2 x 2 STBC 空时分组码 解码 与 均衡

• 本文来源于同门（Jw Lou）总结
• 主要参考文献：MIMO系统中空时译码与频域均衡研究与实现

主要流程介绍（一组数据块示例【即两个OFDM数据块】）

1、四个SISO分析
$$Y_{11}=[X_1{H}_{11},-X_2^{\ast }{H}_{11}]+Z_{11}$$
$$Y_{12}=[X_1{H}_{12},-X_2^{\ast }{H}_{12}]+Z_{12}$$
$$Y_{21}=[X_2{H}_{21},X_1^{\ast }{H}_{21}]+Z_{21}$$
$$Y_{22}=[X_2{H}_{22},X_1^{\ast }{H}_{22}]+Z_{22}$$

1、$Y_{ij}$表示发射天线$i$至接收天线$j$的信号（两个相关时隙）；
2、$X_i$为发送信号（频域）【未STBC编码之前】；
3、$H_{ij}$表示发射天线$i$至接收天线$j$之间的多径信道频域矩阵；
4、$Z_{ij}$表示发射天线$i$至接收天线$j$之间的高斯白噪声，功率为${\sigma }_{ij}^2$;
5、$(\cdot {)}^{\ast }$表示共轭。

2、MISO接收信号
$$R_1=Y_{11}+Y_{21}=[X_1{H}_{11}+X_2{H}_{21},-X_2^{\ast }{H}_{11}+X_1^{\ast }{H}_{21}]+Z_{11}+Z_{21}$$
$$R_2=Y_{12}+Y_{22}=[X_1{H}_{12}+X_2{H}_{22},-X_2^{\ast }{H}_{12}+X_1^{\ast }{H}_{22}]+Z_{12}+Z_{22}$$

$R_i$表示第$i$个接收天线接收到的信号

3、MISO分时隙表示
$$R_{11}=X_1{H}_{11}+X_2{H}_{21}+Z_{11}+Z_{21}$$
$$R_{12}=-X_2^{\ast }{H}_{11}+X_1^{\ast }{H}_{21}+Z_{11}+Z_{21}$$
$$R_{21}=X_1{H}_{12}+X_2{H}_{22}+Z_{12}+Z_{22}$$
$$R_{22}=-X_2^{\ast }{H}_{12}+X_1^{\ast }{H}_{22}+Z_{12}+Z_{22}$$

$R_{ij}$表示第$i$个天线的第$j$个时隙的接收信号。

4、MIMO-STBC解码

$$S_1=R_{11}{H}_{11}^{\ast }+R_{12}^{\ast }{H}_{21}+R_{21}{H}_{12}^{\ast }+R_{22}^{\ast }{H}_{22}$$

$$S_2=R_{11}{H}_{21}^{\ast }-R_{12}^{\ast }{H}_{11}+R_{21}{H}_{22}^{\ast }-R_{22}^{\ast }{H}_{12}$$

$S_i$表示解调得到的发射信号$X_i$

• 进一步可以表示为：
  $$S_1=X_1\sum _{i,j}|H_{ij}{|}^2+(Z_{11}+Z_{21})H_{11}+(Z_{11}^{\ast }+Z_{21}^{\ast })H_{21}+(Z_{12}+Z_{22})H_{12}+(Z_{12}^{\ast }+Z_{22}^{\ast })H_{22}$$
  $$S_2=X_2\sum _{i,j}|H_{ij}{|}^2-(Z_{11}^{\ast }+Z_{21}^{\ast })H_{11}+(Z_{11}+Z_{21})H_{21}^{\ast }-(Z_{12}^{\ast }+Z_{22}^{\ast })H_{12}+(Z_{12}+Z_{22})H_{22}^{\ast }$$

• 进一步化简：

$$S_1=K{X}_1+N_1$$
$$S_2=K{X}_2+N_2$$

1、$K=\sum _{i,j}|H_{ij}{|}^2$;
2、$N_1=(Z_{11}+Z_{21})H_{11}+(Z_{11}^{\ast }+Z_{21}^{\ast })H_{21}+(Z_{12}+Z_{22})H_{12}+(Z_{12}^{\ast }+Z_{22}^{\ast })H_{22}$
3、$N_2=-(Z_{11}^{\ast }+Z_{21}^{\ast })H_{11}+(Z_{11}+Z_{21})H_{21}^{\ast }-(Z_{12}^{\ast }+Z_{22}^{\ast })H_{12}+(Z_{12}+Z_{22})H_{22}^{\ast }$
4、$N_1$与$N_2$为高斯白噪声，功率为${\sigma }_n^2={\sigma }_{n1}^2={\sigma }_{n2}^2=({\sigma }_{11}^2+{\sigma }_{21}^2)(\overline{|H_{11}{|}^2}+\overline{|H_{21}{|}^2})+({\sigma }_{12}^2+{\sigma }_{22}^2)(\overline{|H_{12}{|}^2}+\overline{|H_{22}{|}^2})$

$\overline{(\cdot )}$表示均值

5、MMSE均衡
$$W=\frac{K^{\ast }}{K^2+{\sigma }_n^2}=\frac{(\sum _{i,j}|H_{ij}{|}^2{)}^{\ast }}{(\sum _{i,j}|H_{ij}{|}^2{)}^2+({\sigma }_{11}^2+{\sigma }_{21}^2)(\overline{|H_{11}{|}^2}+\overline{|H_{21}{|}^2})+({\sigma }_{12}^2+{\sigma }_{22}^2)(\overline{|H_{12}{|}^2}+\overline{|H_{22}{|}^2})}$$

$W$即为均衡系数

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