image processing ideal LPF

2023 11 23
1 个回答
更新时间：2023 11 23
3 次查看（30 天）

0 个投票

I made ideal LPF, but dont know it works well
Plz give me advice
close all;
I = rgb2gray(imread('testa.png'));
M = 512;
I = imresize(I, [M, M]);
[m, n] = size(I);
Ip = uint8(zeros(2*m, 2*n));
Ip(1:m,1:n) = double(I);
F = fft2(Ip);
D0 = round(M/10);
u = 0:2*M-1;
v = 0:2*M-1;
[u, v] = meshgrid(u,v);
D = sqrt( (u-((2*M+1)/2)).^2 + (v-((2*M+1)/2)).^2);
H = double(D <= D0);
G = H.*F;
g = ifft2(G);
subplot(2, 2, 1), imshow(I, []), title('Original Image');
subplot(2, 2, 2), imshow(log(1 + abs(fftshift(F))), []), title('Fourier Transform');
subplot(2, 2, 3), imshow(log(1 + abs((H))), []), title('Low-pass Filter');
subplot(2, 2, 4), imshow(real(g), []), title('Filtered Image');

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回答（1 个）

Image Analyst
Image Analyst 2023-11-23
Ran in:

0 个投票

Ran in:
There is no ideal LPF in general. Maybe there is for your particular image though depending on the high frequency noise in your image. To see how well your script/algorithm works, click the green run triangle. I'm attaching some of my demos for you to compare yours to.
Here's what yours does:
close all;
I = rgb2gray(imread('peppers.png'));
M = 512;
I = imresize(I, [M, M]);
[m, n] = size(I);
Ip = uint8(zeros(2*m, 2*n));
Ip(1:m,1:n) = double(I);
F = fft2(Ip);
D0 = round(M/10);
u = 0:2*M-1;
v = 0:2*M-1;
[u, v] = meshgrid(u,v);
D = sqrt( (u-((2*M+1)/2)).^2 + (v-((2*M+1)/2)).^2);
H = double(D <= D0);
G = H.*F;
g = ifft2(G);
subplot(2, 2, 1), imshow(I, []), title('Original Image');
subplot(2, 2, 2), imshow(log(1 + abs(fftshift(F))), []), title('Fourier Transform');
subplot(2, 2, 3), imshow(log(1 + abs((H))), []), title('Low-pass Filter');
subplot(2, 2, 4), imshow(real(g), []), title('Filtered Image');

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