Digital Signal Processing Reference
In-Depth Information
y
(
n
)
L
s
(
n
)
s
(
n
)
L
C
(
z
)
L
L
1
/C
(
z
)
L
equalizer
(a)
q
(
n
)
L
receiver
y
(
n
)
L
s
(
n
)
s
(
n
)
C
(
z
)
L
L
L
G
(
z
)
L
(b)
equalizer
q
(
n
)
L
Figure 4.22
. Examples of
fractionally spaced equalizers
. (a) The channel inverse
1
/C
L
(
z
) is directly used, and (b) a different filter
G
L
(
z
)isused.
Example 4.1: Fractionally spaced FIR equalizer
Let
L
= 2 and assume that the channel is
C
L
(
z
)=1+2
z
−
1
+4
z
−
2
+
z
−
3
+
z
−
5
.
Suppose we choose the equalizer to be the FIR filter
4
z
−
2
G
L
(
z
)=
16
z
−
3
−
.
29
(We will soon explain how to
find
such equalizers.) Then
C
L
(
z
)
G
L
(
z
)=
16
z
+29+58
z
−
1
−
11
z
−
3
−
7
z
−
5
−
4
z
−
7
29
so that
T
(
z
)=[
C
L
(
z
)
G
L
(
z
)]
↓
2
=1
.
In practice the advance operator in
G
L
(
z
) can be eliminated by using
z
−
2
G
L
(
z
).
If we do this then
T
(
z
)=[
z
−
2
C
L
(
z
)
G
L
(
z
)]
↓
2
=
z
−
1
,
and the only distortion is the unit delay.
Note that the FIR channel in the preceding example can be perfectly equalized
with the FIR equalizer
G
L
(
z
)
.
So, what is the general theory behind this? What
are the conditions on the channel which allow this to be done, and how do we
solve for the FIR equalizer in general? The answers are contained in the next
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