Image Processing Reference
In-Depth Information
Fig. 4.13 An example of histogram stretching
Conceptually it might be easier to appreciate the equation if we rearrange Eq. 4.3 :
255
f 2
g(x,y)
=
f 1 ·
f(x,y)
f 1 ·
a
(4.10)
f 1 · f(x,y)
f 1
255
f 2
g(x,y)
=
(4.11)
First the histogram is shifted left so that f 1 is located at 0. Then each value is
multiplied by a factor a so that the maximum value f 2
f 1 becomes equal to 255.
In Fig. 4.13 an example of histogram stretching is illustrated.
If just one pixel has the value 0 and another 255, histogram stretching will not
work, since f 2
255. A solution is modified histogram stretching where small
bins in the histogram are removed by changing their values to those of larger bins.
But if a significant number of pixels with very small and very high values exit, we
still have f 2
f 1 =
255, and hence the histogram (and image) remains the same, see
Fig. 4.15 . A more robust method to improve the histogram (and image) is therefore
to apply histogram equalization .
f 1 =
4.3.2 Histogram Equalization
Histogram equalization is based on non-linear gray-level mapping using a cumula-
tive histogram .
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