Digital Signal Processing Reference
In-Depth Information
and
T
T
1
2
T
1
2
T
=−
A
c
3
=
x
(
t
)
φ
3
(
t
)d
t
=
A
(
−
1)d
t
2
.
−
T
0
In other words,
x
(
t
)
=
0
.
5
A
[
φ
1
(
t
)
− φ
3
(
t
)].
Example 4.3
Show that the set
{
1, cos(
ω
0
t
), cos(2
ω
0
t
), cos(3
ω
0
t
), . . . , sin(
ω
0
t
), sin(2
ω
0
t
),
sin(3
ω
0
t
), . . .
}
, consisting of all possible harmonics of sine and cosine waves
with fundamental frequency of
ω
0
, is an orthogonal set over any interval
t
=
[
t
0
,
t
0
+
T
0
], with duration
T
0
=
2
π/ω
0
.
Solution
It may be noted that the set
{
1, cos(
ω
0
t
), cos(2
ω
0
t
), cos(3
ω
0
t
), . . . , sin(
ω
0
t
),
sin(2
ω
0
t
), sin(3
ω
0
t
), . . .
}
contains three types of functions: 1,
{
cos(
m
ω
0
t
)
}
,
and
{
sin(
n
ω
0
t
)
}
for arbitrary integers
m
,
n
+
+
is the set of positive
integers. We will consider all possible combinations of these functions.
∈
Z
, where
Z
+
Case 1
The following proof shows that functions
{
cos(
m
ω
0
t
),
m
∈
Z
}
are
=
+
T
0
] with
T
0
=
2
π/ω
0
.
orthogonal to each other over interval
t
[
t
0
,
t
0
Equation (4.10) yields
=
t
0
+
T
0
cos(
m
ω
0
t
) cos(
n
ω
0
t
)d
t
cos(
m
ω
0
t
) cos(
n
ω
0
t
)d
t
for any arbitrary
t
0
.
T
0
t
0
Using
the
trigonometric
identity
cos(
m
ω
0
t
) cos(
n
ω
0
t
)
=
(1
/
2)[cos((
m
−
n
)
ω
0
t
)
+
cos((
m
+
n
)
ω
0
t
)], the above integral reduces as follows:
sin(
m
−
n
)
ω
0
t
2(
m
−
n
)
ω
0
+
sin(
m
+
n
)
ω
0
t
2(
m
+
n
)
ω
0
t
0
+
T
0
=
n
m
t
0
cos(
m
ω
0
t
) cos(
n
ω
0
t
)d
t
=
t
2
+
sin 2
m
ω
0
t
4
m
ω
0
t
0
+
T
0
m
=
n
,
T
0
t
0
or
0
m
=
n
cos(
m
ω
0
t
) cos(
n
ω
0
t
)d
t
=
T
0
2
(4.15)
m
=
n
,
T
0
for
m
,
n
∈
Z
+
. Equation (4.15) demonstrates that the functions in the set
+
{
cos(
m
ω
0
t
),
m
∈
Z
}
are mutually orthogonal.
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