For a 6 3 code the generator matrix G is G 1 0 0 1 0 1 0 1

For a (6, 3) code, the generator matrix G is G = [1 0 0 1 0 1 0 1 0 0 1 1 0 0 1 1 1 0] For all eight possible data words, find the corresponding code words, and verify that this code is a single-error correcting code. Table 16.2 shows the eight data words and the corresponding code words found from c = dG. Note that the distance between any two code words is at least 3. Hence, the code can correct at least one error. Figure 16.1 shows a possible encoder for this code, using a three-digit shift register and three modulo-2 adders. Comm system error detecting code for re-transmission of packets and to improve performance. From above, the generator code for this matrix is G= 1 0 0 1 0 1 0 1 0 0 1 1 0 0 1 1 1 0 If the Bit Error Rate (BER) is 1 in 1,000,000 (1E-06) What is the probability that an error in a code word is NOT detected?

Solution

for not detecting error codeSuppose we want to send 2-bit strings. Each codeword contains two copies of the string plus a parity bit. If the bit-string is 01, we send the 5-bit string 01x01, where x is the parity bit. So in this example, the sender transmits the codeword 01101. For this code, there are only four 5-bit codewords: 00000, 01101, 10110, 11011. When the receiver sees any other string, the error is corrected by replacing it with the codeword that has the least Hamming distance to the received word. Suppose that the string 10001 is received. For Hamming distance d, d(10101, 00000) = 3, d(10101, 01101) = 2, d(10101, 10110) = 1, d(10101, 11011) = 3. So the closest codeword to the received string is 10110, so the receiver assumes that this was the original string. The number of errors that can be detected and corrected depends on the Hamming distance