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# 2.1 data represent on cpu

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### 2.1 data represent on cpu

1. 1. Chapter 2.0:Data Representation on CPUFP203 : Computer Organization
2. 2. Topic Cover2.1 Number System (decimal, binary, octal, and hexadecimal)2.2 Arithmetic Operation in number system.2.3 Convert Decimal, Binary, Octal and Hexadecimal Numbers to different bases.2.4 Coding system: Sign and magnitude, 1‟s Complement and 2‟s Complement Binary Coded Decimal (BCD system) ASCII and EBCDIC
3. 3. INTRODUCTION ExamplesReal World Computer Data Input device DataDear Mom: Keyboard 10110010… Digital 10110010… camera
4. 4. 2.1 Number System• Many number system are in use in digital technology.• Most common are: – Decimal, N10 – Binary, N2 – Octal, N8 – Hexadecimal, N16
5. 5. 2.2 ARITHMETIC OPERATION• Arithmetic operation in number system consist of: – Addition Only cover this 2 topics – Subtraction – Multiplication – Division
6. 6. Decimalnumber system • Decimal system is composed of 10 numerals or symbol. • Symbol: 0,1,2,3,4,5,6,7,8,9 10 Symbol • Example: 23410 Multiplier: 103 102 101 100 . 10-1 = 1000 = 100 = 10 =1 . = 0.1
7. 7. Example: 2746.210 This number is came from this calculation: 2 7 4 2 . 2 103 102 101 100 . 10-1 = 1000 = 100 = 10 =1 . = 0.12746.210 = (2x1000) + (7x100) + (4x10) + (2x1) + (2x0.1) = 2000 + 700 + 40 + 2 + 0.2 = 2746.2 Decimal number = Natural Number
8. 8. Arithmetic Operation Decimal + -
9. 9. Decimal AdditionExample:a. 89310 + 32110 = b. 75710 + 24510 = 89310 75710 + 24510 +32110 100210 121410Try this : 73310 + 79910 = ?
10. 10. Decimal SubtractionExample:a. 5410 - 1710 = b. 15710 - 8910 = 5410 15710 - 1710 - 8910 3710 6810Try this : 533310 - 3710 = ?
11. 11. Octalnumber system • Octal system is composed of 8 numerals or symbol. • Symbol: 0,1,2,3,4,5,6,7 8 Symbol • Example: 658 Multiplier: 83 82 81 80 . 8-1 = 512 = 64 =8 =1 . = 0.125
12. 12. Example: 107.158 This number can be convert to decimal value using this calculation: 1 0 7 . 1 5 82 81 80 . 8-1 8-2 = 64 =8 =1 . = 0.1250 = 0.0156107.158 = (1x64)+(0x8)+(7x1)+(1x0.1250)+(5x0.0156) = 64 + 0 + 7 + 0.1250 + 0.078 = 71.20310
13. 13. Arithmetic Operation Octal + -
14. 14. Octal AdditionSekiranya setiap hasil perjumlahan yang melebihi atau samadengan 8 mestilah ditolak dengan 8.Example:a. 1238 + 3218 = b. 4578 + 2458 = 1238 4578 +3218 + 2458 4448 7248Try this : 7338 + 748 = ?
15. 15. Octal SubtractionSekiranya terdapat peminjam, nombor peminjam mestilahdijumlahkan dengan 8.Example:a. 5248 - 1678 = b. 1678 - 248 = 5248 1678 - 1678 - 248 3358 1438Try this : 15238 - 3648 = ?
16. 16. Binarynumber system • Binary system is composed of 2 numerals or symbol. • Symbol: 0,1 2 Symbol • Example: 1012 Multiplier: 25 24 23 22 21 20 = 32 = 16 =8 =4 =2 =1
17. 17. Example: 10.1012 This number can be convert to decimal value using this calculation: 1 0 . 1 0 1 21 20 . 2-1 2-2 2-3 =2 =1 . = 0.5000 = 0.2500 = 0.125010.1012 = (1x2)+(0x1)+(1x0.5)+(0x0.25)+(1x0.125) = 2 + 0 + 0.5 + 0 + 0.125 = 2.62510
18. 18. Arithmetic Operation Binary + -
19. 19. Binary AdditionThe four basic rules for adding binary digits are as follows: 0+0=0 0+1=1 1+0=1 1 + 1 = 0 carry 1Example:110112 + 100012 = 110112+ 100012 1011002Try this : 101112 + 1112 = ?
20. 20. Binary SubtractionThe four basic rules for subtracting binary digits are as follows:- 0-0=0 0 - 1 = 1 borrow 1 1-0=1 1-1=0Example:10012 -102 = 10012 - 102 1112Try this : 1010112 – 11112 =?
21. 21. Binary SubtractionHave previously looked at the subtraction operation. Aquick review.Just like subtraction in any other base 10110 -10010 00100• And when a borrow is needed. Note that the borrow gives us 2 in the current bit position..
22. 22. Example
23. 23. In General• When there is no borrow into the msb position, then the subtrahend in not larger than the minuend and the result is positive and correct.• If a borrow into the msb does occur, then the subtrahend is larger than the minuend.
24. 24. Consider• Now do the operation 4 – 6• Correct difference is -2 or -0010• Different because 2n was brought in and made the operation M- N+2n
25. 25. Hexadecimalnumber system • Hexadecimal system is composed of 16 numerals or symbol. 10 11 12 13 14 15 • Symbol: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F 16 Symbol • Example: 7A16 Multiplier: 163 162 161 160 . 16-1 = 4096 = 256 = 16 =1 . = 0.0626
26. 26. Example: B6F.7C16 This number can be convert to decimal value using this calculation: B 6 F . 7 C 162 161 160 . 16-1 16-2 = 256 = 16 =1 . 0.0625 = 0.0039B6F.7C16 = (11x256) + (6x16) + (15x1) + (7x0.0625) + (12x0.0039) = 2816 + 96 + 15 + 0.4375 + 0.0468 = 2927.484310
27. 27. Arithmetic Operation Hexadecimal + -
28. 28. Hexadecimal AdditionSekiranya setiap hasil perjumlahan yang melebihi atau samadengan 16 mestilah ditolak dengan 16.Example:a. 3316 + 4716 = b. 20D316 + 12BC16 = 3316 20D316 + 4716 + 12BC16 338F16 7A16Try this : DF16 + AB16 = ?
29. 29. Hexadecimal SubtractionNilai yang kecil daripada 16 boleh dipinjam dari sebelah dengannilai 16.Example:a. 4416 - 1716 = b. 20D316 - 12BC16 = 20D316 4416 - 12BC16 - 1716 0E1716 2D16Try this : DF16 - AB16 = ?
30. 30. 2.3: Convert Decimal, Binary, Octal and Hexadecimal Numbers to different bases
31. 31. Convert Binary to Decimal (N2 – N10)Example: 1111012 This number can be convert to decimal value using this calculation: 1 1 1 1 0 1 25 24 23 22 21 20 = 32 = 16 =8 =4 =2 =1 1111012 = (1x32)+(1x16)+(1x8)+(1x4)+(0x2)+(1x1) = 32 + 16 + 8 + 4 + 0 + 1 = 6110 Try this: Convert 1100.10112 to decimal? Convert 100.10112 to decimal?
32. 32. Convert Binary to Octal (N2 - N8)Convert Binary to Octal adalah dengan membahagikan nomborBinary tersebut kepada 3 bit bermula dari sebelah kanan (LSB) LSB 1111012 1 1 1 1 0 1 22 21 20 22 21 20 =4 =2 =1 =4 =2 =1 1111012 = [(1x4)+(1x2)+(1x1)] [(1x4)+(0x2)+(1x1)] = [4 + 2 + 1][ 4 + 0 + 1] = 758 Try this: Convert 110010112 to Octal?
33. 33. Convert Binary to Hexadecimal (N2 – N16)Convert Binary to Hexadecimal adalah dengan membahagikannombor binary kepada 4 bit bermula dari LSB. Sekiranya bit tersebuttidak mencukupi, maka digit „0‟ perlu ditambah pada MSB LSB 01012 0 1 0 1 23 22 21 20 =8 =4 =2 =1 01012 = (0x8)+(1x4)+(0x2)+(1x1) =0+4+0+1 = 516 Try this: Convert 101111012 to Hexadecimal?
34. 34. Convert Decimal to Binary (N10 – N2)Example: Convert 1810 to binary 2 2 18 9 181 0 2 4 1 2 2 0 2 1 0 0 1 1810 = 100102 Try this: Convert 32.20210 to binary? Convert 8910 to binary?
35. 35. Convert Decimal to Octal (N10 – N8)Example: Convert 300.3410 to Octal i. 300 Divide by 8 ii. 0.34 Multiply by 8 8 300 8 37 4 8 4 5 8 0 4 454 0.3410 = 0.34 x 8 = 2.72 ( 2+0.72 ) 0.72 x 8 = 5.76 ( 5+0.76 ) 0.76 x 8 = 6.08 ( 6+0.08 ) 0.08 x 8 = 0.64 ( 0+0.64 ) 0.25605 0.64 x 8 = 5.12 ( 5+0.12 ) 300.3410 = 454.256058Try this: Convert 32.20210 to Octal?
36. 36. Convert Decimal to Hexadecimal (N10 – N16)Example: Convert 2010 to Hexadecimal 16 20 Balance 16 1 4 0 12010= 1416Try this: Convert 343410 to hexadecimal?
37. 37. Convert Octal to Binary (N8 – N2)Convert Octal to Binary adalah dengan menukar setiap digit oktalkepada nilai 3 bit binary nya MSB 4 58 LSB 1 0 0 1 0 1 22 21 20 22 21 20 =4 =2 =1 =4 =2 =1 1001018 = [(1x4)+(0x2)+(0x1)] [(1x4)+(0x2)+(1x1)] = [4 + 0 + 0][ 4 + 0 + 1] = 100 1012 Try this: Convert 110010112 to Octal?
38. 38. Convert Hexadecimal to Binary (N16 – N2) Convert Octal to Binary adalah dengan menukar setiap digit hexadecimal kepada nilai 4 bit binary nya MSB 3 A16 LSB 0 0 1 1 1 0 1 0 23 22 21 20 23 22 21 20 =8 =4 =2 =1 =8 =4 =2 =1 3A16 = [(0x8)+(0x4)+(1x2)+(1x1)][(1x8)+(0x4)+(1x2)+(0x1)] = [0 + 0 + 2 + 1][ 8 + 0 + 2 + 0] = 0011 10102Try this: Convert EFA16 to Binary?
39. 39. 2.4: Coding System Sign and magnitude, 1‟s Complement & 2‟s Complement
40. 40. 8-Bit Binary Number SystemApply what you have learned to the +127 01111111 pos(+)binary number systems. How do you +126 01111110represent negative numbers in this 8-bit +125 01111101binary system?Cut the number system in half. +1 00000001 0 00000000Use 00000001 – 01111111 to indicate -1 11111111positive numbers. -2 11111110Use 10000000 – 11111111 to indicatenegative numbers. -127 10000001 neg(-) -128 10000000Notice that 00000000 is not positive ornegative.
41. 41. Representing Negative Numbers• As there is no third symbol available to store a negative symbol explicitly we must use a bit to show if a number is negative or not. – We name this bit the „Sign Bit‟ – We use the leftmost bit. – If the „Sign Bit‟ is 1 then the number is negative, if it is 0 then it is positive.
42. 42. Sign Bit• What did do you notice about the +127 01111111 pos(+) most significant bit of the binary +126 01111110 numbers? +125 01111101• The MSB is (0) for all positive numbers. +1 00000001• The MSB is (1) for all negative 0 00000000 numbers. -1 11111111 -2 11111110• The MSB is called the sign bit.• In a signed number system, this -127 10000001 allows you to instantly determine -128 10000000 neg(-) whether a number is positive or negative.
43. 43. 1‟s Complement• This is just inverting each bit. flip the 0000010 number. 1 1 11 1 0 1 1‟s compliment of 00000010 is 1111101
44. 44. 2‟S Complement ProcessThe steps in the 2’s Complement processFirst, complement all of the digits in a number. – A digit‟s complement is the number you add to the digit to make it equal to the largest digit in the base (i.e., 1 for binary). In binary language, the complement of 0 is 1, and the complement of 1 is 0.Second, add 1. – Without this step, our number system would have two zeroes (+0 & -0), which no number system has.
45. 45. 2‟s Complement ExamplesExample #1 5 = 00000101 Complement Digits  11111010 +1 Add 1 -5 = 11111011Example #2 -13 = 11110011 Complement Digits  00001100 +1 13 = 00001101
46. 46. Using The 2‟s Compliment ProcessUse the 2‟s complement process to add togetherthe following numbers. POS 9 NEG (-9)+ POS  + 5 + POS  + 5 POS 14 NEG -4 POS 9 NEG (-9)+ NEG  + (-5) + NEG  + (-5) POS 4 NEG -4
47. 47. POS + POS → POS AnswerIf no 2‟s complement is needed, use regular binaryaddition. 9  00001001 + 5  + 00000101 14  00001110
48. 48. POS + NEG → POS AnswerTake the 2‟s complement of the negative number anduse regular binary addition. 9  00001001 + (-5) + 11111011 4  1]00000100 8th Bit = 0: Answer is Positive Disregard 9th Bit 00000101  2’s 11111010 Complement Process +1 11111011
49. 49. POS + NEG → NEG Answer Take the 2‟s complement of the negative number and use regular binary addition. (-9) 11110111 + 5  + 00000101 -4  11111100 8th Bit = 1: Answer is Negative 11111100 00001001 To Check:   2’sPerform 2’sComplement 00000011 11110110 Complement ProcessOn Answer +1 +1 00000100 11110111
50. 50. NEG + NEG → NEG Answer Take the 2‟s complement of both negative numbers and use regular binary addition. 2’s Complement (-9)  11110111 Numbers, See Conversion Process + (-5)  + 11111011 In Previous Slides -14  1]11110010 8th Bit = 1: Answer is Negative Disregard 9th Bit 11110010 To Check: Perform 2’sComplement 00001101On Answer +1 00001110
51. 51. 2.4: Coding System Binary Coded Decimal (BCD System) ASCII and EBCDIC
52. 52. Binary-Coded Decimal (BCD) Four bits per digit Digit Bit pattern Note: the following bit 0 0000 patterns are not used: 1 0001 1010 2 0010 1011 1100 3 0011 1101 4 0100 1110 5 0101 1111 6 0110 7 0111 8 1000 9 1001
53. 53. Example• 709310 = ? (in BCD) 7 0 9 3 0111 0000 1001 0011
54. 54. ASCII• ASCII = American National Standard Code for Information Interchange• 7-bit code• 8th bit is unused (or used for a parity bit)• 27 = 128 codes• Two general types of codes: – 95 are “Graphic” codes (displayable on a console) – 33 are “Control” codes (control features of the console or communications channel)
55. 55. ASCII Chart 000 001 010 011 100 101 110 1110000 NULL DLE 0 @ P ` p0001 SOH DC1 ! 1 A Q a q0010 STX DC2 " 2 B R b r0011 ETX DC3 # 3 C S c s0100 EDT DC4 \$ 4 D T d t0101 ENQ NAK % 5 E U e u0110 ACK SYN & 6 F V f v0111 BEL ETB 7 G W g w1000 BS CAN ( 8 H X h x1001 HT EM ) 9 I Y i y1010 LF SUB * : J Z j z1011 VT ESC + ; K [ k {1100 FF FS , < L l |1101 CR GS - = M ] m }1110 SO RS . > N ^ n ~1111 SI US / ? O _ o DEL
56. 56. 000 001 010 011 100 101 110 1110000 NULL DLE 0 @ P ` p0001 SOH DC1 ! 1 A Q a q0010 STX DC2 " 2 B R b r0011 ETX DC3 # 3 C S c s0100 EDT DC4 Most significant bit \$ 4 D T d t0101 ENQ NAK % 5 E U e u0110 ACK SYN & 6 F V f v0111 BEL ETB 7 G W g w1000 BS CAN ( 8 H X h x1001 HT EM ) 9 I Y i y1010 LF SUB * : J Z j z1011Least significant ESC VT bit + ; K [ k {1100 FF FS , < L l |1101 CR GS - = M ] m }1110 SO RS . > N ^ n ~1111 SI US / ? O _ o DEL
57. 57. e.g., ‘a’ = 1100001 000 001 010 011 100 101 110 1110000 NULL DLE 0 @ P ` p0001 SOH DC1 ! 1 A Q a q0010 STX DC2 " 2 B R b r0011 ETX DC3 # 3 C S c s0100 EDT DC4 \$ 4 D T d t0101 ENQ NAK % 5 E U e u0110 ACK SYN & 6 F V f v0111 BEL ETB 7 G W g w1000 BS CAN ( 8 H X h x1001 HT EM ) 9 I Y i y1010 LF SUB * : J Z j z1011 VT ESC + ; K [ k {1100 FF FS , < L l |1101 CR GS - = M ] m }1110 SO RS . > N ^ n ~1111 SI US / ? O _ o DEL
58. 58. 95 Graphic codes 000 001 010 011 100 101 110 1110000 NULL DLE 0 @ P ` p0001 SOH DC1 ! 1 A Q a q0010 STX DC2 " 2 B R b r0011 ETX DC3 # 3 C S c s0100 EDT DC4 \$ 4 D T d t0101 ENQ NAK % 5 E U e u0110 ACK SYN & 6 F V f v0111 BEL ETB 7 G W g w1000 BS CAN ( 8 H X h x1001 HT EM ) 9 I Y i y1010 LF SUB * : J Z j z1011 VT ESC + ; K [ k {1100 FF FS , < L l |1101 CR GS - = M ] m }1110 SO RS . > N ^ n ~1111 SI US / ? O _ o DEL
59. 59. 33 Control codes 000 001 010 011 100 101 110 1110000 NULL DLE 0 @ P ` p0001 SOH DC1 ! 1 A Q a q0010 STX DC2 " 2 B R b r0011 ETX DC3 # 3 C S c s0100 EDT DC4 \$ 4 D T d t0101 ENQ NAK % 5 E U e u0110 ACK SYN & 6 F V f v0111 BEL ETB 7 G W g w1000 BS CAN ( 8 H X h x1001 HT EM ) 9 I Y i y1010 LF SUB * : J Z j z1011 VT ESC + ; K [ k {1100 FF FS , < L l |1101 CR GS - = M ] m }1110 SO RS . > N ^ n ~1111 SI US / ? O _ o DEL
60. 60. Alphabetic codes 000 001 010 011 100 101 110 1110000 NULL DLE 0 @ P ` p0001 SOH DC1 ! 1 A Q a q0010 STX DC2 " 2 B R b r0011 ETX DC3 # 3 C S c s0100 EDT DC4 \$ 4 D T d t0101 ENQ NAK % 5 E U e u0110 ACK SYN & 6 F V f v0111 BEL ETB 7 G W g w1000 BS CAN ( 8 H X h x1001 HT EM ) 9 I Y i y1010 LF SUB * : J Z j z1011 VT ESC + ; K [ k {1100 FF FS , < L l |1101 CR GS - = M ] m }1110 SO RS . > N ^ n ~1111 SI US / ? O _ o DEL
61. 61. Numeric codes 000 001 010 011 100 101 110 1110000 NULL DLE 0 @ P ` p0001 SOH DC1 ! 1 A Q a q0010 STX DC2 " 2 B R b r0011 ETX DC3 # 3 C S c s0100 EDT DC4 \$ 4 D T d t0101 ENQ NAK % 5 E U e u0110 ACK SYN & 6 F V f v0111 BEL ETB 7 G W g w1000 BS CAN ( 8 H X h x1001 HT EM ) 9 I Y i y1010 LF SUB * : J Z j z1011 VT ESC + ; K [ k {1100 FF FS , < L l |1101 CR GS - = M ] m }1110 SO RS . > N ^ n ~1111 SI US / ? O _ o DEL
62. 62. Punctuation, etc. 000 001 010 011 100 101 110 1110000 NULL DLE 0 @ P ` p0001 SOH DC1 ! 1 A Q a q0010 STX DC2 " 2 B R b r0011 ETX DC3 # 3 C S c s0100 EDT DC4 \$ 4 D T d t0101 ENQ NAK % 5 E U e u0110 ACK SYN & 6 F V f v0111 BEL ETB 7 G W g w1000 BS CAN ( 8 H X h x1001 HT EM ) 9 I Y i y1010 LF SUB * : J Z j z1011 VT ESC + ; K [ k {1100 FF FS , < L l |1101 CR GS - = M ] m }1110 SO RS . > N ^ n ~1111 SI US / ? O _ o DEL
63. 63. The Problem• Representing text strings, such as “Hello, world”, in a computer
64. 64. “Hello, world” Example Binary Hexadecimal DecimalH = 01001000 = 48 = 72e = 01100101 = 65 = 101l = 01101100 = 6C = 108l = 01101100 = 6C = 108o = 01101111 = 6F = 111, = 00101100 = 2C = 44 = 00100000 = 20 = 32w = 01110111 = 77 = 119o = 01100111 = 67 = 103r = 01110010 = 72 = 114l = 01101100 = 6C = 108d = 01100100 = 64 = 100
65. 65. EBCDICExtended BCD Interchange Code(pronounced ebb’-se-dick)• 8-bit code• Developed by IBM• Rarely used today• IBM mainframes only
66. 66. EBCDIC “Extended Binary CodedDecimal Interchange Code” code table
67. 67. EBCDIC “Extended Binary CodedDecimal Interchange Code” code tableExample: MSB LSB1111 1111 1110 1001 1111 0111 1101 0111EBCDIC CODE Z 6 PMessage below are represented in EBCDIC code. What is the message?Please convert by using EBCDIC Code table given:i) 1111 1100 1011 0101 1101 1001 EBCDIC CODE