CSc 28 Data representation
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Binary numbers
• Binary number is simply a number comprised of only 0's and
1's.
• Computers use binary numbers because it's easy for them to
communicate using electrical current -- 0 is off, 1 is on.
• You can express any base 10 (our number system -- where
each digit is between 0 and 9) with binary numbers.
• The need to translate decimal number to binary and binary to
decimal.
• There are many ways in representing decimal number in
binary numbers.
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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 1’s has contribution which depends 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
– Two’s complement notation
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Unsigned Representation
• Represents positive integers.
• Unsigned representation of 157
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
• Addition is simple:
1001 + 0101= 1110
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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 represent negative
numbers
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Representation of negative numbers
• Is a representation of negative numbers possible?
• Unfortunately:
– you can not just stick a negative sign in front of a binary number.
(it does not work like that)
• There are three methods used to represent negative
numbers.
– Signed magnitude notation
– Excess notation
– Two’s 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
position
7
6
5
4
3
2
1
0
Bit pattern
1
0
0
1
1
1
0
1
24
23
22
contribution
-
20
• The magnitude is 0011101 with a value 24+23+22+20= 29
• Then the number represented by 10011101 is –29
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Exercise 1
1. 3710 has 0010 0101 in signed magnitude
notation. Find the signed magnitude of –3710 ?
2. Using the signed magnitude notation find the
8-bit binary representation of the decimal value
2410 and -2410.
3. Find the signed magnitude of –63 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
see whether it is 00000000 or 10000000
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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 The
magnitude.
• 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 fixed value subtracted from it.
For n - bit binary sequences the value subtracted fixed
value is 2(n-1)
– Most significant bit:
0 for negative numbers
1 for positive numbers
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Excess Notation with n bits
1000…0 represent 2n-1 is the decimal value in unsigned
notation.
Decimal value
In unsigned
notation
- 2n-1 =
Decimal value
In excess
notation
Therefore, in excess notation:
1000…0 will represent 0 .
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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, 28-1, the fixed value
-24 + 28-1 = -24 + 128= 104
• then, find the unsigned value of 104
10410 = 01101000 (unsigned notation).
-2410 = 01101000 (excess notation)
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Example (4) 101012
• 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 integer 101012 = 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 unsigned
integers.
• Excess notation is not normally used to represent
integers.
• It is mainly used in floating point representation for
representing fractions (later floating point rep.)
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Your turn…
1.
Find 10011001 is an 8-bit binary sequence.
–
–
2.
Find the decimal value it represents if it was in unsigned
and signed magnitude
Suppose this representation is excess notation, find the
decimal value it represents?
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
unsigned value with a fixed value subtracted from it
For n-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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Two’s Complement Notation
The most used representation for integers.
– All positive numbers begin with 0.
– All negative numbers begin with 1.
– One representation of zero
0 is represented as 0000
using 4-bit binary sequence
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Two’s Complement Notation with 4-bits
Binary pattern
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
1
1
1
1
0
0
0
0
Value in 2’s complement.
1 1
1 0
01
0 0
1 1
1 0
0 1
0 0
1 1
1 0
0 1
0 0
1 1
1 0
0 1
0 0
7
6
5
4
3
2
1
0
-1
-2
-3
-4
-5
-6
-7
-8
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Properties of Two’s Complement
Notation
•
•
•
•
Positive numbers begin with 0
Negative numbers begin with 1
Only one representation of 0, i.e. 0000
Relationship between +n and –n
0100
+4
00010010
+18
1100
-4
11101110
-18
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Advantages of Two’s Complement Notation
• It is easy to add two numbers.
+
0 0 0 1 +1
0 1 0 1 +5
1000 -8
+ 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
• Two’s complement is widely used in ALU
(Arithmetic Logic Unit)
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Evaluating numbers in two’s 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 Two’s Complement
The most significant bit is 1, indicating a negative number
Method 1
(+5 – 25-1 = 5 – 24 = 5-16 = -11)
0101 = +5
Method 2
4
1
-24
3
0
2
1
22
1
0
0
1
20
=
-11
Method 3
Corresponding + number is 01011 = 8 + 2+1 = 11
the result is then –11.
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Two’s complement-summary
• In two’s complement the most significant for an n-bit number
has a contribution of –2(n-1)
• One representation of zero
• All arithmetic operations can be performed by using addition
and 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. Two’s 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
New position
Bit pattern
Contribution
7 6 5 4 3
4 3 2 1 0
1 0 0 1 1.
24
21 20
2 1 0
-1 -2 -3
1 0 1
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 fraction part.
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Floating Point Representation
In decimal notation we can get around this
problem using scientific notation or floating
point 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
Mantissa
Sign
Base
- 159 * 1014
- 15.9 * 1015
- 1.59 * 1016
Exponent
A calculator might display 159 E14
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Floating point format
 M  B E
Sign
Sign
mantissa or
significand
base
Exponent
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exponent
Mantissa
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Floating Point Representation format
Sign
Exponent
Mantissa
• The exponent is biased by a fixed value b, called the
bias
• The mantissa should be normalised, e.g. if the real
mantissa is of the form 1.f then the normalised
mantissa should be f, where f is 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 in biased
127 form
The remaining 23 bits will carry the mantissa normalised to
be between 1 and 2.
i.e. 1<= mantissa < 2
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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
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Mantissa
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Basic Conversion
Converting a decimal number to a floating point number
Steps:
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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IEEE – Example 1
Convert 6.75 to 32 bit IEEE format.
1. The Mantissa. The Integer first.
•
•
•
•
•
6/2 =3r0
3/2 =1r1
= 1102
1/2 =0r1
2. Fraction next.
.75 * 2 = 1.5
= 0.112
.5 * 2 = 1.0
3. put the two parts together… 110.11
Now normalise
1.1011 *
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IEEE – Example 1
Convert 6.75 to 32 bit IEEE format.
1. The Mantissa. The Integer first.
6/2 =3r0
3/2 =1r1
= 1102
1/2 =0r1
2. Fraction next.
.75 * 2 = 1.5
= 0.112
.5 * 2 = 1.0
3. put the two parts together… 110.11
Now normalise
1.1011 *
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IEEE – Example 1
Convert 6.75 to 32 bit IEEE format.
1. The Mantissa. The Integer first.
6/2 =3r0
3/2 =1r1
1/2 =0r1
2. Fraction next.
.75 * 2 = 1.5
.5 * 2 = 1.0
= 1102
= 0.112
3. Put the two parts together… = 110.11
Now normalise
1.1011
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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
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
Complement
00000000
0
0
00000001
00000010
01111110
01111111
10000000
10000001
11111110
11111111
1
2
126
127
-0
-1
-126
-127
1
2
126
127
-128
-127
-2
-1
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Biased
127
Exponent.
-127
{reserved}
-126
-125
-1
0
1
2
127
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 or 10000001 in binary
NOTE: Mantissa always ends up with a value of ‘1’ before the
Dot. This is a waste of storage therefore it is implied but not
actually stored. 1.1000 is stored as .1000
6.75 in 32 bit floating point IEEE representation:0 10000001 10110000000000000000000
sign(1) exponent(8)
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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
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Mantissa
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Example (2)
Which number does the following IEEE single
precision notation represent?
1
1000 0000
0100 0000 0000 0000 0000 000
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
128 –127 = 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
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Single Precision Format
• The exponent is formatted using excess - 127
notation, with an implied base of 2
Example:
Exponent:
10000111
Representation: 135 – 127 = 8
• The stored values 0 and 255 of the exponent are used
to indicate 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 the special 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 two’s complement
are only
-232 and 231 - 1
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Range of numbers
Normalized (positive range; negative is symmetric)
smallest
00000000100000000000000000000000
+2-126× (1+0) = 2-126
largest
01111111011111111111111111111111
+2127× (2-2-23)
0
2-126
Positive underflow
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2127(2-2-23)
Positive overflow
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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
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Mantissa
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IEEE 754 double precision
Biased = 1023
• 11-bit exponent with an excess of 1023
• For example:
If the exponent is -1
• Add 1023 to it. -1+1023 = 1022
• Find the binary representation of 1022
– which is 0111 1111 110
• The exponent field will now hold 0111 1111 110
Represent -1 with an excess of 1023
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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)
Sign
Exponent
Mantissa
• Sign bit
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 is of the form 1.M then the normalised
mantissa should be M where M is 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
• If 8-bits are used, the first bit is always set to 0
• Character representation in ASCII
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ASCII – example
Symbol
decimal
7
8
9
:
;
<
=
>
?
@
A
B
C
55
56
57
58
59
60
61
62
63
64
65
66
67
Binary
00110111
00111000
00111001
00111010
00111011
00111100
00111101
00111110
00111111
01000000
01000001
01000010
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
'H' 'e' 'l' 'l' o' ' ' 'W' 'o' 'r' 'l' 'd' '\0'
• Notation: enclose strings in double quotes
– "Hello world"
• Representation convention: null character defines end
of string
– Null is sometimes written as '\0'
– Its binary representation is the number 0
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Layered View of Representation
Text
string
Information
Information
Character
Data
Information
Data
Information
Bit string
Data
Data
Sequence of
characters
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Working With A Layered
View of Representation
• Represent “SI” at the two layers shown on the previous
slide
• Representation schemes:
– Top layer - Character string to character sequence:
Write each letter separately, enclosed in quotes. End string
with ‘\0’
– Bottom layer - Character to bit-string:
Represent a character using the binary equivalent according
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 don’t care
that all the bits run together
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Your turn
Use the ASCII table to write the ASCII code for the
following:
CIS110
6 = 2*3
(1 space before and after the =)
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Unicode - representation
• ASCII code can represent only 128 = 27 characters.
• It only represents the English Alphabet plus some
control characters.
• Unicode is designed to represent the worldwide
interchange.
• It uses 16 bits and can represents 32,768 characters.
• For compatibility, the first 128 Unicode are the same
as the one of the ASCII.
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Color representation
• Colours can represented using a sequence of bits.
• 256 colors – 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 or bigger than the size of the range.
• That is, find x for 2 x => 256
• 24-bit color – how many possible colors can be
represented?
Hints
16 million possible colors (why 16 millions?)
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24-bits -- the True colour
– 24-bit color is often referred to as the true color.
– Any real-life shade, detected by the naked eye, will
be among the 16 million possible colors.
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Example: 2-bit per pixel
0
4=22
choices
00 (off, off)=white
01 (off, on)=light grey
10 (on, off)=dark grey
11 (on, on)=black
=
(white)
=
(light grey)
0
0
1
=
1
0
=
1
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(dark grey)
(black)
1
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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 information 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-bit code
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 digitise sound waves.
• A sample is the measurement of the amplitude 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 a sample.
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Digitizing Sound
Capture amplitude at
these points
Lose all variation
between data points
Zoomed Low Frequency
Signal
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Summary
• Integer representation
–
–
–
–
Unsigned,
Signed,
Excess notation, and
Two’s 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-point
representation:
–
–
–
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 the representation.
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