Creating phase plot from initial data

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Gavin Seddon 2017-3-30

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评论： Star Strider 2018-3-21

I have a data table with no associated function, is it possible to generate the vector field and phase diagram from this? GS.

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KSSV 2017-3-30

What data you have?

Gavin Seddon 2017-11-15

You are welcome, I have tried the arc tangent with my table and it appears I should define two points and calculate the angle. Also I cannot calculate the gradient since it complains that my matrix is not equal, Possibly because I am using fxy (multivariate). Nevertheless I have been of data now to complete the job. GS.

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Star Strider 2017-3-30

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Use the gradient function to create the derivatives.

See if this works with your data:

x = sin(0:0.5:2*pi*2);                          % Create Data
y = cos(0:0.5:2*pi*2);                          % Create Data
u = gradient(x);                                % Calculate ‘x’ Derivative
v = gradient(y);                                % Calculate ‘y’ Derivative
figure(1)
quiver(x, y, u, v)
grid
axis equal

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Star Strider 2017-4-29

The symfun function allows you to create a symbolic function, then manipulate it symbolically so that it does what you want.

To use it in your numeric MATLAB script, you then have to use the matlabFunction function to create an anonymous function or a function file from it (your decision, so choose whatever is most appropriate). You can then use that function in your numeric MATLAB code. (Print the anonymous function to the Command Window by removing the parenthesis from the end of the matlabFunction call, then copy it and paste it into your numeric calculation script.) Then you can call the function from your script, just as you would any other function.

Please read through the documentation on matlabFunction thoroughly. It has a number of options that can make your task significantly easier.

I don’t know what your symfun function is, so look through the documentation on the odeToVectorField function as well to see if it could be applicable. Use it before you call matlabFunction.

It is best not to do extensive or recursive calculations in the Symbolic Math Toolbox, unless you need its extended-precision calculation ability. It isn’t optimised for such calculations, and is extremely slow, when compared to doing the same calculations in non-symbolic code.

Star Strider 2017-5-17

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I’m not certain where we’re starting from, here. If both ‘fs’ and ‘fx’ are symbolic functions, you can manipulate them in the Symbolic Math Toolbox and then use matlabFunction to convert them to anonymous functions to use in numerical calculations. (The Symbolic Math Toolbox can do one-off numerical calculations. It is extremely inefficient for iterative calculations.)

If you want to manipulate your anonymous function in the Symbolic Math Toolbox, you have to create it new as a symbolic function. (This is a relatively simple copy-paste of everything except the function handle designation. It may be necessary to remove the dot-operators for element-wise calculations.) You can use it as an anonymous function in your symbolic code, however since it exists as a function handle, and not a symbolic object, you cannot do much with it as an anonymous function in the Symbolic Math Toolbox.

To Illustrate —

syms x y 
anon_fcn = @(x) x.^2 + 5*x - 2;                         % Anonymous Function
out = anon_fcn(y);                                      % Substitutes ‘y’
% q1 = diff(anon_fcn);                                    % Throws Error
sym_fcn = symfun(x.^2 + 5*x - 2, x);                    % Creates Symbolic Function
q2 = diff(sym_fcn);                                     % Calculates Derivative

Gavin Seddon 2018-3-21

Hello, I have decided the best way forward is to draw a labelled triangle. Then using my derivative as the absolute hypotenuse Value, then the angle cos = adjacent/hypotenuse. Thank you. GS.