High Pass Butterworth Filter on MATLAB Images

Question

I need to implement a Butterworth high resolution filter in MATLAB to filter images. I implemented one, but it does not seem to work. Here is the code I wrote. Can someone tell me what is wrong?

n=1;
d=50;
A=1.5;
im=imread('imagex.jpg');
h=size(im,1);
w=size(im,2);
[x y]=meshgrid(-floor(w/2):floor(w-1/2),-floor(h/2):floor(h-1/2));
hhp=(1./(d./(x.^2+y.^2).^0.5).^(2*n));
image_2Dfilter=fftshift(fft2(im));
Image_butterworth=image_2Dfilter;
imshow(Image_butterworth);
ifftshow(Image_butterworth);

Answer 1

Firstly, there is no such team, which is called ifftshow. Secondly, you are not filtering anything. All you do is visualize the spectrum of the image.

, , . -, , . , , . , , .

abs. , imshow , , . , , .

. , :

im = imread('cameraman.tif');
figure;
imshow(im);

, - fftshift. double, . , , abs, :

fftim = fftshift(fft2(double(im)));
mag = abs(fftim);
figure;
imshow(mag, []);

, , . log . , , . , 1 , , . , , , log:

figure;
imshow(log(1 + mag), []);

. . . meshgrid . -1, :

[x y]=meshgrid(-floor(w/2):floor(w/2)-1,-floor(h/2):floor(h/2)-1);

, , , . , , . , . :

D(u,v) - , Do - , B - , , . n - . Do d = 50. B = sqrt(2) - 1 , Do, D(u,v) = 1 / sqrt(2) = 0.707, 3 , . , B 1, B = sqrt(2) - 1.

. , . . , fftshift, , , . , uint8, , .

:

%// Your code with meshgrid fix
n=1;
d=50;
h=size(im,1);
w=size(im,2);
fftim = fftshift(fft2(double(im)));
[x y]=meshgrid(-floor(w/2):floor(w/2)-1,-floor(h/2):floor(h/2)-1);
%hhp=(1./(d./(x.^2+y.^2).^0.5).^(2*n));

%%%%%%// New code
B = sqrt(2) - 1; %// Define B
D = sqrt(x.^2 + y.^2); %// Define distance to centre
hhp = 1 ./ (1 + B * ((d ./ D).^(2 * n)));
out_spec_centre = fftim .* hhp;

%// Uncentre spectrum
out_spec = ifftshift(out_spec_centre);

%// Inverse FFT, get real components, and cast
out = uint8(real(ifft2(out_spec)));

%// Show image
imshow(out);

, , :

figure;
imshow(log(1 + abs(out_spec_centre)), []);

:

. , . , .

out , :

! uint8 0 , 255-255. , , ... , [0,1], uint8 255. , , , .... - :

%// Your code with meshgrid fix
n=1;
d=50;
h=size(im,1);
w=size(im,2);
fftim = fftshift(fft2(double(im)));
[x y]=meshgrid(-floor(w/2):floor(w/2)-1,-floor(h/2):floor(h/2)-1);
%hhp=(1./(d./(x.^2+y.^2).^0.5).^(2*n));

%%%%%%// New code
B = sqrt(2) - 1; %// Define B
D = sqrt(x.^2 + y.^2); %// Define distance to centre
hhp = 1 ./ (1 + B * ((d ./ D).^(2 * n)));
out_spec_centre = fftim .* hhp;

%// Uncentre spectrum
out_spec = ifftshift(out_spec_centre);

%// Inverse FFT, get real components
out = real(ifft2(out_spec));

%// Normalize and cast
out = (out - min(out(:))) / (max(out(:)) - min(out(:)));
out = uint8(255*out);

%// Show image
imshow(out);

:

Answer 2

, -

n=1;
D0=50; % change the name for d0, d is usuaally the (u²+v²)⁽1/2)

A=1.5; % normally the amplitude is 1

im=imread('cameraman.jpg');

[M,N]=size(im); % is easy to get the h and w like this

% compute the 2d fourier transform in order to multiply

F=fft2(double(im));

% compute your filter and do the meshgrid for your matrix but it is M*n, and get only the real part

u=0:(M-1);
v=0:(N-1);

idx=find(u>M/2);
u(idx)=u(idx)-M;
idy=find(v>N/2);
v(idy)=v(idy)-N;
[V,U]=meshgrid(v,u);
D=sqrt(U.^2+V.^2);
H =A * (1./(1 + (D0./D).^(2*n)));

% multiply element by element

G=H.*F;
g=real(ifft2(double(G)));
subplot(1,2,1); imshow(im); title('Input image');
subplot(1,2,2); imshow(g,[ ]); title('filtered image');