Extension to Grayscale - Logic-based Nonlinear Image Processing - page 77

Figure 7.5 Stack filter design: counts are aggregated over all threshold levels.

ber of counts ( N 0 and N 1 ) of the corresponding pixel in the ideal image is recorded
as either 0 or 1. The counts are aggregated over every threshold level. As in the bi-
nary case, the optimum filter is found by setting the output value to 0 or 1 according
to whether N 0 or N 1 is larger for each input combination.
However, a little caution is required here as there is no guarantee that this ap-
proach will result in a filter based on a PBF. An example of stack filter design is
shown in Fig. 7.6. The noisy and ideal signals comprising the training set have been
thresholded and a 3-point sliding window is used to compile the table of observa-
tions. The filter output is determined for each input combination depending on
whether N 0 or N 1 is larger.
However, the function arising from this approach cannot be used as it will vio-
late the stacking property which may in turn result in an output signal stack contain-
ing “holes”. Instead of using the non-PBF arising from this process it is necessary
to convert it to a PBF. Ideally the closest PBF, i.e., the one that causes the smallest
increase in error, must be used. Tabus et al. devised a technique to convert a
non-PBF to the closest PBF, resulting in the smallest increase in error. 3,4
The technique is based on the understanding that there are two ways to remove
complemented terms from binary functions:

• The minterms that cause the complemented terms in the non-PBF may be re-

moved from the expression, or

• Further (mirror) minterms may be added to combine with the complemented

terms and remove the necessity for negation.

In the example shown in Fig. 7.6, the problem is caused by the minterm X 0 X 1 X 2 .
The solutio n is either to remove it from the function or include an additional
minterm X 0 X 1 X 2 to combine with the one above and eliminate the negation. Refer-
ring to the table of observations in Fig. 7.6, it can be seen that the cost of the second
option is lower in terms of increase in error so this should be the preferred option.