Question B A 3-bit redundancy-check code (odd parity for columns, even parity fo

Question

Question B

A 3-bit redundancy-check code (odd parity for columns, even parity for rows) will be used to transmit data. In a transmission system data sequences are encoded using 4-bit words (3-bit data andeven parity bit).

a) How many of the bits in the sequence below are redundant bits (i.e. they were added as part of the redundancy-check code before transmission), and how many bits are for the data only?

0 1 0 1 1 1 1 1 0 1 0 1

Redundant Bits:?

Data Bits: ?

b) Give the coding of the data that will be transmitted for the following original data:

coding ( 1 0 1 1 0 0 0 1 1 0 1 0 ) = ?

Explanation / Answer

ANSWER to Question (A) 4 bit word (3 data and one redundancy) we have to check for odd parity. a) 1011100001110100 b)encode data is 1101101011100100 as we know that the encoded word length is 4 and total number of bits encoded is 16 so there are 4 number of words. if we we will write each word as a row of a matrix then it would be as follows 1 1 0 1 1 0 1 0 1 1 1 0 0 1 0 0 now we will look for last column of each row for odd parity check(as last column contains odd parity check bits) Inspecting the rows we found that at 2nd row the word is 1010. As the last bit is '0'(it is the odd parity check bit) even though there are two '1's. so the second word is errored. so the answer is 2. ANSWER to question B A 3-bit redundancy-check code (odd parity for columns, even parity for rows) will be used to transmit data. In a transmission system data sequences are encoded using 4-bit words (3-bit data andeven parity bit). a) encoded data is 010111110101 if we we will write each word as a row of a matrix then it would be as follows 0 1 0 1 1 1 1 1 0 1 0 1 In this last row is odd parity check bits. so there are 4 odd parity check bit. apart from this 4 bits there are remaining 8 bits. so there are two words of 4 bit each. For each word there are 1 parity check bit. so there are 2 number of even parity check bits. Altogether the total no. of parity bits= 4 (odd parity) +2 (even parity) =6 bits so there are 6 bits of redundancy data. the remaining 6 bits are data bits. Redundant Bits: 6 Data Bits: 6 b) coding ( 1 0 1 1 0 0 0 1 1 0 1 0 ) = ? parity check bit(even parity of each rows) 1 0 1 0 1 0 0 1 0 1 1 0 0 1 0 1 1 1 1 1(the last column for odd parity check bit of each column) so the encoded data is 1 0 1 0 1 0 0 1 0 1 1 0 0 1 0 1 1 1 1 1

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