1. Chapter 5
Data representation

2. 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 two’s 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, two’s 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

3. Additional Reading Essential Reading Further Reading
Stalling (2003): Chapter 9
Further Reading
Brookshear (2003): Chapter
Burrell (2004): Chapter 2
Schneider and Gersting (2004): Chapter 4.2

4. Introduction In Chapter 5 In Chapter 3 In Chapter 4:
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

5. 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 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)

6. 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.

7. Example =
= 3*104+4*102+5*100 =
=1*24+1*22+1*20

8. Examples: decimal -- binary
Find the binary representation of
Find the decimal value represented by the following binary representations:

9. Examples: Representing fractional numbers
Find the binary representations of:
0.510 = 0.1
= 0.11
Using only 8 binary digits find the binary representations of:
0.210
0.810

10. Number representation
Representing whole numbers
Representing fractional numbers

11. Integer Representations
Unsigned notation
Signed magnitude notion
Excess notation
Two’s complement notation.

12. Unsigned Representation
Represents positive integers.
Unsigned representation of 157:
Addition is simple: =
position 7 6 5 4 3 2 1
Bit pattern
contribution 27 24 23 22 20

13. Advantages and disadvantages of unsigned notation
One representation of zero
Simple addition
Disadvantages
Negative numbers can not be represented.
The need of different notation to represent negative numbers.

14. 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

15. 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.

16. Example - Suppose 10011101 is a signed magnitude representation.
The sign bit is 1, then the number represented is negative
The magnitude is with a value = 29
Then the number represented by is –29.
position 7 6 5 4 3 2 1
Bit pattern
contribution - 24 23 22 20

17. Exercise 1
3710 has in signed magnitude notation. Find the signed magnitude of –3710 ?
Using the signed magnitude notation find the 8-bit binary representation of the decimal value 2410 and
Find the signed magnitude of –63 using 8-bit binary sequence?

18. 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 = =
To test if a number is 0 or not, the CPU will need to see whether it is or
0 is always performed in programs.
Therefore, having two representations of 0 is inconvenient.

19. Signed-Summary In signed magnitude notation, Advantages:
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.

20. 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

21. Excess Notation with n bits
1000…0 represent 2n-1 is the decimal value in unsigned notation.
Therefore, in excess notation:
1000…0 will represent 0 .
Decimal value
In unsigned
notation
Decimal value
In excess
notation
- 2n-1 =

22. Example (1) - excess to decimal
Find the decimal number represented by in excess notation.
Unsigned value
= = = 15310
Excess value:
excess value = 153 – 27 = 152 – 128 = 25.

23. Example (2) - decimal to excess
Represent the decimal value 24 in 8-bit excess notation.
We first add, 28-1, the fixed value
= = 152
then, find the unsigned value of 152
15210 = (unsigned notation).
= (excess notation)

24. example (3) Represent the decimal value -24 in 8-bit excess notation.
We first add, 28-1, the fixed value
= = 104
then, find the unsigned value of 104
10410 = (unsigned notation).
-2410 = (excess notation)

25. Example (4) (10101) Unsigned Sign Magnitude Excess notation
= = 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 = 510
So the value represented in signed magnitude is -510
Excess notation
As an unsigned binary integer = 2110
subtracting = = 16, we get = 510.
So the value represented in excess notation is 510.

26. 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.).

27. Exercise 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 ?

28. 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.

29. 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.

30. Two’s Complement Notation with 4-bits
Binary pattern Value in 2’s complement.

31. 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.

32. Advantages of Two’s Complement Notation
It is easy to add two numbers. + +
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

33. 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.

34. 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 =
Method 3
Corresponding + number is = = 11
the result is then –11.

35. 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.

36. Exercise
Determine the decimal value represented by in each of the following four systems.
Unsigned notation?
Signed magnitude notation?
Excess notation?
Tow’s complements?

37. Fraction Representation
To represent fraction we need other representations:
Fixed point representation
Floating point representation.

38. Fixed-Point Representation
old position
New position
Bit pattern
Contribution =19.625
Radix-point

39. 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.

40. 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

41. Base Mantissa Exponent Sign
Floating Point
could be represented as
Base
Mantissa
Exponent
Sign
- 159 * 1014
When we deal with very large and very small numbers we often resort to using scientific notation.
Calculators, for example, often represent the results of large calculations this way because there is an insufficient number of digits to show the result normally
159 is called the mantissa 14 the exponent which is the number of places to move the decimal point.
* 1015
* 1016
A calculator might display 159 E14

42. Floating point format  M  B E Sign mantissa or base exponent
significand
Sign
Exponent
Mantissa

43. Floating Point Representation format
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.
Sign
Exponent
Mantissa

44. The number will occupy 32 bits
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

45. 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

46. 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.

47. IEEE – Example 1 Convert 6.75 to 32 bit IEEE format. = 1102 = 0.112
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…
Now normalise * 22
= 1102
= 0.112

48. IEEE – Example 1 Convert 6.75 to 32 bit IEEE format. = 1102 = 0.112
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…
Now normalise * 22
= 1102
= 0.112

49. IEEE – Example 1 Convert 6.75 to 32 bit IEEE format. = 1102 = 0.112
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…
Now normalise * 22
= 1102
= 0.112

50. IEEE Biased 127 Exponent To generate a biased 127 exponent
Take the value of the signed exponent and add 127.
Example.
216 then = and my value for the exponent would be 143 =
So it is simply now an unsigned value ....

51. Possible Representations of an Exponent

52. Why Biased ? The smallest exponent 00000000
Only one exponent zero
The highest exponent is
To increase the exponent by one simply add 1 to the present pattern.

53. Back to the example Our original example revisited…. 1.1011 * 22
Exponent is =129 or 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 is stored .1000
6.75 in 32 bit floating point IEEE representation:-
sign(1) exponent(8) mantissa(23)

54. 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

55. Example (2)
which number does the following IEEE single precision notation represent?
The sign bit is 1, hence it is a negative number.
The exponent is = 12810
It is biased by 127, hence the real exponent is
128 –127 = 1.
The mantissa:
It is normalised, hence the true mantissa is
1.01 =
Finally, the number represented is: x 21 = -2.50 1

56. Single Precision Format
The exponent is formatted using excess-127 notation, with an implied base of 2
Example:
Exponent:
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

57. In comparison
The smallest and largest possible 32-bit integers in two’s complement are only -232 and
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58. Normalized (positive range; negative is symmetric)
Range of numbers
Normalized (positive range; negative is symmetric)
smallest
+2-126× (1+0) = 2-126
largest
+2127× (2-2-23)
2127(2-2-23)
2-126
Positive overflow
Positive underflow
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59. Representation in IEEE 754 double precision format
It uses 64 bits
1 bit sign
11 bit biased exponent
52 bit mantissa
Sign
Exponent
Mantissa

60. 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 = 1022
We then find the binary representation of 1022
Which is
The exponent field will now hold
This means that we just represent -1 with an excess of 1023.

61. IEEE 754 Encoding Single Precision Double Precision Represented Object
Exponent
Fraction
non-zero
+/- denormalized number
1~254
anything
1~2046
+/- floating-point numbers
255
2047
+/- infinity
NaN (Not a Number) 61 61

62. 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.f then the normalised mantissa should be f, where f is a binary sequence.
Sign
Exponent
Mantissa

63. 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.

64. ASCII – example Symbol decimal Binary 7 55 00110111 8 56 00111000 : ;
< =
> ? @ A B C

65. 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 defines end 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'

66. Layered View of Representation
Text string
Information
Data
Sequence of
characters
Information
Data
Character
Bit string

67. 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.
NEED TO PROVIDE PRINTED ASCII TABLES

68. Solution SI
‘S’ ‘I’ ‘\0’
The colors are intended to help you read it; computers don’t care that all the bits run together.

69. exercise
Use the ASCII table to write the ASCII code for the following:
CIS110
6=2*3

70. 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.

71. 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?)

72. 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.

73. Example: 2-bit per pixel=
(white) 1
(dark grey)
(light grey)
(black)
4=22 choices
00 (off, off)=white
01 (off, on)=light grey
10 (on, off)=dark grey
11 (on, on)=black

74. 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.

75. 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

76. Representing Sound Graphically
X axis: time
Y axis: pressure
A: amplitude (volume)
: wavelength (inverse of frequency = 1/)

77. 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.

78. Digitizing Sound Capture amplitude at these points
Lose all variation between data points
Zoomed Low Frequency Signal

79. Summary Integer representation Fraction 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

80. Exercise
Represent +0.8 in the following floating-point representation:
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.