chapter 5 data representation

7/27/2019 Chapter 5 Data Representation

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Chapter 5

Data representation


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Learning outcomes

By the end of this Chapter you will be able to: Explain how integers are represented in computers using:

Unsigned, signed magnitude, excess, and twos complement notations

Explain how fractional numbers are represented in computers Floating point notation (IEEE 754 single format)

Calculate the decimal value represented by a binary sequence in: Unsigned, signed notation, excess, twos complement, and the IEEE 754

notations.

Explain how characters are represented in computers E.g. using ASCII and Unicode

Explain how colours, images, sound and movies are represented


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Additional Reading

Essential Reading

Stalling (2003): Chapter 9 Further Reading

Brookshear (2003): Chapter 1.4 - 1.7

Burrell (2004): Chapter 2 Schneider and Gersting (2004): Chapter 4.2


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Introduction

In Chapter 3 How information is stored

in the main memory, magnetic memory,

and optical memory

In Chapter 4: We studied how information is processed in computers How the CPU executes instructions

In Chapter 5 We will be looking at how data is represented in computers Integer and fractional number representation Characters, colours and sounds representation


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Binary numbers

Binary number is simply a number comprised of only 0's and1's.

Computers use binary numbers because it's easy for them tocommunicate using electrical current -- 0 is off, 1 is on.

You can express any base 10 (our number system -- whereeach digit is between 0 and 9) with binary numbers.

The need to translate decimal number to binary and binary todecimal.

There are many ways in representing decimal number in binarynumbers. (later)


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How does the binary system

work?

It is just like any other system

In decimal system the base is 10

We have 10 possible values 0-9 In binary system the base is 2

We have only two possible values 0 or 1. The same as in any base, 0 digit has non

contribution, where 1s has contribution whichdepends at there position.


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Example

3040510 = 30000 + 400 + 5

= 3*104+4*102+5*100

101012 =10000+100+1

=1*24+1*22+1*20


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Examples: decimal -- binary

Find the binary representation of 12910.

Find the decimal value represented by

the following binary representations: 10000011 10101010


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Examples: Representing

fractional numbers

Find the binary representations of:

0.510 = 0.1

0.7510 = 0.11 Using only 8 binary digits find the binary

representations of:

0.210 0.810


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Number representation

Representing whole numbers

Representing fractional numbers


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Integer Representations

Unsigned notation Signed magnitude notion

Excess notation Twos complement notation.


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Unsigned Representation

Represents positive integers.

Unsigned representation of 157:

Addition is simple:

1 0 0 1 + 0 1 0 1 = 1 1 1 0.

position 7 6 5 4 3 2 1 0

Bit pattern 1 0 0 1 1 1 0 1

contribution 27 24 23 22 20


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Advantages and disadvantages of

unsigned notation

Advantages:

One representation of zero

Simple addition Disadvantages

Negative numbers can not be represented.

The need of different notation to representnegative numbers.


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Representation of negative

numbers

Is a representation of negative numberspossible?

Unfortunately:

you can not just stick a negative sign in front of a binarynumber. (it does not work like that)

There are three methods used to representnegative numbers.

Signed magnitude notation Excess notation notation Twos complement notation


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Signed Magnitude

Representation

Unsigned: - and +are the same.

In signed magnitude

the left-most bit represents the sign of the integer. 0 for positive numbers. 1 for negative numbers.

The remaining bits represent to

magnitude of the numbers.


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Example

Suppose 10011101 is a signed magnitude representation.

The sign bit is 1, then the number represented is negative

The magnitude is 0011101 with a value 24+23+22+20= 29

Then the number represented by 10011101 is29.

position 7 6 5 4 3 2 1 0

Bit pattern 1 0 0 1 1 1 0 1

contribution - 24 23 22 20


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Exercise 1

1. 3710has 0010 0101 in signed magnitude notation. Findthe signed magnitude of3710?

2. Using the signed magnitude notation find the 8-bitbinary representation of the decimal value 2410and -2410.

3. Find the signed magnitude of63 using 8-bit binary

sequence?


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Disadvantage of Signed

Magnitude

Addition and subtractions are difficult.

Signs and magnitude, both have to carry out

the required operation. They are two representations of 0

00000000 = + 010 10000000 = - 010

To test if a number is 0 or not, the CPU will need to seewhether it is 00000000 or10000000. 0 is always performed in programs.

Therefore, having two representations of0 isinconvenient.


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Signed-Summary

In signed magnitude notation, The most significant bit is used to represent the sign. 1 represents negative numbers

0 represents positive numbers. The unsigned value of the remaining bits represent Themagnitude.

Advantages: Represents positive and negative numbers

Disadvantages: two representations of zero, Arithmetic operations are difficult.


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Excess Notation

In excess notation:

The value represented is the unsigned value with a fixedvalue subtracted from it.

Forn-bit binary sequences the value subtracted fixed value is2(n-1).

Most significant bit: 0 for negative numbers 1 for positive numbers


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Excess Notation with n bits

10000 represent 2n-1is the decimal value in

unsigned notation.

Therefore, in excess notation:

10000 will represent 0 .

Decimal value

In unsigned

notation

Decimal value

In excess

notation- 2n-1 =


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Example (1) - excess to decimal

Find the decimal number represented by

10011001 in excess notation.

Unsigned value 100110002 = 27 +24 + 23 +20 = 128 + 16 +8 +1 = 15310 Excess value:

excess value = 153 27 = 152 128 = 25.


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Example (2) - decimal to excess

Represent the decimal value 24 in 8-bit

excess notation.

We first add, 28-1, the fixed value 24 + 28-1 = 24 + 128= 152

then, find the unsigned value of 152

15210 = 10011000 (unsigned notation). 2410 = 10011000 (excess notation)


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example (3)

Represent the decimal value -24 in 8-bit

excess notation.

We first add, 2

8-1

, the fixed value -24 + 28-1 = -24 + 128= 104

then, find the unsigned value of104

10410 = 01101000 (unsigned notation).

-2410 = 01101000 (excess notation)


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Example (4) (10101)

Unsigned 101012 = 16+4+1 = 2110 The value represented in unsigned notation is 21

Sign Magnitude

The sign bit is 1, so the sign is negative The magnitude is the unsigned value 01012 = 510 So the value represented in signed magnitude is -510

Excess notation As an unsigned binary integer101012 = 2110 subtracting 25-1 = 24 = 16, we get 21-16 = 510.

So the value represented in excess notation is 510.


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Advantages of Excess Notation

It can represent positive and negative integers.

There is only one representation for 0.

It is easy to compare two numbers.

When comparing the bits can be treated as unsignedintegers.

Excess notation is not normally used to represent

integers.

It is mainly used in floating point representation forrepresenting fractions (later floating point rep.).


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Exercise 2

1. Find 10011001 is an 8-bit binary sequence. Find the decimal value it represents if it was in unsigned

and signed magnitude.

Suppose this representation is excess notation, findthe decimal value it represents?

2. Using 8-bit binary sequence notation, find the

unsigned, signed magnitude and excess

notation of the decimal value 1110 ?


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Excess notation - Summary

In excess notation, the value represented is the unsignedvalue with a fixed value subtracted from it.

i.e. forn-bit binary sequences the value subtracted is 2(n-1).

Most significant bit:

0 for negative numbers . 1 positive numbers.

Advantages: Only one representation of zero. Easy for comparison.


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Twos Complement Notation

The most used representation for

integers.

All positive numbers begin with 0.All negative numbers begin with 1. One representation of zero

i.e. 0 is represented as 0000 using 4-bit binarysequence.


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Twos Complement Notation with 4-bits

Binary pattern Value in 2s complement.

0 1 1 1 7

0 1 1 0 6

0 1 0 1 5

0 1 0 0 40 0 1 1 3

0 0 1 0 2

0 0 0 1 1

0 0 0 0 0

1 1 1 1 -1

1 1 1 0 -2

1 1 0 1 -3

1 1 0 0 -4

1 0 1 1 -5

1 0 1 0 -6

1 0 0 1 -71 0 0 0 -8


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Properties of Twos Complement

Notation

Positive numbers begin with 0

Negative numbers begin with 1

Only one representation of 0, i.e. 0000 Relationship between +n andn.

0 1 0 0 +4 0 0 0 1 0 0 1 0 +18

1 1 0 0 -4 1 1 1 0 1 1 1 0 -18


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Advantages of Twos

Complement Notation

It is easy to add two numbers.0 0 0 1 +1 1 0 0 0 -8

0 1 0 1 +5 0 1 0 1 +5+ +

0 1 1 0 +6 1 1 0 1 -3

Subtraction can be easily performed.

Multiplication is just a repeated addition. Division is just a repeated subtraction

Twos complement is widely used in ALU


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Evaluating numbers in twos

complement notation

Sign bit = 0, the number is positive. The value is determined in

the usual way.

Sign bit = 1, the number is negative. three methods can be

used:

Method 1 decimal value of(n-1) bits, then subtract 2n-1

Method 2 - 2n-1 is the contribution of the sign bit.

Method 3 Binary rep. of the corresponding positive number.

Let V be its decimal value.

- V is the required value.


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Example- 10101 in Twos

Complement

The most significant bit is 1, hence it is anegative number.

Method 1

0101 = +5 (+5 25-1 = 5 24 = 5-16 = -11) Method 2

4 3 2 1 0

1 0 1 0 1

-24 22 20 = -11

Method 3 Corresponding + number is 01011 = 8 + 2+1 = 11

the result is then11.


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Twos complement-summary

In twos complement the most significant for an n-bit number

has a contribution of2(n-1).

One representation of zero

All arithmetic operations can be performed by using additionand inversion.

The most significant bit: 0 for positive and 1 for negative.

Three methods can the decimal value of a negative number:

Method 1 decimal value of(n-1) bits, then subtract 2n-1

Method 2 - 2n-1 is the contribution of the sign bit.

Method 3 Binary rep. of the corresponding positive number.

Let V be its decimal value.

- V is the required value.


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Exercise - 10001011

Determine the decimal value

represented by 10001011 in each of

the following four systems.1. Unsigned notation?2. Signed magnitude notation?3. Excess notation?

4. Tows complements?


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Fraction Representation

To represent fraction we need other

representations:

Fixed point representation Floating point representation.


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Fixed-Point Representation

old position 7 6 5 4 3 2 1 0

New position 4 3 2 1 0 -1 -2 -3

Bit pattern 1 0 0 1 1 . 1 0 1

Contribution 24 21 20 2-1 2-3 =19.625

Radix-point


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Limitation of Fixed-Point

Representation

To represent large numbers or very

small numbers we need a very long

sequences of bits.

This is because we have to give bits to

both the integer part and the fractionpart.


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Floating Point Representation

In decimal notation we can get around thisproblem using scientific notation orfloatingpoint notation.

Number Scientific notation Floating-point

notation

1,245,000,000,000 1.245 1012 0.1245 1013

0.0000001245 1.245 10-7 0.1245 10-6

-0.0000001245 -1.245 10-7 -0.1245 10-6


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Floating Point

-15900000000000000

could be represented as

- 15.9 * 1015

- 1.59 * 1016A calculator might display 159 E14

- 159 * 1014Sign Exponent

BaseMantissa


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M B E

Sign mantissa or base exponentsignificand

Floating point format

Sign Exponent Mantissa


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format

The exponent is biased by a fixed valueb, called the bias.

The mantissa should be normalised,e.g. if the real mantissa if of the form 1.f

then the normalised mantissa should bef, where fis a binary sequence.



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IEEE 745 Single Precision

The number will occupy 32 bits

The first bit represents the sign of the number;

1= negative 0= positive.

The next 8 bits will specify the exponent stored inbiased 127 form.

The remaining 23 bits will carry the mantissanormalised to be between 1 and 2.

i.e. 1


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Representation in IEEE 754

single precision

sign bit:

0 for positive and,

1 for negative numbers 8 biased exponent by 127

23 bit normalised mantissa



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Basic Conversion

Converting a decimal number to a floating point

number.

1.Take the integer part of the number and generate

the binary equivalent. 2.Take the fractional part and generate a binary

fraction

3.Then place the two parts together and normalise.


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IEEEExample 1

Convert 6.75 to 32 bit IEEE format.

1. The Mantissa. The Integer first. 6 / 2 = 3 r 0

3 / 2 = 1 r 1 1 / 2 = 0 r 1

2. Fraction next.

.75 * 2 = 1.5

.5 * 2 = 1.0 3. put the two parts together 110.11

Now normalise 1.1011 * 22

= 1102

= 0.112


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= 0.112

IEEEExample 1

Convert 6.75 to 32 bit IEEE format.


3 / 2 = 1 r 1 1 / 2 = 0 r 1

2. Fraction next. .75 * 2 = 1.5

.5 * 2 = 1.0

3. put the two parts together 110.11 Now normalise 1.1011 * 22

= 1102


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IEEEExample 1 Convert 6.75 to 32 bit IEEE format.


3 / 2 = 1 r 1 1 / 2 = 0 r 1

2. Fraction next. .75 * 2 = 1.5

.5 * 2 = 1.0

3. put the two parts together 110.11 Now normalise 1.1011 * 22

= 1102

= 0.112


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IEEE Biased 127 Exponent

To generate a biased 127 exponent

Take the value of the signed exponent and add 127.

Example.

216 then 2127+16 = 2143 and my value for the

exponent would be 143 = 100011112

So it is simply now an unsigned value ....


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Possible Representations of an

Exponent

Binary Sign Magnitude 2's

ComplementBiased127

Exponent.

00000000 0 0 -127{reserved}

00000001 1 1 -126

00000010 2 2 -125

01111110 126 126 -1

01111111 127 127 0

10000000 -0 -128 1

10000001 -1 -127 2

11111110 -126 -2 127

11111111 -127 -1 128{reserved}


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Why Biased ?

The smallest exponent 00000000

Only one exponent zero 01111111

The highest exponent is 11111111

To increase the exponent by one simply add 1 to the

present pattern.


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Back to the example

Our original example revisited. 1.1011 * 22

Exponent is 2+127 =129 or10000001 in binary.

NOTE:Mantissa always ends up with a value of 1

before the Dot. This is a waste of storage therefore itis implied but not actually stored. 1.1000 is stored

.1000

6.75 in 32 bit floating point IEEE representation:- 0 1000000110110000000000000000000

sign(1) exponent(8) mantissa(23)


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single precision

sign bit:

0 for positive and,

1 for negative numbers 8 biased exponent by 127

23 bit normalised mantissa



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Example (2)

which number does the following IEEE singleprecision notation represent?

The sign bit is 1, hence it is a negative number.

The exponent is 1000 0000 = 12810 It is biased by 127, hence the real exponent is

128127 = 1.

The mantissa: 0100 0000 0000 0000 0000 000. It is normalised, hence the true mantissa is

1.01 = 1.2510

Finally, the number represented is: -1.25 x 21 = -2.50

1 1000 0000 0100 0000 0000 0000 0000 000


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Single Precision Format

The exponent is formatted using excess-127 notation, with animplied base of 2 Example:

Exponent: 10000111 Representation: 135 127 = 8

The stored values 0 and 255 of the exponent are used toindicate special values, the exponential range is restricted to

2-126 to 2127

The number 0.0 is defined by a mantissa of 0 together with thespecial exponential value 0

The standard allows also values +/- (represented as mantissa

+/-0 and exponent 255 Allows various other special conditions


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In comparison

The smallest and largest possible 32-bit

integers in twos complement are only-

232 and231- 1

2013/7/28 PITT CS 1621 57


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Range of numbers

Normalized (positive range; negative is

symmetric)

00000000100000000000000000000000 +2-126(1+0) = 2-126

01111111011111111111111111111111+2127(2-2-23)

smallest

largest

0 2-1262127(2-2-23)

Positive underflow Positive overflow

2013/7/28 PITT CS 1621 58



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double precision format

It uses 64 bits

1 bit sign

11 bit biased exponent 52 bit mantissa



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IEEE 754 double precision

Biased = 1023

11-bit exponent with an excess of1023.

For example:

If the exponent is -1 we then add 1023to it. -1+1023 = 1022 We then find the binary representation of 1022

Which is 0111 1111 110 The exponent field will now hold 0111 1111 110

This means that we just represent -1 with an excess of1023.


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IEEE 754 Encoding

Single Precision Double Precision Represented Object

Exponent Fraction Exponent Fraction

0 0 0 0 0

0 non-zero 0 non-zero+/- denormalized

number

1~254 anything 1~2046 anything+/- floating-point

numbers

255 0 2047 0 +/- infinity

255 non-zero 2047 non-zero NaN (Not a Number)


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format (summary)

the sign bit represents the sign

0 for positive numbers 1 for negative numbers

The exponent is biased by a fixed value b, called the bias.

The mantissa should be normalised, e.g. if the real

mantissa if of the form 1.fthen the normalised mantissashould be f, where fis a binary sequence.



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Character representation- ASCII

ASCII (American Standard Code for Information Interchange)

It is the scheme used to represent characters.

Each character is represented using 7-bit binary code.

If8-bits are used, the first bit is always set to 0

See (table 5.1 p56, study guide) for character representation inASCII.


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ASCIIexample

Symbol decimal Binary

7 55 00110111

8 56 00111000

9 57 00111001: 58 00111010

; 59 00111011

< 60 00111100

= 61 00111101

> 62 00111110

? 63 00111111

@ 64 01000000

A 65 01000001

B 66 01000010

C 67 01000011


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Character strings

How to represent character strings?

A collection of adjacent words (bit-string units)can store a sequence of letters

Notation: enclose strings in double quotes

"Hello world"

Representation convention: null character definesend of string Null is sometimes written as '\0' Its binary representation is the number 0

'H' 'e' 'l' 'l' o' ' ' 'W' 'o' 'r' 'l' 'd' '\0'


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Layered View of Representation

Text

string

Sequence of

characters

Character

Bit string

Information

Data

Information

Data

Information

Data

Information

Data


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Working With A Layered

View of Representation

Represent SI at the two layers shown on theprevious slide.

Representation schemes:

Top layer - Character string to charactersequence:Write each letter separately, enclosed in quotes. Endstring with \0.

Bottom layer - Character to bit-string:Represent a character using the binary equivalentaccording to the ASCII table provided.


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Solution

SI S I \0

010100110100100000000000

The colors are intended to help you read it; computers dont care that all the bitsrun together.


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exercise

Use the ASCII table to write the ASCII

code for the following:

CIS110 6=2*3


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Unicode - representation

ASCII code can represent only 128 = 27 characters.

It only represents the English Alphabet plus some controlcharacters.

Unicode is designed to represent the worldwideinterchange.

It uses 16 bits and can represents 32,768 characters.

For compatibility, the first 128 Unicode are the same asthe one of the ASCII.


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Colour representation

Colours can represented using a sequence of bits.

256 colours how many bits?

Hint for calculating To figure out how many bits are needed to represent a range

of values, figure out the smallest power of 2 that is equal to orbigger than the size of the range.

That is, find xfor2 x => 256

24-bit colour how many possible colors can berepresented?

Hints 16 million possible colours (why 16 millions?)


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24-bits -- the True colour

24-bit color is often referred to as the truecolour.

Any real-life shade, detected by the nakedeye, will be among the 16 million possiblecolours.


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Example: 2-bit per pixel

4=22 choices

00 (off, off)=white

01 (off, on)=lightgrey

10 (on, off)=darkgrey

11 (on, on)=black

00

= (white)

10

= (dark grey)

10

= (light grey)

11

= (black)


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Image representation

An image can be divided into many tiny squares, called

pixels.

Each pixel has a particular colour.

The quality of the picture depends on two factors: the density of pixels. The length of the word representing colours.

The resolution of an image is the density of pixels.

The higher the resolution the more informationinformation the image contains.


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Bitmap Images

Each individual pixel(pi(x)cture element) in a

graphic stored as a binary number

Pixel: A small area with associated coordinate location

Example: each point below is represented by a 4-bitcode corresponding to 1 of 16 shades of gray


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Representing Sound Graphically

X axis: time

Y axis: pressure

A: amplitude (volume)

: wavelength (inverse of frequency = 1/)


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Sampling

Sampling is a method used to digitisesound waves.

A sample is the measurement of theamplitude at a point in time.

The quality of the sound depends on: The sampling rate, the faster the better

The size of the word used to represent asample.


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Digitizing Sound

Zoomed Low Frequency Signal

Capture amplitude at

these points

Lose all variation

between data points


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Summary

Integer representation

Unsigned, Signed, Excess notation, and Twos complement.

Fraction representation

Floating point (IEEE 754 format ) Single and double precision

Character representation

Colour representation

Sound representation


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Exercise

1. Represent +0.8 in the following floating-pointrepresentation: 1-bit sign

4-bit exponent 6-bit normalised mantissa (significand).2. Convert the value represented back to

decimal.

3. Calculate the relative error of therepresentation.

chapter 5 data representation

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