Curve fitting toolbox giving different answer to custom function?

Hello,

I have data which looks like this when plotted:

A custom equation summing a linear slope and a Gaussian gives me a really nice fit:

However, when I try to implement that as code, the fit fails pretty spectacularly.

% energy = 100-by-1 double
% counts = 100-by-1 double (both cropped from main data)

% if I use the toolbox to fit using 'energy' as the x-axis, it fails. However, if I use xdat it gives the result shown above. 
xdat = linspace(1,length(counts), length(counts))';

% All this I copied from the 'Generate Code' option under 'File'. I thought I had a problem with StartPoint, but they're 

[xData, yData] = prepareCurveData( xdat, counts );

% Set up fittype and options.
ft = fittype( '(m*x+c) + a1*exp(-((x-b1)/c1)^2)', 'independent', 'x', 'dependent', 'y' );
opts = fitoptions( 'Method', 'NonlinearLeastSquares' );
opts.Display = 'Off';
opts.StartPoint = [0.3302 0.2297 0.1139 0.3109 0.2284]; % these change, which I didn't expect when fitting the same data.

% Fit model to data.
[fitresult, gof] = fit( xData, yData, ft, opts );
plot(fitresult)

fitresult = 
     General model:
     fitresult(x) = (m*x+c) + a1*exp(-((x-b1)/c1)^2)
     Coefficients (with 95% confidence bounds):
       a1 =        1042  (1024, 1059)
       b1 =       59.75  (59.66, 59.85)
       c =       311.4  (302.8, 319.9)
       c1 =      -7.268  (-7.415, -7.121)
       m =     -0.9449  (-1.094, -0.7957)

Any clue to where I am failing would be very much appreciated.

[Alex Sha]

Hi, AlexAS, if possible, would you please attache your data file.

[AlexAS]

Hi Alex, I've attached them. BW, Alex

[the cyclist]

Sorry, is this an Alex-only question? Maybe I need to leave.

Answer 1

[the cyclist]

I don't fully understand what is going on here, but if I plot that function over the range 0:100 instead of 0:1, the shape of the curve is replicated:

x = 0 : 1 : 100;

a1 = 1042;
b1 = 59.75;
c = 311.4;
c1 = -7.268;
m = -0.9449;

fitresult = (m*x+c) + a1*exp(-((x-b1)/c1).^2);

figure
plot(x,fitresult)

That makes sense to me, since the exponential term clear has a peak where x == b1. I don't think the data you plotted in your first graph (which peaks around 1.05) is really what was fitted.

[AlexAS]

Thanks for this. I get the same result using your starting points, but it's still not clear to me where I'm going astray. However, I can at least move forward while I figure it out! Cheers, Alex

Answer 2

[Alex Sha]

Hi, AlexAS, I get the global solution as fellow:

Root of Mean Square Error (RMSE): 23.9947153172582

Sum of Squared Residual: 58150.3826787827

Correlation Coef. (R): 0.996437682897764

R-Square: 0.992888055898666

Adjusted R-Square: 0.992742914182312

Determination Coef. (DC): 0.992888055898665

Chi-Square: 98.7754060511867

F-Statistic: 3350.60470214536

Parameter Best Estimate

---------- -------------

a1 1038.49566359074

b1 1.0436517640913

c 387.185685520493

c1 0.036134594114558

m -122.65365053643