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CHAPTER 4 COMPUTER SYSTEM – Von Neumann Model

CHAPTER 4 COMPUTER SYSTEM – Von Neumann Model. Topics Basic components of a computer Instruction processing Examples from the LC-3 simulator Changing the sequence of execution. Basic Components of a Computer System. The basic components are: Central processing unit (CPU) Memory

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CHAPTER 4 COMPUTER SYSTEM – Von Neumann Model

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  1. CHAPTER 4COMPUTER SYSTEM – Von Neumann Model Topics • Basic components of a computer • Instruction processing • Examples from the LC-3 simulator • Changing the sequence of execution S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  2. Basic Components of a Computer System The basic components are: • Central processing unit (CPU) • Memory • Input devices • Output devices A clock controls all the operations performed in a computer system. A computer system is a synchronous machine. S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  3. Von Neumann Model of a Computer System S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  4. Central Processing Unit (CPU) The CPU consists of: • Arithmetic logic unit (ALU) • Control unit (CU) • Small set of storage areas, called registers Arithmetic Logic Unit (ALU) The ALU is the primary processing unit in a computer. It performs all the arithmetic and logic operations. Control Unit (CU) The CU controls all the operations performed by a computer and determines what operations the program calls for and in what order they need to be carried out. S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  5. CPU (Continued) The CPU is responsible for: • The complete processing of information • All the decision-making operations Word Size of a processor: • Number of bits the processor reads from or writes into the memory in one clock cycle • Number of bits normally processed by ALU in one instruction • Also width of registers • LC-3 is a 16 bit processor S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  6. Memory Memory provides the storage space for • the program that is currently being executed • the data that is needed during the execution of the program The memory unit consists of a number of memory locations of the same width. The number of bits that can be stored in a memory location corresponds to the memory width. Each memory location is identified by a unique address. Memory capacity = Total number of memory locations * memory width S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  7. Interface to Memory MAR: Memory Address Register MDR: Memory Data Register S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  8. Block Diagram of a Memory Unit memory width 7 6 5 4 3 2 1 0 0 1 2 . MAR 10 bits . 8 bits MDR . 10-bit memory address . 8-bit data (A9 – A0) . (D7 – D0) . 1023 For this example, memory capacity = 210 x 8 bits = 1 Kbytes (The above figure does not show the control signals needed to control the memory read and memory write operations.) S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  9. Memory - Basic Operations LOAD - Read data from a memory location • Write the address of the location into the MAR. • Send a “read” signal to the memory. • Read the data from MDR. STORE - Write data to a memory location • Write the data to the MDR. • Write the address of the location into the MAR. • Send a “write” signal to the memory. S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  10. Input/Output (I/O) Devices Input devices transfer information from the outside world into the memory of a computer system. Examples: Keyboard, disk drives, scanner, mouse Output devices transfer information from the memory of a computer system into the outside world. Examples: Printers, display units, plotters, disk drives Some devices provide both input and output Examples: disk, network Driver - Program that controls access to a device S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  11. Buses in a Computer System Buses are the physical link between the various components within the computer. The computer employs mainly three types of buses: • Address bus • Data bus • Control bus S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  12. Buses in a Computer System (Continued) Three types of buses in a computer system: • Address bus: Carries the address generated by the CPU to the memory and I/O. • Data bus: Allows the transfer of data between the CPU, memory and I/O. • Control bus: Carries the control signals, generated by the control unit, to the various components in the system in order to ensure proper sequencing of data and instruction movement. S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  13. Block Diagram of a Basic Computer Architecture CPU Clock CU ALU Clock Generator Address Busp Data Bus q Control Bus r Main Memory I/O Unit I/O Bus(represents address, data, & control) Keyboard Printer Secondary Memory S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  14. Instruction The instruction specifies two items: • Opcode: operation to be performed • Operands: data/locations to be used for operation An instruction is encoded as a sequence of bits. Often, but not always, instructions have a fixed length, such as 16 or 32 bits. Instruction Format – The binary representation of an instruction S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  15. Instruction (Continued) Control unit • interprets the instruction • generates sequence of control signals to carry out operation. A computer’s instructions and their formats is known as its Instruction Set Architecture (ISA). S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  16. Instruction Processing Fetch instruction from memory Decode instruction Evaluate address Fetch operands from memory Execute operation Store result S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  17. Instruction Processing: FETCH F D Load the instructionfrom memory into Instruction Register • Copy contents of PC into MAR MAR ← PC • Increment PC so that PC points to the next instruction in sequence PC ← PC + 1 • Send “read” control signal to memory MDR←Memory [MAR] • Copy contents of MDR into IR IR ← MDR EA OP EX S S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  18. Example Instruction: LDR Instruction LDR R2, R3, #6 – Loads (reads) data from memory into the destination register R2 Memory address: Base register + SEXT (offset) Memory address = R3 + 6 Load the contents of the memory location [R3 + 6] to R2 S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  19. Instruction Processing: DECODE F D • First identify the opcode • Identify the operands Example: LDR R2, R3, #6 Opcode: IR[15:12] = 0110 Operands: Offset: IR[5:0] = #6(6 decimal; 000110 binary) Base register: IR[8:6] = 011 = R3 EA OP EX S S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  20. Instruction Processing: EVALUATE ADDRESS F D For instructions that require memory access, compute address used for memory access. Example: LDR R2, R3, #6 Address = SEXT (offset) + Base Register MAR ← 6 + R3 EA OP EX S S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  21. Instruction Processing: FETCH OPERANDS F D Obtain source operands needed to perform operation. Example: LDR R2, R3, #6 Load data from memory (LDR) MDR ← Memory [MAR] EA OP EX S S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  22. Instruction Processing: EXECUTE F D Perform the operation, using the source operands. Example: LDR R2, R3, #6 No operation done in this step EA OP EX S S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  23. Instruction Processing: STORE RESULT F D Write results to destination. Destination can be a register or Memory. Example: LDR R2, R3, #6 Data read from the memory is placed in destination register. R2 ← MDR EA OP EX S S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  24. Changing the Sequence of Instructions In the FETCH phase, we increment PC by 1. Need special instructions called control instructions to change the contents of the PC during the Execute phase. • jumps (unconditional) Always change the PC • branches (conditional) PC is changed only if the specified condition is true S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  25. Examples: LC-3 Control Instructions JMP R2 ; PC ← R2 RET ; PC ← R7 BRzp NEXT ; if the last result was zero or positive then PC ← address corresponding to NEXT S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  26. Instruction Processing Summary Three basic kinds of instructions: • Arithmetic/logic instructions (ADD, AND, …) • Data transfer instructions (LD, ST, …) • Control instructions (JMP, BRnz, …) Six basic phases of instruction processing: F  D  EA  OP  EX  S Not all phases are needed by every instruction Phases may take variable number of machine cycles S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  27. Example Instruction: LDR Instruction LDR R2, R3, #6 ; R2 ← mem [R3 + 6] Opcode: IR[15:12] = 0110 Operands: Base register: R3 given by IR[8:6] Offset: 6 given by IR[5:0] Destination: R2 given by IR[11:9] S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  28. Instruction Processing: LDR R2, R3, #6 Fetch MAR ← PC PC ← PC + 1 MDR←Memory [MAR] IR ← MDR Decode Identify the opcode Identify the operands Execute MAR ← 6 + R3 MDR ← Memory [MAR] R2 ← MDR S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  29. Example Instruction: ADD Instruction ADD R6, R2, R6 ; R6 ← R2 + R6 Opcode: IR[15:12] = 0001 Operands: Source1: R2 given by IR[8:6] Source2: R6 given by IR[2:0] Destination: R6 given by IR[11:9] S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  30. Instruction Processing: ADD R6, R2, R6 Fetch MAR ← PC PC ← PC + 1 MDR←Memory [MAR] IR ← MDR Decode Identify the opcode Identify the operands Execute R6 ← R2 + R6 S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  31. Example Instruction: JMP Instruction JMP R3 ; PC ← R3 Opcode: IR[15:12] = 0001 Operand: Base register R3 given by IR[8:6] S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

  32. Instruction Processing: JMP R3 Fetch MAR ← PC PC ← PC + 1 MDR←Memory [MAR] IR ← MDR Decode Identify the opcode Identify the operand Execute PC ← R3 S. Barua – CPSC 240 sbarua@fullerton.edu http://sbarua.ecs.fullerton.edu

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