Digital Signal Processing Reference
In-Depth Information
g
(
t
)
0.66
3e
−0.2
t
0.33
n
= 100
3
n
= 20
0
n
= 5
−0.33
t
−8
p
−6
p
−4
p
−2
p
0 2
p
4
p
6
p
8
p
10
p
−0.66
−0.99
Fig. 4.10. CT periodic signal
g
(
t
) with fundamental period
T
0
= 2π
0.8
0.85
0.9
0.95
1
1.05
1.1
1.15
1.2
considered in Example 4.8.
Fig. 4.9. Magnified sketch of Fig. 4.8 at
t
= 1.
x
(
t
) has a jump discontinuity at
t
=
t
j
. The reconstructed value for
x
(
t
j
)is
given by
x
(
t
j
)
=
1
+
)
+
x
(
t
j
−
)]
.
2
[
x
(
t
j
(4.42)
=
1isgivenby
For example, the reconstructed value of
w
(
t
) in Fig. 4.8 at
t
w
(1)
=
1
2
[
w
(1
−
)
+
w
(1
+
)]
=
1
1
3
−
2
3
=−
1
6
.
Figure 4.9 is an enlargement of part of Fig. 4.8 at
t
=
1, where it is observed
that the reconstructed signals have a value of
−
1
/
6at
t
2
=
1.
Example 4.8
Consider the periodic signal
g
(
t
) shown in Fig. 4.10. Calculate the CTFS
coefficients.
Solution
Because
T
0
=
2
π
, the fundamental frequency
ω
0
=
1. The dc coefficient
a
0
is
given by
2
π
2
π
−
0
.
2
t
1
T
0
1
2
π
3
2
π
e
−
0
.
2
t
d
t
a
0
=
g
(
t
)d
t
=
3e
=
−
0
.
2
0
T
0
0
=
15
−
0
.
4
π
]
≈
1
.
7079
.
2
π
[1
−
e
The CTFS coefficients
a
n
are given by
2
π
1
T
0
1
2
π
−
0
.
2
t
cos(
nt
)d
t
a
n
=
g
(
t
) cos(
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 cos(
nt
)
+
n
sin(
nt
)
]
2
0
=
or
3
2
π
1
n
2
+
0
.
2
2
[
−
0
.
2e
−
0
.
4
π
a
n
=
+
0
.
2]
0
.
3
(
n
2
+
3
.
4157
−
0
.
4
π
]
=
0
.
2
2
)
π
[1
−
e
≈
25
n
2
.
1
+
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