Blind Equalization in Neural Networks

where

$E\left[x\left(n-k\right)x^{*}\left(n-l\right)x\left(n-m\right)x^{*}\left(n-r\right)\right]=\{\begin{array}{cc}E\left[\left|x{\left(n\right)}^4\right|\right]& k=l=m=r\\ E\left[\left|x{\left(n\right)}^2\right|\right]& k=l\cancel{=}m=r,k=r\cancel{=}l=m\\ {\left|E\left[x^2\left(n\right)\right]\right|}^2& k=m\cancel{=}l=r\\ 0& \text{others}\end{array}$

If k₁ = 0, L = k₂, τ = k₃, τ₁ = k₄, there is

$C_{4y}\left(L,\tau ,{\tau }_1\right)=h_0{h}_{{\tau }_1}^{*}{h}_{\tau }{h}_L^{*}{C}_{4x}\left(0,0,0\right)\left(5.48\right)$

$C_{4y}\left(L,0,{\tau }_1\right)=h_0{h}_{{\tau }_1}^{*}{h}_0{h}_L^{*}{C}_{4x}\left(0,0,0\right)\left(5.49\right)$

$C_{4y}\left(L,0,{\tau }_1\right)h_{\tau }=C_{4y}\left(L,\tau ,{\tau }_1\right)h_0\text{for}0\le {\tau }_1\le L\left(5.50\right)$

So,

$C_{4y}^{*}\left(L,0,{\tau }_1\right)C_{4y}\left(L,0,{\tau }_1\right)h_{\tau }=C_{4y}^{\tau }\left(L,0,{\tau }_1\right)C_{4y}\left(L,\tau ,{\tau }_1\right)h_0\left(0\le {\tau }_1\le L\right)\left(5.51\right)$

${\displaystyle \sum _{{\tau }_1=0}^L{C}_{4y}^{*}\left(L,0,{\tau }_1\right)}{C}_{4y}\left(L,0,{\tau }_1\right)h_{\tau }={\displaystyle \sum _{{\tau }_1=0}^L{C}_{4y}^{*}}\left(L,0,{\tau }_1\right)C_{4y}\left(L,\tau ,{\tau }_1\right)h_0\left(5.52\right)$

$h_{\tau }{\displaystyle \sum _{{\tau }_1=0}^L{C}_{4y}^{*}\left(L,0,{\tau }_1\right)C_{4y}\left(L,0,{\tau }_1\right)=h_0{\displaystyle \sum _{{\tau }_1=0}^L{C}_{4y}^L\left(L,o,{\tau }_1\right)C_{4y}\left(L,\tau ,{\tau }_1\right)}\left(5.53\right)}$

$\begin{array}{l}{h}_{\tau }=\frac{h_0{\displaystyle {\sum }_{{\tau }_1=0}^L{C}_{4y}^{*}\left(L,0,{\tau }_1\right)C_{4y}\left(L,\tau ,{\tau }_1\right)}}{{\displaystyle {\sum }_{{\tau }_1=0}^L{C}_{4y}^{*}\left(L,0,{\tau }_1\right)C_{4y}\left(L,0,{\tau }_1\right)}}\\ =\frac{h_0{\displaystyle {\sum }_{{\tau }_1=0}^L{C}_{4y}^{*}\left(L,0,{\tau }_1\right)C_{4y}(L,}}{}\end{array}$