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CH3_ChannelCoding_4thyear_Part5

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الكلية كلية الهندسة     القسم  الهندسة الكهربائية     المرحلة 4
أستاذ المادة احمد عبد الكاظم حمد الركابي       10/07/2018 21:25:48
Lecture 3/9
Table 12.1
.
13. Systematic cyclic code
Suppose we are given the generator polynomial and the requirement is to encode the message sequence into an systematic cyclic code. That is, the message bits are transmitted in unaltered form, as shown by the following structure
where ( ) are parity bits. Let
Then the code polynomial can be expressed as
Since is a factor of , we have
or
Where is the remainder of i.e.,
As represent shifted to the right by digits, the last digits of this codeword are precisely , and the first digits corresponding to must be parity check digits. Thus, we can state the encoding procedure for a systematic cyclic code as follows:
1. Multiplying the message polynomial by
2. Dividing by to obtain the remainder
3. Adding to to form the code polynomial .
Example 13.1: Consider a (7, 4) cyclic code with generator polynomial . Let data word =(1010) Find the corresponding systematic code word .
Solution: , and
Lecture 3/10
Thus, and ? ?
Hence, the corresponding codeword is
14. Encoder for Cyclic Codes
The encoding operations for generating a cyclic code may be performed by a linear feedback shift register. The first k bits at the output of the encoder are simply the k information bits. These k bits are also clocked simultaneously into the shift register, since switch 1 is in the closed position. After the k information bits are all clocked into the encoder, the positions of the two switches are reversed. At this time, the contents of the shift register are simply the – parity check bits, which correspond to the coefficients of the remainder polynomial . These – bits are clocked out one at a time and sent to the modulator.
Example 14.1: The shift register for encoding the (7,4) cyclic code with generator polynomial is illustrated in Fig.(14.1). Suppose the input message bits are d=(0110). The contents of the shift register are as follows:
Hence, the three parity check bits are 100, which correspond to the code bits , , and . Hence, the corresponding codeword is
Fig.(14.1).

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