c=111,010,100,100,010,000,100) 若采用P=10 的删余矩阵，则刷余后的码率为1/2的ubo码的输出码字为 01 c=(11,00,10,10,01,00,10) c +⊕ 十 001 Branch label:/c (b)分量码的trellis图 图4.7(7,5)Turbo码编码器 4.2.3 A Heuristic Discussion of the Basic Rationale Behind Turbo codes It is well known that in a coded system,the code performance is dominated by low weight code words.With the use of RSC encoders and interleavers,the probability that both encoders produce low weight outputs is very low.This is because that the pseudorandom interleaver permutes the input bits,the two input sequences u and i are almost always 4.9 4-9 c  (111,010,100,100,010,000,100) 若采用 1 0 0 1        P 的删余矩阵，则删余后的码率为1/2的turbo码的输出码字为 c  (11,00,10,10,01,00,10) u 1p c 2 p c s u c c （a） (b) 分量码的 trellis 图 图 4.7 (7, 5) Turbo 码编码器 4.2.3 A Heuristic Discussion of the Basic Rationale Behind Turbo codes It is well known that in a coded system, the code performance is dominated by low weight code words. With the use of RSC encoders and interleavers, the probability that both encoders produce low weight outputs is very low. This is because that the pseudorandom interleaver permutes the input bits, the two input sequences u and u are almost always