Classical transceiver techniques - Signal Processing and Optimization for Transceiver Systems - page 207

0.5 V

− 0.5 V

x x x

−3 A − A A 3 A

Figure 5.25 . Examples showing the application of the mod V operator. The points
±A and ± 3 A correspond to the words in a 2-bit PAM constellation.

What makes the pre-equalization system successful is an important property
of the modulo operator. Given two real numbers u 1 and u 2 , it is readily verified
that

( u 1 + u 2 )mod V = u 1 mod V + u 2 mod V mod V.

(5 . 92)

Using this we will show that for the transceiver in Fig. 5.23, as long as − 0 . 5 V ≤

s ( n ) < 0 . 5 V for all n, the signal s ( n ) at the receiver is exactly equal to s ( n )in

absence of channel noise:

s ( n )= s ( n ) . (5 . 93)

Thus, even though x ( n ) may have any value in the range x ( n ) < 0 . 5 V,

after it passes through the channel H ( z ) and the modulo operator again, the
result is precisely the transmitted symbol s ( n ) from the constellation. When
there is channel noise, this is modified to

− 0 . 5 V ≤

s ( n )=[ s ( n )+ q ( n )] mod V = s ( n )+[ q ( n )mod V ] mod V.

(5 . 94)

Thus, even though the modulo operator is nonlinear, the transceiver reproduces
s ( n ) perfectly at the receiver, except for channel noise. If q ( n )issmallcompared
to the room between the largest PAM symbol and the boundary 0 . 5 V , Eq. (5.94)
canbewrittenas

s ( n )= s ( n )+ q ( n ) . In this case it is as though there is no
modulo operation at all, and the channel has been perfectly pre-equalized!

Proof of Eq. (5.93). Since B ( z ) is as in Eq. (5.89), we see from Fig. 5.23
that

x ( n )= L )+ s ( n ) mod V, − h (1) x ( n − 1) − ... − h ( L ) x ( n −

which implies
x ( n )+ h (1) x ( n

L ) mod V = s ( n ) .

− 1) + ... + h ( L ) x ( n − (5 . 95)

But the output of the channel H ( z ) in Fig. 5.23 is

y ( n )= x ( n )+ h (1) x ( n − 1) + ... + h ( L ) x ( n − L ) .