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2 November 2010 C.Ravindra Murthy 1

Computer Organization

C.Ravindra Murthy

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Data Types Information that a computer is dealing with:

Data

Numeric Data Integer, real

Non-Numeric Data Letters, Symbols

Relationship between data elements Data Structures

Linear Lists, Trees, Queues, etc.,

Programs Instructions

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Data Types: Numeric Data

Representation Non-positional number System

Roman number system

Positional number System Each digit position has a value called a weight associated with it

Examples: Decimal, Octal, Hexadecimal, Binary

Base (or radix) R number Uses R distinct symbols for each digit

Example AR = an-1, an-2, .. a1, a0, a-1, a-m V(AR) = SUM (ak* R

k) for k = -m to n-1 R = 10 Decimal number System

R = 2 Binary

R = 8 Octal

R = 16 Hexadecimal

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Representation Why Positional Number System for Digital

Computers??

Major consideration is the COST and TIME Cost of building hardware

ALU, CPU and Communications

Time to processing

Arithmetic Addition of Numbers Table foraddition Non-positional Number System

Table for addition is infinite Impossible to build, very expensive even if it can be built

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Representation Positional Number System

Table for addition is finite

Physically realizable, but cost wise the smaller thetable size, the less expensive

Binary is favorable to Decimal

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Positive (unsigned) Binary

Numbers Unsigned binary numbers are typically used to represent

computer addresses or other values that are guaranteednot to be negative

An n-bit unsigned binary integer A = an-1, an-2, .. a1, a0 hasa value of

For example1011 = 1 x 23 + 0 x 22 + 1 x 21 + 1 x 20

= 8 + 2 + 1 = 11

An n-bit unsigned binary integer has a range from 0 to 2n - 1

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Octal and Hexadecimal Numbers Octal, base 8 numbers were used in the early days of computing to

represent binary numbers

Octal numbers are made by grouping binary numbers together

three bits at a time Hexadecimal base 16 numbers are representation of choice today

Hexadecimal numbers are made by grouping binary numberstogether four bits at a time

For example:

Octal: 7 2 5 1 7 5 2 2 .

Binary: 1 1 1 0 1 0 1 0 1 0 0 1 1 1 1 1 0 1 0 1 0 0 1 0

Hex: E A 9 F 5 2

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Complements of NumbersTwo types of complements for base R number system:

Rs complement (R-1)s complement

The (R-1)s Complement

Subtract each digit of a number from (R-1)

Examples: 9s complement of 83510 is 164101s complement of 10102 is 01012(bit by bit complement operation)

The Rs ComplementAdd 1 to the low-order digit of its (R-1)s complement

Examples: 10s complement of 83510 is 16410 + 1 = 165102s complement of 10102 is 01012 + 1 = 01102

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Negative (Signed) Binary

NumbersOnes complement format

Negative numbers are represented by a bit-by-bit

complementation of the (positive) magnitude (the process of

negation)

Sign bit interpreted as in sign-magnitude format

Examples (8-bit words):

+42 = 0 00101010

-42 = 1 11010101

Min: - (2 n - 2 -m) = 1111 1111 . 1111 1111

Max: + (2 n - 2 -m) = 0111 1111 . 1111 1111

Zero: - 0 = 1111 1111 . 1111 1111

Zero: +0 = 0000 0000 . 0000 0000

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NumbersTwos complement format

Negative numbers, -X, are represented by the pseudo-

positive number: 2n - |X|

An n-bit unsigned binary integer A = an-1 an-2... a1 a0 has avalue of

For example: 1011 = -1 x 23 + 0 x 22 + 1 x 21 + 1 x 20

= -8 + 2 + 1 = -5

With 2n digits: 2 n-1 -1 positive numbers

2 n -1 negative numbers

Given the representation for +X, the representation for -X is

found by taking the 1s complement of +X and adding 1

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NumbersTwos complement format

Most significant bit is the sign bit.

Number representation is not symmetric.

Only one representation for zero.Easy to negate, add, and subtract numbers.

A little bit trickier for multiply and divide.

Min: - (2 n) = 1000 0000 . 0000 0000

Max: + (2 n - 2 -m) = 0111 1111 . 1111 1111

Zero: = 0000 0000 . 0000 0000

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Signed 2s Complement AdditionAdd the two numbers, including their sign bit, and discard any carryout of left-most(sign) bit

Examples:

6 0 0110 -6= 1 1010+ 9 0 1001 + 9= 0 1001

15 0 1111 3= 0 0011

6 0 0110 -9 1 0111

+ -9 1 0111 + -9 1 0111

-3 1 1101 -18 (1) 0 1110

9 0 1001

+ 9 0 1001

18 1 0010

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Detecting 2s Complement

OverflowWhen adding two's complement numbers, overflowwill only occur if

the numbers being added have the same sign the sign of the result

is different

Ifwe perform the addition

an-1 an-2 ... a1 a0+ bn-1bn-2 b1 b0----------------------------------

= sn-1sn-2 s1 s0

Overflow can be detected as

where cn-1and cn are the carry in and carry out of the most

significant bit.

1! nn ccV

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Signed 2s Complement

SubtractionTo subtract two's complement numbers we first negate the second

number and then add the corresponding bits of both numbers.

Examples:

3 = 0011 -3 = 1101 -3 = 1101 3 = 0011- 2 = 0010 - -2 = 1110 - 2 = 0010 - -2 = 1110

become:

3 = 0011 -3 = 1101 -3 = 1101 3 = 0011+ -2 = 1110 + 2 = 0010 + -2 = 1110 + 2 = 0010

1 = 0001 -1 = 1111 -5 = 1011 5 = 0101

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Floating Point Number

RepresentationThe location of the fractional point is not fixed to a certain location

--> The range of the representable numbers is wide

--> high precision

F = EM

m n e k e k-1 ... e 0 m n-1 m n-2 ... m 0 . m -1 ... m -m

sign exponent mantissa

MantissaSigned fixed point number, either an integer or a fractional number

Exponent

Designates the position of the radix point

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RepresentationDecimal Value:

V = M * R E

Where: M= Mantissa

E= ExponentR= Radix (10)

Example (decimal):

1234.5678

Exponent Mantissa

Sign Value Sign Value

0 4 0 0.12345678

==> 0.12345678 x 10

+4

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RepresentationExample (binary):

+ 1001.11 (= 9.75)

Make a fractional number, counting the number of shifts:

+ .100111 ==> 4 shifts

Exponent Mantissa

Sign Value Sign Value

0 100 0 1001111

Or for a 16-bit numberwith a sign, 5-bit exponent, 10-bit mantissa:

0 00100 1001111000

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Error Detecting Codes ParityParity System: Simplest method for error detectionOne parity bit attached to the information

Even Parity and Odd Parity

EvenP

arityOne bit is attached to the information so that the total number

of 1 bits is an even number

1011001 0

1010010 1 ==> B even = B n-1 (+) B n-2 (+) B 0

Odd ParityOne bit is attached to the information so that the total number

of 1 bits is an odd number

1011001 1

1010010 0 ==> B odd = B n-1 (+) B n-2 (+) B 0 (+) 1

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