Undefined unary operator '.'' for input arguments of type 'function_handle'.

3 visualizzazioni (ultimi 30 giorni)
Hi,
I'm using matlab 2017b and want to find the gradient of a function with two arguments as defined in the mathwork site:
[FX,FY] = gradient(F) returns the x and y components of the two-dimensional numerical gradient of matrix F. The additional output FY corresponds to ∂F/∂y, which are the differences in the y (vertical) direction.
This is my code:
b=@(x,y) (x-1).^2+(x.^2+y).^2;
[dbdX,dbdY] = gradient(b)
The error I got is:
Undefined unary operator '.'' for input arguments of type 'function_handle'.
How do I solve this? Can somebody help me or give me a hint?

Risposta accettata

Robert U
Robert U il 20 Nov 2019
Modificato: Robert U il 20 Nov 2019
Hi Clarisha Nijman,
your b is a function handle to the anonymous function with input arguments x and y. Gradient itself is - as stated in the documentation - the 2-D gradient of a matrix. In order to calculate the matrix you would have to create it.
b=@(x,y) (x-1).^2+(x.^2+y).^2;
[X,Y] = meshgrid(-100:1:100,-100:1:100);
[dbdX,dbdY] = gradient(b(X,Y));
Please, pay attention that gradient needs the grid having equidistant distribution and if spacing is not "one" you have to scale the result.
Using symbolic toolbox you can apply gradient to scalar functions, which would need you to change your function definition.
Kind regards,
Robert
  4 Commenti
Robert U
Robert U il 21 Nov 2019
Modificato: Robert U il 21 Nov 2019
Hi Clarisha Nijman,
you should check your analytical solution.
Chosen resolution has an influence on the numerically obtained result.
% define point to evaluate
xToEval = 1;
yToEval = 0;
% base function
b=@(x,y) (x-1).^2+(x.^2+y).^2;
% analytical solution
dbdXf=@(x,y) 2.*(x-1) + 4.*(x.^2+y).*x;
dbdYf=@(x,y) 2.*(x.^2 + y);
dbdXRef = dbdXf(xToEval,yToEval);
dbdYRef = dbdYf(xToEval,yToEval);
% resolution of grid
resGrid = 0.1;
% define grid
[X,Y] = meshgrid(xToEval-1:resGrid:xToEval+1,yToEval-1:resGrid:yToEval+1);
% gradient on grid
[dbdX,dbdY] = gradient(b(X,Y),resGrid,resGrid);
% point of interest is in the center of the result matrix
dbdX = dbdX((size(X,1)+1)/2,(size(X,1)+1)/2);
dbdY = dbdY((size(Y,1)+1)/2,(size(Y,1)+1)/2);
Kind regards,
Robert
Clarisha Nijman
Clarisha Nijman il 21 Nov 2019
Okee,
I see, you are zooming in on that area by adjusting the meshGrid such that the point of interest in the middle. I was really wondering how to pick the right answer out of the grid.
Now it is clear to me.
Thank you very much,
kind regards,
Clarisha

Accedi per commentare.

Più risposte (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by