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theorie-signal/exercices/tp1.tex
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@ -4,7 +4,7 @@
Théorie du signal --- TP1
\\ \large Décomposition en Série de Fourier
}
\author{}
\author{Alexandre CHEN et Tunui FRANKEN}
\date{Dernière compilation~: \today{} à \currenttime}
\usepackage{../../cours}
@ -46,15 +46,30 @@ Théorie du signal --- TP1
\\
\end{array}
\right.\\
a_n &= \frac{4}{T_0}\left[
a_n &= \frac{4}{T_0}\left(\left[
1 - \frac{2}{T_0}|t|
\frac{\sin(n\omega_0 t)}{n\omega_0}
\right]_0^{T_0/2}
- \int_0^{\sfrac{T_0}{2}} -\frac{2}{T_0}\frac{\sin(n\omega_0 t)}{n\omega_0} \dif t
- \int_0^{\sfrac{T_0}{2}} -\frac{2}{T_0}\frac{\sin(n\omega_0 t)}{n\omega_0}\dif t \right) \\
&= \frac{4}{T_0}\left(
\left[
\left(1 - \frac{2T_0}{T_0 2}\frac{\sin(n\pi)}{\frac{n2\pi}{T_0}}\right) - (1 - 0)
\right]
+ \frac{2}{2n\pi}\left[
\left(
\frac{-\cos(n\pi)}{\frac{2n\pi}{T_0}}
\right) - \frac{1}{\frac{2n\pi}{T_0}}
\right]
\right) \\
&= \frac{4}{T_0}\frac{2}{2n\pi}\left(
\frac{-\cos(n\pi) - 1}{\frac{2n\pi}{T_0}}
\right) \\
&= \frac{2}{(n\pi)^2}(-\cos(n\pi) + 1)
\end{align*}
\begin{align*}
S_x(t) &= a_0 + \sum_{n=0}^{+\infty} a_n\cos(n\omega_0 t) \\
&= \frac{1}{2} + \sum_{n=0}^{+\infty} \frac{2}{(n\pi)^2}(-\cos(n\pi) + 1) \cos(n\omega_0 t)
\end{align*}
\begin{align*}
@ -68,4 +83,71 @@ Théorie du signal --- TP1
\end{align*}
avec $r$ le rapport cyclique tel que $r < 1$ et $T_0 = 0.5s$.
$x_2$ est paire, donc les $b_n$ sont nuls.
\begin{align*}
a_0 &= \frac{1}{T_0}\int_{(T_0)}1\dif t \\
&= \frac{2}{T_0}\int_0^{\sfrac{rT_0}{2}} 1\dif t \\
&= \frac{2}{T_0}[t]_0^{\sfrac{rT_0}{2}} \\
&= \frac{2}{T_0}\frac{rT_0}{2} \\
a_0 &= r
\end{align*}
\begin{align*}
a_n &= \frac{2}{T_0}\int_{(T_0)}1\cdot\cos(n\omega_0 t)\dif t \\
&= \frac{4}{T_0}\left[
\frac{\sin(n\omega_0 t)}{n\omega_0}
\right]_0^{\sfrac{rT_0}{2}} \\
&= \frac{4}{T_0}\frac{\sin(n\pi r)}{n2\pi/T_0} \\
&= \frac{4\sin(n\pi r)}{n2\pi} \\
a_n &= \frac{2\sin(n\pi r)}{n\pi}
\end{align*}
\begin{align*}
S_x(t) &= a_0 + \sum_{n=0}^{+\infty} a_n\cos(n\omega_0 t) \\
&= r + \sum_{n=0}^{+\infty} \frac{2\sin(n\pi r)}{n\pi} \cos(n\omega_0 t)
\end{align*}
\begin{tikzpicture}
\end{tikzpicture}
\subsection{Densité Spectrale de Puissance}
\subsubsection{Pour $x_1$}
\begin{equation*}
c_0 = a_0 = \frac{1}{2}
\end{equation*}
\begin{align*}
c_n &= \frac{1}{2}\frac{2}{(n\pi)^2}(-\cos(n\pi) + 1) \\
&= \frac{-\cos{n\pi} + 1}{(n\pi)^2}
\end{align*}
\begin{align*}
S_{x_1}(f) &= \sum_{n=-\infty}^{+\infty} \left|\frac{-\cos(n\pi) + 1}{(n\pi)^2}\right|^2 \\
&= \sum_{n=-\infty}^{+\infty} \left|\frac{-(-1)^n + 1}{(n\pi)^2}\right|^2 \\
&=
\left\{
\begin{array}{l}
0 \text{ pour $n$ pair} \\
\frac{4}{(n\pi)^2} \text{ pour $n$ impair} \\
\end{array}
\right.
\end{align*}
\subsubsection{Pour $x_2$}
\begin{equation*}
c_0 = a_0 = r
\end{equation*}
\begin{align*}
c_n &= \frac{1}{2}\frac{2\sin(n\pi r)}{n\pi} \\
&= \frac{\sin(n\pi r)}{n\pi}
\end{align*}
\begin{align*}
S_{x_2}(f) &= \sum_{n=-\infty}^{+\infty} \left|\frac{\sin(n\pi r)}{n\pi}\right|^2 \\
\text{pour } r=\frac{1}{2} &\implies
\sum_{n=-\infty}^{+\infty} \left|\frac{\sin(\frac{n\pi}{2})}{n\pi}\right|^2 \\
&= \sum_{n=-\infty}^{+\infty}\frac{1}{(n\pi)^2} \\
\text{pour } r=\frac{1}{3} &\implies
\sum_{n=-\infty}^{+\infty} \left|\frac{\sin(\frac{n\pi}{3})}{n\pi}\right|^2 \\
\text{pour } r=\frac{1}{4} &\implies
\sum_{n=-\infty}^{+\infty} \left|\frac{\sin(\frac{n\pi}{4})}{n\pi}\right|^2 \\
\end{align*}
\end{document}