# Image Enhancement in Spatial Domain

Τεχνίτη Νοημοσύνη και Ρομποτική

6 Νοε 2013 (πριν από 4 χρόνια και 8 μήνες)

64 εμφανίσεις

Image Enhancement in Spatial Domain

Spatial domain process on images can be
described as

g
(
x, y
) =
T
[
f
(
x, y
)]

where
f
(
x,y
) is the input image,
g
(
x,y
) is the
output image,
T

is an operator

T
operates on the neighbors of (
x, y
)

a square or rectangular sub
-
image centered at (
x,y
)
to yield the output g(
x, y
)
.

The simplest form of
T

is the neighborhood
of size 1

1.

g

depends on the value of
f

at (
x, y
) which
is a
gray level transformation

as

s =
T
(r)

where r and s are the gray
-
level of
f(x, y)
and
g(x,
y)
at any point (
x, y
).

Enhancement of any point depends on that point
only.

Point processing

Larger neighborhood provides more
flexibility.

or

filtering

Three basic functions used in image
enhancement

Linear (negative and identity transformation)

s
=

L
-
1
-
r

Logarithmic (log and inverse log)

s
=

c
log
(1
+
r)

Power law (
n
th power and
n
th root transformation)

s
=

cr

or

s
=

c(r+

)

correction

The CRT devices have an intensity
-
to
-
voltage
response which is a power function.

ranges from 1.8 to 2.5

Without

correction
, the monitor output will
become darker than the original input.

Piecewise
-
Linear Transformation Functions

Contrast stretching

The histogram of a digital image with
gray
-
levels in the range [0,

L
-
1] is a
discrete function
h
(
r
k
)

= n
k

where
r
k

is the
k
th level and
n
k

is the number of pixels
having the gray
-
level
r
k
.

A normalized histogram
h
(
r
k
)
=n
k
/n
,
n

is
the total number of pixels in the image.

Histogram equalization

is to find a
transformation
s
=

T
(
r
)

0

r

1

that
satisfies the following conditions:

T
(
r
) is single
-
valued and monotonically
increasing in the interval 0

r

1

0

T
(
r
)

1 for 0

r

1

T
(
r
) is single
-
valued so that its inverse function
exist.

The gray
-
level in an image may be viewed as a
random variable, so we let
p
r
(
r
) and
p
s
(
s
)

denote
the probability density function of random
variables
r

and
s
.

If
p
r
(
r
)

and
T
(
r
)

are known and
T
-
1
(
s
) is single
-
valued and monotonically increased function
then

Assume the transformation function as

where
w i
s a dummy variable,

s = T
(
r
)

is a cumulative distribution function
(CDF) of the random variable
r
.

r
r
dw
w
p
r
T
s
0
)
(
)
(
ds
dr
r
p
s
p
r
s
)
(
)
(

Proof

For discrete case

p
r
(r
k
) = n
k
/n

for
k
= 0,1….
L
-
1

The discrete version of the transformation
function is

s
k
= T
(
r
k
)

=

automatic, without the need for parameter
specifications.

k
j
k
j
j
j
n
n
r
p
0
0
)
(

Given the input image with

p
r
(
r
), and the
specific output image with

p
z
(
z
), find the
transfer function between the

r

and
z
.

Let
s = T
(
r
)

=

Define a random variable z with the
property

G
(
z
)
= = s

From the above equations
G
(
z
)

= T
(
r
) we
have

z = G
-
1
(
s
)

= G
-
1
[
T
(
r
)]

r
r
dw
w
p
0
)
(

z
z
dt
t
p
0
)
(

For discrete case:

From given histogram
p
r
(r
k
), k
=0, 1,….
L
-
1

s
k
=

T
(
r
k
) =

From given histogram
p
z
(
z
i
)
, i
=0, 1,…
L
-
1

v
k
=

G(z
k
)
=

=

s
k

Finally, we have
G
(
z
k
)

= T
(
r
k
), and

z
k
=G
-
1
(
s
k
)
=G
-
1
[
T
(
r
k
)]

k
j
k
j
j
j
n
n
r
p
0
0
)
(

k
i
i
z
z
p
0
)
(
1.
Obtain the histogram of each given image.

2.
Pre
-
compute a mapped level
s
k

for each level
r
k
.

3.
Obtain the transformation function
G

from
given
p(z)

4.
Precompute
z
k

for each value
s
k

using
iterative scheme as follows:

1.
To find
z
k
= G
-
1
(s
k
) = G
-
1
(v
k
)
, however, it may not
exist such

z
k
.

2.
Since we are dealing with integer, the closest we
can get to satisfying
G(z
k
)

s
k
= 0

5.
For each pixel in the original image, if the
value of that pixel is
r
k
,

map this value to its
corresponding levels
s
k
; then map level
s
k

into the final level
z
k
.

Spatial filtering: using a filter kernel
( which is a subimage,
w(x, y)
) to operate
on the image
f(x,y)
.

The response R of the pixel (x, y) after
filtering is

R = w(
-
1,
-
1)f(x
-
1, y
-
1) +w(
-
1, 0)f(x
-
1,
y)+….+w(0, 0)f(x, y)+….+w(1, 0)f(x+1, y)+w(1,
1)f(x+1, y+1)

The output image
g(x,y)

is:

a
a
s
b
b
t
t
y
s
x
f
t
s
w
y
x
g
)
,
(
)
,
(
)
,
(

9
1
9
9
1
1
.......
i
i
i
z
w
z
w
z
w
R

3.6 Smoothing Spatial Filters

Weighted Averaging

a
a
s
b
b
s
a
a
s
b
b
s
t
s
w
t
y
s
x
f
t
s
w
y
x
g
)
,
(
)
,
(
)
,
(
)
,
(

Median Filter

The response is based on ordering (ranking) the
pixels contained in the image area encompassed
by the filter, and then replacing the value of the
center pixel with the value determined by the
ranking result

For certain noise, such as salt
-
and
-
pepper
noise, median filter is effective.

Image averaging

low
-
pass filtering

image
blurring

spatial integration

Image sharpening

high
-
pass filtering

spatial differentiation.

It enhances the edges and the other discontinuities

First order difference is

f/

x = f
(
x+1, y
)

-

f
(
x, y
)

f/

y = f
(
x, y+1
)

-

f
(
x, y
)

Second order difference

2
f/

x
2
= f
(
x+1, y
)

+ f
(
x
-
1, y
)

-

2f
(
x, y
)

Isotropic filter, rotational invariant
-

Laplacian

2
f =

2
f/

x
2
+

2
f/

y
2

2
f = [
f
(
x
+1
,y
)
+f
(
x
-
1
,y
)
+f
(
x, y
+1)
+f
(
x, y
-
1)]
-

4
f(x,y)

Image enhancement

g
(
x,y
)
=f
(
x,y
)

2
f
(
x,y
)

g
(
x,y
)
=
f
(
x,y
) +

2
f
(
x,y
)

f
s
(
x,y
)
=f
(
x,y
)
-

f*
(
x,y
)
,

where f*
(
x,y
)

is the blurred image.

High boost filtering

f
hb
(
x,y
)=
Af
(
x,y
)
-
f*
(
x,y
)=(
A
-
1
)
f
(
x,y
)+
f
(
x,y
)
-
f*
(
x,y
)

=(
A
-
1
)
f

(
x,y
)+
f
s
(
x,y
)

Using Laplacian

f
hb
(
x,y
)=
Af
(
x,y
)

2
f
(
x,y
)

f
hb
(
x,y
)=
Af
(
x,y
)
+

2
f
(
x,y
)

Shapening

rescaling

0~255

t
；如果
t
<0
，把所有

|
t
|

M
：如果

M
>255

p

p
’=(
p
/
M
)x255

p

f
(
x,y
)
=
[
G
x
, G
y
]
=
[

f/

x
,

f/

y

]

f
(
x,y
)
=
mag (

f
)
=
[
G
x
2
, G
y
2
]
1/2

=
[(

f/

x
)
2
+(

f/

y
)
2

]
1/2

Robert operator

G
x
=

f/

x=z
9
-
z
5

G
y
=

f/

y

=z
8
-
z
6

f
(
x,y
)
=
[(
z
9
-
z
5
)
2
, (z
8
-
z
6

)
2

]
1/2

=
|
z
9
-
z
5
|+|z
8
-
z
6
|

Sobel operator

f
(
x,y
)=|
z
7
+
2z
8
+
z
9
)
-
(
z
1
+
2z
2
+
z
3
)|

+|(z
3
+2z
6
+z
9
)
-
(z
1
+2z
4
+z
7
)|

3.8 Combining Spatial Enhancement Methods