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Intensity Transformations (Chapter 3)PowerPoint Presentation

Intensity Transformations (Chapter 3)

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### Intensity Transformations (Chapter 3)

CS474/674 – Prof. Bebis

Point Processing Transformations

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

Negative Image

- O(r,c) = 255-I(r,c)

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).

Thresholding

- Special case of contrast compression

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

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

compress

stretch

Exponential transformation

- Reverse effect of that obtained using logarithmic mapping.

stretch

compress

Histogram Equalization

- A fully automatic gray-level stretching technique.
- Need to talk about image histograms first ...

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)

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

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)

- 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.

Histogram Equalization

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

Histogram Equalization (cont’d)

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

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

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

X(j)

X: # of heads

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

- 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!

Gaussian

fX(a)da

non-decreasing

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:

fX(x)

FX(x)

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:

Special transformation!

Proof:

Histogram Equalization (cont’d)

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

ps(s)

pr(r)

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])

desired

histogram

is known

fR(r)

fZ(z)

fR(a)da

fZ(a)da

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)

- 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
z=G-1(T(r))

fR(a)da

fz(a)da

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.

z=G-1(T(r))=Q(r)

Histogram Specification Example

3 bit

64 x 64 image

input histogram

specified histogram

actual histogram

Histogram Specification (cont’d)

- Histogram specification might yield superior results than histogram equalization.

results of histogram equalization

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 Processing

Idea:

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

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

equalization

3 x 3 neighborhood

global histogram

equalization

Example: Comparison of Standard Deviation Values

σ is useful for estimating image contrast!

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

Question

- 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]

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