Nota Utm

1. 1
Chapter 2
Number Systems
i. Decimal, Binary, Hexadecimal, Octal
ii. Number conversions
iii. Binary Arithmetic

2. 2
Digital Number Systems
• Many number systems are used in digital
electronics.
– Decimal number system
– Binary number system
– Octal number system
– Hexadecimal number system

3. 3
Decimal System
• We use decimal numbers everyday
• Base-10 system
• 10 symbols: 0,1,2,3,4,5,6,7,8,9
• Examples: 23, 5, 7, 88.9, 0.2

4. 4
Decimal System
• The position of each digit in a decimal
number can be assigned a weight
• For example: 2745.214
– 2745.214 is a decimal number
– 2 is a digit, 7 is a digit, 4 is a digit…
2 7 4 5 2 1 4.
103 102 101 100 10-110-210-3
weights

5. 5
Decimal System
2745.21410 =
(2x103) + (7x102) + (4x101) + (5x100) + (2x10-1) + (1x10-2)
+ (4x10-3)
2 7 4 5 2 1 4.
103 102 101 100 10-110-210-3
weights

6. 6
Decimal System
• Most significant digit (MSD)—the digit that
carries the most weight
• Least significant digit (LSD)– the digit that
carries the least weight
2 7 4 5 2 1 4.
103 102 101 100 10-110-210-3
weights

7. 7
Decimal Counting
• 0 1 2 3 4 5 6 7 8 9-----
10 11---------------------------- 19---- 20----
------------------------------- 99---- 100------------
------

8. 8
Binary System
• Difficult to design a system that works with
10 different voltage levels
• Base-2 system
• 2 digits/symbols: 0, 1
• Examples: 0, 1, 01, 111, 101010
• Similar to decimal number, the position of
each digit in a binary number can be
assigned a weight.

9. 9
Binary System
• The position of each digit (bit) in a binary
number can be assigned a weight
• For example: 1011.101
– 1011.101 is a binary number
– 1 is a digit, 0 is a digit, 1 is a digit…
1 0 1 1 1 0 1.
23 22 21 20 2-1 2-2 2-3
weights

10. 10
Binary System
1011.1012=(1x23) + (0x22) + (1x21) + (1x20) +
(1x2-1) + (0x2-2) + (1x2-3)
= 8 + 0 + 2 + 1 + 0.5 + 0 + 0.125
= 11.62510
1 0 1 1 1 0 1.
23 22 21 20 2-1 2-2 2-3
weights

11. 11
Binary counting
• 0 1 10 11 100 101 110 111--------
--------- 11111

12. 12
Hexadecimal System
• Base-16 system
• 16 symbols: 10 numeric digits and 6
alphabetic characters
– 0,1,2,3,4,5,6,7,8,9
– A,B,C,D,E,F
• Compact way of writing binary number
• Widely used in computer and
microprocessor applications

13. 13
Hexadecimal System
• Examples: 1C16 , A8516
• 1CH, A85H
• The position of each digit in a hexadecimal
number can be assigned a weight
• For example: 2AF8.98E
– 2AF8.98E is a hexadecimal number
– 2 is a digit, A is a digit, F is a digit…
2 A F 8 9 8 E.
163 162 161 160 16-116-216-3
weights

14. 14
Hexadecimal Counting
• 0 1 2 3 4 5 6 7 8 9 A B C
D E F 10 11 12-------------------
19 1A---------------------------------------
FF 100 101 102------------------

15. 15
Octal System
• Base-8 system
• 8 digits: 0,1,2,3,4,5,6,7
• Convenient way to express binary numbers
and codes. Use 3 bits of binary

16. 16
Table of Number Systems
DECIMAL BINARY HEXADECIMAL OCTAL
0 0000 0 0
1 0001 1 1
2 0010 2 2
3 0011 3 3
4 0100 4 4
5 0101 5 5
6 0110 6 6
7 0111 7 7
8 1000 8 10
9 1001 9 11
10 1010 A 12
11 1011 B 13
12 1100 C 14
13 1101 D 15
14 1110 E 16
15 1111 F 17

17. 17
Review
• Give the largest decimal number that can be
represented in binary with 8 bits.
• Determine the weight of the 1 in 100002.
• Convert the binary number 10111101.0112 to
decimal. Ans:189.37510

18. 18
Conversions
• Binary Decimal
• Decimal Binary
• Hexadecimal Decimal
• Decimal Hexadecimal
• Binary Hexadecimal
• Hexadecimal Binary
• Octal Decimal
• Decimal Octal
• Binary Octal
• Octal Binary

19. 19
BINARY number DECIMAL number
Method 1: Sum-of-Weights
Binary
Decimal
Given 110012
S1: write the weights
1 1 0 0 1Binary number
weights 2021
222324
S2: write the sum of the products of each
digit with its weight
110012 =(1x24)+(1x23)+(0x22)+(0x21)+(1x20)
=16 + 8 + 1
=2510
Binary
Decimal
?
Answer: 2510

20. 20
Decimal Binary Conversion
• Decimal number binary number
– Method: sum-of-weights
• Decimal whole number binary number
– Method: division-by-2
• Decimal fraction binary number
– Method: multiplication-by-2

21. 21
DECIMAL number BINARY number
Method 1: Sum-of-Weights
Decimal
Binary
Given 2510
Step 1: Find the power of two that fulfills the following:
a. nearest to the given decimal number; and
b. its decimal number is less than or equal to the given decimal number.
22? No, because it is not the nearest. 8 (23) is nearer and still less than 25.
25? No, because its decimal number, 32 is more than 25.
24? Yes, because it is the nearest and its decimal, 16 is less than 25.
Step 2: Subtract the power of two (from Step 1) from the given decimal number.
25-16=9
The result of the subtraction is 9.
Step 3: If the result of the subtraction in Step 2 is 0, go to Step 4. Else, repeat Steps
1 and 2 for the result of the subtraction in Step 2.
Step 4: Write out the binary number based on all the powers of two from Step 1.

22. 22
DECIMAL Number BINARY Number
Method 1: Sum-of-Weights
Given 5110
• Step 1: the power of two which is nearest to 51 but less than
51 is 25
• Step 2: the result of subtraction 51-32=19
• Step 3:
– Repeat Step 1: the power of two nearest to 19 but less than 19 is 24
– Repeat Step 2: the result of subtraction 19-16=3
– Repeat Step 1: the power of two which is nearest to 3 and less than 3 is 21
– Repeat Step 2: the result of subtraction 3-2=1
– Repeat Step 1: the power of two which is nearest to 1 and equal to 1is 20
– Repeat Step 2: the result of subtraction 1-1=0
• Step 4:
1 0 0 1 1Binary number
Weights 24 23 22
21 20
1
25

23. 23
DECIMAL Number BINARY Number
Method 2: Repeated Division-by-2
• Method 1 can convert both
whole numbers and
fractional numbers to binary.
• Method 2 is to convert
whole numbers to binary.
• Given 4510
– Repeat the division until the
quotient is 0.
22
2
45
=
Remainder
11
2
22
=
5
2
11
=
2
2
5
=
1
2
2
=
0
2
1
=
Quotient
1
0
1
1
0
1
LSB
MSB
The binary number is 1011012

24. 24
DECIMAL fraction BINARY fraction
Method 3: Repeated Multiplication-by-2
• Method 3 is to convert decimal fraction to
binary.
• Repeat the multiplication until the fractional
part is all zeros.
• Given 0.3125
0.3125 x 2 = 0.625
0.625 x 2 = 1.25
0.25 x 2 = 0.50
0.50 x 2 = 1.00
Carry
0
1
0
1
MSB
LSB
The binary fraction is
.0101

25. 25
DECIMAL Number HEXADECIMAL Number
Method: Repeated division-by-16
• Repeat the division until the quotient is 0.
• Example: 65010
40
16
650
=
2
16
40
=
0
16
2
=
quotient Remainder
(decimal)
Remainder
(hexadecimal)
LSD
MSD
A
8
2
10
8
2
So, the hexadecimal number is 28A16

26. 26
BINARY Number HEXADECIMAL Number
• Step 1: Break the binary number into 4-bit
groups, starting from LSB.
• Step 2: Replace each 4-bit with the equivalent
hexadecimal number.
• Example: 111111000101101001
00111111000101101001binary
hexadecimal 961F3
The hexadecimal number is 3F16916

27. 27
HEXADECIMAL Number BINARY Number
• Step: Replace each digit of the hexadecimal
number with the equivalent 4-bit binary
number.
• Example: CF8E16
C F 8 EHexadecimal
Binary 1110100011111100
The binary number is 11001111100011102

28. 28
DECIMAL Number OCTAL Number
Method: Repeated division-by-8
• Repeat the division until the quotient is 0.
• Example: 35910
44
8
359
=
5
8
44
=
0
8
5
=
quotient Remainder
LSD
MSD
7
4
5
So, the octal number is 5 4 78

29. 29
BINARY Number OCTAL Number
• Step 1: Break the binary number into 3-bit
groups, starting from LSD.
• Step 2: Replace each 3-bit group with the
equivalent octal number.
• Example: 1101012
110101binary
octal 56
The octal number is 658

30. 30
OCTAL Number BINARY Number
• Step: Replace each octal digit with the
equivalent 3-bit group.
• Example: 138
Octal
Binary
1 3
001 011
So, the binary number is 0010112

31. 31
Exercise
• Fill in the blanks.
Decimal Binary Hexadecimal Octal
1101.0112
10101.112
245.62510
70310
A8516

32. 32
Exercise
• Answers
Decimal Binary Hexadecimal Octal
13.37510 1101.0112 D.616 15.38
21.7510 10101.112 15.C16 25.68
245.62510 11110101.1012 F5.A16 365.58
70310 10101111112 2BF16 12778
269310 1010100001012 A8516 52058

33. Binary Arithmetic
• Binary addition
• Binary subtraction
• Binary multiplication
• Binary division

34. Binary addition
• 0 + 0 = 0 with a carry of 0
• 0 + 1 = 1 with a carry of 0
• 1 + 0 = 1 with a carry of 0
• 1 + 1 = 10 with a carry of 1
• Example: 111 + 11 = ?
111 7
+ 11 +3
1010 10
In decimal…

35. Binary subtraction
• 0 – 0 = 0
• 1 – 1 = 0
• 1 – 0 = 1
• 10 – 1 = 1
• Example: 101 – 11 = ?
101 5
- 11 -3
10 2
In decimal

36. Binary Multiplication
• 0 x 0 = 0
• 0 x 1 = 0
• 1 x 0 = 0
• 1 x 1 = 1
• Example: 101 x 111 = ?
101 5
x 111 x7
101 35
101
101
100011

37. Binary Division
• The procedure is same as decimal division
• Examples: 110 ÷ 11 =?
110
11
0
0
11
10
6
6
0
3
2
In decimal…

38. Review
• Why binary numbers? Why not decimal?
– Digital system understand binary numbers, not
decimal numbers.

39. Review
• Why hexadecimal?
– Shorthand of binary numbers
– Easy to convert to and from binary
• Examples:
– 1 1111 1000 0000 1010 1000 00012
– 1F80A8116

40. 40
Chapter 2
Number Systems
iv. Binary Codes
v. Representation of Negative Numbers
vi. Arithmetic operations

41. 41
Chapter 2
Number Systems
iv. Binary Codes
v. Representation of Negative Numbers
vi. Arithmetic operations

42. Outline
• Binary Codes
– BCD 8421
– Gray
– ASCII
– EBCDIC

43. Code and Encoding
• Code-- Special group of symbols that is used
to represent numbers, letters or words.
• Encoding– The process of converting a
number/letter/word into a code.
Number,
Letter,
Word
encoding Code
Generally,
For example,
Numbers,
Letters,
Words
Straight binary coding
Binary
number

44. Other Codes
• External world is decimal but digital systems
use binary numbers.
– So, conversions are often.
• Other codes are introduced because the
conversion between the code and the decimal
number is easier.

45. BCD Code
• Binary Coded Decimal Code
• Binary Coded Decimal (BCD)—a way to
represent each digit of a decimal number with
its 4-bit binary number.
Example: 87410
8 7 4Decimal
BCD 1000 0111 0100
Therefore, the BCD code for 87410 is 1000 0111 0100
Do you know why?
1000 1111 is not a BCD

46. BCD Code / 8421 Code
• Convert a BCD code to its decimal equivalent.
– Step 1: Break the BCD into 4-bit groups, starting
from LSB
– Step 2: Replace each 4-bit group with its
equivalent decimal
• Example: 0110 1000 0011 1001
6 8 3 9
So, the decimal equivalent is 683910

47. BCD coding vs. straight binary coding
• BCD coding is easier.
Decimal Binary
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
8 1000
9 1001
BCD coding
13710
13710
1 3 7decimal
BCD 0001 0011 0111
The BCD code is 0001 0011 0111
137
2
68
2
=68 1
=34 0
34
2
=17 0
=8 117
2
8
2
=4 0
4
2
=2 0
2
2
=1 0
1
2
=0 1
The straight binary code is 100010012

48. Gray Code
• No weights assigned to
the bit positions
• Only a single bit change
from one code word to
the next in sequence.
• Good— minimize the
chance for error.
DECIMAL BINARY GRAY CODE
0 0000 0000
1 0001 0001
2 0010 0011
3 0011 0010
4 0100 0110
5 0101 0111
6 0110 0101
7 0111 0100
8 1000 1100
9 1001 1101
10 1010 1111
11 1011 1110
12 1100 1010
13 1101 1011
14 1110 1001
15 1111 1000

49. Conversions
BINARY TO GRAY CONVERSION
1---+--->1 ---+--->0---+--->1---+--->1
1 0 1 1 0
GRAY TO BINARY CONVERSION
1 1 0 1 1
1 0 0 1 0
Binary
Gray
Gray
Binary

50. Alphanumeric codes
• Codes that represent numbers and alphabetic
characters (letters).
• At minimum, the code must represent 10
decimal digits (0-9) and 26 letters (A-Z).
• 6 bits are needed in the code that represent
the numbers and letters.
• ASCII is the most common alphanumeric code.

51. ASCII Code
• American Standard Code for International
Interchange
• Used in computers and electronic equipment
processor
1011001

52. ASCII Code
• ASCII code has 128 characters and symbols
• Represented by 7-bit binary code
• Can be considered an 8-bit code with MSB 0.
• The first 32 ASCII characters are non-graphic
commands only for control purposes—The
ASCII Control Characters.
• E.g.: null, line feed, start of text, escape

53. ASCII CODE

54. Character or
Number
ASCII-8 Binary EBCDIC Binary
A 01000001 11000001
E 01000101 11000101
Z 01011010 11101001
0 00110000 11110000
1 00110001 11110001
5 00110101 11110101
EBCDIC ALPHANUMERIC CODE
Extended Binary Coded Decimal Interchange Code
•8-bit character encoding

55. EBCDIC

56. 56
Chapter 2
Number Systems
iii. Binary Codes
iv. Representation of Negative Numbers
v. Arithmetic operations

57. Outline
• Representation of negative numbers
– Sign and magnitude
– 1’s complement
– 2’s complement

58. Representation of negative numbers
• Computer must handle both positive and
negative numbers.
• A signed binary number consists of both sign
and magnitude information.
• 3 types of representation:
– Sign and magnitude (least used)
– 1’s complement
– 2’s complement (most important)

59. Sign and Magnitude
• The sign bit—the left-most bit in a signed
binary number
– A ‘0’ sign bit indicates a positive number
– A ‘1’ sign bit indicates a negative number
• The remaining bits are the magnitude bits.
• Example: Express the decimal number -39 as
an 8-bit number in the sign-magnitude.
3910=1001112= 0010 01112
The sign bit is 1. Therefore, -3910= 1010 01112

60. Sign and Magnitude
0000
0111
0011
1011
1111
1110
1101
1100
1010
1001
1000
0110
0101
0100
0010
0001
+0
+1
+2
+3
+4
+5
+6
+7-0
-1
-2
-3
-4
-5
-6
-7
0 100 = + 4
1 100 = - 4
+
-

61. 1’s complement
• To find the 1’s complement of a given binary
number,
– Change all 1s to 0s and all 0s to 1s.
• Example: Find the 1’s complement of
10110010
1 0 1 1 0 0 1 0
0 1 0 0 1 1 0 1

62. 1’s Complement
0000
0111
0011
1011
1111
1110
1101
1100
1010
1001
1000
0110
0101
0100
0010
0001
+0
+1
+2
+3
+4
+5
+6
+7-7
-6
-5
-4
-3
-2
-1
-0
0 100 = + 4
1 011 = - 4
+
-

63. 2’s complement
• To find the 2’s complement of a given binary number,
– Add 1 to the LSB of the 1’s complement
• Example: Find the 2’s complement of 10110010
A 10110010 Binary number
01001101 1’s complement
+ 1 add 1
01001110 2’s complement
B 10110010
01001110

64. 2’s Complement
0000
0111
0011
1011
1111
1110
1101
1100
1010
1001
1000
0110
0101
0100
0010
0001
+0
+1
+2
+3
+4
+5
+6
+7-8
-7
-6
-5
-4
-3
-2
-1
0 100 = + 4
1 100 = - 4
+
-
• Only one representation for 0
• One more negative number than positive number

65. Comparison
+19 -19
Sign-magnitude
1’s complement
2’s complement
Exercise:
Express the decimal numbers +19 and -19 as an 8-bit
number in the sign-magnitude, 1’s complement and 2’s
complement forms.

66. Comparison
+19 -19
Sign-magnitude 00010011 10010011
1’s complement 00010011 11101100
2’s complement 00010011 11101101
Exercise:
Express the decimal numbers +19 and -19 as an 8-bit
number in the sign-magnitude, 1’s complement and 2’s
complement forms.

67. Comparison

68. Decimal Value of Sign-Magnitude Number
• Convert the magnitude bits to its decimal
value.
• The sign bit determines whether the number
is positive or negative.
• Example: 100101012
00101012 = 24 + 22 + 20
= 16 + 4 + 1
= 21
The sign bit is 1. Therefore, the decimal number is -21.

69. Decimal Value of 1’s Complement Number
• Positive number:
– Sum the weights of all bit position where there are 1s.
• Negative number:
– Step1: Assign negative value to the weight of the sign bit.
– Step 2: Sum all the weights where there are 1s.
– Step 3: Add 1 to the result.
• Example: 111010002
Step 1: -27 26 25 24 23 22 21 20
1 1 1 0 1 0 0 0
Step 2: -128+64+32+8=-24
Step 3: -24+1=-23

70. • Assign negative value to the weight of the sign
bit.
• Sum all the weights where there are 1s.
• Example: 101010102
Step 1: -27 26 25 24 23 22 21 20
1 0 1 0 1 0 1 0
Step 2: -128+32+8+2=-86
Decimal Value of 2’s Complement Number

71. Range of Unsigned Integer Numbers
• The maximum decimal number that can be
represented by n-bit is
– 2n-1
• 8-bit
– The maximum decimal number is 28-1=255

72. Range of SIGNED Integer Numbers
• The smallest decimal number that can be
represented by n-bit is -(2n-1)
• The biggest decimal number that can be
represented by n-bit is +(2n-1-1)
• The range is -(2n-1) to +(2n-1-1).
• Example: 8-bit
– The range is -27 to 27-1 or -128 to +127

73. 73
Chapter 2
Number Systems
iv. Binary Codes
v. Representation of Negative Numbers
vi. Arithmetic operations

74. Outline
• Arithmetic Operations of 2’s Complement
– Addition
– Subtraction
– Overflow
• Simpler addition scheme makes 2’s
Complement the most common choice for
integer number systems within digital systems

75. Addition and Subtraction : 2’s Complement
4
+ 3
7
0100
+ 0011
0111
-4
+ 3
-1
1100
0011
1111
• Example: 4-bit system
• Example: Overflow in 4-bit system
4
- 3
1
0100
+ 1101
10001
Discard
carry
7
+ 4
11
0111
+ 0100
1011
Wrong sign
-4
+ (-3)
-7
1100
+ 1101
11001
Discard
carry
Wrong magnitude
•Does overflow occur for
this case if this is an 8-bit
system?
• Carry in sign different
from carry out sign, a sign
that overflow occurs.

76. 2’s Complement Overflow
Overflow
Overflow when carry in to sign does not equal carry out
Expected 0 1 1 1 Actual
+5 0 1 0 1
+ +3 0 0 1 1
+8 1 0 0 0 -8
Expected 1 0 0 0 Actual
-7 1 0 0 1
+ -2 1 1 1 0
-9 1 0 1 1 1 +7
Overflow
Expected 0 0 0 0 Actual
+5 0 1 0 1
+ +2 0 0 1 0
+7 0 1 1 1 +7
No Overflow
Expected 1 1 1 1 Actual
-3 1 1 0 1
+ -5 1 0 1 1
-8 1 1 0 0 0 -8
No Overflow

77. • Add two positive numbers but get a negative
number
• or two negative numbers but get a positive
number
Overflow Condition

78. ExampleExample
Add two numbers in two’s complement
representation: (-35) + (+20) (-15)
Addition and Subtraction : 2’s Complement

79. ExampleExample
Add two numbers in two’s complement
representation: (-35) + (+20) (-15)
SolutionSolution
Carry 1 1 1
1 1 0 1 1 1 0 1
++ 0 0 0 1 0 1 0 0
----------------------------------
Result 1 1 1 1 0 0 0 1 -15
Addition and Subtraction : 2’s Complement

80. Overflow Condition
ExampleExample
Add two numbers in two’s complement
representation: (+127) + (+3) (+130)

81. Overflow Condition
ExampleExample
Add two numbers in two’s complement
representation: (+127) + (+3) (+130)
SolutionSolution
Carry 1 1 1 1 1 1 1
0 1 1 1 1 1 1 1
++ 0 0 0 0 0 0 1 1
----------------------------------
Result 1 0 0 0 0 0 1 0 --126 (Error)126 (Error)
An overflow has occurred.An overflow has occurred.