Channel equalization with transmitter redundancy - Signal Processing and Optimization for Transceiver Systems - page 248

M samples

x ( n )

L zeros

(a)

n

0 M P 2 P

part of y ( n ) due to 0 th
input block

y ( n )

(b)

n

0 P 2 P

Figure 7.4 . (a) The channel input padded with zeros, and (b) one block of the channel
output. See text.

To recover the M input samples we only need to consider the first M equa-
tions above, and write
⎡ ⎤ ⎡ ⎤ ⎡ ⎤

y ( kP )
y ( kP +1)
.
y ( kP + M − 1)

c (0) 0 ... 0 x ( kP )

x ( kP +1)
.
x ( kP + M − 1)

⎣ ⎦ ⎣

c (1) c (0) ... 0

⎦ ⎣ ⎦

= .

. . .

. . .

c ( M − 1) c ( M − 2) ... c (0)

(7 . 6)

Since c (0) M matrix is nonsingular and can be inverted to yield

the M input samples x ( n ) which in turn equal the M transmitted symbols
s ( kM ) ,...,s ( kM + M

=0the M ×

1) . In practice there is channel noise, and the right-
hand sides of Eqs. (7.5) and (7.6) have a noise term as well, so the recovery of
x ( n ) is not perfect. We will see later that, with noise present, inversion of the
matrix in (7.6) is not the best thing to do. All the P samples of y ( n )inEq.
(7.5) should be used to estimate x ( n ) (equivalently s ( n )) with minimum error,
using the ideas of statistical optimal filtering (Sec. 4.10).
Note that the price paid for the suppression of interblock interference is the
bandwidth expansion factor, or redundancy ratio γ, defined in Eq. (7.2). For
a given channel order L we can choose M to be arbitrarily large ,andmake
γ → 1 . While this appears to be a simple way to avoid bandwidth expansion,
there are some practical issues to keep in mind. For example, the computational
complexity involved in the inversion of (7.5) grows rapidly with M . Secondly,
there is amplification of the ever-present channel noise in the inversion process.
This amplification increases as M is increased, as we shall see in future chapters.
Before we get into these details, let us see what happens if we choose M as small
as possible, namely M =1 .

−

more, that all the elements along any line parallel to the diagonal are identical. Since the
L + 1 nonzero elements in each column form a “band,” we say the matrix in Eq. (7.5) is a
full banded Toeplitz matrix. The matrix in Eq. (7.6) is not full banded because some of the
elements c ( k ) drop out as we move to the right.