IFFT2 as a Symbolic Function

11 次查看（过去 30 天）

John Shackleton 2020-8-10

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/577834-ifft2-as-a-symbolic-function

I have a 2D matrix which I would like to take an explicit derivative with, so it may be helpful to have a functional form of the matrix. I thought performing a 2D FFT and then expressing the matrix in terms of the Fourier basis would be a good idea, but I'm having some difficulties. I'd like that if given parameters of an x, y meshgrid and the matrix f (which is a function of x and y) over the meshgrid, the function would create a function f2 such that f2(x(ii, jj), y(ii, jj)) = f(ii, jj).

xc = -10;
yc = -10;
d = -0.1;
[x, y] = meshgrid(xc:d:10, yc:d:10);
f = sin(x + y);
m = size(f, 1);
n = size(f, 2);
coefs = fft2(f);
omega_x = exp((2*pi.*1i./n).*(0:(n-1)));
omega_y = exp((2*pi.*1i./n).*(0:(m-1)));
omega_y = omega_y';
func = @(x, y) (1/m).*(1/n).*sum( ...
    sum( ...
    (omega_x.^((x-xc)./d).* ...
    (omega_y.^((y-yc)./d))) ...
    .*coefs ...
    ) ...
    );
f2 = zeros(size(f));
for ii = 1:size(f, 1)
    for jj = 1:size(f, 2)
        f2(ii, jj) = func(x(ii, jj), y(ii, jj));
    end
end