638 Higher Engineering Mathematics
interval is thus
2π
p
. Let the ordinates be labelled y 0 ,
y 1 , y 2 , ... y p (note that y 0 = y p ). The trapezoidal rule
states:
Area = (width of interval)
1
2
(first + last ordinate)
+ sum of remaining ordinates
≈
2π
p
1
2
(y 0 + y p ) + y 1 + y 2 + y 3 + · · ·
Since y 0 = y p , then
1
2
(y 0 + y p ) = y 0 = y p
Hence area ≈
2π
p
p
k=1
y k
Mean value =
area
length of base
≈
1
2π
2π
p
p
k=1
y k ≈
1
p
p
k=1
y k
However, a 0 = mean value of f (x) in the range 0 to 2π.
Thus a 0 ≈
1
p
p
k=1
y k
(1)
Similarly, a n = twice the mean value of f (x) cos nx in
the range 0 to 2π,
thus a n ≈
2
p
p
k=1
y k cos nx k
(2)
and b n = twice the mean value of f (x) sin nx in the
range 0 to 2π,
thus b n ≈
2
p
p
k=1
y k sinnx k
(3)
Problem 1. The values of the voltage v volts 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 voltage V against angle θ and
analyse the voltage into its first three constituent
harmonics, each coefficient correct to 2 decimal
places.
80
90
180
360
270
y 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 voltage V against angle θ is shown in
Fig. 70.2. The range 0 to 2π is divided into 12 equal
intervals giving an interval width of
2π
12
, i.e.
π
6
rad
or 30 ◦ . The values of the ordinates y 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 tabulated in the proforma shown in Table 70.1, on
page 639.
From equation (1), a 0 ≈
1
p
p
k=1
y k =
1
12
(212)
= 17.67 (since p = 12)
From equation (2), a n ≈
2
p
p
k=1
y k cos nx k
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), b n ≈
2
p
p
k=1
y k sin nx k
hence
b 1 ≈
2
12
(−278.53) = −46.42
interval is thus
2π
p
. Let the ordinates be labelled y 0 ,
y 1 , y 2 , ... y p (note that y 0 = y p ). The trapezoidal rule
states:
Area = (width of interval)
1
2
(first + last ordinate)
+ sum of remaining ordinates
≈
2π
p
1
2
(y 0 + y p ) + y 1 + y 2 + y 3 + · · ·
Since y 0 = y p , then
1
2
(y 0 + y p ) = y 0 = y p
Hence area ≈
2π
p
p
k=1
y k
Mean value =
area
length of base
≈
1
2π
2π
p
p
k=1
y k ≈
1
p
p
k=1
y k
However, a 0 = mean value of f (x) in the range 0 to 2π.
Thus a 0 ≈
1
p
p
k=1
y k
(1)
Similarly, a n = twice the mean value of f (x) cos nx in
the range 0 to 2π,
thus a n ≈
2
p
p
k=1
y k cos nx k
(2)
and b n = twice the mean value of f (x) sin nx in the
range 0 to 2π,
thus b n ≈
2
p
p
k=1
y k sinnx k
(3)
Problem 1. The values of the voltage v volts 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 voltage V against angle θ and
analyse the voltage into its first three constituent
harmonics, each coefficient correct to 2 decimal
places.
80
90
180
360
270
y 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 voltage V against angle θ is shown in
Fig. 70.2. The range 0 to 2π is divided into 12 equal
intervals giving an interval width of
2π
12
, i.e.
π
6
rad
or 30 ◦ . The values of the ordinates y 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 tabulated in the proforma shown in Table 70.1, on
page 639.
From equation (1), a 0 ≈
1
p
p
k=1
y k =
1
12
(212)
= 17.67 (since p = 12)
From equation (2), a n ≈
2
p
p
k=1
y k cos nx k
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), b n ≈
2
p
p
k=1
y k sin nx k
hence
b 1 ≈
2
12
(−278.53) = −46.42