Digital Signal Processing Reference
In-Depth Information
x
1
(
t
)
x
2
(
t
)
2
2
t
t
01
2
−2
−1
−2
−1
−2
(a)
(b)
Fig. 5.9. CT signals used in
Example 5.12. (a)
x
1
(
t
);
(b)
x
2
(
t
).
Example 5.12
Calculate the Fourier transform of the functions
x
1
(
t
) and
x
2
(
t
) shown in
Fig. 5.9.
Solution
(a) The mathematical expression for the CT function
x
1
(
t
), illustrated in
Fig. 5.9(a), is given by
2
t
−
1
≤
t
≤
1
x
1
(
t
)
=
<
t
≤
2
0
elsewhere.
Since
x
1
(
t
) is an even function, its CTFT is calculated using Eq. (5.40) as
follows:
∞
1
2
X
1
(
ω
)
=
2
x
(
t
) cos(
ω
t
)d
t
=
2
(2
t
) cos(
ω
t
)d
t
+
2
2 cos(
ω
t
)d
t
,
0
0
1
which simplifies to
1
2
t
sin(
ω
t
)
ω
+
1
cos(
ω
t
)
ω
2
sin(
ω
t
)
ω
X
1
(
ω
)
=
4
+
4
0
1
or
sin(
ω
)
ω
+
cos(
ω
)
ω
2
−
1
ω
2
sin(2
ω
)
ω
−
sin(
ω
)
ω
X
1
(
ω
)
=
4
+
4
4
ω
2
=
[
ω
sin(2
ω
)
+
cos(
ω
)
−
1]
.
(5.42)
The above result validates the symmetry property for real-valued, even func-
tions. Property 5.1 states that the CTFT of a real-valued, even function is real
and even. This is indeed the case for
X
1
(
ω
) in Eq. (5.42).
(b) The function
x
2
(
t
), shown in Fig. 5.9(b), is expressed as follows:
#
−
2
−
2
≤
t
≤
1
−
1
≤
t
≤
1
2
t
x
2
(
t
)
=
#
<
t
≤
2
0
elsewhere.
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