chapter 105 a numerical method of harmonic analysis AWS
CHAPTER 105 A NUMERICAL METHOD OF HARMONIC ANALYSIS EXERCISE 368 Page 1088
1. Determine the Fourier series to represent the periodic function given by the table of values, up to and including the third harmonic and each coefficient correct to 2 decimal places. Use 12 ordinates. Angle θ°
30 60
Displacement y 40
θ°
90 120 150 180 210 240 270 300 330 360
43 38 30
y
cos θ
y cos θ
30
40
0.866
34.64
60
43
0.5
90
38
0
120
30
–0.5
150
23
–0.866
180
17
–1
–17
0
210
11
–0.866
–9.53
–0.5
–5.5
240
9
–0.5
–4.5
270
10
0
300
13
0.5
330
21
0.866
18.19
360
32
1
32
0
12
∑y
12
∑y
k
k =1
= 287
k
sin θ
23 17
11
9
10
13 21
y sin θ
cos 2θ
y cos 2θ
sin 2θ
0.5
20
0.5
20
0.866
34.64
0
0
21.50
0.866
37.24
–0.5
–21.5
0.866
37.24
–1
–43
0
0
0
1
38
–1
–38
0
0
0
0
–1
–38
–15.0
0.866
25.98
–0.5
–15
–0.866
–25.98
1
30
0
0
–19.92
0.5
11.5
0.5
11.5
–0.866
–19.92
0
0
1
23
1
17
0
0
–1
–17
0
0
0.5
5.5
0.866
9.53
0
0
–1
–11
0
y sin 2θ
32
cos 3θ
ycos 3θ
sin 3θ
y cos 3θ 40
1
–0.866
–7.79
–0.5
–4.5
0.866
7.79
1
9
0
0
0
–1
–10
–1
–10
0
0
0
0
1
10
6.5
–0.866
–11.26
–0.5
–6.5
–0.866
–11.26
–1
–13
0
0
–0.5
–10.5
0.5
10.5
–0.866
–18.19
0
0
–1
–21
1
32
0
1
32
0
0
cos θ k
k =1
= 46.88
0
12
∑y
k
sin θ k
k =1
= 87.67
12
∑y
k
cos 2θ k
k =1
0
12
∑y
k
sin 2θ k
k =1
= 13.85
=1
12
∑y
k
cos 3θ k
k =1
12
∑y
k
sin 3θ k
k =1
= −2
=3
1 p 1 (287) = 23.92 a0 ≈ ∑ yk = 12 p k =1 2 p ∑ yk cos nxk p k =1
hence, a1 ≈
2 2 2 −0.33 (46.88) = 7.81 , a2 ≈ (1) = 0.17 , a3 ≈ (−2) = 12 12 12
2 p bn ≈ ∑ yk sin nxk p k =1
hence, b1 ≈
2 2 2 2.31 , b3 ≈ (3) = (87.67) = 14.61 , b2 ≈ (13.85) = 0.50 12 12 12
an ≈
∞
Substituting these values into the Fourier series:
f ( x) = a0 + ∑ ( an cos nx + bn sin nx ) n =1
gives:
y = 23.92 + 7.81 cos θ + 0.17 cos 2θ – 0.33 cos 3θ + … 1549
© 2014, John Bird
+ 14.61 sin θ + 2.31 sin 2θ + 0.50 sin 3θ 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
or
2. Determine the Fourier series to represent the periodic function given by the table of values, up to and including the third harmonic and each coefficient correct to 2 decimal places. Use 12 ordinates. Angle θ°
0
Voltage v
θ°
cosθ
30
60
90
120 150 180
210
240 270 300 330
–5.0 –1.5 6.0 12.5 16.0 16.5 15.0 12.5 6.5 –4.0
v cos θ
sin θ
v sin θ
cos 2θ
v cos 2θ
sin 2θ
v sin 2θ
–7.0 –7.5
cos 3θ
0
–5.0
1
–5.0
0
0
1
–5.0
0
0
1
v cos 3θ –5.0
30
–1.5
0.866
–1.299
0.5
–0.75
0.5
–0.75
0.866
–1.299
0
0
v
sin 3θ 0
v cos 3θ 0
1
–1.5
60
6.0
0.5
3.0
0.866
5.196
–0.5
–3.0
0.866
5.196
–1
–6.0
0
0
90
12.5
0
0
1
12.5
–1
–12.5
0
0
0
0
–1
–12.5
120
16.0
–0.5
–8
0.866
13.856
–0.5
–8.0
–0.866
–13.856
1
16.0
0
0
150
16.5
–0.866
–14.29
0.5
8.25
0.5
8.25
–0.866
–14.289
0
0
1
16.5
180
15.0
–1
–15
0
210
12.5
–0.866
–10.83
–0.5
–6.25
240
6.5
–0.5
–3.25
–0.866 –1
1
15.0
0.
0
–1
–15.0
0
0
0.5
6.25
0.866
10.825
0
0
–1
–12.5
–5.629
–0.5
–3.25
0.866
5.629
1
6.5
0
0
4
–1
4.0
0
0
0
0
1
–4.0
270 –4.0
0
300 –7.0
0.5
–3.5
–0.866
6.062
–0.5
3.5
–0.866
6.062
–1
7.0
0
0
330 –7.5
0.866
–6.50
–0.5
3.75
0.5
–3.75
–0866
6.495
0
0
–1
7.5
12
0
0
12
12
12
12
12
12
∑ vk
∑ vk cos θk
∑ vk sin θk
∑ vk cos 2θk
∑ vk sin 2θk
∑ vk cos 3θk
∑v
= 60
= − 64.669
= 40.985
= 0.75
= 4.763
= 3.5
= −6.5
a0 ≈
1 p 1 vk = (60) = 5.00 ∑ p k =1 12
k =1
k =1
2 p an ≈ ∑ vk cos nxk p k =1
k =1
hence, a1 ≈
a3 ≈
2 (3.5) = 0.58 12
bn ≈
2 p ∑ vk sin nxk p k =1
b3 ≈
2 (−6.5) = −1.08 12
hence, b1 ≈
k =1
k =1
k =1
k
sin 3θ k
k =1
2 2 (−64.669) = 0.13 , −10.78 , a2 ≈ (0.75) = 12 12
2 2 (40.985) = 6.83 , b2 ≈ (4.763) = 0.79 , 12 12
1550
© 2014, John Bird
∞
f ( x) = a0 + ∑ ( an cos nx + bn sin nx )
Substituting these values into the Fourier series:
n =1
y = 5.00 – 10.78 cos θ + 0.13 cos 2θ + 0.58 cos 3θ + …
gives:
+ 6.83 sin θ + 0.79 sin 2θ – 1.08 sin 3θ y = 5.00 – 10.78 cos θ + 6.83 sin θ + 0.13 cos 2θ + 0.79 sin 2θ + 0.58 cos 3θ – 1.08 sin 3θ
or
3. Determine the Fourier series to represent the periodic function given by the table of values, up to and including the third harmonic and each coefficient correct to 2 decimal places. Use 12 ordinates. Angle θ°
30
Current i
0
θ°
i
cosθ 0.866
30
0
60
–1.4
0.5
60
90 120 150 180
–1.4 –1.8 –1.9 –1.8
i cos θ 0 –0.7
sinθ 0.5 0.866
i sin θ 0
210
240
270
300
330
360
0
2.2
3.8
3.9
3.5
2.5
–1.3
cos 2θ
i cos 2θ
sin 2θ
i sin 2θ
cos 3θ
i cos 3θ
sin 3θ
i cos 3θ
0.5
0
0.866
0
0
0
1
0
–1.212
–0.5
0.7
0.866
–1.212
–1
1.4
0
0
90
–1.8
0
0
1
–1.8
–1
1.8
0
0
0
0
–1
1.8
120
–1.9
–0.5
0.95
0.866
–1.645
–0.5
0.95
–0.866
1.645
1
–1.9
0
0
–0.9
–1.8
150
–1.8
–0.866
1.56
0.5
0.5
–0.90
–0.866
1.548
0
0
1
180
–1.3
–1
1.3
0
0
1
–1.3
0
0
–1
1.3
0
0
210
0
–0.866
0
–0.5
0
0.5
0
0.866
0
0
0
–1
0
240
2.2
–0.5
–0.866
–1.905
–0.5
–1.1
0.866
1.905
1
2.2
0
0
270
3.8
0
0
–1
–3.8
–1
–3.8
0
0
0
0
1
3.8
300
3.9
0.5
1.95
–0.866
–3.377
–0.5
–1.95
–0.866
–3.377
–1
–3.9
0
0
330
3.5
0.866
3.03
–0.5
–1.75
0.5
1.75
–0.866
–3.031
0
0
–1
–3.5
360
2.5
1
2.5
0
1
2.5
0
0
1
2.5
0
0
12
12
–1.1
0
12
12
12
12
12
∑ yk
∑ yk cos θk
∑ yk sin θk
∑ yk cos 2θk
∑ yk sin 2θk
∑ yk cos 3θk
∑y
= 7.7
= 9.49
= −16.389
= −1.35
= −2.522
= 1.6
= 0.3
k =1
k =1
k =1
k =1
k =1
k =1
k
sin 3θ k
k =1
a0 ≈
1 p 1 yk = (7.7) = 0.64 ∑ p k =1 12
an ≈
2 p 2 2 2 yk cos nxk hence, a1 ≈ (9.49) = 1.58 , a2 ≈ (−1.35) = −0.23 , a3 ≈ (1.6) = 0.27 ∑ p k =1 12 12 12
bn ≈
2 p 2 2 yk sin nxk hence, b1 ≈ (−16.389) = −0.42 , −2.73 , b2 ≈ (−2.522) = ∑ p k =1 12 12 b3 ≈
2 (0.3) = 0.05 12
1551
© 2014, John Bird
∞
Substituting these values into the Fourier series:
f ( x) = a0 + ∑ ( an cos nx + bn sin nx ) n =1
gives:
i = 0.64 + 1.58 cos θ – 0.23 cos 2θ + 0.27 cos 3θ + … – 2.73 sin θ – 0.42 sin 2θ + 0.05 sin 3θ
or
i = 0.64 + 1.58 cos θ – 2.73 sin θ – 0.23 cos 2θ – 0.42 sin 2θ + 0.27 cos 3θ + 0.05 sin 3θ
1552
© 2014, John Bird
EXERCISE 369 Page 1092
1. Without performing calculations, state which harmonics will be present in the waveforms shown below.
(a) Only odd cosine terms are present. This is because the mean value is zero, the function is an even one since it is symmetrical about the vertical axis, and the positive and negative cycles are identical in shape (b) Only even sine terms are present. This is because the mean value is zero, the function is an odd one since it is symmetrical about the origin, and the waveform repeats after half a cycle 2. Analyse the periodic waveform below of displacement y against angle θ into its constituent harmonics, as far as and including the third harmonic, by taking 30° intervals.
1553
© 2014, John Bird
θ°
cos θ
y cos θ
sinθ
y sin θ
cos 2θ
y cos 2θ
sin 2θ
y sin 2θ
cos 3θ
y cos3θ
sin 3θ
30
10
0.866
8.66
0.5
0.5
5
0.866
8.66
0
0
1
y cos 3θ 10
60
–6
0.5
–3.0
0.866
–5.196
–0.5
3
0.866
–5.196
–1
6
0
0
y
5
90
–17
0
0
1
–17
–1
17
0
0
0
0
–1
17
120
–17
–0.5
8.5
0.866
–14.72
–0.5
8.5
–0.866
14.722
1
–17
0
0
150
–13
–0.866
0.5
–6.5
0.5
–6.5
–0.866
11.258
0
0
1
–13
180
–4
–1
1
–4
0
0
–1
4
0
0
210
10
–0.866
–8.66
–0.5
–5.0
0.5
5
0.866
8.66
0
0
–1
–10
240
24
–0.5
–12
–0.866
–20.78
–0.5
–12
0.866
20.784
1
24
0
0
270
33
0
0
–1
–33
–1
–33
0
0
0
0
1
33
300
36
0.5
18
–0.866
–31.18
–0.5
–18
–0.866
–31.176
–1
–36
0
0
330
33
0.866
28.58
–0.5
–16.5
0.5
16.5
–0.866
–28.578
0
0
–1
–33
360
24
1
24
0
1
24
0
1
24
0
0
12
11.26 4
12
0
0
0
12
12
0
12
12
12
∑ yk
∑ yk cos θk
∑ yk sin θk
∑ yk cos 2θk
∑ yk sin 2θk
∑ yk cos 3θk
∑y
= 113
= 79.34
= −144.88
= 5.5
= −0.866
=5
=4
k =1
k =1
k =1
k =1
k =1
k =1
k
sin 3θ k
k =1
1 p 1 (113) = 9.4 a0 ≈ ∑ yk = 12 p k =1 an ≈
2 p 2 2 2 yk cos nxk hence, a1 ≈ (79.34) = 13.2 , a2 ≈ (5.5) = 0.83 0.92 , a3 ≈ (5) = ∑ p k =1 12 12 12
bn ≈
2 p ∑ yk sin nxk p k =1
hence, b1 ≈
2 2 −0.14 , (−144.88) = −24.1 , b2 ≈ (−0.866) = 12 12
b3 ≈
2 (4) = 0.67 12 ∞
Substituting these values into the Fourier series:
f ( x) = a0 + ∑ ( an cos nx + bn sin nx ) n =1
gives:
y = 9.4 + 13.2 cos θ + 0.92 cos 2θ + 0.83 cos 3θ + … – 24.1 sin θ – 0.14 sin 2θ + 0.67 sin 3θ
or
y = 9.4 + 13.2 cos θ – 24.1 sin θ + 0.92 cos 2θ – 0.14 sin 2θ + 0.83 cos 3θ + 0.67 sin 3θ
3. For the waveform of current shown below, state why only a d.c. component and even cosine terms will appear in the Fourier series and determine the series, using π/6 rad intervals, up to and including the sixth harmonic.
1554
© 2014, John Bird
The function is even, thus no sine terms will be present The function repeats itself every half cycle, hence only even terms will be present Hence, the Fourier series will contain a d.c. component and even cosine terms only
θ° 30
1.5
60 90
i cos 2θ
cos 4θ
i cos 4θ
0.5
0.75
–0.5
–0.75
5.5
–0.5
–2.75
–0.5
–2.75
10
–1
–10
1
10
120
5.5
–0.5
–2.75
–0.5
150
1.5
0.5
0.75
180
0
1
0
210
1.5
0.5
240
5.5
–0.5
270
10
–1
–10
300
5.5
–0.5
–2.75
330
1.5
0.5
0.75
–0.5
360
0
1
0
1
12
∑ yk = 48 k =1
a0 ≈
cos 2θ
i
cos 6θ –1
i cos 6θ –1.5
1
5.5
–1
–10
–2.75
1
5.5
–0.5
–0.75
–1
–1.5
1
0
1
0
0.75
–0.5
–0.75
–1
–1.5
–2.75
–0.5
–2.75
1
5.5
1
10
–1
–10
–0.5
–2.75
1
5.5
–0.75
–1
–1.5
1
0
12
0
12
∑ yk cos 4θk = 6
∑ yk cos 2θk = −28
k =1
k =1
12
∑y
k
cos 6θ k = −4
k =1
1 p 1 yk = (48) =4.00 ∑ 12 p k =1
2 p 2 2 2 an ≈ ∑ yk cos nxk hence, a2 ≈ (−28) = −4.67 , a4 ≈ (6) = 1.00 , a3 ≈ (−4) = −0.66 p k =1 12 12 12 ∞
Substituting these values into the Fourier series:
f ( x= ) a0 + ∑ ( an cos nx ) n =1
gives:
i = 4.00 – 4.67 cos 2θ + 1.00 cos 4θ – 0.66 cos 6θ + …
1555
© 2014, John Bird
4. Determine the Fourier series as far as the third harmonic to represent the periodic function y given by the waveform shown. Take 12 intervals when analysing the waveform.
The following values are read from the graph (accuracy will depend on values read) Angle θ°
0
30
60 90
120 150 180 210 240 270 300
330
Voltage v
0
25
50
95
–78
θ°
cosθ
y cos θ
sin θ
75 y sin θ
90 cos 2θ
0 y cos 2θ
–20 –45 –75 sin 2θ
y sin 2θ
–95
cos 3θ
sin 3θ
0
0
1
0
0
1
0
0
0
1
y cos 3θ 0
30
25
0.866
21.65
0.5
12.5
0.5
12.5
0.866
21.65
0
0
60
50
0.5
25
0.866
43.3
–0.5
–25
0.866
43.3
–1
–50
0
0
90
75
0
0
1
75
–1
–75
0
0
0
0
–1
–75
120
95
–0.5
–47.5
0.866
82.27
–0.5
–47.5
–0.866
–82.27
1
95
0
0
150
90
–0.866
–77.94
0.5
45
0.5
45
–0.866
–77.94
0
0
1
90
180
0
–1
0
0
0
1
0
0
210
–20
–0.866
17.32
–0.5
10
0.5
–10.0
0.866
–17.32
240
–45
–0.5
22.5
–0.866
38.97
–0.5
22.5
0.866
270
–75
0
0
–1
75
–1
75
0
300
–95
0.5
–47.5
–0.866
82.27
–0.5
47.5
330
–78
0.866
–67.548
–0.5
39
0.5
–39.0
y
12
0
12
12
12
0
0
y cos 3θ 0
1
25
–1
0
0
0
0
0
–1
20
–38.97
1
–45
0
0
0
0
0
1
–75
–0.866
82.27
–1
95
0
0
–0866
67.548
0
0
–1
78
12
12
12
∑ yk
∑ yk cos θk
∑ yk sin θk
∑ yk cos 2θk
∑ yk sin 2θk
∑ yk cos 3θk
∑y
= 22
= − 154.018
= 503.31
=6
= −1.732
= 95
= 63
k =1
k =1
k =1
k =1
k =1
k =1
k
sin 3θ k
k =1
1 p 1 a0 ≈ ∑ yk = (22) =1.83 p k =1 12 an ≈
2 p ∑ yk cos nxk p k =1
hence, a1 ≈
2 2 2 (−154.018) = 1.0 , a3 ≈ (95) = 15.83 −25.67 , a2 ≈ (6) = 12 12 12
1556
© 2014, John Bird
bn ≈
2 p ∑ yk sin nxk p k =1
b3 ≈
2 (63) = 10.5 12
hence, b1 ≈
2 2 (503.31) = 83.89 , b2 ≈ (−1.732) = −0.29 , 12 12
∞
Substituting these values into the Fourier series:
f ( x) = a0 + ∑ ( an cos nx + bn sin nx ) n =1
gives:
y = 1.83 – 25.67 cos θ + 1.0 cos 2θ + 15.83 cos 3θ + … + 83.89 sin θ – 0.29 sin 2θ + 10.5 sin 3θ
or
y = 1.83 – 25.67 cos θ + 83.89 sin θ + 1.0 cos 2θ – 0.29 sin 2θ + 15.83 cos 3θ + 10.5 sin 3θ
1557
© 2014, John Bird
1. Determine the Fourier series to represent the periodic function given by the table of values, up to and including the third harmonic and each coefficient correct to 2 decimal places. Use 12 ordinates. Angle θ°
30 60
Displacement y 40
θ°
90 120 150 180 210 240 270 300 330 360
43 38 30
y
cos θ
y cos θ
30
40
0.866
34.64
60
43
0.5
90
38
0
120
30
–0.5
150
23
–0.866
180
17
–1
–17
0
210
11
–0.866
–9.53
–0.5
–5.5
240
9
–0.5
–4.5
270
10
0
300
13
0.5
330
21
0.866
18.19
360
32
1
32
0
12
∑y
12
∑y
k
k =1
= 287
k
sin θ
23 17
11
9
10
13 21
y sin θ
cos 2θ
y cos 2θ
sin 2θ
0.5
20
0.5
20
0.866
34.64
0
0
21.50
0.866
37.24
–0.5
–21.5
0.866
37.24
–1
–43
0
0
0
1
38
–1
–38
0
0
0
0
–1
–38
–15.0
0.866
25.98
–0.5
–15
–0.866
–25.98
1
30
0
0
–19.92
0.5
11.5
0.5
11.5
–0.866
–19.92
0
0
1
23
1
17
0
0
–1
–17
0
0
0.5
5.5
0.866
9.53
0
0
–1
–11
0
y sin 2θ
32
cos 3θ
ycos 3θ
sin 3θ
y cos 3θ 40
1
–0.866
–7.79
–0.5
–4.5
0.866
7.79
1
9
0
0
0
–1
–10
–1
–10
0
0
0
0
1
10
6.5
–0.866
–11.26
–0.5
–6.5
–0.866
–11.26
–1
–13
0
0
–0.5
–10.5
0.5
10.5
–0.866
–18.19
0
0
–1
–21
1
32
0
1
32
0
0
cos θ k
k =1
= 46.88
0
12
∑y
k
sin θ k
k =1
= 87.67
12
∑y
k
cos 2θ k
k =1
0
12
∑y
k
sin 2θ k
k =1
= 13.85
=1
12
∑y
k
cos 3θ k
k =1
12
∑y
k
sin 3θ k
k =1
= −2
=3
1 p 1 (287) = 23.92 a0 ≈ ∑ yk = 12 p k =1 2 p ∑ yk cos nxk p k =1
hence, a1 ≈
2 2 2 −0.33 (46.88) = 7.81 , a2 ≈ (1) = 0.17 , a3 ≈ (−2) = 12 12 12
2 p bn ≈ ∑ yk sin nxk p k =1
hence, b1 ≈
2 2 2 2.31 , b3 ≈ (3) = (87.67) = 14.61 , b2 ≈ (13.85) = 0.50 12 12 12
an ≈
∞
Substituting these values into the Fourier series:
f ( x) = a0 + ∑ ( an cos nx + bn sin nx ) n =1
gives:
y = 23.92 + 7.81 cos θ + 0.17 cos 2θ – 0.33 cos 3θ + … 1549
© 2014, John Bird
+ 14.61 sin θ + 2.31 sin 2θ + 0.50 sin 3θ 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
or
2. Determine the Fourier series to represent the periodic function given by the table of values, up to and including the third harmonic and each coefficient correct to 2 decimal places. Use 12 ordinates. Angle θ°
0
Voltage v
θ°
cosθ
30
60
90
120 150 180
210
240 270 300 330
–5.0 –1.5 6.0 12.5 16.0 16.5 15.0 12.5 6.5 –4.0
v cos θ
sin θ
v sin θ
cos 2θ
v cos 2θ
sin 2θ
v sin 2θ
–7.0 –7.5
cos 3θ
0
–5.0
1
–5.0
0
0
1
–5.0
0
0
1
v cos 3θ –5.0
30
–1.5
0.866
–1.299
0.5
–0.75
0.5
–0.75
0.866
–1.299
0
0
v
sin 3θ 0
v cos 3θ 0
1
–1.5
60
6.0
0.5
3.0
0.866
5.196
–0.5
–3.0
0.866
5.196
–1
–6.0
0
0
90
12.5
0
0
1
12.5
–1
–12.5
0
0
0
0
–1
–12.5
120
16.0
–0.5
–8
0.866
13.856
–0.5
–8.0
–0.866
–13.856
1
16.0
0
0
150
16.5
–0.866
–14.29
0.5
8.25
0.5
8.25
–0.866
–14.289
0
0
1
16.5
180
15.0
–1
–15
0
210
12.5
–0.866
–10.83
–0.5
–6.25
240
6.5
–0.5
–3.25
–0.866 –1
1
15.0
0.
0
–1
–15.0
0
0
0.5
6.25
0.866
10.825
0
0
–1
–12.5
–5.629
–0.5
–3.25
0.866
5.629
1
6.5
0
0
4
–1
4.0
0
0
0
0
1
–4.0
270 –4.0
0
300 –7.0
0.5
–3.5
–0.866
6.062
–0.5
3.5
–0.866
6.062
–1
7.0
0
0
330 –7.5
0.866
–6.50
–0.5
3.75
0.5
–3.75
–0866
6.495
0
0
–1
7.5
12
0
0
12
12
12
12
12
12
∑ vk
∑ vk cos θk
∑ vk sin θk
∑ vk cos 2θk
∑ vk sin 2θk
∑ vk cos 3θk
∑v
= 60
= − 64.669
= 40.985
= 0.75
= 4.763
= 3.5
= −6.5
a0 ≈
1 p 1 vk = (60) = 5.00 ∑ p k =1 12
k =1
k =1
2 p an ≈ ∑ vk cos nxk p k =1
k =1
hence, a1 ≈
a3 ≈
2 (3.5) = 0.58 12
bn ≈
2 p ∑ vk sin nxk p k =1
b3 ≈
2 (−6.5) = −1.08 12
hence, b1 ≈
k =1
k =1
k =1
k
sin 3θ k
k =1
2 2 (−64.669) = 0.13 , −10.78 , a2 ≈ (0.75) = 12 12
2 2 (40.985) = 6.83 , b2 ≈ (4.763) = 0.79 , 12 12
1550
© 2014, John Bird
∞
f ( x) = a0 + ∑ ( an cos nx + bn sin nx )
Substituting these values into the Fourier series:
n =1
y = 5.00 – 10.78 cos θ + 0.13 cos 2θ + 0.58 cos 3θ + …
gives:
+ 6.83 sin θ + 0.79 sin 2θ – 1.08 sin 3θ y = 5.00 – 10.78 cos θ + 6.83 sin θ + 0.13 cos 2θ + 0.79 sin 2θ + 0.58 cos 3θ – 1.08 sin 3θ
or
3. Determine the Fourier series to represent the periodic function given by the table of values, up to and including the third harmonic and each coefficient correct to 2 decimal places. Use 12 ordinates. Angle θ°
30
Current i
0
θ°
i
cosθ 0.866
30
0
60
–1.4
0.5
60
90 120 150 180
–1.4 –1.8 –1.9 –1.8
i cos θ 0 –0.7
sinθ 0.5 0.866
i sin θ 0
210
240
270
300
330
360
0
2.2
3.8
3.9
3.5
2.5
–1.3
cos 2θ
i cos 2θ
sin 2θ
i sin 2θ
cos 3θ
i cos 3θ
sin 3θ
i cos 3θ
0.5
0
0.866
0
0
0
1
0
–1.212
–0.5
0.7
0.866
–1.212
–1
1.4
0
0
90
–1.8
0
0
1
–1.8
–1
1.8
0
0
0
0
–1
1.8
120
–1.9
–0.5
0.95
0.866
–1.645
–0.5
0.95
–0.866
1.645
1
–1.9
0
0
–0.9
–1.8
150
–1.8
–0.866
1.56
0.5
0.5
–0.90
–0.866
1.548
0
0
1
180
–1.3
–1
1.3
0
0
1
–1.3
0
0
–1
1.3
0
0
210
0
–0.866
0
–0.5
0
0.5
0
0.866
0
0
0
–1
0
240
2.2
–0.5
–0.866
–1.905
–0.5
–1.1
0.866
1.905
1
2.2
0
0
270
3.8
0
0
–1
–3.8
–1
–3.8
0
0
0
0
1
3.8
300
3.9
0.5
1.95
–0.866
–3.377
–0.5
–1.95
–0.866
–3.377
–1
–3.9
0
0
330
3.5
0.866
3.03
–0.5
–1.75
0.5
1.75
–0.866
–3.031
0
0
–1
–3.5
360
2.5
1
2.5
0
1
2.5
0
0
1
2.5
0
0
12
12
–1.1
0
12
12
12
12
12
∑ yk
∑ yk cos θk
∑ yk sin θk
∑ yk cos 2θk
∑ yk sin 2θk
∑ yk cos 3θk
∑y
= 7.7
= 9.49
= −16.389
= −1.35
= −2.522
= 1.6
= 0.3
k =1
k =1
k =1
k =1
k =1
k =1
k
sin 3θ k
k =1
a0 ≈
1 p 1 yk = (7.7) = 0.64 ∑ p k =1 12
an ≈
2 p 2 2 2 yk cos nxk hence, a1 ≈ (9.49) = 1.58 , a2 ≈ (−1.35) = −0.23 , a3 ≈ (1.6) = 0.27 ∑ p k =1 12 12 12
bn ≈
2 p 2 2 yk sin nxk hence, b1 ≈ (−16.389) = −0.42 , −2.73 , b2 ≈ (−2.522) = ∑ p k =1 12 12 b3 ≈
2 (0.3) = 0.05 12
1551
© 2014, John Bird
∞
Substituting these values into the Fourier series:
f ( x) = a0 + ∑ ( an cos nx + bn sin nx ) n =1
gives:
i = 0.64 + 1.58 cos θ – 0.23 cos 2θ + 0.27 cos 3θ + … – 2.73 sin θ – 0.42 sin 2θ + 0.05 sin 3θ
or
i = 0.64 + 1.58 cos θ – 2.73 sin θ – 0.23 cos 2θ – 0.42 sin 2θ + 0.27 cos 3θ + 0.05 sin 3θ
1552
© 2014, John Bird
EXERCISE 369 Page 1092
1. Without performing calculations, state which harmonics will be present in the waveforms shown below.
(a) Only odd cosine terms are present. This is because the mean value is zero, the function is an even one since it is symmetrical about the vertical axis, and the positive and negative cycles are identical in shape (b) Only even sine terms are present. This is because the mean value is zero, the function is an odd one since it is symmetrical about the origin, and the waveform repeats after half a cycle 2. Analyse the periodic waveform below of displacement y against angle θ into its constituent harmonics, as far as and including the third harmonic, by taking 30° intervals.
1553
© 2014, John Bird
θ°
cos θ
y cos θ
sinθ
y sin θ
cos 2θ
y cos 2θ
sin 2θ
y sin 2θ
cos 3θ
y cos3θ
sin 3θ
30
10
0.866
8.66
0.5
0.5
5
0.866
8.66
0
0
1
y cos 3θ 10
60
–6
0.5
–3.0
0.866
–5.196
–0.5
3
0.866
–5.196
–1
6
0
0
y
5
90
–17
0
0
1
–17
–1
17
0
0
0
0
–1
17
120
–17
–0.5
8.5
0.866
–14.72
–0.5
8.5
–0.866
14.722
1
–17
0
0
150
–13
–0.866
0.5
–6.5
0.5
–6.5
–0.866
11.258
0
0
1
–13
180
–4
–1
1
–4
0
0
–1
4
0
0
210
10
–0.866
–8.66
–0.5
–5.0
0.5
5
0.866
8.66
0
0
–1
–10
240
24
–0.5
–12
–0.866
–20.78
–0.5
–12
0.866
20.784
1
24
0
0
270
33
0
0
–1
–33
–1
–33
0
0
0
0
1
33
300
36
0.5
18
–0.866
–31.18
–0.5
–18
–0.866
–31.176
–1
–36
0
0
330
33
0.866
28.58
–0.5
–16.5
0.5
16.5
–0.866
–28.578
0
0
–1
–33
360
24
1
24
0
1
24
0
1
24
0
0
12
11.26 4
12
0
0
0
12
12
0
12
12
12
∑ yk
∑ yk cos θk
∑ yk sin θk
∑ yk cos 2θk
∑ yk sin 2θk
∑ yk cos 3θk
∑y
= 113
= 79.34
= −144.88
= 5.5
= −0.866
=5
=4
k =1
k =1
k =1
k =1
k =1
k =1
k
sin 3θ k
k =1
1 p 1 (113) = 9.4 a0 ≈ ∑ yk = 12 p k =1 an ≈
2 p 2 2 2 yk cos nxk hence, a1 ≈ (79.34) = 13.2 , a2 ≈ (5.5) = 0.83 0.92 , a3 ≈ (5) = ∑ p k =1 12 12 12
bn ≈
2 p ∑ yk sin nxk p k =1
hence, b1 ≈
2 2 −0.14 , (−144.88) = −24.1 , b2 ≈ (−0.866) = 12 12
b3 ≈
2 (4) = 0.67 12 ∞
Substituting these values into the Fourier series:
f ( x) = a0 + ∑ ( an cos nx + bn sin nx ) n =1
gives:
y = 9.4 + 13.2 cos θ + 0.92 cos 2θ + 0.83 cos 3θ + … – 24.1 sin θ – 0.14 sin 2θ + 0.67 sin 3θ
or
y = 9.4 + 13.2 cos θ – 24.1 sin θ + 0.92 cos 2θ – 0.14 sin 2θ + 0.83 cos 3θ + 0.67 sin 3θ
3. For the waveform of current shown below, state why only a d.c. component and even cosine terms will appear in the Fourier series and determine the series, using π/6 rad intervals, up to and including the sixth harmonic.
1554
© 2014, John Bird
The function is even, thus no sine terms will be present The function repeats itself every half cycle, hence only even terms will be present Hence, the Fourier series will contain a d.c. component and even cosine terms only
θ° 30
1.5
60 90
i cos 2θ
cos 4θ
i cos 4θ
0.5
0.75
–0.5
–0.75
5.5
–0.5
–2.75
–0.5
–2.75
10
–1
–10
1
10
120
5.5
–0.5
–2.75
–0.5
150
1.5
0.5
0.75
180
0
1
0
210
1.5
0.5
240
5.5
–0.5
270
10
–1
–10
300
5.5
–0.5
–2.75
330
1.5
0.5
0.75
–0.5
360
0
1
0
1
12
∑ yk = 48 k =1
a0 ≈
cos 2θ
i
cos 6θ –1
i cos 6θ –1.5
1
5.5
–1
–10
–2.75
1
5.5
–0.5
–0.75
–1
–1.5
1
0
1
0
0.75
–0.5
–0.75
–1
–1.5
–2.75
–0.5
–2.75
1
5.5
1
10
–1
–10
–0.5
–2.75
1
5.5
–0.75
–1
–1.5
1
0
12
0
12
∑ yk cos 4θk = 6
∑ yk cos 2θk = −28
k =1
k =1
12
∑y
k
cos 6θ k = −4
k =1
1 p 1 yk = (48) =4.00 ∑ 12 p k =1
2 p 2 2 2 an ≈ ∑ yk cos nxk hence, a2 ≈ (−28) = −4.67 , a4 ≈ (6) = 1.00 , a3 ≈ (−4) = −0.66 p k =1 12 12 12 ∞
Substituting these values into the Fourier series:
f ( x= ) a0 + ∑ ( an cos nx ) n =1
gives:
i = 4.00 – 4.67 cos 2θ + 1.00 cos 4θ – 0.66 cos 6θ + …
1555
© 2014, John Bird
4. Determine the Fourier series as far as the third harmonic to represent the periodic function y given by the waveform shown. Take 12 intervals when analysing the waveform.
The following values are read from the graph (accuracy will depend on values read) Angle θ°
0
30
60 90
120 150 180 210 240 270 300
330
Voltage v
0
25
50
95
–78
θ°
cosθ
y cos θ
sin θ
75 y sin θ
90 cos 2θ
0 y cos 2θ
–20 –45 –75 sin 2θ
y sin 2θ
–95
cos 3θ
sin 3θ
0
0
1
0
0
1
0
0
0
1
y cos 3θ 0
30
25
0.866
21.65
0.5
12.5
0.5
12.5
0.866
21.65
0
0
60
50
0.5
25
0.866
43.3
–0.5
–25
0.866
43.3
–1
–50
0
0
90
75
0
0
1
75
–1
–75
0
0
0
0
–1
–75
120
95
–0.5
–47.5
0.866
82.27
–0.5
–47.5
–0.866
–82.27
1
95
0
0
150
90
–0.866
–77.94
0.5
45
0.5
45
–0.866
–77.94
0
0
1
90
180
0
–1
0
0
0
1
0
0
210
–20
–0.866
17.32
–0.5
10
0.5
–10.0
0.866
–17.32
240
–45
–0.5
22.5
–0.866
38.97
–0.5
22.5
0.866
270
–75
0
0
–1
75
–1
75
0
300
–95
0.5
–47.5
–0.866
82.27
–0.5
47.5
330
–78
0.866
–67.548
–0.5
39
0.5
–39.0
y
12
0
12
12
12
0
0
y cos 3θ 0
1
25
–1
0
0
0
0
0
–1
20
–38.97
1
–45
0
0
0
0
0
1
–75
–0.866
82.27
–1
95
0
0
–0866
67.548
0
0
–1
78
12
12
12
∑ yk
∑ yk cos θk
∑ yk sin θk
∑ yk cos 2θk
∑ yk sin 2θk
∑ yk cos 3θk
∑y
= 22
= − 154.018
= 503.31
=6
= −1.732
= 95
= 63
k =1
k =1
k =1
k =1
k =1
k =1
k
sin 3θ k
k =1
1 p 1 a0 ≈ ∑ yk = (22) =1.83 p k =1 12 an ≈
2 p ∑ yk cos nxk p k =1
hence, a1 ≈
2 2 2 (−154.018) = 1.0 , a3 ≈ (95) = 15.83 −25.67 , a2 ≈ (6) = 12 12 12
1556
© 2014, John Bird
bn ≈
2 p ∑ yk sin nxk p k =1
b3 ≈
2 (63) = 10.5 12
hence, b1 ≈
2 2 (503.31) = 83.89 , b2 ≈ (−1.732) = −0.29 , 12 12
∞
Substituting these values into the Fourier series:
f ( x) = a0 + ∑ ( an cos nx + bn sin nx ) n =1
gives:
y = 1.83 – 25.67 cos θ + 1.0 cos 2θ + 15.83 cos 3θ + … + 83.89 sin θ – 0.29 sin 2θ + 10.5 sin 3θ
or
y = 1.83 – 25.67 cos θ + 83.89 sin θ + 1.0 cos 2θ – 0.29 sin 2θ + 15.83 cos 3θ + 10.5 sin 3θ
1557
© 2014, John Bird
