Digital Signal Processing Reference
In-Depth Information
⎡
⎤
1000
2100
4210
0421
0042
0004
⎡
⎤
21 10 4 0
10 21 10 4
4 102110
0
⎣
⎦
⎣
⎦
A
†
A
=
A
=
and
.
4
10
21
This result holds for any
M
and
L,
and is a consequence of the full banded
Toeplitz property of
A
,
as we shall show. Imagine
A
K
is only a partial matrix
obtained by dropping rows from
A
. For example if
⎡
⎣
⎤
⎦
1000
2100
4210
0421
0042
⎡
⎣
⎤
⎦
21 10 4 0
10 21 10 4
4 102110
0
A
†
K
A
K
=
A
K
=
then
.
4
10
5
The Toeplitz property is violated by
A
†
K
A
K
because of the last diagonal element.
So
A
†
K
A
K
is Toeplitz for
K
=
M
+
L
but not necessarily so for smaller
K.
Proof of Toeplitz property.
To prove that
A
†
A
is Toeplitz when
A
is full
banded Toeplitz, note that the
m
th column of
A
is the full impulse response
c
(
n
) shifted down by
m.
Thus the (
k, m
)elementof
A
†
A
is
[
A
†
A
]
km
=
n
L
c
∗
(
n − k
)
c
(
n − m
)=
c
(
n
)
c
∗
(
n − k
+
m
)=
r
(
k − m
)
,
n
=0
where
r
(
)=
n
c
(
n
)
c
∗
(
n
−
)
is the autocorrelation of the deterministic sequence
c
(
n
). Using the fact that
r
(
)=
r
∗
(
) we see that
A
†
A
has the form demonstrated below for
M
=4
.
−
⎡
⎤
r
∗
(1)
r
∗
(2)
r
∗
(3)
r
(0)
r
∗
(1)
r
∗
(2)
⎣
r
(1)
r
(0)
⎦
R
=
A
†
A
=
.
(8
.
23)
r
(2)
r
(1)
r
(0)
r
∗
(1)
r
(3)
r
(2)
r
(1)
r
(0)
This matrix is clearly Hermitian and Toeplitz.
Here are some consequences of the fact that
A
†
A
has the form (8.23):
1.
Insensitivity to channel phase
. Given an FIR channel
L
L
c
(
n
)
z
−n
=
c
(0)
z
−
1
z
k
)
,
C
(
z
)=
(1
−
n
=0
k
=1
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