How are 12 outputs being reconciled so that a Homomorphic filter could be applied? Take a look at this paper. It writes in the section 4.1:


  We selected 12 not overlapping filters, to analyze 12 different directions, rotated with respect to 15° each other. A homomorphic filter is then applied to enhance lines and to produce a dark uniform background. Finally a threshold is applied to obtain a binary mask. The threshold value is computed as mean plus standard deviation of the image intensity (see Figure 3(b)).


I could not understand this.

They began the process with a single image. Then, they applied 12 filters to that input image so that they can produce 12 outputs from one image.

How are they selecting one output image from those 12 outputs so that they can apply the homomorphic filter?  I mean, out of those 12 outputs, to which output image they would apply the homomorphic filter?

If they apply homomorphic filter and thresholding to 12 images, when did those 12 images become one image?
From my reading, the authors don't combine the 12 images in a single image: each generates a directional scratch mask, each of them being processed independently. If I read too fast, tell me. Indeed, they first remove a low-pass version, and apply a logarithmic homomorphic filter. But you could combine the binary masks with an OR to get a single image of potential scratches.

However, it would be possible with filters alone to recombine them by addition.

Suppose that you only have two filters $f_1$ and $f_2$, that are applied to an image $i$. $\mathcal{F}$ is the Fourier transform: $I=\mathcal{F}(i)$ and $F_k=\mathcal{F}(f_k)$.

The hypothesis that they do not overlap is related to their frequency behavior, with spectra $F_1$ and $F_2$. Their supports $S_{F_1}$ and $S_{F_2}$ (imagine 2D patches in the full 2D frequency domain $S_F$) do not overlap if their intersection is void: $S_{F_1} \cap S_{F_2}= \emptyset$. What is not explicit is that, often, the filters are also complementary, their union covers the whole frequency domain: $S_{F_1} \cup S_{F_2}= S_F$. Thus, the spectrum $I$ of $i$ is cut into two complementary parts, and $I = F_1\cdot I+F_2\cdot I$. Then, you could recover $i$ with
$$i = \mathcal{F}^{-1}(F_1\cdot I+F_2\cdot I) = \mathcal{F}^{-1}(F_1\cdot I)+\mathcal{F}^{-1}(F_2\cdot I)\,.$$

Equivalently, this amount to directly write, in the original domain:

$$i = f_1\star i + f_2\star i\,.$$

Visually, the frequency domain is a pie, and each filter is a slice. When you have the the slices, you can reproduce the original pie. In practice, people don't use exactly non overlapping filters in frequency, but orthogonal filters in space, that sum to unity to equally cover all frequencies.  This is a little more general as non overlapping filters, as $F_1\cdot F_2 = 0$, thus $f_1\star f_2 = \mathcal{F}^{-1}(F_1\cdot F_2) = 0$,  are a subset of orthogonal filters.