Digital Signal Processing Reference
In-Depth Information
Solution
Since the fundamental period
T
0
=
4, the angular frequency
ω
0
=
2
π/
4
= π/
2.
The exponential CTFS coefficients
D
n
are calculated directly from the definition
as follows:
2
1
T
0
=
1
4
−
j
n
ω
0
t
d
t
−
j
n
ω
0
t
d
t
D
n
=
f
(
t
)e
f
(
t
)e
T
0
−
2
2
2
=
1
4
d
t
−
j
1
4
f
(
t
) co
s
(
n
ω
0
t
)
f
(
t
) s
in(
n
ω
0
t
)
d
t
.
−
2
−
2
even function
odd function
Since the integration of an odd function within the limits [
t
0
,
−
t
0
] is zero,
2
2
=
1
4
=
1
2
D
n
f
(
t
) cos(
n
ω
0
t
)d
t
(3
−
3
t
) cos(
n
ω
0
t
)d
t
,
−
2
0
which simplifies to
2
=
1
2
(3
−
3
t
)
sin(
n
ω
0
t
)
n
ω
0
−
3
cos(
n
ω
0
t
)
(
n
ω
0
)
2
D
n
0
−
sin(2
n
ω
0
)
n
ω
0
−
cos(2
n
ω
0
)
(
n
ω
0
)
2
=
3
2
1
(
n
ω
0
)
2
+
.
Substituting
ω
0
= π/
2, we obtain
=
3
2
−
sin(
n
π
0
)
0
.
5
n
π
−
cos(
n
π
)
(0
.
5
n
π
)
2
1
(0
.
5
n
π
)
2
6
(
n
π
)
2
[1
−
(
−
1)
n
]
D
n
+
=
or
0
n
is even
12
(
n
π
)
2
D
n
=
(4.47)
n
is odd.
In Examples 4.11 and 4.12, the exponential CTFS coefficients can also be
derived from the trigonometric CTFS coefficients calculated in Examples 4.7
and 4.8 using Eq. (4.45).
Example 4.14
Calculate the exponential Fourier series of the signal
x
(
t
) shown in Fig. 4.13.
Solution
The fundamental period
T
0
=
T
, and therefore the angular frequency
ω
0
=
2
π/
T
. The exponential CTFS coefficients are given by
T
/
2
τ/
2
=
1
T
=
1
T
−
j
n
ω
0
t
d
t
−
j
n
ω
0
t
d
t
.
1
e
D
n
x
(
t
)e
−
T
/
2
−τ/
2
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