Polar Code（3）编码实例

前言

在《Polar Code（2）编码原理》中详细阐述了Polar Code编码原理，为更好地理解编码过程，本文将给出一个编码实例。

设码长为\(N=8\)，消息比特数\(K=4\)。列出所有使用到的公式：

\[c_{1}^{N}=u_{1}^{N}{G_{N}}\]

\[{G_{N}}={B_{N}}{F^{\otimes n}}\]

\[{F^{\otimes n}}=F\otimes {F^{\otimes \left( n-1 \right)}}\]

\[F=\left[ \begin{matrix} 1 & 0 \\ 1 & 1 \\ \end{matrix} \right]\]

\[{B_{N}}={R_{N}}\left( {I_{2}}\otimes {B_{N/{2}\;}} \right)\]

\[{B_{2}}={I_{2}}\]

极化信道的可靠性估计

假设对于各二进制删除信道（BEC），已计算出各个分裂信道的巴氏参数\(Z\left( W_{N}^{\left( i \right)} \right)\)。

比特混合

假设通过第一步得到巴氏参数最小的4个子信道序号为“4，6，7，8”，则消息比特序号集合记为

\[A=\left\{ 4,6,7,8 \right\}\]

则固定比特序号集合为

\[{A^{c}}=\left\{ 1,2,3,5 \right\}\]

设信息比特集合为\(\left( {i_{1}},{i_{2}},{i_{3}},{i_{4}} \right)=\left( 1,1,1,1 \right)\)，固定比特集合为\(\left( 0,0,0,0 \right)\)，则最终得到

\[u_{1}^{8}=\left[ 0,0,0,{i_{1}},0,{i_{2}},{i_{3}},{i_{4}} \right]=\left[ 0,0,0,1,0,1,1,1 \right]\]

构造生成矩阵

求排序矩阵BN

递归式：

\[{B_{8}}={R_{8}}\left( {I_{2}}\otimes {B_{4}} \right)\]

\[{B_{4}}={R_{4}}\left( {I_{2}}\otimes {B_{2}} \right)\]

计算：

\[{B_{2}}=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\]

\[{I_{2}}\otimes {B_{2}}=\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right]\]

\({R_{4}}\)由\({I_{4}}\)变换得来，先排\({I_{4}}\)的奇数列，再排\({I_{4}}\)的偶数列：

\[{R_{4}}=\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right]\Leftarrow {I_{4}}=\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right]\]

\[{ {B}_{4}}={ {R}_{4}}\left( { {I}_{2}}\otimes { {B}_{2}} \right)=\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right]\cdot \left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right]\]

\[{I_{2}}\otimes {B_{4}}=\left[ \begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{matrix} \right]\]

\({R_{8}}\)由\({I_{8}}\)变换得来，先排\({I_{8}}\)的奇数列，再排\({I_{8}}\)的偶数列：

\[{R_{8}}=\left[ \begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{matrix} \right]\Leftarrow {I_{8}}=\left[ \begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{matrix} \right]\]

\[\begin{align} & { {B}_{8}}={ {R}_{8}}\left( { {I}_{2}}\otimes { {B}_{4}} \right)=\nonumber \left[ \begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{matrix} \right]\cdot \left[ \begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{matrix} \right] \\ & \begin{matrix} {} & {} & {} & \begin{matrix} {} & {} & = \nonumber \left[ \begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{matrix} \right] \\ \end{matrix} \\ \end{matrix} \\ \end{align}\]

求F的n次克罗内克积

递归式：

\[{F^{\otimes 3}}=F\otimes {F^{\otimes 2}}\]

\[{F^{\otimes 2}}=F\otimes {F^{\otimes 1}}\]

计算：

\[{F^{\otimes 1}}\text{=}F=\left[ \begin{matrix} 1 & 0 \\ 1 & 1 \\ \end{matrix} \right]\]

\[{F^{\otimes 2}}=F\otimes {F^{\otimes 1}}=\left[ \begin{matrix} 1 & 0 \\ 1 & 1 \\ \end{matrix} \right]\otimes {F^{\otimes 1}}=\left[ \begin{matrix} {F^{\otimes 1}} & 0 \\ {F^{\otimes 1}} & {F^{\otimes 1}} \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 1 \\ \end{matrix} \right]\]

\[{F^{\otimes 3}}=F\otimes {F^{\otimes 2}}=\left[ \begin{matrix} 1 & 0 \\ 1 & 1 \\ \end{matrix} \right]\otimes {F^{\otimes 2}}=\left[ \begin{matrix} {F^{\otimes 2}} & 0 \\ {F^{\otimes 2}} & {F^{\otimes 2}} \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \end{matrix} \right]\]

求生成矩阵GN

\[\begin{align} & { {G}_{8}}={ {B}_{8}}{ {F}^{\otimes 3}}=\nonumber \left[ \begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{matrix} \right]\cdot \left[ \begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \end{matrix} \right] \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{=} \nonumber \left[ \begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \end{matrix} \right] \\ \end{align}\]

生成Polar Code

\[\nonumber \begin{align} & c_{1}^{8}=u_{1}^{8}{G_{8}}=\left[ \begin{matrix} 0 & 0 & 0 & 1 & 0 & 1 & 1 & 1 \\ \end{matrix} \right]\cdot \left[ \begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \end{matrix} \right] \\ & \ \ \ \ =\left[ \nonumber \begin{matrix} 0 & 1 & 1 & 0 & 1 & 0 & 0 & 1 \\ \end{matrix} \right] \\ \end{align}\]