Digital Signal Processing Reference
In-Depth Information
passed through a filter (system) with impulse response h(t), then the filter output
sampled as T
S
can be written as,
t
=
T
S
y
(
t
)
=
r
(τ )
h
(
t
−
τ )
d
τ
0
T
S
T
S
(3.39)
=
s
(τ )
h
(
T
S
−
τ )
d
τ
+
n
(τ )
h
(
T
S
−
τ )
d
τ
0
0
=
y
S
(
T
S
)
+
y
n
(
T
S
)
where,
y
S
(
are the signal and noise component of the filter output.
The problem is to select a filter impulse that maximizes the output SNR defined as,
T
S
)
and
y
n
(
T
S
)
y
S
(
T
S
)
E
S
y
n
(
SNR
=
(3.40)
T
S
)
Now,
T
S
T
S
E
y
n
(
T
S
)
=
E
[
n
(
τ
)
n
(
t
)]
h
(
T
S
−
τ
)
h
(
T
S
−
t
)
dtd
τ
0
0
T
S
T
S
=
2
δ
(
t
−
τ
)
h
(
T
S
−
τ
)
h
(
T
S
−
t
)
dtd
τ
(3.41)
0
0
T
S
=
2
h
2
(
T
S
−
t
)
dt
0
Therefore, putting the value of noise variance from Eq. (
3.41
) into Eq. (
3.40
),
T
S
2
s
(τ )
h
(
T
S
−
τ )
d
τ
0
SNR
=
T
S
2
h
2
(
T
S
−
t
)
dt
0
T
S
2
h
(τ )
s
(
t
−
τ )
d
τ
0
=
T
S
2
h
2
(
T
S
−
t
)
dt
0
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