Curve fitting with constraints
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Hi,
I am trying to fit an curve with an exponential function and need it to pass through (0,0) with a gradient of 0 at that point.
I've attached a picture of the data and curve. I am trying to use to the curve fitting toolbox to do this.
Any ideas would be greatly appreciated
Thanks
댓글 수: 8
Torsten
2019년 3월 1일
In your graphics, the x-values are positive (range from 0 to 90) and the y-values are negative( range from 0 to -3). If this is reversed in reality, you can fit the inverse function by the above ansatz.
채택된 답변
Matt J
2019년 3월 1일
편집: Matt J
2019년 3월 1일
Using the 'power1' fit type, I obtain pretty close agreement.
>> cfit = fit(x1(x1>0),y1(x1>0),'power1')
cfit =
General model Power1:
cfit(x) = a*x^b
Coefficients (with 95% confidence bounds):
a = -0.07755 (-0.07888, -0.07622)
b = 3.316 (3.312, 3.32)
>> plot(cfit,x1,y1)
댓글 수: 4
Nicholas Ross
2024년 2월 26일
@Matt J thank you for posting this! I wasn't getting a good result with exp1 or lsqcurvefit and this fit my curve perfectly! Thanks for making my Monday morning
추가 답변 (2개)
Alex Sha
2020년 2월 1일
Hellow, Paul, if adding one more parameter in your fitting function, i.e. y = a*exp(b*x) + c*exp(d*x)+f
Then:
from y(0) = 0, we get: f = -(a+c);
from y'(0) = 0, we get: d=-a*b/c;
It can be easy to get the:
Root of Mean Square Error (RMSE): 214.862401971985
Sum of Squared Residual: 434512996.964381
Correlation Coef. (R): 0.999996411191975
R-Square: 0.999992822396829
Adjusted R-Square: 0.99999282087114
Determination Coef. (DC): 0.999991748082927
Chi-Square: -26589.2854996605
F-Statistic: 569341000.214239
Parameter Initial Value
a -20088.9862898102
b 0.0320322385142336
c 55098.3960866006
댓글 수: 0
John D'Errico
2019년 3월 1일
편집: John D'Errico
2019년 3월 1일
You cannot use the curve fitting toolbox to fit a model that has constraints on it like the ones you have. At least not directly.
Lets look at your model function however. The general model is:
y = a*exp(b*x) + c*exp(d*x)
Your constraints are that
y(0) = 0
y'(0) = 0
What do they mean in terms of the parameters?
y(0)=0 ==> a + b = 0
Or, we can write it as a = -b, essentially allowing us to eliminate b from the model completely. So now we can re-write the model in the form
y = a*exp(b*x) - a*exp(d*x)
Next, consider the slope boundary condition at x=0. Again, since exp(0) is 1, this reduces to:
y'(0) = 0 ==> a*b - a*c = 0
That is a problem however. Why? If we assume a is non-zero (as the problem becomes trivial if a is zero) then we now know that b == c. So we can reduce the model once more, into the form:
y(x) = a*exp(b*x) - a*exp(b*x) = 0
Does that seem problematic? It has a difference of two identical terms. So the ONLY exponential model that is a sum of exactly two exponentials of the form that you want, that also has y(0)=y-(0)=0, is the oddly trivial one:
y(x) = 0
So, can you solve this using the curve fitting toolbox? I suppose you might be able to do so, although I'm not sure how to enter a model that is constant at zero for all x.
I would strongly recommend you choose a model that does allow the required properties to hold, but also allows a fit to exist. Without seeing your data, it is very difficult to know what that model might be. For example, you might be able to use a very simple model of the form:
y(x) = a2*x.^2 + a3*x.^3 + a4*x.^4
etc. Don't go too high in the order of that model, as it will rapidly have numerical difficulties. You may need to consider scaling your data. A virtue of the model form I have suggested is it automatically forces y(0)=y'(0)=0.
My guess is that the data you have has relatively large x. So, how would I scale this problem? I would scale it as:
s = max(abs(x));
Now, solve the problem using the curve fitting toolbox by scaling x and y as:
xs = x/s;
ys = y/s^2;
ft = fittype('a2*x.^2 + a3*x.^3 + a4*x.^4','indep','x');
mdl = fit(xs(:),ys(:),ft);
Don't forget to unscale the coefficients afterwards. If you need a better recommendation for a model, you need to provide the data.
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