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\title{\Huge\bf Réponse fréquentielle des systèmes linéaires invariants}



\author[Y. Ariba]{\Large\bf Yassine Ariba}

\institute[INSA, Toulouse]{\Large Dpt GEI - INSA, Toulouse.}
\date[]{}

\begin{document}
\thispagestyle{empty}

\begin{frame}

\maketitle

\vspace{2\bigskipamount}

\begin{flushright}
Automatique des systèmes linéaires
\end{flushright}

\end{frame}
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\begin{frame}
\frametitle{\Large Pré-requis}

Pour cette leçon, il est nécessaire d'être à l'aise avec les notions de :

\bigskip

\begin{itemize}
\item transformée de Laplace
\item représentation des systèmes LTI par fonction de transfert
\item réponse temporelle
\begin{center}
\includegraphics[width=7cm]{./figures/principe-analyse-temporelle}
\end{center}
\end{itemize}

\bigskip

Il est conseillé d'avoir en tête

\medskip

\begin{itemize}
\item dans le cours d'électronique : régime sinusoïdal
\end{itemize}

 
\end{frame}
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\begin{frame}
\frametitle{Intérêt ?}

\begin{block}{}
Réponse fréquentielle $\Rightarrow$ comment réagit le système en fonction des\\ \hspace{3.3cm} caractéristiques fréquentielles du signal d'entrée ?
\end{block}

\bigskip

\uncover<2->{

Exemple : analyse modale de structures

\bigskip

Mesure de la réponse d'une structure mécanique à une excitation.

\bigskip

\begin{center}
\includegraphics[width=9cm]{./figures/analyse-modale-structures1}
\end{center}


$\Rightarrow$ Objectif : estimer les fréquences de résonance d'une structure mécanique.


}


\end{frame}
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\begin{frame}
\frametitle{Exemple introductif}

Soit un circuit RC

\vspace{1\bigskipamount}


\begin{center}
\includegraphics[width=7cm]{./figures/introduction-circuit-rc}
\end{center}

\vspace{2\bigskipamount}

\uncover<2->{
Régime sinusoïdal :
\begin{equation*}
\left\{\begin{array}{l}
e(t)=e_m\sin(\omega t)\\[3mm]
v(t)=v_m\sin(\omega t+\phi)
\end{array}\right.
\end{equation*}

\bigskip

$\color{structureColor}\blacktriangleright$ Sinusoïde en entrée $\Rightarrow$ sinusoïde en sortie de même pulsation\\[2mm]

\hspace{3.5cm} mais d'amplitude et de phase différentes

 }
 
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\begin{frame}
%\frametitle{Réponse fréquentielle: exemple introductif 1}


Pour $R=1k\Omega$ et $C=200\mu F$, appliquons une tension $e(t)=\sin(8t)$

\vspace{1\bigskipamount}

\includegraphics[width=8cm]{./figures/exp-circuit-rc}

\medskip

 \begin{center}
\includegraphics[trim = 15mm 76mm 10mm 80mm, clip,width=7cm]{./figures/intro-reponse-frequentielle-sinus-circuit-rc}
\end{center}

 
\end{frame}
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\begin{frame}
%\frametitle{Réponse fréquentielle: exemple introductif 1}

Testons les réponses du circuit pour les pulsations $\omega=\{0.8,4,8,40 \}$ en rad/s

\begin{columns}

\column{6cm}

\begin{center}
\includegraphics[trim = 00mm 70mm 00mm 80mm,
clip,width=5.0cm]{./figures/intro-reponse-frequentielle-sinus-circuit-rc-1}
\end{center}

\column{6cm}
\uncover<2->{
\begin{center}
\includegraphics[trim = 00mm 70mm 00mm 80mm,
clip,width=5.0cm]{./figures/intro-reponse-frequentielle-sinus-circuit-rc-2}
\end{center}}


\end{columns}


\begin{columns}

\column{6cm}
\uncover<3->{
\begin{center}
\includegraphics[trim = 00mm 70mm 00mm 80mm,
clip,width=5.0cm]{./figures/intro-reponse-frequentielle-sinus-circuit-rc-3}
\end{center}}

\column{6cm}
\uncover<4->{
\begin{center}
\includegraphics[trim = 00mm 70mm 00mm 80mm,
clip,width=5.0cm]{./figures/intro-reponse-frequentielle-sinus-circuit-rc-4}
\end{center}}

\end{columns}
 
\end{frame}
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\begin{frame}
%\frametitle{Réponse fréquentielle: exemple introductif 1}

\medskip

Selon la pulsation, l'amplitude et la phase changent


\begin{columns}

\column{6cm}
Traçons :
\begin{itemize}
\item le {\bf gain} (en dB) : $\displaystyle\text{gain}_{dB}=20\log \left|\frac{v_m}{e_m}\right|$ 
\item le {\bf déphasage} (en deg) : $\displaystyle\phi = \frac{d}{T}360$
\end{itemize}

\column{4cm}

\includegraphics[trim = 18mm 64mm 10mm 87mm,clip,width=4.0cm]{./figures/intro-reponse-frequentielle-notations}
\end{columns}


\uncover<2->{

\medskip

 \begin{center}
\includegraphics[trim = 00mm 80mm 00mm 85mm, clip,width=8cm]{./figures/intro-reponse-frequentielle-circuit-rc-graph2}
\end{center}}


 
\end{frame}
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\begin{frame}
%\frametitle{Réponse fréquentielle: exemple introductif 1}

\medskip

En répétant l'opération pour toutes les pulsations entre $0.1rad/s$ et $100rad/s$

\vspace{1\bigskipamount}

 \begin{center}
\includegraphics[trim = 15mm 76mm 10mm 85mm, clip,width=8cm]{./figures/intro-reponse-frequentielle-bode-circuit-rc-1}
\end{center}


\bigskip

\hspace{2cm}$\Rightarrow$ Cette représentation graphique est le diagramme de Bode


 
\end{frame}
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\begin{frame}
\frametitle{Cas général des systèmes LTI} 

\begin{block}{}
{\bf Définition :} la réponse fréquentielle d'un système LTI est définie comme l'ensemble des réponses du système en régime permanent à des entrées sinusoïdales en fonction de la pulsation
\end{block}

\vspace{1.5\bigskipamount}

\begin{center}\includegraphics[width=7cm]{./figures/bloc-io-reponse-frequentielle}\end{center}


\vspace{1.5\bigskipamount}

\uncover<2->{
La réponse fréquentielle est caractérisée par

\bigskip

\begin{itemize}
\item son amplification : \quad$\displaystyle \frac{y_0}{u_0} = |F(j\omega)|$\qquad(sans unité)\\[2mm]
\item son déphasage : \quad$\displaystyle 360\frac{d}{T} = \arg F(j\omega)$\qquad(en deg)
\end{itemize}}

\end{frame}
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\begin{frame}<1-| handout:1>
%\frametitle{Cas général des systèmes LTI} 
\small


Calculons la réponse d'une fonction de transfert stable 


\begin{equation*}
F(s)=\frac{N(s)}{(s+p_1)\cdots(s+p_n)}\quad\text{ à l'entrée }\quad \hat{u}(s)= u_0\frac{\omega}{s^2+\omega^2}
\end{equation*}

\bigskip

\uncover<2->{
 La sortie s'écrit :
 \begin{equation*}
 \hat{y}(s)=F(s)u_0\frac{\omega}{s^2+\omega^2} = \frac{A_1}{s+p_1}+\cdots+\frac{A_n}{s+p_n}+\frac{B_1}{s+j\omega}+\frac{B_2}{s-j\omega}
 \end{equation*}}

\bigskip 

\begin{overprint}
   \onslide<3 | handout:0> 
   Quelle est l'expression temporelle  de la fonction $\displaystyle \hat{f}(s)=\frac{1}{s+a}$ ?\\
   \smallskip
   ($\forall t\geq 0$)
   
   \vspace{1\bigskipamount}
   \begin{minipage}{5cm}
\begin{enumerate}
\item[(A)] $\displaystyle f(t) = a\:t$\\[4mm]
\item[(B)] $\displaystyle f(t) = e^{-at}$
\end{enumerate}
   \end{minipage}
   \begin{minipage}{5cm}
\begin{enumerate}
\item[(C)] $\displaystyle f(t) = \sin(a\:t)$\\[4mm]
\item[(D)] $\displaystyle f(t) = t^a$
\end{enumerate}
   \end{minipage}

   \onslide<4 | handout:0>
   Quelle est l'expression temporelle  de la fonction $\displaystyle \hat{f}(s)=\frac{1}{s+a}$ ?\\
   \smallskip
   ($\forall t\geq 0$)
   
   \vspace{1\bigskipamount}
   \begin{minipage}{5cm}
\begin{enumerate}
\item[(A)] $\displaystyle f(t) = a\:t$\\[4mm]
\item[\color{green}{(B)}]{\color{green} $\displaystyle f(t) = e^{-at}$}
\end{enumerate}
   \end{minipage}
   \begin{minipage}{5cm}
\begin{enumerate}
\item[(C)] $\displaystyle f(t) = \sin(a\:t)$\\[4mm]
\item[(D)] $\displaystyle f(t) = t^a$
\end{enumerate}
   \end{minipage}
   \onslide<5-6 | handout:1> 
   En temporel :
   \begin{equation*}
   y(t) = A_1e^{-p_1t}+\cdots+A_2e^{-p_nt}+B_1e^{-j\omega t}+ B_2e^{j\omega t}
   \end{equation*}
   
   \medskip
   
   Pour un système stable, en régime permanent :
 \begin{equation*}
 y_{RP}(t) = B_1e^{-j\omega t}+ B_2e^{j\omega t}\uncover<3->{\qquad\text{ avec }\quad
 {\footnotesize \left\{\begin{array}{l}
 \displaystyle B_1 = F(s)\frac{u_0\omega}{s^2+\omega^2}(s+j\omega)\bigg|_{s=-j\omega} = \frac{u_0F(-j\omega)}{-2j}\\[4mm]
 \displaystyle B_2 = F(s)\frac{u_0\omega}{s^2+\omega^2}(s-j\omega)\bigg|_{s=j\omega} = \frac{u_0F(j\omega)}{2j}
 \end{array}\right.}}
 \end{equation*}

 \end{overprint} 
 
 

\medskip
 
 \uncover<6->{
 \begin{equation*}
 y_{RP}(t) = u_0|F(j\omega)|\bigg(\frac{e^{j\phi}e^{j\omega t}- e^{-j\phi}e^{-j\omega t}}{2j}\bigg)  = \underbrace{u_0|F(j\omega)|}_{y_0}\sin(\omega t+\varphi)
 \end{equation*}}
\end{frame}
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\begin{frame}
\frametitle{Représentations graphiques}


 $F(j\omega)$ est une fonction complexe de la pulsation $\omega$\\[3mm]
 
 
 \begin{itemize}
 \item le diagramme de Bode,\\[1mm]
 \hspace{1.5cm}$\hookrightarrow$ 2 graphes, gain et phase en fonction de $\omega$\\[3mm]
 \item le diagramme de Black-Nichols.\\[1mm]
 \hspace{1.5cm}$\hookrightarrow$ gain en fonction de la phase\\[3mm]
 \item le diagramme de Nyquist,\\[1mm]
 \hspace{1.5cm}$\hookrightarrow$ plan complexe\\[3mm]
 \end{itemize}
 

\uncover<2->{
\begin{columns}

\column{5cm}
\begin{center}
\includegraphics[width=4.5cm]{./figures/plan-complexe}
\end{center}

\column{5cm}
\vspace{-3mm}
\begin{center}
\includegraphics[trim = 37mm 90mm 35mm 95mm,clip,width=5.0cm]{./figures/nyquist-RC}
\end{center}

\end{columns}}


\end{frame}
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\begin{frame}
\frametitle{Exemple}

Oscillations forcées d'un pendule

\vspace{1\bigskipamount}

\begin{columns}

\column{5cm}
\begin{center}
\includegraphics[width=4.5cm]{./figures/intro-reponse-frequentielle-pendule}
\end{center}

\column{5cm}

Fonction de transfert :

\medskip

\begin{equation*}
\frac{\hat{x}(s)}{\hat{z}(s)} = \frac{60}{s^2+3s+60}
\end{equation*}

\medskip

\uncover<2->{
$s\to j\omega$

\begin{equation*}
F(j\omega) = \frac{60}{-\omega^2+3\omega j+60}
\end{equation*}}

\end{columns}

\vspace{2.5\bigskipamount}

\uncover<3->{

gain = $\displaystyle \frac{60}{\sqrt{(60-\omega^2)^2+9\omega^2}}$ \qquad et \qquad déphasage = $\displaystyle \arg\frac{60}{60-\omega^2+3\omega j}$

}



\end{frame}
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\begin{frame}<1-| handout:1>
%\frametitle{Représentations graphiques}


 
 
 \begin{center}
\includegraphics[trim = 20mm 90mm 20mm 95mm, clip,width=8cm]{./figures/intro-reponse-frequentielle-bode-masse-ressort}
\end{center}

\begin{overprint}

   \onslide<1 | handout:1> 
   Tracé facile avec MATLAB :\\
   
   \begin{center}
   \includegraphics[width=5cm]{./figures/matlab_bode}
   \end{center}
   
   \onslide<2 | handout:0> 
Comment se comporte la masse pour une excitation à $\omega=7$ rad/s ?

\medskip

\begin{enumerate}
\item[(A)] la masse et le point A ont la même amplitude d'oscillation\\[3mm]
\item[(B)] la masse ne bouge pas\\[3mm]
\item[(C)] les oscillations de la masse sont plus amples que celle du point A\\[3mm]
\item[(D)] la masse et le point A sont en phase
\end{enumerate}

   \onslide<3 | handout:0>
Comment se comporte la masse pour une excitation à $\omega=7$ rad/s ?

\medskip

\begin{enumerate}
\item[(A)] la masse et le point A ont la même amplitude d'oscillation\\[3mm]
\item[(B)] la masse ne bouge pas\\[3mm]
\item[\color{green}{(C)}] {\color{green}les oscillations de la masse sont plus amples que celle du point A}\\[3mm]
\item[(D)] la masse et le point A sont en phase
\end{enumerate}
 
 \end{overprint} 


\end{frame}
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\begin{frame}
\frametitle{Pour la prochaine fois...}

Reprendre l'exemple du système mécanique avec oscillations forcées

\bigskip

\begin{itemize}
\item Calculer le gain et le déphasage en $\omega=7$ rad/s\\[3mm]
\item Vérifier vos calculs en simulant (Simulink) la réponse à $z(t) = \sin(7t)$\\[3mm]
\item Tracer et interpréter le diagramme de Nyquist correspondant\\
\end{itemize}

\vspace{2\bigskipamount}


\begin{columns}

\column{3cm}

\centering
Des ressources sur moodle 

\medskip

\begin{center}
\includegraphics[width=1.4cm]{./figures/QR_moodleinsa}
\end{center}

\column{3cm}

\centering
Des exemples sur MATLAB Central 

\medskip

\begin{center}
\includegraphics[width=1.4cm]{./figures/Matlab_Logo}
\end{center}

\column{3cm}

Des questions ?

\medskip

{\tt ariba@insa-toulouse.fr}

\end{columns}

 
\end{frame}
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\end{document}