In the process of coding.the information sequence u is divided into data frames of length k.These subseq ences of length k are applied to the k-input terminals of the encoder,producing coded sequence oflength An example: Let n2.=1 and m=2.Consider a rate-1/2(2,1,2)convolutional code which is specified by the following two generator sequences: g"=(101),g=(11) Note:g",g2)可看作编码器的两个冲激响应，由u=6=(I00)得到。冲激响应至多 持续m+1个时间单位，且可写为： g"=(g,g",g）g2=（882,g2,g2） -Let u=()be the input message sequence.Then the two output sequences are c)=u*g) 编码方程（与冲激响应的卷积运算） c2 =u+g) -At the /th time unit,the input is a single bitThe corresponding output is a block of two bits,(cc),which is given by P-28”4g”+g”+中w c"=4+42 →G=4++"型 memory The output codeword is given by c=). Foru=1011100.),c=(11,0L,00,10,01,10,11,.) 5 „ In the process of coding, the information sequence u is divided into data frames of length k. These subsequences of length k are applied to the k-input terminals of the encoder, producing coded sequence of length n. „ An example: Let n=2, k=1 and m=2. Consider a rate-1/2 (2,1,2) convolutional code which is specified by the following two generator sequences: (1) (2) g g = = (101), (111) u (1) c c (2) c Note: (1) (2) g g , 可看作编码器的两个冲激响应，由u = δ = (100.)得到。冲激响应至多 持续m +1个时间单位，且可写为： ( ) ( ) (1) (1) (1) (1) (2) (2) (2) (2) 01 0 1 , , , , , m m g g = = gg g gg g . . - Let 0 1 u = (,) u u " be the input message sequence. Then the two output sequences are (1) (1) (2) (2) * * = ⎫ ⎬ = ⎭ c ug c ug 编码方程 (与冲激响应的卷积运算) - At the lth time unit, the input is a single bit l u . The corresponding output is a block of two bits, (1) (2) (, ) l l c c , which is given by () () () () () 0 11 1 m j jj j j l li i l l lm m i c u g ug u g u g − −− = = = + ++ ∑ " (1) 2 2 1 2 ll l l ll l memory cu u c uu u − − − ⎧ = + ⎪ ⇒ ⎨ =+ + ⎪ ⎩   - The output codeword is given by ( ) (1) (2) (1) (2) 00 11 c = cc cc , ," . For (1011100 ), (11,01,00,10,01,10,11, ) u c = = "