11 23
![\[ \begin{array}{l} \int {\cos ^3 xdx = } \int {\cos ^2 xd\sin x} \\ = \cos ^2 x\sin x - \int {\sin xd(\cos ^2 x)} \\ = \cos ^2 x\sin x - 2\int {\sin x\cos x( - \sin x)dx} \\ = \cos ^2 x\sin x + 2\int {\sin ^2 x\cos xdx} \\ = \cos ^2 x\sin x + 2\int {\sin ^2 xd\sin x} \\ = (1 - sin^2 x)\sin x + 2\int {U^2 dU} \\ = \sin x - \sin ^3 x + \frac{2}{3}\sin ^3 x + c \\ = \sin x - \frac{1}{3}\sin ^3 x + c \\ \end{array} \]](http://lucideye.is-programmer.com/user_files/lucideye/epics/12dc615a6290517d3c8fd121d8a76a5956f2d072.png "\[ \begin{array}{l} \int {\cos ^3 xdx = } \int {\cos ^2 xd\sin x} \\ = \cos ^2 x\sin x - \int {\sin xd(\cos ^2 x)} \\ = \cos ^2 x\sin x - 2\int {\sin x\cos x( - \sin x)dx} \\ = \cos ^2 x\sin x + 2\int {\sin ^2 x\cos xdx} \\ = \cos ^2 x\sin x + 2\int {\sin ^2 xd\sin x} \\ = (1 - sin^2 x)\sin x + 2\int {U^2 dU} \\ = \sin x - \sin ^3 x + \frac{2}{3}\sin ^3 x + c \\ = \sin x - \frac{1}{3}\sin ^3 x + c \\ \end{array} \]")
![$$\[ \left[ {\begin{array}{*{20}c} 1 & 1 & 1 \\ 3 & 1 & { - 1} \\ 5 & { - 2} & 3 \\ \end{array}} \right]\left[ {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array}} \right] = \left[ {\begin{array}{*{20}c} 6 \\ 2 \\ {10} \\ \end{array}} \right] \] $$](http://lucideye.is-programmer.com/user_files/lucideye/epics/b69068a85d924b159368db4c1316c85118c4fd94.png "$$\[ \left[ {\begin{array}{*{20}c} 1 & 1 & 1 \\ 3 & 1 & { - 1} \\ 5 & { - 2} & 3 \\ \end{array}} \right]\left[ {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array}} \right] = \left[ {\begin{array}{*{20}c} 6 \\ 2 \\ {10} \\ \end{array}} \right] \] $$")
![\[ \left[ {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array}} \right] = \left[ {\begin{array}{*{20}c} 1 & 1 & 1 \\ 3 & 1 & { - 1} \\ 5 & { - 2} & 3 \\ \end{array}} \right]^{ - 1} \left[ {\begin{array}{*{20}c} 6 \\ 2 \\ {10} \\ \end{array}} \right] \]](http://lucideye.is-programmer.com/user_files/lucideye/epics/5c8a994303dfa4b09e743c17b10346c7e39b5e49.png "\[ \left[ {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array}} \right] = \left[ {\begin{array}{*{20}c} 1 & 1 & 1 \\ 3 & 1 & { - 1} \\ 5 & { - 2} & 3 \\ \end{array}} \right]^{ - 1} \left[ {\begin{array}{*{20}c} 6 \\ 2 \\ {10} \\ \end{array}} \right] \]")
![\[ \left[ {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array}} \right] = \left[ {\begin{array}{*{20}c} { - 0.0417} & {0.2083} & {0.0833} \\ {0.5833} & {0.0833} & { - 0.1667} \\ {0.4583} & { - 0.2917} & {0.0833} \\ \end{array}} \right]\left[ {\begin{array}{*{20}c} 6 \\ 2 \\ {10} \\ \end{array}} \right] \]](http://lucideye.is-programmer.com/user_files/lucideye/epics/233271eccaea525e709c5ef2fb034d43960be712.png "\[ \left[ {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array}} \right] = \left[ {\begin{array}{*{20}c} { - 0.0417} & {0.2083} & {0.0833} \\ {0.5833} & {0.0833} & { - 0.1667} \\ {0.4583} & { - 0.2917} & {0.0833} \\ \end{array}} \right]\left[ {\begin{array}{*{20}c} 6 \\ 2 \\ {10} \\ \end{array}} \right] \]")
![\[ \left[ {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array}} \right] = \left[ {\begin{array}{*{20}c} 1 \\ 2 \\ 3 \\ \end{array}} \right] \]](http://lucideye.is-programmer.com/user_files/lucideye/epics/2ee3cfc2ff96ec70ff50b0308fb3586c583db743.png "\[ \left[ {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array}} \right] = \left[ {\begin{array}{*{20}c} 1 \\ 2 \\ 3 \\ \end{array}} \right] \]")
11 4
由于G=SH，We do not need to factorize the overall channel matrix. 这里的overall channel matrix指的是G,而接收端滤波矩阵是F=SH,这里是否矛盾?
反馈矩阵B, 经过反馈矩阵后是B-1a,再经过信道后是GB-1a=SHB-1a;然后再经过接收方的滤波器F,得到的是:
SHSHB-1a
=HB-1a (因为Ｓ是酉阵特性)
=HH-1diag(T)a
=diag(T)a
其中diag(T)仅仅是一个功率因子.
现在,假如H是对角阵,那么T也是一个对角阵.这种分解还会存在吗？
当MIMO 2x2的情况下MT=2,MR=2，N=64，那么信道矩阵
H=![$\left[ {\begin{array}{*{20}c} {{\bf{{\rm H}}}_{1,1} } & {{\bf{{\rm H}}}_{2,1} } \\ {{\bf{{\rm H}}}_{1,2} } & {{\bf{{\rm H}}}_{2,2} } \\ \end{array}} \right]$](http://lucideye.is-programmer.com/user_files/lucideye/epics/9e3080fc28a975fb1cbe3968c3116c868b8b388b.png "$\left[ {\begin{array}{*{20}c} {{\bf{{\rm H}}}_{1,1} } & {{\bf{{\rm H}}}_{2,1} } \\ {{\bf{{\rm H}}}_{1,2} } & {{\bf{{\rm H}}}_{2,2} } \\ \end{array}} \right]$")
同时128x128的ICI系数矩阵等于S=diag[S1,S2]，因为其中每一个子块都是酉阵，所以S也是酉阵。
对于G=SH，仅仅分解H,那么H=FHT等价于T=FH。
接收方滤波矩阵D=FSH，反馈矩阵B=diag-1(T)T，那么发送信号为B-1a，经过信道后是SHB-1a
经过接收方滤波之后是DSHB-1a= FSHSHB-1a= FSHSHB-1a= TB-1a
可见，MIMO情况下和SISO情况下，反馈矩阵B的求法是一样的，接收端滤波矩阵稍有区别(这是因为cholesky分解和QR分解造成的不同方案)。，但反馈矩阵一定是一个三角阵。
同时，这次分解中，H不是一个对角阵了。
![\begin{array}{l} \left[ {\begin{array}{*{20}c} 1 & 2 & { - 1} \\ 3 & 4 & { - 2} \\ 5 & { - 4} & 1 \\ \end{array}} \right]\begin{array}{*{20}c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \\ \to \left[ {\begin{array}{*{20}c} 1 & 2 & { - 1} \\ 0 & { - 2} & 1 \\ 0 & { - 14} & 6 \\ \end{array}} \right]\begin{array}{*{20}c} 1 & 0 & 0 \\ { - 3} & 1 & 0 \\ { - 5} & 0 & 1 \\ \end{array} \to \left[ {\begin{array}{*{20}c} 1 & 2 & { - 1} \\ 0 & { - 2} & 1 \\ 0 & 0 & { - 1} \\ \end{array}} \right]\begin{array}{*{20}c} 1 & 0 & 0 \\ { - 3} & 1 & 0 \\ {16} & { - 7} & 1 \\ \end{array} \\ \to \left[ {\begin{array}{*{20}c} 1 & 2 & 0 \\ 0 & { - 2} & 0 \\ 0 & 0 & { - 1} \\ \end{array}} \right]\begin{array}{*{20}c} { - 15} & 7 & { - 1} \\ {13} & { - 6} & 1 \\ {16} & { - 7} & 1 \\ \end{array} \to \left[ {\begin{array}{*{20}c} 1 & 0 & 0 \\ 0 & { - 2} & 0 \\ 0 & 0 & { - 1} \\ \end{array}} \right]\begin{array}{*{20}c} { - 2} & 1 & 0 \\ {13} & { - 6} & 1 \\ {16} & { - 7} & 1 \\ \end{array} \\ \to \left[ {\begin{array}{*{20}c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}} \right]\begin{array}{*{20}c} { - 2} & 1 & 0 \\ { - 13/2} & 3 & { - 1/2} \\ { - 16} & 7 & { - 1} \\ \end{array} \\ \end{array}](http://lucideye.is-programmer.com/user_files/lucideye/epics/c39f4363a291c1f5494331e55d95d41106ed6332.png "\begin{array}{l} \left[ {\begin{array}{*{20}c} 1 & 2 & { - 1} \\ 3 & 4 & { - 2} \\ 5 & { - 4} & 1 \\ \end{array}} \right]\begin{array}{*{20}c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \\ \to \left[ {\begin{array}{*{20}c} 1 & 2 & { - 1} \\ 0 & { - 2} & 1 \\ 0 & { - 14} & 6 \\ \end{array}} \right]\begin{array}{*{20}c} 1 & 0 & 0 \\ { - 3} & 1 & 0 \\ { - 5} & 0 & 1 \\ \end{array} \to \left[ {\begin{array}{*{20}c} 1 & 2 & { - 1} \\ 0 & { - 2} & 1 \\ 0 & 0 & { - 1} \\ \end{array}} \right]\begin{array}{*{20}c} 1 & 0 & 0 \\ { - 3} & 1 & 0 \\ {16} & { - 7} & 1 \\ \end{array} \\ \to \left[ {\begin{array}{*{20}c} 1 & 2 & 0 \\ 0 & { - 2} & 0 \\ 0 & 0 & { - 1} \\ \end{array}} \right]\begin{array}{*{20}c} { - 15} & 7 & { - 1} \\ {13} & { - 6} & 1 \\ {16} & { - 7} & 1 \\ \end{array} \to \left[ {\begin{array}{*{20}c} 1 & 0 & 0 \\ 0 & { - 2} & 0 \\ 0 & 0 & { - 1} \\ \end{array}} \right]\begin{array}{*{20}c} { - 2} & 1 & 0 \\ {13} & { - 6} & 1 \\ {16} & { - 7} & 1 \\ \end{array} \\ \to \left[ {\begin{array}{*{20}c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}} \right]\begin{array}{*{20}c} { - 2} & 1 & 0 \\ { - 13/2} & 3 & { - 1/2} \\ { - 16} & 7 & { - 1} \\ \end{array} \\ \end{array}")
有什么区别吗？
，发现