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Architecture des ordinateurs : memoires
objectif général : Connaître les composants qui stockent les programmes et les données dans un ordinateur
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objectif général : comprendre comment l'ordinateur communique avec son environnement
objectifs spécifiques :
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- comprendre la scrutation ou interrogation
- comprendre les requêtes d'interruption
- comprendre l'accès direct à la mémoire ou DMA
- connaître le fonctionnement des bus
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3. The RBS is an advanced antenna system capable of beamforming and massive MIMO to enhance coverage, capacity, and performance.
4. Additional modules like the capacity plug-in unit, radio modules, and fiber distribution unit that integrate with the system are also described.
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- CRC Press is publishing the book as an imprint of the Taylor & Francis Group.
- Copyright ownership is held by Taylor & Francis Group, LLC and no claim is made to original U.S. government works.
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1326472.ppt
- 1. Le Système Processeur David Saint-Mellion
- 2. Un Système de Traitement • Gérer une application • Couplé par le Module E-S à l ’application • Le Programme exécuté par le processeur
- 4. Le Module d’échange Dialoguer avec l'extérieur • E-S TOR • E-S Communication • Timers : compter, gérer le temps • E-S analogiques
- 5. Le Processeur • Maître du bus • Initialisation « Reset » : le µP charge le vecteur de RESET • µP a besoin d ’énergie et d ’une horloge
- 6. Le bloc Mémoire • Mémoire Mortes: Lire des données • Mémoire Vive : Écrire et Lire des Données • Le Vecteur de Reset et le Boot en Mémoire Morte
- 7. Décodage adresse Le µP gère un ensemble de positions mémoires Il faut positionner les composants dans l’espace Mémoire Le Décodage d ’adresse élabore les « CS » des composants 0000 h FFFFh 64 Ko Composant @ Début Taille
- 8. Exemple • µP 16 lignes @ • Composant 1 : 14 bornes @ • Composant 2 : 15 bornes @ Pour qu’un composant soit sélection, il faut CS=0 Composant 1 : il faut A15=0 et A14=0 Composant 2 : il faut A15=1 Pour accéder au premier registre d ’un composant, il faut : CS=0 et la valeur «0» sur les bornes @ du composant /CS /CS OU A14 A15 A15 =1 15 14 16 Comp 1 Comp 2
- 9. Espace mémoire du µP 64 Ko Plan Mémoire @ début composant 1 A15 à A14 «0» et A13 à A0 à «0» - 0000 h @ début composant 2 A15 «1» et A14 à A0 à «0» - 8000h @ fin composant 1 A15 à A14 «0» et A13 à A0 à «1» - 3FFFh @ fin composant 2 A15 «1» et A14 à A0 à «1» - FFFFh 0000h Composant 1 3FFFh FFFFh Composant 2 8000h Plan Mémoire