Chapter 2 Data Representation on CPU (part 1)

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This topic introduces the numbering systems: decimal, binary, octal and hexadecimal. The topic covers the conversion between numbering systems, binary arithmetic, one's complement, two's complement, signed number and coding system. This topic also covers the digital logic components.

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Chapter 2 Data Representation on CPU (part 1)

  1. 1. Chapter 2: DATA REPRESENTATION IN COMPUTER MEMORY SUMMARY: This topic introduces the numbering systems: decimal, binary, octal and hexadecimal. The topic covers the conversion between numbering systems, binary arithmetic, one's complement, two's complement, signed number and coding system. This topic also covers the digital logic components. CLO 2:apply appropriate method to solve arithmetic problem in numbering system (C3). RTA: (08 : 08)
  2. 2. 2.1 Understand data representation on CPU. 2.1.1 Define decimal, binary, octal and hexadecimal number. 2.1.2 Perform arithmetic operation (addition and subtraction) in different number bases. 2.1.3 Convert decimal, binary, octal and hexadecimal numbers to different bases and vice-versa
  3. 3. INTRODUCTION  The binary system and decimal system is most important in digital system.  Decimal - Universally used to represent quantities outside a digital system.  Its means, there will be situations decimal values must be converted to binary values before entered to digital system.  Example : Calculator / Computer
  4. 4. DECIMAL NUMBERING SYSTEM  Decimal system is composed of 10 numerals or symbols.  These 10 sysmbols are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.  Using these symbols as digits of a number, it can express any quantity.
  5. 5.  Base number = 10  Basic number = 0,1,2,3,4,5,6,7,8,9 23410 Basic number Base number
  6. 6. Positional Values (weights) 2 7 4 6 . 2 102 103 101 100 10-1 Positional values (weights) MSD LSD 2746.210 is from calculation below: 2746.2 = (2x103 ) + (7x102 ) + (4x101 ) + (6x100 ) + (2x10-1 ) = 2000 + 700 + 40 +6 +0.2 = 2746.2 Most significant digit Least significant digit
  7. 7. BINARY NUMBERING SYSTEM  Define Binary numbers  Binary numbers representing number in which only digits 0 or 1.  ADDITION BINARY NUMBERS Basic binary addition rule : 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 10 1 + 1 + 1 = 11  Example : 101 + 101 = 1010 1011 + 1011 = ?
  8. 8.  Ex 1: 110112 + 100012 = 1011002  Ex 2:  101112 + 1112 = ________ Exercise
  9. 9. Subtraction Four conditions in binary subtraction  0 - 0 = 0  0 - 1 = 1 borrow 1  1 - 0 = 1  1 - 1 = 0  10 - 1 = 1  If a 10 being borrow a 1, what’s left with that 10 is a 1
  10. 10.  Ex 1: 10012 – 102 = 1112  Ex 2:  1010112 – 11112 =__________
  11. 11. Conversions of Binary Numbers Binary to Decimal conversions  Example : 1 1 0 1 12  24 + 23 + 22 + 21 + 20 = 16 + 8 + 2 + 1 = 2710
  12. 12. Decimal to Binary conversions  Convert 2510 to binary number Exercise: Convert 3010 to binary number
  13. 13. OCTAL NUMBERING SYSTEM  The octal number system has a base of eight, meaning that it has eight possible digits: 0,1,2,3,4,5,6 and 7.  The digit positions in an octal number have weights as follows : 84 83 82 81 80 •. 8-1 8-2 8-3
  14. 14.  Ex: 5248 – 1678 = 3358 1678 – 248 = _________ Octal number - (Subtraction - Pengurangan)
  15. 15. Octal –to-decimal conversion Convert 3728 to decimal number 3728 = 3 x (8 2 ) + 7 x (8 1 ) + 2 x (8 0 ) = (3 x 64) + (7x 8) + (2 x 1) = 25010
  16. 16. Octal number - Addition (Penambahan)  Ex: 1238 + 3218 4448  Ex: 4578 + 2458
  17. 17.  Decimal integer can be converted to octal by using the same repeated-division method with a division factor of 8. Decimal-to-Octal Conversion
  18. 18. Octal –to- Binary conversion
  19. 19. Binary to Octal conversion
  20. 20.  The hexadecimal number system uses base 16.  It has 16 possible digit symbols.  It uses the digits 0 through 9 plus the letters A, B, C, D, E and F as the 16 digit symbols. HEXADECIMAL NUMBERING SYSTEM 7A16 Basic number Base number
  21. 21. Hexadecimal number - Addition (Penambahan)  Ex: 3 316  + 4 716  Ex: 2 0 D 316 + 1 2 B C16 7 A16
  22. 22. Hexadecimal number - Subtraction (Pengurangan)  Ex: 4 416 - 1 716  Ex: 3 2 5 516 - 3 1 8 216 2 D16
  23. 23. Hexadecimal-To-Decimal Conversion  A hexadecimal number can be converted to its decimal equivalent by using the fact that each hex digit position has a weight that is a power of 16.
  24. 24.  Ex 1: 1416 = (1 x 161 ) + (4 x 160 ) = 16 + 4 = 2010
  25. 25.  Ex 2: ABC16 = (10 x 162 ) + (11 x 161 ) = (12 x 160 ) = 2560 + 176 + 12 = 274810
  26. 26.  Decimal to hex conversion can be done using repeated division by 16.  Ex: Convert 2010 to hex Decimal-To-Hexadecimal Conversion 16 20 16 1 4 2010 = 1416
  27. 27.  Like the octal number system, the hexadecimal number system is used primarily as a “shorthand” method for representing binary numbers.  It is a relatively simple matter to convert a hex number to binary .  Each hex digit is converted to its four-bit binary equivalent. Hexadecimal-to-Binary Conversion
  28. 28.  Ex: 111010012 = __________ 00101101001111002 =___________
  29. 29.  The binary number is grouped into groups of four bits, and each group is converted to its equivalent hex digit.  Zero are added, as needed to complete a four-bit group.  Ex: 1012 = 0101 = 516 Binary-to-Hexadecimal Conversion
  30. 30. Summary Hexadecimal Decimal Binary 0 0 0000 1 1 0001 2 2 0010 3 3 0011 4 4 0100 5 5 0101 6 6 0110 7 7 0111
  31. 31. Summary Hexadecimal Decimal Binary 8 8 1000 9 9 1001 A 10 1010 B 11 1011 C 12 1100 D 13 1101 E 14 1110 F 15 1111
  32. 32. 2.1.4 Describe the coding system a. Sign and magnitude b. 1’s Complement and 2’s Complement c. Binary Coded Decimal (BCD system) d. ASCII and EBCDIC
  33. 33. Describe the coding system Sign and Magnitude Positive sign, ( 0 ) or Negative sign ( 1 ) Magnitude = Size Or value
  34. 34. Describe the coding system Sign and Magnitude Example 1 : Represent -9 in sign and magnitude. -9 sign magnitude 1 1001 11001 -9 in sign &magnitude value is 11001
  35. 35. Describe the coding system Sign and Magnitude Example 2 : Represent 25 in sign and magnitude. +25 sign magnitude 0 11001 011001 25 in sign & magnitude value is 011001
  36. 36. One’s Complements and Two’s Complements  One’s Complements  One’s complements is used in binary number.  The one’s complement of a binary number is obtained by changing each 0 to 1 and 1 to a 0.  Only change negative number  In other words, change each bit in the number to its complement.
  37. 37.  Exp:  10011001 – original binary number  01100110 – complement each bit to form 1’s complement  Thus, we say that the 1’s complement of 10011001 is 01100110.
  38. 38.  Exp:  Convert -2710 to 1’s complement a) -2710 = 001002  Convert -4510 to 1’s complement -------------
  39. 39.  Two’s Complement  The 2’s complement of a binary number is formed by taking the 1’s complement of the number and adding 1 to the least-significant-bit (LSB) position.  Exp: 101101 binary equivalent of 45 010010 complement each bit to form 1’s complement + 1 add 1 010011 2’s complement of original binary number Two’s complement = One’s Complement + 1
  40. 40. Exercise 1. Convert the number below to 1’s complement 2’s complement ------------------- ------------------- -101110012 0100 0110 0100 0111 -5768 01000 0001 01000 0010 -124516 -45
  41. 41. Addition in 1’s complement  Exp 1 : 810 + (-310) 1000 8 change to binary number + 1100 -3 change to 1’s complement 10100 1 add carry to LSB 0101  Exercise 2 : 510 + (-210) ----------------
  42. 42.  Exp 1: 810 + (-310) 1100 -3 change to 1’s complement + 1 1101 -3 change to 2’s complement + 1000 8 change to binary number 10101 Addition in 2’s complement This carry is disregarded, the result is 0101 (sum=5)
  43. 43.  Exp 2: -810 + 310 0111 -8 change to 1’s complement + 1 1000 -8 change to 2’s complement + 0011 3 change to binary number 1011 negative sign bit
  44. 44.  Only binary number which have –ve sign need to change to 1’s complement. If the number is decimal number, change the number to binary number.  The –ve number that already change to 1’s complement, means that the number already change to +ve. So that, the subtraction process have change to addition process.  Overflow bit in addition process need to carry to LSB and add with the number. Subtraction in 1’s complement
  45. 45.  Exp 1: 2510 – 1310 Step 1 : Convert 2510 and -1310 to binary number. 2510 = 110012 1310 = 11012 Step 2 : Change 1310 = 1101 to 1’s complement 1310 = 11012 change to 1’s complement = 10010 11001 25 change to binary number + 10010 -13 change to 1’s complement 101011 + 1 add 1 01100 total=12
  46. 46.  Change the number are given to binary number.  For each –ve binary number, we must change to 1’s complement (change 0 to 1 and 1 to 0).  Then, add the number with 1. Subtraction in 2’s complement
  47. 47.  Exp 1: 2510 – 1310 Step 1 : Convert 2510 and -1310 to binary number. 2510 = 110012 1310 = 11012 Step 2 : Change -1310 to 2’s complement -1310 = 011012 change to 1’s complement = 10010 10010 change to 2’s complement = 10011 11001 25 change to binary + 10010 -13 change to 2’s complement 101100 total=12 Carry disregard
  48. 48.  Solve this arithmetic with 2’s complement. i. 428 – 158 ii. 101110 - 56910 Exercise
  49. 49. Signed Number  Addition of Signed Number  Addition number same sign  Exp: +4 + (+8) = +12 +4 00000100 +8 00001000 +12 00001100
  50. 50.  Addition number different sign  Exp 1: (-4) + (+8) = +4 -4 11111100 + (+8) 00001000 +4 1 00000100 This carry is disregarded 00000100 +4 11111011 1’s complements + 1 11111100 2’s complements
  51. 51.  Exp 1: (-12) + (+5) = -7 -12 11110100 + +5 00000101 -7 11111001 00001100 +12 11110011 1’s complements + 1 11110100 2’s complements Negatif sign bit
  52. 52. Subtraction of Signed Number Positive Number (-) Negative Number  Exp: +10 – (-5) = +10 + (+5) = + 15 +10 00001010 + 5 00000101 +15 00001111
  53. 53. Negative Number (-) Positive Number  Exp: -10 – (+5) = - 15 - 10 11110110 11110110 - (+5) 00000101 + 11111011 -15 111110001
  54. 54. Negative Number (-) Negative Number  Exp: -10 – (-5) = -10 + (5) = - 5 -10 11110110 - (- 5) + 00000101 - 5 11111011
  55. 55. BCD Code  BCD – Binary Coded Decimal  It’s contain BCD 8421, 2421, 3321, 5421, 5311, 4221 and etc.  The common used is BCD 8421 and BCD 2421.
  56. 56. 4 bit BCD Code Desimal 5421 5311 4221 3321 2421 8421 7421 0 0000 0000 0000 0000 0000 0000 0000 1 0001 0001 0001 0001 0001 0001 0001 2 0010 0011 0010 0010 0010 0010 0010 3 0011 0100 0011 0011 0011 0011 0011 4 0100 0101 1000 0101 0100 0100 0100 5 1000 1000 0111 1010 1011 0101 0101 6 1001 1001 1100 1100 1100 0110 0110 7 1010 1011 1101 1101 1101 0111 1000 8 1011 1100 1110 1110 1110 1000 1001 9 1100 1101 1111 1111 1111 1001 1010
  57. 57. Binary-Coded-Decimal Code  If each digit of a decimal number is represented by its binary equivalent, the result is a code called binary-coded-decimal.  Decimal digit can be as large as 9, four bits are required to code each digit (the binary code for 9 is 1001)
  58. 58.  Exp: 87410 8 7 4 (decimal) 1000 0111 0100 (BCD)
  59. 59. BCD 8421 Code – to – Binary Number  Exp: Convert 1001 0110BCD 8421 to binary number. Step 1: Change BCD 8421 code to decimal number. 1001 0110 9 6 Step 2 : Change decimal number to binary number. 1001 0110BCD 8421 = 11000002
  60. 60.  Exp: Convert 10010102 to BCD 8421 code. Step 1: Change binary number to decimal number. 10010102 = 7410 Step 2: Change decimal number to BCD 8421 code. 10010102 = 01110111BCD 8421 Binary Number – to – BCD 8421 Code
  61. 61.  The most widely used alphanumeric code is the American Standard Code for Information Interchange (ASCII).  The ASCII code is a seven-bit code and so it has 27 = 128 possible code groups. ASCII Code
  62. 62. MSB LSB Binary 000 001 010 011 100 101 110 111 Binary Hex 0 1 2 3 4 5 6 7 0000 0 Nul Del sp 0 @ P p 0001 1 Soh Dc1 1 1 A Q a q 0010 2 Stx Dc2 “ 2 B R b r 0011 3 Etx Dc3 # 3 C S c s 0100 4 Eot Dc4 $ 4 D T d t 0101 5 End Nak % 5 E U e u 0110 6 Ack Syn & 6 F V f v 0111 7 Bel Etb ‘ 7 G W g w 1000 8 Bs Can ( 8 H X h x 1001 9 HT Em ) 9 I Y i y 1010 A LF Sub . : J Z j z 1011 B VT Esc + ; K k 1100 C FF FS , < L l 1101 D CR GS - = M m 1110 E SO RS . > N n 1111 F SI US / ? O o ASCII code
  63. 63.  Exp: An operator is typing in a BASIC program at the keyboard of a certain microcomputer. The computer converts each keystroke into its ASCII code and stores the code as a byte in memory. Determine the binary strings that will be entered into memory when operator types in the following BASIC statement: GOTO 25
  64. 64.  Solution: Locate each character (including the space) and record ASCII code. G 01000111 O 01001111 T 01010100 O 01001111 (space) 00100000 2 00110010 5 00110101 *0 was added to the leftmost bit of each ASCII code because the Codes must be stored as bytes (eight bits).
  65. 65. Exercise : 1.The following message encode in ASCII code. What the meaning of this code ?  a) 54 4F 4C 4F 45 47  b) 48 45 4C 4C 4F  c) 41 50 41 4B 48 41 42 41 52
  66. 66. EBCDIC  Stand for Extended Binary Coded Decimal Interchange Code.  was first used on the IBM 360 computer, which was presented to the market in 1964.  Used with large computer such as mainframe.

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