Consider the following parity protocol. The data consists of m bits, and the che
ID: 3620209 • Letter: C
Question
Consider the following parity protocol. The data consists of m bits,and the checkbits consist of m parity bits (I know, its wasteful)
Label the bits by their position, i.e., first data bit is b0, the
second bit b1, the last data bit bm-1, then bm is the first parity
bit, bm+1 the second parity bit, and bm+m-1 the last parity bit.
Assume each parity bit coversthe previous m bits (regardless
of whether they are data or parity bits. E.g. but bm (the first
parity bit) covers data bits b0 ... bm-1, and bit bm+1 (the second
parity bit) covers data bits b1 .. bm-1 AND the parity bit bm, etc.
What is the hamming distance of this scheme?
Explanation / Answer
Dear user, Let us suppose, m = 4 for simplicity and let the data bits be 1101 So the code word with the parity bits calculated as per the scheme will be 11011110 They are represented as: b0 b1 b2 b3 b4 b5 b6 b7 1 1 0 1 1 1 1 0 The hamming distance depends on the number of bits that are changed in the data word. Since the parity bits depend on the values of data and parity bits, the hamming distance changes. If 1 bit is changed, that means error occurred in 1 bit then 1 data bit and two parity bits that are changed by the data bit are also changed. For example, if the first bit of the data word is changed and become as 0101, The code word becomes 01010010 So the hamming distance of 11011110 and 01010010 is 3 Similarly as the number of bits change in the dataword, the hamming distance changes. In general, if only 1 bit changes, the minimum hamming distance becomes m – 1 And if the maximum of total m bits are change, then the hamming distance is 2m – 1