mohatta2015.pdf

MORE DETAILS 77
To obtain the unity-gain max SINR solution directly, we want the coefficient in
front of s 2 to be 1. This is achieved by setting w 2 = 1/c, so that
The impairment power is given by
«2 = s 2 + {d/c + wic)si + (ωιηι + (l/c)ri2). (4-31)
Ñ 2 = (c 2 + a 2 )w\ + 2dw! + {d 2 + a 2 )/c 2 . (4.32)
Setting the derivative w.r.t. w\ to zero and solving for wi gives
Wl
= 72-^2- ( 4·33 )
For the MMSE solution, we start with (4.30). Substituting models for τ\ and
r 2 , the error e 2 = z 2 — s 2 can be modeled as
e 2 = {w 2 c - l)s 2 + {w 2 d + wic)si + (wini + w 2 n 2 ). (4-34)
The MSE is the power in e 2 , which is
E 2 = (w 2 c - l) 2 + (w 2 d + wie) 2 + (w 2 + w 2 2)a 2 . (4.35)
To find the MMSE weights, we take the derivative of E 2 w.r.t. w\ and set it to
zero. We do the same w.r.t. w 2 . This gives two equations in two unknowns, which
can be written in matrix form as
where
o _ Γ c 2 + d 2 + σ 2 cd
"■-[ cd c 2 +d 2 + a 2
The solution to this set of equations is
Rw = h, (4.36)
[Z\], and h=[0]. (4.37)
= -c 2 d cjc 2 + d 2 +a 2 )
Wl
(c 2 + d 2 + σ 2 ) 2 - c 2 d 2 W2
' {c 2 +d 2 +σ 2 ) 2 -c 2 d 2, [ '
and the resulting MSE is given by
MSE = w^Rw - 2wr h + x = ! - ^ I f 2) t ^ ,
(4.39)
where superscript "'Ρ' denotes transpose (turns a column vector into a row vector).
The minimum ISI solution can be obtained by setting σ 2 to zero.
The elements in R have a special interpretation. The diagonal elements are the
average received sample power, i.e., the average of r\ or r\. Specifically, the average
received signal power is the desired signal power (c 2 + d 2 ) and the average noise
power (σ 2 ). The off-diagonal elements are the average of the product of adjacent
received values, i.e., the average of r\r 2 . Specifically, using the models,
E{nr 2 } = E{(cs 2 +dsi+n 2 )(csi+dso + ni)} (4.40)
= E{c s 2 s\ + cds 2 SQ + cs 2 ni + dcs x
+ d 2 siSo + ds\ni + n 2 cs\ + n 2 ds n + n 2 n\). (4.41)