Higher Engineering Mathematics

688 FOURIER SERIES

Table 73.2

Ordinate θ i sinθ isinθ sin 3θ isin 3θ sin 5θ isin 5θ

y 1 30 2 0.5 1 12 0.5 1

y 2 60 7 0.866 6.06 00 −0.866 −6.06

y 3 90 10 110 − 1 − 10 110

y 4 120 7 0.866 6.06 00 −0.866 −6.06

y 5 150 2 0.5 1 12 0.5 1

y 6 180 0 000000

y 7 210 − 2 −0.5 1 − 12 −0.5 1

y 8 240 − 7 −0.866 6.06 00 0.866 −6.06

y 9 270 − 10 − 1101 − 10 − 110

y 10 300 − 7 −0.866 6.06 00 0.866 −6.06

y 11 330 − 2 −0.5 1 − 12 −0.5 1

y 12 360 0 000000

∑^12

k= 1

yksinθk= 48. 24

∑^12

k= 1

yksin 3θk=− 12

∑^12

k= 1

yksin 5θk=− 0. 24

Fourier series has only odd sine terms present, i.e.

i=b 1 sinθ+b 3 sin 3θ+b 5 sin 5θ+···

A proforma, similar to Table 73.1, but without
the ‘cosine terms’ columns and without the ‘even
sine terms’ columns is shown in Table 73.2 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 as shown in Fig. 73.5.
From equation (3), Section 73.2,

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

Thus the Fourier series for currentiis given by:

i= 8 .04 sinθ− 2 .00 sin 3θ− 0 .04 sin 5θ

Now try the following exercise.

Exercise 247 Further problems on a num-

erical method of harmonic analysis

1. Without performing calculations, state which
   harmonics will be present in the waveforms
   shown in Fig. 73.6.

[

(a) only odd cosine terms present

(b) only even sine terms present

]

f(t)

4

0

− 4

2 ππ−π− 2 π 4 π t

(a)

2 π

−π 0

π

y

10

− 10

(b)

x

Figure 73.6

2. Analyse the periodic waveform of displace-
   mentyagainst angleθin Fig. 73.7(a) into
   its constituent harmonics as far as and
   including the third harmonic, by taking 30◦
   intervals.
   ⎡

⎣

y= 9. 4 + 13 .2 cosθ− 24 .1 sinθ

+ 0 .92 cos 2θ− 0 .14 sin 2θ

+ 0 .83 cos 3θ+ 0 .67 sin 3θ

⎤

⎦