fitting data with power function

2011 6 26
3 个回答
更新时间：2018 4 8
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Hi,
I need to curve fit to data, which i had meassured. I must use method of least squares and for fitting i must use a „power function“ y= a*x^b ((ftype=fittype('power1')); I need to find the coefficients a and b. But i dont know how to do in Matlab, i try to write some m-scripts but it doesnt works.
Example of meassured data:
X=[-90; -80; -70; -60; -50; -40; -30; -20; -10;0; 10; 20; 30; 40; 50;60; 70; 80; 90]
Y=[-2.8691; -2.4261; -2.0042; -1.7497; -1.3936; -1.1392; -0.9356; -0.7321; -0.5286; 0; 0.3872; 0.6416; 0.8451; 1.0995; 1.3031; 1.6083;1.9136; 2.3207; 2.524]
I will be grateful for any idea or any help. Thank you very much.

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safak
safak 2018-4-8
did you check the command window, coefficients may be printed out there

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回答（3 个）

bym
bym 2011-6-26

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Plotting your data it does not seem to fit a power function and seems linear. You can use the backslash operator to do least squares
doc mldivide
or in the plot window, use tools>basic fitting to fit a linear equation

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Jana
Jana 2011-6-27
This data was only example, i have a set of data and each data I will fit with power function, I find coefficient a and b, than I can compare data,this is the reason to power function.
bym
bym 2011-6-27
you can use the backslash to find the coefficients, you will just have to transform your data using logarithms.
Jana
Jana 2011-6-27
But i cannot use logarithmus, because i have a zero defined in my data, i try to use a logaritmus, but it doesnt work for zero, do you know how to write a script using logaritmus,and backslash for data with zero?
bym
bym 2011-6-27
if you have zeros in your data, then maybe a different approach is warranted. If you can post some 'real' data then maybe a solution can be found

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Jana
Jana 2011-6-28
编辑：Walter Roberson 2018-4-8

0 个投票

Data : X=[-90; -80; -70; -60; -50; -40; -30; -20; -10;0; 10; 20; 30; 40; 50;60; 70; 80; 90]
Y=[-2.8691; -2.4261; -2.0042; -1.7497; -1.3936; -1.1392; -0.9356; -0.7321; -0.5286; 0; 0.3872; 0.6416; 0.8451; 1.0995; 1.3031; 1.6083;1.9136; 2.3207; 2.524]
other measurement:
x= =[-90; -80; -70; -60; -50; -40; -30; -20; -10;0; 10; 20; 30; 40; 50;60; 70; 80; 90]
y=[-2.8182 -2.6147 -2.4621 -2.1059 -1.8515 -1.6481 -1.2409 -0.9865 -0.7831 -0.0707 0.5398 0.8961 1.1504 1.4048 1.6083 1.9136 2.117 2.3206 2.5751]]
value -.0707 is something like offset, it must be deducted from all other data in y.
other measurement:
X=[-90; -80; -70; -60; -50; -40; -30; -20; -10;0; 10; 20; 30; 40; 50;60; 70; 80; 90]
Y= [-3.5306 -3.2762 -3.0218 -2.6656 -2.3604 -2.0042 -1.5462 -1.1392 -0.7831 -0.0707 0.7434 1.0995 1.6084 2.0663 2.5242 2.9821 3.2874 3.5418 3.9488]
value -0,0707 is ofsfset too, and it must be dedicated from other values in y.
i use this algoritm, but its ignore zero, so its wrong way !
close all
clear all
clc
x=[-90; -80; -70; -60; -50; -40; -30; -20; -10; 10; 20; 30; 40; 50;60; 70; 80; 90]';
y=[-2.8691; -2.4261; -2.0042; -1.7497; -1.3936; -1.1392; -0.9356; -0.7321; -0.5286; 0.3872; 0.6416; 0.8451; 1.0995; 1.3031; 1.6083;1.9136; 2.3207; 2.524]';
plot(x,y,'r-*')
lnx = log(abs(x));
lny = log(abs(y));
A = [lnx;ones(size(x))];
v = inv(A*A')*A*lny';
a = v(1);
b = exp(v(2));
hold on
u = 0:1:100;
plot(u,b*u.^a);
u = -100:1:0;
plot(u,-b*(-u).^a);
hold off
a
b
legend('namerena data', 'prolozeni dat',2)
Matt Fig
Matt Fig 2011-6-28

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If I use this data and plot as you show, it looks near linear. Why would you think this should be a power law relation?
x=[-90; -80; -70; -60; -50; -40; -30; -20; -10; 10; 20; 30; 40; 50;60; 70; 80; 90]';
y=[-2.8691; -2.4261; -2.0042; -1.7497; -1.3936; -1.1392; -0.9356; -0.7321; -0.5286; 0.3872; 0.6416; 0.8451; 1.0995; 1.3031; 1.6083;1.9136; 2.3207; 2.524]';
plot(x,y,'r-*')
pp = polyfit(x,y,1); % Fit a line to the data
yp = polyval(pp,x);
hold on
plot(x,yp,'b') % Plot the line in blue.

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Jana
Jana 2011-6-30
I need to compare coefficients,how can I have a coefficients from polyfit and polyval?
Because the comparison is the result of my thesis..

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