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Intensity Transformations (Chapter 3). CS474/674 – Prof. Bebis. Spatial Domain Methods. f(x,y). g(x,y). Point Processing. g(x,y). f(x,y). Area/Mask Processing. Point Processing Transformations. Convert a given pixel value to a new pixel value based on some predefined function. .

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Intensity transformations chapter 3

Intensity Transformations (Chapter 3)

CS474/674 – Prof. Bebis

Spatial domain methods
Spatial Domain Methods









Point processing transformations
Point Processing Transformations

  • Convert a given pixel value to a new pixel value based on some predefined function.

Negative image
Negative Image

  • O(r,c) = 255-I(r,c)

Contrast stretching or compression
Contrast Stretching or Compression

  • Stretch gray-level ranges where

    we desire more information

    (slope > 1).

  • Compress gray-level ranges that

    are of little interest

    (0 < slope < 1).


  • Special case of contrast compression

Intensity level slicing
Intensity Level Slicing

  • Highlight specific ranges of gray-levels only.

Same as double


Bit level slicing
Bit-level Slicing

  • Highlighting the contribution made by a specific bit.

  • For pgm images, each pixel is represented by 8 bits.

  • Each bit-plane is a binary image

Logarithmic transformation
Logarithmic transformation

  • Enhance details in the darker regions of an image at the expense of detail in brighter regions.



Exponential transformation
Exponential transformation

  • Reverse effect of that obtained using logarithmic mapping.



Histogram equalization
Histogram Equalization

  • A fully automatic gray-level stretching technique.

  • Need to talk about image histograms first ...

Image histograms
Image Histograms

  • An image histogram is a plot of the gray-level frequencies (i.e., the number of pixels in the image that have that gray level).

Image histograms cont d
Image Histograms (cont’d)

  • Divide frequencies by total number of pixels to represent as probabilities.

Properties of image histograms
Properties of Image Histograms

  • Histograms clustered at the low end correspond to dark images.

  • Histograms clustered at the high end correspond to bright images.

Properties of image histograms cont d
Properties of Image Histograms (cont’d)

  • Histograms with small spread correspond to low contrastimages (i.e., mostly dark, mostly bright, or mostly gray).

  • Histograms with wide spread correspond to high contrastimages.

Properties of image histograms cont d1
Properties of Image Histograms (cont’d)

High contrast

Low contrast

Histogram equalization1
Histogram Equalization

  • The main idea is to redistribute the gray-level values uniformly.

Histogram equalization cont d
Histogram Equalization (cont’d)

  • In practice, the equalized histogram might not be completely flat.

Probability definitions
Probability - Definitions

  • Random experiment: an experiment whose result is not certain in advance (e.g., throwing a die)

  • Outcome: the result of a random experiment

  • Sample space: the set of all possible outcomes (e.g., {1,2,3,4,5,6})

  • Event: a subset of the sample space (e.g., obtain an odd number in the experiment of throwing a die = {1,3,5})

Random variables review
Random Variables - Review

  • A function that assigns a real number to random experiment outcomes (i.e., helps to reduce space of possible outcomes)


X: # of heads

Random variables example
Random Variables - Example

  • Consider the experiment of throwing a pair of dice

  • Define the r.v. X=“sum of dice”

  • X=x corresponds to the event

Probability density function
Probability density function

  • The probability density function(pdf) is a real-valued function fX(x) describing the density of probability at each point in the sample space.

  • In the discrete case, this is just a histogram!


Probability distribution function




Probability distribution function

  • The integral of fX(x) defines the probability distribution functionFX(x) (i.e., cumulative probability)

  • In the discrete case, simply take the sum:



Random variable transformations
Random Variable Transformations

  • Suppose Y=T(X)

    • e.g., Y=X+1

  • If we know fX(x), can we find fY(y)?

  • Yes - it can be shown that:

Transformations of r v example
Transformations of r.v. - Example

μ=0, σ=1


μ=1, σ=1

Histogram equalization cont d1
Histogram Equalization (cont’d)

The intensity levels can be viewed as a random variable in [0,1]



Histogram equalization cont d2
Histogram Equalization (cont’d)

For PGM images:

L=256 (graylevels)

k=0,1,2, …, L-1 (possible graylevels)

rk=k/(L-1) (normalized graylevel in [0, 1])

Histogram equalization cont d3
Histogram Equalization (cont’d)

then, de-normalize:

sk x (L-1)

Histogram equalization example
Histogram Equalization Example

3 bit

64 x 64 image

input histogram

equalized histogram

Histogram equalization examples
Histogram Equalization Examples

original images and histograms

equalized images and histograms

Histogram specification matching




is known





Histogram Specification (Matching)

  • Histogram equalization yields a uniform pdf only.

  • What if we want to obtain a histogram other than uniform?

so, Q(r)=G-1(T(r))

Histogram specification cont d
Histogram Specification (cont’d)

  • fS(s) and fV(v) are uniform

  • G(z) can be computed by specifying fZ(z) but I2 and I’2 are unknown!

  • z=G-1(v) requires that v is a r.v. with uniform pdf

  • IDEA: use z=G-1(s) instead of z=G-1(v)

    • s is a r.v. with uniform pdf

  • The desired transformation is




Histogram specification cont d1
Histogram Specification (cont’d)

  • Comments

    • We do not need to apply T( ) and G-1( ) separately!

    • Combine them: Q=G-1T, thus, z=Q(r)

    • Histogram specification assumes that we know G-1 (not always easy to find).

    • G( ) must be in [0,1] and must be non-decreasing.


Histogram specification example
Histogram Specification Example

3 bit

64 x 64 image

input histogram

specified histogram

actual histogram

Histogram specification cont d2
Histogram Specification (cont’d)

  • Histogram specification might yield superior results than histogram equalization.

results of histogram equalization

Histogram specification cont d3
Histogram Specification (cont’d)

results of histogram specification



Local histogram processing
Local Histogram Processing

  • Histogram equalization/specification are global methods.

    • The intensity transformation is computed using pixels from the entire image.

  • Global transformations are not appropriate for enhancing little details in an image.

    • The number of pixels in these areas might be very small, contributing very little to the computation of the transformation.

Local histogram processing1
Local Histogram Processing


Define a transformation function based on the intensity distribution in a neighborhood of every pixel in the image!

Local histogram processing cont d
Local Histogram Processing (cont’d)

1. Define a neighborhood and move its center from pixel to pixel.

2. At each location, the histogram of the points in the neighborhood is computed. Obtain histogram equalization or histogram specification transformation.

3. Map the intensity of the pixel centered in the neighborhood.

4. Move to the next location and repeat the procedure.

Local histogram processing example
Local Histogram Processing: Example

local histogram


3 x 3 neighborhood

global histogram


Histogram statistics
Histogram Statistics


(average intensity)

n-th moment

around mean


(2nd moment)

Example comparison of standard deviation values
Example: Comparison of Standard Deviation Values

σ is useful for estimating image contrast!

Using histogram statistics for image enhancement
Using Histogram Statistics for Image Enhancement

  • Useful when parts of the image might contain hidden features.

Task: enhance dark

areas without changing

bright areas.

Idea: Find dark, low contrast

areas using local statistics.

Using histogram statistics for image enhancement example
Using Histogram Statistics for Image Enhancement: Example


  • Intensity operations can yield pixel values outside of the range [0 – 255].

  • You should convert values back to the range [0 – 255] to ensure that the image is displayed properly.

  • How would you find the following mapping?

    [fmin – fmax]  [ 0 – 255]