Curve fitting: Getting very poor results on seemingly simple data set ("fit" function)
显示 更早的评论
Hello everyone,
I will be the first to admit I am a complete newbie with the "fit" function and "cfun" data type, and find the documentation available on it much less than user friendly, so please refrain from directing me to it - I have been through it front to back and have not found a solution.
I am going through a dataset in which I am analyzing hundreds of discrete blocks, each of which has an essentially periodic behaviour. I need to find the amplitude of this periodic behaviour. I have included an example.
My problem is this. I used a very clear sample dataset containing 13 samples (one of the best I have) with a very obvious cosine behaviour. I set the fit function up and ran it, and got parameters that give me almost a straight line. I tried using repmat thinking maybe it was my sample size, and this does almost nothing.
I manually solved the periodic fit and have included the plot with the original data, the empirically derived fit, and the results from the 'fit' function below. I have also provided what I believe to be the relevant chunk of code, but I could be wrong. Please let me know what else if anything I need to include. As I have to perform this operation on literally hundreds or thousands of these blocks, this is not something I can feasibly do manually, so if someone can direct me as to what I'm doing wrong, I would be infinitely grateful.
Thank you kindly for your time.
Here is what cfun spits out:
cfun =
General model Sin1:
cfun(x) = a1*sin(b1*x+c1)
Coefficients (with 95% confidence bounds):
a1 = 0.8323 (-22.99, 24.66)
b1 = 0.004459 (-0.9294, 0.9383)
c1 = 1.936 (-72.97, 76.84)
And here is the relevant snippet:
%%%%%%%%%%%%%%%%%
% CURVE FITTING %
%%%%%%%%%%%%%%%%%
x = 2*pi.*((0:num_phases-1)./(num_phases-1));
num_phases2 = num_phases*10;
x2 = 2*pi.*((0:num_phases2-1)./(num_phases2-1));
x = x';
x2 = x2';
y = results(working_block*num_phases+1:working_block*num_phases+num_phases,1);
y2 = repmat(y,10,1);
cfun = fit(x,y,'sin1');
test = 0.025*cos(2*pi.*((0:num_phases-1)./(num_phases-1)))+0.775;
figure(1);
plot(2*pi.*((0:num_phases-1)./(num_phases-1)),results(working_block*num_phases+1:working_block*num_phases+num_phases,1)); hold on;
plot(2*pi.*((0:num_phases-1)./(num_phases-1)),test,'-r');
plot(cfun,'-g');
xlabel('Block');
ylabel('Correlation');
legend('Original Data','0.025cos(2pi*x)+0.775','Fitted Data');
采纳的回答
Your model without an intercept term doesn't fit the data pattern worth a hoot...
With no intercept there's no offset to shift an initial zero value of sin(0) and the amplitude of the sin/cos has to be about +/-0.025 to fit the range of the data.
Try something more like
g = fittype( @(a,b,c,d,x) a.*sin(b.*x+c)+d )
With that I get a reasonable fit. Of course, w/ four parameters and only 13 data points you're using up a sizable fraction of the available DOF...
ADDENDUM:
Came to see if had been any followup and noticed what hadn't actually seen before...
I see in your legend text you don't have a phase angle in the equation--altho you don't show it, in that case you mis-specified the model. As the tool showed you, the model you fitted was
cfun(x) = a1*sin(b1*x+c1)
whereas you wanted (apparently)
cfun(x) = a1*sin(b1*x)+c1
based on the text in
legend('Original Data','0.025cos(2pi*x)+0.775','Fitted Data');
IOW, you had a misplaced ')' in the function definition...
2 个评论
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Interpolation 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
