Th`ese Marouan BOUALI - Sites personnels de TELECOM ParisTech

96 4. A Variational approach for the destriping issue
along strong edges. In this section, we overcome this drawback and introduce a new variational
model that satisfies the requirements of optimal destriping.
In chapter 3, we attributed the limitations of standard destriping techniques to the simplistic
assumptions made on the stripe noise. Equalization based methods (moment matching,
histogram matching and OFOV techniques) assume that the stripe noise can be modelled
with a gain and offset between the detectors of the sensor. The presence of strong nonlinearities
on MODIS detectors responses, discredits this assumption. A simple and more
realistic assumption lies in the directional aspect of stripe noise. For instance, if we denote
by n the stripe noise, it is resonable to assume that :
∣ ∀(x, y) ∈ Ω,
∂n(x, y)
∣∣∣ ∣ ∂x ∣ ≪ ∂n(x, y)
∂y ∣ (4.85)
The integration of the previous inequality over the entire image domain Ω results in :
∫
∫
∂n(x, y)
∣ ∂x ∣ ≪ ∂n(x, y)
∣ ∂y ∣ (4.86)
Ω
which translates a realistic characteristic of stripe noise when written as :
Ω
TV x (n) ≪ TV y (n) (4.87)
where TV x and TV y are respectively horizontal and vertical variations of the noise n.
The L 1 norm being an edge-preserving norm can be introduced in a variational model
where the fidelity term takes into account the stripe free horizontal information while
the regularization term only smoothes the result along the vertical axis. We propose a
destriping variational model based on the minimization of the following energy :
E(u) =TV x (u − I s )+λT V y (u) (4.88)
We anticipate the non-differentiability of this functional at points where ∂u
= 0 by introducing a small parameter ɛ in (4.88) which becomes :
∂u
∂y
∂x = ∂Is
∂x
and
∫
E ɛ (u) =
Ω
√ (∂(u ) − Is ) 2 ∫
+ ɛ 2 + λ
∂x
Ω
√ (∂u ) 2
+ ɛ 2 (4.89)
The model (4.89) will be refered to hereafter as the Unidirectional Variational Destriping
Model (UVDM). Its Euler-Lagrange equation is given by :
− ∂
∂x
⎛
⎜
⎝
∂(u−I s)
∂x
√ (
∂(u−Is)
∂x
⎞
⎛
⎟
) 2 ⎠ − λ ∂ ⎜
∂y ⎝
+ ɛ 2
√ (
∂u
∂y
∂u
∂y
∂y
⎞
⎟
) 2 ⎠ =0 (4.90)
+ ɛ 2