Work on td1

theorie-signal/exercices/td1.tex

@@ -1,12 +1,17 @@
 \documentclass[a4paper,french,11pt]{article}
 
-\title{Théorie du signal --- TD1 \\ Décomposition en Série de Fourier}
+\title{
+Théorie du signal --- TD1
+\\
+\large Décomposition en Série de Fourier
+}
 \author{}
 \date{Dernière compilation~: \today{} à \currenttime}
 
 \usepackage{../../cours}
 \usepackage{enumitem}
 \usepackage{xfrac}
+\usepackage{tikz}
 
 \begin{document}
 
@@ -18,29 +23,59 @@
     \begin{align*}
         x(t) =
         \left\{
-        \begin{array}{l}
-            - 1 \,\forall\, t \in [-\frac{T_0}{2};0] \\
-            1 \,\forall\, t \in [0;\frac{T_0}{2}] \\
+        \begin{array}{ll}
+            -1 \quad \forall\, t \in [-\frac{T_0}{2};0] \\
+            1 \quad \forall\, t \in [0;\frac{T_0}{2}] \\
         \end{array}
         \right.
     \end{align*}
 
     \subsection{Tracer le signal $x(t)$}
 
+        \begin{center}
+        \begin{tikzpicture}
+
+            \draw[help lines, dashed] (-7,-2) grid (7,2);
+            \draw[-latex] (-7,0) -- (7,0) node[below]{$t$};
+            \draw[-latex] (0,-2) -- (0,2) node[left]{$x(t)$};
+
+            \foreach \i in {-6, -4, -2, 0, 2, 4, 6}{
+                \draw[very thick, teal]
+                    plot[domain=\i-1:\i]({\x}, {-1})
+                    plot[domain=\i:\i+1]({\x}, {1})
+                    ;
+            }
+
+        \end{tikzpicture}
+        \end{center}
+
     \subsection{Calculer les coefficients de Fourier réels $a_0, a_n, b_n$ du signal $x(t)$}
 
         $x$ est impaire donc $a_0 = a_n = 0$.
-        \begin{align*}
-            b_n &= \frac{2}{T_0} \int_{(T_0)} x(t) \sin(n\omega_0 t) \dif t \\
-                &= \frac{4}{T_0} \int_0^{T_0/2} 1 \sin(n\omega_0 t) \dif t \\
-                &= \frac{4}{T_0} \left[\frac{-\cos(n\omega_0 t)}{n\omega_0}\right]_0^{T_0/2} \\
-                &= \frac{4}{T_0} \left(\frac{-\cos(n2\pi f_0 \frac{T_0}{2}) + 1}{n2\pi f_0}\right) \\
-                &= 2 \left(\frac{-\cos(n\pi) + 1}{n\pi}\right) \\
-                &= \frac{2}{n\pi} (-\cos(n\pi) + 1) \\
-                &= \frac{2}{n\pi} (-(-1)^n + 1) \\
-            b_n &= \boxed{\frac{2}{n\pi} (1 -(-1)^n)} \\\\
-            x(t) = \sum_{n=1}^{+\infty} \frac{2}{n\pi}(1 - (-1)^n) \sin(2n\pi f_0 t)
-        \end{align*}
+
+        \begin{tabularx}{\linewidth}{XX}
+            {
+            \begin{equation*}
+                \text{formule~:} \quad b_n = \frac{2}{T_0} \int_{(T_0)} x(t) \sin(n\omega_0 t) \dif t
+            \end{equation*}
+            \vspace{4cm}
+            \begin{equation*}
+                x(t) = \sum_{n=1}^{+\infty} \frac{2}{n\pi}(1 - (-1)^n) \sin(n\omega_0 t)
+            \end{equation*}
+            } &
+            {
+            \begin{align*}
+                b_n &= \frac{4}{T_0} \int_0^{T_0/2} 1 \sin(n\omega_0 t) \dif t \\
+                    &= \frac{4}{T_0} \left[\frac{-\cos(n\omega_0 t)}{n\omega_0}\right]_0^{T_0/2} \\
+                    &= \frac{4}{T_0} \left(\frac{-\cos(n2\pi f_0 \frac{T_0}{2}) + 1}{n2\pi f_0}\right) \\
+                    &= 2 \left(\frac{-\cos(n\pi) + 1}{n\pi}\right) \\
+                    &= \frac{2}{n\pi} (-\cos(n\pi) + 1) \\
+                    &= \frac{2}{n\pi} (-(-1)^n + 1) \\
+                b_n &= \boxed{\frac{2}{n\pi} (1 -(-1)^n)}
+            \end{align*}
+            } \\
+        \end{tabularx}
+
 
     \subsection{Tracer la DSP du signal $x(t)$}