Post on 05-Jan-2016
description
Least Square Method
Curve Fitting
21 mxmy Linear Least Square Method
nnn Rmxmy 21
22121
221
2 222 nnnnnnn Rxmmymyxmxmy
N
nn
N
nnnnnnn Rxmmymyxmxmy
1
2
12121
221
2 222
02221 1
21
21
1
1
2
N
n
N
nn
N
nnnn
N
nn
xmyxxmm
R
022211
22
1
2
N
nn
N
nn
N
nn
xyNmm
R
2
11
2
1111
N
nn
N
n
N
nn
N
nn
N
nnn
xxN
yxyxNm
n
2
11
2
111 1
2
2
N
nn
N
n
N
nn
N
nnn
N
n
N
nn
xxN
xyxyxm
n
n
M
iiimxy
1
21 mxmy
m M ในกรณี�ที่��มี�ค่�าพารามี�เตอร� จำ�านวน ต�ว
εmXy
mXymXyεε
TTN
nnR
1
2
N
nn
N
nnnnnnn Rxmmymyxmxmy
1
2
12121
221
2 222
02 T1
2
XmXym
N
n
R
0 mXyX T
0 mXXyX TT
yXXXm TT 1
Example 1 on page 362
21 pTpk
T]115 119 138 141[k
T]1500 1000 500 390[T
2
1
21
]1 [ p
pT
pTpk
y'XX'Xp 1
% Linear1.mk =[141 138 119 115]';T = [390 500 1000 1500]';X = [T,[1 1 1 1]'];y = k;p =inv(X'*X)*X'*y
Non Linear System of Equation
0,...,,
0,...,,
0,...,,
21
212
211
nn
n
n
xxxf
xxxf
xxxf
0... ,,1,
2
,,21,2
1
,,11,1,1,
n
ikinin
ikii
ikiiikik
x
fxx
x
fxx
x
fxxff
n
ikin
iki
iki
n
ikin
iki
ikiik
x
fx
x
fx
x
fx
x
fx
x
fx
x
fxf
,
1,2
,1,2
1
,1,1
,,
2
,,2
1
,,1,
...
...
sym
...
...
...
,
2
,
1
,
,2
2
,2
2
,2
,1
2
,1
1
,1
n
ininin
n
iii
n
iii
x
f
x
f
x
f
x
f
x
f
x
fx
f
x
f
x
f
Z
= Jacobian Matrix
inii
Ti
iniiT
i
iniiT
i
fff
xxx
xxx
,,2,1
1,1,21,11
,,2,1
...
...
...
F
x
x
1 iiT
i xZxZF
iT
ii xZFZx
11
Example
03
1
3
112
27
10
9
2
3
52
03
2
3
11
27
8
9
2
3
42
2
212121
2
212121
xxxxxx
xxxxxx
2
,2
1
,2
2
,1
1
,1
x
f
x
f
x
f
x
f
ii
ii
Z
2121212
,2
2121211
,2
2121212
,1
2121211
,1
3
1
3
11
3
22
27
20
27
10
9
2
3
52
3
1
3
11
3
22
9
4
27
10
9
2
3
52
3
2
3
11
3
4
27
16
27
8
9
2
3
42
3
2
3
11
3
2
9
4
27
8
9
2
3
42
xxxxxxx
f
xxxxxxx
f
xxxxxxx
f
xxxxxxx
f
i
i
i
i
clear; c1 = 1; c2 = 1; %First Guess NONLINEAR1.m
for i=1:10
X = [c1;c2];
f1 = 2*(4/3-2/9*c1-8/27*c2)*(c1+c2)+(1-1/3*c1-2/3*c2)^2;
f2 = 2*(5/3-2/9*c1-10/27*c2)*(c1+2*c2)+(1+1/3*c1+1/3*c2)^2;
F = [f1;f2];
df1c1 = 2*(4/3-2/9*c1-8/27*c2)- 4/9*(c1+c2)- 2/3*(1-1/3*c1-2/3*c2);
df1c2 = 2*(4/3-2/9*c1-8/27*c2)- 16/27*(c1+c2)- 4/3*(1-1/3*c1-2/3*c2);
df2c1 = 2*(5/3-2/9*c1-10/27*c2)- 4/9*(c1+2*c2)+ 2/3*(1+1/3*c1+1/3*c2);
df2c2 = 4*(5/3-2/9*c1-10/27*c2)- 20/27*(c1+2*c2)+ 2/3*(1+1/3*c1+1/3*c2);
Z = [df1c1,df1c2;df2c1,df2c2];
W = -F+Z*X;
Xnew = inv(Z)*W;
c1 =Xnew(1);c2= Xnew(2);
fprintf(' %g %g %g\n',i,c1,c2);
end
Non Linear Least Square εmXy
02 T1
2
m
XmXy
m
N
n
R
0
T
mXym
X
Numerically
mXymXyεε
TTN
nnR
1
2
other
*mm
m
*mX*mXy
Expand {y} by Taylor
m
*mXH
*mXyHHH*mm T1T
*mmH*mX y
*mmHH*mXH TT y
Example : Stress Relaxation
54321 expexp at-aat-ay = a
Time Force
.000001, 3.083900
.016670, 1.647600
.033330, 1.176900
.050000, .935450
.066670, .784570
.083330, .681970
.100000, .609550
Example : Stress Relaxation
54321 expexp at-aat-ay = a
2
,2
1
,2
2
,1
1
,1
x
f
x
f
x
f
x
f
ii
ii
Z
54321 a
f
a
f
a
f
a
f
a
f iiiiiH
*mXyHHH*mm T1T
m
*mXH
54321 expexp at-aat-ay = a
54321 a
f
a
f
a
f
a
f
a
f iiiiiH
t-aa
fi2
1
exp
1 5
= a
fi
t-ata= a
fi21
2
exp
t-a=a
fi4
3
exp
t-ata= a
fi43
4
exp
Stess.m
% Parameters Estimation% Using Stress Relaxation Data% Five parametersclear; clc; clf;Data= ...[ 1, .000001, 3.083900 2, .016670, 1.647600 3, .033330, 1.176900 4, .050000, .935450 5, .066670, .784570 6, .083330, .681970 7, .100000, .609550 8, .116670, .555230 9, .133330, .506950 10, .150000, .470740 20, .316670, .289690 30, .483330, .223300 40, .650000, .181050 50, .816670, .156910 60, .983330, .138810 70, 1.150000, .126740 80, 1.316700, .114670 90, 1.483300, .108630 100, 1.650000, .102600 110, 1.816700, .090530 120, 1.983300, .084490 130, 2.150000, .084490 140, 2.316700, .078460 150, 2.483300, .078460 160, 2.650000, .078460 170, 2.816700, .072420 180, 2.983300, .072420 190, 3.150000, .066390 200, 3.316700, .066390 220, 3.650000, .060350 240, 3.983300, .060350 260, 4.316700, .054320 280, 4.650000, .060350 300, 4.983300, .048280 350, 5.816700, .048280 400, 6.650000, .048280 500, 8.316700, .042250 600, 9.983300, .036210];
t = Data(:,2);y = Data(:,3)*1000/224.8089;y = y/y(1);plot(t,y,'ro');legend('Data','Fit');xlabel('Time');ylabel('Force');title ('Non Linear Regression');axis([min(t) round(max(t)) 0 1]);hold on;Plothandle = plot(t,y,'-','EraseMode','xor');
a =[10;10;10;1;1]; % First Guess of three parametersda =1; i=0;while da>1e-20 ym = a(1)*exp(-a(2)*t)+a(3)*exp(-a(4)*t)+a(5); dyda1 = exp(-a(2)*t); dyda2 = -a(1)*t.*exp(-a(2)*t); dyda3 = exp(-a(4)*t); dyda4 = -a(3)*t.*exp(-a(4)*t); dyda5 = 1+zeros(size(t));
Jacob = [dyda1,dyda2,dyda3,dyda4,dyda5]; anew = a +0.1*inv(Jacob'*Jacob)*Jacob'*(y-ym); da = norm(anew-a); a = anew; i=i+1; fprintf( 'Iter = %g Par = %g %g %g %g%g\n',i,a(1),a(2),a(3),a(4),a(5)); set (Plothandle,'ydata',ym); drawnow; hold off;end
ybar = mean(y); SSr = sum((ym-ybar).^2);Syy = sum((y-ybar).^2); R2 = SSr/Syy;fprintf( ' Parameter = %g %g %g %g %g\n',a(1),a(2),a(3),a(4),a(5));fprintf( ' R_square = %g \n',R2);