Digital Signal Processing Reference
In-Depth Information
y
(
t
)
r
(
n
)
r
(
n
)
2
j
ω
1
= s
(
n
)
s
(
n
)
H
(
j
ω
*
)
H
(
j
ω)
A
(
e
)
D/C
C/D
T
T
digital filter
channel
matched filter
Figure 5.7
. Reconstruction of
s
(
n
) from
r
(
n
)
.
Since these relationships will be encountered frequently, we shall summa-
rize them as a lemma. A deeper discussion in the setting of
2
and
L
2
spaces
(Appendix A) is given in Sec. 5.4.
Lemma 5.3
. Given a function
h
(
t
) with Fourier transform
H
(
jω
)
,
the fol-
lowing statements are equivalent:
♠
1. The set of functions
is linearly independent, that is, there does
not exist a nonzero sequence
c
(
n
)
{
h
(
t
−
nT
)
}
2
such that
n
c
(
n
)
h
(
t
∈
−
nT
)=0for
all
t.
H
j
ω
+
2
πk
T
∞
2
>
0 for all
ω.
2.
k
=
−∞
3. There exists an alias-free(
T
) band
such that
H
(
jω
)isnonzeroevery-
where in it. That is, the channel
H
(
jω
)has
su
cient passband width
(though not necessarily as a single contiguous piece) for sending symbols
at the rate 1
/T.
A
4. There does
not
exist a region
R
of frequencies
θ
a
<ω<θ
b
such that
H
(
jω
)vanisheseverywherein
R
and in
all its shifted versions
R
+2
πk/T
(see Fig. 5.8 for a demonstration).
♦
Proof.
Assume the functions
{h
(
t − nT
)
}
are linearly dependent. Then
there exists a nonzero sequence
c
(
n
)
∈
2
such that
c
(
n
)
h
(
t
−
nT
)=0
for all
t.
Taking Fourier transforms, this implies
C
(
e
jωT
)
H
(
jω
) = 0 for all
ω
, which implies
C
(
e
jωT
)
H
j
ω
+
2
πk
T
∞
2
2
=0
(5
.
10)
k
=
−∞
for all
ω.
Since
c
(
n
)
∈
2
is a nonzero sequence,
C
(
e
jωT
)isnonzeroin
some region
ω
a
<ω<ω
b
.
In this region, therefore, the summation in the
preceding left-hand side must vanish identically.
Search WWH ::
Custom Search