Digital Signal Processing Reference
In-Depth Information
Substituting
ω
0
= π/
2, we obtain
(
−
3)
sin(
n
π
)
0
.
5
n
π
−
3
cos(
n
π
)
(0
.
5
n
π
)
2
1
(0
.
5
n
π
)
2
a
n
=
+
3
(
−
1)
n
(0
.
5
n
π
)
2
1
(0
.
5
n
π
)
2
12
(
n
π
)
2
[1
−
(
−
1)
n
]
=
3
0
−
+
=
or
0
n
is even
a
n
=
24
(
n
π
)
2
n
is odd.
The CTFS representation of
f
(
t
)isgivenby
∞
24
(
n
π
)
2
f
(
t
)
=
cos(0
.
5
n
π
t
)
n
=
1
,
3
,
5
,
=
24
π
2
+
1
+
1
cos(0
.
5
π
t
)
9
cos(1
.
5
π
t
)
25
cos(2
.
5
π
t
)
+
.
Example 4.10
Calculate the CTFS coefficients for the following signal:
4
t
+
π
4
10
t
+
π
3
x
(
t
)
=
3
+
cos
+
sin
.
Solution
The fundamental period of cos(4
t
+
π/
4) is given by
T
1
=
π/
2, while the
fundamental period of sin(10
t
+
π/
3) is given by
T
2
=
π/
5. Since the ratio
T
1
T
2
=
5
2
is a rational number, Proposition 1.2 states that
x
(
t
) is periodic with a fun-
damental period of
π
. The fundamental frequency
ω
0
is therefore given by
ω
0
=
2.
Since
x
(
t
) is a linear combination of sinusoidal functions, the CTFS coef-
ficients can be calculated directly by expanding the sine and cosine terms as
follows:
=
2
π/
T
0
π
4
π
4
π
3
x
(
t
)
=
3
+
cos(4
t
) cos
−
sin(4
t
) sin
+
sin(10
t
) cos
π
3
+
cos(10
t
) sin
.
Substituting the values of sin(
π/
4), cos(
π/
4), sin(
π/
3), and cos(
π/
3), we obtain
√
√
√
+
1
3
2
x
(
t
)
=
3
+
2
cos(4
t
)
−
2
sin(4
t
)
2
sin(10
t
)
+
cos(10
t
)
.
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