ECE 178: HW #4 Solutions
Anindya Sarkar & Emre Sargin
Q1
.
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.
Q2
.
a) If the least significant bit of the intensity value is 1 then the intensity value will decrease
by 1.
Otherwise the intensity value will remain the same.
Note that, in the former one,
intensity value was odd at the beginning and it became even at the end.
The number of
pixels having different gray level values would decrease since the odd intensity values are all
zero in the histogram. The contribution of each odd intensity value will be added to the left
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 Winter '08
 MANJUNATH
 Byte, sK, IMAX, intensity value, histogram equalization transformation

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