Question

a. Suppose the information content of a packet is the bit pattern 1110101010101111 and an even parity scheme is being used. What would the value of the checksum field (sequence of parity bits) be for the case of a two-dimensional parity scheme? Your answer should be such that a minimum-length checksum field is used.

b. With the same bit-pattern and your derived checksum, prove that the 2D scheme will not be able to detect all burst errors?

Answer 1

Solution

Part 1

1110101010101111 is divided to 4 bit codes : 1110 , 1010 ,1010 , 1111

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

For checksum to be length minimum ,checksum field should be a 4x4 matrix with two dimensional even parity bit.

So we create a matrix of RxC dimensions and add one even parity bit at the end of each row and column.

Even Parity

If total number of 1 is even, parity bit is 0

If total number of 1 is odd, parity bit is 1

Part 2

A burst error  is a contiguous sequence of symbols, received over a communication channel, such that the first and last symbols have error and no contiguous subsequence of m exists which is correctly received symbols within the error burst.

The correct sequence without code is shown below:

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

Imagine a burst errors involving the highligted bits(in green) above so that the bits are changed from 0 to 1 and 1 to 0.

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

This error cannot be detected by the 2D parity scheme using the row and column parity bits.