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This document introduces problems in various image processing topics
Typology: Exercises
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1 1
(^2 )
r (^) z
Pr(r) (^) Pz(z)
a) Find the pixel transformation y = g(r) such that after transformation the image has a flat PDF, i.e. which accomplishes histogram equalisation. Assume continuous variables r,y.
b) It is desired to find a transformation z = f (r) such that the transformed image will have the PDF of Pz (z) shown above right. Assume continuous quantities and determine the transformation function z = f (r).
b) Do histogram equalisation on the following image which has 8 discrete pixel levels (0 - 7), transforming it into a histogram equalised image also with 8 discrete grey levels in the range (0-7).
1 1 1 1 1 1 1 1 0 2 5 5 5 5 2 0 0 3 2 6 7 2 3 0 0 3 3 2 2 3 3 0 0 2 3 2 2 3 3 0 0 3 2 4 4 2 4 0 0 2 6 4 4 4 2 0 1 1 1 1 1 1 1 1
b) Histogram equalisation is applied to the coin image below (A) and the result is the image (B). Why does the image look so bad, with few details visible on the coin?
c) A different algorithm is applied to (A) and produces the image (C) which enhances both the background and the coin. Describe in detail how would write such an algorithm. (Hint: Use the image histogram to select parts of the image on and off the coin).
A−Original Image B−Histogram Equalised
C−After applying algorithm
Fig 2. Image enhancement of coin image.
a) Correlate the image ’I’ with the filter ’F’ above and compute the output image (assume ’same’ correlation with zeros outside the input image and round down the output pixel values to the nearest integer)
b) Apply a 3 by 3 median filter to the same image ’I’ to produce an 3 by 3 output image, again assume zeros outside of the image.
c) The result in part b demonstrates one general advantage and one general disadvantage of the median filter as compared to the mean filter.
d) Show that correlating the Laplace operator ’L’ with the image ’I’ is equivalent (except for a proportional factor) to locally subtracting a five point local mean from each original value of the image.
i) The imaginary part of F(u,v) is zero at all u,v
ii) F(u,v) is purely real and positive for all u,v.
iii) F(0,0) = 0
iv) F(u,v) has circular symmetry
v) The real part of F(u,v) is zero at all u,v
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