在 Matlab 问题中使用 FFT 的 2D 反卷积 [英] 2D Deconvolution using FFT in Matlab Problems
问题描述
我用我在 matlab 中创建的图像与我也在 matlab 中定义的 2D 高斯函数进行了卷积，现在我正在尝试对结果矩阵进行解卷积，以查看是否使用 fft2 和 ifft2 命令获得了 2D 高斯函数.然而，我得到的矩阵是不正确的(据我所知).这是我迄今为止所做的代码:
I have convoluted an image I created in matlab with a 2D Gaussian function which I have also defined in matlab and now I am trying to deconvolve the resultant matrix to see if I get the 2D Gaussian function back using the fft2 and ifft2 commands. However the matrix I get as a result is incorrect (to my knowledge). Here is the code for what I have done thus far:
% 输入图像 (img) [300x300 数组] 的代码
% Code for input image (img) [300x300 array]
N = 100;
t = linspace(0,2*pi,50);
r = (N-10)/2;
circle = poly2mask(r*cos(t)+N/2+0.5, r*sin(t)+N/2+0.5,N,N);
img = repmat(circle,3,3);
% 二维高斯函数代码，c = 0 sig = 1/64 (Z) [300x300 数组]
% Code for 2D Gaussian Function with c = 0 sig = 1/64 (Z) [300x300 array]
x = linspace(-3,3,300);
y = x';
[X Y] = meshgrid(x,y);
Z = exp(-((X.^2)+(Y.^2))/(2*1/64));
% Code for 2D Convolution of img with Z (C) [599x599 array]
% Code for 2D Convolution of img with Z (C) [599x599 array]
C = conv2(img,Z);
% 我已经使用 img 和 C 的横截面轮廓向量测试了这个卷积是正确的，并且得到的 x-y 图是我对卷积的期望.
% I have tested that this convolution is correct using cross section profile vectors for img and C and the resulting x-y plots are what i expect from the convolution.
% 根据我对卷积的了解，该算法在傅立叶空间中用作乘法器，因此通过将我的输出(卷积图像)的傅立叶变换除以我的输入(img)，我应该得到点扩散函数(Z -2D 高斯函数)在逆傅里叶变换后通过除法应用于该结果.
% From my knowledge of convolution, the algorithm works as a multiplier in Fourier space, therefore by dividing the Fourier transform of my output (convoluted image) by my input (img) I should get back the point spread function (Z - 2D Gaussian function) after the inverse Fourier transform is applied to this result by division.
% 尝试二维反卷积的代码
% Code for attempted 2D deconvolution
Fimg = fft2(img,599,599);
添加了 % 零填充以将结果增加到 599x599 数组
% zero padding added to increase result to 599x599 array
FC = fft2(C);
R = FC/Fimg;
% 我现在收到此错误提示:警告:矩阵接近单数或严重缩放.结果可能不准确.RCOND = 2.551432e-22
% I now get this error prompt: Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 2.551432e-22
iFR = ifft2(R);
我期待 iFR 接近 Z，但我得到了完全不同的东西.它可能是具有复数值的 Z 的近似值，但我似乎无法检查它，因为我不知道如何在 matlab 中绘制 3D 复矩阵.因此，如果有人能告诉我我的答案是正确还是错误，以及如何使这种反卷积起作用?我将不胜感激.
I'm expecting iFR to be close to Z but I'm getting something completely different. It may be an approximation of Z with complex values but I can't seem to check it since I don't know how to plot a 3D complex matrix in matlab. So if anyone can tell me whether my answer is correct or incorrect and how to get this deconvolution to work? I'd be much appreciated.
推荐答案
R = FC/Fimg 需要 R = FC./Fimg; 你需要做按元素划分.
R = FC/Fimg needs to be R = FC./Fimg; You need to do division element-wise.
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