Custom exponential fit doesn't seem to work

Hey everyone,

I was stumbling in a bit of a problem and I hope one of you might be able to help me out. I have some data points (it's about 250ish) and am trying to fit an exponentially decreasing function with non-zero constant term to it, i.e. . However, no matter what I tried so far (different fitting functions, methods, starting points), I always end up with something that looks a lot like a constant function and basically resembles the data points not at all. This is the code I've been using so far:

% Here comes a previous part of the code, which extracts the data in
% question from a larger array and ends up giving you some x- and y-values.
% Those will  be attached to the question.
x0 = [1 -2 12500]; 
fitfun = fittype( @(a,b,c,x) a*exp(b*x)+c );
[fitted_curve,gof] = fit(Data(:,1),Data(:,2),fitfun,'StartPoint',x0)

Unfortunately, when I plot the results, I end up with the following:

That obviously can't be the best fit possible.I've already looked at the fit-Function itself, but didn't get any useful information regarding my problem from the doc. Even though I'm almost certain it's just some trivial error, I fail to spot it. So thanks in advance for anyone trying to help me out here.

Cheers!

[Alex Sha]

The start values your given/guessed are not appropriate。Refer to the result below:

Sum Squared Error (SSE): 31299597.4141903
Root of Mean Square Error (RMSE): 353.128292963896
Correlation Coef. (R): 0.992075938598263
R-Square: 0.984214667945624

Parameter	Best Estimate      	Std. Deviation      	Confidence Bounds[95%] 
---------	-------------      	--------------      	--------------------------------
a        	4.76448778913171E18	1.06042570762486E-5 	[4.76448778913171E18, 4.76448778913171E18]
b        	-0.0445919965391538	24.8230867373737    	[-48.935539282319, 48.8463552892407]
c        	12317.7442468735   	4.75238610300809E-14	[12317.7442468735, 12317.7442468735]

[Jannis Korn]

@Alex Sha Thanks for the info! Is there a way to automatically find more appropriate initial guesses? This is part of a code processing a bunch of different sets of data, so trying varying initial guesses for each of them is not exactly what I planned to do

Answer 1

[Matt J]

Is there a way to automatically find more appropriate initial guesses?

Use fminspleas from the File Exchange. Then you only need an initial guess for 'b'.

https://www.mathworks.com/matlabcentral/fileexchange/10093-fminspleas?s_tid=FX_rc2_behav

load Data

x=Data(:,1); y=Data(:,2);


funlist={@(b,x)exp(b.*x),1};

warning off
[b,ac]=fminspleas(funlist,0,x,y,-2,0)

b = -0.0386

ac = 2×1

1.0e+16 * 4.9888 0.0000

warning on

f=@(x) ac(1)*exp(b*x)+ac(2);

plot(x,y,'x',x,f(x))

[Jannis Korn]

Thanks a lot for your comprehensive answer!

Answer 2

[Steven Lord]

Consider specifying the Normalize name-value pair argument in your fit call to normalize the X data. In my example below I'll use the normalize function instead of that argument, but it's the same general idea.

format longg
x = (750:10:800)';

Your x data is so large in magnitude that the exponential overflows to Inf.

plainExp = exp(x)

plainExp = 6×1

Inf Inf Inf Inf Inf Inf

Even if we cut the magnitude by a factor of 10, the values of the exponential are still really large in magnitude.

expOfOneTenthX = exp(x/10)

expOfOneTenthX = 6×1

1.0e+00 * 3.733241996799e+32 1.01480038811389e+33 2.75851345452317e+33 7.49841699699012e+33 2.03828106651267e+34 5.54062238439351e+34

But if we normalize the x data, the values are much smaller:

centeredAndScaledX = normalize(x)

centeredAndScaledX = 6×1

-1.33630620956212 -0.801783725737273 -0.267261241912424 0.267261241912424 0.801783725737273 1.33630620956212

and so are the values of the exponential. These can be "nicer" to work with.

centedAndScaledExp = exp(centeredAndScaledX)

centedAndScaledExp = 6×1

0.26281466013124 0.448528198864534 0.765473071693191 1.30638168340533 2.22951422570875 3.80496278061747

Answer 3

[Matt J]

It would be better to parametrize the function as

xmin=min(x)

f=@(x)a*exp(-b.*(x-xmin))+c;

The fit is better conditioned that way, and you don't get such insanely large values for a and c. Your original approach works much better with this change, as you can see below.

load Data; x=Data(:,1); y=Data(:,2);


xmin=min(x);
fitfun = fittype( @(a,b,c,x) a*exp(b*(x-xmin))+c );

[fitted_curve,gof] = fit(Data(:,1),Data(:,2),fitfun,'StartPoint',[max(y) -2 min(y)])

fitted_curve = 
     General model:
     fitted_curve(x) = a*exp(b*(x-xmin))+c
     Coefficients (with 95% confidence bounds):
       a =   1.424e+04  (1.396e+04, 1.452e+04)
       b =    -0.04459  (-0.04604, -0.04314)
       c =   1.232e+04  (1.226e+04, 1.237e+04)

gof = struct with fields:
           sse: 3.1300e+07
       rsquare: 0.9842
           dfe: 248
    adjrsquare: 0.9841
          rmse: 355.2577

plot(fitted_curve,x,y)

[Matt J]

You don't really even need a StartPoint guess (although you do need positivity bounds):

[fitted_curve,gof] = fit(Data(:,1),Data(:,2),fitfun,'Lower',[0,0,0])

Warning: Start point not provided, choosing random start point.

fitted_curve = 
     General model:
     fitted_curve(x) = a*exp(-b*(x-xmin))+c
     Coefficients (with 95% confidence bounds):
       a =   1.424e+04  (1.396e+04, 1.452e+04)
       b =     0.04459  (0.04314, 0.04604)
       c =   1.232e+04  (1.226e+04, 1.237e+04)

gof = struct with fields:
           sse: 3.1300e+07
       rsquare: 0.9842
           dfe: 248
    adjrsquare: 0.9841
          rmse: 355.2577

plot(fitted_curve,x,y)