4.intensity transformations

Digital Image ProcessingIII. Intensity transformations

Hamid LagaInstitut Telecom / Telecom Lille1

[email protected]

http://www.img.cs.titech.ac.jp/~hamid/

In the previous lecture

• Colors• The results of the interaction of light with the surface as observed by a

sensor from a viewing point(the surface absorbs some light, depending on the material properties)

• Image• a 2D function which encodes the color at each location of the image plane

• Digital image• a 2D matrix, each entry is called a pixel, it can have a single value

(grayscale) or a vector of 3 values (color)

• Color spaces• There are many color spaces, from now we will use the RGB color space.

• If we need to use another color space I will mention it explicitly

2

In the this lecture

• Basic processing operations that can be done on an image•

3

Intensity transformations

• Outline

– Basic intensity transformations(Image negatives, log transformations, power-law or Gamma transformations)

– Image histogram(Definitions, histogram equalization, local histogram processing, histogram statistics for image enhancement)

– Your first TP (to be done in Matlab)(Introduction to Matlab, Image manipulation with Matlab, …)

– Summary

4

Some definitions

• Spatial domain• The image plane itself

• Processing in the spatial domain deals with direct manipulation of the image pixels

• Other domains• Frequency domain

• Consider the image as a 2D signal and apply Fourier transform

• Process the image in the transformed domain

• Apply the inverse transform to return into the spatial domain.

• Types of transformations that can be applied to an image• Intensity transformations (this lecture)

• Operate on single pixels

• Spatial filtering (next lecture)• Working on a neighborhood of every pixel

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Basics

• Intensity transformation and spatial filtering

• f : input image

• g: output image

• T: an operator applied to the image f

• In the discrete case• f and g are 2D matrices

of size WxH

• T is a matrix of size wxhwhich is much smaller than WxH

• The value of each pixel of g is the resultof applying the operator T at the correspondinglocation in f.

• If wxh = 1x1 intensity transform

• Otherwise, it’s spatial filtering.

6

),(),( yxfTyxg

Basic intensity transforms

• Definition

• g(x, y) = T (f(x, y))

• Examples

• Image negatives

• Log transformations

• Power-law (Gamma) transformations

• In the following

• Each image is represented with a 2D matrix of L intensity levels. That is each

pixel takes a value between 0 and L-1

7


• Image negatives

• Reversing the intensity levels

g(x, y) = L – 1 – f(x, y)

– It produces the equivalent of a photographic negatives

8

Input image Its negative


• Image negatives

• Can be used to enhance white or gray details embedded in dark regions

especially when the black areas are dominant in size

9

Input image Its negative



g(x, y) = c . Log (1 + f(x, y))

– c is a constant and f(x, y) >= 0.

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Input image Its log transform c=1



– Maps a narrow range of low intensity values into a wider range of output

values (spreading)

– The opposite is true for the higher values of input levels (compressing)

11 x

Log(

1 +

x)



– Maps a narrow range of low intensity values into a wider range of output

values (spreading)

– The opposite is true for the higher values of input levels (compressing)

12

Input imagePixel values range from 0 to 106

Its log transform with c=1Values range from 0 to 6.2



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

– c and are positive constants.

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


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Original image and results of applyingGamma transformations with

c=1 and Gamma = 0.6, 0.4, and 0.3(increasing the dynamic range of

low intensities)



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


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Original image and results of applyingGamma transformations with c=1 and Gamma = 3, 4, and 5

(increasing the dynamic range of highintensities)


• Piecewise linear transformation functions

– Piecewise linear functions can be arbitrary complex

more flexibility in the design of the transformation

16

Piecewise linear transformations

• Contrast stretching

– Low contrast images can result from poor illumination

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Low contrast imageReduce

intensity

ReduceintensityIncrease

intensity

Piecewise linear transformations

• Contrast stretching

– Low contrast images can result from poor illumination

18

Low contrast image Contrast stretching

Intensity transformations

• Outline

– Basic intensity transformations(Image negatives, log transformations, power-law or Gamma transformations, piecewise-linear transformation functions)

– Image histogram(Definitions, histogram equalization, histogram matching)

– Your first TP (to be done in Matlab)(Introduction to Matlab, Image manipulation with Matlab, …)

– Summary

19

Image histogram - definitions

• Histogram– The histogram function is defined over all possible

intensity levels. – For each intensity level, its value is equal to the number of

the pixels with that intensity.

• Applications

– Image enhancement, compression, segmentation

– Very popular tool for realtime image processing

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Image histogram - examples

• Consider the following 5x5 image

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1 8 4 3 4

1 1 1 7 8

8 8 3 3 1

2 2 1 5 2

1 1 8 5 2

Image histogram - example

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1 8 4 3 4

1 1 1 7 8

8 8 3 3 1

2 2 1 5 2

1 1 8 5 2

1 2 3 4 5 6 7 8


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Graph of the histogram function

Original image


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Original image


25


Original image


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Original image


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Uniform distribution

Image histogram - definitions

• Normalized histogram– The normalised histogram function is the histogram function divided

by the total number of the pixels of the image

where n is the number of pixels in the image, and r is an intensity level.

– It is the frequency of appearance of each intensity level (color) in the image

– It gives a measure of how likely is for a pixel to have a certain intensity.

That is, it gives the probability of occurrence the intensity.– The sum of the normalised histogram function over the range of all

intensities is 1.

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n

rhrp

)()(

Cumulative Distribution Function (CDF)

• If h is the normalized histogram of an image, the CDF of h is

define as:

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Histogram h CDF of the normalized h

i

j

h ihiCDF0

)()(

Histogram equalization

• Goal• Increase the global contrast of an image by design a transformation (mapping) T

that that makes pixel intensities uniformly distributed over the whole range [0, L-1]

• Input• An image f of intensity levels in [0, L-1]

• Intensity levels can be viewed as random variables

• Let’s hf be the normalized histogram of f. hf is also the probability distribution function which we denote by Pf

• We denote by r intensity values of f

• Output• An image g = T(f) such that the intensities of g are uniformly over the range [0, L-

1].

• This will improve the contrast by spreading out the most frequent intensity values

• We denote by s = T(r).

30

Review of probabilities

• Uniform probability distribution• A probability distribution P is uniform if all events have the same probability of

occurrence.

• If N is the number of events then P(x) = 1 / N for every event x.

31

Intensities

Fre

qu

en

cy o

f

occ

urr

en

ce

0 1

Ideal equalized

histogram function


• We want to compute• An image g = T(f) such that the intensities of g are better distributed on the

histogram hg

• hg as a PDF, the intensities of g should follow a uniform distribution.

• Requirements for a good mapping T

• Condition A: T is required to be a monotonically increasing function in the interval [0, L-1]

• That is if x < y T(x) <= T(y)

• Avoids reversing intensities

• Condition B: x [0, L-1] T(x) [0, L-1] • This guarantees that the range of output intensities are the same as the range of input

intensities.

• Condition C: You may require strict monotonicity• x < y T(x) < T(y)

• This guarantees the mapping (or transformation )T is one-to-one.

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• Example of good mapping T

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Input intensity

Ou

tpu

t in

ten

sity

Input intensity

Ou

tpu

t in

ten

sity

Conditions A, B, and C are satisfied

Conditions A and B are satisfied, but not C


• Consider the following transformation

• In the discrete case

– That is, we add the values of the normalised histogram function from 1 to k to find where the intensity will be mapped.

• This is the cumulative distribution function

• Short quiz:• Prove that the transformation T defined as above satisfies the condition A

and condition B.

• Does it satisfy condition C?

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r

f dwwpLrTs0

)()1()(

r

f rhLrTs0

)()1()(


• Solution to the Quiz

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• Interesting property

• If we apply the transformation T to the image f we get an image g of

histogram hg such that

• This is a uniform distribution !!!

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1

1)(],1,1[

LshLs g

Histogram equalization – Algorithm

• Input

• A discrete image f of L levels of intensity

• Algorithm

• Step1: Compute the normalized histogram hf of input image f

• Step 2: Compute the CDF of hf, denoted CDFhf

• Step 3: Find a transformation T: [0, L-1] [0, L-1] such that

• Step 4: Apply the transformation on each pixel of the input image f

• Propert ies

• The procedure is fully automatic and very fast

• Suitable for real-time applications

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r

q

hf rCDFLqhLrT0

)()1()()1()(

Histogram equalization – Example

• Do histogram equalization on the 5x5 image with integer intensities in the range between one and eight

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1 8 4 3 4

1 1 1 7 8

8 8 3 3 1

2 2 1 5 2

1 1 8 5 2



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1 8 4 3 4

1 1 1 7 8

8 8 3 3 1

2 2 1 5 2

1 1 8 5 2

20.0)(

04.0)(

00.0)(

08.0)(

08.0)(

12.0)(

16.0)(

32.0)(

8

7

6

5

4

3

2

1

rp

rp

rp

rp

rp

rp

rp

rp

)(

)(

)(

)(

68.008.012.016.032.0)(

60.012.016.032.0)(

48.016.032.0)(

32.0)(

8

7

6

5

4

3

2

1

rT

rT

rT

rT

rT

rT

rT

rT

Intensity transformation functionNormalised

histogram function



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20.0)(

04.0)(

00.0)(

08.0)(

08.0)(

12.0)(

16.0)(

32.0)(

8

7

6

5

4

3

2

1

rp

rp

rp

rp

rp

rp

rp

rp

Intensity transformation function

Normalised histogram function

00.1)(

80.0)(

76.0)(

76.0)(

68.0)(

60.0)(

48.0)(

32.0)(

8

7

6

5

4

3

2

1

rT

rT

rT

rT

rT

rT

rT

rT The 32% of the pixels have intensity r1. We expect them to cover 32% of the possible intensities.

The 48% of the pixels have intensity r2 or less. We expect them to cover 48% of the possible intensities.

The 60% of the pixels have intensity r3 or less. We expect them to cover 60% of the possible intensities.

……………………………


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42


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Histogram equalization of color images

• The above procedure is for grayscale images

• For color images

• Apply separately the procedure to each of the three channels

• will yield in a dramatic change in the color balance

• NOT RECOMMENED unless you have a good reason for doing that.

• Recommended method

• Convert the image into another color space such as the Lab color space OR

HSL/HSV color space

• Apply the histogram equalization on the Luminance channel

• Convert back into the RGB color space.

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Lab color space

• Lab color space is designed to approximate the human vision

• The L component matches the human perception of lightness

• The Lab space is much larger than the gamut of computer displays.

• Conversions

• Will be covered in a special lecture on Colors.

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Histogram matching

• Histogram equalization tries to make the intensity distribution uniform. – Because of the irregular initial distribution, usually this is not possible.– Sometimes histogram equalization introduces artefacts,

(see the example with the photograph of the moon).

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Intensities

Fre

qu

en

cy o

f

occ

urr

en

ce

0 1

Ideal equalized

histogram function

Histogram matching

• Instead– We can specify the function we want for the histogram

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Intensities

Fre

qu

en

cy o

f

occ

urr

en

ce

0 1

hf: initial (source)histogram

P: specified (target)histogram

Find a transformation T such that the normalized

histogram of T(f) matches the probability distribution P.

P can be the normalized histogram of another image g,

then T transfers the histogram of g to f.

Histogram matching

• Goal

• Given an image f with a normalized histogram hf

• Given a probability distribution P

• Find a transformation T such that the normalized histogram of T(f)

matches the probability distribution P.

• Remarks

• If P is a uniform distribution then T is the same as the histogram

equalization.

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Histogram matching - procedure

• Input

• A source image f and a target distribution function P

• P can be the normalized histogram of another image g

• Procedure

• Calculate the histogram hf of f

• Calculate the CDFf of f and the CDFP of P

• For each gray level x [0, L-1], find the gray level y for which

• CDFf(x) = CDFP(y)

• We denote this mapping by M

• Can be implemented as a lookup table

• Apply the mapping M to all pixels of the input image f

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Histogram matching - examples

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Histogram matching - examples

• Undoing with histogram equalization

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Summary

• What we have seen

• Basic intensity transformations applied to individual pixels

• Histogram

• A very powerful tool in image processing

• A connection to probability and statistics

• Histogram equalization

• Histogram matching

• Enables the matching of the histogram of one image to the histogram of another

image

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Next

• Spatial filtering, convolution, and enhancement

– Fundamentals(Kernel, convolution, filtering)

– Smoothing (Linear / non-linear filters)

– Sharpening (The Laplacian - 2nd order derivatives, the gradient - first order derivatives)

– Image pyramids (The Gaussian pyramid)

– Filtering in the frequency domain (just a short introduction)

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Technology

4.intensity transformations