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CHAPTER 4 COMPUTER SYSTEM – Von Neumann Model

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1 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 S. Barua – CPSC

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

3 Von Neumann Model of a Computer System
S. Barua – CPSC

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

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

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

7 Interface to Memory MAR: Memory Address Register
MDR: Memory Data Register S. Barua – CPSC

8 Block Diagram of a Memory Unit
memory width 1 2 . MAR bits bits MDR 10-bit memory address 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

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. Send a “write” signal to the memory. S. Barua – CPSC

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

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

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

13 Block Diagram of a Basic Computer Architecture
CPU Clock CU ALU Clock Generator Address Bus p 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

14 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

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

16 Instruction Processing
Fetch instruction from memory Decode instruction Evaluate address Fetch operands from memory Execute operation Store result S. Barua – CPSC

17 Instruction Processing: FETCH
Load the instruction from 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 D EA OP EX S S. Barua – CPSC

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

19 Instruction Processing: DECODE
F 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; binary) Base register: IR[8:6] = 011 = R3 D EA OP EX S S. Barua – CPSC

20 Instruction Processing: EVALUATE ADDRESS
F 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 D EA OP EX S S. Barua – CPSC

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

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

23 Instruction Processing: STORE RESULT
F 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 D EA OP EX S S. Barua – CPSC

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

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

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

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

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 R2 ← MDR S. Barua – CPSC

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

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

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

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


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