Computer Organization and Architecture Tutorial 2 Kenneth Lee.

Computer Organization and Architecture

Tutorial 2

Kenneth Lee

One instruction has 20 bits8 bits for operation code (opcode)12 bits for the address of operand

Binary, Octal, Decimal, Hexadecimal Binary:

Base two Only 2 numbers: 0, 1 0+1=1; 1+1=10

Octal: Base eight 8 numbers: 0, 1, 2, 3, 4, 5, 6, 7 1+2=3; 7+1=10

Decimal: Base ten 9 numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 1+2=3; 9+1=10

Hexadecimal Base sixteen 16 numbers: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F 1+2=3; 9+1=A; E+1=F; F+1=10

• 2.2.a

Load M(2) (Load the contents of memory address 2 to accumulator )

• 2.2.b Fetch: access the memory to fetch the instruction

Execute: access the memory address 2 to load the data

This program will store the absolute value of the contents of memory address 0FA into memory address 0FB

Text book: p 53

(Text book p 56)

Text book p 64

•Memory to processor•Processor to memory•I/O to processor•Processor to I/O•I/O to or from memory

(Text book p 67-68)

•Reduces propagation delay because each bus can be shorter•Reduces bottleneck effects

(Text book p 70-71)

MAR (Memory address register)Specifies the address in the memory for the next read or write

MBR (Memory buffer register)Contains the data to be written into memory or receives the data read from memory

Two registers used most frequentlyAll memory address related operation need go through MARAll data transfer related operation need go through MBR

12 bit address

16 bit memory

16 bit bus

16 bitinstruction

Ps. a 16-bit bus need one cycle to transfer a 8-bit instruction