©brooks/cole, 2003 chapter 5 computer organization
TRANSCRIPT
©Brooks/Cole, 2003
Chapter 5
ComputerOrganization
©Brooks/Cole, 2003
Figure 5-1
Computer hardware (subsystems)
©Brooks/Cole, 2003
CENTRALCENTRALPROCESSINGPROCESSING
UNITUNIT(CPU)(CPU)
CENTRALCENTRALPROCESSINGPROCESSING
UNITUNIT(CPU)(CPU)
5.15.1
©Brooks/Cole, 2003
Figure 5-2
CPU
©Brooks/Cole, 2003
MAIN MEMORYMAIN MEMORYMAIN MEMORYMAIN MEMORY
5.25.2
©Brooks/Cole, 2003
Table 5.1 Memory unitsTable 5.1 Memory units
UnitUnit------------kilobyte
megabytegigabyteterabytepetabyteexabyte
Exact Number of bytesExact Number of bytes------------------------
210 bytes220 bytes230 bytes240 bytes250 bytes260 bytes
ApproximationApproximation------------103 bytes106 bytes109 bytes1012 bytes1015 bytes1018 bytes
©Brooks/Cole, 2003
Figure 5-3
Main memory
©Brooks/Cole, 2003
Memory addresses are defined usingMemory addresses are defined usingunsigned binary integers. unsigned binary integers.
Note:Note:
©Brooks/Cole, 2003
Example 1Example 1
A computer has 32 MB (megabytes) of memory. How many bits are needed to address any single byte in memory?
SolutionSolution
The memory address space is 32 MB, or 2The memory address space is 32 MB, or 22525 (2 (255 x x 222020). This means you need). This means you needloglog22 2 22525 or 25 bits, to address each byte. or 25 bits, to address each byte.
©Brooks/Cole, 2003
Example 2Example 2
A computer has 128 MB of memory. Each word in this computer is 8 bytes. How many bits are needed to address any single word in memory?
SolutionSolution
The memory address space is 128 MB, which The memory address space is 128 MB, which means 2means 22727. However, each word is 8 (2. However, each word is 8 (233) bytes, ) bytes, which means that you have 2which means that you have 22424 words. This words. This means you need logmeans you need log22 2 22424 or 24 bits, to address or 24 bits, to address
each word.each word.
©Brooks/Cole, 2003
Figure 5-4
Memory hierarchy
©Brooks/Cole, 2003
Figure 5-5
Cache
©Brooks/Cole, 2003
INPUT / OUTPUTINPUT / OUTPUTINPUT / OUTPUTINPUT / OUTPUT
5.35.3
©Brooks/Cole, 2003
Figure 5-6
Physical layout of a magnetic disk
©Brooks/Cole, 2003
Figure 5-7
Surface organization of a disk
©Brooks/Cole, 2003
Figure 5-8
Mechanical configuration of a tape
©Brooks/Cole, 2003
Figure 5-9
Surface organization of a tape
©Brooks/Cole, 2003
Table 5.2 CD-ROM speedsTable 5.2 CD-ROM speeds
SpeedSpeed------------
1x2x4x6x8x12x16x24x32x40x
Data RateData Rate------------------------
153,600 bytes per second307,200 bytes per second614,400 bytes per second921,600 bytes per second1,228,800 bytes per second1,843,200 bytes per second 2,457,600 bytes per second3,688,400 bytes per second 4,915,200 bytes per second6,144,000 bytes per second
ApproximationApproximation------------150 KB/s300 KB/s600 KB/s900 KB/s1.2 MB/s1.8 MB/s2.4 MB/s3.6 MB/s4.8 MB/s6 MB/s
©Brooks/Cole, 2003
Figure 5-12
Making a CD-R
©Brooks/Cole, 2003
Figure 5-13
Making a CD-RW
©Brooks/Cole, 2003
Table 5.3 DVD capacitiesTable 5.3 DVD capacities
FeatureFeature---------------------------------single-sided, single-layersingle-sided, dual-layer
double-sided, single-layerdouble-sided, dual-layer
CapacityCapacity------------
4.7 GB8.5 GB9.4 GB17 GB
©Brooks/Cole, 2003
SUBSYSTEMSUBSYSTEMINTERCONNECTIONINTERCONNECTION
SUBSYSTEMSUBSYSTEMINTERCONNECTIONINTERCONNECTION
5.45.4
©Brooks/Cole, 2003
Figure 5-14
Connecting CPU and memory using three buses
©Brooks/Cole, 2003
Figure 5-15
Connecting I/O devices to the buses
©Brooks/Cole, 2003
PROGRAMPROGRAMEXECUTIONEXECUTIONPROGRAMPROGRAM
EXECUTIONEXECUTION
5.55.5
©Brooks/Cole, 2003
Figure 5-21
Steps of a cycle
©Brooks/Cole, 2003
Figure 5-22
Contents of memory and register before execution
©Brooks/Cole, 2003
Figure 5-23.a
Contents of memory and registers after each cycle
©Brooks/Cole, 2003
Figure 5-23.b
Contents of memory and registers after each cycle
©Brooks/Cole, 2003
Figure 5-23.c
Contents of memory and registers after each cycle
©Brooks/Cole, 2003
Figure 5-23.d
Contents of memory and registers after each cycle