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Chapter 5
Data representation
1
Learning outcomes

By the end of this Chapter you will be able to:
•
Explain how integers are represented in computers using:
•
Explain how fractional numbers are represented in computers
•
•
Unsigned, signed magnitude, excess, and two’s complement notations
Floating point notation (IEEE 754 single format)
•
Calculate the decimal value represented by a binary sequence in:
•
Explain how characters are represented in computers
•
•
•
Unsigned, signed notation, excess, two’s complement, and the IEEE 754
notations.
E.g. using ASCII and Unicode
Explain how colours, images, sound and movies are represented
2
Additional Reading

Essential Reading

Further Reading
• Stalling (2003): Chapter 9
• Brookshear (2003): Chapter 1.4 - 1.7
• Burrell (2004): Chapter 2
• Schneider and Gersting (2004): Chapter 4.2
3
Introduction


In Chapter 3
•
•
•
•
in the main memory,
magnetic memory,
and optical memory
In Chapter 4:
•
•

How information is stored
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
4
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. (later)
5
How does the binary system
work?

It is just like any other system
In decimal system the base is 10

In binary system the base is 2

• We have 10 possible values 0-9
• 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.
6
Example


3040510 = 30000 + 400 + 5
= 3*104+4*102+5*100
101012 =10000+100+1
=1*24+1*22+1*20
7
Examples: decimal -- binary


Find the binary representation of 12910.
Find the decimal value represented by
the following binary representations:
• 10000011
• 10101010
8
Examples: Representing
fractional numbers

Find the binary representations of:

Using only 8 binary digits find the binary
representations of:
• 0.510 = 0.1
• 0.7510 = 0.11
• 0.210
• 0.810
9
Number representation


Representing whole numbers
Representing fractional numbers
10
Integer Representations
• Unsigned notation
• Signed magnitude notion
• Excess notation
• Two’s complement notation.
11
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
24
23
22
contribution
27
Addition is simple:
20
1 0 0 1 + 0 1 0 1 = 1 1 1 0.
12
Advantages and disadvantages of
unsigned notation

Advantages:

Disadvantages
• One representation of zero
• Simple addition
• Negative numbers can not be represented.
• The need of different notation to represent
negative numbers.
13
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 notation
• Two’s complement notation
14
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.
15
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.
16
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?
17
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.
0 is always performed in programs.
•
Therefore, having two representations of 0 is
inconvenient.
18
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:

Disadvantages:
• Represents positive and negative numbers
• two representations of zero,
• Arithmetic operations are difficult.
19
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
20
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 .
21
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.
22
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)
23
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)
24
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 integer 101012 = 2110
subtracting 25-1 = 24 = 16, we get 21-16 = 510.
So the value represented in excess notation is 510.
25
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.).
26
Exercise 2
1.
•
•
2.
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, 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 ?
27
Excess notation - Summary

In excess notation, the value represented is the unsigned
value with a fixed value subtracted from it.
•


i.e. 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.
28
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
• i.e.
0 is represented as 0000 using 4-bit binary
sequence.
29
Two’s Complement Notation with 4-bits
Binary pattern
0 1 1 1
0 1 1 0
0 1 01
0 1 0 0
0 0 1 1
0 0 1 0
0 0 0 1
0 0 0 0
1 1 1 1
1 1 1 0
1 1 0 1
1 1 0 0
1 0 1 1
1 0 1 0
1 0 0 1
1 0 0 0
Value in 2’s complement.
7
6
5
4
3
2
1
0
-1
-2
-3
-4
-5
-6
-7
-8
30
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
31
Advantages of Two’s
Complement Notation

It is easy to add two numbers.
0 0 0 1 +1
+ 0 1 0 1 +5
1 0 0 0 -8
0+1 0 1 +5
0110
1 1 0 1 -3
+6
• 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
32
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.
33
Example- 10101 in Two’s
Complement



The most significant bit is 1, hence it is a
negative number.
Method 1
• 0101 =
(+5 – 25-1 = 5 – 24 = 5-16 = -11)
Method 2
4
1
-24

+5
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.
34
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.
35
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. Tow’s complements?
36
Fraction Representation

To represent fraction we need other
representations:
• Fixed point representation
• Floating point representation.
37
Fixed-Point Representation
old position
New position
Bit pattern
Contribution
7 6 5 4 3 2 1 0
4 3 2 1 0 -1 -2 -3
1 0 0 1 1. 1 0 1
24
21 20 2-1
2-3 =19.625
Radix-point
38
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.
39
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
40
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
41
Floating point format
 M  B E
Sign
Sign
mantissa or
significand
Exponent
base
exponent
Mantissa
42
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 if of the form 1.f
then the normalised mantissa should be
f, where f is a binary sequence.
43
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
44
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
45
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.
46

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.
= 0.112
.75 * 2 = 1.5
.5 * 2 = 1.0
3. put the two parts together… 110.11
Now normalise
1.1011 *
22
47
IEEE – Example 1
 Convert 6.75 to 32 bit IEEE format.




1.
The Mantissa. The Integer first.
6/2 =3r0
3 / 2 = 1 r 1 = 1102
1/2
=0r1

2. Fraction next.

.75 * 2 = 1.5
.5 * 2 = 1.0



= 0.112
3. put the two parts together…
Now normalise
110.11
1.1011 * 22
48
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
.5 * 2 = 1.0



= 0.112
3. put the two parts together…
Now normalise
110.11
1.1011 * 22
49
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 ....
50
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
Biased
127
Exponent.
-127
{reserved}
-126
-125
-1
0
1
2
127
128
{reserved}
51
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.
52
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
.1000
6.75 in 32 bit floating point IEEE representation:0 10000001 10110000000000000000000
sign(1) exponent(8)
mantissa(23)
53
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
54
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
55
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
56
In comparison

The smallest and largest possible 32-bit
integers in two’s complement are only -232 and
231 - 1
2017/5/4
PITT CS 1621
57
57
Range of numbers
 Normalized (positive range; negative is
symmetric)
smallest
00000000100000000000000000000000
+2-126× (1+0) = 2-126
largest
01111111011111111111111111111111
+2127× (2-2-23)
0
2127(2-2-23)
2-126
Positive overflow
Positive underflow
2017/5/4
PITT CS 1621
58
58
Representation in IEEE 754
double precision format

It uses 64 bits
• 1 bit sign
• 11 bit biased exponent
• 52 bit mantissa
Sign
Exponent
Mantissa
59
IEEE 754 double precision
Biased = 1023


11-bit exponent with an excess of 1023.
For example:
•
If the exponent is -1
• we then add 1023 to 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 of
1023.
60
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)
61
61
Floating Point Representation
format (summary)
Sign



Exponent
Mantissa
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.f then the normalised mantissa
should be f, where f is a binary sequence.
62
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

See (table 5.1 p56, study guide) for character representation in
ASCII.
63
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
64
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

Representation convention: null character defines
end of string
• "Hello world"
• Null is sometimes written as '\0'
• Its binary representation is the number 0
65
Layered View of Representation
Text
string
Information
Information
Character
Data
Information
Data
Information
Bit string
Data
Data
Sequence of
characters
66
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.
67
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.
68
exercise

Use the ASCII table to write the ASCII
code for the following:
• CIS110
• 6=2*3
69
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.
70
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 or
bigger than the size of the range.
That is, find x for 2 x => 256
24-bit colour – how many possible colors can be
represented?
•
Hints
• 16 million possible colours (why 16 millions?)
71
24-bits -- the True colour
• 24-bit color is often referred to as the true
•
colour.
Any real-life shade, detected by the naked
eye, will be among the 16 million possible
colours.
72
Example: 2-bit per pixel

4=22 choices
• 00 (off, off)=white
• 01 (off, on)=light
0
grey
10 (on, off)=dark
grey
11 (on, on)=black
0
•
•
=
(white)
=
(light grey)
0
1
=
1
0
=
1
(dark grey)
(black)
1
73
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
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:
•
•
•
2.
3.
1-bit sign
4-bit exponent
6-bit normalised mantissa (significand).
Convert the value represented back to
decimal.
Calculate the relative error of the
representation.
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