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[VC] Finish fourier transform sec
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@@ -15,9 +15,32 @@ More general:
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\[
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F(k) = \sum_{n = 0}^{N - 1} x_n e^{-i 2\pi kn \div N} \text{ for } k \in \{ 0, 1, \ldots, N - 1 \}
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\]
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% TODO: Formula to compute the matrix
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For image:
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For images at location $(u, v)$, or $F = Uf$, with $U$ Fourier Mat.:
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\[
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F(u, v) = \sum_{x = 0}^{N - 1}\sum_{y = 0}^{M - 1} I(x, y) \cdot e^{-i2\pi\left( \frac{ux}{N} + \frac{vy}{M} \right)}
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\]
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% TODO: Formula to compute the matrix
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For continuous time, we get:
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\[
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\cF[g(x, y)](u, v) = \int \int_{\R^2} g(x, y) e^{-2i\pi (ux + vy)} \dx x \dx y
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\]
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Inverted by negating exp and swapping $g$ and $F$ (or $F$ and $I$)
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\shorttheorem[Convolution] $\cF[f * g] = F \cdot G$, $\cF[f \cdot g] = F * G$, with $F, G$ the fourier transforms
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\shortremark[Filtering] Can be achieved: $h * f = \cF^{-1}[H \cdot F]$
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\shortremark[Dirac Delta] $\delta(x) = \begin{cases}
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\8 & \text{ if }x = 0 \\
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0 & \text{otherwise}
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\end{cases}$,
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following prop for sampling to alleviate issue of $0$ everywhere after fourier:
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\[
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\int_{-\8}^{\8} f \cdot \delta(x - a) \dx = f(a)
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\]
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\shortremark[Restoration] To restore image, inverse kernel $\tilde{h}(x) = \cF^{-1}\left[ \frac{1}{H(x)} \right]$, with $H(x) = \cF[h(x)]$
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and $h(x)$ degradation filter.
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\shortexample For motion blur: $h(x, y) = \frac{1}{2l} [\theta(x + l) - \theta(x -l)]\delta(y)$. $H(x) = \frac{\sin(2\pi ul)}{2\pi ul}$
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@@ -12,6 +12,7 @@
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\renewcommand{\remarkShortNamingEN}{Rem}
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\renewcommand{\lemmaShortNamingEN}{Lem}
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\renewcommand{\theoremShortNamingEN}{Thm}
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\renewcommand{\exampleShortNamingEN}{Ex}
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\renewcommand{\descriptorNameDisplay}[1]{\textbf{#1}}
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\setupCheatSheet[0.5cm]{Visual Computing}
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