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    <p>Let <strong>f</strong> be the original image, <strong>g</strong> be the filtered one, and <strong>h</strong> the filter applied to <strong>f</strong>, so that:</p> <pre><code>f * h = g </code></pre> <p>Passing that to the frequency domain:</p> <pre><code>F.H = G, so H = G/F </code></pre> <p>The problem is that inverting F is VERY sensitive to noise.</p> <p>How to implement that in MATLAB:</p> <pre><code>close all; f = imread('cameraman.tif'); [x,y] = size(f); figure,imshow(f); h = fspecial('motion', 20, 40); % abitrary filter just for testing the algorithm F = fft2(f); H = fft2(h,x,y); G = F.*H; g = ifft2(G); % the filtered image figure, imshow(g/max(g(:))); % Inverting the original image epsilon = 10^(-10); small_values = find(abs(F)&lt;epsilon); F(small_values) = epsilon; F_i = ones(x,y)./F; H_calculated = G.*F_i; h_calculated = ifft2(H_calculated); % remove really small values to try to infer the original size of h r = sum(h_calculated,1)&lt;epsilon; c = sum(h_calculated,2)&lt;epsilon; h_real = h_calculated(~r,~c); % Calculate error % redo the filtering with the found filter figure,g_comp = ifft2(fft2(f).*fft2(h_real,x,y)); imshow(g_comp/max(g_comp(:))); rmse = sqrt(mean(mean((double(g_comp) - double(g)).^2,2),1)) </code></pre> <p><strong>edit</strong>: Just to explain the epsilon part:</p> <p>It can be that some values in F are zero, or very close to zero. If we try to invert F with these small values, we would have problems with infinity. The easy way to solve that is to truncate every value in F that is smaller than an arbitrarily small limit, epsilon on the code.</p> <p>Mathematically, what was done is this:</p> <pre><code>For all F &lt; epsilon, F = epsilon </code></pre>
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    1. COIt's also very sensitive to spectral nulls.
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    2. COYep, I know, that's why I'm truncating it. To me the real problem with this approach is noise. But it's definitely better than nothing at all...
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