Digital Signal Processing Reference
In-Depth Information
where
r
(0) =
n
|
|
2
= energy of the channel. Note therefore that the
c
(
n
)
quantity
L
A
2
M
|c
(
n
)
|
2
=
r
(0) =
(8
.
24)
n
=0
and depends only on the channel energy
r
(0) and not on
M.
Example 8.2: Channels with zeros outside the unit circle
Consider the channel
C
(
z
)=1+
z
−
1
+0
.
31
z
−
2
+0
.
03
z
−
3
This has order
L
= 3 and the three zeros are
inside
the unit circle:
z
1
=
−
0
.
2
.
Choose
M
=8sothat
P
=
M
+
L
=11
.
Thenthesizeof
A
K
can be
K
=8
,
9
,
10, or 11
.
Calculations show:
0
.
3
,z
2
=
−
0
.
5
,
and
z
3
=
−
A
K
2
/M
K
8
2
.
37
9
2
.
05
10
2
.
02
11
2
.
02
The noise gain therefore decreases only slightly as we increase the size of
A
K
. Now consider the channel
C
rev
(
z
)=0
.
03 + 0
.
31
z
−
1
+
z
−
2
+
z
−
3
,
which is the time-reversed version of
C
(
z
). This has all the zeros
outside
the unit circle. Calculations show:
A
K
2
/M
8
.
03
K
10
13
×
10
8
9
.
30
×
10
3
.
03
×
10
3
11
2
.
02
A
K
2
is very large
for
K
=
M, M
+1
,
and
M
+2
.
But for the full size matrix
A
K
with
K
= 11,
the quantity
For
C
rev
(
z
)
,
since the zeros are outside the unit circle,
A
K
2
is identical for
C
(
z
)and
C
rev
(
z
) as expected.
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