640 Higher Engineering Mathematics
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
(a n cos nx + b n sin nx)
gives: v = 17.67 + 69.66 cos θ − 6.50 cos 2θ
− 8.17 cos 3θ +· · · −46.42 sin θ
+ 4.91 sin 2θ + 9.17 sin 3θ +· · ·
(4)
Note that in equation (4), (−46.42 sin θ + 69.66 cosθ)
comprises the fundamental, (4.91 sin 2θ − 6.50 cos 2θ)
comprises the second harmonic and (9.17 sin 3θ −
8.17 cos 3θ) comprises the third harmonic.
It is shown in Chapter 17 that:
a sin ω t + b cos ω t = R sin(ω t + α)
where a = R cos α, b = R sin α, R =
√
a 2 + b 2 and
α = tan −1 b
a
For the fundamental, R =
(−46.42) 2 + (69.66) 2
= 83.71
If
a = R cos α, then cos α =
a
R
=
−46.42
83.71
which is negative,
and if b = R sin α, then sin α =
b
R
=
69.66
83.71
which is positive.
The only quadrant where cos α is negative and sin α is
positive is the second quadrant.
Hence α = tan
−1 b
a
= tan
−1 69.66
−46.42
= 123.68
◦ or 2.16 rad
Thus (−46.42 sin θ + 69.66 cosθ )
= 83.71 sin(θ + 2.16)
By a similar method it may be shown that the second
harmonic
(4.91 sin 2θ − 6.50 cos 2θ)= 8.15 sin(2θ − 0.92)
and the third harmonic
(9.17sin3θ −8.17cos3θ)=12.28sin(3θ −0.73)
Hence equation (4) may be re-written as:
v = 17.67 + 83.71 sin(θ + 2.16)
+ 8.15 sin(2θ − 0.92)
+ 12.28 sin(3θ − 0.73) volts
which is the form used in Chapter 25 with complex
waveforms.
Now try the following exercise
Exercise 234 Further problems on
numerical harmonic analysis
Determine the Fourier series to represent the periodic functions given by the tables of values in
Problems 1 to 3, up to and including the third harmonic and each coefficient correct to 2 decimal
places. Use 12 ordinates in each case.
1. Angle θ ◦
30 60 90 120 150 180
Displacement y 40 43 38 30 23 17
Angle θ
◦
210 240 270 300 330 360
Displacement y 11 9 10 13 21 32
⎡
⎣
y = 23.92 + 7.81 cos θ + 14.61 sin θ
+ 0.17 cos 2θ + 2.31 sin 2θ
− 0.33 cos 3θ + 0.50 sin 3θ
⎤
⎦
2. Angle θ ◦
0
30 60 90 120 150
Voltage v −5.0 −1.5 6.0 12.5 16.0 16.5
Angle θ ◦ 180 210 240 270 300 330
Voltage v 15.0 12.5 6.5 −4.0 −7.0 −7.5
⎡
⎣
v = 5.00 − 10.78 cos θ + 6.83 sin θ
− 1.96 cos 2θ + 0.80 sin 2θ
+ 0.58 cos 3θ − 1.08 sin 3θ
⎤
⎦
3.
Angle θ ◦ 30 60 90 120 150 180
Current i 0 −1.4 −1.8 −1.9 −1.8 −1.3
Angle θ ◦ 210 240 270 300 330 360
Current i 0 2.2 3.8 3.9 3.5 2.5
⎡
⎣
i = 0.64 + 1.58 cosθ − 2.73 sin θ
− 0.23 cos 2θ − 0.42 sin 2θ
+ 0.27 cos 3θ + 0.05 sin 3θ
⎤
⎦
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
(a n cos nx + b n sin nx)
gives: v = 17.67 + 69.66 cos θ − 6.50 cos 2θ
− 8.17 cos 3θ +· · · −46.42 sin θ
+ 4.91 sin 2θ + 9.17 sin 3θ +· · ·
(4)
Note that in equation (4), (−46.42 sin θ + 69.66 cosθ)
comprises the fundamental, (4.91 sin 2θ − 6.50 cos 2θ)
comprises the second harmonic and (9.17 sin 3θ −
8.17 cos 3θ) comprises the third harmonic.
It is shown in Chapter 17 that:
a sin ω t + b cos ω t = R sin(ω t + α)
where a = R cos α, b = R sin α, R =
√
a 2 + b 2 and
α = tan −1 b
a
For the fundamental, R =
(−46.42) 2 + (69.66) 2
= 83.71
If
a = R cos α, then cos α =
a
R
=
−46.42
83.71
which is negative,
and if b = R sin α, then sin α =
b
R
=
69.66
83.71
which is positive.
The only quadrant where cos α is negative and sin α is
positive is the second quadrant.
Hence α = tan
−1 b
a
= tan
−1 69.66
−46.42
= 123.68
◦ or 2.16 rad
Thus (−46.42 sin θ + 69.66 cosθ )
= 83.71 sin(θ + 2.16)
By a similar method it may be shown that the second
harmonic
(4.91 sin 2θ − 6.50 cos 2θ)= 8.15 sin(2θ − 0.92)
and the third harmonic
(9.17sin3θ −8.17cos3θ)=12.28sin(3θ −0.73)
Hence equation (4) may be re-written as:
v = 17.67 + 83.71 sin(θ + 2.16)
+ 8.15 sin(2θ − 0.92)
+ 12.28 sin(3θ − 0.73) volts
which is the form used in Chapter 25 with complex
waveforms.
Now try the following exercise
Exercise 234 Further problems on
numerical harmonic analysis
Determine the Fourier series to represent the periodic functions given by the tables of values in
Problems 1 to 3, up to and including the third harmonic and each coefficient correct to 2 decimal
places. Use 12 ordinates in each case.
1. Angle θ ◦
30 60 90 120 150 180
Displacement y 40 43 38 30 23 17
Angle θ
◦
210 240 270 300 330 360
Displacement y 11 9 10 13 21 32
⎡
⎣
y = 23.92 + 7.81 cos θ + 14.61 sin θ
+ 0.17 cos 2θ + 2.31 sin 2θ
− 0.33 cos 3θ + 0.50 sin 3θ
⎤
⎦
2. Angle θ ◦
0
30 60 90 120 150
Voltage v −5.0 −1.5 6.0 12.5 16.0 16.5
Angle θ ◦ 180 210 240 270 300 330
Voltage v 15.0 12.5 6.5 −4.0 −7.0 −7.5
⎡
⎣
v = 5.00 − 10.78 cos θ + 6.83 sin θ
− 1.96 cos 2θ + 0.80 sin 2θ
+ 0.58 cos 3θ − 1.08 sin 3θ
⎤
⎦
3.
Angle θ ◦ 30 60 90 120 150 180
Current i 0 −1.4 −1.8 −1.9 −1.8 −1.3
Angle θ ◦ 210 240 270 300 330 360
Current i 0 2.2 3.8 3.9 3.5 2.5
⎡
⎣
i = 0.64 + 1.58 cosθ − 2.73 sin θ
− 0.23 cos 2θ − 0.42 sin 2θ
+ 0.27 cos 3θ + 0.05 sin 3θ
⎤
⎦