ICT12 3. Channel coding

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$Some fractional functional inequalities - UPC$

23.1. IntroductionSourceHelloworld!Sourceencoderu···1001···FeedbackchannelfChannelencoderx···1001011···ModulatorBinarydigitalChannelcodingis requiredto protectinformationNoiseReceiverHelloworld!Sourcedecoder···1001···x′Channeldecoderg···1000011···ychannelDemodulatorNoisy communications channel. I(S, R) = H(S) − H(S|R)againstthe effectsof noisein thechannel.
3For definiteness, let us assume that the source encoding alphabet andthe channel alphabet are both B = {0,1}. Later (Section 3.3) we will considera more general setup.Encoding. The channel encoder breaks the bit stream coming from thesource encoder into blocks u of k bits. Then, for each such informationblock, it produces a block x = f(u) of n > k bits, which is sent throughthe channel (f: B k → B n is assumed 1-to-1). The difference r = n − k iscalled the redundancy of the channel and the fact that x has r more bitsthan u is what will be used as protection against the channel errors.We will use C to denote the set of codewords and we will say that it is acode of type [n, k]. The rate of the code is defined as k/n.Since there are 2 k possible information blocks u, we also have 2 k codewordsx of length n (so |C| = 2 k ). Thus the proportion of codewords (orcode density) with respect to binary words of length n is 2 k /2 n =1/2 n−k = 1/2 r .
Page 1: ICT123. Channel coding25.9.2012 SXD
Page 6 and 7: 65.2. Two decoding criteriaFirst so
Page 8 and 9: 83.3. General block codesThe Hammin
Page 10: 10CodersA coder of type [n, k] is a
Page 14 and 15: 143.4. Minimum distance decodingNot
Page 16 and 17: 16D 6D 5x 6D 1D 7x 1x 7D 2x 2D 3The
Page 18: 18Remark. From the computational po

ICT12 3. Channel coding

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