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2.11 Error Codes
It is impossible to avoid the interference of noise, which causes errors in the
received binary data at other systems. The bits of the data may change (either 0 to
1 or 1 to 0) during transmission. Therefore error detection and correction code play
an important role in the transmission of data from one source to another. There are
four method for detect errors:
2.11.1 Parity Method for Error Detection
There are two parity methods, even and odd. In the even parity method, the value
of the bit is chosen so that the total number of 1s in the transmitted signal,
including the parity bit, is even. Similarly, with odd parity, the value of the bit is
chosen so that the total number of 1s is odd.
Even parity (ep): makes the total no. of 1?s even
Odd parity (op): makes the total no. of 1?s odd
Example: Assign the proper even parity bit to the following code groups:
(a) 1010 (b) 111000 (c) 101101 (d) 1000111001001 (e) 101101011111
Sol.
Make the parity bit either 1 or 0 as necessary to make the total number of 1s
even. The parity bit will be the left-most bit (color).
(a) 01010 (b) 1111000 (c) 0101101 (d) 0100011100101 (e) 110110101111135
Example: An odd parity system receives the following code groups: 10110, 11010,
110011, 110101110100, and 1100010101010. Determine which groups, if any,
are in error.
Sol. Since odd parity is required, any group with an even number of 1s is
incorrect. The following groups are in error: 110011 and 1100010101010.
2.11.2 Checksum Method
Checksums are similar to parity bits except, the number of bits in the sums is larger
than parity and the result is always constrained to be zero. That means if the
checksum is zero, an error is detected. A checksum of a message is an arithmetic
sum of codewords of a certain length. The sum is stated by means of 1’s
complement and stored or transferred as a code extension of the actual code word.
At the receiver, a new checksum is calculated by receiving the bit sequence from
the transmitter.
Checksum of messages = M1 + M2 + M3 + M4 + … = 0 0 0 0 0
Example: If k = 4 and n =8, find the checksum of four segments: (10110011
10101011 01011010 11010101), along with each transmitted message, the
checksum of all the messages are also transmitted.
Sol. K=4, n = 8
10110011 10110011
10101011 10101011
01011110 01011110
1 1
01011111 01011111
01011010 01011010
10111001 10111001
11010101 11010101
10001110 10001110
1 1
Sum: 10001111 10001111
Checksum: 01110000 01110000
Sum: 11111111
Complement = 00000000
Conclusion = Accept data
At sender side At receiver side36
2.11.3 Cyclic Redundancy Check (CRC)
In cyclic redundancy code (CRC), the transmitted bit sequence is:
TX RX
Series data + CRC data data + CRC data
The transmitted CRC is compared with the RX CRC and if they match then there
are no errors. If they do not match then error is there.
Example: Determine the transmitted CRC for the following byte of data (D) and
generator code (G). Verify that the remainder is 0. D: 11010011, G: 1010
Sol.
Since the generator code has four data bits, add four 0s (blue) to the data byte. The
appended data (D ) is37
Example: During transmission, an error occurs in the second bit from the left in
the appended data byte generated in Example above. The received data is
D’ = 100100110100 Apply the CRC process to the received data to detect the error
using the same generator code (1010)