从相位/大小的Matlab逆FFT仅

Question

所以我有这个形象'我'。我用F = fft2（I）来获得二维傅里叶变换。重建它，我可以去ifft2（F）。

问题是，我需要从a）幅度和b）F相位分量重建此图像。我怎样才能分离傅里叶变换的这两个分量，然后从每个分量重建图像？

我试过abs（）和angle（）函数来获得幅度和相位，但是第一个阶段不能正确重构。

帮助？

Answer 1

您需要一个与F和0相位相同的矩阵，以及另一个与F相同的相位和均匀的幅度。正如你注意到的abs给你的大小。为了得到均匀一致的相位矩阵，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

为时已晚把另一个回答这个职位，但...反正

@ zhilevan，你可以使用我写的代码使用mtrw的回答是：

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');