilde P\ol A^T_{d11}-V^T\ol A^T_{d12}-\tilde M^T+\tilde N+\ol \tau \tilde Y^T+\textcolor[rgb]{0.66,0.14,0.44}{\tilde P U_2^TH_d^TD_d^TG_d^TU_2-V^TU_1^TH_d^TD_d^TG_d^TU_2}$;\\
$\tilde\Phi_{22}=-\tilde N-\tilde N^T-(1-d)\tilde Q+\ol\tau \tilde Z$;\\
$\tilde\Phi_{31}=\tilde\Phi_{31}^T=\ol\tau\lambda (\ol A_{11}\tilde P-\ol A_{12}V)+\textcolor[rgb]{0.66,0.14,0.44}{U_2^TGDHU_2\tilde P-U_2^TGDHU_1V^T}$;\\
$\tilde\Phi_{32}=\tilde\Phi_{32}^T=\ol\tau\lambda (\ol A_{d11}\tilde P-\ol A_{d12}V)+\textcolor[rgb]{0.66,0.14,0.44}{U_2^TG_dD_dH_dU_2\tilde P-U_2^TG_dD_dH_dU_1V}$;\\
$\tilde\Phi_{33}=-\ol\tau\lambda \tilde P$
\end{itemize}
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\begin{itemize}
\vspace{-10mm}\item[\begin{footnotesize}\ding{226}\end{footnotesize}]\xiaoerhao\li\sihao\textcolor[rgb]{0.66,0.14,0.44}{Proof of inequality (\ref{eq.uncer}):2}\\
\xiaosihao As the inequality (\ref{eq.uncer}) is symmetric matrix,we
rewrite the colored terms in two new terms like follows:
\small\bea &&\Omega_2=\lt[\ba{cc}U^T_2G&U^T_2G_d\\0&0\\\ol\tau\lambda U^T_2G&\ol\tau\lambda U^T_2G_d\ea\rt]\nm \\
&&\Omega_3=\lt[\ba{ccc}HU_2P-HU_1V&0\\0&H_dU_2P-H_dU_1V \ea\rt]\nm\\
&&\ol D=\lt[\ba{cc}D&0\\0&D_d\ea\rt]\eea\\[-20mm]

\small\bea
&&\Omega_2\ol D\lt[\ba{cc}\Omega_3\quad 0\ea\rt]\nm\hspace*{\fill}\\
=&&\tilde PU_2^TH^TD^TG^TU_2+U_2^TGDHU_2^T\tilde P-V^TU_1^TH^TD^TG^TU_2+U_2^TGDHU_1V\quad\textcolor[rgb]{0,0,1}{\hei(terms~in~\tilde\Phi_{11})}\nm\\
+&&\tilde P U_2^TH_d^TD_d^TG_d^TU_2-V^TU_1^TH_d^TD_d^TG_d^TU_2\quad\textcolor[rgb]{0,0,1}{\hei(terms~in~\tilde\Phi_{21})}\nm\\
+&&U_2^TGDHU_2\tilde
P-U_2^TGDHU_1V^T\quad\textcolor[rgb]{0,0,1}{\hei(terms~in~\tilde\Phi_{31})}\nm\\
+&&U_2^TG_dD_dH_dU_2\tilde P-U_2^TG_dD_dH_dU_1V \quad\textcolor[rgb]{0,0,1}{\hei(terms~in~\tilde\Phi_{32})}\nm\eea\\[-15mm]

\end{itemize}
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\begin{itemize}
\item[\begin{footnotesize}\HandPencilLeft\end{footnotesize}]\xiaoerhao\li\sihao\textcolor[rgb]{0.66,0.14,0.44}{Colored in the matrix $\Omega_1$}\\
\quad\quad$ \Omega_1= \lt[\ba{ccc}\textcolor[rgb]{0,0,1}{\hei\Omega_{111}}~~\textcolor[rgb]{0,0,1}{\hei\Omega^T_{121}}~~\Omega^T_{131}\\
 \Omega_{121}~~\textcolor[rgb]{1,0,0}{\hei\Omega_{122}}~~\Omega^T_{132}\\
\textcolor[rgb]{0,0,1}{\hei\Omega_{131}}~~\textcolor[rgb]{0,0,1}{\hei\Omega_{132}}~~\textcolor[rgb]{1,0,0}{\hei\Omega_{133}}\ea\rt]\nm$\\[-10mm]
\sihao\item[\begin{footnotesize}\PlusCenterOpen\end{footnotesize}]\sihao\li\textcolor[rgb]{0.66,0.14,0.44}{The red terms don't have  uncertain terms}\\[-30mm]
\item[\begin{footnotesize}\PlusCenterOpen \end{footnotesize}]\sihao\li\textcolor[rgb]{0.66,0.14,0.44}{The blue terms have a half part of uncertain terms}\\[-40mm]
\item[\begin{footnotesize}\PlusCenterOpen \end{footnotesize}]\sihao\li\textcolor[rgb]{0.66,0.14,0.44}{The blue terms have another half part of uncertain terms}\\[-15mm]
\color{black} Then, (\ref{eq.uncer}) can be easily
rewritten~!\\[-45mm]
\sanhao\bea\Omega_1+\Omega_2\ol D\lt[\Omega_3\quad
0\rt]+\lt[\Omega_3\quad 0\rt]^T\ol D^T\Omega_2^T<0\nm\eea

\end{itemize}
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%                                           Numerical Simulation parameters d lambda & tao              %
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\begin{itemize}
\item[\begin{footnotesize}\HandRightUp  \end{footnotesize}]\erhao \hei \textcolor[rgb]{0.00,0.00,1.00}{Numerical
Simulation}\\[5mm]
\sihao Consider the system with parameters as follows:\\[5mm]
 \small \hei $A = \lt[\ba{cc}-4&1.5\\4&-2\ea\rt]$,\quad$A_d =
\lt[\ba{cc}0.7&0.8\\0.6&0.9\ea\rt]$, \quad$B =
\lt[\ba{c}1\\0.5\ea\rt]$;\\[5mm]
for simply,there are only two uncertaintities\quad $\Delta A = \theta_1A_1$\quad$\Delta A_d = \beta_1A_{d1}$;\\[5mm]
where\quad $A_1 = \lt[\ba{cc}1&0\\0&0\ea\rt]$,\quad$A_{d1} =
\lt[\ba{cc}0.2&0\\0&0\ea\rt]$\\[5mm]
it can easily obtain: \\[5mm]$G = \lt[\ba{c}1\\0\ea\rt]$,$H = \lt[\ba{cc}1\quad 0\ea\rt]$\\[5mm]
$G_d = \lt[\ba{c}1\\0\ea\rt]$,$H = \lt[\ba{cc}0.2\quad 0\ea\rt]$
\end{itemize}
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%                                           Numerical Simulation parameters d lambda & tao              %
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\begin{itemize}
\item[\HandPencilLeft] \sihao \hei \textcolor[rgb]{0.00,0.00,1.00}{The parameters of d, $\lambda$, $\&$ $\bar{\tau}$}
    \begin{itemize}
    \sihao
    \vspace{-8mm}
    \item [$\star$]\hei \textcolor[rgb]{0.00,0.00,0.00}{When $\lambda=0$, the condition is delay independent;}
    \vspace{-8mm}
    \item [$\star$]\hei \textcolor[rgb]{0.00,0.00,0.00}{When $\lambda\neq 0$ inequalities and are nonlinear:   }
        \begin{itemize}
        \vspace{-10mm}
        \item [$\blacktriangleright$] \xiaosihao\textcolor[rgb]{0.00,0.00,0.00}{the LMIs can also be solved using LMI toolbox by fixing $\lambda$ first}
        \vspace{-11mm}
        \item [$\blacktriangleright$] \xiaosihao\textcolor[rgb]{0.00,0.00,0.00}{then searching for feasible solutions using different
values of $\lambda$.}
        \end{itemize}
    \end{itemize}
\end{itemize}
%% 绘制表格
\vspace{0.2em} \centering
the value of \textcolor[rgb]{0.00,0.00,1.00}{\sihao\hei $\bar{\tau}$} with \textcolor[rgb]{0.00,0.00,1.00}{d} and \textcolor[rgb]{0.00,0.00,1.00}{$\lambda$} changed\\
\vspace{3mm}
\begin{minipage}[c]{168mm}
\centering \fontsize{11pt}{13pt}\selectfont
\begin{tabular}{cc||c c c c c c c c c c}
\hline
              & &    &    &    &    & $\lambda$\\
\cline{3-12}  & & 0.1& 0.2& 0.3& 0.4& 0.5& 0.6& 0.7& 0.8& 0.9&1.0\\
\hline
  & 0  & \color{blue}{7.0612}& 4.0350& 3.0432& 2.5587& 2.2729& 2.0827& 1.9447& 1.8377& 1.7504& 1.6763\\
  & 0.1& 6.9989& 3.9939& 3.0094& 2.5284& 2.2443& 2.0552& 1.9178& 1.8114& 1.7246& 1.6511\\
  & 0.2& 6.9252& 3.9455& 2.9700& 2.4934& 2.2119& 2.0243& 1.8881& 1.7827& 1.6969& 1.6245\\
  & 0.3& 6.8362& 3.8876& 2.9234& 2.4526& 2.1746& 1.9895& 1.8552& 1.7514& 1.6673& 1.5966\\
  & 0.4& 6.7263& 3.8170& 2.8673& 2.4045& 2.1314& 1.9498& 1.8185& 1.7173& 1.6356& 1.5674\\
 d& 0.5& 6.5865& 3.7285& 2.7986& 2.3466& \color{red}{2.0806\footnote[2]{\color{red}{used in the following simulation}}}& 1.9043& 1.7773& 1.6800& 1.6020& 1.5373\\
  & 0.6& 6.4015& 3.6141& 2.7118& 2.2755& 2.0201& 1.8517& 1.7311& 1.6395& 1.5668& 1.5070\\
  & 0.7& 6.1426& 3.4590& 2.5985& 2.1860& 1.9466& 1.7902& 1.6793& 1.5960& 1.5307& 1.4777\\
  & 0.8& 5.7467& 3.2340& 2.4425& 2.0692& 1.8557& 1.7181& 1.6219& 1.5508& 1.4957& 1.4517\\
  & 0.9& 5.0319& 2.8662& 2.2102& 1.9097& 1.7415& 1.6349& 1.5615& 1.5079& 1.4670& \color{blue}{1.4348}\\
\hline
\end{tabular}\label{Tab:2.1}
\end{minipage}
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%                                           Numerical Simulation
%                                              States                                       %
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%\vspace{15mm}
\begin{itemize}
\item [$\star$]\xiaoerhao \hei \textcolor[rgb]{0.00,0.00,1.00}{Simulation Result: 1}
\end{itemize}
    \centering \sihao \hei \textcolor[rgb]{0.00,0.00,0.00}{The system's states [x1,x2]}
          \begin{center}
          \includegraphics[width=148mm]{graph/states.pdf}\\
          \end{center}
          \wuhao\textcolor[rgb]{0.00,0.00,0.00}{Fig 1.Stats [x1,x2] with $d=0.5,\lambda=0.5,\bar{\tau}=2.0806$}\\[-10mm]
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%                                              Amplified States Graph                                      %
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\begin{slide}
\begin{itemize}
\item [$\star$]\xiaoerhao\hei\textcolor[rgb]{0.00,0.00,1.00}{Simulation Result: 2}
\end{itemize}
    \centering \sihao \hei \textcolor[rgb]{0.00,0.00,0.00}{Amplified States Graph:[x1,x2]nearby X-Axis}
          \begin{center}
          \includegraphics[width=148mm]{graph/statesG.pdf}\\[-5mm]
          \end{center}
          \xiaosihao\textcolor[rgb]{0.00,0.00,0.00}{Fig 2.Amplified States Graph:[x1,x2]with
          $d=0.5,\lambda=0.5,\bar{\tau}=2.0806$}\\[-10mm]
\end{slide}

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%                                               Phase
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\begin{slide}
\begin{itemize}
\item [$\star$]\textcolor[rgb]{0.00,0.00,1.00}{Simulation Result: 3}
\end{itemize}
    \centering \sihao \hei \textcolor[rgb]{0.00,0.00,0.00}{The phase of [x1,x2] }
          \begin{center}
          \includegraphics[width=148mm]{graph/phase.pdf}\\[-10mm]
          \end{center}
\centering \wuhao\textcolor[rgb]{0.00,0.00,0.00}{Fig 3.Phase of states [x1,x2] with $d=0.5,\lambda=0.5,\bar{\tau}=2.0806$}\\
\end{slide}

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%                                             Amplified  Phase
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\begin{slide}
\begin{itemize}
\item [$\star$]\textcolor[rgb]{0.00,0.00,1.00}{Simulation Result: 4}
\end{itemize}
    \centering \sihao \hei \textcolor[rgb]{0.00,0.00,0.00}{The Amplified phase graph of [x1,x2] }
          \begin{center}
          \includegraphics[width=148mm]{graph/phaseG.pdf}\\[-10mm]
          \end{center}
          \centering \wuhao\textcolor[rgb]{0.00,0.00,0.00}{Fig 4.Amplified Phase of [x1,x2] with
          $d=0.5,\lambda=0.5,\bar{\tau}=2.0806$}\\
\end{slide}
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%                                             Output of Controller
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\begin{slide}
\begin{itemize}
\item [$\star$]\textcolor[rgb]{0.00,0.00,1.00}{Simulation Result: 5}
\end{itemize}
    \centering \sihao \hei \textcolor[rgb]{0.00,0.00,0.00}{The Output of Sliding Mode Controller}
          \begin{center}
          \includegraphics[width=148mm]{graph/output.pdf}\\[-10mm]
          \end{center}
          \centering \wuhao\textcolor[rgb]{0.00,0.00,0.00}{Fig 5.The output of the sliding Mode controller with
          $d=0.5,\lambda=0.5,\bar{\tau}=2.0806$}\\
\end{slide}


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%                                             Parameters of Adaptive
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\begin{itemize}
\item [$\star$]\textcolor[rgb]{0.00,0.00,1.00}{Simulation Result: 6}
\end{itemize}
    \centering \sihao \hei \textcolor[rgb]{0.00,0.00,0.00}{The parameters of the adaptation laws:\quad$\hat{\theta}$,\quad$\hat{\beta}$,\quad$\hat{c}$ and
    $\hat{k}$}\footnote[3]{\wuhao\hei\textcolor[rgb]{1.00,0.00,0.00}{The parameters\quad$\hat{\theta}$\quad and \quad$\hat{\beta}$ are required to be less than
    1}}
          \begin{center}
          \includegraphics[width=148mm]{graph/parameters.pdf}\\[-10mm]
          \end{center}
          \centering \wuhao\textcolor[rgb]{0.00,0.00,0.00}{Fig 6.The parameters of system with
          $d=0.5,\lambda=0.5,\bar{\tau}=2.0806$}\\
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%                                                  Thanks                                               %
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\vspace{-8mm} \yihao\centering\textcolor[rgb]{0.00,0.00,0.00}{\Huge
Thank You !}
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