Fixes

README.md

@@ -26,7 +26,7 @@ Make sure to compile with xelatex (required by fontspec).
 2. After you launched the docker daemon (started Docker Desktop) you can simply run execute the `./build_latex_local.sh` file to generate the pdf files. They are located in a newly created subfolder called `build`.  
 (Most probably you have to change the execution permissions with `chmod +x build_latex_local.sh` before you are able to run the script for the first time.)
 
-### Native LaTeX compiler
+### Native LaTeX Compiler
 
 If you want to use your own native installed LaTeX compiler we recommend to use [TeXLive](https://www.tug.org/texlive/) which integrates well within visual studio code.
 
@@ -80,3 +80,7 @@ The LaTeX template of this summary can be found [here](https://github.com/MeierT
 ## License
 
 Distributed under the MIT License. See [LICENSE](LICENSE) for more information.
+
+## Disclaimer
+
+All information, images, and materials presented here are sourced from the official lecture slides of the course "Control Systems II" taught by Prof. Dr. Emilio Frazzoli (Spring Semester 2024) and the textbook authored by Gioele Zardini. This course summary is intended solely for educational and academic purposes. Any other use or distribution is strictly prohibited.

src/BSB/reach_observability.tex

@@ -1,7 +1,7 @@
 % Please ignore the mess I've created here, I don't feel comfortable either.
 \begin{center}
     \begin{tikzcd}[ampersand replacement=\&]
-        \substack{\text{unstable} \\ \text{modes}}      \&[2em] \substack{\text{all} \\ \text{modes}} \&[-3em] \\[-2em]
+        \substack{\text{unstable} \\ \text{modes} \\ \in{} \mathcal{R} \\ \in{} \mathcal{O}}      \&[2em] \substack{\text{all} \\ \text{modes}} \&[-3em] \\[-2em]
         \& \mathcal{C} \arrow[d, teal, xshift=-0.5ex,
         "{\color{teal}\begin{matrix}                        %chktex 18
             \mathsf{CT}: & \text{always}                \\
@@ -10,7 +10,7 @@
         "'                                                  %chktex 18
         ] \& {\footnotesize\begin{cases}
                     \mathsf{CT:} & \mathbf{x}_0 = 0 \to~\mathbf{x_f} \\
-                    \mathsf{DT:} & \mathbf{x_f} \to~0
+                    \mathsf{DT:} & \mathbf{x}_0 = \mathbf{x_f} \to~0
                 \end{cases}}
         \\[2em]
         \mathcal{S}                                     \& \mathcal{R} \arrow[l]\arrow[u, xshift=0.5ex] \&

src/images/digit_controller_design.drawio

@@ -1,6 +1,6 @@
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@@ -48,17 +48,6 @@
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@@ -81,9 +70,20 @@
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src/sections/LQG.tex

@@ -151,7 +151,7 @@ \subsection{Transfer Function}
 
 \textbf{Remarks}
 \begin{itemize}
-    \item In $G_{yr}^{cl}$, only the closed-loop poles of the controller are retained (observer and system are cancelled) and open-loop zeros are retained.
+    \item In $G_{yr}^{cl}$, only the closed-loop poles of the controller (observer and system are cancelled) and open-loop zeros are retained.
     \item $G_{uy}$, has the same order as the plant and the number of closed-loop poles is doubled (all stable)
 \end{itemize}
 

src/sections/LQR.tex

@@ -13,14 +13,14 @@ \subsection{Cost Functional}
           \begin{itemize}
               \item is symmetric i.e. $\mathbf{Q}=\mathbf{Q}^{\mathsf{T}}$
               \item is positive semidefinite i.e. $\mathbf{x}^{\mathsf{T}} \mathbf{Qx} \geq 0$ or $\mathrm{eig}(\mathbf{Q}) \geq 0$
+              % TODO: Why must it be sqrt (below)? Compare with LQE technical conditions
               \item must give positive weights to unstable modes to penalize them in the cost function (the pair $\mathbf{A}, \sqrt{\mathbf{Q}}$ is detectable)
           \end{itemize}
     \item $\mathbf{R}$ 
     \begin{itemize}
         \item must be positive definite i.e. $\mathbf{u}^{\mathsf{T}} \mathbf{Ru} >0\; \forall \mathbf{u}\neq \mathbf{0}$ or $\mathrm{eig}(\mathbf{R}) > 0$ (otherwise not every control effort is penalized).
-        \item is symmetric (see derivation in H2-Sythesis Section~\ref{h2_synth})
+        \item is symmetric (see derivation in H2-Synthesis Section~\ref{h2_synth})
     \end{itemize}
-          % TODO: Why must Q but not R be symmetric? -> I think this should be true. -> From H2 Synthesis: R = D^TD which yields a symmetric matrix
     \item The pair $\mathbf{A,B}$ must be stabilizable
     \item In general, there is an additional cross-coupling term $\mathbf{x}^{\mathsf{T}}\mathbf{Nu}$ in $J$ which is often neglected i.e.\ equal to $0$
 \end{itemize}
@@ -168,12 +168,20 @@ \subsection{Symmetric Root Locus}
 
 \begin{itemize}
     \item $2n$ branches, where $n$ is the size of $\mathbf{A}$
+    \begin{itemize}
+        \item like in standard RL, every branch starts in an OL pole
+        \item but asymptotic behavior different from standard RL
+    \end{itemize}
     \item symmetric to the real and imaginary axis
     \item LQR closed-loop poles are all in the LHP
     \item $\rho \rightarrow \infty$ (expensive control):\newline
           CL poles approach stable OL poles and the mirror-images of the unstable OL poles (all on LHP)
     \item $\rho \rightarrow 0$ (cheap control):\newline
           CL poles approach MP OL zeros and the mirror-images of the NMP OL zeros or go to infinity along the LHP asymptotes (all on LHP)
+    \item note that for $\rho \rightarrow \infty$ there are no asymptotes to $\infty$ but only convergence towards (mirror) OL poles
+    \begin{itemize}
+        \item for $\rho \rightarrow 0$ there is also convergence to $\infty$
+    \end{itemize}
 \end{itemize}
 
 \begin{examplesection}[Example: Symmetric Root Locus]