How do I extract the exponent from a semilogy plot?

If I use semilogy to plot some data such that the plot appears as a straight line, how can I extract the exponent of this line? Can I use polyfit to get this or is there a different way?

Answer 1

[Stephan]

Hi,

i think you should use lsqcurvefit (requires Optimization Toolbox) or fit (requires Curve Fitting Toolbox) - depending on the toolboxes you have access to. I would expect an exponential function if there is a straight line in the semilogy plot. Due to this polyfit is not the correct choice. I recommend to work through this examples:

lsqcurvefit:

https://de.mathworks.com/help/optim/ug/lsqcurvefit.html#buuu2mv-1

fit:

https://de.mathworks.com/help/curvefit/fit.html#bto2vuv-9

In case of lsqcurvefit the resulting elements of solutiuon x contain the needed exponent. In case of fit you get the values with the coeffvalues function.Therefore see also Fit Postprocessing.

Best regards

Stephan

[Janna Hinchliff]

Hi, thanks for your help, I'm having a go with fit but struggling to get it to work. I have a very small data set , with

x = [2.3561 2.4831 2.6233];
y = [0.2452 1.4632 1.6916];

which I'm trying to fit to the curve a*exp(-b*x). The fit isn't very good but I need to look into how good it is so I need to fit something of this form to it. I tried to use

    expfit = fittype('a*exp(-b*x)','dependent',{'y'},'independent',{'x'},'coefficients',{'a','b'})
    fitEa = fit(x',y',expfit)

but this gave me the error

Function 'subsindex' is not defined for values of class 'fittype'. How can I do this correctly?

[Stephan]

I dont get an error with your fittype - which Matlab Release do you use?

x = [2.3561 2.4831 2.6233];
y = [0.2452 1.4632 1.6916];
expfit = fittype('a*exp(-b*x)','dependent',{'y'},'independent',{'x'},...
    'coefficients',{'a','b'})
f = fit(x',y',expfit);
fvals = f(x);

format long
coeffs = coeffvalues(f)
names = coeffnames(f)
format short

subplot(2,1,1)
plot(f,x,y)
legend('Location','southeast')

subplot(2,1,2)
semilogy(x,fvals)
ylim([0 2])

[Janna Hinchliff]

No worries, I sorted it - I had another variable stored as fit from a different part of the code! Thanks for your help!

[Stephan]

No problem - if this was useful for you please accept my answer.