¿What distance does Matlab use in the function "fft2"?

Is it possible to use Chebyshev's distance?

Here you can see the definition of the distance, it is a particular case of the Minkowski distance, when real number p (greater than or equal to one) has infinity.

I try to determine this with empirical way, but i obtain a strange result...

%We generate the sinusoidal grid of 50 cycles   
N = 256 ;
x = 1:N ;
y = 1:N ;
[X,Y] = meshgrid(x,y) ;
I=sin(2*pi*(50*X+50*Y*0)/256)
t=figure; imagesc(I); colormap gray; axis image;
title('Sinusoidal grid of 50 cycles')
exportgraphics(t,'sinusoidal_grid_of_50_cycles.jpg')
%Transform apply the fourier transform,
% we are left with the real part and remove
% the core value
A=real(fftshift(fft2(I)));
A(floor(256/2+1),floor(256/2+1))=0;
t=figure; imagesc(A); colormap gray; axis image;
title('FFT2 of Sinusoidal grid of 50 cycles')
exportgraphics(t,'FFT2_of_sinusoidal_grid_of_50_cycles.jpg')
for i=1:256
for j=1:256
if A(i,j) > 0.1
i 
j
end
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%distance between points to (129,129)     %
%(129,79) (129,179)                       %
% 50 pixels for all p>1                   %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 	Rotation example number 1         %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%We generate the sinusoidal grid of 50 cycles 
N = 256 ;
x = 1:N ;
y = 1:N ;
[X,Y] = meshgrid(x,y) ;
J = sin(2*pi*(50*X+50*Y)/256)
t=figure; imagesc(J); colormap gray; axis image;
title('Rotation example number 1')
exportgraphics(t,'rotation_example_number_1.jpg')
%J=imrotate(I,45,'bilinear','crop')
h=size(J,1);
w=size(J,2);
J=real(fftshift(fft2(J)));
J(floor(h/2+1),floor(w/2+1))=0;
t=figure; imagesc(J); colormap gray; axis image;
title('FFT2 of Rotation example number 1')
exportgraphics(t,'FFT2_of_rotation_example_number_1.jpg')
for i=1:256
for j=1:256
if J(i,j) > 0.1
i 
j
end
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%distance between points to (129,129)         %
%(79,79) (179,179)                            %   
% 70.71 long pixels (euclidean distance) p=2  % 
% 100 long pixels (Manhattan distance) p=1    %
% 50 long pixels p->\infty                    %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 	Rotation example number 2         %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%We generate the sinusoidal grid of 50 cycles 
N = 256 ;
x = 1:N ;
y = 1:N ;
[X,Y] = meshgrid(x,y) ;

I=sin(2*pi*(50*X+25*Y)/256)
t=figure; imagesc(I); colormap gray; axis image;
title('Rotation example number 2')
exportgraphics(t,'rotation_example_number_2.jpg')
%Transform apply the fourier transform,
% we are left with the real part and remove
% the core value
A=real(fftshift(fft2(I)));
A(floor(256/2+1),floor(256/2+1))=0;
t=figure; imagesc(A); colormap gray; axis image;
title('FFT2 of Rotation example number 2')
exportgraphics(t,'FFT2_of_rotation_example_number_2.jpg')
for i=1:256
    for j=1:256
        if A(i,j) > 0.1
            i
            j
        end
    end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%distance between points to (129,129)         %
%(104,79) (154,179)                           %
% 55.9 long pixels (euclidean distance) p=2   %
% 75  long pixels (Manhattan distance)  p=1   %
% 50 long pixels p->\infty                    %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%sqrt((154-129).^2+(179-129).^2)
%sqrt((104-129).^2+(79-129).^2)

%abs(154-129)+abs(179-129)
%abs(154-129)+abs(179-129)


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 	Rotation example number 3         %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%We generate the sinusoidal grid of 50 cycles 
N = 256 ;
x = 1:N ;
y = 1:N ;
[X,Y] = meshgrid(x,y) ;
% I'm not sure if this function is correct,
I=sin(2*pi*(50*X+8*Y)/256)
t=figure; imagesc(I); colormap gray; axis image;
title('Rotation example number 3')
exportgraphics(t,'rotation_example_number_3.jpg')
%Transform apply the fourier transform,
% we are left with the real part and remove
% the core value
A=real(fftshift(fft2(I)));
A(floor(256/2+1),floor(256/2+1))=0;
t=figure; imagesc(A); colormap gray; axis image;
title('FFT2 of Rotation example number 3');
exportgraphics(t,'FFT2_of_rotation_example_number_3.jpg')
for i=1:256
    for j=1:256
        if A(i,j) > 0.1
            i
            j
        end
    end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%distance between points to (129,129)         %
%(121,79) (137,179)                           %
% 50.63 long pixels (euclidean distance) p=2  %
% 58  long pixels (Manhattan distance)  p=1   %
% 50 long pixels p->\infty                    %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


%sqrt((121-129).^2+(79-129).^2)
%sqrt((137-129).^2+(179-129).^2)

%abs(121-129)+abs(79-129)
%abs(137-129)+abs(179-129)



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Answer: if there is nothing wrong it seems %
%       that the distance used is the       % 
%       Minkowski distance p -> infty       %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

[Rena Berman]

(Answers Dev) Restored edit

Answer 1

[Matt J]

fft2 implements the Discrete Fourier Transform, not the continuous one. Therefore, the choice of distance is irrelevant.

[Walter Roberson]

Distance does not apply even to continuous Fourier Transform as far as I can see.

[Matt J]

@Walter Roberson The infinite integral required to compute a continuous Fourier Transform will not generally converge with respect to all distance metrics. For instance, sinc(t) is not absolutely integrable, but it is square integrable.