1 / 64

Logic Design

Logic Design. Number Systems Operation. 1- Decimal Numbers. 2- Binary Numbers. 3- Octal Numbers. 4- Hexadecimal Numbers. Decimal Numbers. In the decimal number system each of the ten digits (10 digits), 0 through 9 (0, 1, 2, 3, 4, 5, 6, 7, 8, and 9).

choleman
Download Presentation

Logic Design

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Logic Design

  2. Number Systems Operation 1- Decimal Numbers. 2- Binary Numbers. 3- Octal Numbers. 4- Hexadecimal Numbers.

  3. Decimal Numbers • In the decimal number system each of the ten digits (10 digits), 0 through 9 (0, 1, 2, 3, 4, 5, 6, 7, 8, and 9). • Decimal weight 10 4 10 3 10 2 101 10 0 . 10 -1 10 -2 10 -3 • i.e (345) 10 • 300+40+5=102 *3+101 *4+100 *5=345= (345) 10

  4. Decimal Numbers • i.e (23.5) 10 • 2*101 + 3*100 + 5*10 -1 = 20+3+0.5=23.5 Where 10 0 =1 • Any number to power 0 equal 1

  5. Binary Numbers The binary number system its two digits a base - two system. The two binary digits (bits) are 1 and 0 (1,0). Binary weight 23 22 21 2 0 Weight value 8 4 2 1

  6. Binary – to – Decimal Conversion • Binary number 1101101 • 1 1 0 1 1 0 1 • 2 6 2 5 2 4 2 3 2 2 2 1 2 0 • = 26 *1 + 25 * 1 + 24 *0 + 23 *1 + 22 *1 + 21*0+20*1 • = 64+32+0+8+4+0+1=96+13=109  (109) 10

  7. Decimal – to – Binary Conversion • Example: • Number (58) 10 • 2 58 • 2 29 0 • 2 14 1 • 2 7 0 • 2 3 1 • 2 1 1 • 0 1 (58) 10 = (111010) 2

  8. Octal Numbers • The octal number system is composed of eight digits, which are 0, 1, 2, 3, 4, 5, 6, and 7 • To count above 7, begin another column and start over: 10, 11, 12, 13, 14, 15, 16, and 17. 20, 21, 22, 23, 24, 25, 26, and 27. 30, 31, … … … … 37.

  9. Octal – to – Decimal Conversion Weight … … 83 8 2 8 1 8 0 Example: (2374) 8= 2*83 +3*82 +7*81 +4*80 = 2*512+3*64+7*8+4*1 = 1024+192+56+4 = (1276) 10

  10. Decimal – to – Octal Conversion (359) 10 8 359 8 44 7 8 5 4 0 5 (359) 10 = (547) 8

  11. Octal – to – Binary Conversion • Octal digit can be represented by a 3-bit binary number. • Octal digit binary 0 1 2 3 4 5 6 7 000 001 010 011 100 101 110 111 • Examples: (25) 8 (2 5) 8 (010 101) (010101) 8

  12. Binary – to – Octal Conversion • Conversion binary number to octal number is start with right – most group of three bits and moving from right to left. • Examples: (110101) 2 110101 (6 5) 8 (65) 8

  13. Hexadecimal Numbers The hexadecimal number system has a base of sixteen; it is composed of 16 digits and alphabetic character.

  14. Binary – to – Hexadecimal conversion • 4-bit groups, starting at the right-most bit. • Example: (1100101001010111)2 1100101001010111 C A 5 7 (1100101001010111)2 = (CA57) 16

  15. Hexadecimal – to – Binary Conversion • Example: (10A4) 16 1 0 A 4 0001 0000 1010 0100 (10A4) 16= (1000010100100) 2

  16. Example • Convert the number 235 from decimal to binary 2 235 2 117 1 2 58 1 2 29 0 2 14 1 2 7 0 2 3 1 2 1 1 2 0 1 (235) 10 = (11101011) 2

  17. Example • Convert the number 412 from octal to binary 4 12 100001010 (412) 8 = (100001010) 2

  18. Example • Convert the number 11100011 from binary to octal 011100011 3 4 3 (11100011) 2 = (343) 8

  19. Example • Convert 1000101 from binary to decimal = 26 *1 + 25 * 0 + 24 *0 + 23 *0 + 22 *1 + 21*0+20*1 = 64 + 0 + 0 + 0 + 4 + 0 + 1 = 69 (1000101) 2 = (69) 10

  20. Example • Convert 1110110101 from binary to hexadecimal • 001110110101 • 3 B 5 • (1110110101) 2 = (3B5) 16

  21. Example • Convert 671 from octal to hexadecimal • 5 7 1 • 101111001 • 000101111001 1 7 8 (571) 8 = (178) 10

  22. Example Convert 2CA from hexadecimal to Octal 2 C A 001011001010 001011001010 1 3 1 2 (2CA) 16 = (1312) 8

  23. Homework 1 • Convert 246 from octal to decimal. Hint: you can divided the number to 8 until you get the number zero or you can convert the number to binary and from binary to decimal

  24. Homework 2 Convert E9B from hexadecimal to decimal. Hint or you can convert the number to binary and from binary to decimal

  25. Math Operations 1. Adding Operation . 2. subtraction operation. Note: multiplication is a repeated adding operation and division operation is a repeated subtraction operation

  26. Adding in Octal Numbering System 1 1 Carry ( 5 6 4 ) 8 ( 4 2 7 ) 8 • (1 2 1 3 ) 8 1 Carry ( 3 7 1 ) 8 ( 3 1 2 ) 8 ( 7 0 3 ) 8 ( 2 3 5 ) 8 ( 4 3 2 ) 8 ( 6 6 7 ) 8

  27. Adding in Hexadecimal Numbering System ( 1 7 4 ) 16 ( A 5 5 ) 16 ( B C 9 ) 16 1 1 Carry ( F 8 B ) 16 ( 2 8 7 ) 16 (1 2 1 2 ) 16 1 Carry ( 4 5 A ) 16 ( 6 E 2 ) 16 ( B 3 C ) 16

  28. Adding in Binary Numbering System 1111 1 Carry ( 110111) 2 ( 101111) 2 (1100110 ) 2 ( 1001) 2 ( 0100) 2 (1101 ) 2 11 Carry ( 100110) 2 ( 010011) 2 ( 1110 01 ) 2

  29. Subtraction in Octal Numbering System 312 ( 4 2 6 ) 8 ( 2 6 3 ) 8 ( 1 4 3 ) 8 7 51015 ( 6 0 5 ) 8 ( 3 6 7 ) 8 ( 2 1 6 ) 8 ( 6 4 5 ) 8 ( 1 3 2 ) 8 ( 5 1 3 ) 8

  30. Subtraction in Hexadecimal Numbering System C 13 ( F D 3 ) 16 ( A 6 8 ) 16 ( 5 6 B ) 16 17 8 7 11 ( 9 8 1 ) 16 ( 4 A 4 ) 16 ( 4 D D ) 16 ( F 9 5 ) 16 ( B 6 4 ) 16 ( 4 3 1 ) 16

  31. Subtraction in Binary Numbering System 0 10 ( 11001) 2 ( 10101 )2 (00100)2 1 0 1010 ( 11001) 2 ( 10110 )2 (00011)2 ( 110111)2 ( 100100)2 ( 010011 )2

  32. Subtraction in Binary Numbering System • Subtraction using 1’s • M= 01010100 • N= 01000100 • Find = M-N • Converting N to 1’s Complement N=10111011 • Convert the subtraction operation to adding. • Is there carry in extra bit, the carry will be added to the number. ( 01010100)2 ( 10111011)2 ( 100001111 )2 1 ( 00010000 ) 2

  33. Subtraction in Binary Numbering System M= 01010100 N= 01000100 Find = N-M ( 01000100)2 ( 10101011)2 (11101111)2 No Carry

  34. Subtraction in Binary Numbering System ( 1001)2 ( 1010)2 ( 10011 )2 The carry discarded it is mean it is positive number Subtraction using 2’s • M=1001 • N=0110 • Find = M-N • Converting N to 2’s complement • The 1’s complement of 0110 = 1001 • Adding 1 to 1001 = 1010

  35. Subtraction in Binary Numbering System ( 0110)2 ( 0111)2 ( 01101 )2 There is no carry bit, this means it is negative number to see the number convert the result to 2’complent. Subtraction using 2’s • M=1001 • N=0110 • Find = N-M • Converting M to 2’s complement • The 1’s complement of 1001 = 0110 • Adding 1 to 0110 = 0111

  36. AND Gate

  37. OR Gate

  38. Inverter (NOT) Gate

  39. NAND Gate

  40. NOR Gate

  41. XOR Gate

  42. XNOR Gate

  43. Half Adder The half adder adds two binary digits and produces two outputs as sum and carry; XOR is applied to both inputs to produce sum and AND gate is applied to both inputs to produce carry.

  44. Full Adder A full - adder is a combinational circuit that forms the arithmetic sum of three input bits. It consists of three inputs and two outputs.

  45. Boolean Algebra • Boolean algebra is a mathematical system for the manipulation of variables that can have one of two values. • In formal logic, these values are “true” and “false.” • In digital systems, these values are “on” and “off,” 1 and 0, or “high” and “low.” • Boolean expressions are created by performing operations on Boolean variables. • Common Boolean operators include AND, OR, and NOT.

  46. 3.2 Boolean Algebra • A Boolean operator can be completely described using a truth table. • The truth table for the Boolean operators AND and OR are shown at the right. • The AND operator is also known as a Boolean product. The OR operator is the Boolean sum.

  47. 3.2 Boolean Algebra • The truth table for the Boolean NOT operator is shown at the right. • The NOT operation is most often designated by an overbar. It is sometimes indicated by a prime mark ( ‘ ) or an “elbow” ().

  48. 3.2 Boolean Algebra • A Boolean function has: • At least one Boolean variable, • At least one Boolean operator, and • At least one input from the set {0,1}. • It produces an output that is also a member of the set {0,1}.

  49. 3.2 Boolean Algebra • The truth table for the Boolean function: is shown at the right.

  50. 3.2 Boolean Algebra • As with common arithmetic, Boolean operations have rules of precedence. • The NOT operator has highest priority, followed by AND and then OR. • This is how we chose the (shaded) function subparts in our table.

More Related