6.30 Noisy Channel Coding Theorem

225

Received Codeword Probability Decoded Bit

000 (1 − pe)3 0 001 pe(1 − pe)2 0 010 pe(1 − pe)2 0 011 pe 3

2 (1 − pe) 1 100 pe(1 − pe)2 0 101 pe 2 (1 − pe) 1 110 pe 2 (1 − pe) 1 111 pe 1 Table 6.1: In this example, the transmitter encodes 0 as 000. The channel creates an error (changing a 0 into a 1) with probability pe. The first column lists all possible received data words and the second the probability of each data word being received. The last column shows the results of the majority-vote decoder. When the decoder produces 0, it successfully corrected the errors introduced by the channel (if there were any; the top row corresponds to the case in which no errors occurred). The error probability of the decoders is the sum of the probabilities when the decoder produces 1.

Exercise 6.26

(Solution on p. 257.)

Demonstrate mathematically that this claim is indeed true. Is 3pe 2 (1 − pe) + pe 3 ≤ pe?

6.26 Block Channel Coding34

Because of the higher datarate imposed by the channel coder, the probability of bit error occurring in the digital channel increases relative to the value obtained when no channel coding is used. The bit interval duration must be reduced by K N in comparison to the no-channel-coding situation, which means the energy per bit Eb goes down by the same amount. The bit interval must decrease by a factor of three if the transmitter is to keep up with the data stream, as illustrated in Figure 6.21.

Point of Interest: It is unlikely that the transmitter’s power could be increased to compensate.

Such is the sometimes-unfriendly nature of the real world. Because of this reduction, the error probability pe of the digital channel goes up. The question thus becomes does channel coding really help: Is the effective error probability lower with channel coding even though the error probability for each transmitted bit is larger? The answer is no: Using a repetition code for channel coding cannot ultimately reduce the probability that a data bit is received in error. The ultimate reason is the repetition code’s inefficiency: transmitting one data bit for every three transmitted is too inefficient for the amount of error correction provided.

Exercise 6.27 (Solution on p. 257.)

Using MATLAB, calculate the probability a bit is received incorrectly with a three-fold repetition code. Show that when the energy per bit Eb is reduced by 1/3 that this probability is larger than the no-coding probability of error. The repetition code (p. 224) represents a special case of what is known as block channel coding. For every K bits that enter the block channel coder, it inserts an additional N − K error-correction bits to produce a block of N bits for transmission. We use the notation (N , K) to represent a given block code’s parameters. In the three-fold repetition code (p. 224), K = 1 and N = 3. A block code’s coding efficiency E equals the ratio K/N , and quantifies the overhead introduced by channel coding. The rate at which bits must be

34 This content is available online at http://cnx.org/content/m0094/2.15/.