Image Enhancement in Spatial Domain
Dr. Sanjay Kumar Singh
Associate Professor, Department of Communication Engineering,
School of Electronics Engineering, Vellore Institute of Technology,
Vellore ‐ 632014, Tamil Nadu [India]
Fall 2022‐23
Image Enhancement
• Image Enhancement refers to accentuation, or sharpening, of
image features such as edges, boundaries, or contrast to make a
graphic display more useful for analysis & display.
• Image enhancement includes gray level and contrast
manipulation,
noise
reduction,
edge crispening
and
sharpening,
filtering,
interpolation
and
magnification,
pseudocoloring, and so on.
• Image enhancement remains a very important process because of
its usefulness in virtually all image processing applications.
Image Enhancement
Point Operations
Spatial Operations
Transform Operations
• Contrast stretching
• Noise smoothing
• Linear filtering
• Noise clipping
• Median filtering
• Root filtering
• window slicing
• Unsharp masking
• Homomorphic filtering
• Histogram modeling
• Low-pass, bandpass
high-pass filtering
• Zooming
Point Operations
• Zero memory operations
u 0, L 
f ()
v 0, L 
Contrast
Contrast Equation?

Many different equations for contrast exist
Examples:

Michalson’s equation for contrast

Basic Grey Level Transformations

3 most common gray level transformation:

Linear


Logarithmic


Negative/Identity
Log/Inverse log
Power law

nth power/nth root
Logarithmic Transformations
Maps
narrow range of input levels => wider range of
output values
Inverse log transformation does opposite transformation
The general form of the log transformation is
New pixel value
Old pixel value
s = c * log(1 + r)
Log transformation of Fourier transform shows more detail
s = log(1 + r)
Power Law Transformations
Power
law transformations have the form
Power
γ
s=c*r
Old pixel value
New pixel value
Constant
Map
narrow range of
dark input values into
wider range of output
values or vice versa
Varying
γ gives a whole
family of curves
Power Law Example
Original
Magnetic
Resonance
(MR) image of fractured
human spine
s = r 0.6
s = r 0.4
Different
power values
highlight different
details
Contrast Stretching
• Low contrast images occur often due to poor or nonuniform lighting conditions or due to nonlinearity or small
dynamic range of the imaging sensor.
• The gray scale intervals where pixels occur most frequently
would be stretched most to improve the overall visibility
of a scene.
u,

v    u  a   va ,
  u  b  v ,
b

0ua
aub
bu L
v

vb

For dark region stretch    a  L 3
2
midregion stretch β >1, b  3 L
bright region stretch γ 1
va
0

a
b
L
u
Clipping and Thresholding
Clipping
• A special case of contrast stretching where α  γ  0
• Useful for noise reduction when the input signal is known
to lie in the range  a, b  .


v
Image
Histogram
v
u
a
b
u
v
u
Thresholding
• Special case of clipping where a  b  t
• Goal is to segment an image into object and background.
• Eliminates unimportant shading variation. Hence, result is
always a binary image.
v
v
u
u
Digital Negative
• A negative image can be obtained by reverse scaling of the
gray levels according to the transformation, v  L  u
• Producing negative prints of images.
v
L
0
L
u
Intensity Level Slicing
• These transformations permit segmentation of certain gray
level regions from the rest of the image. This technique is
useful when different features of an image are contained in
different gray levels.
v
v
L
a b
Without background
 L,
v
0,
u
aub
otherwise
45
a b L
With background
 L,
v
 u,
u
aub
otherwise
Histogram Processing
 Modify an image so that its histogram has a desired shape.
Histogram: Discrete function
p(rk )  nk
where,
rk
: k th gray level
k
: 0,1, 2,..., L -1
nk : No. of pixels in the image having gray level rk
Histogram: Continuous function
 rk  rk 1   0
Histogram
pixels
130
36
36
22
0
Image 16x14 = 224 pixels
1
2
3
level
p  rk   nk
Example:
rk
p(0)  18
p(1)  6
p(2)  2
p(3)  p (4)  p (5)  p(6)  0
p(7)  10
Normalized Histogram: Discrete Function
nk
;
p  rk  
n
k  0,1, 2,..., L  1
where,
rk
: k th gray level
nk : No. of pixels with k thgray level in the image
n
: total No. of pixels in the image
p  rk  : probability of occurrence of gray level rk
Example:
nk
p  rk  
n
rk
18 1
p(0) 

36 2
2
1
p(2) 

36 18
10 5
p(7) 

36 18
6 1
p(1) 

36 6
p(3)  p(4)  p(5)  p(6)  0
f
f
x
x
Very Different Images May Have Same Histogram!
Histogram reflects the pixel intensity distribution,
not the spatial distribution!
Histogram Modeling
• Histogram‐modeling techniques modify an image
contrast and brightness so that its histogram has a
desired shape and range.
• Histogram modeling has been found to be a
powerful technique for image enhancement.
• Common Techniques: Histogram Stretching, Histogram
Shrinking, Histogram Sliding, Histogram Equalization
The histogram of an image represents the relative frequency of
occurrence of the various gray levels in the image.
Stretching
 S MAX  S MIN 
I '  x, y   
  I  x, y   I MIN   S MIN
 I MAX  I MIN 
I MAX : Largest gray level in the image I  x, y 
I MIN : Smallest gray level in the image I  x, y 
S MAX : Maximum possible gray level value
(For an 8‐bit image this is 255)
S MIN : Minimum possible gray level value
(For an 8‐bit image this is 0)
Original Image
Image after
contrast
stretching
Histogram of
original Image
Histogram
after
contrast
stretching
Example: Stretching
 S MAX  S MIN 
I '  x, y   
  I  x, y   I MIN   S MIN
 I MAX  I MIN 
 210  10 
y
 x  50  10

100  50 
Shrinking
 S MAX  S MIN 
I '  x, y   
  I  x, y   I MIN   S MIN
 I MAX  I MIN 
I MAX : Largest gray level in the image I  x, y 
I MIN : Smallest gray level in the image I  x, y 
S MAX : Maximum desired in the compressed histogram
S MIN : Minimum desired in the compressed histogram
Original Image
Histogram after
Shrinking to the
range [50, 175]
Histogram of
original Image
Histogram
of shrinked
Image
Sliding
S  x, y   I  x, y   Offset
Offset: Amount to slide the histogram
Original Image
Histogram of
original Image
Positive value
after histogram
Sliding
Histogram of the
Image after Sliding
Histogram Equalization
Aim:
 To improve the contrast of an image
 To transform an image in such a way that the
transformed image has a nearly uniform distribution of
pixel values
Equalized Histogram
Original Histogram
r
0
s  T (r )
1
s
0
1
In practice, the equalized histogram might not be completely
flat.
Transformation:
 Assume r has been normalized to the interval [0,1], with r = 0
representing black and r = 1 representing white
s  T (r ) ; 0  r  1
 The transformation function satisfies the following conditions:
(a). T  r  : (i). Monotonically increasing
i.e.
d
T  r   0 : Positive Slope
dr
(ii). Single Valued: if s  T  r  then, r  T 1  s 
(b). 0  T (r )  1 for 0  r  1
s
1
T(r)
sk=T(rk)
0
rk
1
r
 The requirement in (a) that T(r) be single valued is needed to
guarantee that the inverse transformation will exist, and the
monotonicity condition preserves the increasing order from
black to white in the output image.
 Condition (b) guarantees that the output gray levels will be in
the same range as the input levels.
The inverse transformation from s back to r is denoted
s  T 1 (r ) where 0  r  1
Note: There are some cases that even if T(r) satisfies
conditions (a) and (b), it is possible that the corresponding
inverse T 1 ( s ) may fail to be single valued.
PDF vs CDF
 PDF (probability density function) is the probability of
each element
 CDF (cumulative distribution function) is summation of
the probability of the element that value less than or
equal this element
p  x
P  x
CDF
PDF
1/6
1
x
1/6
x
PDF vs CDF
f
f
a . Histogram
b  . Cumulative histogram
x
x
Modified histogram
Linear cumulative histogram
by modification
Relationship between PDF and CDF
 PDF can be found from the following equation
d [ P( x)]
p ( x) 
 P( x)
dx
 CDF can be found from the following equation
P( x  xn ) 
xn

x x
0
p ( x)dx
Idea of Histogram Equalization:
 The gray levels in an image may be viewed as random variables
in the interval [0,1].
 Let pr  r  denote the pdf of random variable r and ps  s 
denote the pdf of random variable s; if pr  r  and T(r) are
known and T 1  s  satisfies condition (a), the formula should be
dr
p s ( s )  pr ( r )
ds
r
s  T (r )  ( L  1)  pr ( w)dw
0
By Leibniz’s rule that the derivative of a definite integral w.r.t.
its upper limit is simply the integrand evaluated at that limit.
ds dT (r )

dr
dr
r

d 
 ( L  1)   pr ( w)dw  ( L  1) pr (r )
dr  0

Substituting into the first equation
dr
ps ( s )  pr (r )
ds
1
1
, 0  s 1
 pr (r )

( L  1) pr (r ) L  1
The probability of occurrence of gray level rk in an image is
approximated by
nk
pk (rk ) 
n
k  0,1,2,..., L  1
k
sk  T (rk )  ( L  1) pr (rj )
k
nj
j 0
n
 ( L  1)
j 0
; k  0,1, 2,..., L  1
(a).
(b).
FIGURE (a). An arbitrary PDF
(b). Result of applying all transformation to all intensity
levels r. The resulting intensities s, have a uniform PDF,
independent of the form of PDF of the r's.
Example:
p r 
r
Solution:
Steps for Histogram Equalization
1. Normalize the gray value to make the range between 0
and 1.
2. Find pdf values of individual gray value (normalized).
3. Find the CDF of normalized gray value.
4. Perform r to s mapping.
Example: An 8‐level 64  64 image has a gray level distribution, nk
shown in the table below, rk indicates a gray level in the image.
(a). Draw the histogram of the original Image
(b). Draw the histogram based on histogram equalization.
Show all steps.
Steps:
Step 1: Calculate total no. of pixels
n  64  64  4096 pixels
Step 2: Find
nk
pr  rk  
n
k
 
Step 3: Calculate sk  T  rk    L  1  pr rj ;
j 0
Step 4: Map gray level
Step 5: Assign o/p level sk
Step 6: Calculate ps  sk 
Step 7: Plot graph
k  0,1, 2,..., 7
Normalized
Histogram
k
 sk  T  rk    L  1  pr  rj ; k  0,1, 2,..., 7
j 0
0
s0  T  r0   7 pr  r0   7  0.19  1.33
 1
j 0
1
s1  T  r1   7 pr  rj   7  pr  r0   pr  r0    7  0.19  0.25  3.08  3
j 0
2
s2  T  r2   7 pr  rj   s1  7  pr  r2    3.08  7  0.21  4.55
 5
j 0
3
s3  T  r3   7 pr  rj   s2  7  pr  r3    4.55  7  0.16  5.67
 6
j 0
4
s4  T  r4   7 pr  rj   s3  7  pr  r4    5.67  7  0.08  6.23
 6
s5  T  r5   s4  7  0.06  6.65
 7 
s6  T  r6   s5  7  0.03  6.86
 7 
s7  T  r7   s6  7  0.02  7.00
 7 
j 0
Transfer function: CDF
Assigning o/p level sk
Mapping gray level
Equalized Histogram
Histogram Matching (Specification)
Objective:
To convert the image so that it has a particular histogram that
can be arbitrarily specified.
Input Image
Reference Image
Resulted Image
Input Image
Histogram
Equalization
Cumulative
Distribution
Mapping
Function
/Look‐up
Table
Reference
Image
Histogram
Equalization
Cumulative
Distribution
Steps: 1. Equalize the histogram of the input image.
2. Equalize the specified histogram.
3. Relate the two equalized histograms.
Modified
Image
r
s  T (r )  ( L  1)  pr ( w)dw
0
z
G ( z )  ( L  1)  p z (t )dt  s
0
z  G 1 ( s )  G 1[T (r )]
p z (z ) is the desired PDF
k
k
nj
j 0
j 0
n
sk  T (rk )  ( L  1) pr (rj )  ( L  1)
, k  0,1, 2,..., L  1
k
vk  G ( z k )  ( L  1) p z ( zi ) sk , k  0,1,2,..., L  1
i 0
1
z k  G [T (rk )], k  0,1,2,..., L  1
(b).
(a).
(a). Graphical interpretation of
mapping from rk to sk via T(r)
(b). Mapping of zq to its corres‐
ponding value vq via G(z).
(c).
(c). Inverse mapping from sk to
its corresponding value of zk .
Steps for histogram matching:
– Obtain the histogram of the given image, T(r)
– Precompute a mapped level s k for each level rk
– Obtain the transformation function G from the given p z (z )
– Precompute z k for each value of s k
– Map rk to its corresponding level s k ; then map level s k into
the final level z k .
(a). Histogram of the 3‐bit image (b). Specified histogram (c). Transformation
function obtained from the specified histogram (d). Result of performing
histogram specification, Compare (b) and (d).
pr  rk 
Input Image
sk
pz  zq 
Reference Image
vq
i
sk
vq
j
Mapping/ Look‐up Table
Histogram Statistics (Global)
Mean (average intensity):
L 1
1
m   ri p ( ri ) 
MN
i0
M 1 N 1
  f ( x, y )
x0 y 0
Variance (2nd moment):

2
L 1
1
2
  2 ( r )   ( ri  m ) p ( ri ) 
MN
i0
M 1 N 1
   f ( x, y )  m 
x0 y 0
nth moment around mean:
L 1
n (r )   (ri  m) n p (ri )
i 0
2
Example
σ is useful for estimating image contrast!
  14.3
  31.6
  49.2
Local Enhancement
Pixel-to-pixel translation
Nonoverlapping region
Histogram using a local neighborhood,
for example 33 neighborhood
Local Histogram Statistics
– rs ,t is the gray level at coordinates (s, t) in the neighborhood
–
p(rs ,t ) is the neighborhood normalized histogram component
–
average local intensity (mean):
mS xy 
r
s ,t
( s ,t )S xy
p(rs ,t )
where, sxy denotes a neighborhood
–
local variance

2
S xy

 [r
( s ,t )S xy
s ,t
 mS xy ] p(rs ,t )
2
Global vs Local Histogram
Enhancement using
Arithmetic/Logic Operations
AND
OR
NOT
Subtraction
Addition
Multiplication
Division
(a). Original Image
(b). Gaussian Noise
FIGURE: Image Addition
(a) + (b)
(a)  2
(a). Original Image
FIGURES: Image
Multiplication &
Division
(a)  2
Image subtraction and Change Detection
 The difference between two “similar” images is computed as
g ( x, y )  f1 ( x, y )  f 2 ( x, y )