Fitting two Parameters to an equation with data

Hello, I've been stumped trying to solve this problem for a while now and was hoping someone here might be able to help. For background info I'm not completely new to MATLAB, I have a good understanding of the fundamentals and I've developed scripts and functions to run things like Newton-Rhapson method etc...

I have a set of data:

x=[0;69.18248;78.63425;83.30743;86.66291;87.06997]; t=[0;1;2;3;4;5];

And I want to fit an equation to this data. I have the equation shown below:

x= 1 - 0.5584*exp(-a*t) - 0.4416*exp(-b*t)

As you can see, I have the x data points for each t value. I do not, however, know the constant parameters a and b. I wish to find a value for them that will best fit this data. I've looked around for a solution and some say it might be possible using fminsearch but I don't understand how to set up the functions required to do so. Any help is much appreciated.

[Matt J]

Your equation can't possibly give a good fit to the data you've shown. Your equation gives values less than 1 for all t, but your given x data are very much greater than 1.

[R]

I was told the value for a was approximately 0.02-0.04 and that b was around 0.0003. I'm just trying to confirm these two values. I know that a should be 100-1000 times larger than b. Also, using your code I get an error:

"Subscripted assignment dimension mismatch.

Error in fminsearch (line 191) fv(:,1) = funfcn(x,varargin{:});"

[Yao Li]

I've tried the curve fitting tool. It seems Matt is right. Your equation cannot fit the data well

[R]

Okay guys, I'm sorry. I was given the wrong data points. Those x values should be /100 as they are percentages...

Answer 1

[Matt J]

 y=1-x;
 fun=@(ab) abs( .5584*exp(-ab(1)*t)+.4416*exp(-ab(2)*t)  - y);

 ab=fminsearch(fun,[0,0])

 a=ab(1);
 b=ab(2);

[Matt J]

Sorry, should be norm() instead of abs().

[R]

Alright, thanks a lot for the help! You've solved my problem.

Answer 2

[Yao Li]

General model:
     f(x) = 1 - 0.5584*exp(-a*x) - 0.4416*exp(-b*x)
Coefficients (with 95% confidence bounds):
       a =       16.92  (-2.71e+06, 2.71e+06)
       b =      0.3107  (0.2453, 0.3762)

Goodness of fit:
  SSE: 0.002167
  R-square: 0.9962
  Adjusted R-square: 0.9952
  RMSE: 0.02328

[mht6]

Could you elaborate on how you achieved/calculated the goodness of fit? It would be nice to apply similar information to data I am fitting

[Matt J]

By requesting two or more outputs from fit(), see for example,

https://www.mathworks.com/help/curvefit/fit.html#bto2vuv-7