add filtres numériques
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@@ -19,7 +19,7 @@ Les inconvénients des filtres récursifs :
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- Ils possèdent une phase non-linéaire
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- Ils possèdent une phase non-linéaire
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- Ils peuvent avoir une instabilité numérique
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- Ils peuvent avoir une instabilité numérique
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Fonction de transfert en $z$ :
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***Fonction de transfert en $z$ :***
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$$F(z) = \frac{Y(z)}{U(z)} = \frac{b_{0}+b_{1}z^{-1}+\dots+b_{n}z^{-1}}{1+a_{1}z^{-1}+\dots+a_{n}z^{-n}}$$
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$$F(z) = \frac{Y(z)}{U(z)} = \frac{b_{0}+b_{1}z^{-1}+\dots+b_{n}z^{-1}}{1+a_{1}z^{-1}+\dots+a_{n}z^{-n}}$$
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Numérisation par conversion DA et AD d'un filtre analogique :
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Numérisation par conversion DA et AD d'un filtre analogique :
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$$F(z) = (1-z^{-1})Z \left\{ L^{-1}\left[ \frac{F(s)}{s} \right] \right\} $$
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$$F(z) = (1-z^{-1})Z \left\{ L^{-1}\left[ \frac{F(s)}{s} \right] \right\} $$
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@@ -75,8 +75,231 @@ b_{1} = \frac{2\left( d_{0}-\frac{4d_{2}}{T_{s^{2}}} \right)}{C}
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b_{2} = \frac{d_{0}-\frac{2d_{1}}{T_{s}}+\frac{4d_{2}}{T_{s^{2}}}}{C}
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b_{2} = \frac{d_{0}-\frac{2d_{1}}{T_{s}}+\frac{4d_{2}}{T_{s^{2}}}}{C}
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\end{gather}$$
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\end{gather}$$
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# Filtre numérique non-récursif (FIR)
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Les filtres numériques non-récursifs (FIR) sont des filtres numériques qui n'utilisent pas de rétroactions.
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Ils sont basés sur des équations aux différences.
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Les filtres non-récursifs sont utilisés pour des applications où la phase du filtre est critique.
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Ils sont utilisés dans le traitement du signal et la conception de filtres.
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Les intérêts des filtres non-récursifs :
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- Les fonctions de transfert ne possèdent que des coefficients $b_i$. ($\forall a_{i} = 0$)
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- Ils possèdent une phase linéaire exacte
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- Ils sont stables
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Les inconvénients des filtres non-récursifs :
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- Ils sont plus coûteux en calcul
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- Ils nécessitent plus de matériel
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Coefficients symétrique: pairs $b_{n-i} = b_{i}$, impairs $b_{n-i} = -b_{i}$, $b_{n/2}=0$ <br>
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**Symétrie paire:**
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$$H(\mathbf{j}\omega)
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= \sum^{n}_{i=o}b_{i}e^{-\frac{\mathbf{j}i\omega}{f_{s}}}
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= e^{\frac{-\mathbf{j}n\omega}{2f_{s}}} \sum^{n/2}_{i=0}b_{i}\cos\left[ \frac{\left( i-\frac{n}{2} \right)\omega}{f_{s}} \right]$$
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**Symétrie impaire:**
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$$H(\mathbf{j}\omega)=\mathbf{j}e^{\frac{-\mathbf{j}n\omega}{2f_{s}}}\sum^{n/2}_{i=0}b_{i}\sin\left( \frac{(i-\frac{n}{2})\omega}{f_{s}} \right)$$
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Les filtres FIR à coefficients symétriques ont une phase linéaire.
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$$
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\text{Symétrie paire : }\varphi(\omega)=-\frac{n\omega}{2f_{s}}
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\qquad \qquad
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\text{Symétrie impaire : } \varphi(\omega)= -\frac{n\omega}{2f_{s}}+\frac{\pi}{2}
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$$
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$$\text{Temps de propagation : }t_{gr}=-\frac{d\varphi}{d\omega}=\frac{n}{2f_{s}}=\frac{n}{2}T_{s}$$
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Il y a donc 4 types de filtres FIR à phase linéaire : n resp. symétrie paire/impaire
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***Fonction de transfert en $z$ :***
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$$F(z) = \frac{Y(z)}{U(z)} = b_{0}+b_{1}z^{-1}+\dots+b_{n}z^{-1}$$
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***Équation aux différences :***
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$$y[k]= b_{n}u[k-n]+ \dots + b_{1}u[k-1]+b_{0}u[k]$$
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***Réponse en fréquence :***
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$$z=\exp(\mathbf{j}\omega\cdot T_{s})$$
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## Filtres FIR à symétrie paire :
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$$b_{n-k} = b_{k}$$
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Exemple : FIR à symétrie paire d'ordre 6
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$$\begin{align}
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F(z) &= \frac{Y(z)}{U(z)} \\
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&= 1+2z^{-1}+3z^{-2}+4z^{-3}+3z^{-4}+2z^{-5}+z^{-6} \\
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&= \frac{z^{6}+2z^{5}+3z^{4}+4z^{3}+3z^{2}+2z^{1}+1}{z^{6}}
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\end{align}$$
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Les filtres FIR à symétrie paire sont toujours de type passe-bas ou coupe-bande. <br>
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Ils possèdent autant de pôles que l'order du filtre. Ces derniers sont à (0,0) sur le cercle unité. <br>
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Ils possèdent autant de zéros que l'ordre du filtre. Ces derniers sont sur le cercle unité. <br>
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## Filtres FIR à symétrie impaire :
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$$b_{n-k} = -b_{k}$$
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Exemple : FIR à symétrie impaire d'ordre 6
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$$\begin{align}
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F(z) &= \frac{Y(z)}{U(z)} \\
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&= -1-2z^{-1}-3z^{-2}+0z^{-3}+3z^{-4}+2z^{-5}+z^{-6} \\
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&= \frac{z^{6}+2z^{5}+3z^{4}+0z^{3}-3z^{2}-2z^{1}-1}{z^{8}}
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\end{align}$$
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Les filtres FIR à symétrie paire sont toujours de type passe-haut ou passe-bande. <br>
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Ils possèdent autant de pôles que l'ordre du filtre. Ces derniers sont à (0,0) sur le cercle unité. <br>
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Ils possèdent autant de zéros que l'ordre du filtre. <br>
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# Filtres numériques FIR idéaux
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$$\omega_{c_{d}} = 2\pi \frac{f_{c_{a}}}{f_{s}}$$
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## Filtre passe-bas idéal
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$$H(\mathbf{j}\omega)=\begin{cases}
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1 &\text{ si } |\omega| \leq \omega_{c} \\
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0 &\text{ si } |\omega| > \omega_{c}
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\end{cases} $$
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La transformée de Fourier inverse donne la réponse impulsionnelle d’un filtre passe-bas idéal.
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$$h[n] = \begin{cases}
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\frac{\omega_{c}}{\pi} \frac{\sin((n-N_{s})\omega_{c})}{(n-N_{s})\omega_{c}} &\text{ si } 0\leq n \leq N \\
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\\
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\frac{\omega_{c}}{\pi} &\text{ si } n=N_{s} \\
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\\
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0 &\text{ sinon}
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\end{cases}$$
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On tronque à $N=ordre$ du filtre. $N_{s}=\frac{N}{2}$ <br>
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Ces coefficients ne sont rien d’autre que les coefficients de la fonction de transfert en *z* du filtre passe-bas idéal.
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## Filtre passe-haut idéal
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$$H(\mathbf{j}\omega)=\begin{cases}
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0 &\text{ si } |\omega| \leq \omega_{c} \\
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1 &\text{ si } |\omega| > \omega_{c}
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\end{cases}$$
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La transformée de Fourier inverse donne la réponse impulsionnelle d’un filtre passe-bas idéal.
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$$h[n] = \begin{cases}
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-\frac{\omega_{c}}{\pi} \frac{\sin((n-N_{s})\omega_{c})}{(n-N_{s})\omega_{c}} &\text{ si } 0\leq n \leq N \\
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\\
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1-\frac{\omega_{c}}{\pi} &\text{ si } n=N_{s} \\
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\\
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0 &\text{ sinon}
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\end{cases}$$
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On tronque à $N=ordre$ du filtre. $N_{s}=\frac{N}{2}$ <br>
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Ces coefficients ne sont rien d’autre que les coefficients de la fonction de transfert en *z* du filtre passe-haut idéal.
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## Filtre passe-bande idéal
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$$H(\mathbf{j}\omega)=\begin{cases}
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1 &\text{ si } \omega_{c_{1}} \leq \omega \leq \omega_{c_{2}} \\
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0 &\text{ si } \omega \leq \omega_{c_{1}} \quad et \quad \omega\geq \omega_{c_{2}}
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\end{cases}$$
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La transformée de Fourrier inverse donne la réponse impulsionnelle d’un filtre passe-bande idéal.
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$$h[n] = \begin{cases}
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\frac{\omega_{c_{2}}}{\pi} \frac{\sin((n-N_{s})\omega_{c_{2}})}{(n-N_{s})\omega_{c_{2}}}-\frac{\omega_{c_{1}}}{\pi} \frac{\sin((n-N_{s})\omega_{c_{1}})}{(n-N_{s})\omega_{c_{1}}} &\text{ si } 0\leq n \leq N \\
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\\
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\frac{\omega_{c_{2}}-\omega_{c_{1}}}{\pi} &\text{ si } n=N_{s}\\
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\\
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0 &\text{ sinon}
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\end{cases}$$
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On tronque à $N=ordre$ du filtre. $N_{s}=\frac{N}{2}$ <br>
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Ces coefficients ne sont rien d’autre que les coefficients de la fonction de transfert en *z* du filtre passe-bande idéal.
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## Filtre coupe-bande idéal
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$$H(\mathbf{j}\omega)=\begin{cases}
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0 &\text{ si } \omega_{c_{1}} \leq \omega \leq \omega_{c_{2}} \\
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1 &\text{ si } \omega \leq \omega_{c_{1}} \quad et \quad \omega\geq \omega_{c_{2}}
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\end{cases}$$
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La transformée de Fourrier inverse donne la réponse impulsionnelle d’un filtre coupe-bande idéal.
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$$h[n] = \begin{cases}
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\frac{\omega_{c_{1}}}{\pi} \frac{\sin((n-N_{s})\omega_{c_{1}})}{(n-N_{s})\omega_{c_{1}}}
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-\frac{\omega_{c_{2}}}{\pi} \frac{\sin((n-N_{s})\omega_{c_{2}})}{(n-N_{s})\omega_{c_{2}}}
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&\text{ si } 0\leq n \leq N \\
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\\
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1-\frac{\omega_{c_{2}}-\omega_{c_{1}}}{\pi} &\text{ si } n=N_{s}\\
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\\
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0 &\text{ sinon}
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\end{cases}$$
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On tronque à $N=ordre$ du filtre. $N_{s}=\frac{N}{2}$ <br>
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Ces coefficients ne sont rien d’autre que les coefficients de la fonction de transfert en *z* du filtre coupe-bande idéal.
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# Synthèse par fenêtrage
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Le point de départ de la synthèse des filtres FIR est donné par la considération des réponses impulsionnelles des filtres idéaux.
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Comme celles-ci sont infiniment longues et donc non réalisables, on va se contenter d’une approximation de leurs réponses.
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## Méthode de fenêtrage
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- **Fenêtre rectangulaire :** $$W[n] = \begin{cases}
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1 &\text{ si } 0\leq n \leq N \\
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0 &\text{ sinon}
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\end{cases}$$
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- **Fenêtre triangulaire (Bartlett) :** $$W[n] = \begin{cases}
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2 \frac{n}{N} &\text{ si } 0\leq n \leq \frac{N}{2} \\
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2-2 \frac{n}{N} &\text{ si } \frac{N}{2}\leq n \leq N \\
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0 &\text{ sinon}
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\end{cases}$$
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- **Fenêtre sinusoïdale (Hann) :** $$W[n] = \begin{cases}
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0,5\left( 1-\cos\left( \frac{2\pi n}{N} \right) \right) &\text{ si } 0\leq n\leq N \\
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0 &\text{ sinon}
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\end{cases}$$
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- **Fenêtre de Hamming :** $$W[n] = \begin{cases}
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0,54-0,46\cos\left( \frac{2\pi n}{N} \right) &\text{ si } 0\leq n\leq N \\
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0 &\text{ sinon}
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\end{cases}$$
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- **Fenêtre de Blackman :** $$W[n] = \begin{cases}
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0,42-0,5\cos\left( \frac{2\pi n}{N} \right)+0,08\cos\left( \frac{4\pi n}{N} \right) &\text{ si } 0\leq n\leq N \\
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0 &\text{ sinon}
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\end{cases}$$
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- **Fenêtre de Kaiser-Bessel :** $$W[n] = \begin{cases}
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\frac{I_{0}\left( \beta\sqrt{1-\left( \frac{1-n}{N_{s}} \right)^{2}} \right)}{I_{0}(\beta)} &\text{ si } 0\leq n\leq N \\
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0 &\text{ sinon}
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\end{cases}$$
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$I_{0}$ : Fonction de Bessel modifiée d’ordre 0 <br>
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$\beta$ : Paramètre de forme
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***Caractéristiques spectrales de quelques fenêtres***
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| Fenêtres | Largeur du 1er lobe | Atténuation du 1er lobe [dB] | Décroissance des lobes suivants | Ondulation Rp [dB] | Atténuation max [dB] | Bande de transition<br>[rad/sec]<br>$\Delta\omega$ |
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|---------------------|---------------------|------------------------------|---------------------------------|--------------------|----------------------|----------------------------------------------------|
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| Rectangulaire | $4\pi/N$ | $-13$ | $20 dB/dec$ | $0,74$ | $21$ | $1,8\pi/N$ |
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| Triangulaire | $8\pi/N$ | $-27$ | $40 dB/dec$ | $0$ | $25$ | $6,1\pi/N$ |
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| Cosinus | $8\pi/N$ | $-32$ | $60 dB/dec$ | $0,055$ | $44$ | $6,2\pi/N$ |
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| Hamming | $8\pi/N$ | $-43$ | $20 dB/dec$ | $0,014$ | $53$ | $6,6\pi/N$ |
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| Blackman | $12\pi/N$ | $-58$ | $60 dB/dec$ | $0,0017$ | $74$ | $11\pi/N$ |
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| Kaiser $\beta=4,54$ | $7,2\pi/N$ | $-30$ | $20 dB/dec$ | $0,025$ | $50$ | $5,8\pi/N$ |
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| Kaiser $\beta=5,66$ | $8,4\pi/N$ | $-42$ | $20 dB/dec$ | $0,009$ | $60$ | $7,2\pi/N$ |
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## Comment calculer son filtre
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$H(\mathbf{j}\omega)$ : Gain <br>
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$\delta_{1}$ : L'erreur de gain tolérable <br>
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$\delta_{2}$ : L'erreur d'atténuation tolérable <br>
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$\omega_{p}$ : Fréquence de pulsation de passage <br>
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$\omega_{a}$ : Fréquence de pulsation d'arrêt <br>
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1. En fonction du gabarit, déterminer le type de filtre désiré (`passe-bas`, `passe-haut`, `passe-bande`, `coupe-bande`)
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2. Calculer $f_{c} \to \omega_{c}$ et $\Delta f \to \Delta\omega$
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$$\begin{gather}
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f_{c} = \frac{f_{a}+f_{p}}{2} \qquad & \qquad \Delta f = f_{a}-f_{p} \\
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\omega_{c} = 2\pi \frac{f_{c}}{f_{s}} \quad & \quad \Delta\omega = \frac{2\pi\Delta f}{f_{s}}
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\end{gather}$$
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3. Choisir une fenêtre $W[n]$ satisfaisante du point de vue de l’atténuation et du reste du cahier des charges.(voir tableau ci-dessus)
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4. Calculer l'ordre du filtre à l'aide de la bande de transition du filtre choisit. Arrondir au nombre pair supérieur.
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5. Calculer la réponse impulsionnelle $h_{f}[n]$ du filtre avec $h[n]*W[n]$
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6. Calculer les coefficients $b_{n}$ du filtre avec $h_{f}[n]$
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<details>
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<summary>Démarche selon les slides</summary>
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1. Connaissant le gabarit du filtre désiré, choisir le filtre idéal correspondant
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2. Calculer les pulsations caractéristiques se situant au centre des bandes de transitions
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3. Rechercher la réponse impulsionnelle $h[n]$ du filtre idéal, si celle-ci n’est pas connue, on peut la calculer par transformation de Fourier inverse
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4. Choisir une fenêtre $W[n]$ satisfaisante du point de vue de l’atténuation
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5. Connaissant la largeur de la bande de transition , calculer l’ordre du filtre $N$
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6. Calculer les coefficients du filtre en multipliant la réponse impulsionnelle par la fenêtre choisie
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</details>
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Reference in new issue
Block a user