homework 3 of modeling final
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Course: Model Analysis
and S stem Identification
SCHOOL OF
CIVILENGINEERING
Topic : Homework 3
Professor : LUO Shaoxiang
Student : Chinny Lyheang
Student’s ID : 15129120
Year : 2015-2016
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Model Analysis and System Identification Page 3
subplot(2,1,2) plot(f(1:N/2+1),real(Gxy)); hold off % co-spectrum hold off; xlabel('Frequency (Hz)'); ylabel('Real (Gxy)')
Gyy1=20*log10(sqrt(8/3)*abs(Gyy1(1:N/2+1)/fs/N)); Gyy2=20*log10(sqrt(8/3)*abs(Gyy2(1:N/2+1)/fs/N));
Graph of Signal 1 and Signal 2
According the graph above we compare the signal 1 and signal 2 we got 4 peaks of each
signal means that we got 4 frequencies of each signals
Signal 1
Peak 1
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Peak 1 has its amplitude 74.5712 y and its frequencies at 5.125 x Hz
Peak 2
Peak 2 has its amplitude 52.4889 y and its frequencies at 7.34375 x Hz
Peak 3
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Peak 3 has its amplitude 111.6818 y and its frequencies at 21.13 x Hz
Peak 4
Peak 4 has its amplitude 103.8797 y and its frequencies at 34.16 x Hz
Signal 2
Peak 1
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Peak 1 has its amplitude 58.0195 y and its frequencies at 5.125 x Hz
Peak 2
Peak 2 has its amplitude 48.8263 y and its frequencies at 7.34375 x Hz
Peak 3
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Peak 3 has its amplitude 99.5411 y and its frequencies at 21.13 x Hz
Peak 4
Peak 4 has its amplitude 102.8247 y and its frequencies at 34.16 x Hz
The reason why we take them as the natural frequencies because after we see the graph of each
signal and compare both signal together we can observe they have the peak (high amplitude) at
the same frequencies and for other peaks just only white noise or unnecessary noise so we can
ignore it that we take only 4 natural frequencies of each signal.
2. The modal damping at each selected natural frequency. (Use half-power bandwidth
method to get these damping values)
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Graph of Signal 1 and Signal 2
According the graph of Signal 1 and Signal 2 we got 4 peak and got 4 frequencies of each
signal and it means that we can find the Damping Ratio of each frequencies by using
half-power bandwidth method :
2
b a b a
b a r
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For Signal 1 we have:
Peak 1 we have its amplitude ( ) 74.5712r and 5.125r Hz
( ) 74.571252.73
2 2
r
74.5712 ( 52.7298) 21.8414
( ) ( ) 74.5712 21.8414 96.4126a r b r
So ( )a r , it ranges [-100.2623;-83.0595] & its frequency a range [5.03125;5.0625] and
( )b r ranges [-82.5596;-97.572] & its frequency b range [5.1875;5.21875]
By using interpolation, we got:
5.03125 96.4126 ( 100.2623)
5.0625 5.03125 83.0595 ( 100.2623)
a
96.4126 ( 100.2623)
5.0625 5.03125 5.03125 5.038283.0595 ( 100.2623)
a Hz
5.1875 96.4126 ( 82.5596)
5.21875 5.1875 97.572 ( 82.5596)
b
96.4126 ( 82.5596)
5.21875 5.1875 5.1875 5.216397.572 ( 82.5596)b
11
5.2163 5.03820.01737
2 2 5.125
b a b a
b a r
Peak 2 we have its amplitude ( ) 52.4889r and 7.34375r Hz
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( ) 52.488937.1153
2 2
r
52.4889 ( 37.1153) 15.3736
( ) ( ) 52.4889 15.3736 67.8625a r b r
So ( )a r , it ranges [-78.265; -66.4147] & its frequency a range [7.25;7.28125] and ( )b r
ranges [-57.7025; -71.3505] & its frequencyb
range [7.40625;7.4375]
By using interpolation, we got:
7.25 67.8625 ( 78.265)
7.28125 7.25 66.4147 ( 78.265)
a
67.8625 ( 78.265)
7.28125 7.25 7.25 7.277466.4147 ( 78.265)
a
7.40625 67.8625 ( 57.7025)
7.4375 7.40625 71.3505 ( 57.7025)
b
67.8625 ( 57.7025)
7.4375 7.40625 7.40625 7.429571.3505 ( 57.7025)
b
12
7.4295 7.2774
0.010352 2 7.34375
b a b a
b a r
Peak 3 we have its amplitude ( ) 111.6818r and 21.13r Hz
( ) 111.681878.9709
2 2
r
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111.6818 ( 78.9709) 32.7109
( ) ( ) 111.6818 32.7109 144.3927a r b r
So ( )a r , it ranges [-149.3354; -137.7442] & its frequency a range [20.9375;20.96875] and
( )b r ranges [-143.50196; -144.74281] & its frequency b range [21.40625;21.4375]
By using interpolation, we got:
20.9375 144.3927 ( 149.3354)
20.96875 20.9375 137.7442 ( 149.3354)
a
144.3927 ( 149.3354)
20.96875 20.9375 20.9375 20.9508137.7442 ( 149.3354)
a
21.40625 144.3927 ( 143.50196)
21.4375 21.40625 144.74281 ( 143.50196)
b
144.3927 ( 143.50196)
21.4375 21.40625 21.40625 21.4286144.74281 ( 143.50196)
b
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Model Analysis and System Identification Page 13
So ( )a r , it ranges [-135.1446; -129.9976] & its frequency a range [33.8125;33.84375] and
( )b r ranges [-132.98959; -134.92528] & its frequency b range [34.625;34.65625]
By using interpolation, we got:
33.8125 134.3054 ( 135.1446)33.84375 33.8125 129.9976 ( 135.1446)
a
134.3054 ( 135.1446)
33.84375 33.8125 33.8125 33.8176129.9976 ( 135.1446)
a
34.625 134.3054 ( 132.98959)
34.65625 34.625 134.92528 ( 132.98959)
b
134.3054 ( 132.98959)
34.65625 34.625 34.625 34.6462134.92528 ( 132.98959)
b
14
34.6462 33.8176 0.012132 2 34.15625
b a b a
b a r
For Signal 2 we have:
Peak 1 we have its amplitude ( ) 58.0195r and 5.125r Hz
( ) 58.019541.0259
2 2
r
58.0195 ( 41.0259) 16.9936
( ) ( ) 58.0195 16.9936 75.0131a r b r
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So ( )a r , it ranges [-83.0553; -65.8801] & its frequency a range [5.03125;5.0625] and
( )b r ranges [-67.9777; -84.5827] & its frequency b range [5.1875;5.21875]
By using interpolation, we got:
5.03125 75.0131 ( 83.0553)5.0625 5.03125 65.8801 ( 83.0553)
a
75.0131 ( 83.0553)
5.0625 5.03125 5.03125 5.045965.8801 ( 83.0553)
a Hz
5.1875 75.0131 ( 67.9777)
5.21875 5.1875 84.5827 ( 67.9777)
b
75.0131 ( 67.9777)
5.21875 5.1875 5.1875 5.200784.5827 ( 67.9777)
b
21
5.2007 5.0459 0.01512 2 5.125
b a b a
b a r
Peak 2 we have its amplitude ( ) 48.8263r and 7.34375r Hz
( ) 48.826334.5254
2 2
r
48.8263 ( 34.5254) 14.3009
( ) ( ) 48.8263 14.3009 63.1272a r b r
So ( )a r , it ranges [-63.2043; -54.6305] & its frequency a range [7.28125;7.3125] and
( )b r ranges [-53.9510; -67.5159] & its frequency b range [7.40625;7.4375]
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By using interpolation, we got:
7.28125 63.1272 ( 63.2043)
7.3125 7.28125 54.6305 ( 63.2043)
a
63.1272 ( 63.2043)
7.3125 7.28125 7.28125 7.281554.6305 ( 63.2043)a
7.40625 63.1272 ( 53.951)
7.4375 7.40625 67.5159 ( 53.951)
b
63.1272 ( 53.951)
7.4375 7.40625 7.40625 7.427367.5159 ( 53.951)
b
22
7.4273 7.28150.0099
2 2 7.34375
b a b a
b a r
Peak 3 we have its amplitude ( ) 99.5411r
and 21.13r Hz
( ) 99.541170.3862
2 2
r
99.5411 ( 70.3862) 29.1549
( ) ( ) 99.5411 29.1549 128.696a r b r
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So ( )a r , it ranges [-131.4669; -128.187] & its frequency a range [20.90625;20.9375] and
( )b r ranges [-126.9502; -130.5633] & its frequency b range [21.28125;21.3125]
By using interpolation, we got:
20.9625 128.696 ( 131.4669)
20.9375 20.9625 128.187 ( 131.4669)
a
128.696 ( 131.4669)
20.9375 20.9625 20.9625 20.9413128.187 ( 131.4669)
a
21.28125 128.696 ( 126.9502)
21.3125 21.28125 130.5633 ( 126.9502)
b
128.696 ( 126.9502)
21.3125 21.28125 21.28125 21.2963130.5633 ( 126.9502)
b
23
21.2963 20.94130.0084
2 2 21.13
b a b a
b a r
Peak 4 we have its amplitude ( ) 102.8247r and 34.16r Hz
( ) 102.824772.708
2 2
r
102.8247 ( 72.708) 30.1167
( ) ( ) 102.8247 30.1167 132.9414a r b r
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So ( )a r , it ranges [-135.5201; -130.0136] & its frequency a range [33.8125;33.84375] and
( )b r ranges [-132.54441; -135.76804] & its frequency b range [34.65625;34.6875]
By using interpolation, we got:
33.8125 132.9414 ( 135.5201)
33.84375 33.8125 130.0136 ( 135.5201)
a
132.9414 ( 135.5201)
33.84375 33.8125 33.8125 33.8271130.0136 ( 135.5201)
a
34.65625 132.9414 ( 132.54441)
34.6875 34.65625 135.76804 ( 132.54441)
b
132.9414 ( 132.54441)
34.6875 34.65625 34.65625 34.66135.76804 ( 132.54441)
b
24
34.66 33.8271
0.01222 2 34.16
b a b a
b a r
So the Damping Ratio of the both signals is:
11 21verage,peak1
0.01737 0.01510.016235
2 2a
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12 22verage,peak2
0.01035 0.00990.010125
2 2a
13 23verage,peak3
0.0113 0.00840.00985
2 2a
14 24verage,peak4
0.01213 0.01220.012165
2 2a
3. The phase angle between the two signals at the selected natural frequencies.
According to the graph above we can get the value of the amplitude and frequencies of
the peak. It means that we can obtain the value of the phase angle between the two signal.
For peak 1 : Amplitude = -66.3 , Frequency = 5.125 Hz and Phase Angle = -
1.0852 (rad)
For peak 2 : Amplitude = -50.66 , Frequency = 7.344 Hz and Phase Angle =
0.8817 (rad)
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Signal 1
Amplitude of Peak 1 = -74.57
Amplitude of Peak 2 =-52.49
Amplitude of Peak 3 = -111.7
Amplitude of Peak 4 = -103.9
Signal 2
Amplitude of Peak 1 = -58.02
Amplitude of Peak 2 =-48.83
Amplitude of Peak 3 = -99.54
Amplitude of Peak 4 = -102.8
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