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