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EXAMPLE 17.7 Find the Fourier transform of x t e at u t , a > 0. Plot X f for 1 < f < 1. From (22a), the Fourier transform of x t is 1 1 X f e at e j2ft dt a j2f 0
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X f is a complex function of a real variable. Its magnitude and phase angle, jX f j and X f , respectively, shown in Figs. 17-13(a) and (b), are given by 1 jX f j p 2 a 42 f 2 and X f tan 1 2f =a 25a
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CHAP. 17]
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FOURIER METHOD OF WAVEFORM ANALYSIS
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Fig. 17-13
Alternatively, X f may be shown by its real and imaginary parts, Re X f and Im X f , as in Figs. 17-14(a) and (b). Re X f a a2 42 f 2 2f Im X f 2 a 42 f 2 26a 26b
Fig. 17-14
EXAMPLE 17.8
Find the Fourier transform of the square pulse  1 for T < t < T x t 0 otherwise
From (22a), T X f
e j2ft dt
1 h j2f iT sin 2fT e T j2f f
27
Because x t is even, X f is real.
The transform pairs are plotted in Figs. 17-15(a) and (b) for T 1 s. 2
EXAMPLE 17.9
Find the Fourier transform of x t eat u t ; a > 0. 0 X f
eat e j2ft dt
1 a j2f
28
FOURIER METHOD OF WAVEFORM ANALYSIS
[CHAP. 17
Fig. 17-15
EXAMPLE 17.10 Find the inverse Fourier transform of X f 2a= a2 42 f 2 , a > 0. By partial fraction expansion we have X f 1 1 a j2f a j2f 29
The inverse of each term in (29) may be derived from (24) and (28) so that x t e at u t eat u t e ajtj See Fig. 17-16. for all t
Fig. 17-16
PROPERTIES OF THE FOURIER TRANSFORM
Some properties of the Fourier transform are listed in Table 17-1. Several commonly used transform pairs are given in Table 17-2.
CONTINUOUS SPECTRUM
jX f j2 , as de ned in Section 17.9, is called the energy density or the spectrum of the waveform x t . Unlike the periodic functions, the energy content of a nonperiodic waveform x t at each frequency is zero. However, the energy content within a frequency band from f1 to f2 is
CHAP. 17]
FOURIER METHOD OF WAVEFORM ANALYSIS
Table 17-1 Fourier Transform Properties 1 1 Time Domain x t X f e j2ft dt Frequency Domain X f x t e j2ft dt
1 1
1. 2. 3. 4. 5. 6. 7. 8. 9.
x t real x t even, x t x t x t , odd, x t x t X t 1 x 0
X f X f X f X f X f X f x f 1 X f df X 0
x t dt 1 X f =a jaj 1 dX f j2 df
y t x at y t tx t y t x t y t x t t0
Y f
Y f
Y f X f Y f e j2ft0 X f
Table 17-2 x t 1. 2. 3. 4. e at u t ; a > 0 e ajtj ; a > 0 te at u t ; a > 0 exp t2 = 2
Fourier Transform Pairs X f 1 a j2f 2a a2 42 f 2 1 a j2f 2  exp f 2  2
7. 8. 9. 10.
1  t sin 2f0 t cos 2f0 t
 f 1  f f0  f f0 2j  f f0  f f0 2
FOURIER METHOD OF WAVEFORM ANALYSIS
[CHAP. 17
f2 W 2
jx f j2 df
30
EXAMPLE 17.11 Find the spectrum of x t e at u t eat u t , a > 0, shown in Fig. 17-17.
Fig. 17-17 We have x t x1 t x2 t . Since x1 t e at u t and x2 t eat u t , X1 f Then from which 1 a j2f X2 f 1 a j2f
X f X1 f X2 f jX f j2
j4f a2 42 f 2
162 f 2 a2 42 f 2 2 y1 t e jatj
EXAMPLE 17.12 Find and compare the energy contents W1 y2 t e at u t eat u t , a > 0, within the band 0 to 1 Hz. Let a 200. From Examples 17.10 and 17.11, jY1 f j2 4a2 a2 42 f 2 2 and jY2 f j2
162 f 2 a2 42 f 2 2
Within 0 < f < 1 Hz, the spectra and energies may be approximated by jY1 f j2 % 4=a2 10 4 J=Hz jY2 f 2 j % 10 7 f 2 and and W1 2 10 4 J 200 mJ W2 % 0
The preceding results agree with the observation that most of the energy in y1 t is near the low-frequency region in contrast to y2 t .
Solved Problems
17.1 Find the trigonometric Fourier series for the square wave shown in Fig. 17-18 and plot the line spectrum.
In the interval 0 < !t < , f t V; and for  < !t < 2, f t V. The average value of the wave is zero; hence, a0 =2 0. The cosine coe cients are obtained by writing the evaluation integral with the functions inserted as follows:
CHAP. 17]
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