Your SlideShare is downloading. ×
Ch1.number systems
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×

Saving this for later?

Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime - even offline.

Text the download link to your phone

Standard text messaging rates apply

Ch1.number systems

760
views

Published on

Lgic Design

Lgic Design

Published in: Technology, Education

1 Comment
0 Likes
Statistics
Notes
  • Be the first to like this

No Downloads
Views
Total Views
760
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
62
Comments
1
Likes
0
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. LOGIC DESIGNDr. Mahmoud Abo_elfetouh
  • 2. Course objectives• This course provides you with a basic understanding of what digital devices are, how they operate, and how they can be designed to perform useful functions.• The course is intended to give you an understanding of Binary systems, Boolean algebra, digital design techniques, logic gates, logic minimization, standard combinational circuits, sequential circuits, flip-flops, synthesis of synchronous sequential circuits, and arithmetic circuits.
  • 3. ContentsWeek No. Topic Lecture Practical Total 1 Number Systems and Codes 3 2 5 2 Number Systems and Codes 2 5 3 Boolean Algebra and Logic 3 2 5 Simplification 3 Minimization Techniques- 4 2 5 Karnaugh Map 3 Minimization Techniques- 5 2 5 Karnaugh Map 3 6 Logic Gates 3 2 5 7 Arithmetic Circuits-Adders 3 2 5 8 Mid- Term Exam
  • 4. ContentsWeek No. Topic Lecture Practical Total Arithmetic Circuits -Subtracter 9 3 2 5 10 Combinational Circuits 2 5 3 11 Combinational Circuits 3 2 5 12 Flip-Flops 3 2 5 Flip-Flops 13 3 2 5 14 Counters - Registers 3 2 5 15 Memory Devices 3 2 5 16 Final- Term Exam
  • 5. TextbookLogic and Computer Design Fundamentals, 4thEdition by M. Morris Mano and Charles R. Kime,Prentice Hall, 2008
  • 6. Chapter 1Number Systems
  • 7. 1. Number Systems Location in course textbook Chapt. 1ITEC 1011 Introduction to Information Technologies
  • 8. Common Number Systems Used by Used in System Base Symbols humans? computers? Decimal 10 0, 1, … 9 Yes No Binary 2 0, 1 No Yes Octal 8 0, 1, … 7 No No Hexa- 16 0, 1, … 9, No No decimal A, B, … FITEC 1011 Introduction to Information Technologies
  • 9. Quantities/Counting (1 of 3) Hexa- Decimal Binary Octal decimal 0 0 0 0 1 1 1 1 2 10 2 2 3 11 3 3 4 100 4 4 5 101 5 5 6 110 6 6 7 111 7 7 p. 33ITEC 1011 Introduction to Information Technologies
  • 10. Quantities/Counting (2 of 3) Hexa- Decimal Binary Octal decimal 8 1000 10 8 9 1001 11 9 10 1010 12 A 11 1011 13 B 12 1100 14 C 13 1101 15 D 14 1110 16 E 15 1111 17 FITEC 1011 Introduction to Information Technologies
  • 11. Quantities/Counting (3 of 3) Hexa- Decimal Binary Octal decimal 16 10000 20 10 17 10001 21 11 18 10010 22 12 19 10011 23 13 20 10100 24 14 21 10101 25 15 22 10110 26 16 23 10111 27 17 Etc.ITEC 1011 Introduction to Information Technologies
  • 12. Conversion Among Bases • The possibilities: Decimal Octal Binary Hexadecimal pp. 40-46ITEC 1011 Introduction to Information Technologies
  • 13. Quick Example 2510 = 110012 = 318 = 1916 BaseITEC 1011 Introduction to Information Technologies
  • 14. Decimal to Decimal (just for fun) Decimal Octal Binary Hexadecimal Next slide…ITEC 1011 Introduction to Information Technologies
  • 15. Weight 12510 => 5 x 100 = 5 2 x 101 = 20 1 x 102 = 100 125 BaseITEC 1011 Introduction to Information Technologies
  • 16. Binary to Decimal Decimal Octal Binary HexadecimalITEC 1011 Introduction to Information Technologies
  • 17. 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 resultsITEC 1011 Introduction to Information Technologies
  • 18. Example Bit “0” 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 4310ITEC 1011 Introduction to Information Technologies
  • 19. Octal to Decimal Decimal Octal Binary HexadecimalITEC 1011 Introduction to Information Technologies
  • 20. 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 resultsITEC 1011 Introduction to Information Technologies
  • 21. Example 7248 => 4 x 80 = 4 2 x 81 = 16 7 x 82 = 448 46810ITEC 1011 Introduction to Information Technologies
  • 22. Hexadecimal to Decimal Decimal Octal Binary HexadecimalITEC 1011 Introduction to Information Technologies
  • 23. 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 resultsITEC 1011 Introduction to Information Technologies
  • 24. Example ABC16 => C x 160 = 12 x 1 = 12 B x 161 = 11 x 16 = 176 A x 162 = 10 x 256 = 2560 274810ITEC 1011 Introduction to Information Technologies
  • 25. Decimal to Binary Decimal Octal Binary HexadecimalITEC 1011 Introduction to Information Technologies
  • 26. 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.ITEC 1011 Introduction to Information Technologies
  • 27. Example 12510 = ?2 2 125 2 62 1 2 31 0 2 15 1 2 7 1 2 3 1 2 1 1 0 1 12510 = 11111012ITEC 1011 Introduction to Information Technologies
  • 28. Decimal to Octal Decimal Octal Binary HexadecimalITEC 1011 Introduction to Information Technologies
  • 29. Decimal to Octal • Technique – Divide by 8 – Keep track of the remainderITEC 1011 Introduction to Information Technologies
  • 30. Example 123410 = ?8 8 1234 8 154 2 8 19 2 8 2 3 0 2 123410 = 23228ITEC 1011 Introduction to Information Technologies
  • 31. Decimal to Hexadecimal Decimal Octal Binary HexadecimalITEC 1011 Introduction to Information Technologies
  • 32. Decimal to Hexadecimal • Technique – Divide by 16 – Keep track of the remainderITEC 1011 Introduction to Information Technologies
  • 33. Example 123410 = ?16 16 1234 16 77 2 16 4 13 = D 0 4 123410 = 4D216ITEC 1011 Introduction to Information Technologies
  • 34. Octal to Binary Decimal Octal Binary HexadecimalITEC 1011 Introduction to Information Technologies
  • 35. Octal to Binary • Technique – Convert each octal digit to a 3-bit equivalent binary representationITEC 1011 Introduction to Information Technologies
  • 36. Example 7058 = ?2 7 0 5 111 000 101 7058 = 1110001012ITEC 1011 Introduction to Information Technologies
  • 37. Hexadecimal to Binary Decimal Octal Binary HexadecimalITEC 1011 Introduction to Information Technologies
  • 38. Hexadecimal to Binary • Technique – Convert each hexadecimal digit to a 4-bit equivalent binary representationITEC 1011 Introduction to Information Technologies
  • 39. Example 10AF16 = ?2 1 0 A F 0001 0000 1010 1111 10AF16 = 00010000101011112ITEC 1011 Introduction to Information Technologies
  • 40. Binary to Octal Decimal Octal Binary HexadecimalITEC 1011 Introduction to Information Technologies
  • 41. Binary to Octal • Technique – Group bits in threes, starting on right – Convert to octal digitsITEC 1011 Introduction to Information Technologies
  • 42. Example 10110101112 = ?8 1 011 010 111 1 3 2 7 10110101112 = 13278ITEC 1011 Introduction to Information Technologies
  • 43. Binary to Hexadecimal Decimal Octal Binary HexadecimalITEC 1011 Introduction to Information Technologies
  • 44. Binary to Hexadecimal • Technique – Group bits in fours, starting on right – Convert to hexadecimal digitsITEC 1011 Introduction to Information Technologies
  • 45. Example 10101110112 = ?16 10 1011 1011 2 B B 10101110112 = 2BB16ITEC 1011 Introduction to Information Technologies
  • 46. Octal to Hexadecimal Decimal Octal Binary HexadecimalITEC 1011 Introduction to Information Technologies
  • 47. Octal to Hexadecimal • Technique – Use binary as an intermediaryITEC 1011 Introduction to Information Technologies
  • 48. Example 10768 = ?16 1 0 7 6 001 000 111 110 2 3 E 10768 = 23E16ITEC 1011 Introduction to Information Technologies
  • 49. Hexadecimal to Octal Decimal Octal Binary HexadecimalITEC 1011 Introduction to Information Technologies
  • 50. Hexadecimal to Octal • Technique – Use binary as an intermediaryITEC 1011 Introduction to Information Technologies
  • 51. Example 1F0C16 = ?8 1 F 0 C 0001 1111 0000 1100 1 7 4 1 4 1F0C16 = 174148ITEC 1011 Introduction to Information Technologies
  • 52. Exercise – Convert ... Hexa- Decimal Binary Octal decimal 33 1110101 703 1AF Don’t use a calculator! Skip answer AnswerITEC 1011 Introduction to Information Technologies
  • 53. Exercise – Convert … Answer Hexa- Decimal Binary Octal decimal 33 100001 41 21 117 1110101 165 75 451 111000011 703 1C3 431 110101111 657 1AFITEC 1011 Introduction to Information Technologies
  • 54. Common Powers (1 of 2) • Base 10 Power Preface Symbol Value 10-12 pico p .000000000001 10-9 nano n .000000001 10-6 micro .000001 10-3 milli m .001 103 kilo k 1000 106 mega M 1000000 109 giga G 1000000000 1012 tera T 1000000000000ITEC 1011 Introduction to Information Technologies
  • 55. Common Powers (2 of 2) • Base 2 Power Preface Symbol Value 210 kilo k 1024 220 mega M 1048576 230 Giga G 1073741824 • What is the value of “k”, “M”, and “G”? • In computing, particularly w.r.t. memory, the base-2 interpretation generally appliesITEC 1011 Introduction to Information Technologies
  • 56. Example In the lab… 1. Double click on My Computer 2. Right click on C: 3. Click on Properties / 230 =ITEC 1011 Introduction to Information Technologies
  • 57. Exercise – Free Space • Determine the “free space” on all drives on a machine in the lab Free space Drive Bytes GB A: C: D: E: etc.ITEC 1011 Introduction to Information Technologies
  • 58. Review – multiplying powers • For common bases, add powers ab ac = ab+c 26 210 = 216 = 65,536 or… 26 210 = 64 210 = 64kITEC 1011 Introduction to Information Technologies
  • 59. Fractions • Decimal to decimal (just for fun) 3.14 => 4 x 10-2 = 0.04 1 x 10-1 = 0.1 3 x 100 = 3 3.14 pp. 46-50ITEC 1011 Introduction to Information Technologies
  • 60. Fractions • Binary to 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 pp. 46-50ITEC 1011 Introduction to Information Technologies
  • 61. Fractions • Decimal to binary .14579 x 2 3.14579 0.29158 x 2 0.58316 x 2 1.16632 x 2 0.33264 x 2 0.66528 x 2 1.33056 11.001001... etc. p. 50ITEC 1011 Introduction to Information Technologies
  • 62. Fractions • Octal to decimal 15.42 => 2 x 8-2 = 0.03125 4 x 8-1 = 0.5 5 x 80 = 5.0 1 x 81 = 8.0 13.53125 pp. 46-50ITEC 1011 Introduction to Information Technologies
  • 63. Fractions • Decimal to octal .14 x 8 3.14 1.12 x 8 0.96 x 8 7.68 x 8 5.44 x 8 3.52 x 8 4.16 3.107534... etc. p. 50ITEC 1011 Introduction to Information Technologies
  • 64. Fractions • Hexadecimal to decimal 2B.84 => 4 x 16-2 = 0.015625 8 x 16-1 = 0.5 B x 160 = 11.0 2 x 161 = 32.0 43.515625 pp. 46-50ITEC 1011 Introduction to Information Technologies
  • 65. Fractions • Decimal to Hexadecima .1 x 16 3.1 1.6 x 16 9.6 x 16 9.6 x 16 9.6 x 16 9.6 x 16 9.6 3.199999... etc. p. 50ITEC 1011 Introduction to Information Technologies
  • 66. Exercise – Convert ... Hexa- Decimal Binary Octal decimal 29.8 101.1101 3.07 C.82 Don’t use a calculator! Skip answer AnswerITEC 1011 Introduction to Information Technologies
  • 67. Exercise – Convert … Answer Hexa- Decimal Binary Octal 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.82ITEC 1011 Introduction to Information Technologies
  • 68. Binary Addition (1 of 2) • Two 1-bit values A B A+B 0 0 0 0 1 1 1 0 1 1 1 10 “two” pp. 36-38ITEC 1011 Introduction to Information Technologies
  • 69. Binary Addition (2 of 2) • Two n-bit values – Add individual bits – Propagate carries – E.g., 1 1 10101 21 + 11001 + 25 101110 46ITEC 1011 Introduction to Information Technologies
  • 70. Multiplication (1 of 3) • Decimal (just for fun) 35 x 105 175 000 35 3675 pp. 39ITEC 1011 Introduction to Information Technologies
  • 71. Multiplication (2 of 3) • Binary, two 1-bit values A B A B 0 0 0 0 1 0 1 0 0 1 1 1ITEC 1011 Introduction to Information Technologies
  • 72. Multiplication (3 of 3) • Binary, two n-bit values – As with decimal values – E.g., 1110 x 1011 1110 1110 0000 1110 10011010ITEC 1011 Introduction to Information Technologies
  • 73. Binary Subtraction (1 of 2) • Two 1-bit values A B A- B 0 0 0 Borrow 1 0 1 1 1 0 1 1 1 0 pp. 36-38ITEC 1011 Introduction to Information Technologies
  • 74. Binary Subtraction (2 of 2) • Two n-bit values – Subtract individual bits – Propagate borrows – E.g., 10 11001 0 25 - 10101 - 21 00100 4ITEC 1011 Introduction to Information Technologies
  • 75. Binary Subtraction (2 of 2) • Two n-bit values – Subtract individual bits – Propagate borrows – E.g., 10 10 0 10001 11001 25 - 10101 - 10101 - 21 00100 00100 4ITEC 1011 Introduction to Information Technologies
  • 76. Subtraction with Complements • Complements are used for simplifying the subtraction operations. • There are two types of complements for each base-r system: the rs complement and the (r — l)s complement. • 2s complement and 1s complement for binary numbers, and the 10s complement and 9s com-plement for decimal numbers.ITEC 1011 Introduction to Information Technologies
  • 77. 9s complement • The 9s complement of a decimal number is obtained by subtracting each digit from 9. • • The 9s complement of 546700 is: • 999999 - 546700 = 453299. • The 9s complement of 012398 is: • 999999 - 012398 = 987601.ITEC 1011 Introduction to Information Technologies
  • 78. Binary numbers, the 1s complement • The 1s complement of a binary number is formed by changing 1s to 0s and 0s to 1s. • Examples: • The 1s complement of 1011000 is • 0100111 • The 1s complement of 0101101 is • 1010010ITEC 1011 Introduction to Information Technologies
  • 79. 10s complement • The 10s complement can be formed by leaving all least significant 0s un-changed, subtracting the first nonzero least significant digit from 10, and subtracting all higher significant digits from 9. • The 10s complement of 012398 is 987602. • The 10s complement of 246700 is 753300.ITEC 1011 Introduction to Information Technologies
  • 80. 2s complement • The 2s complement can be formed by leaving all least significant 0s and the first 1 unchanged, and replacing 1s with 0s and 0s with 1s in all other higher significant digits. • The 2s complement of 1101100 is • 0010100 • The 2s complement of 0110111 is • 1001001ITEC 1011 Introduction to Information Technologies
  • 81. Subtraction with Complements • The subtraction of two n-digit unsigned numbers M — N in base r can be done as follows: • Add M to the rs complement of N. • If M ≥ N, the sum will produce an end carry, r n, which is discarded; what is left is the result M - N. • If M < N, the sum does not produce an end carry. To obtain the answer in a familiar form, take the rs complement of the sum and place a negative sign in front.ITEC 1011 Introduction to Information Technologies
  • 82. Examples to illustrate the procedure • Given the two binary numbers; • X = 1010100 and Y = 1000011, • perform the subtraction: • (a) X — Y • (b) Y — X • using 2s complements.ITEC 1011 Introduction to Information Technologies
  • 83. X — Y= 1010100 —1000011 • X = 1010100 • 2s complement of Y = + 0111101 • Sum = 10010001 • Discard end carry = -10000000 • Answer: X — Y = 0010001 •ITEC 1011 Introduction to Information Technologies
  • 84. Y — X= 1000011 — 1010100 • Y = 1000011 • 2s complement of X = + 0101100 • Sum = 1101111 • There is no end carry. • Answer: • Y - X = -(2s complement of 1101111) • = - 0010001ITEC 1011 Introduction to Information Technologies
  • 85. Thank you Next topicITEC 1011 Introduction to Information Technologies
  • 86. Binary Codes • Binary codes are codes which are represented in binary system with modification from the original ones. • Binary codes are classified as: – Weighted Binary Systems – Non Weighted CodesITEC 1011 Introduction to Information Technologies
  • 87. Weighted Binary Systems • Weighted binary codes are those which obey the positional weighting principles, • Each position of the number represents a specific weight. • The codes 8421, 2421, 5421, and 5211 are weighted binary codes.ITEC 1011 Introduction to Information Technologies
  • 88. Weighted Binary SystemsITEC 1011 Introduction to Information Technologies
  • 89. 8421 Code/BCD Code • The BCD (Binary Coded Decimal) is a straight assignment of the binary equivalent. • It is possible to assign weights to the binary bits according to their positions. • The weights in the BCD code are 8,4,2,1. • Example: The bit assignment 1001, can be seen by its weights to represent the decimal 9 because: • 1x8+0x4+0x2+1x1 = 9 • Ex. number 12 is represented in BCD as [0001 0010]ITEC 1011 Introduction to Information Technologies
  • 90. 2421 Code • 2421 Code This is a weighted code, its weights are 2, 4, 2 and 1. • A decimal number is represented in 4-bit form and the total four bits weight is 2 + 4 + 2 + 1 = 9. • Hence the 2421 code represents the decimal numbers from 0 to 9.ITEC 1011 Introduction to Information Technologies
  • 91. 5211 Code • 5211 Code This is a weighted code, its weights are 5, 2, 1 and 1. • A decimal number is represented in 4-bit form and the total four bits weight is 5 + 2 + 1 + 1 = 9. • Hence the 5211 code represents the decimal numbers from 0 to 9.ITEC 1011 Introduction to Information Technologies
  • 92. Reflective Code • Reflective Code A code is said to be reflective when code for 9 is complement for the code for 0, and so is for 8 and 1 codes, 7 and 2, 6 and 3, 5 and 4. • Codes 2421, 5211, and excess-3 are reflective, whereas the 8421 code is not.ITEC 1011 Introduction to Information Technologies
  • 93. Sequential Codes • Sequential Codes A code is said to be sequential when two subsequent codes, seen as numbers in binary representation, differ by one. • This greatly aids mathematical manipulation of data. • The 8421 and Excess-3 codes are sequential, whereas the 2421 and 5211 codes are not.ITEC 1011 Introduction to Information Technologies
  • 94. Excess-3 Code • 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). • Example: 1000 of 8421 = 1011 in Excess-3ITEC 1011 Introduction to Information Technologies
  • 95. Error Detecting and Correction Codes • For reliable transmission and storage of digital data, error detection and correction is required.ITEC 1011 Introduction to Information Technologies
  • 96. Error Detecting Codes • When data is transmitted from one point to another there are chances that data may get corrupted. • To detect these data errors, we use special codes, which are error detection codes.ITEC 1011 Introduction to Information Technologies
  • 97. Parity check • In parity codes, every binary message is checked if they have even number of ones or even number of zeros. • Based on this information an additional bit is appended to the original data. • At the receiver side, once again parity is calculated and matched with the received parity, and if they match, data is ok, otherwise data is corrupt. • There are two types of parity: Even parity and Odd ParityITEC 1011 Introduction to Information Technologies
  • 98. Parity • There are two types of parity: • Even parity: Checks if there is an even number of ones; if so, parity bit is zero. When the number of ones is odd then parity bit is set to 1. Message Even parity code xyz xyz p 000 000 0 001 001 1 011 011 0ITEC 1011 Introduction to Information Technologies
  • 99. Parity • Odd Parity: Checks if there is an odd number of ones; if so, parity bit is zero. When number of ones is even then parity bit is set to 1. • • Message Odd parity code xyz xyz p 000 000 1 001 001 0 011 011 1ITEC 1011 Introduction to Information Technologies
  • 100. Alphanumeric Codes • The binary codes that can be used to represent all the letters of the alphabet, numbers and mathematical symbols, punctuation marks, are known as alphanumeric codes or character codes. • These codes enable us to interface the input-output devices like the keyboard, printers, video displays with the computer.ITEC 1011 Introduction to Information Technologies
  • 101. ASCII Code • ASCII Code ASCII stands for American Standard Code for Information Interchange. • It has become a world standard alphanumeric code for microcomputers and computers. • It is a 7-bit code representing 27 = 128 different characters. • These characters represent 26 upper case letters (A to Z), 26 lowercase letters (a to z), 10 numbers (0 to 9), 33 special characters and symbols and 33 control characters. •ITEC 1011 Introduction to Information Technologies
  • 102. ASCII Code • The 7-bit code is divided into two portions, The leftmost 3 bits portion is called zone bits and the 4-bit portion on the right is called numeric bits. Character 7-bit ASCII A 100 0001 B 100 0010 3 011 0011ITEC 1011 Introduction to Information Technologies
  • 103. ASCII Code • An 8-bit version of ASCII code is known as ASCII-8. • The 8-bit version can represent a maximum of 256 characters.ITEC 1011 Introduction to Information Technologies
  • 104. EBCDIC Code • EBCDIC Code EBCDIC stands for Extended Binary Coded Decimal Interchange. • It is mainly used with large computer systems like mainframes. • EBCDIC is an 8-bit code and thus accommodates up to 256 characters. • An EBCDIC code is divided into two portions: 4 zone bits (on the left) and 4 numeric bits (on the right).ITEC 1011 Introduction to Information Technologies