Page 530 - Fiber Optic Communications Fund

Digital Signal Processing 511

The adaptive equalizer has 2K + 1 adjustable complex coefficients. The coefficients W[k] can be adjusted so
that the mean square error is minimum,
J
= 0 (11.70)
W[k]
and
J
= 0. (11.71)
∗
W [k]
Using Eqs. (11.67) and (11.69) in Eq. (11.70), we find
J ∗ ∗
= < −x [n]y[n − k]+ y[n − k]̂x [n] >
W[k]
∗
=− < y[n − k]e [n] >= 0. (11.72)
∗
∗
Note that W[k] and W [k] are independent variables and, therefore, ̂x [n]∕W[k]= 0. From Eq. (11.71), we
obtain
J ∗
=− < y [n − k]e[n] >= 0, (11.73)
∗
W [k]
which is nothing but the complex conjugate of Eq. (11.72).
The tap weights W[−K], W[−K + 1], … , W[K] are optimum when the cost function J is minimum. To find
the optimum tap weights, we follow an iterative procedure. Initially, tap weights are chosen arbitrarily as
(0) (0) (0) (0)
W =[W [−K], W [−K + 1], … , W [K]], (11.74)
where ‘(0)’ stands for the zeroth iteration. To update the tap weights for the next iteration, we need to move
in a vector space of 2K + 1 dimensions such that we are closer to a minimum of the cost function J.The
gradient vector is defined as
G =[g[−K], g[−K + 1], … , g[K]], (11.75)
J ∗
g[k]= 2 =−2 < y [n − k]e[n] >. (11.76)
∗
W [k]
(0)
At the starting point, we have the tap weight vector W (0) and the gradient vector G . From Eq. (11.71), we
see that J is minimum when g[k] is zero. But at the starting point, g[k] may not be zero. Iteratively, we need
to find W[k] such that g[k] is close to zero. The tap weight vector for the next iteration should be chosen in
(0)
(0)
a direction opposite to G . This is because, if we move in the direction of G , J would be maximized. So,
the tap weights for the next iteration are chosen as
Δ (0)
(0)
(1)
W = W − G (11.77)
2
or
W[k] (1) = W[k] (0) − Δ g[k] (0)
2
∗
= W[k] (0) +Δ < y [n − k]e[n] >, (11.78)
where Δ is a step-size parameter and the factor 1∕2 in Eq. (11.77) is introduced for convenience. The con-
vergence of the iterative procedure depends on the value of Δ chosen. In practice, it is difficult to evaluate
the expectation operator of Eq. (11.78), which requires knowledge of the channel response H[n]. Instead, the