hwsol-0405

hwsol-0405 - ECE 178 W04 HOMEWORK #5 SOLUTIONS TA: Evan...

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ECE 178 W04 HOMEWORK #5 SOLUTIONS TA: Evan Ruzanski Problem 3.3 The transformations required to produce the individual bit planes are nothing more than mappings of the truth table for eight binary variables. In this truth table, the values of the 7 th bit are 0 for byte values 0 to 127, and 1 for byte values 128 to 255, thus giving the transformation mentioned in the problem statement. Note that the given transformed values of either 0 or 255 simply indicate a binary image for the 7 th bit plane. Any other two values would have been equally valid, though less conventional. Continuing with the truth table concept, the transformation required to produce an image of the 6 th bit plane outputs a 0 for byte values in the range [0, 63], a 1 for byte values in the range [64, 127], a 0 for byte values in the range [128, 191], and a 1 for byte values in the range [192, 255]. Similarly, the transformation for the 5 th bit plane alternates between eight ranges of byte values, the transformation for the 4 th bit plane alternates between 16 ranges, and so on. Finally, the output of the transformation for the 0 th bit plane alternates between 0 and 255 depending if the byte values are even or odd. Thus, this transformation alternates between 128 byte value ranges, which explains why an image of the 0 th bit plane is usually the busiest looking of all the bit plane images. Problem 3.5 All that histogram equalization does is re-map histogram components on the intensity scale. To obtain a uniform (flat) histogram would require in general that pixel intensities be actually redistributed so that there are L groups of nL = pixels with the same intensity, where L
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hwsol-0405 - ECE 178 W04 HOMEWORK #5 SOLUTIONS TA: Evan...

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