Logic System Design Ktu Module1. Number Systems introduction, Logic Gates introduction

1. NUMBER SYSTEMS
Decimal
Binary
Octal
Hexadecimal
MODULE 1
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2. 2
Number systems
COMMON NUMBER
SYSTEMS
System Base Symbols
Used by
humans?
Used in
computers?
Decimal 10 0, 1, … 9 Yes No
Binary 2 0, 1 No Yes
Octal 8 0, 1, … 7 No No
Hexa-
decimal
16 0, 1, … 9,
A, B, … F
No No
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3. 3
Number systems
QUANTITIES/COUNTING (1
OF 2)
Decimal Binary Octal
Hexa-
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
Decimal Binary Octal
Hexa-
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 F
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4. 4
Number systems
QUANTITIES/COUNTING (2
OF 2)
Decimal Binary Octal
Hexa-
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
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5. 5
Number systems
CONVERSION AMONG
BASES
The possibilities:
Hexadecimal
Decimal Octal
Binary
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6. 6
Number systems
QUICK EXAMPLE
2510 = 110012 = 318 = 1916
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7. 7
Number systems
DECIMAL TO DECIMAL
Hexadecimal
Decimal Octal
Binary
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8. 8
Number systems
12510 => 5 x 100 = 5
2 x 101 = 20
1 x 102 = 100
125
Base
Weight
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1 2 5
0
1
2 Position
Weights
100
101
102

9. 9
Number systems
BINARY TO DECIMAL
Hexadecimal
Decimal Octal
Binary
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10. 10
Number systems
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 results
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11. 11
Number systems
EXAMPLE
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
4310
Bit “0”
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12. 12
Number systems
OCTAL TO DECIMAL
Hexadecimal
Decimal Octal
Binary
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13. 13
Number systems
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 results
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14. 14
Number systems
EXAMPLE
7248 => 4 x 80 = 4
2 x 81 = 16
7 x 82 = 448
46810
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15. 15
Number systems
HEXADECIMAL TO DECIMAL
Hexadecimal
Decimal Octal
Binary
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16. 16
Number systems
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 results
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17. 17
Number systems
EXAMPLE
ABC16 => C x 160 = 12 x 1 = 12
B x 161 = 11 x 16 = 176
A x 162 = 10 x 256 = 2560
274810
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18. 18
Number systems
DECIMAL TO BINARY
Hexadecimal
Decimal Octal
Binary
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19. 19
Number systems
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.
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20. EXAMPLE
12510 = ?2
2 125
62 1
2
31 0
2
15 1
2
7 1
2
3 1
2
1 1
2
0 1
12510 = 11111012

21. 21
Number systems
OCTAL TO BINARY
Hexadecimal
Decimal Octal
Binary
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22. 22
Number systems
OCTAL TO BINARY
Technique
Convert each octal digit to a 3-bit equivalent binary
representation
EXAMPLE
7058 = ?2 7 0 5
111 000 101
7058 = 1110001012
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23. 23
Number systems
HEXADECIMAL TO BINARY
Hexadecimal
Decimal Octal
Binary
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24. 24
Number systems
HEXADECIMAL TO BINARY
Technique
Convert each hexadecimal digit to a 4-bit equivalent binary
representation
EXAMPLE
10AF16 = ?2
1 0 A F
0001 0000 1010 1111
10AF16 = 00010000101011112
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25. 25
Number systems
DECIMAL TO OCTAL
Hexadecimal
Decimal Octal
Binary
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26. 26
Number systems
DECIMAL TO OCTAL
Technique
 Divide by 8
 Keep track of the remainder
EXAMPLE
123410 = ?8
8 1234
154 2
8
19 2
8
2 3
8
0 2
123410 = 23228
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27. 27
Number systems
DECIMAL TO HEXADECIMAL
Hexadecimal
Decimal Octal
Binary
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28. 28
Number systems
DECIMAL TO HEXADECIMAL
Technique
 Divide by 16
 Keep track of the remainder
123410 = ?16
123410 = 4D216
16 1234
77 2
16
4 13 = D
16
0 4
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29. 29
Number systems
BINARY TO OCTAL
Hexadecimal
Decimal Octal
Binary
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30. 30
Number systems
BINARY TO OCTAL
Technique
Group bits in threes, starting on right
Convert to octal digits
10110101112 = ?8
1 011 010 111
1 3 2 7
10110101112 = 13278
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31. 31
Number systems
BINARY TO HEXADECIMAL
Hexadecimal
Decimal Octal
Binary
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32. 32
Number systems
BINARY TO HEXADECIMAL
Technique
 Group bits in fours, starting on right
 Convert to hexadecimal digits
10101110112 = ?16
10 1011 1011
2 B B
10101110112 = 2BB16
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33. 33
Number systems
OCTAL TO HEXADECIMAL
Hexadecimal
Decimal Octal
Binary
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34. 34
Number systems
OCTAL TO HEXADECIMAL
Technique
 Use binary as an intermediary
10768 = ?16
1 0 7 6
001 000 111 110
2 3 E
10768 = 23E16
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35. 35
Number systems
HEXADECIMAL TO OCTAL
Hexadecimal
Decimal Octal
Binary
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36. 36
Number systems
HEXADECIMAL TO OCTAL
Technique
Use binary as an intermediary
1F0C16 = ?8
1 F 0 C
0001 1111 0000 1100
1 7 4 1 4
1F0C16 = 174148
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37. 37
Number systems
EXERCISE
Decimal Binary Octal
Hexa-
decimal
33
1110101
703
1AF
Skip answer Answer
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38. 38
Number systems
EXERCISE – CONVERT …
Decimal Binary Octal
Hexa-
decimal
33 100001 41 21
117 1110101 165 75
451 111000011 703 1C3
431 110101111 657 1AF
Answer
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39. 39
FRACTIONS
Decimal to decimal
3.14 => 4 x 10-2 = 0.04
1 x 10-1 = 0.1
3 x 100 = 3
3.14
Number systems LOGIC SYSTEM DESIGN||MODULE 1||DR. BIBIN VINCENT||ASSOCIATE PROFESSOR||KMEA

40. 40
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
Number systems LOGIC SYSTEM DESIGN||MODULE 1||DR. BIBIN VINCENT||ASSOCIATE PROFESSOR||KMEA

41. 41
FRACTIONS
Decimal to binary
3.14579
.14579
x 2
0.29158
x 2
0.58316
x 2
1.16632
x 2
0.33264
x 2
0.66528
x 2
1.33056
etc.
11.001001...
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42. 42
EXERCISE
Decimal Binary Octal
Hexa-
decimal
29.8
101.1101
3.07
C.82
Skip answer Answer
Number systems LOGIC SYSTEM DESIGN||MODULE 1||DR. BIBIN VINCENT||ASSOCIATE PROFESSOR||KMEA

43. 43
Number systems
EXERCISE
Decimal Binary Octal
Hexa-
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.82
Answer
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44. BINARY NUMBER SYSTEM
1’s and 2’s compliment
Signed number
representation
Binary arithmetic
Binary subtraction
MODULE 1
44
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45. 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”
Number systems 40
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46. 46
Number systems
BINARY ADDITION (2 OF 2)
Two n-bit values
Add individual bits
Propagate carries
E.g.,
10101 21
+ 11001 + 25
101110 46
1
1
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47. 47
Binary Subtraction
Number systems LOGIC SYSTEM DESIGN||MODULE 1||DR. BIBIN VINCENT||ASSOCIATE PROFESSOR||KMEA

48. 48
Number systems
Binary Subtraction
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49. 49
Number systems
Binary Multiplication Binary Division
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50. 50
Number systems LOGIC SYSTEM DESIGN||MODULE 1||DR. BIBIN VINCENT||ASSOCIATE PROFESSOR||KMEA

51. 51
Number systems LOGIC SYSTEM DESIGN||MODULE 1||DR. BIBIN VINCENT||ASSOCIATE PROFESSOR||KMEA

52. 52
Number systems
REPRESENTING SIGNED (POSITIVE AND
NEGATIVE) NUMBERS
Ex. 4-bit signed magnitude
1 bit for sign
3 bits for magnitude
1111
0111
7
1110
0110
6
1101
0101
5
1100
0100
4
1011
0011
3
1010
0010
2
1001
0001
1
1000
0000
0
N
N 

DECIM
AL
SIGN BIT
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53. 53
Number systems
Ex 1. Find the sign/mag representation of -
610
Step1: find binary representation using 8 bits
610 = 000001102
Step2: if the number is a negative number flip
left most bit
10000110
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54. 54
Number systems
Ex 2. Find the signmag representation of -3610
Step 1: find binary representation using 8 bits
3610 = 001001002
Step 2: if the number is a negative number flip left
most bit
10100100
So: -3610 = 101001002 (in 8-bit sign/magnitude
form) LOGIC SYSTEM DESIGN||MODULE 1||DR. BIBIN VINCENT||ASSOCIATE PROFESSOR||KMEA

55. 55
Number systems
Ex 3. Find the signmag representation of 7010
Step 1: find binary representation using 8 bits
7010 = 010001102
Step 2: if the number is a negative number flip left
most bit
01000110 (no flipping, since it is +ve)
So: 7010 = 010001102 (in 8-bit sign/magnitude
form) LOGIC SYSTEM DESIGN||MODULE 1||DR. BIBIN VINCENT||ASSOCIATE PROFESSOR||KMEA

56. 56
Number systems
What is this signmag number?
100000002
The machine will think of it as - 0 , which is a non
valid value.
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57. 57
Number systems
TWO’S COMPLEMENT
REPRESENTATION
Ex. Find the two’s complement representation
of –610
Step1: find binary representation in 8 bits
610 = 000001102
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58. 58
Number systems
Step 2: Complement the entire positive number, and then add
one
00000110
(complemented) -> 11111001
(add one) -> + 1
11111010
So: -610 = 111110102 (in 2's complement form)
TWO’S COMPLEMENT
REPRESENTATION
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59. 59
Number systems
Alternative method for step 2
Scan binary representation from right
too left, find first one bit, from low-order
(right) end, and complement the remaining
pattern to the left.
00000110
(left complemented) --> 11111010
TWO’S COMPLEMENT
REPRESENTATION
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60. 60
Number systems
Ex 2: Find the Two’s Complement of -7610
Step 1: Find the 8 bit binary representation of the positive value.
7610 = 010011002
Step 2:
Find first one bit, from low-order (right) end, and complement the pattern to the left.
01001100
(left complemented) -> 10110100
So: -7610 = 101101002 (in 2's complement form, using any of above methods)
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61. 61
Number systems
Ex 3: Find the Two’s Complement of 7210
Step 1: Find the 8 bit binary representation of the
positive value.
7210 = 010010002
Step 2:
Since number is positive do
nothing.
So: 7210 = 010010002 (in 2's complement form,
using any of above methods)
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62. 62
Number systems
Binary subtraction using 2’s
compliment
To subtract +7 from +3, we add the code for -7, 1001, to that of +3,
0011:
0011 (+3)
+1001 (-7)
1100 (-4)
The result, 1100, is the code for -4, the result of
subtracting +7 from +3.
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63. BINARY CODES
Grey Code
BCD Code
Excess -3 Code
Parity and Hamming Code
ASCII Code
MODULE 1
63
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64. 64
Number systems
BCD
The BCD is simply the 4 bit
representation of the decimal digit.
For multiple digit base 10 numbers,
each symbol is represented by its
BCD digit
What happened to 6 digits not used?
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65. 65
Would it be easy for you if you can replace
a decimal number with an individual
binary code?
Such as 00011001 = 1910
The 8421 code is a type of BCD to do
that.
BCD code provides an excellent interface
to binary systems:
Keypad inputs
Digital readouts
Number systems
BCD
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66. 66
ex1: dec-to-BCD
(a) 35
(b) 98
(c) 170
(d) 2469
ex2: BCD-to-dec
(a) 10000110
(b) 001101010001
(c) 1001010001110000
Decimal
Digit
0 1 2 3 4 5 6 7 8 9
BCD 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001
Let’s crack these…
Note: 1010, 1011, 1100, 1101, 1110, and 1111 are INVALID CODE!
BCD
Number systems LOGIC SYSTEM DESIGN||MODULE 1||DR. BIBIN VINCENT||ASSOCIATE PROFESSOR||KMEA

67. 67
Number systems
BCD OPERATION
Consider the following BCD operation
Decimal: Add 4 + 1
Covert to binary 0 1 0 0
And 0 0 0 1
Getting 0 1 0 1
Which is still a BCD representation of a
decimal digit
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68. 68
Number systems
A second example
 3 0 0 1 1
 +3 0 0 1 1
Getting 6 or 0 1 1 0
And in range and a BCD digit
representation
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69. 69
Number systems
AND NOW
Consider 5 + 5
5 0 1 0 1
+5 0 1 0 1
giving 1 0 1 0 which is binary 10 but not a BCD
digit!
What to do?
Try adding 6 !!
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70. 70
Number systems
ADDING 6
Had 1010 and want to add 6 or 0110
 so 1 0 1 0
 plus 6 0 1 1 0
Giving 1 0 0 0 0
Or a carry out to the next binary digit, or if the
binary in BCD, the next BCD digit.
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71. 71
Number systems
ANOTHER CARRY EXAMPLE
Add 7 + 6
 have 7 0 1 1 1
 plus 6 0 1 1 0
Giving 1 1 0 1 and again out of range
Adding 6 0 1 1 0
Giving 1 0 0 1 1 so a 1 carries out to the next BCD
digit
FINAL BCD answer 0001 0011 or 1310
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72. 72
Number systems
MULTIBIT BCD
Add the BCD for 417 to 195
Would expect to get 612
BCD setup - start with Least Significant Digit
 0 1 0 0 0 0 0 1 0 1 1 1
 0 0 0 1 1 0 0 1 0 1 0 1
 1 1 0 0
Adding 6 0 1 1 0
Gives 1 0 0 1 0
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73. 73
Number systems
CONTINUING MULTIBIT
Had a carry to the 2nd BCD digit position
 1
 0 1 0 0 0 0 0 1 done
 0 0 0 1 1 0 0 1 0 0 1 0
 1 0 1 1
Again must add 6 0 1 1 0
Giving 1 0 0 0 1
And another carry
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74. 74
Number systems
STILL CONTINUING MULTIBIT
Had a carry to the 3rd BCD digit position
 1
 0 1 0 0 done done
 0 0 0 1 0 0 0 1 0 0 1 0
 0 1 1 0
And answer is 0110 0001 0010 or the BCD for the base
10 number 612
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75. 75
Number systems
GRAY CODES
Gray code. The reflected binary code
(RBC), also known as Gray code after
Frank Gray, is a binary numeral system
where two successive values differ in
only one bit (binary digit). The reflected
binary code was originally designed to
prevent spurious output from
electromechanical switches.
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76. 76
The Gray code is unweighted and is not an
arithmetic code.
There are no specific weights assigned to the bit
positions.
Important: the Gray code exhibits only a
single bit change from one code word to the
next in sequence.
Number systems
GRAY CODES
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77. 77
GRAY CODES
Number systems LOGIC SYSTEM DESIGN||MODULE 1||DR. BIBIN VINCENT||ASSOCIATE PROFESSOR||KMEA

78. 78
Number systems
Binary-to-Gray code conversion
The MSB in the Gray code is the same as corresponding
MSB in the binary number.
Going from left to right, add each adjacent pair of
binary code bits to get the next Gray code bit. Discard
carries.
ex: convert 101102 to Gray code
1 + 0 + 1 + 1 + 0 binary
1 1 1 0 1 Gray
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79. 79
Number systems
Gray-to-Binary Conversion
The MSB in the binary code is the same as the
corresponding bit in the Gray code.
Add each binary code bit generated to the Gray code bit in
the next adjacent position. Discard carries.
ex: convert the Gray code word 11011 to binary
1 1 0 1 1 Gray
+ + + +
1 0 0 1 0 Binary
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80. 80
Number systems
The Excess-3 (XS-3) BCD code does not use the principle of positional
weights into consideration while converting the decimal numbers to
4-bit BCD system. Therefore, we can say that this code is a non-
weighted BCD code.
The function of XS-3 code is to transform the decimal numbers into
their corresponding 4-bit BCD code.
In this code, the decimal number is transformed to the 4-bit BCD code
by first adding 3 to all the digits of the number and then converting
the excess digits, so obtained, into their corresponding 8421 BCD
code. Therefore, we can say that the XS-3 code is strongly related with
8421 BCD code in its functioning.
The Excess-3 (XS-3) Code
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81. 81
Number systems
Decimal digits Excess-3 BCD code
0 0011
1 0100
2 0101
3 0110
4 0111
5 1000
6 1001
7 1010
8 1011
9 1100
The Excess-3 (XS-3) Code
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82. 82
Number systems
ASCII CODES
ASCII stands for American Standard Code for
Information Interchange
The code uses 7 bits to encode 128 unique characters
It has the word length 7 and codes decimal digits, the characters
of the latin alphabet as well as special character. From the 128
possible binary words are 32 pseudo-words and/or control
characters.
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Number systems
ASCII CODES
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Number systems
ERROR DETECTION / CORRECTING CODES
Why might we need Error
detection/correction?
Even & Odd Parity
 Error detection
Hamming code
 Used for error detection & error correction
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Number systems
PARITY BITS
ASCII – 7 bit code (hex 00 to 7F)
Could use “8th” bit for parity bit:
X1011010
 Even parity: make total number of “1” bits is even
01011010
Odd parity: make total number of “1” bits odd
11011010
If a parity bit is added to a bit stream, then there is a basis to check
for bit(s) being corrupted.
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Number systems
Suppose you receive a binary bit word
“0101” and you know you are using an odd
parity.
Is the binary word errored?
The answer is yes:
 There are 2 1-bit, which is an even number
 We are using an odd parity
 So there must have an error.
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Number systems
A single bit is appended to each data chunk
makes the number of 1 bits even/odd
Example: even parity
(1)1000000
(0)1111101
(1)1001001
Example: odd parity
(0)1000000
(1)1111101
(0)1001001
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Number systems
Suppose you are using an odd parity.
What should the binary word “1010” look
like after you add the parity bit?
Answer:
There is an even number of 1-bits.
 So we need to add another 1-bit
Our new word will look like “10101”.
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Number systems
Hamming Code (7,4) [1 bit error correction]
A Hamming code is a linear error-correcting code named after its
inventor, Richard Hamming. Hamming codes can detect up to two bit
errors, and correct single-bit errors. This method of error correction is best
suited for situations in which randomly occurring errors are likely, not for
errors that come in bursts.
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Number systems
1 2 3 4 5 6 7
P1 P2 D3 P4 D5 D6 D7
The format of a Hamming code is:
•Here 1,2,4 are the parity bits and 3,5,6,7 are the
Information/Data bits
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Example: Encode the data bits 1101 into a 7-bit even parity Hamming code.
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