Transcript Vibrating Beam Modeling Results Prime (Group 7) Abby, Jacob, TJ, Leo

Vibrating Beam Modeling
Results
Prime (Group 7)
Abby, Jacob, TJ, Leo
The model and data
We are attempting to use the ordinary
differential equation model:
d 2 y (t )
dy (t )
C
 Ky(t )  0
2
d (t )
dt
c
k
C , K
m
m
This is assuming that the beam behaves
like an under damped harmonic
oscillator
-5
6
Two-parameter model estimation
x 10
experimental data
model displacement
4
displacement
2
0
-2
-4
-6
-8
1
1.5
2
2.5
C = 0.68439333
standard error (C) = 0.00897737
95 % confidence interval :
( 0.66643859, 0.70234806 )
3
3.5
time (s)
4
4.5
5
5.5
6
K = 1525.693399
standard error (K) = 0.350108
95% confidence interval :
( 1524.993183, 1526.393614 )
σ2 = 1.0559e-010
-5
Two-parameter model estimation
x 10
experimental data
model displacement
4
displacement
2
0
-2
-4
-6
1.25
1.3
1.35
1.4
1.45
time (s)
1.5
1.55
1.6
1.65
• notice the model only appears to estimate one frequency of
the three that the data appears to contain.
The Optimization Algorithm
The data vector is massaged by truncating
at the max and adding the average back
into it 5000 times
A loop using fminsearch adds random
vectors to C and K. The values that
produce the lowest least-squares cost are
kept.
-5
6
Experimental Data and Model
x 10
4
displacement (m)
2
0
-2
-4
-6
-8
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
time (s)
C1 = -0.64450000
standard error (C1) = 0.00908483
95% confidence interval :
( -0.66266967, -0.62633033 )
K1 = -1530.600000
standard error (K1) = 0.354825
95% confidence interval :
(-1531.309651, -1529.890349 )
σ12 = 4.056e-011
compared to
cost1=1.9064e-007
σ2 = 1.0559e-010
cost = 5.1366e-007
Residual Plots
Old residual plot
New residual plot
•Ideally the plot should be random about zero
•The occurrence of the diagonal pattern implies that
there may be a better fit model
QQ Distribution
-5
6
-5
QQ Plot of Sample Data versus Standard Normal
x 10
4
QQ Plot of Sample Data versus Standard Normal
x 10
3
4
2
Quantiles of Input Sample
Quantiles of Input Sample
2
0
-2
1
0
-1
-4
-2
-6
-8
-4
-3
-3
-2
-1
0
1
Standard Normal Quantiles
First QQ plot
2
3
4
-4
-4
-3
-2
-1
0
1
2
Standard Normal Quantiles
Second QQ plot
The QQ or normal probability plot shows that the function doesn’t
have a normal distribution.
3
4
Diagnostic Plots
First C, K values
Second C, K values
•Exhibits non-constant variance
The ODE Model
The model appears to generally mimic the
behavior of the system
It seems to only capture one out of three
frequencies displayed
It shows residuals that are not evenly
distributed or random about zero
An improved model should be attempted
PDE Model
Next, we attempted to fit the PDE model to the
data
y 2
y
 2
0  t 2 dx  0  t dx  0 x 2 Mdx  0 fdx
l
l
l
l

2 y
3 y 
M   yI ( x) 2  cI ( x) 2 
x
 xt 

where   a function which satisfies the bc' s
dy
M
y (t ,0)  0,
(t ,0)  0,
(t , l )  0, M(t , l )  0
dx
x
The PDE Model
Initial parameter guesses
-4
1
x 10
Displacement (m)
Model
Data
0
-1
0
0.5
1
1.5
2
2.5
Time (s)
3
3.5
4
4.5
PDE Model
Initial parameter guess
-5
x 10
Model
Data
8
6
Displacement (m)
4
2
0
-2
-4
-6
-8
0.3
0.35
0.4
0.45
Time (s)
0.5
0.55
•The model catches the first two, but misses the third
frequency
0.6
PDE Model
Second Parameter Guess
-5
10
x 10
Model
Data
8
6
Displacement (m)
4
2
0
-2
-4
-6
-8
0
0.5
1
1.5
2
2.5
Time (s)
3
3.5
4
4.5
5
PDE Model
Second Parameter Guess
-5
x 10
Model
Data
6
4
Displacement (m)
2
0
-2
-4
-6
-8
0.08
0.1
0.12
0.14
0.16
0.18
Time (s)
0.2
0.22
0.24
0.26
0.28
Further Improvements
Optimize the parameter selection
Use statistical analysis for the PDE model
of our data
Attempt alternate statistical analysis
techniques
Collect multiple data sets and improve
laboratory settings
Research current literature for more
accurate models