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Number system
The document provides an extensive overview of number systems, detailing conversion techniques among decimal, binary, octal, and hexadecimal bases. It also covers arithmetic operations in binary, including addition and multiplication, as well as representations of signed numbers using different methods like sign-magnitude, 1's complement, and 2's complement. Additional topics include binary-coded decimal (BCD), Gray code, ASCII coding, and an introduction to Boolean algebra and logic gates.
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Number system
- 1.
- 2. NUMBER SYSTEM BASE Symbol Intro…
- 3. Common Number Systems System Base Symbols Used by humans? Used in computers? Decimal 10 0, 1, … 9 Yes No Binary 2 0, 1 No Yes Octal 8 0, 1, … 7 No No Hexa-decimal 16 0, 1, … 9, A, B, … F No No
- 4. Conversion Among Bases The possibilities: Decimal Octal Hexadecimal Binary
- 5. Quick Example 2510= 110012 = 318 = 1916 Base/Radix
- 6. Decimal to Decimal(just for fun) Decimal Octal Hexadecimal Binary
- 7. 12510 => 5x 100= 5 2 x 101= 20 1 x 102= 100 125 Base Weight
- 8. Binary to Decimal Hexadecimal Decimal Octal Binary
- 9. Binary to Decimal Technique Multiply each bit by 2n, where n is the “weight” of the bit The weight is the position of the bit, starting from 0 on the right Add the results
- 10. Example 1010112 =>1 x 20 = 1 1 x 21 = 2 0 x 22 = 0 1 x 23 = 8 0 x 24 = 0 1 x 25 = 32 4310 Bit “0”
- 11. Octal to Decimal Decimal Octal Hexadecimal Binary
- 12. Octal to Decimal Technique Multiply each bit by 8n, where n is the “weight” of the bit The weight is the position of the bit, starting from 0 on the right Add the results
- 13. Example 7248 =>4 x 80 = 4 2 x 81 = 16 7 x 82 = 448 46810
- 14. Hexadecimal to Decimal Decimal Octal Hexadecimal Binary
- 15. Hexadecimal to Decimal Technique Multiply each bit by 16n, where n is the “weight” of the bit The weight is the position of the bit, starting from 0 on the right Add the results
- 16. Example ABC16 =>C x 160 = 12 x 1 = 12 B x 161 = 11 x 16 = 176 A x 162 = 10 x 256 = 2560 274810
- 17. Decimal to Binary Decimal Octal Hexadecimal Binary
- 18. Decimal to Binary Technique Divide by two, keep track of the remainder First remainder is bit 0 (LSB, least-significant bit) Second remainder is bit 1 Etc.
- 19. Example 12510 =?2 2 125 2 6 2 1 2 3 1 0 2 1 5 1 2 7 1 2 3 1 2 1 1 0 1 12510 = 11111012
- 20. Octal to Binary Decimal Octal Hexadecimal Binary
- 21. Octal to Binary Technique Convert each octal digit to a 3-bit equivalent binary representation
- 22. Example 7058 =?2 7 0 5 111 000 101 7058 = 1110001012
- 23. Hexadecimal to Binary Decimal Octal Hexadecimal Binary
- 24. Hexadecimal to Binary Technique Convert each hexadecimal digit to a 4-bit equivalent binary representation
- 25. Example 10AF16 =?2 1 0 A F 0001 0000 1010 1111 10AF16 = 00010000101011112
- 26. Decimal to Octal Decimal Octal Hexadecimal Binary
- 27. Decimal to Octal Technique Divide by 8 Keep track of the remainder
- 28. Example 123410 =?8 8 1234 8 154 2 19 2 8 2 3 8 0 2 123410 = 23228
- 29. Decimal to Hexadecimal Decimal Octal Hexadecimal Binary
- 30. Decimal to Hexadecimal Technique Divide by 16 Keep track of the remainder
- 31. Example 123410 =?16 123410 = 4D216 16 1234 77 2 16 4 13 = D 16 0 4
- 32. Binary to Octal Decimal Octal Hexadecimal Binary
- 33. Binary to Octal Technique Group bits in threes, starting on right Convert to octal digits
- 34. Example 10110101112 =?8 1 011 010 111 1 3 2 7 10110101112 = 13278
- 35. Binary to Hexadecimal Decimal Octal Hexadecimal Binary
- 36. Binary to Hexadecimal Technique Group bits in fours, starting on right Convert to hexadecimal digits
- 37. Example 10101110112 =?16 10 1011 1011 2 B B 10101110112 = 2BB16
- 38. Octal to Hexadecimal Decimal Octal Hexadecimal Binary
- 39. Octal to Hexadecimal Technique Use binary as an intermediary
- 40. Example 10768 =?16 1 0 7 6 001 000 111 110 2 3 E 10768 = 23E16
- 41. Hexadecimal to Octal Decimal Octal Hexadecimal Binary
- 42. Hexadecimal to Octal Technique Use binary as an intermediary
- 43. Example 1F0C16 =?8 1 F 0 C 0001 1111 0000 1100 1F0C16 = 174148 1 7 4 1 4
- 44. Exercise – Convert... Decimal Binary Octal Don’t use a calculator! Hexa-decimal 33 1110101 703 1AF Skip answer Answer
- 45. Exercise – Convert… Answer Decimal Binary Octal Hexa-decimal 33 100001 41 21 117 1110101 165 75 451 111000011 703 1C3 431 110101111 657 1AF
- 46. Common Powers (1of 2) Base 10 Power Preface Symbol 10-12 pico p 10-9 nano n 10-6 micro 10-3 milli m 103 kilo k 106 mega M 109 giga G 1012 tera T Value .000000000001 .000000001 .000001 .001 1000 1000000 1000000000 1000000000000
- 47. Common Powers (2of 2) Base 2 Power Preface Symbol 210 kilo k 220 mega M 230 Giga G Value 1024 1048576 1073741824 • What is the value of “k”, “M”, and “G”? • In computing, particularly w.r.t. memory, the base-2 interpretation generally applies
- 48. Review – multiplyingpowers For common bases, add powers ab ac = ab+c 26 210 = 216 = 65,536 or… 26 210 = 64 210 = 64k
- 49. Binary Addition (1of 2) Two 1-bit values A B A + B 0 0 0 0 1 1 1 0 1 1 1 10 “two”
- 50. Binary Addition (2of 2) Two n-bit values Add individual bits Propagate carries E.g., 10101 21 + 11001 + 25 101110 46
- 51. Multiplication (1 of3) Decimal (just for fun) 35 x 105 175 000 35 3675
- 52. Multiplication (2 of3) Binary, two 1-bit values A B A B 0 0 0 0 1 0 1 0 0 1 1 1
- 53. Multiplication (3 of3) Binary, two n-bit values As with decimal values E.g., 1110 x 1011 1110 1110 0000 1110 10011010
- 54. Fractions Decimalto decimal (just for fun) 3.14 => 4 x 10-2 = 0.04 1 x 10-1 = 0.1 3 x 100 = 3 3.14
- 55. Fractions Binaryto decimal 10.1011 => 1 x 2-4 = 0.0625 1 x 2-3 = 0.125 0 x 2-2 = 0.0 1 x 2-1 = 0.5 0 x 20 = 0.0 1 x 21 = 2.0 2.6875
- 56. Fractions Decimalto binary 3.14579 .14579 x 2 0.29158 x 2 0.58316 x 2 1.16632 x 2 0.33264 x 2 0.66528 x 2 1.33056 etc. 11.001001...
- 57. Exercise – Convert... Decimal Binary Octal Don’t use a calculator! Hexa-decimal 29.8 101.1101 3.07 C.82
- 58. Exercise – Convert… Answer Decimal Binary Octal Hexa-decimal 29.8 11101.110011… 35.63… 1D.CC… 5.8125 101.1101 5.64 5.D 3.109375 11.000111 3.07 3.1C 12.5078125 1100.10000010 14.404 C.82
- 59.
- 60. Sign-Magnitude Notation Thesimplest form of representation that employs a sign bit, the leftmost significant bit. For an N-bit word, the rightmost N-1 bits hold the magnitude of the integer. Thus, 00010010 = +18 10010010 = -18
- 61. Drawbacks of Sign-Magnitude Notation Addition and subtraction require a consideration of both the signs of the numbers, and their relative magnitudes, in order to carry out the requested operation.
- 62. 1’s Complement Notation Positive integers are represented in the same way as sign-magnitude notation. A negative integer is represented by the 1's complement of the positive integer in sign-magnitude notation. The ones complement of a number is obtained by complementing each one of the bits, i.e., a 1 is replaced by a 0, and a 0 is replaced by a 1. 18 (Base 10) = 00010010 -18 = 1's complement of 18 = 11101101
- 63. 2’s Complement Notation The 2's complement representation of positive integers is the same as in sign-magnitude representation. A negative number is represented by the 2's complement of the positive integer with the same magnitude. 1. Perform the 1's complement operation. 2. Treating the result as an unsigned binary integer, add 1.
- 64. 2’s Complement Example 18 = 00010010 1's complement = 11101101 +1 ________ 2’S of 18 = 11101110 = -18d
- 65. Binary Addition andSubtration The simplest implementation is one in which the numbers involved can be treated as unsigned integers for purposes of addition. 1’s complement Addition 2’s complement Addition
- 66. 1’s complement Addition Consider the Subtrahend & Minuend. Calculate the 1’s complement of the Subtrahend number. Add it to the Minuend If the resulting is producing any carry then add the carry to the LSB of the Result the sign is same as of the Minuend. If there is no carry then again take the 1’s complement of the result and place minus sign before the result. This is the Require Result.
- 67. Example ex. +10001 Minuend -6 0110 Subtrahend 1’s complement of 6 = 1001 Add : 0001 Minuend 1001 1’s Complement of Subtrahend Result =1010 (No carry) Again take the 1’ s Complement of 1010 = - 0101 Answer = > -5
- 68. Ones Complement EndAround Carry Perform the operation -2 + -4 1’s complement of -2 = 1101 1’s complement of -4 = 1011 ------- 1)1000 --> 1 -------- 1001 = -6
- 69. 2’s complement Addition Calculate the 2’s complement of the Subtrahend number. Add the 2’s complement to the Minuend Number If the resultant has a carry discard it & this is the final Ans. Else if there is no carry calculate the 2’s complement of the Resultant. This is the final Answer.
- 70. 2’s Complement Addition : a)5 - 7 Binary equivalent of 7 = 0111 1’s complement of 7 = 1000 2’s complement of 7 = 1001 Now add 5 = 0101 + (-)7 = 1001 1110 = -2 In 2’s Complement addition, the Carry Out of the most significant bit is ignored!
- 71. 8421 BCD Code In the 8421 Binary Coded Decimal (BCD) representation each decimal digit is converted to its 4-bit pure binary equivalent. This coding is an example of a binary coded (each decimal number maps to four bits) weighted (each bit represents a number: 1, 2, 4, etc.) code. 57dec = 0101 0111bcd
- 72. Questions Q.What is the decimal number for the following 8421 BCD Code 0001 1001 0111 0010bcd? A. 1972 Q. What is the 8421 BCD Code for the decimal number 421? A. 0100 0010 0001
- 73. Gray Code Gray coding is used for its speed & freedom from errors. In BCD or 8421 BCD when counting from 7 (0111) to 8 (1000) requires 4 bits to be changed simultaneously. If this does not happen then various numbers could be momentarily generated during the transition so creating spurious numbers which could be read. Gray coding avoids this since only one bit changes between subsequent numbers. Two simple rules. 1. Start with all 0s. 2. Proceed by changing the least significant bit (lsb) which will bring about a new state.
- 74. Gray Code Continued Decimal Gray Code 0 0000 1 0001 2 0011 3 0010 4 0110 5 0111 6 0101 7 0100 8 1100 9 1101 10 1111 11 1110 12 1010 13 1011 14 1001 15 1000
- 75. Excess-3 Code Excess-3 is a non weighted code used to express decimal numbers. The code derives its name from the fact that each binary code is the corresponding 8421 code plus 0011(3). 1000 of 8421 = 1011 in Excess-3
- 76. ASCII Code ASCII: American Standard Code for Information Interchange The standard ASCII code defines 128 character codes (from 0 to 127), of which, the first 32 are control codes (non-printable), and the other 96 are representable characters. In addition to the 128 standard ASCII codes there are other 128 that are known as extended ASCII, and that are platform-dependent.
- 77.
- 78.
- 79. Boolean function •Boolean function: Mapping from Boolean variables to a Boolean value. • Truth table: Represents relationship between a Boolean function and its binary variables. It enumerates all possible combinations of arguments and the corresponding function values.
- 80. Boolean function andlogic diagram • Boolean algebra: Deals with binary variables and logic operations operating on those variables. • Logic diagram: Composed of graphic symbols for logic gates. A simple circuit sketch that represents inputs and outputs of Boolean functions.
- 81. Gates Referto the hardware to implement Boolean operators. The most basic gates are
- 82. Boolean function andtruth table
- 83. Basic identities ofboolean algebra • A Boolean algebra is a closed algebraic system containing set of elements and the operators. • The . (dot) operator is known as logical And. • The + (plus) which refer to logical OR.
- 84. Basic Identities ofBoolean Algebra (1) x + 0 = x (2) x · 0 = 0 (3) x + 1 = 1 (4) x · 1 = x (5) x + x = x (6) x · x = x (7) x + x’ = 1 (8) x · x’ = 0 (9) x + y = y + x (10) xy = yx (11) x + ( y + z ) = ( x + y ) + z (12) x (yz) = (xy) z X I/P O/P = X + I/P 0 0 0 = > X 1 0 1 => X X I/P O/P = X . I/P 0 0 0 1 0 0 X I/P O/P = X . I/P 0 1 0 => X 1 1 1 => X
- 85. Basic Identities ofBoolean Algebra (DeMorgan’s Theorem) (15) ( x + y )’ = x’ y’ (16) ( xy )’ = x’ + y’ (17) (x’)’ = x
- 86. Function Minimization usingBoolean Algebra Examples: (a) a + ab = a(1+b)=a (b) a(a + b) = a.a +ab=a+ab=a(1+b)=a. (c) a + a'b = (a + a')(a + b)=1(a + b) =a+b (d) a(a' + b) = a. a' +ab=0+ab=ab
- 87. Try F= abc + abc’ + a’c
- 88. The other typeof question Show that; 1- ab + ab' = a 2- (a + b)(a + b') = a 1- ab + ab' = a(b+b') = a.1=a 2- (a + b)(a + b') = a.a +a.b' +a.b+b.b' = a + a.b' +a.b + 0 = a + a.(b' +b) + 0 = a + a.1 + 0 = a + a = a
- 89. More Examples Show that; (a) ab + ab'c = ab + ac (b) (a + b)(a + b' + c) = a + bc (a) ab + ab'c = a(b + b'c) = a((b+b').(b+c))=a(b+c)=ab+ac (b) (a + b)(a + b' + c) = (a.a + a.b' + a.c + ab +b.b' +bc) = (a + ab’ + a.c + a.b + 0 + b.c) =( a(1+b’ + c + b)+ b.c) = ( a+ b.c)
- 90. Principle of Duality The dual of a statement S is obtained by interchanging . and +; 0 and 1. Dual of (a*1)*(0+a’) = 0 is (a+0)+(1*a’) = 1 Dual of any theorem in a Boolean Algebra is also a theorem. This is called the Principle of Duality.
- 91. DeMorgan's Theorem (a)(a + b)' = a'b' (b) (ab)' = a' + b' Generalized DeMorgan's Theorem (a) (a + b + … z)' = a'b' … z' (b) (a.b … z)' = a' + b' + … z‘
- 92. DeMorgan's Theorem F = ab + c’d’ F’ = ?? F = ab + c’d’ + b’d F’ = ?? Show that: (a + b.c)' = a'.b' + a'.c'
- 93. More DeMorgan's example Show that: (a(b + z(x + a')))' =a' + b' (z' + x') Sol: (a(b + z(x + a')))' = a' + (b + z(x + a'))' = a' + b' (z(x + a'))' = a' + b' (z' + (x + a')') = a' + b' (z' + x'(a')') = a' + b' (z' + x'a) =a‘+b' z' + b'x'a =(a‘+ b'x'a) + b' z' =(a‘+ b'x‘)(a +a‘) + b' z' = a‘+ b'x‘+ b' z‘ = a' + b' (z' + x')
- 94. More Examples (a(b+ c) + a'b)'=b'(a' + c') ab + a'c + bc = ab + a'c (a + b)(a' + c)(b + c) = (a + b)(a' + c)