Continuous-time Fourier transform - Continuous and Discrete Time Signals and Systems - page 215

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.