Number system utm notes
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The document discusses various number systems used in digital electronics including decimal, binary, hexadecimal, and octal number systems. It provides details on how decimal, binary, and hexadecimal numbers are represented and converted between number systems. Various methods for converting between decimal, binary, hexadecimal, and octal numbers are presented including the sum-of-weights method and division/multiplication methods. The use of binary coded decimal codes for easier conversion between decimal and binary numbers is also covered.
Number system utm notes
- 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.