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Transcript of Homework 9
President University Erwin Sitompul SMI 10/1
Lecture 10
System Modeling and Identification
Dr.-Ing. Erwin SitompulPresident University
http://zitompul.wordpress.com
2 0 1 4
President University Erwin Sitompul SMI 10/2
Homework 9Chapter 6 Identification from Step Response
Time Percent Value MethodDetermine the approximation of the model in the last example, if after examining the t/t table, the model order is chosen to be 4 instead of 5.
President University Erwin Sitompul SMI 10/3
t/τ Table
5 values of ti/τ are to be located for n = 4
10 1.8t
30 2.7t
50 3.6t
70 4.8t
90 6.7t
405.97
6.7
296.04
4.8
23.56.53
3.6
186.67
2.7
126.67
1.8
avg
(5.97 6.04 6.53 6.67 6.67)6.38
5
4
50( )
(6.38 1)G s
s
Result:
Solution to Homework 9Chapter 6 Identification from Step Response
President University Erwin Sitompul SMI 10/4
Solution to Homework 9Chapter 6 Identification from Step Response
Step Response
Time (sec)
0 5 10 15 20 25 30 35 40 45 50 55 60 65 700
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
: 5th order approximation: 4th order approximation
President University Erwin Sitompul SMI 10/5
Least Squares MethodsChapter 6 Least Squares Methods
The Least Squares Methods are based on the minimization of squares of errors.
The errors are defined as the difference between the measured value and the estimated value of the process output, or between y(k) and y(k).
There are two version of the methods: batch version and recursive version.
^
President University Erwin Sitompul SMI 10/6
Least Squares MethodsChapter 6 Least Squares Methods
Consider the discrete-time transfer function in the form of:1
11
1
( )1
mm
nn
b z b zG z
a z a z
The aim of Least Squares (LS) Methods is to identify the parameters a1, ..., an, b1, ..., bm from the knowledge of process inputs u(k) and process output y(k).
As described by the transfer function above, the relation of process inputs and process outputs is:
1 1( ) ( 1) ( ) ( 1) ( )n my k a y k a y k n b u k b u k m
President University Erwin Sitompul SMI 10/7
Least Squares MethodsChapter 6 Least Squares Methods
This relation can be written in matrix notation as:T( ) ( )y k km θ
where: T
1 1n ma a b bθ
T ( ) ( 1) ( ) ( 1) ( )k y k y k n u k u k m m
• Vector of Parameters
• Vector of Measured Data
Hence, the identification problem in this case is how to find θ based on the actual process output y(k) and the measured data from the past m(k).
President University Erwin Sitompul SMI 10/8
Least Squares MethodsChapter 6 Least Squares Methods
Assuming that the measurement was done for k times, with the condition k ≥ n + m, then k equations can be constructed as:
1 1(1) (0) (1 ) (0) (1 )n my a y a y n b u b u m
1 1(2) (1) (2 ) (1) (2 )n my a y a y n b u b u m
1 1( ) ( 1) ( ) ( 1) ( )n my k a y k a y k n b u k b u k m
or:T
T
T
(1)(1)
(2) (2)
( ) ( )
y
y
y k k
m
mθ
m
Y Mθ
President University Erwin Sitompul SMI 10/9
If M is nonsingular, then the direct solution can be calculated as:1θ M Y
In this method, error is minimized as a linear function of the parameter vector.
The disadvantage of this solution is, that error can be abruptly larger for t > k.
Least Squares MethodsChapter 6 Least Squares Methods
• Least Error (LE) Method, Batch Version
President University Erwin Sitompul SMI 10/10
Least Squares MethodsChapter 6 Least Squares Methods
A better way to calculate the parameter estimate θ is to find the parameter set that will minimize the sum of squares of errors between the measured outputs y(k) and the model outputs y(k) = mT(k)θ
^
2T
1
( ) ( )i
k
J y k k
θ m θ
T Y Mθ Y Mθ
T T T T T T Y Y Y Mθ θ M Y θ M Mθ
The extreme of J with respect to θ is found when:
0
dJ
d
θ
θ
President University Erwin Sitompul SMI 10/11
The derivation of J(θ) with respect to θ can be calculated as:
Least Squares MethodsChapter 6 Least Squares Methods
T TTd d
d d Ax A x A
x xif A symmetric
T T T0 2dJ
d
θM Y M Y M Mθ
θ
T T T T T TJ θ Y Y Y Mθ θ M Y θ M Mθ
TTd
d x Ax A A x
x2 Ax
0
T T2 2M Y M Mθ
1T Tθ M M M Y• Least Squares (LS) Method,
Batch Version
President University Erwin Sitompul SMI 10/12
Performing the “Second Derivative Test”,
Least Squares MethodsChapter 6 Least Squares Methods
Second Derivative Test• If f ’(x) = 0 and f ”(x) > 0 then f has a local minimum at x• If f ’(x) = 0 and f ”(x) < 0 then f has a local maximum at x• If f ’(x) = 0 and f ”(x) = 0 then no conclusion can be drawn
T T T0 2dJ
d
θM Y M Y M Mθ
θ
2T
2 2d J
d
θM M
θAlways positive definite
1T T θ M M M Y is a solution that will minimize the
squares of errors
President University Erwin Sitompul SMI 10/13
In order to guarantee that MTM is invertible, the number of row of M must be at least equal to the number of its column, which is again the number of parameters to be identified.
More row of M increase the accuracy of the calculation. In other words, the number of data row does not have to be the same as the sum of the order of numerator and denominator of the model to be identified.
If possible, rows with any value assumed to be zero (because no measurement data exist) should not be used.
Least Squares MethodsChapter 6 Least Squares Methods
President University Erwin Sitompul SMI 10/14
The parameters of a model with the structure of:
Example: Least Squares MethodsChapter 6 Least Squares Methods
22
1 1 21 2
( )1
b zG z
a z a z
are to be identified out of the following measurement data:0 1 2 3 4 5 6
( ) 0.5 0.8 0.2 0.8 0.7 0.8 0.7
( ) 0 0 1 1.4 0.6 5.4 17.2
k
u k
y k
Perform the batch version of the Least Squares Methods to find out a1, a2, and b2.
Hint: n + m = 2 + 1 At least 3 measurements must be available/ utilized.Hint: If possible, avoid to many zeros due to unavailable data for u(k) = 0 and y(k) = 0, k < 0.
President University Erwin Sitompul SMI 10/15
Using the least allowable data, from k = 2 to k = 4, the matrices Y and M can be constructed as:
(2)
(3)
(4)
y
y
y
Y
0 1 2 3 4 5 6
( ) 0.5 0.8 0.2 0.8 0.7 0.8 0.7
( ) 0 0 1 1.4 0.6 5.4 17.2
k
u k
y k
T
T
T
(2)
(3)
(4)
m
M m
m
1
1.4
0.6
(1) (0) (0)
(2) (1) (1)
(3) (2) (2)
y y u
y y u
y y u
0 0 0.5
1 0 0.8
1.4 1 0.2
Example: Least Squares MethodsChapter 6 Least Squares Methods
1
2
2
a
a
b
θ
President University Erwin Sitompul SMI 10/16
Example: Least Squares MethodsChapter 6 Least Squares Methods
1T
3.56 5.62 3.20
5.62 9.93 5.28
3.20 5.28 4
M M
T
2.24
0.6
0.5
M Y
1T Tθ M M M Y
3
4
2
1
2
2
a
a
b
2
1 1 2
2 ( )
1 3 4
zG z
z z
President University Erwin Sitompul SMI 10/17
Homework 10Chapter 6 Least Squares Methods
Redo the example, utilizing as many data as possible. Does your result differ from the result given in the slide? What could be the reason for that? Which result is more accurate?
President University Erwin Sitompul SMI 10/18
Homework 10AChapter 6 Least Squares Methods
Redo the example, utilizing least allowable data, if the structure of the model is chosen to be
11
2 1 21 2
( )1
b zG z
a z a z
After you found the three parameters a1, a2, and b1, for G2(z), use Matlab/Simulink to calculate the response of both G1(z) and G2(z) if they are given the sequence of input as given before.
2
1 1 2
2( )
1 3 4
zG z
z z
Compare y(k) from Slide 10/15 with y1(k) and y2(k) from the outputs of the transfer functions G1(z) and G2(z). Give analysis and conclusions.
0 1 2 3 4 5 6
( ) 0.5 0.8 0.2 0.8 0.7 0.8 0.7
k
u k
Odd-numbered Student-ID
11
2 11
( )1
b zG z
a z
Even-numbered Student-ID