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Tiêu đề Computer Systems Organization
Trường học University of Computer Science
Chuyên ngành Computer Systems Organization
Thể loại Tài liệu
Thành phố Hanoi
Định dạng
Số trang 43
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ALU input registerALU output register ALU ALU input bus Figure 2-2.. public class Interp {static int PC; // program counter holds address of next instrstatic int AC; // the accumulator,

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COMPUTER SYSTEMS

ORGANIZATION

1

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Central processing unit (CPU)

Figure 2-1 The organization of a simple computer with one

CPU and two I/O devices.

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ALU input register

ALU output register ALU

ALU input bus

Figure 2-2 The data path of a typical von Neumann machine.

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public class Interp {

static int PC; // program counter holds address of next instrstatic int AC; // the accumulator, a register for doing arithmeticstatic int instr; // a holding register for the current instructionstatic int instr3type; // the instruction type (opcode)

static int data3loc; // the address of the data, or−1 if none

static int data; // holds the current operand

static boolean run3bit = true; // a bit that can be turned off to halt the machinepublic static void interpret(int memory[ ], int starting3address) {

// This procedure interprets programs for a simple machine with instructions having// one memory operand The machine has a register AC (accumulator), used for// arithmetic The ADD instruction adds am integer in memory to the AC, for example// The interpreter keeps running until the run bit is turned off by the HALT instruction.// The state of a process running on this machine consists of the memory, the

// program counter, the run bit, and the AC The input parameters consist of

// of the memory image and the starting address

PC = starting3address;

while (run3bit) {

instr = memory[PC]; // fetch next instruction into instr

PC = PC + 1; // increment program counter

instr3type = get3instr3type(instr); // determine instruction type

data3loc = find3data(instr, instr3type); // locate data (−1 if none)

if (data3loc >= 0) // if data3loc is−1, there is no operand

data = memory[data3loc]; // fetch the data

execute(instr3type, data); //execute instruction

}

}

private static int get3instr3type(int addr) { }

private static int find3data(int instr, int type) { }

private static void execute(int type, int data){ }

}

Figure 2-3 An interpreter for a simple computer (written in Java).

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Instructionexecutionunit

Writebackunit

Figure 2-4 (a) A five-stage pipeline (b) The state of each

stage as a function of time Nine clock cycles are illustrated.

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Instructionexecutionunit

Writebackunit

Instructiondecodeunit

Operandfetchunit

Instructionexecutionunit

Writebackunit

Figure 2-5 (a) Dual five-stage pipelines with a common

in-struction fetch unit.

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S2 S3 S5Instruction

decodeunit

Operandfetchunit

LOAD Writeback

ALUALU

Figure 2-6 A superscalar processor with five functional units.

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(a) (b)

CPU

Sharedmemory

BusCPU CPU CPU

Local memories

CPU

Sharedmemory

BusCPU CPU CPU

Figure 2-8 (a) A single-bus multiprocessor (b) A

multicom-puter with local memories.

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1234567

012345

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Figure 2-10 Number of bits per cell for some historically

in-teresting commercial computers.

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Address Big endian Address

1 5 9 13

2 6 10 14

3 7 11 15

32-bit word

Little endian

Byte 3

(b)

7 11 15

2 6 10 14

1 5 9 13

0 4 8 12

32-bit word

Figure 2-11 (a) Big endian memory (b) Little endian memory.

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Figure 2-12 (a) A personnel record for a big endian machine.

(b) The same record for a little endian machine (c) The result

of transferring the record from a big endian to a little endian.

(d) The result of byte-swapping (c).

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A

C1

A

C1

1000

0

1

(c)

ErrorA

B

C

1100

0

Figure 2-14 (a) Encoding of 1100 (b) Even parity added (c) Error in AC.

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Memory word 1111000010101110 0

1 6

1 7

0 8

0 9

0 10

0 11

0 12

1 13

0 14

1 15

1 16

0 17

1 18

1 19

1 20

0 21

Parity bits

Figure 2-15 Construction of the Hamming code for the

memory word 1111000010101110 by adding 5 check bits to the

16 data bits.

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Bus

Main memory CPU

Figure 2-16 The cache is logically between the CPU and

main memory Physically, there are several possible places it

could be located.

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4-MB memory chip

Connector

Figure 2-17 A single inline memory module (SIMM) holding

32 MB Two of the chips control the SIMM.

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Main memory Cache

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of armmotion

Diskarm

Read/writehead

Preamble

E C

1 sect

or

Figure 2-19 A portion of a disk track Two sectors are illustrated.

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Surface 2

Surface 1

Surface 0

Read/write head (1 per surface)

Direction of arm motion Surface 3

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P16-19 Strip 12 Strip 17 Strip 18

Strip 12 P16-12 Strip 13 Strip 14

Strip 4

Strip 9 Strip 5 Strip 10 Strip 6

RAID level 2

Strip 11 Strip 7

Strip 8

Strip 4

Strip 0

Strip 9 Strip 5 Strip 1

Strip 10 Strip 6 Strip 2

Strip 11 Strip 7 Strip 3

Strip 8 Strip 4 Strip 0

Strip 9 Strip 5 Strip 1

Strip 10 Strip 6 Strip 2

Strip 11 Strip 7 Strip 3

Bit 1 Bit 2 Bit 3 Bit 4 Bit 5 Bit 6

RAID level 1

Strip 10 Strip 6 Strip 2

Strip 11 Strip 7 Strip 3

Strip 8

Strip 4

Strip 0

Strip 9 Strip 5 Strip 1

P8-11 Strip 6 Strip 2

Strip 10 P4-7 Strip 3

Strip 19 Strip 15

Strip 11 Strip 7 P0-3

Figure 2-23 RAID levels 0 through 5 Backup and parity

drives are shown shaded.

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Spiral groove

Pit Land

2K block of user data

Figure 2-24 Recording structure of a Compact Disc or CD-ROM.

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Frames of 588 bits,each containing

24 data bytes

Symbols of

14 bits each

42 Symbols make 1 frame

98 Frames make 1 sector

Figure 2-25 Logical data layout on a CD-ROM.

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Printed label

Protective lacquerReflective gold layer

layer

SubstrateDirection

Infraredlaserdiode

Dark spot in thedye layer burned

by laser whenwriting1.2 mm

Dye

Polycarbonate

Figure 2-26 Cross section of a CD-R disk and laser (not to

scale) A silver CD-ROM has a similar structure, except

without the dye layer and with a pitted aluminum layer instead

of a gold layer.

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Polycarbonate substrate 1

Polycarbonate substrate 2

Semireflective layer

Semireflective layer

Aluminum reflector

Aluminum reflector

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Keyboard disk driveFloppy

Harddisk drive

Harddiskcontroller

Floppydiskcontroller

Keyboardcontroller

VideocontrollerMemory

CPU

Bus

Figure 2-29 Logical structure of a simple personal computer.

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Networkcontroller

Videocontroller

Printercontroller

Sound

ISAbridge

SCSI

scanner

MainmemorySCSI

bus

PCI bus

ISA buscache

Figure 2-30 A typical modern PC with a PCI bus and an ISA

bus The modem and sound card are ISA devices; the SCSI

controller is a PCI device.

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(a) (b)

Electron gun

Grid Screen Spot on screen

Vacuum Vertical

deflection

plate

Horizontal scan

Vertical retrace Horizontal retrace

Figure 2-31 (a) Cross section of a CRT (b) CRT scanning pattern.

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Figure 2-32 (a) The construction of an LCD screen (b) The

grooves on the rear and front plates are perpendicular to one

another.

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Video RAM

Main memory

Video board A2B2C2

Figure 2-33 Terminal output on a personal computer.

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KeyboardSome signals:

Protective ground (1)Transmit (2)

Receive (3)Request to send (4)Clear to send (5)Data set ready (6)Common return (7)Carrier detect (8)Data terminal ready (20)

TerminalABC

Figure 2-34 Connection of an RS-232-C terminal to a

com-puter The numbers in parentheses in the list of signals are the pin numbers.

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Pointer controlled by mouse

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(a) (b)

letter ‘‘A’’ printed with 24 overlapping needles.

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Laser Rotating octagonal

mirror

Drum sprayed and charged

Light beam strikes drum

Toner

Scraper Discharger

Drum

Blank

paper

Heated rollers

Stacked output

Figure 2-37 Operation of a laser printer.

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(a) (b) (c) (d) (e) (f)

Figure 2-38 Halftone dots for various gray scale ranges (a)

0–6 (b) 14–20 (c) 28–34 (d) 56–62 (e) 105–111 (f) 161–167.

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Low amplitude

High frequency frequency Low

Phase change

01001011000100 over a telephone line bit by bit (a)

Two-level signal (b) Amplitude modulation (c) Frequency

modu-lation (d) Phase modumodu-lation.

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Digitalbit pipe

carrier'sinternalnetwork

exchange

Figure 2-40 ISDN for home use.

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