CODEWITHSUNDEEP
matlabequationcurvecurve-fitting
user238469
user238469

Reputation:

How can I find equation of a plot connecting data points in Matlab?

I have various plots (with hold on) as show in the following figure: enter image description here

I would like to know how to find equations of these six curves in Matlab. Thanks.

Upvotes: 9

Views: 33409

Answers (5)

user238469
user238469

Reputation:

I found interactive fitting tool in Matlab simple and helpful, though somewhat limited in scope:

enter image description here

Upvotes: 3

Mesop
Mesop

Reputation: 5263

There is a way to extract information with the current figure handle (gcf) from you graph.

For example, you can get the series that were plotted in a graph:

% Some figure is created and data are plotted on it
figure;
hold on;
A = [ 1 2 3 4 5 7] % Dummy data
B = A.*A % Some other dummy data
plot(A,B);
plot(A.*3,B-1);

% Those three lines of code will get you series that were plotted on your graph
lh=findall(gcf,'type','line'); % Extract the plotted line from the figure handle
xp=get(lh,'xdata'); % Extract the Xs
yp=get(lh,'ydata'); % Extract the Ys

There must be other informations that you can get from the "findall(gcf,...)" methods.

Upvotes: 0

Jack_of_All_Trades
Jack_of_All_Trades

Reputation: 11498

The advice, though there might be better answer, from me is: try to see the rate of increase in the curve. For example, cubic is more representative than quadratic if the rate of increase seems fast and find the polynomial and compute the deviation error. For irregular curves, you might try spline fitting. I guess there is also a toolbox in matlab for spline fitting.

Upvotes: 0

Diego
Diego

Reputation: 18379

Does rxns stand for reactions? In that case, your curves are most likely exponential. An exponential function has the form: y = a*exp(b * x) . In your case, y is the width of mixing zone, and x is the time in years. Now, all you need to do is run exponential regression in Matlab to find the optimal values of parameters a and b, and you'll have your equations.

Upvotes: 0

richard
richard

Reputation: 605

The graph above seems to be linear interpolation. Given vectors X and Y of data, where X contains the arguments and Y the function points, you could do 

f = interp1(X, Y, x)

to get the linearly interpolated value f(x). For example if the data is 

X = [0 1 2 3  4  5];
Y = [0 1 4 9 16 25];

then 

y = interp1(X, Y, 1.5)

should give you a very rough approximation to 1.5^2. interp1 will match the graph exactly, but you might be interested in fancier curve-fitting operations, like spline approximations etc.

Upvotes: 0

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