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Th`ese Marouan BOUALI - Sites personnels de TELECOM ParisTech

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92 4. A Variational approach for the <strong>de</strong>striping issue<br />

(FC) algorithm [Frankot and Chellappa, 1988], which provi<strong>de</strong>s a solution by consi<strong>de</strong>ring<br />

the Fourier transform of the gradient field G components. Let us <strong>de</strong>note by (ξ x ,ξ y ) the<br />

frequency components as in [Bracewell, 1986] and recall the differentiation properties of<br />

the Fourier transform :<br />

F<br />

⎧⎪ ( ∂u<br />

( ∂x)<br />

) = j.ξx F(u)<br />

⎨ F ∂u<br />

∂y<br />

= j.ξ y F(u)<br />

( )<br />

F ∂ 2 u<br />

= ξ 2 (4.73)<br />

xF(u)<br />

⎪ ⎩<br />

∂ 2 x<br />

F<br />

(<br />

∂ 2 u<br />

∂ 2 y<br />

)<br />

= ξyF(u)<br />

2<br />

where j is the imaginary unit √ −1. The fourier transform of the Poisson equation (4.72)<br />

can be written as :<br />

(<br />

ξ<br />

2<br />

x + ξy<br />

2 )<br />

F(u) =−jξx F(G x ) − jξ y F(G y ) (4.74)<br />

Denoting u F , G F x and G F y the fourier transform of u, G x and G y , the Frankot-Chellapa<br />

reconstruction algorithm gives :<br />

u F = −jξ xG F x − jξ y G F y<br />

ξ 2 x + ξ 2 y<br />

(4.75)<br />

As we will see, gradient field integration problems and their gradient-based variational<br />

formulation offer an interesting perspective on the <strong>de</strong>striping issue. In fact, compared to<br />

other restoration-oriented variational mo<strong>de</strong>ls, the energy functional (4.70) clearly separates<br />

the information related to vertical gradient and horizontal gradient. Going further<br />

in this gradient-based reasoning, we make the following remark :<br />

If the stripe noise is additive, it mostly affects the vertical gradient of the striped image<br />

Let us the consi<strong>de</strong>r the following image formation mo<strong>de</strong>l :<br />

I s = I + n (4.76)<br />

where I is the stripe free true image and n is the stripe noise. The linear operator K is<br />

consi<strong>de</strong>red to be the i<strong>de</strong>ntity. Let us assume that the unidirectional signature of the stripe<br />

noise n translates on its horizontal gradient as :<br />

∂n(x, y)<br />

∂x<br />

≈ 0 (4.77)<br />

The partial <strong>de</strong>rivatives along the x and y-axis of the image formation mo<strong>de</strong>l (4.76) :<br />

∂I s<br />

∂x = ∂I<br />

∂x + ∂n<br />

∂x<br />

∂I s<br />

∂y = ∂I<br />

∂y + ∂n<br />

∂y<br />

(4.78)

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