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
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
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
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
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
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…
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
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
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
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
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
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
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.
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
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
