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CHAP. 17]

FOURIER METHOD OF WAVEFORM ANALYSIS

Fig. 17-18

an ¼

1 

¼0

435

Fig. 17-19

ð 

ð 2 V cos n!t dð!tÞ þ

0



(   2 )  V 1 1 sin n!t  sin n!t ðVÞ cos n!t dð!tÞ ¼  n n 0 

for all n

Thus, the series contains no cosine terms. Proceeding with the evaluation integral for the sine terms, ð  ð 2 1  V sin n!t dð!tÞ þ ðVÞ sin n!t dð!tÞ bn ¼  0  (   2 ) V 1 1 ¼  cos n!t þ cos n!t  n n 0  ¼

V 2V ð cos n þ cos 0 þ cos n2  cos nÞ ¼ ð1  cos nÞ n n

Then bn ¼ 4V=n for n ¼ 1; 3; 5; . . . ; and bn ¼ 0 for n ¼ 2; 4; 6; . . . . f ðtÞ ¼

The series for the square wave is

4V 4V 4V sin !t þ sin 3!t þ sin 5!t þ     3 5

The line spectrum for this series is shown in Fig. 17-19. This series contains only odd-harmonic sine terms, as could have been anticipated by examination of the waveform for symmetry. Since the wave in Fig. 17-18 is odd, its series contains only sine terms; and since it also has half-wave symmetry, only odd harmonics are present.

17.2

Find the trigonometric Fourier series for the triangular wave shown in Fig. 17-20 and plot the line spectrum. The wave is an even function, since f ðtÞ ¼ f ðtÞ, and if its average value, V=2, is subtracted, it also has half-wave symmetry, that is, f ðtÞ ¼ f ðt þ Þ. For  < !t < 0, f ðtÞ ¼ V þ ðV=Þ!t; and for 0 < !t < , f ðtÞ ¼ V  ðV=Þ!t. Since even waveforms have only cosine terms, all bn ¼ 0. For n  1, ð ð 1 0 1  ½V þ ðV=Þ!t cos n!t dð!tÞ þ ½V  ðV=Þ!t cos n!t dð!tÞ an ¼    0 ð   ð0 ð V !t !t cos n!t dð!tÞ þ ¼ cos n!t dð!tÞ  cos n!t dð!tÞ     0  ( 0   ) V 1 !t 1 !t sin n!t sin n!t ¼ 2 cos n!t þ  2 cos n!t þ  n  n2 n  0 ¼

V 2V ½cos 0  cosðnÞ  cos n þ cos 0 ¼ 2 2 ð1  cos nÞ 2 n2  n

As predicted from half-wave symmetry, the series contains only odd terms, since an ¼ 0 for n ¼ 2; 4; 6; . . . . For n ¼ 1; 3; 5; . . . ; an ¼ 4V=2 n2 . Then the required Fourier series is f ðtÞ ¼

V 4V 4V 4V þ cos !t þ cos 3!t þ cos 5!t þ    2 2 ð3Þ2 ð5Þ2

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An  
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