image processing ideal LPF

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');

Antworten (1)

Image Analyst
Image Analyst am 23 Nov. 2023

0 Stimmen

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');

Kategorien

Mehr zu Read, Write, and Modify Image finden Sie in Hilfe-Center und File Exchange

Tags

Gefragt:

am 23 Nov. 2023

Beantwortet:

am 23 Nov. 2023

Community Treasure Hunt

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

Start Hunting!

Translated by