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@@ -39,5 +39,6 @@
 \input{lectures/2020-05-25.tex}
 \input{lectures/2020-05-27.tex}
 \input{lectures/2020-06-03.tex}
+\input{lectures/2020-06-04.tex}
 
 \end{document}
diff --git a/lectures/2020-06-04.tex b/lectures/2020-06-04.tex
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+\chapter{Appendix: Discretization of an analog system}
+\newlecture{Sergio Savaresi}{04/06/2020}
+
+\missingfigure{Fig1}
+
+\subsection*{A to D converter}
+
+\missingfigure{Fig2}
+
+\begin{description}
+    \item[Time discretization] $\Delta T$ is the sampling time
+    \item[Amplitude discretization] Number of levels used for discretization, e.g. \emph{10-bit discretization} uses $2^{10}$ levels of amplitude
+\end{description}
+
+An high quality of A/D converter:
+\begin{itemize}
+    \item Can use small $\Delta T$
+    \item High number of levels (16-bits)
+\end{itemize}
+
+
+\subsection*{D to A converter}
+
+\missingfigure{Fig3}
+
+If $\Delta T$ is sufficiently small, the step-wise analog signal is very similar to a smooth analog signal.
+
+\missingfigure{Fig4}
+
+What is the model of the Digital Perspective?
+
+\missingfigure{Fig5}
+
+\begin{itemize}
+    \item We make B.B. system identification from measured data: directly estimate a discrete-time model
+    \item We have a physical W.B. model (continuous time), we need to discretize it
+\end{itemize}
+
+The most used approach is the State-Space Transformation.
+
+\[
+    S: \begin{cases}
+        \dot{x} = Ax + bu \\
+        y = Cx + (Du)
+    \end{cases}
+    \qquad
+    \text{sampling time $\Delta T$}
+    \qquad
+    S: \begin{cases}
+        x(t+1) = Fx(t) + Gu(t) \\
+        y(t) = Hu(t) + (Du(t))
+    \end{cases}
+\]
+
+Transformation formulas:
+\begin{align*}
+    F &= e^{A\Delta T} \\
+    G &= \int_0^{\Delta T} e^{A\delta}B\, d\delta \\
+    H &= C \\
+    D &= D
+\end{align*}
+
+\begin{remark}
+    How the poles of the continuous time system are transformed?
+
+    Can be proved that the eigenvalues (poles) follow the \emph{sampling transformation rule}.
+    \[
+        z = e^{s\Delta T} \qquad \lambda_F = e^{\lambda_A \Delta T}
+    \]
+
+    \missingfigure{Fig6}
+
+    How the zeros in of $S$ in continuous time are transformed into zeros in discrete time?
+
+    Unfortunately there is no simple rule like the poles. We can only say:
+    \[
+        G(s) = \frac{\text{polynomial in $s$ with $h$ zeros}}{\text{polynomial in $s$ with $k$ poles}} \qquad \text{if $G(s)$ is strictly proper: } k > h
+    \]
+    \[
+        G(z) = \frac{\text{polynomial in $z$ with $k-1$ zeros}}{\text{polynomial in $z$ with $k$ poles}} \qquad \text{$G(z)$ with relative degree 1}
+    \]
+
+    We have new $k-h-1$ zeros that are generated by the discretization.
+    They are called \emph{hidden zeros}.
+
+    Unfortunately these hidden zeros are frequently outside the unit circle, which means that $G(z)$ is not minimum phase even if $G(s)$ is minimum phase.
+
+    We need for instance GMVC to design the control system.
+\end{remark}
+
+Another simple discretization technique frequently used is the discretization of time-derivative $\dot{x}$.
+
+\begin{align*}
+    \text{\textbf{eulero backward}} &\qquad \dot{x} \approx \frac{x(t)-x(t-1)}{\Delta T} = \frac{x(t)-z^{-1}x(t)}{\Delta T} = \frac{z-1}{z\Delta T} x(t) \\
+    \text{\textbf{eulero forward}} &\qquad \dot{x} \approx \frac{x(t+1)-x(t)}{\Delta T} = \frac{zx(t)-x(t)}{\Delta T} = \frac{z-1}{\Delta T} x(t)
+\end{align*}
+
+General formula
+\[
+    \dot{x}(t) = \left[ \frac{z-1}{\Delta T} \frac{1}{\alpha z + (1-\alpha)} \right]x(t) \qquad \text{with } 0 \le \alpha \le 1
+\]
+\begin{itemize}
+    \item if $\alpha = 0$ it's Eulero Forward
+    \item if $\alpha = 1$ it's Eulero Backward
+    \item if $\alpha = \frac{1}{2}$ it's Tustin method
+\end{itemize}
+
+The critical choice is $\Delta T$ (sampling time).
+The general intuitive rule is: the smaller $\Delta T$, the better.
+
+\missingfigure{Fig7}
+
+If $\Delta T$ is smaller, $\omega_S$ larger
+
+\missingfigure{Fig8}
+
+\[
+    \Delta T \rightarrow f_S = \frac{1}{\Delta T} \qquad \omega_S = \frac{2\pi}{\Delta T} \qquad f_N = \frac{1}{2} f_S \qquad \omega_N = \frac{1}{2} \omega_S
+\]
+
+Hidden problems of a too-small $\Delta T$:
+\begin{itemize}
+    \item Sampling devices (A/D and D/A) cost
+    \item Computational cost: update an algorithm every $1 \mu s$ is much heavier than every $1 ms$
+    \item Cost of memory (if data logging is needed)
+    \item Numerical precision \emph{cost} (hidden computational cost)
+\end{itemize}
+
+\missingfigure{Fig9}
+
+If $\Delta T$ is very small (tends to zero), we squeeze all the poles very closed to $(1,0)$. We need very high numerical precision (use a lot of digits) to avoid instability.
+
+Rule of thumb of control engineers: $f_S$ is between 10 and 20 times the system bandwidth we are interested in.
+
+\missingfigure{Fig10}
+
+\begin{remark}[Another way of managing the choice of $\Delta T$ w.r.t. the aliasing problem]
+    \missingfigure{Fig11}
+
+    The classical way to deal with aliasing is to use anti-alias analog filters
+
+    \missingfigure{Fig12}
+
+    For example, if A/D is at $1KHz$ ($\Delta T = 1ms$)
+    \missingfigure{Fig13}
+
+    \paragraph{Full Digital Approach} without analog anti-alias filter
+
+    \missingfigure{Fig14}
+\end{remark}