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Answer to: Is there an easier way to count fringes? I synthesized an image to resemble your image with x = linspace(0, 1, 256); [x,y] = meshgrid(x,x); image = sin((7*x + 2*y) * 2*pi) ./ (0.1 + (x - 0.5).^2 + (y - 0.6).^2); imagesc(image); Then I can get the vector the 2D FFT A = abs(fft2(image)); A = A(1:end/2, 1:end/2) [~, imax] = max(abs(A(:))) % 899 % ...


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If the watermark is additive and static and, the content is smooth, and the vide long enough you could compute the watermark that minimizes $\sum ||S(F_i - W)||^2$, for the frames $F_i$ and a fixed watermark $W$. The solution $W$ is expected to converge to the watermark. But it is always better to use the original video.


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