EEN 6113
DIGITAL LOGIC DESIGN
2
1
NUMBER SYSTEMS AND CODES
2
Learning objectives

Review the decimal number system

Convert from binary to decimal and from decimal to binary

Convert from octal to decimal and from decimal to octal

Convert from hex to decimal and from decimal to hex

Convert from hex to binary and from binary to hex

Perform binary addition, subtraction, multiplication and division on two binary
numbers

Add and subtract hexadecimal numbers

Determine the 1’s and 2’s complements of a binary number

Manipulate signed binary numbers using the 2’s complement system
3
Learning objectives

Express decimal numbers in Binary Coded Decimal (BCD) form

Add BCD numbers

Interpret the American Standard Code for Information Interchange (ASCII)

Convert from binary to gray code and from gray code to binary

Explain how to use parity method to detect code errors

Assign proper parity bit to a code

Describe the Hamming code and know how to correct a single-bit error

Determine if there is an error in a code based on the parity bit
4
Decimal Numbers
 The decimal numbering system has 10 types of digits
 Consist of numbers 0 through 9
 The decimal numbering system has a base of 10 with
each position weighted by a factor of 10 ….
103
102
101
100
10-1
10-2
10-3
10-4
Decimal point
 Example:-
18.510 = 1 x 101 + 8 x 100 + 5 x 10-1
10-5
5
Binary Numbers

The binary numbering system has 2 types of digits

Consist of numbers 0 and 1 only

The binary numbering system has a base of 2 with
each position weighted by a factor of 2
23

22
21
20
2-1
2-2
2-3
2-4
2-5
Example:110112 = 1 x 24 + 1 x 23 + 0 x 22 + 1 x 21 + 1 x 20
6
Binary to Decimal Conversion
 Convert binary to decimal by summing the positions that
contain a 1:
 An example with a greater number of bits:
Ronald Tocci/Neal Widmer/Gregory Moss
Digital Systems: Principles and Applications, 10e
Copyright ©2007 by Pearson Education, Inc.
Columbus, OH 43235
All rights reserved.
7
Exercises
Convert the following binary numbers to decimal without using a
calculator. Use calculator to verify your answers.
a) 100112
(19)
b) 0011102
(14)
c) 0101.110 2
(5.75)
8
Decimal to Binary Conversion
Repeated Division (Method 1)
•
•
•
•
Divide the decimal number by 2.
Write the remainder after each division
until a quotient of zero is obtained.
The first remainder is the
Least Significant Bit (LSB) and
it represents the lowest power of 2.
The last is the Most Significant Bit (MSB)
and it represents the highest power of 2.
Copyright ©2007 by Pearson Education, Inc.
Columbus, OH 43235
All rights reserved.
9
Decimal to Binary Conversion
 Convert 3710 to binary:
1001012
Copyright ©2007 by Pearson Education, Inc.
Columbus, OH 43235
All rights reserved.
10
Exercises
Convert the following decimal numbers to binary with
and without using a calculator.
a) 5210
(1101002)
b) 10810
(11011002)
11
Decimal Fractions to Binary Conversion
Radix Multiply Technique


The technique consists of the following steps:
1.
Successively multiply the given fraction by the required base,
noting the integer portion of the product at each step. Use the
fractional part of the product as the multiplicand for subsequent
steps. Stop when the fraction either reaches 0 or recurs/repeats.
2.
Collect the integer digits at each step from first to last and arrange
them left to right
Note: If the radix multiplication process doesn’t converge to 0,
it’s not possible to represent a decimal fraction in binary exactly.
Accuracy, then, depends on the number of bits used to
represent the fraction.
12
Decimal Fractions to Binary Conversion
Radix Multiply Technique
(0.250)10 =(?)2
0.25
X 2
0.50
X 2
1.00
(0.25)10 =(0.01)2
(0.345)10 =(?)2
0.345
X 2
0.690
X 2
1.380
X 2
0.760
X 2
1.520
X 2
1.040
X 2
0.080
(0.345)10 =(0.010110)2
The fraction may
never reach 0;
stop when the
required number
of the fraction
digits is obtained;
the fraction will
not be accurate
13
Decimal Fractions to Binary Conversion
 Example:
(25.45)10 =(?)2
For the integer of 2510
For the fraction of 0.4510
0.45
X 2
0.90
X 2
1.80
X 2
1.60
X 2
1.20
X 2
0.40
X 2
0.80
X 2
1.60
Repeated division by 2
yields the whole number
while repeated
multiplication by 2 of the
fraction yields the binary
fraction
Repeats
(25.45)10 =(11001.011100)2
14
Exercises
Convert the following decimal numbers to binary.
a) 52.7510
(110100.112)
b) 108.2510
(1101100.012)
15
Octal Numbers
 The octal numbering system has 8 types of digits
 Consist of numbers 0 through 7
 The octal numbering system has a base of 8 with
each position weighted by a factor of 8 ….
83
82
 Example:-
81
80
8-1
8-2
8-3
8-4
2748 = 2 x 82 + 7 x 81 + 4 x 80 = 18810
8-5
16
Decimal → Octal
Example: Convert 15010 into Octal.
8
150
8
18
6
8
2
2
0
2
Remainder
15010 =2268
17
Exercises
Perform the following conversion
a)
Convert to Octal :35910
(5478)
b)
Convert to decimal: 2748
(250 )
c)
Convert to decimal: 24.68
(20.75 )
10
10
18
Binary → Octal
Example: Convert 0001011102 to octal
Method: Form 3 binary bit, convert accordingly with max
conversion =7
0001011102
000
101
1102
0
5
6
Answer =0568
Beware !
Avoid incorrect
grouping
19
Binary → Octal (Fractions)
Example: Convert 010110.0111002 to Octal
010 110 . 011 1002
2
6
3
Answer = 26.348
4
20
Exercises
Convert the following binary numbers to octal.
a) 110012
(318)
b) 1010.0112
(12.38)
21
Hexadecimal Numbers
 Hexadecimal allows convenient handling of long
binary strings, using groups of 4 bits—Base 16
 The hexadecimal numbering system has a base of 16
with each position weighted by a factor of 16 ….
163
162
161
160
16-1
16-2
Hexadecimal point
 Example:-
16-3
16-4
16-5
22
Hexadecimal Number System
Relationships between
hexadecimal, decimal, and
binary numbers.
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Columbus, OH 43235
All rights reserved.
23
Hex → Decimal
The hexadecimal numbering system uses the numerals 0 through 9
and the letters A through F.
Counting : 0 1 2 3 4 5 6 7 8 9 A B C D E F
Example : Convert 4D16 to decimal
1010
= ( 4 x 161) + (D x 160)
= (4 x 16) + ( 13 x 1)
=7710
1510
24
Decimal → Hex
Convert 42310 to hex:
Method: Use repeated division
Copyright ©2007 by Pearson Education, Inc.
Columbus, OH 43235
All rights reserved.
25
Exercise
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Columbus, OH 43235
All rights reserved.
26
Exercise - solutions
Copyright ©2007 by Pearson Education, Inc.
Columbus, OH 43235
All rights reserved.
27
Decimal → Hex (Fractions)
Convert 53.3910 to Hex
53 /16 = 3 , remainder = 5
3 / 16 = 0 , remainder = 3 (MSB)
0.39 x 16 = 6.24
Answer = 35.63D7H
0.24 x 16 = 3.84
0.84 x 16 = 13.44
0.44 x 16 = 7.04
Continue to reduce error
28
Hex → Binary
 Leading zeros can be added to the left of the MSB to fill out
the last group.
29
Binary → Hex
 It is easy to convert between base 16 and base 2, because 16 = 24 .
 Convert from binary to hex by grouping bits in four starting with the
LSB.
 Each group is then converted to the hex equivalent
 The binary number is grouped into groups of four bits & each is converted to its
equivalent hex digit.
Answer = 3A616 can be written as 3A6H
Copyright ©2007 by Pearson Education, Inc.
Columbus, OH 43235
All rights reserved.
30
Intermediate conversion Technique
➢ Convert 15010 to binary
Method 1 : Using conventional Division method
Method 2 : Intermediate method
Decimal
Octal
Binary
8 150
8
18
6
8 2
2
0
2
2268
010 010
1102
31
Intermediate conversion Technique
 Convert 37810 to a 16-bit binary number
Decimal
Hex
Binary
= 1010
= 0111
= 0001
Answer = 17A16 = 0001 0111 10102
32
Exercises
Convert the following hexadecimal numbers.
a) 1B16=( ? )2
(110112)
b) 7CA16=(
(199410)
? )10
33
Final Exam
34
Final Exam
35
Homework
36
Arithmetic - Binary Addition
Decimal , Binary Addition
8
0
0
1
1
1
+6
+0
+1
+1
+ 1
14
0
1
10
11
Carry out to next stage
One bit
adder
sum
One bit
adder
sum
carry
37
Arithmetic - Binary Addition
ADD 47 and 29 in binary
11111
47
+ 29
101111
+ 011101
76
1 001100
Carry bits
Begin with LSB and work towards the MSB, writing down
(memories) carries as they are generated.
38
Arithmetic - Hexadecimal Addition
ADD 2CH and 48H
Use the following steps to perform
hexadecimal addition:
1. Add one column at a time.
2. Convert to decimal and add the
numbers.
3. If the result of step two is 16 or larger
subtract the result from 16 and carry 1
to the next column.
4. If the result of step two is less than 16,
convert the number to hexadecimal.
1
2C
C=12
+48
12+8=20
74
20-16=4
Carry 1 (representing 16)
remain 4
39
Exercises
40
Exercises-solution
41
Binary subtraction

In subtracting, we have become accustomed to “Borrowing” from other
column to find a result.

In Decimal numbers, the base is 10, which means for subtraction, the
borrowed number is 10.

In Octal numbers, the base is 8, which means for subtraction, the borrowed
number is 8.

In Hex numbers, the base is 16, which means for subtraction, the borrowed
number is 16.
42
Binary subtraction

However, for reasons of efficiency, computers rarely perform subtractions in this
manner; instead, these operations are typically performed by means of
complement techniques.

There are two complement techniques;
1.One’s complement
2.Two’s complement
The method of two’s complement arithmetic is commonly used
in computers to handle negative numbers.
43
Representation of Negative Numbers

There are three forms in which signed integer numbers can be represented in
binary:

Sign-magnitude; 1’s complement; 2’s complement.

The method of two’s complement arithmetic is commonly used in computers to
handle negative numbers

With sign magnitude, the MSB is used to flag a negative number. So for example
with 4-bit numbers we would have 0011 = 3 and 1011 = −3. This is simple to
see, but is not good for doing arithmetic.
44
Representation of Negative Numbers
With 1’s complement

The negative numbers are the 1’s complements of the corresponding positive
numbers. For example, using eight bits, the decimal number (-25) is expressed as
the 1’s complement of (+25) (00011001) as 11100110

Note: in the 1’s complement form, a negative number is the 1’s complement of the
corresponding positive number
With 2’s complement

The negative numbers are the 2’s complements of the corresponding positive
numbers. For example, using eight bits, the decimal number (-25) is expressed as
the 1’s complement of (+25) (00011001) as 11100111

Note: in the 2’s complement form, a negative number is the 2’s complement of the
corresponding positive number
45
Example
Express the decimal number -39 as an 8-bit number in the sign-magnitude, 1’s
complement, and 2’s complement forms.

8-bit number of +39 is 00100111.

In sign-magnitude form, -39 is produced by changing the sign bit to a 1 and
leaving the magnitude bits as they are. The number is 10100111.

In the 1’s complement form, -39 is produced by taking the 1’s complement of
+39 (00100111). The number is 11011000.

In the 2’s complement form, -39 is produced by taking the 2’s complement of
+39 (00100111). The number is 11011001.
46
One’s Complement
Inverting each bit
(changing each 0→1, and each 1→0)
Example:
0101102 → to form One’s complement
→ 1010012
0 1 0 1 1 0
inverting
1 0 1 0 0 1
First Complement
Two’s Complement
Method = One’s Complement + 1
Example: Find the 2’s Complement of 28 in binary (8 bits)
Step 1 : Find binary equivalent 2810 = 000111002
Step 2 : Find 1’s Complement number = 111000112
Step 3 : add 1 to get 2’s Complement
11100011
= One’s Complement
+1
11100100
2’s Complement of 28D
= One’s Complement + 1
47
Substrate 47 from 123 using 2’s Complement 48
binary arithmetic
123
01111011
-(47)
-(00101111)
?
?
Convert to 2’s
Complement
directly
Solve for 4710 : +00101111
1’s complement : 11010000
2’s complement:
11010000
+1
11010001
Now, you can perform subtraction
in addition form.
01111011
+11010001
101001100
Carry-- ignored
=76
+ve sign
49
Example:- Calculate 12 – 3 and
3 – 12 in 5-bit word
Note:12D = 01100
12
3
-3
-12
3D = 00011
Find 2’s complement for 3
Find 2’s complement for 12
00011 → 11100 + 1 =11101
01100→ 10011 + 1 =10100
01100
00011
+11101
+10100
101001
=+9
Ignore carry
+ve sign
10111
(for prove, take the 2’s
complement: 01001) = +9
=-9
(2’s complement form)
-ve sign
50
Subtract 23H from 81H

First convert the Hexa into binary

It is important to hold an equal number of places for numbers being subtracted.
81
10000001
-23
-00100011
?
?
10000001
+11011100
1 01011101
carry
Convert to 1’s
Complement
1 01011101
End
round
carry
+1
01011110
answer
0101
1110
= 5EH
51
Exercises
Calculate the following assuming a two’s complement
number system and n = 8 (where MSB is the sign bit).
Check your results by decimal arithmetic.
a) 8 − 3
b) − 25 − 19
c)
− 120 + 30
52
Overflow condition
53
Solutions
54
Solutions
55
Binary Multiplication
The multiplication of binary numbers is done in the same manner as the
multiplication of decimal numbers. The process is actually simpler since the
multiplier digits are either 0 or 1.
Perform binary multiplication of 910 x 1110
0101(5) Multiplicand
x 0011(3) Multiplier
0101
0101
0000
+ 0000
0001111 = (15)
56
Binary Division

Compare Dividend (X) and Y (Divisor).
a) If X >= Y, the quotient bit is 1 and perform the subtraction X-Y.
b) If X < Y, the quotient bit is 0 and do not perform any subtractions.

4. 3. Shift Y one bit to the right and go to step 2.
0111
Divisor (Y)
111
110101
- 111
1100
- 111
Quotient
Dividend (X)
1011
- 111
100
Remainder
57
Binary Division
11010
Divisor (Y)
1000
11010010
Quotient
Dividend (X)
- 1000
-
1010
1000
-
1001
1000
10
Remainder
58
Exercises
Perform the following binary division.
a) 110 ÷ 11
(102)
59
Binary codes : Binary Coded Decimal
The BCD system is employed by computer systems to encode the decimal
number into its equivalent binary number.
 This is generally accomplished by encoding each digit of the decimal number into
its equivalent binary sequence.
 The main advantage of BCD system is that it is a fast and efficient system to
convert the decimal numbers into binary numbers as compared to the pure
binary system.
 The 4-bit BCD system is usually employed by the computer systems to represent
and process numerical data only. In the 4-bit BCD system, each digit of the
decimal number is encoded to its corresponding 4-bit binary sequence. The two
most popular 4-bit BCD systems are:
1. Weighted 4-bit BCD code
2. Excess-3 (XS-3) BCD code

60
Weighted 4-bit BCD code

Using 4 bits to define BCD, but use only 0 –9 digits.

The weighted 4-bit BCD code is more commonly known as 8421 weighted code.

It is called weighted code because it encodes the decimal system into binary
system by using the concept of positional weighting into consideration.

In this code, each decimal digit is encoded into its 4-bit binary number in which
the bits from left to right have the weights 8, 4, 2, and 1, respectively.
Decimal
BCD
0
1
2
3
4
5
6
7
8
9
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
61
8421 BCD
Example: Convert 47910 into 8421 BCD
4 7 9
0100
0111
1001
Answer = 0100 0111 1001BCD
Compare with pure binary conversion
47910 = 0001 1101 11112
62
BCD Addition

Addition of BCD digits is similar to adding 4-bit unsigned binary numbers except
a correction must be made if a result exceeds 1001 by adding 6 to the digit.
Decimal to BCD
+
(5)10
(9)10
(14)10
+
0101
1001
+
1110
0110
0001 0100
(1)
(4)
Since sum exceed
1001, we add 0110
to the answer
63
BCD Addition
Decimal to BCD
+
(56)10
(69)10
+
0101 0110
0110 1001
+
1011 1111
0110 0110
(125)10
0001 0010 0101
(1)
(2)
(5)
Since sum exceed
1001 for both BCD,
we add 0110 to
each sum.
64
Excess-3 BCD Code

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.

This code can be derived from the natural binary BCD (8-4-2-1) code by adding
3 to each coded number.

Therefore, we can say that the XS-3 code is strongly related with 8421 BCD code
in its functioning.
65
BCD & Excess-3
Decimal
BCD
EXCESS-3
0
0000
0011
1
0001
0100
2
0010
0101
3
0011
0110
4
0100
0111
5
0101
1000
6
0110
1001
7
0111
1010
8
1000
1011
9
1001
1100
66
Excess-3 BCD Code
Example: Convert the decimal number 85 to XS-3 BCD code.

Add 3 to each digit of the given decimal number as:
8+3=11 and
5+3=8

The corresponding 4-bit 8421 BCD representation of the decimal digit 11 is 1011.

The corresponding 4-bit 8421 BCD representation of the decimal digit 8 is 1000.

Therefore, the XS-3 BCD representation of the decimal number 85 is 1011 1000.
67
ASCII Code

4-bit BCD systems are inadequate for representing and handling non-numeric data.
For this purpose, 6-bit BCD and 8-BCD systems have been developed.

The 6-bit BCD systems can handle numeric as well as non-numeric data but with few
special characters.

The 8-bit BCD systems were developed to overcome the limitations of 6-bit BCD
systems, which can handle numeric as well as nonnumeric data with almost all the
special characters such as +, -, *, /, @, $, etc. Therefore, the various codes under the
category of 8-bit BCD systems are also known as alphanumeric codes.

The ASCII code is pronounced as ASKEE.

Initially, this code was developed as a 7-bit BCD code to handle 128 characters but
later it was modified to an 8-bit code. 8th bit is used as ‘extended ASCII’ to code
graphical shapes, mathematical symbols, etc. or as check bit
ACSII Code
68
ACSII Code
69
70
Gray Code

This is a variable weighted code and is cyclic. This means that it is arranged so that
every transition from one value to the next value involves only one bit change.

To convert binary to gray code, bring down the most significant digit of the given
binary number, because, the first digit or most significant digit of the gray code number
is same as the binary number.

To obtain the successive gray coded bits to produce the equivalent gray coded number
for the given binary, add the first bit or the most significant digit of binary to the second
one and write down the result next to the first bit of grey code, add the second binary
bit to third one and write down the result next to the second bit of grey code, follow this
operation until the last binary bit and write down the results based on EX-OR logic to
produce the equivalent grey coded binary.
71
Binary to Gray Code

Convert 10110 in binary code to Gray code.
+
+
+
+
Binary
1
0
1
1
0
Gray
1
1
1
0
1
72
Gray Code to Binary

Gray
Convert 11011 in Gray code to Binary code.
1
1
+
Binary
1
0
+
0
1
+
+
0
1
1
0
➢ To convert grey code to binary, bring
down the most siginificant digit of the
given grey code number, because, the
first digit or the most siginificant digit
of the grey code number is same as
the binary number.
➢ To obtain the successive second binary
bit, perform the EX-OR operation
between the first bit or most
siginificant digit of binary to the
second bit of the given grey code.
73
Parity Bit for Error Detection

Many systems use a parity bit as a means for bit error detection.

Any group of bits contain either even or odd number of 1s.

A parity bit is attached to a group of bits to make the total number of 1s in a group
always even or always odd.

An even parity bit makes the total number of 1’s even, and an odd parity bit makes the
total to be odd.

The parity bit can be attached to the code

A given system operates with even or odd parity, but not both.

If a system operates with even parity, a check is made on each group of bits received to
make sure the total number of 1s in that group is even. If there is an odd number of 1s,
an error has occurred.
74
Parity Bit for Error Detection
Used for single bit error detection
8-bits
7-bits
Parity bit
generator
Parity = X
8-bits
Transmission
medium
1
011 0100
0 100 1011
Parity bit
checker
Output
Parity = X’
Noise
Even parity
Total no. of 1’s (including parity bit)
is an even number
7-bits
Odd parity
Total no. of 1’s (including parity bit)
is an odd number
0
011 0100
1 100 1011
75
Hamming Code (Distance-3)

The most common types of error-correcting codes used in RAM are based on the codes
devised by R. W. Hamming. In the Hamming code, k parity bits are added to an n-bit
data word, forming a new word of n + k bits. The bit positions are numbered in
sequence from 1 to n + k.

Those positions numbered with powers of two are reserved for the parity bits. The
remaining bits are the data bits. The code can be used with words of any length.

Parity bit position = 2𝑛

Exclusive-OR gate can be used to generate
and check parity bit
76
Hamming Code (Distance-3)
Example:-
Consider, for example, the 7-bit data word 1011001.

The number of data bits = 7

The number of parity bits to be included = 4

The total number of bits = 11

The parity bits are placed at positions corresponding to power of 2, which corresponds
to the position 1, 2, 4, and 8.
11
10
9
8
7
6
5
4
3
2
1
1
0
1
P8
1
0
0
P4
1
P2
P1
77
Hamming Code (Distance-3)
a. Parity bit 1 covers all the bits positions whose binary representation includes a 1 in the
least significant position (1, 3, 5, 7, 9, 11, etc).
(1011)
(1001)
11
10
9
1
0
1
(0111)
8
+
(0101)
7
6
5
1
0
0
(0011)
4
2
1
1
P1
+
+
+
3
(0001)
For XOR gate, count
the number of 1s, if it’s
even, then the results
is zero, if it’s odd, then
the results is 1.
0
b. Parity bit 2 covers all the bits positions whose binary representation includes a 1 in the
second position from the least significant bit (2, 3, 6, 7, 10, 11, etc).
(1011)
(1010)
11
10
9
1
0
1
+
8
+
(0111)
(0110)
7
6
5
1
0
0
+
4
+
(0011)
(0010)
3
2
1
P2
1
1
78
Hamming Code (Distance-3)
c. Parity bit 4 covers all the bits positions whose binary representation includes a 1 in the
third position from the least significant bit (4–7, 12–15, 20–23, etc).
11
10
9
1
0
1
8
(0111)
(0100)
(0101)
(0100)
7
6
5
4
3
1
0
0
P4
1
+
+
2
1
1
d. Parity bit 8 covers all the bits positions whose binary representation includes a 1 in the
fourth position from the least significant bit bits (8–15, 24–31, 40–47, etc).
(1011)
(1010)
(1001)
(1000)
11
10
9
8
7
6
5
1
0
1
P8
1
0
0
+
+
0
4
3
1
2
1
79
Hamming Code (Distance-3)
Thus, the data transferred will be
11
10
9
8
7
6
5
4
3
2
1
1
0
1
0
1
0
0
1
1
1
0
Where
P1 = 0, P2 = 1, P4 = 1, P8 = 0
80
Hamming Code (Distance-3)
Error Detection and Correction:-
When the 11 bits are read from memory, they are checked again for errors.
The parity of the word is checked over the same groups of bits, including their
parity bits. The four check bits are evaluated as follows:
C1 = XOR of bits (1, 3, 5, 7, 9, 11) = 0
C2 = XOR of bits (2, 3, 6, 7, 10, 11) = 0
C=C8C4C2C1 = 0000 = 0 (No error)
C4 = XOR of bits (4, 5, 6, 7) = 0
C8 = XOR of bits (8, 9, 10, 11) = 0
11
10
9
8
7
6
5
4
3
2
1
1
0
1
0
1
0
0
1
1
1
0
81
Hamming Code (Distance-3)
Error Detection and Correction:-

A 0 check bit designates an even parity over the checked bits, and a 1 designates an
odd parity. Since the bits were written with even parity, the result C = 0 indicates that
no error has occurred.

However, if C≠0, the 4-bit binary number formed by the check bits gives the position of
the incorrect bit if only a single bit is in error.
82
Hamming Code (Distance-3)
Suppose in the previous example, the 6th bit is changed from 0 to 1 during data
transmission, then it gives new parity values in the binary number.
The four check bits are evaluated as follows:
C1 = XOR of bits (1, 3, 5, 7, 9, 11) = 0
C2 = XOR of bits (2, 3, 6, 7, 10, 11) = 1
C=C8C4C2C1 = 0110 = 6 (Error in bit 6)
C4 = XOR of bits (4, 5, 6, 7) = 1
C8 = XOR of bits (8, 9, 10, 11) = 0
11
10
9
8
7
6
5
4
3
2
1
1
0
1
0
1
1
0
1
1
1
0
Bit Error
83
References

Ronald J. Tocci, Neal S. Widmer and Gregory L. Moss, “Digital
Systems - Principles and Applications”, 11th Edition, PrenticeHall, 2007.

S. Brown and Z. Vranesic, “Fundamentals of Digital Logic with
VHDL Design”, 2nd ed., McGraw-Hill, 2005

Alan B. Marcovitz, “Introduction to Logic Design”, 3rd Edition,
McGraw-Hill, 2009.

Donald D. Givone, “Digital Principles and Design”, McGraw-Hill,
2003.