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
.