From d57830f883af0030076a47539d488e3990b65308 Mon Sep 17 00:00:00 2001 From: "flyingscorpio@arch-desktop" Date: Sun, 21 Nov 2021 14:16:24 +0100 Subject: [PATCH] Work on td1 --- theorie-signal/exercices/td1.tex | 65 ++++++++++++++++++++++++-------- 1 file changed, 50 insertions(+), 15 deletions(-) diff --git a/theorie-signal/exercices/td1.tex b/theorie-signal/exercices/td1.tex index 268ce49..37ecba5 100644 --- a/theorie-signal/exercices/td1.tex +++ b/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 $x(t)$}