Higher Engineering Mathematics, Sixth Edition

642 Higher Engineering Mathematics

2180 2120 260

290 230 30 60 90 120 150

180 360

y 5

y (^7) y
8 y^9
y 10
y 11
y 1 y 2 y 3 y 4 240 300
 8
25
5
10
i
210
2150 0 210 270 330
Figure 70.5
With referenceto Fig.70.5,thefollowingcharacteristics
are noted:
(i) The mean value is zero since the area above the
θ axis is equal to the area below it. Thus the
constant term, or d.c. component,a 0 =0.
(ii) Since the waveform is symmetrical about the ori-
gin the functioniis odd, which means that there
are no cosine terms present in the Fourier series.
(iii) The waveform is of the formf(θ )=−f(θ+π)
which means that only odd harmonics are
present.
Investigating waveform characteristics has thus saved
unnecessary calculations and in this case the Fourier
series has only odd sine terms present, i.e.
i=b 1 sinθ+b 3 sin3θ+b 5 sin5θ+···
A proforma, similar to Table 70.1, but without the
‘cosine terms’ columns and without the ‘even sine
terms’ columns is shown in Table 70.2 up to, and
including, the fifth harmonic, from which the Fourier
coefficientsb 1 ,b 3 andb 5 can be determined. Twelve
co-ordinates are chosen and labelledy 1 ,y 2 ,y 3 ,...y 12
asshowninFig.70.5
From equation (3), Section 70.2,
bn=
2
p
∑p
k= 1
iksinnθk,wherep= 12
Hence b 1 ≈
2
12
( 48. 24 )= 8. 04 ,
b 3 ≈
2
12
(− 12 )=− 2. 00 ,
and b 5 ≈
2
12
(− 0. 24 )=− 0. 04
Table 70.2
Ordinate θ i sinθ isinθ sin3θ isin3θ sin5θ isin5θ
y 1 30 2 0.5 1 1 2 0.5 1
y 2 60 7 0.866 6.06 0 0 −0.866 −6.06
y 3 90 10 1 10 − 1 − 10 1 10
y 4 120 7 0.866 6.06 0 0 −0.866 −6.06
y 5 150 2 0.5 1 1 2 0.5 1
y 6 180 0 0 0 0 0 0 0
y 7 210 − 2 −0.5 1 − 1 2 −0.5 1
y 8 240 − 7 −0.866 6.06 0 0 0.866 −6.06
y 9 270 − 10 − 1 10 1 − 10 − 1 10
y 10 300 − 7 −0.866 6.06 0 0 0.866 −6.06
y 11 330 − 2 −0.5 1 − 1 2 −0.5 1
y 12 360 0 0 0 0 0 0 0
∑^12
k= 1
yksinθk= 48. 24
∑^12
k= 1
yksin3θk=− 12
∑^12
k= 1
yksin5θk=− 0. 24