seeing that you can't correct these errors.) So, the probability that you fail to correct an error, P, is simply the probability that more than one bit error occurs. Knowing this, you can easily open statistics books and figure this out to be (or you can take my word that this Pis) p=P(j.n) (6.18) 6.3.3 The Performance of Linear Block Codes In linear block codes you take a chunk of k bits and map it into a chunk of n bits. In the words of all famous engineering textbooks, let me just say "it can be shown that" Chapter Six the number of bits in error, in each chunk of incoming n bits, that can always be corrected is t, where t is the number you calculate from the equation (6.19) Lx_ refers to the integer you get by rounding X down to the nearest integer, and dmin is the number you get by looking at all the channel coder outputs and counting the number of 1s in the channel coder output with the fewest 1s (excluding the output (000000)). For example, consider the channel coder described by Figure 6.10. In this channel coder, the output with the fewest number of 1s (not counting (000000)) is the output (000111). This has three 1s in it, SO dmin = 3. Then, com- puting t, we get 3-1 = This tells us that, for the channel coder in Figure 6.10, you can build a channel decoder and always correct all 1-bit errors. You might be able to correct a few 2-bit errors as well, but t = 1 tells us that the channel decoder cannot always correct all 2-bit errors. The probability that the linear block coder will not correct an error, P, is well approximated by the probability that more than t errors occur in the incoming chunk of n bits. So, using a wee bit of statistics, that means mathematically that p=[M(in) (6.20) Example 6.7 If, in the modulator-channel-demodulator part of the communication system, the probability that a bit is in error is 1%, what is: 1. the Pm for a rate 3/4 parity check bit decoder? 2. the P for a rate 5/10 linear block decoder (assume t = 2)? Solution: For the parity check bit decoder, turning to equation (6.1