Higher Engineering Mathematics

684 FOURIER SERIES

Mean value=

area

length of base

≈

1

2 π

(

2 π

p

)∑p

k= 1

yk≈

1

p

∑p

k= 1

yk

However,a 0 = mean value off(x) in the range

0to2π

Thus a 0 ≈

1

p

∑p

k= 1

yk (1)

Similarly,an=twice the mean value off(x) cosnx

in the range 0 to 2π,

thus an≈

2

p

∑p

k= 1

ykcosnxk (2)

andbn=twice the mean value off(x) sinnxin the

range 0 to 2π,

thus bn≈

2

p

∑p

k= 1

yksinnxk (3)

Problem 1. The values of the voltagevvolts at

different moments in a cycle are given by:

θ◦(degrees) V(volts)

30 62

60 35

90 − 38

120 − 64

150 − 63

180 − 52

210 − 28

240 24

270 80

300 96

330 90

360 70

Draw the graph of voltageVagainst angleθand

analyse the voltage into its first three constituent

harmonics, each coefficient correct to 2 decimal

places.

The graph of voltageVagainst angleθis shown in
Fig. 73.2. The range 0 to 2πis divided into 12 equal

intervals giving an interval width of

2 π

12

, i.e.

π

6

rad

or 30◦. The values of the ordinatesy 1 ,y 2 ,y 3 ,...are

62, 35,−38,...from the given table of values. If a

larger number of intervals are used, results having

y 1

y 2

y 3 y 4 y 5 y 6

y 7

y 8

y 9 y 11 y 12

y 10

270 360 degrees

90 180

80

60

40

20

0

− 20

− 40

− 60

− 80

Voltage (volts)

θ

Figure 73.2

a greater accuracy are achieved. The data is tabulated

in the proforma shown in Table 73.1, on page 685.

From equation (1),a 0 ≈

1

p

∑p

k= 1

yk=

1

12

(212)

= 17 .67 (sincep=12)

From equation (2),an≈

2

p

∑p

k= 1

ykcosnxk

hence a 1 ≈

2

12

(417.94)= 69. 66

a 2 ≈

2

12

(−39)=− 6. 50

and a 3 ≈

2

12

(−49)=− 8. 17

From equation (3),bn≈

2

p

∑p

k= 1

yksinnxk

hence b 1 ≈

2

12

(− 278 .53)=− 46. 42

b 2 ≈

2

12

(29.43)= 4. 91

and b 3 ≈

2

12

(55)= 9. 17

Substituting these values into the Fourier series:

f(x)=a 0 +

∑∞

n= 1

(ancosnx+bnsinnx)

gives: v= 17. 67 + 69 .66 cosθ− 6 .50 cos 2θ

− 8 .17 cos 3θ+··· − 46 .42 sinθ

+ 4 .91 sin 2θ+ 9 .17 sin 3θ+··· (4)