Curve Fitting Matlab

1. Unit:
Curve-Fitting
Introduction:
In experiments we get lot of random given data (x,y). It is difficult to fit the curve passing
through these data points. So we need to find some polynomial function by which approximate
curve can be drawn. This curve may not pass through all the point but errors will be less.
Least square method is a technique which gives best possible polynomial like Y c + c (x). by
which errors will be less.
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1
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n
n
i i
i
Error y y

 
 
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n
i
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1
2
Curve fitting is the process of constructing a curve, or mathematical function that has the best
fit to a series of data points, possibly subject to constraints. Fitted curves can be used as an aid
for data visualization, to infer values of a function where no data are available, and to
summarize the relationships among two or more variables

2. A first degree polynomial equation:
Yn c + c (x).
A first degree polynomial equation is
an exact fit through any two points
with distinct x coordinates.
A second degree polynomial: Yn c + c (x) + c (x)2
.
Fitting a Line (first degree polynomial) equation:
n= number of data points
∑
∑ ∑
∑ .
∑
OR
Determinants:
d=det d1= d2=
Constants: c0=d1/d & c1=d2/d
Fitting a Line : + (x).
Problem: Using least square method Fit a line: Yn c + c (x)
X 1 2 3 4
Y 2 6 12 20
Solution: n= number of data points=4
∑
∑ ∑
∑ .
∑
X Y
1 2 1 2
2 3 4 12
3 12 9 36
4 20 16 80
∑ =10 ∑ .
40 ∑ 30 ∑ 130

3. OR
Determinants:
d=det
1
1 2
=20 d1= Dd2= =120
Constants: C0=d1/d = ‐ 5 and C1=d2/d=6
Best fitted line: + (X). ANSWER: + (X).
Line Flowchart

4. % LINE FITTING Yf=C0+C1(X)
clc
clear all
%-------------Direct input of Data-----------
x=[1 2 3 4];
y=[2 6 12 20];
n=length(x)
%--------------OR-User UInput of data-----
% n=input('Enter n=');
% for i=1:n
% fprintf('Enter x(%d)=',i);
% x(i)=input('');
% fprintf('Enter y(%d)=',i);
% y(i)=input('');
% end
%--------------------------------------------
s1=sum(x); s2=sum(x.^2); s3=sum(y);
s4=sum(x.*y);
%--------------------------------------------
d=[n s1;
s1 s2];
d1=[s3 s1;
s4 s2];
d2=[n s3;
s1 s4];
d=det(d); d1=det(d1); d2=det(d2);
%--------------------------------------------
c0=d1/d; c1=d2/d;
fprintf('BEST FIT ,Yn=%0.3f+%0.3f(X)n',c0,c1)
%--------------------------------------------
yn=c0+c1*x;
plot(x,y,'*',x,yn)
%--------------------------------------------

5. Solved Problems
Fit the line yn=c0+c1*x
x=[1 2 3 4];
y=[2 6 12 20];
‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
X Y X2
XY
‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
1 2 1 2
2 6 4 12
3 12 9 36
4 20 16 80
‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
10 40 30 130
Calculate d d1 d2 c 0 c1 by you
BEST FIT ,Yn=‐5.000+6.000(X)
Fit the line yn=c0+c1*x
L=a0+a1T
x=[ 20 30 40 50 60 70];
y=[800.3 800.4 800.6 800.7 800.9 801];
‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
X Y X2
X.*Y
‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
20 800 400 16006
30 800 900 24012
40 801 1600 32024
50 801 2500 40035
60 801 3600 48054
70 801 4900 56070
‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
270 4804 13900 216201
‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
Calculate d d1 d2 c 0 c1 by you

6. BEST FIT ,Yn=799.994+0.015(X)
Fitting a second degree polynomial:
+ (X) + (X)2
∑ ∑
∑ ∑ ∑
∑ ∑ ∑
0
1
2
∑
∑
∑
.
1 2
1 2 3
2 3 4
0
1
2
5
6
7
D det
1 2
1 2 3
2 3 4
D1 det
5 1 2
6 2 3
7 3 4
D2 Det
5 2
1 6 3
2 7 4
D3 Det
1 5
1 2 6
2 3 7
Constants: C0=d1/d , C1=d2/d and C2=d3/d
Problem: Using least square method fit a second degree polynomial
+ (X) + (X)2
X 1 2 3 4 5 6 7 8 9
Y 2 6 7 8 10 11 11 10 9
Solution:
∑ ∑
∑ ∑ ∑
∑ ∑ ∑
∑
∑
∑
=
0
1
2

7. Table calculatins to be done by you
X Y X2
X3
X4
XY XY2
S1 S5 S2 S3 S4 S6 S7
. 45 74 286 2025 15333 421 2771
On making table and solving
S1 45, S2 286, S3 2025, S4 15333, s5 74 , s6 421, s7 2771,
D det 166320 D1 det ‐154440
D2 Det D3 Det ‐44460
Constants:
C0=D1/D = ‐0.9286 C1=D2/D= 3. C2=D3/D= ‐0.2673
Best fitted parabola: . .+3.523(X) 0.267(X)2

8. Program
% Quadratic FITTING Yf=C0+C1(X)+C2(x^2);
% Quadratic FITTING Yf=C0+C1(X)+C2(x^2);
clc
clear all
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
x=[1 2 3 4 5 6 7 8 9];
y=[2 6 7 8 10 11 11 10 9];
n=9;
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
x2=x.^2; x3=x.^3; x4=x.^4; xy=x.*y; x2y=(x.^2).*y;
s1=sum(x); s2=sum(x2); s3=sum(x3); s4=sum(x4); s5=sum(y); s6=sum(xy); s7=sum(x2y);
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
d=[n s1 s2;
s1 s2 s3
s2 s3 s4];
d1=[s5 s1 s2;
s6 s2 s3
s7 s3 s4];
d2=[n s5 s2;
s1 s6 s3
s2 s7 s4];
d3=[n s1 s5;
s1 s2 s6
s2 s3 s7];
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
d=det(d); d1=det(d1); d2=det(d2); d3=det(d3); c0=d1/d; c1=d2/d; c2=d3/d;
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
fprintf('BEST FIT ,Yn=%0.3f+%0.3f(X)+%0.3f(x^2)n',c0,c1,c2)
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
yn=c0+c1*x+c2*x2;
plot(x,y,'*',x,yn,'r')
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
Option
for i=1:n
fprintf('%0.0f %0.0f %0.0f %0.0f %0.0f %0.0f %0.0f
n',x(i),y(i),x2(i),x3(i),x4(i),xy(i),x2y(i))
end
fprintf('‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
n');
fprintf('%0.0f %0.0f %0.0f %0.0f %0.0f %0.0f %0.0f n',s1,
s5,s2,s3,s4,s6,s7)

9. Solved Problem
Fit the following quadratic/2nd
degree/ parabolic curve + (X) + (X)2
x=[‐4 ‐3 ‐2 ‐1 0 1 2 3 4 5];
y=[21 12 4 1 2 7 15 30 45 67];
‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
X Y x2 x3 x4 xy x2y
‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
‐4 21 16 ‐64 256 ‐84 336
‐3 12 9 ‐27 81 ‐36 108
‐2 4 4 ‐8 16 ‐8 16
‐1 1 1 ‐1 1 ‐1 1
0 2 0 0 0 0 0
1 7 1 1 1 7 7
2 15 4 8 16 30 60
3 30 9 27 81 90 270
4 45 16 64 256 180 720
5 67 25 125 625 335 1675
‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
5 204 85 125 1333 513 319
‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
Remaining d d1 d2 d3 c0 c1 c3 …to be done by user

10. Flowchart: Curve Fitting (Parabola)
s1=sum(x); s2=sum(x.^2); s3=sum(x.^3); s4=sum(x.^4);
s5=sum(y); s6=sum(x.*y); s7=sum((x.^2).*y);
Calculation of Determinents d, d1, d2
c0=d1/d; c1=d2/d; c2=d3/d;
Start
Break
Read x(i), y(i)
For ( i=1 to n)
Read n
yf=c0+c1*x+c2*x.^2
Print C0, C1, C2
plot(x,y)

11. Numerical Power
For the power equation: PV^gamma
=C
P=[0.5 1 1.5 2 2.5 3];
V=[1.62 1 0.75 0.62 0.52 0.46];
Solution: Log(P)+ gamma*log(V)=log(C)
Log(P)= log(C) - gamma *log(V)
Y = C0 + C1 (X)
C0=log(c) C1=-gamma Y=Log(P) X=Log(V)
-------------------------------------------------------------------
V P X Y X^2 XY
-------------------------------------------------------------------
1.62 0.5 0.482426 -0.693147 0.232735 -0.334392
1.0 1.0 0. 0. 0. 0.
0.75 1.5 -0.287682 0.405465 0.082761 -0.116645
0.62 2. -0.478036 0.693147 0.228518 -0.331349
0.52 2.5 -0.653926 0.916291 0.427620 -0.599187
0.46 3. -0.776529 1.098612 0.602997 -0.853104
-----------------------------------------------------------
S1= -1.7137 S3=2.4204 S2=1.5746 S4=-2.2347
d = 6.5109 d1 =-0.0185 d2 = -9.2602
c0 =-0.0028 c1 = -1.4223
c = exp(c0)= 0.9972 gamma =-c1=1.4223
PV^1.422=0.997165

12. Program Power
clc
clear all
%‐‐‐‐‐‐‐‐‐‐‐‐‐Power Equation‐‐‐‐‐‐‐‐‐‐‐
%PV^Gamma=C
p=[0.5 1.0 1.5 2.0 2.5 3.0];
v=[1.62 1.0 0.75 0.62 0.52 0.46];
n=6;
y=log10(p);
x=log10(v);
xy=x.*y;
x2=x.^2;
s1=sum(x); s2=sum(x2); s3=sum(y); s4=sum(xy);
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
d=n*s2‐s1*s1;
d1=s3*s2‐s4*s1;
d2=n*s4‐s1*s3;
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
c0=d1/d; c1=d2/d;
gamma=‐c1;
c=exp(c0);
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
fprintf('PV^%0.3f=%fn',gamma,c)
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
%plot(p,v);
for i=1:n
fprintf('%f %f %f %f %f %f
n',v(i),p(i),x(i),y(i),x2(i),xy(i));
end

13. Numerical Exponential
For the Exponential equation: y=aebx
x=[2 4 6 8];
y=[25 38 56 84];
Solution
y=aebx
ln(y)=ln(a)+bx ln(e)
Y =C0 +C1(x)
yn=ln(y) C0=ln(a) C1=b
------------------------------------------------------
x y yn x^2 x*yn
------------------------------------------------------
2. 25. 3.218876 4. 6.437752
4. 38. 3.637586 16. 14.550345
6. 56. 4.025352 36. 24.152110
8. 84. 4.430817 64. 35.446534
-----------------------------------------------------
S1=20 s3= 15.3126 s2= 120 s4=80.5867
d = 80 d1 = 225.7808 d2 = 16.0944
c0=2.8223 c1=0.2012
a=exp(c0)= 16.8148 b = c1=0.2012
y=16.815 e (0.2012x)

14. Program Exponential
clc
clear all
%‐‐‐‐‐‐‐‐‐‐‐‐‐Exponential‐‐‐‐‐‐‐‐‐‐‐
%y=ae^bx
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
x=[2 4 6 8];
y=[25 38 56 84];
n=length(x);
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
y1=log(y);
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
x2=x.^2; xy1=x.*y1;
s1=sum(x); s2=sum(x2); s3=sum(y1); s4=sum(xy1);
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
d=n*s2‐s1*s1;
d1=s3*s2‐s4*s1;
d2=n*s4‐s1*s3;
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
c0=d1/d; c1=d2/d;
a=exp(c0);
b=c1;
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
fprintf('y=%0.3f e^(%0.3fx)n',a,b)
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
for i=1:n
fprintf('%f %f %f %f %f n',x(i),y(i),y1(i),x2(i),xy1(i));
end

15. Solver :% POLYFIT
clear all
x=[19 25 30 36 40 45 50];
y=[76 77 79 80 82 83 85];
d=1; %d=1 (line) d=2(parabola)
c=polyfit(x,y,d)
% yn = polyval(c, x);
yn=c(2)+c(1)*x;
plot(x,y,'*',x,yn)