從相位/大小的Matlab逆FFT僅

Question

所以我有這個形象'我'。我用F = fft2（I）來獲得二維傅里葉變換。重建它，我可以去ifft2（F）。

問題是，我需要從a）幅度和b）F相位分量重建此圖像。我怎樣才能分離傅里葉變換的這兩個分量，然後從每個分量重建圖像？

我試過abs（）和angle（）函數來獲得幅度和相位，但是第一個階段不能正確重構。

幫助？

Answer 1

您需要一個與F和0相位相同的矩陣，以及另一個與F相同的相位和均勻的幅度。正如你注意到的abs給你的大小。爲了得到均勻一致的相位矩陣，需要使用angle來獲得相位，然後將相位分離回實部和虛部。

> 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

爲時已晚把另一個回答這個職位，但...反正

@ zhilevan，你可以使用我寫的代碼使用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');