Question

所以我有这个'我'的形象。我用F = fft2（I）得到2D傅里叶变换。为了重建它，我可以去ifft2（F）。

问题是，我需要仅从a）幅度和b）F的相位分量重建这个图像。我如何分离傅立叶变换的这两个分量，然后从每个变换图像？

我尝试了abs（）和angle（）函数来获得幅度和相位，但第一阶段不会正确重建。

帮助？

Answer 1

您需要一个幅度与F和0相同的矩阵，另一个矩阵与F具有相同的相位并且幅度均匀。如您所述，abs为您提供了重要性。要获得均匀幅度相同的相位矩阵，您需要使用angle来获得相位，然后将相位分离回实部和虚部。

> F_Mag = abs(F); %# has same magnitude as F, 0 phase
> F_Phase = cos(angle(F)) + j*(sin(angle(F)); %# has magnitude 1, same phase as F
> I_Mag = ifft2(F_Mag);
> I_Phase = ifft2(F_Phase);

Answer 2

现在为这篇文章提出另一个答案为时已晚，但是......无论如何

@ zhilevan，你可以使用我用mtrw的答案编写的代码：

image = rgb2gray(imread('pillsetc.png')); 
subplot(131),imshow(image),title('original image');
set(gcf, 'Position', get(0, 'ScreenSize')); % maximize the figure window
%:::::::::::::::::::::
F = fft2(double(image));
F_Mag = abs(F); % has the same magnitude as image, 0 phase 
F_Phase = exp(1i*angle(F)); % has magnitude 1, same phase as image
% OR: F_Phase = cos(angle(F)) + 1i*(sin(angle(F)));
%:::::::::::::::::::::
% reconstruction
I_Mag = log(abs(ifft2(F_Mag*exp(i*0)))+1);
I_Phase = ifft2(F_Phase);
%:::::::::::::::::::::
% Calculate limits for plotting
% To display the images properly using imshow, the color range
% of the plot must the minimum and maximum values in the data.
I_Mag_min = min(min(abs(I_Mag)));
I_Mag_max = max(max(abs(I_Mag)));

I_Phase_min = min(min(abs(I_Phase)));
I_Phase_max = max(max(abs(I_Phase)));
%:::::::::::::::::::::
% Display reconstructed images
% because the magnitude and phase were switched, the image will be complex.
% This means that the magnitude of the image must be taken in order to
% produce a viewable 2-D image.
subplot(132),imshow(abs(I_Mag),[I_Mag_min I_Mag_max]), colormap gray 
title('reconstructed image only by Magnitude');
subplot(133),imshow(abs(I_Phase),[I_Phase_min I_Phase_max]), colormap gray 
title('reconstructed image only by Phase');

仅来自相位/幅度的Matlab逆FFT

2 个答案: