B. P. Lathi, Zhi Ding - Modern Digital and Analog Communication Systems-Oxford University Press (2009)

362 PRINCIPLES OF DIGITAL DATA TRANSMISSION
of equations by taking the inverse of the matrix P r
P - 1
C = r Po
Example 7.3
For the received pulse P r (t) in Fig. 7.22b, let
P r [0] = 1
P r D ] = -0.3
P r [- 1] = -0.2
Pr[2] = 0.1
P r [-2] = 0.05
Design a three-tap (N = l ) equalizer.
Substituting the foregoing values in Eq. (7.53), we obtain
[
0
] [
1 -0.2 0.05
] [
C- 1
1 = -0.3 1 -0.2 C()
0 0.1 -0.3 1 CJ
]
(7.54)
Solution of this set yields c - 1 = 0.210, co = 1.13, and c1 = 0.318. This tap setting assures
us that po[0] = 1 and po[-1] =po[l] = 0. The ideal output p 0 (t) is sketched in Fig. 7.22c.
Note that the equalizer determined from Eq. (7.53) can guarantee only the zero ISi condition
of Eq. (7.52). In other words, ISi is zero only for k = 0, ±1, ... , ±N. In fact, for k
outside this range, it is quite common that the samples p 0 (kTb) -::/= 0, indicating some residual
ISL For instance, consider the equalizer problem in Example 7 .3. The samples of the equalized
pulse has zero ISi for k = - 1, 0, 1. However, from
Po[k] = L CnPr[k - n]
n=-N
we can see that the three-tap zero-forcing equalizer parameters will lead to
p 0 [-3] = 0.010 p 0 [-2] = 0.0145 p 0 [2] = 0.0176
p 0 [3] = 0.03 18 p 0 [k] = 0 k = 0, ±1, ±4, ...
It is therefore clear that not all the TSI has been removed because of these four nonzero samples
of the equalizer output pulse. In fact, because we only have 2N + 1 (N = 1 in Example 7.3)
parameters in the equalizer, it is impossible to force p 0 [k] = 0 for all k unless N = oo. This
means that we will not be able to design a practical finite tap equalizer that achieves perfect
zero ISL Still, when N is sufficiently large, then typically the residual nonzero sample values
will be small, indicating that most of the ISi has been suppressed.