Higher Engineering Mathematics, Sixth Edition

638 Higher Engineering Mathematics

interval is thus

2 π

p

. Let the ordinates be labelledy 0 ,
y 1 ,y 2 ,...yp(note thaty 0 =yp). The trapezoidal rule
states:
Area=(width of interval)

[

1

2

(first+last ordinate)

+sum of remaining ordinates

]

≈

2 π

p

[

1

2

(y 0 +yp)+y 1 +y 2 +y 3 +···

]

Sincey 0 =yp,then

1

2

(y 0 +yp)=y 0 =yp

Hence area≈

2 π

p

∑p

k= 1

yk

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)intherange0 to 2π.

Thus a 0 ≈

1

p

∑p

k= 1

yk (1)

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

therange0to2π,

thus an≈

2

p

∑p

k= 1

ykcosnxk (2)

andbn=twice the mean value of f(x)sinnxin the

range0to2π,

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

θ◦(degrees) V(volts)

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.

80

90 180

y^270360

7

y 1

y 2

y 3 y 4 y 5 y 6

y 8

y 9 y 11 y 12

y 10

 degrees

Voltage (volts)

60

40

20

220

240

260

280

0

Figure 70.2

The graph of voltageV against angleθis shown in

Fig. 70.2. 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

a greater accuracy are achieved. The data is tab-

ulated in the proforma shown in Table 70.1, on

page 639.

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