1. Monday, October 11 Assignment(s) due:Assignment #7: IEEE FLOATING POINT FORMATS EMULATOR PROGRESS WORKSHEET 1 Quiz(zes) due:Quiz #7: Debug Note:Quiz #8: Floating Point Formats is due next Monday Tonight is the cutoff date for Assignment #5 (Complement Arithmetic) Next Monday is the cutoff date for all Debug and Floating Point assignments and quizzes(these topics will be covered on the midterm) Any assignments turned in to me by Thursday afternoon will be available in the lab on Monday afternoon Please note:The midterm is next Monday (at the beginning of class) - an 8.5 in. * 11 in. cheat sheet will be allowed We will review for the midterm tonight

2. Conversion to/from floating point formats: - decimal number to hexadecimal floating point format: - convert decimal number to binary - convert to normalized scientific form (varies) - convert exponent to its excess-N form - add sign bit - convert binary number to hexadecimal - hexadecimal floating point format to decimal value - convert hexadecimal number to binary - determine sign - determine exponent - determine fraction from its normalized form - convert to its non-scientific binary form - convert to decimal

3. - a problem using the same format as Lab #9 - 32 bits - 1 bit for the sign bit - 7 bits for the exponent - using excess-64 - using base 16 for the exponent - 24 bits for the normalized fraction - radix point to the left of the most significant bit - round-off because of rounding - how would 25.125 10 be stored in hexadecimal?

4. Step 1:Convert 25.125 10 to binary

5. 25.125 10 = 11001.001 2 Step 2:Normalize the binary number

6. Step 1:Convert 25.125 10 to binary 25.125 10 = 11001.001 2 Step 2:Normalize the binary number.00011001001 * (16 2 ) 10

7. Step 1:Convert 25.125 10 to binary 25.125 10 = 11001.001 2 Step 2:Normalize the binary number.00011001001 * (16 2 ) 10 Step 3:Calculate the exponent

8. Step 1:Convert 25.125 10 to binary 25.125 10 = 11001.001 2 Step 2:Normalize the binary number.00011001001 * (16 2 ) 10 Step 3:Calculate the exponent 2 + 64 = 66 10 = 1000010 2

9. Step 1:Convert 25.125 10 to binary 25.125 10 = 11001.001 2 Step 2:Normalize the binary number.00011001001 * (16 2 ) 10 Step 3:Calculate the exponent 2 + 64 = 66 10 = 1000010 2 Step 4:Put the number together 0 1000010 000110010010000000000000

10. Step 1:Convert 25.125 10 to binary 25.125 10 = 11001.001 2 Step 2:Normalize the binary number.00011001001 * (16 2 ) 10 Step 3:Calculate the exponent 2 + 64 = 66 10 = 1000010 2 Step 4:Put the number together 0 1000010 000110010010000000000000 In hexadecimal: 0100 0010 0001 1001 0010 0000 0000 0000 4 2 1 9 2 0 0 0

11. - the same problem using the same format as Lab #9, part 2 (IEEE format) - 32 bits - 1 bit for the sign bit - 8 bits for the exponent - using excess-127 - using base 2 for the exponent - 24 bits for the normalized fraction - radix point to the right of the most significant bit - the most significant bit is "hidden" - round-off because of rounding - how would 25.125 10 be stored in hexadecimal?

12. Step 1:Convert 25.125 10 to binary

13. 25.125 10 = 11001.001 2 Step 2:Normalize the binary number

14. Step 1:Convert 25.125 10 to binary 25.125 10 = 11001.001 2 Step 2:Normalize the binary number 1.1001001 * (2 4 ) 10

15. Step 1:Convert 25.125 10 to binary 25.125 10 = 11001.001 2 Step 3:Calculate the exponent Step 2:Normalize the binary number 1.1001001 * (2 4 ) 10

16. Step 1:Convert 25.125 10 to binary 25.125 10 = 11001.001 2 Step 3:Calculate the exponent 4 +127 = 131 10 = 10000011 2 Step 2:Normalize the binary number 1.1001001 * (2 4 ) 10

17. Step 1:Convert 25.125 10 to binary 25.125 10 = 11001.001 2 Step 4:Put the number together 0 10000011 10010010000000000000000 Step 2:Normalize the binary number 1.1001001 * (2 4 ) 10 Step 3:Calculate the exponent 4 +127 = 131 10 = 10000011 2

18. Step 1:Convert 25.125 10 to binary 25.125 10 = 11001.001 2 Step 4:Put the number together 0 10000011 10010010000000000000000 In hexadecimal 0100 0001 1100 1001 0000 0000 0000 0000 4 1 C 9 0 0 0 0 Step 2:Normalize the binary number 1.1001001 * (2 4 ) 10 Step 3:Calculate the exponent 4 +127 = 131 10 = 10000011 2

19. - using the IEEE standard short real - 32 bits - 1 bit for the sign bit - 8 bits for the exponent - using excess-127 - using base 2 for the exponent - 23 bits for the normalized fraction - radix point to the right of the most significant bit - the most significant bit is hidden - round-off because of rounding - how would 414E0000 16 be represented as a decimal number?

20. Step 1:Convert 414E0000 16 to binary 414E0000 16 = 01000001010011100000... 2 or 0 10000010 10011100000... 2

21. Step 1:Convert 414E0000 16 to binary 414E0000 16 = 01000001010011100000... 2 or 0 10000010 10011100000... 2 Step 2:The exponent is 130 10 or 3 (130 - 127)

22. Step 1:Convert 414E0000 16 to binary 414E0000 16 = 01000001010011100000... 2 or 0 10000010 10011100000... 2 Step 2:The exponent is 130 10 or 3 (130 - 127) Step 3:The fraction is 1.10011100000... 2

23. Step 1:Convert 414E0000 16 to binary 414E0000 16 = 01000001010011100000... 2 or 0 10000010 10011100000... 2 Step 2:The exponent is 130 10 or 3 (130 - 127) Step 3:The fraction is 1.10011100000... 2 Step 4:The normalized form is 1.100111 2 * (2 3 ) 10

24. Step 1:Convert 414E0000 16 to binary 414E0000 16 = 01000001010011100000... 2 or 0 10000010 10011100000... 2 Step 2:The exponent is 130 10 or 3 (130 - 127) Step 3:The fraction is 1.10011100000... 2 Step 4:The normalized form is 1.100111 2 * (2 3 ) 10 Step 5:In binary 1100.111 2

25. Step 1:Convert 414E0000 16 to binary 414E0000 16 = 01000001010011100000... 2 or 0 10000010 10011100000... 2 Step 2:The exponent is 130 10 or 3 (130 - 127) Step 3:The fraction is 1.10011100000... 2 Step 4:The normalized form is 1.100111 2 * (2 3 ) 10 Step 5:In binary 1100.111 Step 6:In decimal 12.875

27. Data Table N: 18 X: 19

28. 20: 400 21: 820 22: 518 23: 019 24: 119 25: 418 26: 319 27: 518 28: 730 29: 623 30: 900 Data Table N: 18 X: 19

29. 20: 400 21: 820 22: 518 23: 019 24: 119 25: 418 26: 319 27: 518 28: 730 29: 623 30: 900 If the input cards are: 013 052 070 005 041 What is the output? Data Table N: 18 X: 19

31. 016 005 013 007

35. 6. Given the following information in DEBUG: -u 100, 10A 21F0:0100BB2001MOV BX, 120 21F0:0103BE0200MOV SI, 2 21F0:01060307MOV AX, [BX] 21F0:01080300ADD AX, [BX + SI] 21F0:010A 034002ADD AX, [BX + SI + 02] 21F0:010D894004MOV [BX + SI +04], AX 21F0:0110CD20INT 20 -e 120 12 03 24 00 10 FF01200121012201230124012501260127 AXBXSI

36. 6. Given the following information in DEBUG: -u 100, 10A 21F0:0100BB2001MOV BX, 120 21F0:0103BE0200MOV SI, 2 21F0:01060307MOV AX, [BX] 21F0:01080300ADD AX, [BX + SI] 21F0:010A 034002ADD AX, [BX + SI + 02] 21F0:010D894004MOV [BX + SI +04], AX 21F0:0110CD20INT 20 -e 120 12 03 24 00 10 FF01200121012201230124012501260127 AXBXSI 1203240010FF

37. 6. Given the following information in DEBUG: -u 100, 10A 21F0:0100BB2001MOV BX, 120 21F0:0103BE0200MOV SI, 2 21F0:01060307MOV AX, [BX] 21F0:01080300ADD AX, [BX + SI] 21F0:010A 034002ADD AX, [BX + SI + 02] 21F0:010D894004MOV [BX + SI +04], AX 21F0:0110CD20INT 20 -e 120 12 03 24 00 10 FF01200121012201230124012501260127 AXBXSI 0120 1203240010FF

38. 6. Given the following information in DEBUG: -u 100, 10A 21F0:0100BB2001MOV BX, 120 21F0:0103BE0200MOV SI, 2 21F0:01060307MOV AX, [BX] 21F0:01080300ADD AX, [BX + SI] 21F0:010A 034002ADD AX, [BX + SI + 02] 21F0:010D894004MOV [BX + SI +04], AX 21F0:0110CD20INT 20 -e 120 12 03 24 00 10 FF01200121012201230124012501260127 AXBXSI 01200002 1203240010FF

39. 6. Given the following information in DEBUG: -u 100, 10A 21F0:0100BB2001MOV BX, 120 21F0:0103BE0200MOV SI, 2 21F0:01060307MOV AX, [BX] 21F0:01080300ADD AX, [BX + SI] 21F0:010A 034002ADD AX, [BX + SI + 02] 21F0:010D894004MOV [BX + SI +04], AX 21F0:0110CD20INT 20 -e 120 12 03 24 00 10 FF01200121012201230124012501260127 AXBXSI 031201200002 1203240010FF

40. 6. Given the following information in DEBUG: -u 100, 10A 21F0:0100BB2001MOV BX, 120 21F0:0103BE0200MOV SI, 2 21F0:01060307MOV AX, [BX] 21F0:01080300ADD AX, [BX + SI] 21F0:010A 034002ADD AX, [BX + SI + 02] 21F0:010D894004MOV [BX + SI +04], AX 21F0:0110CD20INT 20 -e 120 12 03 24 00 10 FF01200121012201230124012501260127 AXBXSI 031201200002 0336 1203240010FF

41. 6. Given the following information in DEBUG: -u 100, 10A 21F0:0100BB2001MOV BX, 120 21F0:0103BE0200MOV SI, 2 21F0:01060307MOV AX, [BX] 21F0:01080300ADD AX, [BX + SI] 21F0:010A 034002ADD AX, [BX + SI + 02] 21F0:010D894004MOV [BX + SI +04], AX 21F0:0110CD20INT 20 -e 120 12 03 24 00 10 FF01200121012201230124012501260127 AXBXSI 031201200002 0336 0246 1203240010FF

42. 6. Given the following information in DEBUG: -u 100, 10A 21F0:0100BB2001MOV BX, 120 21F0:0103BE0200MOV SI, 2 21F0:01060307MOV AX, [BX] 21F0:01080300ADD AX, [BX + SI] 21F0:010A 034002ADD AX, [BX + SI + 02] 21F0:010D894004MOV [BX + SI +04], AX 21F0:0110CD20INT 20 -e 120 12 03 24 00 10 FF01200121012201230124012501260127 AXBXSI 031201200002 0336 0246 1203240010FF4602

44. More practice: Given the following information in DEBUG: -u 100 21F0:0100MOVAX,ABCD 21F0:0103MOVBX,0120 21F0:0106MOVCX,0004 21F0:0109SUBAX,[BX] 21F0:010BADDBX,0002 21F0:010ELOOP109 21F0:0110INT 20 -d 120,12F EB 02 46 00 EB 38 5A 01 - 10 FF 01 02 A3 FF 88 04 Trace through the program above and indicate below the contents of the registers and memory location just before the INT 20 instruction is executed. AX = BX =CX =

45. The number 1.275 10 = ? 2 a)1.010001 b)1.01 c)1.1001101 d)1.1

46. If the number 1.010001 normalized using the IEEE standard (using radix 2 to normalize, radix point to the right of the most significant bit, and the most significant bit being hidden) it will result in which fraction and exponent? a).1010001 and 1 b).1010001 and 0 c).010001 and 1 d).010001 and 0

48. For the binary number 1 10000101 01010001000000000000000: Using excess-127, what is the exponent in base 10? 1.-6 2.0 3.4 4.6

49. For the binary number 1 10000101 01010001000000000000000: With an exponent of 6 and using radix 2, what is the binary normalized form? 1.-.010100001 * (2 6 ) 10 2.-.010100000101 * (2 -6 ) 10 3.-1.01010001 * (2 6 ) 10 4.-1.101010001 * (2 6 ) 10

50. If the binary normalized form is -1.01010001 * (2 6 ) 10 : The number in binary is: 1.-1010100.01 2.-101010001.1 3.-101010001 4.-.00000101010001

51. If the binary number is -1010100.01: The number in decimal is: 1.-168.25 2.-84.25 3.-84.5 4.-168.1

52. Vocabulary: addressDS (data segment) ASCII codeIP (instruction pointer) bitLOOP instruction bytememory compilation/translationregister

53. Noise and Distortion: -occurs every time a signal is transmittted - may be random - white noise (a background hiss) - impulse noise (amplitude peaks that block out data) - external, usually from equipment - internal, signal from another channel - echoes on the line - amplitude changes - circuit failures

54. Transmission in the presence of noise:

55. Error Detection - error-detecting Codes - add redundant bits to each unit of data - receiver knows when an error has occurred - receiver can ask for retransmission

56. - to detect single-bit errors - often used with 7-bit ASCII code characters - add one parity bit so number of 1-bits in the word is always even (even parity) or odd (odd parity) - works about 70% of the time - e.g. to transmit the character ‘A’ (65 10 => 41 16 => 1000001 2 ) 1010000

57. - to detect single-bit errors - often used with 7-bit ASCII code characters - add one parity bit so number of 1-bits in the word is always even (even parity) or odd (odd parity) - works about 70% of the time - e.g. to transmit the character ‘A’ (65 10 => 41 16 => 1000001 2 ) - using odd parity: 1010000 10100001

58. - to detect single-bit errors - often used with 7-bit ASCII code characters - add one parity bit so number of 1-bits in the word is always even (even parity) or odd (odd parity) - works about 70% of the time - e.g. to transmit the character ‘A’ (65 10 => 41 16 => 1000001 2 ) - using odd parity: - using even parity: 1010000 1010000 1010000 1 0

59. Error Detection and Correction - error-correcting Codes - add more redundant bits to each unit of data - receiver can detect and sometimes correct the error -Hamming Codes - use a “block parity” method of adding parity bits - use more than 1 parity bit - allow for detection and correction of single-bit errors

60. - Hamming algorithm: - start numbering the bits at 1 (from the left) bit 1bit 2bit 3bit 4bit 5

61. - Hamming algorithm: - start numbering the bits at 1 (from the left) - all bits whose position is a power of 2 are parity bits - i.e. bits 1, 2, 4, 8, 16 etc. parity bit bit 1bit 2bit 3bit 4bit 5 parity bit parity bit

62. - Hamming algorithm: - start numbering the bits at 1 (from the left) - all bits whose position is a power of 2 are parity bits - i.e. bits 1, 2, 4, 8, 16 etc. - then fill in the data bits parity bit data bit bit 1bit 2bit 3bit 4bit 5 data bit parity bit parity bit

63. - Hamming algorithm: - start numbering the bits at 1 (from the left) - all bits whose position is a power of 2 are parity bits - i.e. bits 1, 2, 4, 8, 16 etc. - then fill in the data bits - so a code with 2 data bits - would require: parity bit data bit bit 1bit 2bit 3bit 4bit 5 data bit parity bit parity bit

64. parity bit data bit bit 1bit 2bit 3bit 4bit 5 data bit parity bit parity bit - Hamming algorithm continued: - each data bit is checked by those parity bits before it whose bit positions add up to its position - data bit 3 is checked by parity bit 1 and parity bit 2

65. parity bit data bit bit 1bit 2bit 3bit 4bit 5 data bit parity bit parity bit - Hamming algorithm continued: - each data bit is checked by those parity bits before it whose bit positions add up to its position - data bit 3 is checked by parity bit 1 and parity bit 2 - data bit 5 is checked by parity bit 1 and parity bit 4

66. parity bit data bit bit 1bit 2bit 3bit 4bit 5 data bit parity bit parity bit - Hamming algorithm continued: - or- parity bit 1 checks data bits 3 and 5 - parity bit 2 checks data bit 3 - parity bit 4 checks data bit 5 parity bit data bit bit 1bit 2bit 3bit 4bit 5 data bit parity bit parity bit parity bit data bit bit 1bit 2bit 3bit 4bit 5 data bit parity bit parity bit

67. parity bit data bit bit 1bit 2bit 3bit 4bit 5 data bit parity bit parity bit - Hamming algorithm continued: - each parity bit forms a group of itself and the data bits it checks - 3 groups are formed - there will be an overlap of data bits - each group must add up to an even number of 1’s (or none) for even parity - or an odd number of 1’s for odd parity - groups: - bits 1, 3 and 5 - bits 2 and 3 - bits 4 and 5

68. - each group contains only 1 parity bit - groups: Group 1:bit 1 Group 2:bit 2 Group 3:bit 4 parity bit data bit bit 1bit 2bit 3bit 4bit 5 data bit parity bit parity bit

69. - each group contains only 1 parity bit - groups: Group 1:bit 1 Group 2:bit 2 Group 3:bit 4 - data bit 3 is checked by parity bit 1 and parity bit 2 - (1 + 2 = 3 or 3 = 1 + 2) parity bit data bit bit 1bit 2bit 3bit 4bit 5 data bit parity bit parity bit

70. - each group contains only 1 parity bit - groups: Group 1:bit 1bit 3 Group 2:bit 2bit 3 Group 3:bit 4 - data bit 3 is checked by parity bit 1 and parity bit 2 - (1 + 2 = 3 or 3 = 1 + 2) parity bit data bit bit 1bit 2bit 3bit 4bit 5 data bit parity bit parity bit

71. - each group contains only 1 parity bit - groups: Group 1:bit 1bit 3 Group 2:bit 2bit 3 Group 3:bit 4 - data bit 3 is checked by parity bit 1 and parity bit 2 - (1 + 2 = 3 or 3 = 1 + 2) - data bit 5 is checked by parity bit 1 and parity bit 4 - (1 + 4 = 5 or 5 = 1 + 4) parity bit data bit bit 1bit 2bit 3bit 4bit 5 data bit parity bit parity bit

72. - each group contains only 1 parity bit - groups: Group 1:bit 1bit 3 bit 5 Group 2:bit 2bit 3 Group 3:bit 4bit 5 - data bit 3 is checked by parity bit 1 and parity bit 2 - (1 + 2 = 3 or 3 = 1 + 2) - data bit 5 is checked by parity bit 1 and parity bit 4 - (1 + 4 = 5 or 5 = 1 + 4) parity bit data bit bit 1bit 2bit 3bit 4bit 5 data bit parity bit parity bit

73. 0 1 1 0 0 0 P1P4P2D3D5 P1 P2 P4 D3 D5 11 Legal codewords for the 2-data-bit Hamming code (even parity):

74. 0 1 1 0 0 0 P1P4P2D3D5 P1 P2 P4 D3 D5 11 Parity BitData Bits P1 D3, D5 P2 D3 P4 D5 Legal codewords for the 2-data-bit Hamming code (even parity):

75. 0 1 1 0 0 0 P1P4P2D3D5 P1 P2 P4 D3 D5 11 Parity BitData Bits P1 D3, D5 P2 D3 P4 D5 Legal codewords for the 2-data-bit Hamming code (even parity):

76. 0 1 1 0 0 0 P1P4P2D3D5 P1 P2 P4 D3 D5 11 Parity BitData Bits P1 D3, D5 P2 D3 P4 D5 0 Legal codewords for the 2-data-bit Hamming code (even parity):

77. 0 1 1 0 0 0 P1P4P2D3D5 P1 P2 P4 D3 D5 11 Parity BitData Bits P1 D3, D5 P2 D3 P4 D5 0 Legal codewords for the 2-data-bit Hamming code (even parity):

78. 0 1 1 0 0 0 P1P4P2D3D5 P1 P2 P4 D3 D5 11 Parity BitData Bits P1 D3, D5 P2 D3 P4 D5 00 Legal codewords for the 2-data-bit Hamming code (even parity):

79. 0 1 1 0 0 0 P1P4P2D3D5 P1 P2 P4 D3 D5 11 Parity BitData Bits P1 D3, D5 P2 D3 P4 D5 00 Legal codewords for the 2-data-bit Hamming code (even parity):

80. 0 1 1 0 0 0 P1P4P2D3D5 P1 P2 P4 D3 D5 11 Parity BitData Bits P1 D3, D5 P2 D3 P4 D5 000 Legal codewords for the 2-data-bit Hamming code (even parity):

81. 0 1 1 0 0 0 P1P4P2D3D5 P1 P2 P4 D3 D5 11 Parity BitData Bits P1 D3, D5 P2 D3 P4 D5 000 Legal codewords for the 2-data-bit Hamming code (even parity):

82. 0 1 1 0 0 0 P1P4P2D3D5 P1 P2 P4 D3 D5 11 Parity BitData Bits P1 D3, D5 P2 D3 P4 D5 000 1 Legal codewords for the 2-data-bit Hamming code (even parity):

83. 0 1 1 0 0 0 P1P4P2D3D5 P1 P2 P4 D3 D5 11 Parity BitData Bits P1 D3, D5 P2 D3 P4 D5 000 10 Legal codewords for the 2-data-bit Hamming code (even parity):

84. 0 1 1 0 0 0 P1P4P2D3D5 P1 P2 P4 D3 D5 11 Parity BitData Bits P1 D3, D5 P2 D3 P4 D5 000 101 Legal codewords for the 2-data-bit Hamming code (even parity):

85. 0 1 1 0 0 0 P1P4P2D3D5 P1 P2 P4 D3 D5 11 Parity BitData Bits P1 D3, D5 P2 D3 P4 D5 000 101 110 Legal codewords for the 2-data-bit Hamming code (even parity):

86. 0 1 1 0 0 0 P1P4P2D3D5 P1 P2 P4 D3 D5 11 Parity BitData Bits P1 D3, D5 P2 D3 P4 D5 000 101 110 110 Legal codewords for the 2-data-bit Hamming code (even parity):

87. - to transmit the data number 01 - we use the Hamming code word - if we transmit - there is an error in a parity bit - the receiver checks the groups: 10 P1P2P4D3D5 011 10 P1P2P4D3D5 111

88. - to transmit the data number 01 - we use the Hamming code word - if we transmit - there is an error in a parity bit - the receiver checks the groups:p1, d3, d5ok 10 P1P2P4D3D5 011 10 P1P2P4D3D5 111

89. - to transmit the data number 01 - we use the Hamming code word - if we transmit - there is an error in a parity bit - the receiver checks the groups:p1, d3, d5ok p2, d3no 10 P1P2P4D3D5 011 10 P1P2P4D3D5 111

90. - to transmit the data number 01 - we use the Hamming code word - if we transmit - there is an error in a parity bit - the receiver checks the groups:p1, d3, d5ok p2, d3no p4, d5ok 10 P1P2P4D3D5 011 10 P1P2P4D3D5 111

91. - to transmit the data number 01 - we use the Hamming code word - if we transmit - there is an error in a parity bit - the receiver checks the groups:p1, d3, d5ok p2, d3no p4, d5ok - only one group is wrong - the error is in the parity bit for that group 10 P1P2P4D3D5 011 10 P1P2P4D3D5 111

92. - to transmit the data number 01 - we use the Hamming code word - if we transmit - there is an error in a data bit - the receiver checks the groups:p1, d3, d5no p2, d3no p4, d5ok - two groups are wrong - the error is in the common bit 10 P1P2P4D3D5 011 11 P1P2P4D3D5 011

93. 10 P1P2P4P8 3 5 6 7 9 10 11 00001 7-bit character ‘A’ using the Hamming code and odd parity: - 65 10 => 41 16 => 1000001 2

94. 10 P1P2P4P8 3 5 6 7 9 10 11 00001 groups:p1, p2, p4, p8, 7-bit character ‘A’ using the Hamming code and odd parity: - 65 10 => 41 16 => 1000001 2

95. 7-bit character ‘A’ using the Hamming code and odd parity: - 65 10 => 41 16 => 1000001 2 10 P1P2P4P8 3 5 6 7 9 10 11 00001 groups:p1, 3 p2, 3 p4 p8

96. 7-bit character ‘A’ using the Hamming code and odd parity: - 65 10 => 41 16 => 1000001 2 10 P1P2P4P8 3 5 6 7 9 10 11 00001 groups:p1, 3, 5 p2, 3 p4, 5 p8

97. 7-bit character ‘A’ using the Hamming code and odd parity: - 65 10 => 41 16 => 1000001 2 10 P1P2P4P8 3 5 6 7 9 10 11 00001 groups:p1, 3, 5 p2, 3, 6 p4, 5, 6 p8

98. 7-bit character ‘A’ using the Hamming code and odd parity: - 65 10 => 41 16 => 1000001 2 10 P1P2P4P8 3 5 6 7 9 10 11 00001 groups:p1, 3, 5, 7 p2, 3, 6, 7 p4, 5, 6, 7 p8

99. 7-bit character ‘A’ using the Hamming code and odd parity: - 65 10 => 41 16 => 1000001 2 10 P1P2P4P8 3 5 6 7 9 10 11 00001 groups:p1, 3, 5, 7, 9 p2, 3, 6, 7 p4, 5, 6, 7 p8, 9

100. 7-bit character ‘A’ using the Hamming code and odd parity: - 65 10 => 41 16 => 1000001 2 10 P1P2P4P8 3 5 6 7 9 10 11 00001 groups:p1, 3, 5, 7, 9 p2, 3, 6, 7, 10 p4, 5, 6, 7 p8, 9, 10

101. 7-bit character ‘A’ using the Hamming code and odd parity: - 65 10 => 41 16 => 1000001 2 10 P1P2P4P8 3 5 6 7 9 10 11 00001 groups:p1, 3, 5, 7, 9, 11 p2, 3, 6, 7, 10, 11 p4, 5, 6, 7 p8, 9, 10, 11

102. 7-bit character ‘A’ using the Hamming code and odd parity: - 65 10 => 41 16 => 1000001 2 10 P1P2P4P8 3 5 6 7 9 10 11 00001 10 P1P2P4P8 3 5 6 7 9 10 11 00001 groups:p1, 3, 5, 7, 9, 11 p2, 3, 6, 7, 10, 11 p4, 5, 6, 7 p8, 9, 10, 11

103. 7-bit character ‘A’ using the Hamming code and odd parity: - 65 10 => 41 16 => 1000001 2 10 P1P2P4P8 3 5 6 7 9 10 11 00001 10 P1P2P4P8 3 5 6 7 9 10 11 000011 groups:p1, 3, 5, 7, 9, 11 p2, 3, 6, 7, 10, 11 p4, 5, 6, 7 p8, 9, 10, 11

104. 7-bit character ‘A’ using the Hamming code and odd parity: - 65 10 => 41 16 => 1000001 2 10 P1P2P4P8 3 5 6 7 9 10 11 00001 10 P1P2P4P8 3 5 6 7 9 10 11 0000111 groups:p1, 3, 5, 7, 9, 11 p2, 3, 6, 7, 10, 11 p4, 5, 6, 7 p8, 9, 10, 11

105. 7-bit character ‘A’ using the Hamming code and odd parity: - 65 10 => 41 16 => 1000001 2 10 P1P2P4P8 3 5 6 7 9 10 11 00001 10 P1P2P4P8 3 5 6 7 9 10 11 00001111 groups:p1, 3, 5, 7, 9, 11 p2, 3, 6, 7, 10, 11 p4, 5, 6, 7 p8, 9, 10, 11

106. 7-bit character ‘A’ using the Hamming code and odd parity: - 65 10 => 41 16 => 1000001 2 10 P1P2P4P8 3 5 6 7 9 10 11 00001 10 P1P2P4P8 3 5 6 7 9 10 11 000011110 groups:p1, 3, 5, 7, 9, 11 p2, 3, 6, 7, 10, 11 p4, 5, 6, 7 p8, 9, 10, 11

107. 7-bit character ‘A’ using the Hamming code and odd parity: - 65 10 => 41 16 => 1000001 2 10 P1P2P4P8 3 5 6 7 9 10 11 00001 10 P1P2P4P8 3 5 6 7 9 10 11 010011110 - what if we transmitted: 1110 groups:p1, 3, 5, 7, 9, 11Check 1, skip 1 p2, 3, 6, 7, 10, 11Check 2, skip 2 p4, 5, 6, 7Check 4, skip 4 p8, 9, 10, 11Check 8, skip 8 Or... Starting at the parity bit:

108. 7-bit character ‘A’ using the Hamming code and odd parity: - 65 10 => 41 16 => 1000001 2 10 P1P2P4P8 3 5 6 7 9 10 11 00001 10 P1P2P4P8 3 5 6 7 9 10 11 010011110 groups:p1, 3, 5, 7, 9, 11Check 1, skip 1 p2, 3, 6, 7, 10, 11Check 2, skip 2 p4, 5, 6, 7Check 4, skip 4 p8, 9, 10, 11Check 8, skip 8 - what if we transmitted: 1110 Or... Starting at the parity bit: Reconstitute the groups, and check

109. From Lab #10: Identify the parity bits and the data bits: Complete the table for the 11-bit Hamming codeword: Group Parity Bit Data Bits 1P1 2P2 3P4 4P8 Using even parity, which bit (if any) is in error: Bit# 1234567891011 a)0 1 1 1 0 0 0 0 1 1 0 b)011100110 1 0 c)110111100 1 1

110. Monday, October 11 Assignment(s) due:Assignment #7: IEEE FLOATING POINT FORMATS EMULATOR PROGRESS WORKSHEET 1 Quiz(zes) due:Quiz #7: Debug Note:Quiz #8: Floating Point Formats is due next Monday Tonight is the cutoff date for Assignment #5 (Complement Arithmetic) Next Monday is the cutoff date for all Debug and Floating Point assignments (these topics will be covered on the midterm) Any assignments turned in to me by Thursday afternoon will be available in the lab on Monday afternoon Please note:The midterm is next Monday (at the beginning of class) - an 8.5 in. * 11 in. cheat sheet will be allowed We will review for the midterm tonight