We’re rewarding the question askers & reputations are being recalculated! Read more.

# Tag Info

Actually the down sampling has no role here. It is all based on a real simple equation: $$I = A + B$$ It is always enough to keep 2 terms of the 3 to restore completely and perfectly the information. So let's look on this: $${I}_{0} = \left( \left( {I}_{0} \downarrow \right) \uparrow \right) + {R}_{0}$$ So if we keep ${R}_{0}$ and we have $\left( ... 2 Because wo want to get the centroid of the image(a block/patch) by the intensity. m00:p = q = 0,sum the intensity matrix. m10:p =1,q = 0,sum of the x-direction. m01:p = 0,q = 1,sum of the y-direction. (m10/m00,m01/m00) is the centroid. 1 A very simple example on a$2\times 2$image $$I_0=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ with (very crude Gaussian) low-pass: $$g=1/4\begin{bmatrix}1&1\\1&1\end{bmatrix}$$ yields a downsampled$I_1$after filtering, with only one pixel (out of 4): $$I_1=\begin{bmatrix}(a+b+c+d/4)\end{bmatrix}$$ It can be upsampled as:$\$U(I_1) = I_1^\...