Example: Let us see if the two codes we defined in Table 10.1 and Table 10.2 bel

Question

Example:
Let us see if the two codes we defined in Table 10.1 and Table 10.2 belong to the class of linear block codes.
1. The scheme in Table 10.1 is a linear block code because the result of XORing any codeword with any other codeword is a valid codeword. For example, the XORing of the second and third codewords creates the fourth one.

2. The scheme in Table 10.2 is also a linear block code. We can create all four codewords by XORing two other codewords.


Example 10.12
Let us look at some transmission scenarios. Assume the sender sends the dataword 1011. The codeword created from this dataword is 10111, which is sent to the receiver. We examine five cases:
1. No error occurs; the received codeword is 10111. The syndrome is O. The dataword 1011 is created.

2. One single-bit error changes aI' The received codeword is 10011. The syndrome is 1. No dataword is created.

3. One single-bit error changes roo The received codeword is 10110. The syndrome is 1. No dataword is created. Note that although none of the dataword bits are corrupted, no dataword is created because the code is not sophisticated enough to show the position of the corrupted bit.

4. An error changes ro and a second error changes a3' The received codeword is 00110. The syndrome is O. The dataword 0011 is created at the receiver. Note that here the dataword is wrongly created due to the syndrome value. The simple parity-check decoder cannot detect an even number of errors. The errors cancel each other out and give the syndrome a value of O.

5. Three bits-a3, az, and aI-are changed by errors. The received codeword is 01011. The syndrome is 1. The dataword is not created. This shows that the simple parity check, guaranteed to detect one single error, can also find any odd number of errors.

QUESTIONS
1. Using Example 10.10, p. 278.
Show how the XOR operation is applied to data in Table10.1 and Table 10.2.


2. Using Example 10.12, p. 280.
Analyze every case and explain the result. Indicate the cases when the error
is missed.

Explanation / Answer

1) Consider Table 10.1 dataword (000,011,101,110)
Then perform XOR operation on 000 and 011
       1 0 1
XOR 0 1 1
---------------
       1 1 0
---------------
Therefore we perform XOR operation on second(011) and third(101) datawords then we get fourth(110).So the table 10.1 is a linear block code.

Consider Table 10.2 dataword (00000,01011,10101,11110)
Then perform XOR operation on 000 and 011
       0 1 0 1 1
XOR 1 0 1 0 1
-------------------
       1 1 1 1 0
--------------------
Therefore we perform XOR operation on second(01011) and third(101) datawords then we get fourth(11110).So the table 10.2 is also a linear block code.

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