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# Formules
## Pour les expressions de la forme $ay'+by=0$
$ay'=-by$
$\frac{y'}{y}=\frac{-b}{a}$
$ln|y|=\int\frac{-b}{a}+c$
$e^{ln|y|}=e^{\int\frac{-b}{a}+c}$
$|y|=e^{\int\frac{-b}{a}+c}$
$|y|=\pm e^c+e^{\int\frac{-b}{a}}$
$|y|=Ke^{\int\frac{-b}{a}}$
## Pour les expressions de la forme $ay'+by=c$
$y=y_p+y_h$ avec $y_h$ solution de $ay'+by=0$
Méthode générale MVC: Méthode de variation de la constante
$y_h=Ke^{-\int\frac{b}{a}}$
$y_p=K(x)e^{-\int\frac{b}{a}}$
$ay'_p+by_p=a(K'e^{-\int\frac{b}{a}}+K(\frac{-b}{a})e^{-\int\frac{b}{a}})+bKe^{-\int\frac{b}{a}}$
$K=\int\frac{c}{a}e^{\int\frac{b}{a}}$
## Pour les expressions d'ordre 2, de la forme $ay''+by'+cy=d$
$a,b,c\in\mathbb{R}$; $d$ est une fonction
$y_h$: Solution de $ay''+by'+cy=0$
Équation caractéristique: $ax^2+bx+c=0$
Si $\Delta>0$: 2 racines réelles $r_1,r_2$. -> $y_h=Ae^{r_1x}+Be^{r_2x}$
Si $\Delta = 0$: 1 racine réelle $r$. -> $y_h=(Ax+B)e^{rx}$
Si $\Delta < 0$: 2 racines complexes conjuguées $a\pm ib$ -> $y_h=e^{ax}(A\cos{(bx)}+B\sin{(bx)})$
$y_p$: On cherche $y_p$ sous une forme similaire à celle de $d$.
Recette n°1: Si $d$ est un polynome, $y_p$ est un polynome de même degré
Exemple: $y''+y=x^2+3$
$y_p=ax^2+bx+c$
$y_p''+y_p=2a+ax^2+bx+c=x^2+3+0x$
$\begin{cases}a=1\\b=0\\2a+c=3\implies c=1\end{cases}$
Recette n°2: $d=a\cos{(\alpha x)}+b\sin{(\alpha x)} \implies y_p=A\cos{(\alpha x)}+B\sin{(\alpha x)}$
Exemple: $y''+y+y=\sin{(2x)}$
$y_p=A\cos{(2x)}+B\sin{(2x)}$
$y_p''+y_p'+y_p=-4y_p+y_p'+y_p$
$=-3y_p+y_p'=3A\cos{(2x)}-3B\sin{(2x)}-2A\sin{(2x)}+2B\cos{(2x)}$
$=(-3A+2B)\cos{(2x)}+(-3B+2A)\sin{(2x)}=\sin{(2x)}$
$\cases{-3A+2B=0\\-3B+2A=1}$
Recette n°3: $d(x)=P(x)e^{\alpha x}\implies y_p=Q(x)e^{\alpha x}$
Avec $d°Q=d°P$ si $\alpha$ n'est pas solution de l'équation caractéristique.
Sinon, $d°Q=d°P+1$ si $\alpha$ est solution.
Sinon, $d°Q=d°P+2$ si $\alpha$ est solution double ($\Delta=0$).
Exemple: $y''-y'-2y=te^t$
$\alpha=1$; Éq caractéristique: $x^2-x-2=0$.
$\alpha$ n'est pas solution de l'éq caractéristique.
$y_p=(at+b)e^t$
$y_p'=ae^{t}+(at+b)e^t=(at+b+a)e^t$
$y_p''=(at+b+2a)e^t$
$y_p''-y_p'-2y_p=((at+b+2a)e^t)-(ae^{t}+(at+b)e^t=(at+b+a)e^t)-2((at+b)e^t)$
$y_p''-y_p'-2y_p=(-2at-2b+a)e^t=te^t$
$-2at-2b+a=t$
$\cases{-2a=1\implies a=-\frac{1}{2} \\ a-2b=0\implies 2b=-\frac{1}{2}\implies b=\frac{1}{4}}$
$y_p=(-\frac{1}{2}t-\frac{1}{4})e^t$
$y_h$: $x^2-x-2=0$
$\Delta=1+8=9$
$x=\frac{1\pm3}{2}=\cases{2\\-1}$
$y_h=Ae^{r_1t}+Be^{r_2t}=Ae^{2t}+Be^{-t}$
$y=y_h+y_p=Ae^{2t}+Be^{-t}+(-\frac{1}{2}t-\frac{1}{4})e^t$
$\cases{A+B=2\\2A-B=1}$
$3A=3\implies A=1 \implies B=1$
# Exercices
## Exercice 1:
$(1+t^2)y'(t)+4ty(t)=0$
$a=1+t^2$
$b=4t$
$y=Ke^{-\int\frac{b}{a}}=Ke^{-\int\frac{4t}{1+t^2}}=Ke^{-2\int\frac{2t}{1+t^2}}$
$y=Ke^{-2ln(1+t^2)}=\boxed{\frac{K}{(1+t^2)^2}}$
## Exercice 2:
$t^2y'(t)+y(t)=0$
$a=t^2$
$b=1$
$y=Ke^{-\int\frac{b}{a}}=Ke^{-\int\frac{1}{t^2}}=\boxed{Ke^{\frac{1}{t}}}$
## Exercice 3:
$y'(t)+2ty(t)=e^{t-t^2}$
$y_h=Ke^{-\int\frac{b}{a}}=Ke^{-\int2t}=\boxed{Ke^{-t^2}}$
$y_p=K(t)e^{-t^2}$
$K=\int\frac{c}{a}e^{\int\frac{b}{a}}=\int\frac{c}{a}e^{t2}=\int\frac{e^{t-t^2}}{1}e^{t2}=\int e^t=\boxed{e^t}$
$y_p=e^te^{-t^2}=\boxed{e^{t-t^2}}$
$y=y_p+y_h=\boxed{e^{t-t^2}+Ke^{-t^2}}$
## Exercice 4:
$\begin{cases}y'(t)-2y(t)=te^t \\ y(0)=4\end{cases}$
$y_h=Ke^{-\int\frac{b}{a}}=Ke^{-\int\frac{2}{1}}=\boxed{Ke^{2t}}$
$y_p=K(t)e^{2t}$
$K(t)=\int\frac{c}{a}e^{\int\frac{b}{a}}=\int\frac{te^t}{1}e^{-2t}=\int te^{-t}$
$=[-e^{-t}t]-\int-e^{-t}1$
$=[-te^{-t}-e^{-t}]$
$K(t)=(-1-t)e^{-t}$
$y_p=(-1-t)e^{-t}e^{2t}$
$y_p=\boxed{(-1-t)e^t}$
$y=y_p+y_h=(-1-t)e^t+Ke^{2t}$
Avec $y(0)=4=-1+K$, donc $K=5$
$\boxed{y=(-1-t)e^t+5e^{2t}}$
## Exercice 5:
Voir exemple de recette 3 dans ## Pour les expressions d'ordre 2, de la forme $ay''+by'+cy=d$.
## Exercice 6:
$\cases{y''(t)+y(t)=0 \\ y(0)=1 \\ y'(0)=0}$
$y=y_h$
Équation caractéristique: $x^2+1=0$.
$\Delta=b^2-4ac=0-4<0$
$x=\frac{-b\pm\sqrt{-\Delta}}{2a}=\frac{0\pm i\sqrt{4}}{2}=\pm i$
$y_h=e^{ax}(A\cos{(bx)}+B\sin{(bx)})=e^0(A\cos{x}+B\sin{x})$
$y=A\cos{x}+B\sin{x}$
$y(0)=A+0=1\implies A=1$
$y'=-A\sin{x}+B\cos{x}$
$y'(0)=B=0\implies B=0$
$\boxed{y=\cos{x}}$
## Exercice 8:
$\cases{x'(t)=y(t)+2\\y'(t)=x(t)+z(t)+e^t\\z'(t)=x(t)+y(t)+z(t)}$
$$
\begin{pmatrix}
x \\ y \\ z
\end{pmatrix}'
= \begin{pmatrix}
0 & 1 & 0 \\
1 & 0 & 1 \\
1 & 1 & 1
\end{pmatrix}
\begin{pmatrix}x \\ y \\ z\end{pmatrix}
+ \begin{pmatrix}2 \\ e^t \\ 0\end{pmatrix}
$$
$$
X_A(x)=\begin{vmatrix}
-X & 1 & 0 \\
1 & -X & 1 \\
1 & 1 & 1-X
\end{vmatrix}
$$
<u>Déterminant d'une matrice carrée:</u>
$\begin{vmatrix}a&b\\c&c\end{vmatrix}=ab-cd$
$\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}\begin{matrix}a&b\\d&e\\g&h\end{matrix}=aei+bfg+cdh-(gec+hfa+idb)$
$\begin{vmatrix}2&2&3\\0&1&0\\7&7&0\end{vmatrix}=\begin{vmatrix}0&2&3\\-1&1&0\\0&7&0\end{vmatrix}=0+0-21-0=-21$
Développement / à C3:
$det\,A=(-1)^{1+3}\times3\times\begin{vsmallmatrix}0&1\\7&7\end{vsmallmatrix}+(-1)^{2+3}\times0\times\begin{vsmallmatrix}2&2\\7&7\end{vsmallmatrix}+(-1)^{3+5}\times0\times\begin{vsmallmatrix}2&2\\0&1\end{vsmallmatrix}$
$C_i\leftarrow C_i+\sum{\alpha_jC_j}$
$X'=AX+B$
$X=\begin{pmatrix}x\\y\\z\end{pmatrix}$
$$
\begin{pmatrix}
x \\ y \\ z
\end{pmatrix}'
= \begin{pmatrix}
0 & 1 & 0 \\
1 & 0 & 1 \\
1 & 1 & 1
\end{pmatrix}
\begin{pmatrix}x \\ y \\ z\end{pmatrix}
+ \begin{pmatrix}2 \\ e^t \\ 0\end{pmatrix}
$$
$A=PDP^{-1} \rightarrow$ Diagonale
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# Intégration par parties (IPP)
Soit $( f, g \in \mathcal{C}^n([a,b]) )$.
---
## (IP1)
$$
(\mathcal{L}(f))^{(n)}(t) = (-t)^n \, \mathcal{L}(x^n f(x))(t)
$$
---
## (IP2)
Si \( f \) est de classe \( \mathcal{C}^n \) avec \( f^{(n)} \in \mathcal{L}^1(\mathbb{R}^+) \), alors :
$$
\mathcal{L}(f^{(n)})(p)
= p^n \mathcal{L}(f)(p)
- p^{n-1} f(0)
- p^{n-2} f'(0)
- \dots
- f^{(n-1)}(0)
$$
---
## (IP3)
Si \( f \) est la primitive de \( g \) qui s'annule en \( 0 \), alors :
$$
\mathcal{L}(f)(p) = \frac{\mathcal{L}(g)(p)}{p}
$$

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Netflix
Why
Démocratiser la culture cinématographique en permettant à quiconque d'avoir accès aux films et séries produits dans le monde entier
How
En développant une plateforme de streaming facilement accessible sur tout appareil et navigateurs connus.
What
Un abonnement donnant accès au catalogue des milliers de films et de séries.
---
Télépartout ("**Où vous voulez, Quand vous voulez**")
Why
Permettre l'accès au voyage rapide, efficace et durable au plus grand nombre
How
En proposant un service de transport instantané d'un point A à un point B, accessible à tous, sur et efficace.
What
Un service de transport instantané.
---
Fermez les yeux deux secondes. Imaginez lendroit idéal pour avoir une idée brillante… du calme, de lespace, du silence. Maintenant, rouvrez-les : vous êtes en open space, avec 12 conversations autour de vous. Les uns sont en réunions commerciales, les autres sont au téléphone avec le support technique. Les potentielles idées sont noyées dans le brouhaha avant de voir le jour. Toujours aussi inspiré ?

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```
%%

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@ -1,3 +1,9 @@
Refs: Refs:
- Ken Robinson's Ted Talk. - Ken Robinson's Ted Talk.
Torseur cinématique
$\begin{vmatrix}\omega_x & V_x \\ \omega_y & V_y \\ \omega_z & V_z\end{vmatrix}_M$
Torseur statique
$\begin{vmatrix}X_{2\rightarrow1} & L_{2\rightarrow1} \\ Y_{2\rightarrow1} & M_{2\rightarrow1} \\ Z_{2\rightarrow1} & N_{2\rightarrow1}\end{vmatrix}_M$

Binary file not shown.

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@ -0,0 +1,48 @@
> [!NOTE] À retenir
> Le `<système>` doit permettre à un `<acteur>` de `<réaliser une activité>` dans `<un contexte précis>`.
Exemple de reformulation du besoin client:
Un pote: "Il me faut un logiciel qui puisse me permettre de savoir quelles cartes Pokémon je possède"
$$\downarrow$$
Le logiciel `PokéCard` doit permettre à `mon pote` de `gérer sa collection de cartes Pokémon` dans `le but de faciliter ses échanges avec ses amis`.
# Exercice 1: Sujet
- Nous désirons réaliser le logiciel embarqué dun Guichet Automatique Bancaire, plus communément appelé GAB. Compte-tenu du temps imparti, on étudiera plus particulièrement la fonction essentielle du GAB, à savoir le retrait dargent.
- Le client :
- Peut consulter létat de son compte, ou bien retirer de largent.
- Pour y parvenir, il doit procéder classiquement en insérant sa carte, saisissant son code, …
Le `Firmware du GAB` doit permettre à `un client de la banque` de `Consulter ses comptes et retirer de l'argent` dans `le but de gérer ses comptes à tout moment`.
# Exercice 2 : Cycle de vie et parties prenantes
• Lister les différentes phases de vie, puis les parties prenantes du GAB :
Le GAB doit permettre à un client de pouvoir consulter létat de son compte et retirer de largent via lutilisation de sa carte bancaire 24h/24 et 7j/7.
Phases de vie:
```mermaid
flowchart TD
Conception --> Développement --> Déploiement --> Utilisation --> Maintenance --> f["Fin de vie"] --> Conception
```
Analyse: Nous
Conception: Nous
Développement: Nous
Déploiement: Technicien
Utilisation: Le client de la Banque ou l'employé de la Banque
Maintenance: Développeur, technicien ou employé de la Banque
Administration: Le Banquier
Fin de vie: Technicien
![[Pasted image 20251016101653.png]]
![[Pasted image 20251016102557.png]]
Story board du GAB
![[Drawing 2025-10-22 16.25.26.excalidraw|1000]]
![[Drawing 2025-10-24 16.11.20.excalidraw]]
![[Pasted image 20251024162613.png]]

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@ -0,0 +1,39 @@
```mermaid
flowchart TD
%% Tâche racine
A[Commander un média]
%% Séquence principale (>>)
A --> B[Sélectionner média]
B --> C[Gérer panier]
C --> D[Saisir infos de livraison]
D --> E[Payer la commande]
E --> F[Consulter confirmation]
%% Détail : Sélectionner média
B --> B1[Consulter catalogue livrable]
B1 --> B2[Rechercher / filtrer médias]
B2 --> B3[Consulter fiche média]
B3 --> B4[Ajouter au panier]
%% Détail : Gérer panier
C --> C1[Afficher panier]
C1 --> C2{Modifier quantités ?}
C2 -->|Oui| C3[Modifier quantités]
C3 --> C1
C2 -->|Non| C4[Valider panier]
%% Détail : Saisir informations de livraison
D --> D1[Choisir / saisir adresse]
D1 --> D2[Choisir mode de livraison]
%% Détail : Payer la commande
E --> E1[Choisir moyen de paiement]
E1 --> E2[Saisir infos de paiement]
E2 --> E3[Confirmer paiement]
%% Détail : Consulter confirmation
F --> F1[Afficher récapitulatif commande]
F --> F2[Notifier le client]
```

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@ -0,0 +1,250 @@
---
excalidraw-plugin: parsed
tags: [excalidraw]
---
==⚠ Switch to EXCALIDRAW VIEW in the MORE OPTIONS menu of this document. ⚠== You can decompress Drawing data with the command palette: 'Decompress current Excalidraw file'. For more info check in plugin settings under 'Saving'
# Excalidraw Data
## Text Elements
Consulter mon compte ^dq2fLe8a
M'authentifier ^w6hHgMiJ
Retirer de l'argent ^J7bLKppY
Consulter mon compte ^7maHwnO3
Retirer de l'argent ^IkhcTle2
Finaliser le retrait ^L8Y8fYKg
Insérer ma carte ^LWwBxuxb
Saisir mon code secret ^PUg0UWko
Demander débit/crédit ^hDo3IXK4
Afficher le solde ^eYURCrFL
Saisir le montant désiré ^uPdbaLrF
Valider l'opération ^SJwxfB5j
Choisir reçu ou pas ^OJM3hO87
Récupérer la carte ^xraUtgvx
M'authentifier ^JcvSxwa7
Modifier mon code en cas d'erreur ^XrEMnHpp
Lire le message d'erreur ^TGuB8GvS
Lire nombre de saisies ^z8SW6Mfm
Saisir mon code secret ^S0p2f3CS
Annuler pour récupérer ma carte ^T5Qhd3Ey
Choisir une option ^xNFetUou
Choisir l'opération ^5YgD1319
Demander consultation ^2qMwKXRl
Demander retrait ^DjxUY2Va
Demander annulation ^ezQJIbal
Choisir la composition de billets ^xFmjq6Gu
Récupérer les billets ^wdRDVzus
Récupérer le reçu ^5IjQ1jaL
%%
## Drawing
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```
%%

View file

@ -0,0 +1,205 @@
---
excalidraw-plugin: parsed
tags: [excalidraw]
---
==⚠ Switch to EXCALIDRAW VIEW in the MORE OPTIONS menu of this document. ⚠== You can decompress Drawing data with the command palette: 'Decompress current Excalidraw file'. For more info check in plugin settings under 'Saving'
# Excalidraw Data
## Text Elements
Intéraction IHM ^JL3lZ3Ac
Action Abstraite ^DokVQs5u
Application ^OAtOQ7US
Retirer de l'argent ^zNQsAatp
Choisir le montant ^lkMNBwPK
Finaliser le retrait ^C8w1O6qU
Saisir le montant ^EUrWYyVH
Choisir les types
de billets ^pJJRgvQ0
Lire le message
d'erreur ^bMznOVkr
[ ] ^2E35mM4V
Reçu ^7ZbTnnxd
Choisir si on
veut un reçu ^0Mu05Pst
Récupérer le
reçu ^JjhTY8Eg
Récupérer la
carte ^9ua8upCL
Récupérer les billets ^P1Fl41vY
%%
## Drawing
```compressed-json
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@ -28,125 +28,337 @@ Débit de sortie ^dib1MOgr
Eau rejetée ^8x4VsbRU Eau rejetée ^8x4VsbRU
A0 ^FsZLri6B
Niveau de trop plein ^C5fiaciL
Débit d'entrée ^uuEp5upz
Débit de sortie ^1eN59Qg0
A1 ^Gyyl46vZ
A2 ^F5xI5pJI
A3 ^lfmgjN2U
A4 ^vwqU500X
A5 ^VYYkkAl1
Récupérer ^Xbej2nLA
Eau de pluie dans une goutière ^Jt547y0Z
Mesurer le niveau ^EjdX1W4k
Refouler ^DzErYKAM
Eau rejetée ^7IE4kxSj
Filtrer ^JGflqsCz
Stocker ^xuiOu561
Redistribuer ^8q4dCxlg
Eau non potable pour le réseau domestique ^zBFBwZR9
A6 ^aYnOWuqf
Aquaboost ^LT8RAa4s
Flotteur ^beEKGCel
Tube flexible ^KYC4hOG9
Filtre mécanique
ou biologique ^YM75itYY
Cuve
bac fermé ^IWpnqjgO
Répartiteur + pompe ^ptVLECb9
Clapet ^jheQh6eO
%% %%
## Drawing ## Drawing
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``` ```
%% %%

View file

@ -0,0 +1,88 @@
---
excalidraw-plugin: parsed
tags: [excalidraw]
---
==⚠ Switch to EXCALIDRAW VIEW in the MORE OPTIONS menu of this document. ⚠== You can decompress Drawing data with the command palette: 'Decompress current Excalidraw file'. For more info check in plugin settings under 'Saving'
# Excalidraw Data
## Text Elements
ESTIA ^IRZgx43A
Consommation d'énergie ^mCSo72dN
Produit ^2OpPqrEr
Permettre aux usagers de l'ESTIA d'ajuster leur consommation d'électricité en fonction de la production d'énergie verte dans le bâtiment ^M721uhmn
%%
## Drawing
```compressed-json
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```
%%

View file

@ -0,0 +1,130 @@
---
excalidraw-plugin: parsed
tags: [excalidraw]
---
==⚠ Switch to EXCALIDRAW VIEW in the MORE OPTIONS menu of this document. ⚠== You can decompress Drawing data with the command palette: 'Decompress current Excalidraw file'. For more info check in plugin settings under 'Saving'
# Excalidraw Data
## Text Elements
Produit ^cbrAFxAW
Capteur conso ^Uo8fK0CF
Usagers ^vqU1BC7w
Centrale de production d'énergie verte ^jYmefdmW
PC portable ^lFE1lvrH
Réseau informatique interne ^xOotpSeN
FC1 ^mrhfmwHz
FC2 ^APOTOavi
FC3 ^a0RrfkVU
FC4 ^6JgXWFFZ
FC5 ^efKSMbMW
FP1 ^EKD15DAV
FC6 ^I6GIVEOs
%%
## Drawing
```compressed-json
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```
%%

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@ -0,0 +1,18 @@
# Bête à corne
![[BeteACorne]]
# Diagramme Pieuvre
![[DiagrammePieuvre|1000]]
| Contrainte | Description |
| ---------- | ----------------------------------------------------------------- |
| FC1 | Récupérer les données en temps réel de production d'énergie verte |
| FC2 | Récupérer les données en temps réel de consommation |
| FC3 | Déterminer la demande (nombre de pc branchés) |
| FC4 | Envoyer les infos de dispo |
| FC5 | Respecter la vie privée |
| FC6 | Savoir si le PC est branché |
| FP1 | Permettre à l'usager de connaitre la dispo d'énergie verte |
# SADT
![[SADT|1000]]

View file

@ -0,0 +1,164 @@
---
excalidraw-plugin: parsed
tags: [excalidraw]
---
==⚠ Switch to EXCALIDRAW VIEW in the MORE OPTIONS menu of this document. ⚠== You can decompress Drawing data with the command palette: 'Decompress current Excalidraw file'. For more info check in plugin settings under 'Saving'
# Excalidraw Data
## Text Elements
Dispositif Alerte Production ^44maUcuh
"Branchez!" ^KqZPJqyA
"Débranchez!" ^wUma6ya6
Récupérer les données ^0tUbNWCL
Données de conso ^KpJA04DZ
Données de prod ^HYLi1eir
Électricité? ^IEtzJVnb
Temps réel ^E8J2R1C0
Protocol Réseau ^OK7R80cq
Nombre de pc à l"ESTIA ^sRIiotoT
Identifier le nombre branchés ^nNyZdIz1
Évaluer ^JHbwPkNC
Disponibilités ^fg2aiW0M
Besoins ^93veLaFG
Alerter ^aYP8rLeN
%%
## Drawing
```compressed-json
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```
%%

View file

@ -0,0 +1,118 @@
---
excalidraw-plugin: parsed
tags: [excalidraw]
---
==⚠ Switch to EXCALIDRAW VIEW in the MORE OPTIONS menu of this document. ⚠== You can decompress Drawing data with the command palette: 'Decompress current Excalidraw file'. For more info check in plugin settings under 'Saving'
# Excalidraw Data
## Text Elements
A0 ^CEnNGifV
Connaitre les disponibilités en énergie verte ^p043ZfmS
Données de conso ^qhR9wwJS
Données de prod ^bz9k3Qw1
Nombre de pc branchés ^qBVSE3Su
Alerte ^GXLgWf3Y
Dispositif Alerte Production ^Z6I7UPFM
Électricité? ^AYxZqMFg
Temps réel ^JB1aFyy5
Protocol Réseau ^2u3G4Av5
%%
## Drawing
```compressed-json
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L1QqSOOzAGBDQ6QW9j7CKnoEyS5EPtjsr2VAp65Jc3UPqHCCcl2obYIAA===
```
%%

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<!DOCTYPE html>
<html lang="fr">
<head>
<!-- Métadonnées de la page -->
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<meta name="description" content="Page web simple en HTML5 respectant les standards du W3C.">
<title>Page HTML5 — Démonstration</title>
</head>
<body>
<!-- En-tête sémantique de la page -->
<header>
<h1>Exemple de page HTML5 conforme au W3C</h1>
<p>Réalisation dans le cadre du module <strong>Développement Web Statique</strong></p>
</header>
<!-- Contenu principal -->
<main>
<!-- Section dintroduction -->
<section>
<h2>Présentation</h2>
<p>
Cette page a été réalisée pour illustrer lutilisation des éléments
<strong>sémantiques</strong> en HTML5 tout en respectant les standards du <abbr
title="World Wide Web Consortium">W3C</abbr>.
Elle contient une image, une liste, un tableau, un lien et du texte, comme demandé dans lexercice.
</p>
<!-- Insertion dune image locale ou distante -->
<figure>
<img src="https://upload.wikimedia.org/wikipedia/commons/6/61/HTML5_logo_and_wordmark.svg"
alt="Logo HTML5">
<figcaption>Logo officiel de HTML5.</figcaption>
</figure>
</section>
<!-- Section sur les éléments sémantiques -->
<section>
<h2>Les éléments sémantiques HTML5</h2>
<p>
HTML5 introduit plusieurs balises sémantiques telles que :
</p>
<!-- Liste non ordonnée -->
<ul>
<li><code>&lt;header&gt;</code> en-tête du document</li>
<li><code>&lt;nav&gt;</code> zone de navigation</li>
<li><code>&lt;article&gt;</code> contenu autonome</li>
<li><code>&lt;section&gt;</code> division thématique</li>
<li><code>&lt;footer&gt;</code> bas de page</li>
</ul>
</section>
<!-- Section sur les standards W3C -->
<section>
<h2>Les standards du W3C</h2>
<p>
Le <strong>W3C</strong> (World Wide Web Consortium) est une organisation qui définit
les <em>normes du web</em>. Ces règles assurent que les sites soient accessibles, compatibles et
bien structurés.
</p>
<p>
Pour plus dinformations, consultez le site officiel :
<!-- Lien externe -->
<a href="https://www.w3.org/" target="_blank" rel="noopener noreferrer">W3C.org</a>.
</p>
</section>
<!-- Section avec tableau -->
<section>
<h2>Comparaison entre HTML et HTML5</h2>
<table>
<thead>
<tr>
<th>Critère</th>
<th>HTML</th>
<th>HTML5</th>
</tr>
</thead>
<tbody>
<tr>
<td>Sémantique</td>
<td>Peu développée</td>
<td>Balises explicites (header, footer, etc.)</td>
</tr>
<tr>
<td>Support multimédia</td>
<td>Utilise des plugins (Flash, etc.)</td>
<td>Intégré (audio, vidéo)</td>
</tr>
<tr>
<td>Compatibilité</td>
<td>Ancienne norme</td>
<td>Standard W3C actuel</td>
</tr>
</tbody>
</table>
</section>
</main>
<!-- Pied de page -->
<footer>
<p>&copy; 2025 - Équipe Persistent Lagoons</p>
<p>Page validée selon les standards du <a href="https://validator.w3.org/" target="_blank"
rel="noopener noreferrer">W3C Validator</a>.</p>
</footer>
</body>
</html>

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https://www.cnbc.com/2025/11/18/cloudflare-down-outage-traffic-spike-x-chatgpt.html
Intro: The article
Context: What is CloudFlare?
What happened and what was affected
Market value drop 5%
Service provider outages: AWS, Azure, and The CrowdStrike Incident (July 2024)
Debate question: Should critical infrastructures be completely independent from service providers to operate ?

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@ -91,3 +91,42 @@ Do you think music is important?
Notes Notes
![[Pasted image 20251007151529.png]] ![[Pasted image 20251007151529.png]]
![[Pasted image 20251014140304.png]]
A few years ago, I bought and read "The Art of War" by Sun Tzu. It is a very old book written by a Chinese general, and it contains a couple hundreds of battle tips. It is organized in a few chapters, 9 if I recall correctly, and it is written in the form of a list of quotes such as "If you know your enemy, and you know yourself, then you shall not fear the outcome of a hundred battles". I bought this book because I had heard about it from a YouTuber, who praised it as one of the best books he had ever read. Even though the contents of this book should be completely obsolete in the era of autonomous drones and nuclear warheads, it is still very adequate in a variety of contexts. I would say this book has radically changed my way of approaching most issues.
Notes from teacher:
- "Relevant" would be better suited than "adequate" in this context.
- Don't hesitate to go a little bit off topic.
---
Tue. 21 / 10 / 2025
Section 1:
Questions 1-5
1. 9h30
2. Helendale
3. Central Street
4. 792
5. 8h55
Questions 6-10
6. $1.80
7. 7h30
8. $7.15
9. Commuter
10. Afternoon
Section 2:
Café
Book reservation
Location of rooms
1. Quilt shop:
2. Handicrafts Museum:
3. School House:
Section 3:
1. 1
2. 3
3. 2

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@ -216,3 +216,130 @@ Escucha a dos personas y marca la información que se refiere a Marta, a Antonio
| 7. Quiere conocer una cultura diferente.<br> | x | x | | 7. Quiere conocer una cultura diferente.<br> | x | x |
| 8. Busca naturaleza.<br> | x | x | | 8. Busca naturaleza.<br> | x | x |
| 9. Tiene pocos días de vacaciones. | x | | | 9. Tiene pocos días de vacaciones. | x | |
---
# Martes, 14 de octubre 2025
![[Mi casa.pdf]]
![[Pasted image 20251014162711.png]]
¿Donde esta el espejo? El espejo esta detras de la persona con la camiseta amarillo y es azul y gris. Espejo ->$11$
Armario -> 8
El sofa -> 1 El sofa esta en la isquina izquierda y es de color azul.
La alfombra -> 9
el ordenador -> 14 El ordenador esta sobre una mesa y es amarillo.
la cama -> 12
la librería -> 7 La libreria esta al lado de la alfombra.
el sillón -> 15
la nevera -> 3
la silla -> 2 La silla esta a la izquierda de la nevera y es verde.
el cuadro -> 4 El cuadro esta a la derecha de la nevera, es un cuadro de dos montañas.
la lavadora -> 10 La lavadora esta a la izquierda de la cama y es gris y roja.
el televisor -> 5 La tele esta en la cima de la maleta violeta.
la mesa -> 16
la estantería -> 6
la lámpara -> 13
1. Las tiendas estan abiertas todavia.
2. La mujer que esta al lado de Carlos, es la directora.
3. Madrid es la capital de España y esta en el centro.
4. Ellos estan en la discoteca, pues hoy es fiesta.
5. La silla que esta a la izquierda de la puerta es de madera.
6. Este libro esta muy bien, es muy interesante.
7. Hace mucho tiempo que tengo mi coche, es muy viejo, pero esta muy nuevo.
8. El mercado no esta lejos de aqui.
9. El hombre que es alto y moreno es mi hermano.
10. Los servicios estan a la derecha.
11. Este ejercicio es muy dificil.
12. La boda de mi hermano es en el restaurante Miramar.
13. No hay prisa, es ponto todavia?
14. El mar esta a pocos kilometros de aqui.
15. La lentejas estan muy sosas, ¿no les has echando sal?
16. ¿Ya estas listo? Pues vamonos, es tarde.
17. Se caso con una viuda que es muy rica.
18. Mi habitacion es grane y siempre esta desrdenada.
19. ---
20. Madrid esta bien para visitar museos.
21. El examen esta en aula 223.
22. ---
23. ¿Estais contentos con el hotel?
Si, es un hotel limpio y barato.
---
Martes, 21 de Octubre 2025
El piso de Julian es un atico de 55$m^2$, esta en el centro, tiene mucha luz, un habitacion, una cocina americana, un acensor. Da a una calle y un mercado.
Tus hermanas estan cansadas, ¿no?
**mi** hermana Eva no esta cansada, pero Luisa si. **su** marido es de Canada y tienen dos hijos de 4 y 5 años. Son muy guapos.
¿Y en que lengua hablan con **sus** hijos?
En ingles y en español.
¡Que suerte tienen **tus** sobrinos! Yo tengo 35 años y hablo muy mal ingles...
![[Pasted image 20251021164403.png]]
![[Pasted image 20251021164411.png]]
8. a
9. b
10. a
11. b
---
# Martes, 18 de noviembre 2025
![[Pasted image 20251118162202.png]]
Escribe los verbos anteriores en el lugar correspondiente y completa la
conjugación.
| Hacer | Poner | Salir | Traer | Caer |
| ------- | ------- | ----- | ------ | ----- |
| Hago | Pongo | Salgo | Traigo | Caigo |
| haces | pones | | | |
| hace | pone | | | |
| hacemos | ponemos | | | |
| hacéis | ponéis | | | |
| hacen | ponen | | | |
| Ver | Saber | Dar | Conocer | Valer |
| --- | ----- | --- | ------- | ----- |
| Veo | Se | Doy | Conozco | Valgo |
Ver la tele: algunas veces por la noche
Salir con los amigos: 2 o 3 vezes por semana
Ir la cine: Una vez por semana
discoteca: Nunca, no le gustan
leer un libro: Todo los dias
deporte: todos los fines de semana
conciertos: casi nunca, son muy caros
viajar: Una vez al año
jugar videojuegos: Muchas veces
---
# Martes, 25 de noviembre 2025
Completa con gustar o encantar (cuando está indicado)
1. A Juan le gusta jugar a la ajedrez.
2. A mis hijos les gusta mucho el chocolate.
3. A los niños les env (encantar) jugar en el parque.
4. A nosotros nos gusta la tortilla de patata.
5. ¿A ti te gustan las películas de Javier Bardem?
6. A vosotros no os gusta jugar a tenis.
7. A ellos les gusta la música clásica.
8. A ti te encanta (encantar) las películas de terror.
9. A nosotros no nos gusta el arte moderno.
10. ¿A ti te gustan los animales?
11. A nosotros nos encanta (encantar) viajar por el mundo.
12. A ellos no les gustan nada los libros.
13. A María les gustan los deportes acuáticos.
14. A mi hermano le gusta mucho ver partidos de fútbol en la tele.
15. ¿A vosotros os gusta cocinar?
16. A ti te encantan (encantar) las flores, ¿verdad?
17. A ellos les encanta (encantar) escuchar música clásica por las mañanas.

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_(Cinématique → Statique → Cinétique → Dynamique → Équilibre Dynamique)_
## I. CINÉMATIQUE
### Repères et variables
Deux repères :
- $R_0(O,\vec{x_0},\vec{y_0},\vec{z_0})$ fixe et galiléen
- $R_1(O,\vec{x_1},\vec{y_1},\vec{z_1})$ lié au solide
Angle $\theta$ en radians. Vitesse angulaire $\dot{\theta} = \dfrac{d\theta}{dt}$ en rad s$^{-1}$. Accélération angulaire $\ddot{\theta} = \dfrac{d^2\theta}{dt^2}$ en rad s$^{-2}$.
### Changement de base dans le plan $x_0y_0$
$$
\begin{cases}
\vec{x_1} = \cos(\theta)\,\vec{x_0} + \sin(\theta)\,\vec{y_0} \\
\vec{y_1} = -\sin(\theta)\,\vec{x_0} + \cos(\theta)\,\vec{y_0}
\end{cases}
$$
### Dérivation dans une base mobile (formule de Poisson)
Vecteur rotation
$$
\vec{\Omega}_{1/0} = \dot{\theta}\,\vec{z}
$$
Formule
$$
\left(\dfrac{d\vec{u}}{dt}\right)_{R_0} = \left(\dfrac{d\vec{u}}{dt}\right)_{R_1} + \vec{\Omega}_{1/0} \wedge \vec{u}
$$
Applications aux axes
$$
\begin{cases}
\dfrac{d\vec{x_1}}{dt}\Big|_0 = \dot{\theta}\,\vec{y_1} \\
\dfrac{d\vec{y_1}}{dt}\Big|_0 = -\dot{\theta}\,\vec{x_1}
\end{cases}
$$
### Vitesse dun point
$$
\vec{V_B} = \vec{V_A} + \vec{AB} \wedge \vec{\Omega}_{1/0}
$$
### Accélération dun point
$$
\vec{\Gamma_B} = \vec{\Gamma_A} + \vec{AB} \wedge \dot{\vec{\Omega}}_{1/0} + \vec{\Omega}_{1/0} \wedge \big(\vec{\Omega}_{1/0} \wedge \vec{AB}\big)
$$
Cas rotation autour dun axe à distance $R$
$$
\vec{\Gamma_B} = R\big(\ddot{\theta}\,\vec{y_1} - \dot{\theta}^{2}\,\vec{x_1}\big)
$$
---
## II. STATIQUE
### Torseur daction mécanique au point $A$
$$
\mathcal{T} =
\begin{Bmatrix}
\vec{F} \\
\vec{M_A}
\end{Bmatrix}_A
$$
### Changement de point
$$
\vec{M_B} = \vec{M_A} + \vec{BA} \wedge \vec{F}
$$
### Torseurs usuels
Poids appliqué en $G$
$$
\mathcal{T}_P =
\begin{Bmatrix}
- m g\,\vec{y_0} \\
\vec{0}_G
\end{Bmatrix}
$$
Liaison pivot au point $O$
$$
\mathcal{T}_{\text{pivot}} =
\begin{Bmatrix}
F_x\,\vec{x_1} + F_y\,\vec{y_1} \\
\vec{0}_O
\end{Bmatrix}
$$
### Somme des actions extérieures au point $O$
$$
\sum \mathcal{T}_{\text{ext}} =
\begin{Bmatrix}
\sum \vec{F_i} \\
\sum \vec{M_{O,i}}
\end{Bmatrix}_O
$$
---
## III. CINÉTIQUE
### Masse et volume
$$
m = \rho V
$$
### Moment dinertie
Définition
$$
I = \int r^2\,dm
$$
Usuels
$$
I_{\text{disque}} = \tfrac{1}{2} m R^2
$$
$$
I_{\text{barre, axe extrémité}} = \tfrac{1}{3} m L^2
$$
### Moment cinétique
$$
\vec{H_G} = I\,\vec{\Omega}_{1/0} = I\,\dot{\theta}\,\vec{z}
$$
### Torseur cinétique
Au point $G$
$$
\mathcal{T}_{ci(1/0)} =
\begin{Bmatrix}
m\,\vec{V_G} \\
\vec{H_G}
\end{Bmatrix}_G
$$
Transport au point $O$
$$
\vec{H_O} = \vec{H_G} + \vec{OG} \wedge \big(m\,\vec{V_G}\big)
$$
---
## IV. DYNAMIQUE
### Principe fondamental de la dynamique
Forme torseur
$$
\sum \mathcal{T}_{\text{ext}} = \mathcal{T}_{dy(1/0)}
$$
### Torseur dynamique au point $G$
$$
\mathcal{T}_{dy(1/0)} =
\begin{Bmatrix}
m\,\vec{\Gamma_G} \\
I\,\ddot{\theta}\,\vec{z}
\end{Bmatrix}_G
$$
---
## V. ÉQUILIBRE DYNAMIQUE
### Mise en équations par projections
Point de départ
$$
\sum \mathcal{T}_{\text{ext}} = \mathcal{T}_{dy}
$$
Projections typiques pour une rotation de rayon $R$
$$
\begin{cases}
\sum F_{x_1} = - m R\,\dot{\theta}^{2} \\
\sum F_{y_1} = m R\,\ddot{\theta} \\
\sum M_{z} = I\,\ddot{\theta}
\end{cases}
$$
### Petits mouvements
Approximations pour petits angles
$$
\sin(\theta) \approx \theta \\
\cos(\theta) \approx 1
$$
Équation linéarisée
$$
\ddot{\theta} + \omega_0^{2}\,\theta = 0
$$
Pulsation propre selon le système
$$
\omega_0 = \sqrt{\dfrac{k}{I}} \quad \text{ou} \quad \omega_0 = \sqrt{\dfrac{g}{L}}
$$
---
## VI. RÉCAP DES SYMBOLES
$\theta$ angle en rad
$\dot{\theta}$ vitesse angulaire en rad s$^{-1}$
$\ddot{\theta}$ accélération angulaire en rad s$^{-2}$
$\vec{\Omega}_{1/0}$ vecteur rotation en rad s$^{-1}$
$\vec{V}$ vitesse en m s$^{-1}$
$\vec{\Gamma}$ accélération en m s$^{-2}$
$m$ masse en kg
$I$ moment dinertie en kg m$^{2}$
$\vec{F}$ force en N
$\vec{M}$ moment en N m
$\wedge$ produit vectoriel

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@ -0,0 +1,76 @@
![[ImplantationAteliers-TD.pdf]]
# A. Analyse de lexistant
## 1- Pour un lot darticle P1 calculer les indicateurs :
### (i) Ratio de fluidité (temps à valeur ajoutée / temps de défilement).
M3 -> M6 = 10m = 2min
M6 -> M2 = 12.5m = 2.5min
M2 -> M7 = 15m = 3min
M7 -> M8 = 3m = 0.6min
M8 -> M9 = 0.5m = 0.1min
Temps de VA = $1+0.5+2+0.7+1.2+0.2=5.6min$
Temps de défilement = $2+2.5+3+0.6+0.1=8.2$
Ratio de fluidité: $\frac{\text{Temps de VA}}{\text{Temps de défilement}}=\frac{5.6}{8.2}\approx0.6829268293$
### (ii) Ratio de linéarité Rzz (distance nominale/distance parcourue).
# B. Proposition dune implantation de type "Flow shop".
## 3- Quel est lintérêt dune telle implantation ?
Moins de temps de défilement -> meilleur ratio de fluidité
## 4- Calculer à laide de la méthode des rangs moyens, la meilleure implantation pour les machines M1 à M9, en prenant en compte les articles P1, P2, P3 et P4 (références les plus fabriquées sur la ligne).
| Produit | Gamme | Séquence |
| ------- | -------------------------------- | ------------- |
| P1 | M3 → M6 → M2 → M7 → M8 → M9 | 6 transitions |
| P2 | M1 → M4 → M2 → M7 → M8 → M9 | 6 transitions |
| P3 | M1 → M5 → M4 → M7 → M2 → M8 → M9 | 7 transitions |
| P4 | M1 → M5 → M4 → M6 → M2 → M8 → M9 | 7 transitions |
| De → À | Nombre total |
| ------- | ------------ |
| M1 → M4 | 2 |
| M1 → M5 | 2 |
| M3 → M6 | 1 |
| M4 → M2 | 2 |
| M4 → M7 | 2 |
| M5 → M4 | 2 |
| M6 → M2 | 2 |
| M7 → M2 | 1 |
| M7 → M8 | 3 |
| M8 → M9 | 4 |
| M2 → M7 | 1 |
| M2 → M8 | 2 |
| Machine | Positions moyennes | Rang moyen |
| ------- | --------------------- | ---------- |
| M1 | (1, 1, 1, 1) | 1,0 |
| M3 | (1) | 1,0 |
| M5 | (2, 2) | 2,0 |
| M4 | (2, 2, 3, 3) | 2,5 |
| M6 | (2, 4) | 3,0 |
| M2 | (3, 3, 5, 5) | 4,0 |
| M7 | (4, 4, 4) | 4,0 |
| M8 | (5, 5, 6, 6, 6, 6, 6) | ≈ 5,8 |
| M9 | (6, 6, 7, 7) | 6,5 |
| Rang | Machine |
| ---- | ------- |
| 1 | M1 / M3 |
| 2 | M5 |
| 3 | M4 |
| 4 | M6 |
| 5 | M2 |
| 6 | M7 |
| 7 | M8 |
| 8 | M9 |
> **M3 → M1 → M5 → M4 → M6 → M2 → M7 → M8 → M9**
## 5- Quest-ce qui se passe si on ajoute à cet analyse larticle P5 ? • Gamme simplifiée de P5 : M3 →M5 →M6 →M2
Ça change pas grand chose car P5 suit déjà l'ordre du flowshop établi.
> ***M3*** → M1 → ***M5*** → M4 → ***M6*** → ***M2*** → M7 → M8 → M9
En revanche on peut inverser M1 et M3 pour optimiser encore un peu plus la chaine
> M1 → ***M3*** → ***M5*** → M4 → ***M6*** → ***M2*** → M7 → M8 → M9
## 6- Proposez un plan dimplantation et lanalyse de déroulement en cohérence avec les résultats obtenus avec la méthode des rangs moyens (résultat exercice 4).

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@ -0,0 +1,34 @@
# dependencies (bun install)
node_modules
# output
out
dist
*.tgz
# code coverage
coverage
*.lcov
# logs
logs
_.log
report.[0-9]_.[0-9]_.[0-9]_.[0-9]_.json
# dotenv environment variable files
.env
.env.development.local
.env.test.local
.env.production.local
.env.local
# caches
.eslintcache
.cache
*.tsbuildinfo
# IntelliJ based IDEs
.idea
# Finder (MacOS) folder config
.DS_Store

View file

@ -0,0 +1,15 @@
# project
To install dependencies:
```bash
bun install
```
To run:
```bash
bun run index.ts
```
This project was created using `bun init` in bun v1.3.2. [Bun](https://bun.com) is a fast all-in-one JavaScript runtime.

View file

@ -0,0 +1,26 @@
{
"lockfileVersion": 1,
"configVersion": 1,
"workspaces": {
"": {
"name": "project",
"devDependencies": {
"@types/bun": "latest",
},
"peerDependencies": {
"typescript": "^5",
},
},
},
"packages": {
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}
}

View file

@ -0,0 +1,14 @@
import { serve } from "bun";
import films from "./sgv-2025-all-the_red_helicopters/catalogues/films.html";
import series from "./sgv-2025-all-the_red_helicopters/catalogues/series.html";
serve({
port: 3000,
routes: {
"/": Response.redirect("/films"),
"/films": films,
"/films.html": films,
"/series": series,
"/series.html": series,
},
});

View file

@ -0,0 +1,12 @@
{
"name": "project",
"module": "index.ts",
"type": "module",
"private": true,
"devDependencies": {
"@types/bun": "latest"
},
"peerDependencies": {
"typescript": "^5"
}
}

@ -0,0 +1 @@
Subproject commit 4c9886a4e0971d3636db1386ee592845c7213375

View file

@ -0,0 +1,29 @@
{
"compilerOptions": {
// Environment setup & latest features
"lib": ["ESNext"],
"target": "ESNext",
"module": "Preserve",
"moduleDetection": "force",
"jsx": "react-jsx",
"allowJs": true,
// Bundler mode
"moduleResolution": "bundler",
"allowImportingTsExtensions": true,
"verbatimModuleSyntax": true,
"noEmit": true,
// Best practices
"strict": true,
"skipLibCheck": true,
"noFallthroughCasesInSwitch": true,
"noUncheckedIndexedAccess": true,
"noImplicitOverride": true,
// Some stricter flags (disabled by default)
"noUnusedLocals": false,
"noUnusedParameters": false,
"noPropertyAccessFromIndexSignature": false
}
}

View file

@ -32,3 +32,100 @@ $V^-=\frac{\frac{V_s}{R_1}}{\frac{1}{R_1}+\frac{1}{R_2}}$
Millman en $y$: Millman en $y$:
![[millmanEnY]] ![[millmanEnY]]
$V_y=\frac{\frac{V_A}{R_A}+\frac{V_B}{R_B}+\frac{V_C}{R_C}}{\frac{1}{R_A}+\frac{1}{R_B}+\frac{1}{RC}}$ $V_y=\frac{\frac{V_A}{R_A}+\frac{V_B}{R_B}+\frac{V_C}{R_C}}{\frac{1}{R_A}+\frac{1}{R_B}+\frac{1}{RC}}$
---
![[Pasted image 20251015114956.png]]
| $f(Hz)$ | $0$ | $f_0$ | $500$ | $1k$ | $3k$ |
| -------- | ------------ | ---------------------------------- | ------- | ------- | -------- |
| $\|T\|$ | $\|T_0\|=10$ | $\frac{\|T_0\|}{\sqrt2}\approx7.1$ | $0.032$ | $0.016$ | $0.0053$ |
| $G_{dB}$ | $20$ | 17 | $-30$ | $-36$ | $-46$ |
Avec $G_{dB}=20log( |T| )$.
# TD2.
![[Pasted image 20251015120304.png]]
1.
Millman en $V^-$:
$V^-=\frac{\frac{V_i}{R_1+Z_C}+\frac{V_0}{R_2}}{\frac{1}{R_1+Z_C}+\frac{1}{R_2}}$
Il y a CR sur la borne $-$, c'est donc un comportement linéaire:
$V^+=V^-=0=\frac{\frac{V_i}{R_1+Z_C}+\frac{V_0}{R_2}}{\frac{1}{R_1+Z_C}+\frac{1}{R_2}}$.
$0=\frac{\frac{V_i}{R_1+Z_C}+\frac{V_0}{R_2}}{\frac{1}{R_1+Z_C}+\frac{1}{R_2}}=\frac{V_i}{R_1+\frac{1}{jC\omega}}+\frac{V_0}{R_2}$
$H(j\omega)=\frac{V_0}{V_i}$
---
$\implies \cases{G_{dB}=20log(|H(j\omega)|)=20log(|A_{V_0}|)+20log(\sqrt{\frac{1}{1+(\frac{\omega}{\omega_0})^2}}) \\ \Phi= arg(A_{V_0})+arg(\frac{1}{1+j\frac{\omega}{\omega_0}})=\pi-arg(1+j\frac{\omega}{\omega_0})=\pi-\arctan{(\frac{\omega}{\omega_0})}}$
---
![[Pasted image 20251120105841.png]]
1.
$V^+=\frac{\frac{+15V}{1.6k\Omega}+\frac{-15V}{1.4k\Omega}}{\frac{1}{1.6k\Omega}+\frac{1}{1.4k\Omega}}=\frac{\frac{+15V}{1.6k\Omega}+\frac{-15V}{1.4k\Omega}}{\frac{1}{1.6k\Omega}+\frac{1}{1.4k\Omega}}=\frac{\frac{+15V}{1.6k\Omega}+\frac{-15V}{1.4k\Omega}}{\frac{3}{2.24}}=(\frac{+15V}{1.6k\Omega}+\frac{-15V}{1.4k\Omega})\times\frac{2.24}{3}=\frac{+15V \times 2.24}{1.6k\Omega \times 3}+\frac{-15V \times 2.24}{1.4k\Omega \times 3}=-1V$
2.
Avec $V^+=-1V$
Si $V_{in}<-1V$ alors $V_{out}=+V_{Sat}$
Si $V_{in}>-1V$ alors $V_{out}=-V_{Sat}$
3.
```functionplot
---
title:
xLabel:
yLabel:
bounds: [-10,10,-10,10]
disableZoom: false
grid: true
---
f(x)=-1
Vin(x)=5sin(x)
```
4.
```functionplot
---
title: Fonction de transfert
xLabel: Vin
yLabel: Vout
bounds: [-20,20,-20,20]
disableZoom: false
grid: true
---
f(x)=1000x+1000
g(x)=15+(sqrt(-x-1))*0
h(x)=-15+(sqrt(x+1))*0
```
![[Pasted image 20251120114534.png]]
1.
$V^+=\frac{\frac{0}{R_1}+\frac{V_{out}}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}}=\frac{V_{out}}{1+\frac{R_2}{R_1}}=\frac{R_1V_{out}}{R_1+R_2}$ avec $V_{out}=\pm V_{Sat}$
2.
Avec $R_1=1K\Omega$, calculer $R_2$ pour $V^+=\frac{1}{3}V_{out}$
$\frac{1}{3}V_{out}=\frac{R_1V_{out}}{R_1+R_2}$
$\frac{1}{3}=\frac{R_1}{R_1+R_2}$
$R_1+R_2=3R_1$
$\boxed{R_2=2R_1=2K\Omega}$
3.
Avec $V_{out}=\pm V_{Sat}=\pm15V$
On a $\cases{V_{Max}^+=\frac{1}{3}\times15=5V \\ V_{Min}^+=\frac{1}{3}\times(-15)=-5V}$
4.
AOP en régime saturé, d'où: $\cases{V^+>V^-\implies V_{out}=+V_{Sat} \\ V^+<V^-\implies V_{out}=-V_{Sat}}$
Si $V_{out}=+V_{Sat}$, alors $V^+=V^+_{Max}$
$Si V_{out}=-V_{Sat}$, alors $V^+=V^+_{Min}$
Si $V_{in}<-5V$ alors $V_{out}=+V_{Sat}$
Si $V_{in}>5V$ alors $V_{out}=-V_{Sat}$
![[Pasted image 20251120121241.png]]

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@ -53,3 +53,268 @@ C2 de $(0100001)_{BS}$ -> $(1011111)_{BS}$
## Exercice SB4 ## Exercice SB4
$(568)_{10}$ = ($0101$ $0110$ $1000$)$_{DCB}$ $(568)_{10}$ = ($0101$ $0110$ $1000$)$_{DCB}$
---
Soit un mot binaire codé sur $10$ bits.
$(Val_{Max})_2=0b1111111111$
$(Val_{Max})_{10}=2^{10}-1=1023$.
$(Val_{Min})_2=0b0000000000$
$(Val_{Max})_{10}=0$.
Pour un convertisseur analogique-numérique, qui encode $0$ à $5V$ sur 10 bits,
614 serait obtenu pour environ $3V$.
$4V$ donnerait $818.4$, soit $0b1100110010$.
| A | B | A Et B | A Ou B |
| :-: | :-: | :----: | :----: |
| 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |
Calculer:
1) $x+xy=x(1+y)=x$.
2) $x(x+y)=x$.
3) $(x+\overline{y})y=xy$
4) $(x+y)(x+z)=x+xz+yx+yz=x+yz$
5) $(x+y)(x+\overline{y})=x+x\overline{y}+yx+y\overline{y}=x$
# TD 2.
![[Pasted image 20251013170054.png]]
$A=\overline{x}yz+x\overline{y}z+xy\overline{z}+xyz$
$=\overline{x}yz+x\overline{y}z+xy\overline{z}+xyz+xyz+xyz$
$=yz(\overline{x}+x)+xz(\overline{y}+y)+xy(\overline{z}+z)$
$=yz+xz+xy$
$B=xy+\overline{x}y\overline{z}+yz$
$=xy(z+\overline{z})+\overline{x}y\overline{z}+yz(x+\overline{x})$
$=xyz+xy\overline{z}+\overline{x}y\overline{z}+xyz+\overline{x}yz$
$=xy+\overline{x}y=y(x+\overline{x})=y$
![[Pasted image 20251013170215.png]]
$C=(x+z)(\overline{x}+y)$
$=x\overline{x}+xy+z\overline{x}+zy$
$=0+xy+z\overline{x}+zy$
$=xy(z+\overline{z})+\overline{x}z(y+\overline{y})+yz(x+\overline{x})$
$=xyz+xy\overline{z}+\overline{x}yz+\overline{x}\overline{y}z+xyz+\overline{x}yz$
$=xyz+xy\overline{z}+\overline{x}yz+\overline{x}\overline{y}z$
$=xy(z+\overline{z})+\overline{x}z(y+\overline{y})$
$=xy+\overline{x}z$
$D=(x\overline{y}+z)(x+\overline{y})z$
$=(xx\overline{y}+x\overline{y}\overline{y}+xz+\overline{y}z)z$
$=(x\overline{y}+x\overline{y}+xz+\overline{y}z)z$
$=x\overline{y}z+x\overline{y}z+xzz+\overline{y}zz$
$=x\overline{y}z+xz+\overline{y}z$
$=z(x+\overline{y}+x\overline{y})$
$=z(x+\overline{y})$
Théorème de Morgan
$\overline{a•b}=\overline{a}+\overline{b}$
$\overline{a+b}=\overline{a}•\overline{b}$
![[Pasted image 20251015091234.png]]
![[Pasted image 20251015094501.png]]
1.
a)
Table de vérité de $F$ et $G$:
| $x$ | $y$ | $F(x,y)$ | $G(x,y)$ |
| --- | --- | -------- | -------- |
| 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 |
b)
Par De Morgan, $F(x,y)=x+\overline{x}y$
$F(x,y)=\overline{\overline{x+\overline{x}y}}$
$F(x,y)=\overline{\overline{x}•\overline{\overline{x}y}}$
$F(x,y)=\overline{\overline{x}•(\overline{\overline{x}}+\overline{y})}$
$F(x,y)=\overline{\overline{x}•(x+\overline{y})}$
$F(x,y)=\overline{\overline{x}x•\overline{x}\overline{y}}$
$F(x,y)=\overline{\overline{x}x}+\overline{\overline{x}\overline{y}}$
$F(x,y)=\overline{\overline{x}\overline{y}}$
$F(x,y)=\overline{\overline{x}}+\overline{\overline{y}}$
$F(x,y)=x+y$
# TD3
![[Pasted image 20251015102351.png]]
![[Pasted image 20251015102357.png]]
$A=\overline{b}$
![[Pasted image 20251015102532.png]]
$B=\overline{a}$
![[Pasted image 20251015102540.png]]
$C=\overline{c}$
![[Pasted image 20251015102551.png]]
$D=\overline{a}\overline{c}+\overline{a}\overline{b}$
![[Pasted image 20251015102611.png]]
$E=c$
![[Pasted image 20251015102624.png]]
$F=1$
![[Pasted image 20251015102631.png]]
$G=d$
![[Pasted image 20251015102641.png]]
$H=\overline{c}d+ab$
![[Pasted image 20251015102647.png]]
$I=\overline{c}$ j
![[Pasted image 20251015110635.png]]
1.
| a/bc | 00 | 01 | 11 | 10 |
| :--: | --- | --- | --- | --- |
| 0 | 0 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 | 1 |
$F=a\overline{c}+bc$
---
# Circuits combinatoires
![[CI-EEA21EA5 - Elec_Num - CH4 - Circuits combinatoires - TD.pdf]]
## Exercice CC1. Comparateur binaire 1 bit
1. À partir de la table de vérité dun comparateur binaire dun bit, obtenir les équations logiques des sorties en fonction des entrées.
| A | B | Égalité | A>B | A<B |
| --- | --- | ------- | --- | --- |
| 0 | 0 | 1 | 0 | 0 |
| 0 | 1 | 0 | 0 | 1 |
| 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 | 0 |
$F_{Égalité}=AB+\overline{AB}$
$F_{A>B}=\overline{A}B$
$F_{A>B}=A\overline{B}$
2. Dessiner le circuit électronique du comparateur.
![[Drawing 2025-11-19 09.29.12.excalidraw|1000]]
## Exercice CC5. Étude de circuits à base de multiplexeurs
### 5.1 Premier cas d'étude
1.
| Entrée | A | B | $F_1$ |
| ------ | --- | --- | ----- |
| 0 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 2 | 1 | 0 | 1 |
| 3 | 1 | 1 | 0 |
$F_1=\overline{B}$
| Entrée | C | D | $F_2$ |
| ------ | --- | --- | ----- |
| 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | e |
| 2 | 1 | 0 | e |
| 3 | 1 | 1 | 1 |
| e/CD | 00 | 01 | 11 | 10 |
| ---- | --- | --- | --- | --- |
| 0 | 0 | 0 | 1 | 0 |
| 1 | 0 | 1 | 1 | 1 |
$F_2=CD+e(C+D)$
2.
$S=\overline{\overline{F_1}F_2}$
$S=\overline{\overline{\overline{B}}(CD+eC+eD)}$
3.
$S=\overline{B(CD+eC+eD)}$
$S=\overline{BCD+BeC+BeD}$
$S=\overline{BCD}•\overline{BeC}•\overline{BeD}$
## Exercice CC8. Machine à café & thé
3 entrées: $C$, $T$ et $J$.
4 sorties: $L$ (Led rouge), $Sc$ (Sortie café), $St$ (Sortie thé) et $B$ (bip sonore)
| $C$ | $T$ | $J$ | - | $S_c$ | $S_t$ | $L$ | $B$ |
| --- | --- | --- | --- | ----- | ----- | --- | --- |
| 0 | 0 | 0 | - | 0 | 0 | 1 | 0 |
| 0 | 0 | 1 | - | 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | - | 0 | 0 | 1 | 0 |
| 0 | 1 | 1 | - | 0 | 1 | 0 | 0 |
| 1 | 0 | 0 | - | 0 | 0 | 1 | 0 |
| 1 | 0 | 1 | - | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | - | 0 | 0 | 1 | 0 |
| 1 | 1 | 1 | - | 0 | 0 | 1 | 1 |
$B=CTJ$
$L=\overline{J}+CT$
$S_c=C\overline{T}J$
$S_t=CT\overline{J}$
3.
![[Drawing 2025-11-21 10.25.20.excalidraw]]
4.
$L=\overline{\overline{L}}=\overline{\overline{\overline{J}+CT}}=\overline{J+\overline{CT}}$
## Exercice CC11. Étude dun circuit combinatoire
1.
| a | b | c | d | - | x | y | z |
| --- | --- | --- | --- | --- | --- | --- | --- |
| 0 | 0 | 0 | 0 | - | X | X | X |
| 0 | 0 | 0 | 1 | - | X | X | X |
| 0 | 0 | 1 | 0 | - | X | X | X |
| 0 | 0 | 1 | 1 | - | X | X | X |
| 0 | 1 | 0 | 0 | - | X | X | X |
| 0 | 1 | 0 | 1 | - | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | - | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 | - | 1 | 0 | 1 |
| 1 | 0 | 0 | 0 | - | X | X | X |
| 1 | 0 | 0 | 1 | - | 1 | 0 | 0 |
| 1 | 0 | 1 | 0 | - | X | X | X |
| 1 | 0 | 1 | 1 | - | X | X | X |
| 1 | 1 | 0 | 0 | - | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 | - | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | - | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | - | 1 | 0 | 1 |
$X=C+\overline{B}$
| ab/cd | 00 | 01 | 11 | 10 |
| ----- | --- | --- | --- | --- |
| 00 | X | X | X | X |
| 01 | X | 0 | 1 | 1 |
| 11 | 0 | 0 | 1 | 1 |
| 10 | X | 1 | X | X |
$Y=B\overline{C}$
| ab/cd | 00 | 01 | 11 | 10 |
| ----- | --- | --- | --- | --- |
| 00 | X | X | X | X |
| 01 | X | 1 | 0 | 0 |
| 11 | 1 | 1 | 0 | 0 |
| 10 | 1 | 0 | X | X |
$Z=C+BD$
| ab/cd | 00 | 01 | 11 | 10 |
| ----- | --- | --- | --- | --- |
| 00 | X | X | X | X |
| 01 | X | 1 | 1 | 1 |
| 11 | 0 | 1 | 1 | 1 |
| 10 | X | 0 | X | X |
2.
$Y=\overline{X}$
3.
![[Screenshot 2025-11-27 at 10.05.18.png]]
---

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@ -20,3 +20,64 @@
- $V_{R_2}=-R_2I=V_s-V^-\implies I=\frac{-(V_s-V^-)}{R_2}$ - $V_{R_2}=-R_2I=V_s-V^-\implies I=\frac{-(V_s-V^-)}{R_2}$
D'où $\frac{V_e-V^-}{R_1}=\frac{-(V_s-V^-)}{R_2}$ et $V^+=V^-=0$ donc $\frac{V_e}{R_1}=\frac{-V_s}{R_2}$ et $\boxed{\frac{V_s}{V_e}=-\frac{R_2}{R_1}}$ D'où $\frac{V_e-V^-}{R_1}=\frac{-(V_s-V^-)}{R_2}$ et $V^+=V^-=0$ donc $\frac{V_e}{R_1}=\frac{-V_s}{R_2}$ et $\boxed{\frac{V_s}{V_e}=-\frac{R_2}{R_1}}$
# Exercice ARL3.
![[Pasted image 20251013141437.png]]
1.
D'après Millman, $V^-=\frac{\frac{V_{e_3}}{R_1}+\frac{V_{e_2}}{R_2}+\frac{V_{e_1}}{R_3}+\frac{V_{out}}{R_4}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}}$
2.
Pour $R_1=R_2=R_3=R_4=1k\Omega$,
$V^-=\frac{\frac{V_{e_3}}{1k\Omega}+\frac{V_{e_2}}{1k\Omega}+\frac{V_{e_1}}{1k\Omega}+\frac{V_{out}}{1k\Omega}}{\frac{1}{1k\Omega}+\frac{1}{1k\Omega}+\frac{1}{1k\Omega}+\frac{1}{1k\Omega}}=\frac{\frac{V_{e_1}+V_{e_2}+V_{e_3}+V_{out}}{1k\Omega}}{\frac{4}{1k\Omega}}=\frac{V_{e_1}+V_{e_2}+V_{e_3}+V_{out}}{4}$
Puisque l'AOP est idéal, $V^-=V^+$, et $V^+=0$ donc $V^-=0$,
ainsi, $0=\frac{V_{e_1}+V_{e_2}+V_{e_3}+V_{out}}{4}$
$V_{out}=-V_{e_1}-V_{e_2}-V_{e_3}$.
3.
$V_{R_1}=V_{e_3}-V^-=R_1i_1$.
Donc, $i_1=\frac{V_{e_3}-V^-}{R_1}=\frac{V_{e_3}}{R_1}$
$V_{R_2}=V_{e_2}-V^-=R_2i_2$.
Donc, $i_2=\frac{V_{e_2}-V^-}{R_2}=\frac{V_{e_2}}{R_2}$
$V_{R_3}=V_{e_1}-V^-=R_3i_3$.
Donc, $i_3=\frac{V_{e_1}-V^-}{R_3}=\frac{V_{e_1}}{R_3}$
$i=\frac{V^--V_{out}}{R_4}=\frac{-V_{out}}{R_4}$
D'après la loi des noeuds, $i=i_1+i_2+i_3$, donc $\frac{-V_{out}}{R_4}=\frac{V_{e_3}}{R_1}+\frac{V_{e_2}}{R_2}+\frac{V_{e_1}}{R_3}$
En posant $R_1=R_2=R_3=R_4=R$, On obtient $V_{out}=-(V_{e_1}+V_{e_2}+V_{e_3})$
![[Pasted image 20251013141455.png]]
1.
D'après Millman, $V^-=\frac{\frac{V_{e_3}}{R_1}+\frac{V_{e_2}}{R_2}+\frac{V_{out}}{R_4}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_4}}$ .
2.
$V^+=\frac{R_5}{R_3+R_5}V_{e_1}$.
3.
$V^-=V^+=\boxed{\frac{R_5}{R_3+R_5}V_{e_1}=\frac{\frac{V_{e_3}}{R_1}+\frac{V_{e_2}}{R_2}+\frac{V_{out}}{R_4}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_4}}}$
Puisque $R_1=R_2=R_3=R_4=R_5=R$,
$\frac{R}{2R}V_{e_1}=\frac{\frac{V_{e_3}}{R}+\frac{V_{e_2}}{R}+\frac{V_{out}}{R}}{\frac{1}{R}+\frac{1}{R}+\frac{1}{R}}$
$\frac{1}{2}V_{e_1}=\frac{\frac{V_{e_3}}{R}+\frac{V_{e_2}}{R}+\frac{V_{out}}{R}}{\frac{1}{R}+\frac{1}{R}+\frac{1}{R}}=\frac{\frac{V_{e_3}+V_{e_2}+V_{out}}{R}}{\frac{3}{R}}=\frac{V_{e_3}+V_{e_2}+V_{out}}{3}$
$V_{e_1}=2\frac{V_{e_3}+V_{e_2}+V_{out}}{3}=\frac{2}{3}(V_{e_3}+V_{e_2}+V_{out})$
$\frac{2}{3}V_{out}=V_{e_1}-\frac{2}{3}(V_{e_3}+V_{e_2})$
$V_{out}=\frac{3}{2}V_{e_1}-(V_{e_3}+V_{e_2})$.
![[Pasted image 20251013151316.png]]
Millman en $V^-$: $V^-=\frac{\frac{V_1'}{R_1}+\frac{V_S}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}}$ <input type="checkbox" checked="true" >
Millman en $V^+$: $V^-=\frac{\frac{V_2'}{R_1}+\frac{0}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}}=\frac{\frac{V_2'}{R_1}}{\frac{1}{R_1}+\frac{1}{R_2}}=\frac{R_2V_2'}{R_1+R_2}$
On a $V^+=V^-$ car on a une CR sur la borne $\boxed{-}$ (AOP en régime linéaire)
$\frac{\frac{V_1'}{R_1}+\frac{V_S}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}}=\frac{R_2V_2'}{R_1+R_2}$
$\frac{\frac{R_1V_S+R_2V_1'}{R_1R_2}}{\frac{1}{R_1}+\frac{1}{R_2}}=\frac{R_2V_2'}{R_1+R_2}$
$R_1V_S=R_2(V_2'-V_1')$
$\boxed{V_S=\color{red}\boxed{\color{white}\frac{R_2}{R_1}}\color{white}(V_2'-V_1')}$
$\color{red}A_0$

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@ -0,0 +1,141 @@
---
excalidraw-plugin: parsed
tags: [excalidraw]
---
==⚠ Switch to EXCALIDRAW VIEW in the MORE OPTIONS menu of this document. ⚠== You can decompress Drawing data with the command palette: 'Decompress current Excalidraw file'. For more info check in plugin settings under 'Saving'
# Excalidraw Data
## Text Elements
C ^VK6UFB0x
T ^KaGsYAan
J ^ywdJkR1H
B ^uSKDShqa
L ^ZNsCKQM8
## Embedded Files
17ce83cf4d5573db8067f3a00f02fcbb3738261f: $$S_c$$
6854b8de3ffeb5e38d690c7a1627d9c4fb04d38a: $$S_t$$
%%
## Drawing
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```
%%

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@ -0,0 +1,104 @@
---
excalidraw-plugin: parsed
tags: [excalidraw]
---
==⚠ Switch to EXCALIDRAW VIEW in the MORE OPTIONS menu of this document. ⚠== You can decompress Drawing data with the command palette: 'Decompress current Excalidraw file'. For more info check in plugin settings under 'Saving'
# Excalidraw Data
## Text Elements
%%
## Drawing
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```
%%

View file

@ -0,0 +1,162 @@
---
excalidraw-plugin: parsed
tags: [excalidraw]
---
==⚠ Switch to EXCALIDRAW VIEW in the MORE OPTIONS menu of this document. ⚠== You can decompress Drawing data with the command palette: 'Decompress current Excalidraw file'. For more info check in plugin settings under 'Saving'
# Excalidraw Data
## Text Elements
A ^j3PgQ4Bk
B ^v2mWPzuo
A>B ^FeGx8OHk
A<B ^VjmF8Woc
A=B ^t6GA4GJY
A ^Jms7fQlJ
_ ^KcFVm05B
B ^Ne4b1rgr
_ ^0d7S8LXo
AB ^G8P2UFeK
_ ^hjkP00Aw
AB ^AkKHnMxP
_ ^xoZewSKo
%%
## Drawing
```compressed-json
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```
%%

View file

@ -0,0 +1,52 @@
---
excalidraw-plugin: parsed
tags: [excalidraw]
---
==⚠ Switch to EXCALIDRAW VIEW in the MORE OPTIONS menu of this document. ⚠== You can decompress Drawing data with the command palette: 'Decompress current Excalidraw file'. For more info check in plugin settings under 'Saving'
# Excalidraw Data
## Text Elements
%%
## Drawing
```compressed-json
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A/7CXopiWP4o2PW43jWIE2SXvWk3FLYSSQmernIAU8TneUi3g3/fANNhomdP0wyPRM9AAHlSQAR0qXYAFVJFladamIegyJ2SoRlJABNDg3O8+AZv8rUgoeHZtB6Hoa3SxFh2+X4PQSh5DgSBndh2e8+hSjKso9HK8tQY6wQhKE0AV1CKuVqqEeDXE6uNxrBRaiRGXa1lV05blCz6oUKiGkapXGj15UVLaZrtKsl4tJah7WhrNumior++vxJD+/KP
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```
%%

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