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  V 1 !t V sin n!t 2 2 cos n 1 cos n!t n 2 n2  n 0
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FOURIER METHOD OF WAVEFORM ANALYSIS
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When n is even, cos n 1 0 and an 0. When n is odd, an 2V= 2 n2 . bn 1   V= !t sin n!t d !t
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The bn coe cients are
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  V 1 !t V V cos n!t cos n 1 n 1 sin n!t n n n 2 n2 0
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Then the required Fourier series is f t V 2V 2V 2V cos 3!t cos 5!t 2 cos !t 4  3 2 5 2 V V V sin 2!t sin 3!t sin !t  2 3
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The even-harmonic amplitudes are given directly by jbn j, since therep are no even-harmonic cosine terms. However, the odd-harmonic amplitudes must be computed using cn a2 b2 . Thus, n n c1 q 2V=2 2 V= 2 V 0:377 c3 V 0:109 c5 V 0:064
The line spectrum is shown in Fig. 17-25.
Find the trigonometric Fourier series for the half-wave-recti ed sine wave shown in Fig. 17-26 and sketch the line spectrum.
The wave shows no symmetry, and we therefore expect the series to contain both sine and cosine terms. Since the average value is not obtainable by inspection, we evaluate a0 for use in the term a0 =2. 1  V 2V V sin !t d !t cos !t  a0 0  0   Next we determine an : an 1  V sin !t cos n!t d !t  0   V n sin !t sin n!t cos n!t cos !t  V cos n 1  n2 1  1 n2 0
With n even, an 2V= 1 n2 ; and with n odd, an 0. However, this expression is indeterminate for n 1, and therefore we must integrate separately for a1 . 1  V 1 V sin !t cos !t d !t sin 2!t d !t 0 a1  0  02 Now we evaluate bn : bn 1   V sin !t sin n!t d !t
  V n sin !t cos n!t sin n!t cos !t  0  n2 1 0
FOURIER METHOD OF WAVEFORM ANALYSIS
[CHAP. 17
Here again the expression is indeterminate for n 1, and b1 is evaluated separately.   1  V !t sin 2!t  V V sin2 !t d !t b1  0  2 4 2 0 Then the required Fourier series is   V  2 2 2 1 sin !t cos 2!t cos 4!t cos 6!t f t  2 3 15 35 The spectrum, Fig. 17-27, shows the strong fundamental term in the series and the rapidly decreasing amplitudes of the higher harmonics.
Fig. 17-26
Fig. 17-27
Find the trigonometric Fourier series for the half-wave-recti ed sine wave shown in Fig. 17-28, where the vertical axis is shifted from its position in Fig. 17-26.
The function is described in the interval  < !t < 0 by f t V sin !t. The average value is the same as that in Problem 17.5, that is, 1 a0 V=. For the coe cients an , we have 2 an 1  0 V sin !t cos n!t d !t
V 1 cos n  1 n2
Fig. 17-28 For n even, an 2V= 1 n2 ; and for n odd, an 0, except that n 1 must be examined separately. 1 0 a1 V sin !t cos !t d !t 0   For the coe cients bn , we obtain bn except for n 1. b1 Thus, the series is f t 1  1  0 V sin !t sin n!t d !t 0
V sin2 !t d !t
  V  2 2 2 1 sin !t cos 2!t cos 4!t cos 6!t  2 3 15 35
CHAP. 17]
FOURIER METHOD OF WAVEFORM ANALYSIS
This series is identical to that of Problem 17.5, except for the fundamental term, which has a negative coe cient in this series. The spectrum would obviously be identical to that of Fig. 17-27. Another Method When the sine wave V sin !t is subtracted from the graph of Fig. 17.26, the graph of Fig. 17-28 results.
Obtain the trigonometric Fourier series for the repeating rectangular pulse shown in Fig. 17-29 and plot the line spectrum.
Fig. 17-29
Fig. 17-30
With the vertical axis positioned as shown, the wave is even and the series will contain only cosine terms and a constant term. In the period from  to  used for the evaluation integrals, the function is zero except from =6 to =6. 1 =6 V 1 =6 2V n an sin a0 V d !t V cos n!t d !t  =6 3  =6 n 6 p p Since sin n=6 1=2, 3=2; 1; 3=2; 1=2; 0; 1=2; . . . for n 1; 2; 3; 4; 5; 6; 7; . . . , respectively, the series is " p   p     V 2V 1 1 3 1 3 1 f t cos !t cos 2!t 1 cos 3!t cos 4!t 6  2 3 2 2 2 4 #     1 1 1 1 cos 5!t cos 7!t 2 5 2 7 or f t
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