1
$\begingroup$

I want to minimize the functional of teh Blind Deconvolution model as given in: Total Variation Blind Deconvolution by Chan and Wong.

Their model is given by:

$$ z = h \ast u + \eta $$

Where $ \ast $ is the convolution operator, $ h $ is the blurring kernel, $ u $ is the sharp noiseless image and $ \eta $ is additive white gaussian noise (AWGN).

The functional to be minimized is given by (Assuming the Blurring Kernel is known):

$$ \underset{u}{\min} f \left( u \right ) = \underset{u}{\min} \left \{ \frac{1}{2} {\left \| h \ast u - z \right \|}^{2}_{L_{2}} + \alpha \int \left | \nabla u \right | dx dy \right \} $$

Where $ \alpha $ is the smoothing term.

Yet, in Blind Deconvolution the Kernel isn't known and the minimization functional is given by:

$$ \underset{u}{\min} f \left( u \right ) = \underset{u}{\min} \left \{ \frac{1}{2} {\left \| h \ast u - z \right \|}^{2}_{L_{2}} + {\alpha}_{1} \int \left | \nabla u \right | dx dy + {\alpha}_{2} \int \left | \nabla h \right | dx dy \right \} $$

Now, The Euler Lagrange equations I calculated and given in the article are:

$$\begin{align} \frac{\delta L}{\delta h} & = \left ( u \ast h - z \right ) \ast u \left( -x, -y \right ) - {\alpha}_{2} \nabla \cdot \left( \frac{\nabla h}{\left | \nabla h \right |} \right) \\ \frac{\delta L}{\delta u} & = \left ( u \ast h - z \right ) \ast h \left( -x, -y \right ) - {\alpha}_{1} \nabla \cdot \left( \frac{\nabla u}{\left | \nabla u \right |} \right) \end{align}$$

What I'm not sure about is how can I use it to solve the problem.
At the article they suggest the alternating method, namely once solve for $ h $ and then for $ u $.

Yet I don't see how to write in in MATLAB code (Or any other pseudo code).
It should be some kind of a Gradient Descent step.

$\endgroup$
3
$\begingroup$

The blind deconvolution should be a minimization on both $u$ and $h$:

$$ \underset{u,h}{\min} f \left( u, h \right ) = \underset{u,h}{\min} \left \{ \frac{1}{2} {\left \| h \ast u - z \right \|}^{2}_{L_{2}} + {\alpha}_{1} \int \left | \nabla u \right | dx dy + {\alpha}_{2} \int \left | \nabla h \right | dx dy \right \} $$

The standard way to do this is to alternately hold one variable fixed and minimize on the other, ie.

\begin{cases} u^{k+1} = \underset{u}{\text{argmin}} f(u,h^k) \\ h^{k+1} = \underset{h}{\text{argmin}} f(u^k,h) \\ \end{cases}

Back when Chan and Wong wrote their paper the way to minimize total variation was to write out the Euler-Lagrange equations which leads to those complicated and slow to solve PDEs. Now there are faster and easier ways to solve each of those subproblems (eg Split-Bregman) which will simplify things for you.

A bit of searching led to me to this 2014 paper: http://www.cvg.unibe.ch/dperrone/tvdb/ which has more information as well as publicly-available MATLAB code. I'm not familiar with their work but it's not easy to publish in CVPR so it's probably worth a look...

$\endgroup$
  • $\begingroup$ Thank You for your answer. 2 Questions: 1. What if I want to implement the straight way, using Gradient Descent, I don't understand how create the corrects dimensions of the signals (At least in the iteration solving for the Blur Kernel). 2. Can I use MATLAB's fminunc function for the minimization? Thank You. $\endgroup$ – Royi Dec 22 '14 at 22:45
  • $\begingroup$ Since you have a TV-penalty on h gradient descent won't work since the TV functional is not differentiable. There is some research on smoothing out the TV-term using a parameter (usually called epsilon) that goes under a square root somewhere but the newer Bregman stuff is easier to implement. I don't know how fminunc works but I suspect it won't be a good choice for this difficult problem. After a first pass over the linked CVPR paper I suggest taking a look at their research and code. $\endgroup$ – dranxo Dec 22 '14 at 23:43
  • $\begingroup$ Hi, I looked at their paper and they do use Gradient Descent (Am I wrong?). They use it alternatively (Holding the Kernel, calculating the Image and vice versa). $\endgroup$ – Royi Dec 23 '14 at 6:28
  • $\begingroup$ Huh, yeah I see it now in alg. 1. In the following paragraph they remark that they only do one iteration - I don't know how that changes things. Also, they have alpha_2 = 0 which simplifies things a bit. In any event, computing the div*(nabla/|nabla|) (the curvature of a level curve) isn't easy. The linked paper has a function gradTVcc(f,epsilon) in their code that handles the computation. $\endgroup$ – dranxo Dec 23 '14 at 21:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.