仅来自相位/幅度的Matlab逆FFT

jiusto 发布于 2018-12-06 fft 最后更新 2018-12-06 22:59 5 浏览

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)相位分量重建这幅图像。我怎样才能将傅里叶变换的这两个分量分开,然后从每个分量重建图像? 我尝试了abs()和angle()函数来获得幅度和相位,但第一阶段不能正确重构。 帮帮我?

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esaepe

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对这篇文章提出另一个答案为时已晚,但是......无论如何 @ zhilevan,您可以使用我用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');

hipsum

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您需要一个与F和0相同幅度的矩阵,以及与F具有相同相位并且幅度均匀的另一个矩阵。正如您所指出的,abs为您提供了幅度。要获得均匀幅度相同的相位矩阵,您需要使用angle来获取相位,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);

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