Digital Signal Processing Reference
In-Depth Information
Fig. 4.11. Periodic signal
f
(
t
)
considered in Example 4.9.
f
(
t
)
3
t
−4
−2
0 2
4
−3
Similarly, the CTFS coefficients
b
n
are given by
2
π
1
T
0
1
2
π
−
0
.
2
t
sin(
nt
)d
t
b
n
=
g
(
t
) sin(
n
ω
0
t
)d
t
=
3e
T
0
0
3
2
π
1
n
2
+
0
.
2
2
[e
−
0
.
2
t
−
0
.
2 sin(
nt
)
−
n
cos(
nt
)
]
2
0
=
or
3
2
π
1
n
2
+
0
.
2
2
[
−
n
e
3
n
(
n
2
+
0
.
2
2
)
π
[1
−
e
≈
17
.
0787
n
1
−
0
.
4
π
−
0
.
4
π
]
b
n
=
+
n
]
=
25
n
2
.
+
The trigonometric CTFS representation of
g
(
t
) is therefore given by
∞
∞
3
.
4157
1
17
.
0787
1
g
(
t
)
=
1
.
7079
+
cos(
nt
)
+
25
n
2
n
sin(
nt
)
.
+
25
n
2
+
n
=
1
n
=
1
Example 4.9
Consider the periodic signal
f
(
t
) as shown in Fig. 4.11. Calculate the CTFS
coefficients.
Solution
Because
T
0
=
4, the fundamental frequency
ω
0
=
π/
2. Since
f
(
t
) is zero-mean,
the dc coefficient
a
0
=
0. Also, since
f
(
t
) is an even function,
b
n
=
0 for all
n
.
The CTFS coefficients
a
n
are given by
2
2
=
2
4
=
4
4
f
(
t
) cos(
n
ω
0
t
)
−
3
t
) cos(
n
ω
0
t
)d
t
a
n
d
t
(3
−
2
0
even function
2
3
t
)
sin(
n
ω
0
t
)
n
ω
0
3
cos(
n
ω
0
t
)
(
n
ω
0
)
2
=
(3
−
−
.
0
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