How to fit a polynomial to a step function??

Question

Swati Jain 2019-3-28

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/452948-how-to-fit-a-polynomial-to-a-step-function

编辑： Swati Jain 2019-3-31

Hi,

I have a noisy step from a scnner result. Is there any way to fit a higher order polynimial to step ? I am not sure which polynomial should work. Attached is the data and figure of my step.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Walter Roberson 2019-3-28

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/452948-how-to-fit-a-polynomial-to-a-step-function#answer_367809

No finite polynomial can possibly fit that -- not unless perhaps you swapped axes. What you have is more like a sigmoid function.

What might be acceptable is to fit it as a ratio of polynomials. If you use cftool and select Rational, and ask for numerator order 2 and denominator order 4, the fit is not terrible. (You might have to change the degree and change back again in order to get a fit that does not have singularities.)

2 个评论
显示无隐藏无

Walter Roberson 2019-3-31

That status is not an error. It says that fmincon found a place it was willing to live with. It would have preferred that the difference between adjacent locations was smaller, but it has reached the minimum step size that you configured in (by default because you pass no options) and would need to take smaller steps to reduce the slope more. You can configure options to permit it to take smaller steps, but probably it would not help.

You appear to be fitting a phased gaussian. Those are notoriously difficult to fit well. When you fit with guassians, one of the common tendencies is for the fitting to push towards a narrow peak with a high phase, which is a adaptation to fit a noise spike.

Swati Jain 2019-3-31

编辑：Swati Jain 2019-3-31

在 MATLAB Online 中打开

Sorry, my bad. I got my mistake. I will have to convolve gaussain with a perfect step and optimize it for minimum error with the measured point cloud. I could solve it with two methods. Both are giving almost same cofficients. But problem is error value (objective value for least square) is too large (around 274.9005). I tried no. of different initial values but every time I got same result. I tired to optimize with two different methods, both of them are giving approximately same cofficients.

My Method1 is as follows:

% Creat a Perfect Step
h=ones(1,numel(xf)); %height=1mm
h(1:5890)=-1.20;
h(5891:end)=-0.15;
figure; 
plot(xf,h,'.-r','linewidth',1.5,'markersize',12);
grid on
set(gca, 'FontName', 'Arial');
set(gca, 'FontSize', 30);
xlabel('x(mm)'); ylabel('y(mm)'); zlabel('z(mm)');
axis equal
% initial parameters
mu= -0.97; 
% mu= -0.970; 
sigma=0.011/2.355; 
A=1/17460; 
xff=xf(1800:9800);
xff=xff';
z_Tr1=z_Tr1(1800:9800);
z_Tr1=z_Tr1';
p0=[A, mu, sigma];
yp=@(p) conv(h,(p(1)/(p(3)*sqrt(2*pi)))*exp(-((xff-p(2)).^2)/(2*p(3)^2)),'same');
objective = @(p) sum(((yp(p)-z_Tr1)).^2);
disp(['Initial Objective: ' num2str(objective(p0))])
popt = fmincon(objective,p0);
figure;
plot(xff,z_Tr1,'.k');
hold on
plot(xff,yp(p0),'.-b');
hold on
plot(xff,yp(popt),'gs')
legend('measured','initial predicted','optimal predicted')
ylabel('y')
xlabel('x')
disp(['Final Objective: ' num2str(objective(popt))])
disp(['Optimal parameters: ' num2str(popt)])

Method2:

f = @(F,xff) conv(h,(F(1)/(F(3)*sqrt(2*pi)))*exp(-((xff-F(2)).^2)/(2*F(3)^2)),'same');
F_fitted = nlinfit(xff, z_Tr1,f,[A mu sigma]);
% Display fitted coefficients
disp(['F = ',num2str(F_fitted)])
% Plot the data and fit
figure(1)
plot(xff,z_Tr1,'*',xff,f(F_fitted,xff),'.-k');
legend('data','fit')