Review of Analog Signal Processing Basics - Digital Signal Processing: Fundamentals and Applications - page 787

Solution:
Applying Equation (B.21) and using Euler's formula, we have

Z N 0:25

x t e jut dt ¼ X u ¼ 10e ju dt

N

0:25

e j0:25u e j0:25u
ju

0:25
0:25 ¼ 10

e jut
ju

¼ 10

¼ 10 cos ð0:25uÞ j sin ð0:25uÞ ½cos ð0:25uÞ þ j sin ð0:25uÞ
ju

¼ 5 sin ð0:25uÞ
0:25u

where the amplitude spectrum is expressed as

sin ð0:25uÞ
0:25u

jX ðuÞj ¼ 5

Using u ¼ 2pf , we can express the spectrum in terms of Hz as

sin ð0:5p f Þ
0:5pf

jX ðf Þj ¼ 5

The amplitude spectrum is shown in Figure B.9 . Note that the first null point is at u ¼ 2p=0:5 ¼ 4p rad/sec, and
the spectrum is symmetric.

FIGURE B.9

Amplitude spectrum for Example B.5.

EXAMPLE B.6
Let xðtÞ be an exponential function given by

( 10e 2t

x t ¼ 10e 2t u t ¼

t 0

0 t < 0

Find its Fourier transform.