where

E[ x(nk)x*(nl)x(nm)x*(nr) ]={ E[ | x(n)4 | ]k=l=m=rE[ | x(n)2 | ]k=l=m=r,k=r=l=m| E[ x2(n) ] |2k=m=l=r0others

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

C4y(L,τ,τ1)=h0hτ1*hτhL*C4x(0,0,0)(5.48)

C4y(L,0,τ1)=h0hτ1*h0hL*C4x(0,0,0)(5.49)

C4y(L,0,τ1)hτ=C4y(L,τ,τ1)h0for0τ1L(5.50)

So,

C4y*(L,0,τ1)C4y(L,0,τ1)hτ=C4yτ(L,0,τ1)C4y(L,τ,τ1)h0(0τ1L)(5.51)

τ1=0LC4y*(L,0,τ1)C4y(L,0,τ1)hτ=τ1=0LC4y*(L,0,τ1)C4y(L,τ,τ1)h0(5.52)

hττ1=0LC4y*(L,0,τ1)C4y(L,0,τ1)=h0τ1=0LC4yL(L,o,τ1)C4y(L,τ,τ1)(5.53)

hτ=h0τ1=0LC4y*(L,0,τ1)C4y(L,τ,τ1)τ1=0LC4y*(L,0,τ1)C4y(L,0,τ1)=h0τ1=0LC4y*(L,0,τ1)C4y(L,

Get Blind Equalization in Neural Networks now with O’Reilly online learning.

O’Reilly members experience live online training, plus books, videos, and digital content from 200+ publishers.