homework 3 of modeling final

8/18/2019 Homework 3 of Modeling Final

1/21

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


2/21


3/21


4/21

Civil Engineering School

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


5/21



Peak 1 has its amplitude 74.5712 y and its frequencies at 5.125 x Hz

Peak 2


Peak 3


6/21




Peak 4


Signal 2

Peak 1


7/21




Peak 2


Peak 3


8/21




Peak 4


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)


9/21



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


10/21



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



11/21



( ) 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]


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


( ) 111.681878.9709

2 2

r


12/21



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]


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


13/21


14/21






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:


( ) 58.019541.0259

2 2

r

58.0195 ( 41.0259) 16.9936

( ) ( ) 58.0195 16.9936 75.0131a r b r


15/21






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


( ) 48.826334.5254

2 2

r

48.8263 ( 34.5254) 14.3009

( ) ( ) 48.8263 14.3009 63.1272a r b r




16/21




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


17/21






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


( ) 102.824772.708

2 2

r

102.8247 ( 72.708) 30.1167

( ) ( ) 102.8247 30.1167 132.9414a r b r


18/21






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


19/21



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)


20/21


21/21



Signal 1

Amplitude of Peak 1 = -74.57

Amplitude of Peak 2 =-52.49



Signal 2


Amplitude of Peak 2 =-48.83



homework 3 of modeling final

Documents