zen barcode ssrs FOURIER METHOD OF WAVEFORM ANALYSIS in Software

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FOURIER METHOD OF WAVEFORM ANALYSIS
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Fig. 17-18   V cos n!t d !t
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Fig. 17-19 2 (   2 )  V 1 1 sin n!t sin n!t V cos n!t d !t  n n 0 
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Thus, the series contains no cosine terms. Proceeding with the evaluation integral for the sine terms,   2 1  V sin n!t d !t V sin n!t d !t bn  0  (   2 ) V 1 1 cos n!t cos n!t  n n 0  V 2V cos n cos 0 cos n2 cos n 1 cos n n n The series for the square wave is
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Then bn 4V=n for n 1; 3; 5; . . . ; and bn 0 for n 2; 4; 6; . . . . f t
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4V 4V 4V sin !t sin 3!t sin 5!t  3 5
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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.
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Find the trigonometric Fourier series for the triangular wave shown in Fig. 17-20 and plot the line spectrum.
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The wave is an even function, since f t f t , and if its average value, V=2, is subtracted, it also has half-wave symmetry, that is, f t f t  . For  < !t < 0, f t V V= !t; and for 0 < !t < , f t V V= !t. Since even waveforms have only cosine terms, all bn 0. For n ! 1, 1 0 1  V V= !t cos n!t d !t V V= !t cos n!t d !t an    0    0  V !t !t cos n!t d !t cos n!t d !t cos n!t d !t     0  ( 0   ) V 1 !t 1 !t sin n!t sin n!t 2 cos n!t 2 cos n!t  n  n2 n  0 V 2V cos 0 cos n cos n cos 0 2 2 1 cos n 2 n2  n
As predicted from half-wave symmetry, the series contains only odd terms, since an 0 for n 2; 4; 6; . . . . For n 1; 3; 5; . . . ; an 4V=2 n2 . Then the required Fourier series is f t V 4V 4V 4V cos !t cos 3!t cos 5!t 2 2 3 2 5 2
FOURIER METHOD OF WAVEFORM ANALYSIS
[CHAP. 17
The coe cients decrease as 1=n2 , and thus the series converges more rapidly than that of Problem 17.1. This fact is evident from the line spectrum shown in Fig. 17-21.
Fig. 17-20
Fig. 17-21
Find the trigonometric Fourier series for the sawtooth wave shown in Fig. 17-22 and plot the line spectrum.
By inspection, the waveform is odd (and therefore has average value zero). Consequently the series will contain only sine terms. A single expression, f t V= !t, describes the wave over the period from  to , and we will use these limits on our evaluation integral for bn .   1  V 1 !t 2V cos n!t cos n V= !t sin n!t d !t 2 2 sin n!t bn   n n  n  As cos n is 1 for even n and 1 for odd n, the signs of the coe cients alternate. The required series is f t 2V fsin !t 1 sin 2!t 1 sin 3!t 1 sin 4!t g 2 3 4 
The coe cients decrease as 1=n, and thus the series converges slowly, as shown by the spectrum in Fig. 17-23. Except for the shift in the origin and the average term, this waveform is the same as in Fig. 17-8; compare the two spectra.
Fig. 17-22
Fig. 17-23
Find the trigonometric Fourier series for the waveform shown in Fig. 17-24 and sketch the line spectrum.
In the interval 0 < !t < , f t V= !t; and for  < !t < 2, f t 0. By inspection, the average value of the wave is V=4. Since the wave is neither even nor odd, the series will contain both sine and cosine terms. For n > 0, we have 1   V= !t cos n!t d !t
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