Digital Signal Processing Reference
In-Depth Information
Example 4.7
Consider the function
w
(
t
)
=
Ev
x
(
t
)
−
a
0
}
shown in Fig. 4.6(b). Express
w
(
t
)
as a trigonometric CTFS.
Solution
From inspection, we see
w
(
t
) is even. Therefore,
b
n
=
0 for all
n
. Since
w
(
t
)
is periodic with a fundamental period
T
0
=
3,
ω
0
=
2
π/
3. The area enclosed
by one period of
w
(
t
), say
t
=
[
−
1
,
2], is given by 2(1
/
3)
+
1(
−
2
/
3)
=
0.
Function
w
(
t
) is, therefore, zero-mean, which imples that
a
0
=
0.
The value of
a
n
is calculated as follows:
1
.
5
1
.
5
=
2
3
=
4
3
a
n
w
(
t
) cos(
n
ω
0
t
)d
t
w
(
t
) cos(
n
ω
0
t
)d
t
,
−
1
.
5
0
which simplifies to
1
1
.
5
=
4
3
1
3
cos(
n
ω
0
t
)d
t
−
4
2
3
cos(
n
ω
0
t
)d
t
a
n
3
0
1
1
.
5
1
=
4
9
sin(
n
ω
0
t
)
n
ω
0
−
8
9
sin(
n
ω
0
t
)
n
ω
0
,
0
1
or
=
4
9
sin(
n
ω
0
)
n
ω
0
−
8
9
sin(1
.
5
n
ω
0
)
n
ω
0
+
8
9
sin(
n
ω
0
)
n
ω
0
a
n
=
4
3
sin(
n
ω
0
)
n
ω
0
−
8
9
sin(1
.
5
n
ω
0
)
n
ω
0
.
Substituting
ω
0
=
2
π/
3, we obtain
2
n
π
2
n
π
3
a
n
=
sin
,
leading to the CTFS representation
∞
2
n
π
2
n
π
3
2
n
π
3
w
(
t
)
=
sin
cos
t
,
(4.41)
n
=
1
which is same as the even component Ev
{
x
(
t
)
−
a
0
}
in Eq. (4.26) in
Example 4.6. The CTFS coefficients
a
n
are plotted in Fig. 4.7.
From Example 4.7, we observe that a rectangular pulse train
w
(
t
)
−
a
0
}
, as shown in Fig. 4.6(b), has a CTFS representation that includes a lin-
ear combination of an infinite number of cosine functions. A question that
arises is why an infinite number of cosine functions are needed. The answer
=
Ev
x
(
t
)
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