使用fft2在Matlab中对两个图像进行线性卷积 [英] Linear convolution of two images in Matlab using fft2

问题描述

我想拍摄两张图像并使用2D FFT在Matlab中将它们组合在一起，而无需使用 conv2 函数。但是，我不确定如何正确填充矩阵并为卷积做好准备。

I would like to take two images and convolve them together in Matlab using the 2D FFT without recourse to the conv2 function. However, I am uncertain with respect to how the matrices should be properly padded and prepared for the convolution.

数学运算如下：

A * B = C

在上面，*是卷积运算符（维基百科链接）。

In the above, * is the convolution operator (Wikipedia link).

以下Matlab程序显示填充和不填充之间的区别矩阵。我怀疑没有填充矩阵导致循环卷积，但我想执行线性卷积而没有别名。

The following Matlab program shows the difference between padding and not padding the matrices. I suspect that not padding the matrices results in a circular convolution, but I would like to perform a linear convolution without aliasing.

如果我这样做填充两个矩阵，然后如何截断卷积的输出，使 C 与 A 和 B 的大小相同？

If I do pad the two matrices, then how do I truncate the output of the convolution so that C is the same size as A and B?

A = rgb2gray(im2double(imread('1.png'))); % input A
B = rgb2gray(im2double(imread('2.png'))); % kernel B

figure;
imagesc(A); colormap gray;
title ('A')

figure;
imagesc(B); colormap gray;
title ('B')

[m,n] = size(A);
mm = 2*m - 1;
nn = 2*n - 1;

C = (ifft2(fft2(A,mm,nn).* fft2(B,mm,nn)));

figure;
imagesc(C); colormap gray;
title ('C with padding')

C0 = (ifft2(fft2(A).* fft2(B)));

figure;
imagesc(C0); colormap gray;
title ('C without padding')

这是程序的输出：

推荐答案

如果没有填充，结果将等同于循环卷积。对于线性卷积，在卷积2个图像（2D信号）A * B时，完整输出的大小 Ma + Mb-1 x Na + Nb-1 ，其中 Ma x Na，Mb x Nb 图像A和B的大小。

Without padding the result will be equivalent to circular convolution as you point out. For linear convolution, in convolving 2 images (2D signals) A*B the full output will be of size Ma+Mb-1 x Na+Nb-1, where Ma x Na, Mb x Nb the sizes of images A and B resp.

填充到预期大小后，通过 ifft2 进行乘法和转换，可以保持中心部分得到的图像（通常对应于A和B中最大的一个）。

After padding to the expected size, multiplying and transforming back, via ifft2, you can keep the central part of the resulting image (usually corresponding to the largest one of A and B).

A = double(imread('cameraman.tif'))./255; % image
B = fspecial('gaussian', [15 15], 2); % some 2D filter function

[m,n] = size(A);
[mb,nb] = size(B); 
% output size 
mm = m + mb - 1;
nn = n + nb - 1;

% pad, multiply and transform back
C = ifft2(fft2(A,mm,nn).* fft2(B,mm,nn));

% padding constants (for output of size == size(A))
padC_m = ceil((mb-1)./2);
padC_n = ceil((nb-1)./2);

% frequency-domain convolution result
D = C(padC_m+1:m+padC_m, padC_n+1:n+padC_n); 
figure; imshow(D,[]);

现在，使用 conv2D <将上述内容与空间域卷积进行比较/ code>

Now, compare the above with doing spatial-domain convolution, using conv2D

 % space-domain convolution result
 F = conv2(A,B,'same');
 figure; imshow(F,[]);

结果在视觉上是相同的，两者之间的总误差（由于四舍五入）在 e-10。

Results are visually the same, and total error between the two (due to rounding) on the order of e-10.

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