CHAP. 17]

437

FOURIER METHOD OF WAVEFORM ANALYSIS

When n is even, cos n  1 ¼ 0 and an ¼ 0. When n is odd, an ¼ 2V=ð2 n2 Þ. bn ¼

1 

ð ðV=Þ!t sin n!t dð!tÞ ¼ 0

The bn coeﬃcients are

  V 1 !t V V cos n!t ðcos nÞ ¼ ð1Þnþ1 sin n!t  ¼ n n n 2 n2 0

Then the required Fourier series is f ðtÞ ¼

V 2V 2V 2V cos 3!t  cos 5!t      2 cos !t  4  ð3Þ2 ð5Þ2 V V V sin 2!t þ sin 3!t     þ sin !t   2 3

Fig. 17-24

Fig. 17-25

The even-harmonic amplitudes are given directly by jbn j, since therepare no even-harmonic cosine terms. ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ However, the odd-harmonic amplitudes must be computed using cn ¼ a2n þ b2n . Thus, c1 ¼

qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ð2V=2 Þ2 þ ðV=Þ2 ¼ Vð0:377Þ

c3 ¼ Vð0:109Þ

c5 ¼ Vð0:064Þ

The line spectrum is shown in Fig. 17-25.

17.5

Find the trigonometric Fourier series for the half-wave-rectiﬁed 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. ð 1  V 2V V sin !t dð!tÞ ¼ ½ cos !t0 ¼ a0 ¼  0   Next we determine an : ð 1  V sin !t cos n!t dð!tÞ  0   V n sin !t sin n!t  cos n!t cos !t  V ¼ ðcos n þ 1Þ ¼  n2 þ 1 ð1  n2 Þ 0

an ¼

With n even, an ¼ 2V=ð1  n2 Þ; and with n odd, an ¼ 0. However, this expression is indeterminate for n ¼ 1, and therefore we must integrate separately for a1 . ð ð 1  V 1 V sin !t cos !t dð!tÞ ¼ sin 2!t dð!tÞ ¼ 0 a1 ¼  0  02 Now we evaluate bn : bn ¼

1 

ð V sin !t sin n!t dð!tÞ ¼ 0

  V n sin !t cos n!t  sin n!t cos !t  ¼0  n2 þ 1 0

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An