Digital Signal Processing Reference
In-Depth Information
Comparing the above equation with the CTFS expansion,
∞
x
(
t
)
=
a
0
+
(
a
n
cos(
n
ω
0
t
)
+
b
n
sin(
n
ω
0
t
))
,
n
=
1
with
ω
0
=
2, we obtain
√
√
3
2
=−
√
=
1
a
0
=
3
,
a
2
=
2
,
a
5
=
,
b
2
2
,
and
b
5
2
.
=
The CTFS coefficients
a
n
and
b
n
, for values of
n
other than
n
0, 2, and 5, are
all zeros.
Example 4.11
A periodic signal is represented by the following CTFS:
∞
x
(
t
)
=
2
π
1
2
m
+
1
sin(4
π
(2
m
+
1)
t
)
.
m
=
0
(i) From the CTFS representation, determine the fundamental period
T
0
of
x
(
t
).
(ii) Comment on the symmetry properties of
x
(
t
).
(iii) Plot the function to verify if your answers to (i) and (ii) are correct.
Solution
(i) The CTFS representation is obtained by expanding the summation as follows:
∞
=
2
π
1
2
m
+
x
(
t
)
1
sin(4
π
(2
m
+
1)
t
)
m
=
0
=
2
π
sin(4
π
t
)
+
1
3
sin(12
π
t
)
+
1
5
sin(20
π
t
)
+
1
7
sin(28
π
t
)
+
.
Note that the signal
x
(
t
) contains the fundamental component sin(4
π
t
) and its
higher-order harmonics. Hence, the fundamental frequency is
ω
0
=
4
π
with
the fundamental period given by
T
0
1
/
2
.
(ii) Because the CTFS contains only sine terms,
x
(
t
) must be odd based on
property (3) on page 156.
(iii) It is generally difficult to evaluate the function
x
(
t
) manually. We use a
M
ATLAB
function
ictfs.m
(provided in the accompanying CD) to calculate
x
(
t
). The function, reconstructed using the first 1000 CTFS coefficients, is plot-
ted in Fig. 4.12 for
=
2
π/
4
π
=
1
.
It is observed that the function is a rectangular
pulse train with a fundamental period of 0
.
5. It is also observed that the function
is odd.
−
≤
t
≤
1
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