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Chapter 3 Intensity Transformations and Spatial Filtering

Spatial domain refers to the image plane itself, and image

processing methods in this category are based on direct
manipulation of pixels in an image.

Two principal categories of spatial processing are intensity

transformations and spatial filtering.

Intensity transformations operate on single pixels of an image for

the purpose of contrast manipulation and image thresholding.

Spatial filtering deals with performing operations, such as image

sharpening, by working in a neighbourhood of every pixel in an
image.

3.1 Background

The Basics of Intensity Transformations and Spatial Filtering

Generally, spatial domain techniques are more efficient

computationally and require less processing resources to
implement.

The spatial domain processes can be denoted by the expression

g ( x, y ) = T [ f ( x, y )] (3.1-1)

where f(x, y) is the input image, g(x, y) is the output image, and T
is an operator on f defined over a neighbourhood of point (x, y).
The operator can apply to a single image or to a set of images.
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Typically, the neighbourhood is rectangular, centered on (x, y),

and much smaller than the image.

Example: Suppose that the neighbourhood is a square of size

3×3 and the operator T is defined as “compute the
average intensity of the neighbourhood.”
At an arbitrary location in an image, say (10, 15), the
output g(10, 15) is computed as the sum of f(10, 15)
and its 8-neighbourhood is divided by 9.
The origin of the neighbourhood is then moved to the
next location and the procedure is repeated to generate
the next value of the output image g.
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The smallest possible neighbourhood is of size 1×1.

Example:
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3.2 Some Basic Intensity Transformation Functions

Intensity transformations are among the simplest of all image

processing techniques. We use the following expression to
indicate a transformation
s = T (r )
where T is a transformation that maps a pixel value r into a pixel
value s.

Image Negatives

The negative of an image with intensity levels in the range [0, L − 1]

is obtained by using the negative transformation shown in Figure 3.3,
which is given by
s = L −1− r (3.2-1)
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Example:

The negative transformation can be used to enhance white or gray

detail embedded in dark regions of an image.

Log Transformations

The general form of the log transformations is

s = c log(1 + r ) (3.2-2)
where c is a constant, and r ≥ 0 .

The log transformation maps a narrow range of low intensity

values in the input into a wider range of output levels. We use the
transformation of this type to expend the values of dark pixels in
an image while compress the higher-level values.

The opposite is true of the inverse log transformation.

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Example:

Figure 3.5(a) shows a Fourier spectrum with values in the range

0 to 1.5 ×106 .

Figure 3.5(b) shows the result of applying (3.2-2) to the spectrum

values, which will rescale the values to a range of 0 to 6.2, and
displaying the results with an 8-bit system.

Power-Law (Gamma) Transformations

Power-law transformations have the basic form

s = cr γ (3.2-3)
where c and γ are positive constants.

A variety of devices used for image capture, printing, and display

according to a power-law. By convention, the exponent in the
power-law equation is referred to as gamma.
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Unlike the log function, changing the value of γ will obtain a

family of possible transformations. As shown in Figure 3.6, the
curves generated with values of γ > 1 have exactly the opposite
effect as those generated with values of γ < 1 .

The process used to correct these power-law response phenomena

is called gamma correction.
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Example:

Gamma correction is important if displaying an image accurately

on a computer screen is of concern.

Gamma correction has become increasingly important as the use

of digital images over the Internet has increased.

In addition to gamma correction, power-law transformations are

very useful for general-purpose contrast manipulation.
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Example 3.1: Contrast enhancement using power-law transformations.

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Example 3.2: Another illustration of power-law transformations.

Figure 3.9(a) shows the opposite problem of Figure 3.8(a).

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Piecewise-Linear Transformation Functions

A complementary approach to the abovementioned methods is to

use piecewise linear functions.

Contrast stretching

One of the simplest piecewise linear functions is a contrast-

stretching transformation.

Contrast-stretching transformation is a process that expands the

range of intensity levels in an image so that it spans the full
intensity range of the recording medium or display device.

Example:
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Intensity-level slicing

Highlighting a specific range of intensities in an image often is of

interest. The process, often called intensity-level slicing, can be
implemented in several ways, though basic themes are mostly used.

One approach is to display in one value all the values in the range
of interest and in another all other intensities, as shown in Figure
3.11 (a).

Another approach is based on the transformation in Figure 3.11(b),

which brightens (or darkens) the desired range of intensities but
leaves all other intensities levels in the image unchanged.
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Example 3.3: Intensity-level slicing

Figure 3.12 (b) shows the result of using a transformation of the

form in Figure 3.11 (a), with the selected band near the top of the
scale, because the range of interest is brighter than the background.

Figure 3.12 (c) shows the result of using the transformation in

Figure 3.11 (b) in which a band of intensities in the mid-gray
region around the mean intensity was set to black, while all other
intensities were unchanged.
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Bit-plane slicing

Instead of highlighting intensity-level ranges, we could highlight

the contribution made to total image appearance by specific bits.

Figure 3.13 shows an 8-bit image, which can be considered as

being composed of eight 1-bit planes, with plane 1 containing the
lowest-order bit of all pixels in the image and plane 8 all the
highest-order bits.

Example:
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Note that each bit plane is a binary image.

For example, all pixels in the border have values 1 1 0 0 0 0 1 0,

which is the binary representation of decimal 194. Those values
can be viewed in Figure 3.14 (b) through (i).

Decomposing an image into its bit planes is useful for analyzing

the relative importance of each bit in the image.

Example:
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3.3 Histogram Processing

The histogram of a digital image with intensity levels in the range

[0, L − 1] is a discrete function h(rk ) = nk , where rk is the kth
intensity value and nk is the number of pixels in the image with
intensity rk .

It is common practice to normalize a histogram by diving each of

its components by the total number of pixels in the image, denoted
by MN, where M and N are the row and column dimensions of the
image.

A normalized histogram is given by

nk
p(rk ) = , for k = 0,1, 2,..., L − 1.
MN
p(rk ) can be seen as an estimate of the probability of occurrence
of intensity level rk in an image. The sum of all components of a
normalized histogram is equal to 1.

Histograms are the basic for numerous spatial domain processing

techniques.

Example:

Figure 3.16, which is the pollen image of Figure 3.10 shown in

four basic intensity characteristics: dark, light, low contrast, and
high contrast, shows the histograms corresponding to these image.

The vertical axis corresponds to value of h(rk ) = nk or

p(rk ) = nk / MN if the values are normalized.
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Histogram Equalization

We consider the continuous intensity values and let the variable r

denote the intensities of an image. We assume that r is in the
range [0, L − 1] .

We focus on transformations (intensity mappings) of the form

s = T (r ) 0 ≤ r ≤ L − 1 (3.3-1)
that produce an output intensity level s for every pixel in the
input image having intensity r. Assume that

(a) T (r ) is a monotonically increasing function in the

interval 0 ≤ r ≤ L − 1 , and

(b) 0 ≤ T (r ) ≤ L − 1 for 0 ≤ r ≤ L − 1 .

In some formations to be discussed later, we use the inverse

r = T −1 ( s) 0 ≤ s ≤ L − 1 (3.3-2)
in which case we change condition (a) to
(a ') T (r ) is a strictly monotonically increasing function in
the interval 0 ≤ r ≤ L − 1 .

Figure 3.17 (a) shows a function that satisfies conditions (a) and
(b).
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From Figure 3.17 (a), we can see that it is possible for multiple
values to map to a single value and still satisfy these two
conditions, (a) and (b). That is, a monotonic transformation
function can perform a one-to-one or many-to-one mapping,
which is perfectly fine when mapping from r to s.

However, there will be a problem if we want to recover the values

of r uniquely from the mapped values.
As Figure 3.17 (b) shows, requiring that T (r ) be strictly
monotonic guarantees that the inverse mappings will be single
valued. This is a theoretical requirement that allows us to derive
some important histogram processing techniques.

The intensity levels in an image may be viewed as random

variables in the interval [0, L − 1] . A fundamental descriptor of a
random variable is its probability density function (PDF).

Let pr (r ) and ps ( s) denote the probability density functions of r

and s. A fundamental result from basic probability theory is that if
pr (r ) and T (r ) are known, and T (r ) is continuous and
differentiable over the range of values of interest, then the PDF of
the transformed variable s can be obtained using the formula
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dr
ps ( s) = pr (r) (3.3-3)
ds
A transformation function of particular importance in image
processing has the form
r
s = T (r) = ( L − 1) ∫ pr (ω )dω (3.3-4)
0

where ω is a dummy variable of integration.

The right side of (3.3-4) is recognized as the cumulative distribution

function of random variable r . Since PDFs always are positive, the
transformation function of (3.3-4) satisfies condition (a) because the
area under the function cannot decreases as r increases.

When the upper limit in (3.3-4) is r = ( L − 1) , the integral

evaluates to 1 (the area under a PDF curve always is 1), so the
maximum value of s is ( L − 1) and condition (b) satisfies as well.

Using (3.3-3) and recalling the Leibniz’s rule that saying the
derivative of a definite integral with respect to its upper limit is
the integrand evaluated at the limit, we have
ds dT (r )
=
dr dr
d  r
= ( L − 1) ∫ pr (ω )dω 
dr  0  (3.3-5)
= ( L − 1) pr (r )

Substituting this result for dr / ds in (3.3-3), yields

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dr
ps ( s) = pr (r )
ds
1
= pr (r )
( L − 1) pr (r )
(3.3-6)
1
= 0 ≤ s ≤ L −1
L −1
which shows the that ps ( s) always is uniform, independently of
the form of pr (r ) .

Example 3.4: Illustration of (3.3-4) and (3.3.6)

Suppose that the continuous intensity values in an image have

the PDF

 2r
 for 0 ≤ r ≤ L − 1
pr (r ) =  ( L − 1)
2

0
 otherwise
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From (3.3-4),
r
s = T (r ) = ( L − 1) ∫ pr (ω )dω (3.3-4)
0
r
2 r2
ω dω =
( L − 1) ∫0
=
L −1

Consider an image in which L = 10 , and suppose that a pixel at

( x, y ) in the input image has intensity r = 3 . Then, the pixel at
( x, y ) in the new image is s = T (r ) = r 2 / 9 = 1 .

We can versify that the PDF of the intensities in the new image
is uniform by substituting pr (r ) into (3.3-6) and using the facts
that s = r / ( L − 1) , r is nonnegative, and L > 1 :
2

dr
ps ( s) = pr (r ) (3.3-6)
ds
−1
2r  ds 
=
( L − 1)2  dr 
−1
2r  d r2 
=  dr L − 1
( L − 1) 2  
2r ( L − 1) 1
= =
( L − 1)2 2r L −1
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For discrete values, we deal with probabilities (histogram values)

and summations instead of probability density functions and
integrals.

The probability of occurrence of intensity level rk in a digital

image is approximated by
nk
pr (rk ) = k = 0,1, 2,..., L − 1 (3.3-7)
MN
where MN is the total number of pixels in the image, nk is the
number of pixels having intensity rk , and L is the number of
possible intensity levels in the image.

The discrete form of the transformation in

r
s = T (r ) = ( L − 1) ∫ pr (ω )dω (3.3-4)
0
is
k
sk = T (rk ) = ( L − 1)∑ pr (rj )
j =0

( L − 1) k
∑ nj
(3.3-8)
= k = 0,1, 2,..., L − 1
MN j =0

The transformation (mapping) T (rk ) in (3.3-8) is called a histogram

equalization transformation.
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Example 3.5: A simple illustration of history equalization.

Suppose that a 3-bit image ( L = 8 ) of size 64 × 64 pixels

( MN = 4096 ) has the intensity distribution shown in Table 3.1.

The histogram of our hypothetical image is sketched in Figure

3.19 (a).

By using (3.3-8), we can obtain values of the histogram equalization

function:
0
s0 = T (r0 ) = 7∑ pr (rj ) = 7 pr (r0 ) = 1.33 ,
j =0
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1
s1 = T (r1 ) = 7∑ pr (rj ) = 7 pr (r0 ) + 7 pr (r1 ) = 3.08 ,
j =0

s2 = 4.55 , s3 = 5.67 , s4 = 6.23 , s5 = 6.65 , s6 = 6.86 , and

s7 = 7.00 . This function is shown in Figure 3.19 (b).

Then, we round them to the nearest integers:

s0 = 1.33 → 1 s1 = 3.08 → 3 s2 = 4.55 → 5 s3 = 5.67 → 6
s4 = 6.23 → 6 s5 = 6.65 → 7 s6 = 6.86 → 7 s7 = 7.00 → 7
which are the values of the equalized histogram.

Observe that there are only five distinct levels:

s0 → 1 : 790 pixels
s1 → 3 : 1023 pixels
s2 → 5 : 850 pixels
s3 → 6 : 985 (656+329) pixels
s5 → 7 : 448 (245+122+81) pixels
Total: 4096

Dividing these numbers by MN = 4096 would yield the

equalized histogram shown in Figure 3.19 (c).

Since a histogram is an approximation to probability density

function, and no new allowed intensity levels are created in the
process, perfectly flat histograms are rare in practical
applications of histogram equalization.

Therefore, in general, it cannot be proved that discrete histogram

equalization results in a uniform histogram.
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Given an image, the process of histogram equalization consists

simply of implementing
( L − 1) k
sk = ∑ nj ,
MN j =0 (3.3-8)

which is based on information that can be extracted directly from

the given image, without the need for further parameter
specifications.

The inverse transformation from s back to r is denoted by

rk = T −1 ( sk ) k = 0,1, 2,..., L − 1 (3.3-9)

Although the inverse transformation is not used in the histogram

equalization, it plays a central role in the histogram-matching
scheme.

Example 3.6: Histogram equalization

The left column in Figure 3.20 shows the four images from
Figure 3.16.

The center column in Figure 3.20 shows the result of performing

histogram equalization on each of the images in left.

The histogram equalization did not have much effect on the

fourth image because the intensities of this image already span
the full intensity scale.
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Figure 3.21 shows the transformation functions used to

generate the equalized images in Figure 3.20.