chapter 5 data representation
TRANSCRIPT
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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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Floating Point Representation
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.
Sign Exponent Mantissa
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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
Sign Exponent 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.
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
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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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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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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
Sign Exponent 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
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Representation in IEEE 754
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Representation in IEEE 754
double precision format
It uses 64 bits
1 bit sign
11 bit biased exponent 52 bit mantissa
Sign Exponent 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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Floating Point Representation
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.
Sign Exponent Mantissa
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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.