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1. Data is transmitted in 256-byte blocks or packets. Eight sequential packets are transmitted. The system BER is 4:10000. On average, how many errors can be transmitted in this transmission?

2. Determine the 4 Hamming bits for the 8-bit number 11010110.

Answer :

Final answer:

The average number of errors that can be transmitted in this transmission is approximately 6.5536 errors. The 4 Hamming bits for the 8-bit number 11010110 are H1 = 0, H2 = 1, H3 = 1, and H4 = 0.

Explanation:

Transmission Errors:

In this scenario, data is transmitted in 256-byte blocks or packets. Eight sequential packets are transmitted. The system BER is 4:10000.

The BER ratio of 4:10000 means that for every 4 bits transmitted, there is an average of 1 error. To calculate the average number of errors in this transmission, we can use the formula:

Average number of errors = (Total number of bits transmitted) x (BER ratio)

Since each packet contains 256 bytes, and each byte contains 8 bits, the total number of bits transmitted is:

Total number of bits transmitted = (Number of packets) x (Number of bytes per packet) x (Number of bits per byte)

Substituting the given values:

Total number of bits transmitted = 8 x 256 x 8 = 16,384 bits

Now, we can calculate the average number of errors:

Average number of errors = 16,384 x (4/10000) = 6.5536 errors

Hamming Bits:

To determine the 4 Hamming bits for the 8-bit number 11010110, we can use the following steps:

  1. Assign positions to the bits starting from 1, skipping the positions for the Hamming bits. In this case, the positions are: 1, 2, 3, H1, 4, H2, 5, 6, H3, 7, 8, H4.
  2. Calculate the values of the Hamming bits using the parity of the data bits. The Hamming bits are calculated as follows:
  3. H1 = parity of positions 3, 5, 7
  4. H2 = parity of positions 3, 6, 7
  5. H3 = parity of positions 5, 6, 7
  6. H4 = parity of positions 1, 2, 3, 4, 5, 6, 7
  7. Replace the Hamming bit positions with the calculated values. In this case, the 4 Hamming bits are:
  8. H1 = 0 (even parity of 1, 0, 1)
  9. H2 = 1 (odd parity of 1, 0, 1)
  10. H3 = 1 (odd parity of 0, 1, 1)
  11. H4 = 0 (even parity of 1, 1, 0, 1, 0, 1, 1)

Learn more about transmission errors and hamming bits here:

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