Digital Image Processing - Image Enhancement

Unit III Image Enhancement

Two mark Questions with Answers

1. What is a mask? A Mask is a small two-dimensional array, in which the value of

the mask coefficient determines the nature of the process, such as image sharpening.

The enhancement technique based on this type of approach is referred to as mask processing.

2. How can an image negative be obtained?

The negative of an image with gray levels in the range [0, L-1] is obtained by using the negative transformation, which is given by the expression.

s = L-1- r, where„s‟ is output pixel, „r‟ is input pixel

3. What is the difference between contrast stretching and compression of dynamic range?

Contrast Stretching

Produce higher contrast than the original by Darkening the levels below m in the original image. Brightening the levels above m in the original image.

Compression of dynamic range

It compresses the dynamic range of images with large variations in pixel values

Example of image with dynamic range: Fourier spectrum image

It can have intensity range from 0 to 106 or higher. We can‟t see the significant degree of detail as it will be lost

in the display.

The contrast stretching increases the dynamic range of the gray levels

4. What is a histogram?

Histogram of a digital image with gray levels in the range [0,L-1] is

a discrete function.

h(rk) = nk Where,

rk : the kth gray level nk : the number of pixels in the image having gray level rk h(rk) : histogram of a digital image with gray levels rk

5. What is meant by histogram equalization?

It is a technique used to obtain linear histogram. It is also known as histogram linearization. Condition for uniform histogram is Ps(s) = 1

(or) The histogram equalization is an approach to enhance a given

image. The approach is to design a transformation T(.) such that the gray values in the output is uniformly distributed in [0, 1].

6. How can histogram equalization be applied locally?

Histogram processing methods are global processing, in the sense

that pixels are modified by a transformation function based on the gray-level content of an entire image.

Sometimes, we may need to enhance details over small areas in an image, which is called a local enhancement.

7. What is Image Enhancement?

Image enhancement is a technique to process an image so that the

result is more suitable than the original image for specific applications.

8. In local Histogram processing, why are non-overlapping regions used?

It‟s used to reduce computation is to utilize nonoverlapping regions, but it usually produces an undesirable checkerboard effect.

9. What is meant by histogram matching or histogram specification?

Histogram equalization yields an image whose pixels are (in theory) uniformly distributed among all gray levels. Sometimes, this may not be desirable. Instead, we may want a transformation that yields an output image with a pre-specified histogram. This technique is called histogram specification.

10. How can noise reduction be accomplished using image averaging?

Consider a noisy image g(x,y) formed by the addition of noise (x,y) to an original image f(x,y).

g(x,y) = f(x,y) + (x,y) if noise has zero mean and be uncorrelated then it can be shown that if

),( yxg

Then,

),(2

),(2 , yxyxg

if „K‟ increase, it indicates that the variability (noise) of the pixel at each location (x,y) decreases.

11. Differentiate between linear and nonlinear spatial filters.

s.no. Linear spatial filter Non-linear spatial filter

1. 2.

Response is a sum of products of the filter co-efficient. 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).

They do not explicitly use co-efficients in the sum-of-products. R = w1z1 + w2z2 + … +w9z9 9 = ∑ wizi i=1

1( , ) ( , )

1

Kg x y g x y

iK i

12 2( , ) ( , )g x y x yK

12. What is image Negatives?

The negative of an image with gray levels in the range [0, L-1] is

obtained by using the negative transformation, which is given by the expression.

s = L-1- r, Where s is output pixel, r is input pixel

13. Differentiate between Correlation and Convolution with specific reference to an image and a filter mask.

Convolution in frequency domain reduces the multiplication in the x

domain

The correlation of 2 continuous functions f(x) and g(x) is defined by

14. Define derivative filter.

For a function f (x, y), the gradient f at co-ordinate (x, y) is defined as the vector

15. What is the principal difficulty with the smoothing method with reference to edges and sharp details?

Median filtering is a powerful smoothing technique that does not blur

the edges significantly . Max/min filtering is used where the max or min value of the

neighbourhood gray levels replaces the candidatepel .

Shrinking and expansion are useful operations especially in two tone images.

16. What is the basic characteristic of a high pass filter mask coefficients? The basic strategy behind weighting the center point the highest and

then reducing the value of the coefficients as a function of increasing distance from the origin is simply an attempt to reduce blurring in the smoothing process.

17. What is the effect of averaging with reference to detail in an image?

An important application of image averaging is in the field of astronomy, where imaging with very low light levels is routine, causing sensor noise frequently to render single images virtually useless for analysis.

18. Outline a simple procedure to produce an enhanced image using a fourier transform and a filter transfer function.

Frequency domain techniques are based on modifying the Fourier

transform of an image.

19. How can blurring or smoothing process be explained in the frequency domain?

Smoothing Filters are used for blurring and for noise reduction

� Blurring is used for removal of small details prior to object extraction. � bridging of small gaps in lines or curves. � Smoothing Linear Filters (Averaging Filters) � replace the average value defined by the filter mask. � have the undesirable effect of blur edges

20. How can image sharpening be achieved by a high pass filtering process

in the frequency domain? Image sharpening deals with enhancing detail information in an image. The detail information is typically contained in the high spatial

frequency components of the image. Therefore, most of the techniques contain some form of high pass filtering.

High pass filtering can be done in both the spatial and frequency domain.

– Spatial domain: using convolution mask (e.g. enhancement filter). – Frequency domain: using multiplication mask.

21. What is homomorphic filtering?

Homomorphic filtering is a generalized technique for signal and

image processing, involving a nonlinear mapping to a different domain in which linear filter techniques are applied, followed by mapping back to the original domain.

22. Write the application of sharpening filters.

The applications of sharpening filters are as follows, i. Electronic printing and medical imaging to industrial application

ii. Autonomous target detection in smart weapons.

23. What do you mean by point processing? Image enhancement at any Point in an image depends only on the

gray level at that point is often referred to as Point processing.

24. Define high boost filter. High boost filtered image is defined as

HBF= A (original image)-LPF = (A-1) original image + original image –LPF HBF= (A-1) original image +HPF

25. Name the different types of derivative filters.

The different types of derivative filters are

i. Perwitt operators ii. Roberts cross gradient operators

iii. Sobel operators.

Twelve mark Questions

1. What is image enhancement? Explain Contrast stretching and compression of dynamic range.

Image enhancement is a technique to process an image so that the result is more suitable than the original image for specific applications.

The suitableness is up to each application. A method which is quite useful for enhancing an image may

not necessarily be the best approach for enhancing another images

Image enhancement widely used in computer graphics. It is the sub areas of image processing.

Enhancement approaches: 1. Spatial domain 2. Frequency domain

1) Spatial Domain : (image plane)

Techniques are based on direct manipulation of pixels in an image.

2) Frequency Domain : Techniques are based on modifying the Fourier transform of an

image. There are some enhancement techniques based on various combinations of methods from these two categories.

Contrast Stretching Low contrast images occur often due to poor or nonuniform

lighting conditions or due to nonlinearity or small dynamic range of the image sensor.

Expands the range of intensity levels in an image so that it spans

the full intensity range of the recording medium or display device.

The figure shows a typical contrast stretching transformation

FIG : Contrast Stretching Transformation

For , Dark region stretch 1,3

La ;

Mid region stretch 2

1,3

Lb ;

Bright region stretch 1

Which can be expressed as,

, 0

( ) ,

( ) ,

a

b

u for u a

u a V for a u b

u b V for b u L

The slope of the transformation is chosen greater than unity in the region of stretch. The parameters a & b can be obtained by examining the histogram of the image. For example, the gray scale intervals where pixels occur most frequently would be stretched most to improve the overall visibility of the scene. Produce Higher contrast than the original image: By darkening the levels below „m‟ in the original image. By Brightening the levels above „m‟ in the original image.

Compression of dynamic range.

Sometimes the dynamic range of the image data may be very large.

For example, the dynamic range of a typical unitarily transformed image is so large that only a few pixels are visible.

The dynamic range can be compressed via the logarithmic transformation.

log (1 )10

C u

Where „C‟ is a scaling constant and „u‟ is a Gray levels.

This transformation enhance the small magnitude pixels compared to those pixels with large magnitudes.

2. Explain histogram equalization and histogram specification. How can

they be applied for local enhancement?

Histogram Processing Histogram of a digital image with gray levels in the range [0,L-1] is a discrete function

h(rk) = nk Where

rk : the kth gray level nk : the number of pixels in the image having gray

level rk h(rk) : histogram of a digital image with gray levels rk

Histogram Equalization Histogram EQUALization

Aim: To “equalize” the histogram, to “flatten”, “distrubute as uniform as possible”.

● As the low-contrast image's histogram is narrow and centred towards the middle of the gray scale, by distributing the histogram to a wider range will improve the quality of the image.

● Adjust probability density function of the original histogram so that the probabilities spread equally The histogram equalization is an approach to enhance a given image. The approach is to design a transformation T(.) such that the gray values in the output is uniformly distributed in [0, 1]. Let us assume for the moment that the input image to be enhanced has continuous gray values, with r = 0 representing black and r = 1 representing white. We need to design a gray value transformation s = T(r), based on the histogram of the input image, which will enhance the image.

As before, we assume that: (1) T(r) is a monotonically increasing function for 0≤r≤1 (preserves order from black to white). (2) T(r) maps [0,1] into [0,1] (preserves the range of allowed Gray values). Let us denote the inverse transformation by r = T -1(s) . We assume that the inverse transformation also satisfies the above two conditions. We consider the gray values in the input image and output image as random variables in the interval [0, 1]. Let pin(r) and pout(s) denote the probability density of the Gray values in the input and output images. If pin(r) and T(r) are known, and r = T -1(s) satisfies condition 1, we can write (result from probability theory):

( ) ( )1( )

drp s p rout in ds r T s

One way to enhance the image is to design a transformation T(.) such that the gray values in the output is uniformly distributed in [0, 1], i.e. pout (s) = 1, 0≤s≤1 . In terms of histograms, the output image will have all gray values in “equal proportion”. This technique is called histogram equalization.

Next we derive the gray values in the output is uniformly distributed in [0, 1]. ·Consider the transformation

( ) ( ) 0 1,0r

s T r p w dw rin

Note that this is the cumulative distribution function (CDF) of pin (r) and satisfies the previous two conditions. From the previous equation and using the fundamental theorem of calculus,

( )ds p rindr

Therefore, the output histogram is given by

1( ) ( ) 1 1, 0 11( )( ) 1( )

p s p r sr T sout in p rin r T s

The output probability density function is uniform, regardless of the input. Thus, using a transformation function equal to the CDF of input gray values r, we can obtain an image with uniform gray values. This usually results in an enhanced image, with an increase in the dynamic range of pixel values.

How to implement histogram equalization? Step 1:For images with discrete gray values, compute:

( )nkp r

in nk 0 1r

k 0 1k L

L: Total number of gray levels nk: Number of pixels with gray value rk

n: Total number of pixels in the image Step 2: Based on CDF, compute the discrete version of the previous transformation :

( ) ( )0

ks T r p r

in jk k j

0 1k L

Example: Consider an 8-level 64 x 64 image with gray values (0, 1, …,7). The normalized gray values are (0, 1/7, 2/7, …, 1). The normalized histogram is given below:

NB: The gray values in output are also (0, 1/7, 2/7, …, 1). Notice that there are only five distinct gray levels --- (1/7, 3/7,5/7, 6/7, 1) in the output image. We will relabel them as (s0,s1, …, s4 ). With this transformation, the output image will have histogram

Histogram Specification (Histogram Matching) Histogram equalization yields an image whose pixels are (in theory) uniformly distributed among all gray levels. Sometimes, this may not be desirable. Instead, we may want a transformation that yields an output image with a pre-specified histogram. This technique is called histogram specification. Given Information

(1) Input image from which we can compute its histogram .

(2) Desired histogram.

Goal

Derive a point operation, H(r), that maps the input image into an output image that has the user-specified histogram.

Again, we will assume, for the moment, continuous-gray values.

Approach of derivation

Step1: Equalize the levels of the original image Step2: Specify the desired pdf and obtain the transformation function Step3: Apply the inverse transformation function to the levels obtained in step 1

Histogram equalization has a disadvantage which is that it can generate only one type of output image.

With Histogram Specification, we can specify the shape of the histogram that we wish the output image to have.

It doesn‟t have to be a uniform histogram Consider the continuous domain , Let pr(r) denote continuous probability density function of gray-level of input image, r Let pz(z) denote desired (specified) continuous probability density function of gray-level of output image, z Let s be a random variable with the property Histogram equalization Where w is a dummy variable of integration Next, we define a random variable z with the property Histogram equalization Where t is a dummy variable of integration

Thus, s = T(r) = G(z)

Therefore, z must satisfy the condition, z = G-1(s) = G-1[T(r)] Assume G-1 exists and satisfies the condition (a) and (b) We can map an input gray level r to output gray level z

r

r dw)w(p)r(Ts0

sdt)t(p)z(g

z

z 0

Procedure Conclusion:

1. Obtain the transformation function T(r) by calculating the histogram equalization of the input image

( ) ( )0

rs T r p w dwr

2. Obtain the transformation function G(z) by calculating histogram equalization of the desired density function

( ) ( )0

zG z p t dt sz

3. Obtain the inversed transformation function G-1

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

4. Obtain the output image by applying the processed gray-level from the inversed transformation function to all the pixels in the input image

Histogram specification is a trial-and-error process There are no rules for specifying histograms, and one must resort to

analysis on a case-by-case basis for any given enhancement task.

Local Enhancement Histogram processing methods are global processing, in the sense

that pixels are modified by a transformation function based on the gray-level content of an entire image.

Sometimes, we may need to enhance details over small areas in

an image, which is called a local enhancement. The image pre-processing may be used for different goals. For example for manual or automatic image processing. So we

have developed another image enhancement procedure, the local histogram equalization.

The main idea is to take into account histogram distribution over local window and combine it with global histogram distribution. We have used nonlinear histogram equalization for combination of local and global histogram.

a) Original image (slightly blurred to reduce noise) b) global histogram equalization (enhance noise & slightly increase

contrast but the construction is not changed) c) local histogram equalization using 7x7 neighborhood (reveals the small

squares inside larger ones of the original image.

Define a square or rectangular neighborhood and move the center of this area from pixel to pixel.

At each location, the histogram of the points in the neighborhood is computed and either histogram equalization or histogram specification transformation function is obtained.

Another approach used to reduce computation is to utilize nonoverlapping regions, but it usually produces an undesirable checkerboard effect.

Explain the result in c) Basically, the original image consists of many small squares inside the

larger dark ones. However, the small squares were too close in gray level to the larger

ones, and their sizes were too small to influence global histogram equalization significantly.

So, when we use the local enhancement technique, it reveals the small areas.

Note also the finer noise texture is resulted by the local processing using relatively small neighborhoods.

3. Explain image subtraction technique and its application in mask mode

radiography.

Image subtraction technique The difference between two images f(x,y) and h(x,y) are expressed as,

G(x,y)= f(x,y) – h(x,y)

Is obtained by computing the difference between all pairs of corresponding pixels from f and h. The key usefulness of subtraction is the enhancement of difference between images.

In many imaging applications it is desired to compare two

complicated busy images. A simple but powerful method is to align the two images and subtract

them. The difference image is then enhance. For example, the missing

components on a circuit board can be detected by subtracting its image from that of a properly assembled board.

Another application is imaging of the blood vessels and arteries in a

body. The blood stream is injected with radio – opaque dye and X-ray images are taken before and after the injection. The difference of the two images yields a clear display of the blood flow paths.

Other applications of change detection are in security monitoring

systems, automated inspection of printed circuits and so on.

Mask mode radiography

One of the most commercially successful and beneficial uses of image subtraction is in the area of medical imaging called mask mode radiography .

h(x,y) is the mask, an X-ray image of a region of a patient‟s body

captured by an intensified TV camera (instead of traditional X-ray film) located opposite an X-ray source

f(x,y) is an X-ray image taken after injection a contrast medium into the patient‟s bloodstream

images are captured at TV rates, so the doctor can see how the medium propagates through the various arteries in the area being observed (the effect of subtraction) in a movie showing mode.

Note We may have to adjust the gray-scale of the subtracted image to be [0,

255] (if 8-bit is used) first, find the minimum gray value of the subtracted image second, find the maximum gray value of the subtracted image set the minimum value to be zero and the maximum to be 255 while the rest are adjusted according to the interval [0, 255], by timing each value with 255/max

Subtraction is also used in segmentation of moving pictures to track the changes after subtract the sequenced images, what is left should be the

moving elements in the image, plus noise

4. Explain image averaging.

Consider a noisy image g(x,y) formed by the addition of noise (x,y) to an original image f(x,y)

if noise has zero mean and be uncorrelated then it can be shown that if

= image formed by averaging K different noisy images If the noise is uncorrelated and has zero expectation, then

, , ,g x y f x y x y

1 1 1

, , ,0 0 0

M M Mg x y f x y x yi

i i i

, , ,g x y f x y x y

, ,E g x y f x y

),( yxg

1( , ) ( , )1

Kg x y g x y

iK i

)},({ yxgE

= expected value of g (output after averaging) = original image f(x,y)

),(2

),(2 , yxyxg

= variances of g and if K increase, it indicates that the variability (noise) of the pixel at each location (x,y) decreases. (or) Assume n(x,y) a white noise with mean=0, and variance If we have a set of noisy images The noise variance in the average image is Note: the images gi(x,y) (noisy images) must be registered (aligned) in

order to avoid the introduction of blurring and other artifacts in the output image.

12 2,, x yg x y M

2 2( , )E n x y

( , )g x yi

1( , ) ( , )1

Mg x y g x yave iM i

21 1 12 2( , ) ( , )

21 1

M ME n x y E n x y

i iM MMi i

5. What are smoothing filters? Explain low pass spatial filtering and

median filtering.

Smoothing is fundamentally a low pass operation in the frequency domain.

Spatial Filtering Spatial filters are designed to highlight or suppress specific features in an image based on their spatial frequency.. Filtering is performed by using convolution windows.

Used to enhance the appearance of an image It is based on concept of image texture It highlight or suppress specific features in an image based on their

spatial frequency

use filter (can also be called as mask/kernel/template or window) the values in a filter subimage are referred to as coefficients, rather than

pixel. our focus will be on masks of odd sizes, e.g. 3x3, 5x5,…

Spatial Filtering Process

simply move the filter mask from point to point in an image. at each point (x,y), the response of the filter at that point is calculated

using a predefined relationship.

Spatial Filtering (Masking)

Therefore, R= w1z1 + w2z2 + ….. +w9z9

The operation is similar to Convolution. Hence the masks are also called convolution masks.

• Non linear operations such as finding median may also be done on a neighborhood.

• Near the edges parts of the masks may lie beyond the image boundary. • To avoid this either a smaller filtered image is accepted. • Or zeros are padded along the image boundary.

...1 1 2 2

R w z w z w zmn mnmn

w zi ii i

Median Filters replaces the value of a pixel by the median of the gray levels in the

neighborhood of that pixel (the original value of the pixel is included in the computation of the median)

Quite popular because for certain types of random noise (impulse noise salt and pepper noise) , they provide excellent noise-reduction capabilities, with considering less blurring than linear smoothing filters of similar size.

Forces the points with distinct gray levels to be more like their neighbors.

Isolated clusters of pixels that are light or dark with respect to their neighbors, and whose area is less than n2/2 (one-half the filter area), are eliminated by an n x n median filter.

Eliminated = forced to have the value equal the median intensity of the neighbors.

larger clusters are affected considerably less

Median Filters

Excellent at noise removal, without the smoothing effects that can

occur with other smoothing filters Particularly good when salt and pepper noise is present

Max filter is good for pepper noise and min is good for salt noise.

The gradient of the image intensity at each point, gives the direction of the largest possible increase from light to dark and the rate of change in that direction. The result therefore shows how "abruptly" or "smoothly" the image changes at that point, and therefore how likely it is that that part of the image represents an edge, as well as how that edge is likely to be oriented. In practice, the magnitude (likelihood of an edge) calculation is more reliable and easier to interpret than the direction calculation.

Mathematically, the gradient of a two-variable function (here the image intensity function) is at each image point a 2D vector with the components given by the derivatives in the horizontal and vertical directions.

The Sobel operator represents a rather inaccurate approximation of the image gradient, but is still of sufficient quality to be of practical use in many applications

Sobel operators :

What are sharpening filters? Explain Derivative filters.

Sharpening filters are used to enhance the edges of objects and adjust the contrast and the shade characteristics. In combination with threshold they can be used as edge detectors. Sharpening or high-pass filters let high frequencies pass and reduce the lower frequencies and are extremely sensitive to shut noise.

To construct a high-pass filter the kernel coefficients should be set positive near the center of the kernel and in the outer periphery negative.

The sharpening filters are divided into the following groups:

High Pass Filters (Uni Crisp) Laplacian of Gaussian / Mexican Hat filters. Unsharp Masking High Boost filtering Difference of Gaussians

The applications of sharpening filters are as follows,

i. Electronic printing and medical imaging to industrial application ii. Autonomous target detection in smart weapons.

Derivative filters For a function f (x, y), the gradient f at co-ordinate (x, y) is defined as the vector

The strength of the response of a derivative operator is proportional to the degree of discontinuity of the image at the point at which the operator is applied.

thus, image differentiation enhances edges and other discontinuities (noise) Deemphasizes area with slowly varying gray-level values.

First-order derivative

a basic definition of the first-order derivative of a one-dimensional function f(x) is the difference

Second-order derivative similarly, we define the second-order derivative of a one-dimensional

function f(x) is the difference First and Second-order derivative of f(x,y) when we consider an image function of two variables, f(x,y), at which

time we will dealing with partial derivatives along the two spatial axes.

Gradient operator ( , ) ( , ) ( , )

ff x y f x y f x y

x y x y

Laplacian operator

(linear operator)

( 1) ( )f

f x f xx

2( 1) ( 1) 2 ( )

2f

f x f x f xx

2 2( , ) ( , )22 2

f x y f x yf

x y

Effect of Laplacian Operator as it is a derivative operator,

it highlights gray-level discontinuities in an image it deemphasizes regions with slowly varying gray levels

tends to produce images that have grayish edge lines and other discontinuities, all superimposed

on a dark, featureless background.

The gradient of an image f(x,y) at location (x,y) is the vector

The gradient vector points are in the direction of maximum rate of change of f at (x,y)

In edge detection an important quantity is the magnitude of this vector (gradient) and is denoted as ∆f.

∆f = mag (∆f) = [Gx2+Gy2] ½

The direction of gradient vector also is an important quantity.

α(x,y) = tan-1(Gy/Gx)

6. Explain Low pass filtering in frequency domain. Discuss using an ideal

filter. • The basic model for filtering in the frequency domain

( , ) ( , ) ( , )G u v H u v F u v

Where, F(u,v): the Fourier transform of the image to be smoothed

H(u,v): a filter transfer function • Smoothing is fundamentally a lowpass operation in the frequency

domain. • There are several standard forms of lowpass filters (LPF).

– Ideal lowpass filter – Butterworth lowpass filter – Gaussian lowpass filter

Ideal Lowpass Filters (ILPFs) • The simplest lowpass filter is a filter that “cuts off” all high-frequency

components of the Fourier transform that are at a distance greater than a specified distance D0 from the origin of the transform.

• The transfer function of an ideal lowpass filter

1 if ( , ) 0

( , )0 if ( , )

0

D u v DH u v

D u v D

Where,

D(u,v) : the distance from point (u,v) to the center of their frequency

rectangle (M/2, N/2)

12 2 2( , ) ( / 2) ( / 2)D u v u M v N

Fig: a) Perspective plot of an ideal low pass filter transfer function

Fig : b) Filter displayed as an image

Fig : c) Filter radial cross section

LPF is a type of “nonphysical” filters and can‟t be realized with electronic components and is not very practical.

7. Explain low pass filtering in frequency domain. Differentiate between using ideal filter and Butterworth filter for low pass filtering.

• The basic model for filtering in the frequency domain

( , ) ( , ) ( , )G u v H u v F u v

Where, F(u,v): the Fourier transform of the image to be smoothed

H(u,v): a filter transfer function • Smoothing is fundamentally a lowpass operation in the frequency

domain.

• There are several standard forms of lowpass filters (LPF). – Ideal lowpass filter – Butterworth lowpass filter – Gaussian lowpass filter

Ideal Lowpass Filters (ILPFs) • The simplest low pass filter is a filter that “cuts off” all high-frequency

components of the Fourier transform that are at a distance greater than a specified distance D0 from the origin of the transform.

• The transfer function of an ideal lowpass filter

1 if ( , ) 0

( , )0 if ( , )

0

D u v DH u v

D u v D

Where,

D(u,v) : the distance from point (u,v) to the center of their frequency

rectangle (M/2, N/2)

12 2 2( , ) ( / 2) ( / 2)D u v u M v N

Fig: a) Perspective plot of an ideal low pass filter transfer function

Fig : b) Filter displayed as an image

Fig : c) Filter radial cross section

LPF is a type of “nonphysical” filters and can‟t be realized with electronic components and is not very practical.

The drawback of this filter function is a ringing effect which occurs

along the edges of filtered real domain image.

The drawback of this filter function is a ringing effect which occurs along the edges of the filtered real domain image.

Butterworth low pass filter

The BLPF may be viewed as a transition between ILPF and GLPF, BLPF of order 2 is a good compromise between effective low pass filtering and acceptable ringing characteristics.

• The transfer function of a Butterworth lowpass filter of order n with

cutoff frequency at distance D0 from the origin is defined as:

1( , )2

1 ( , ) /0

H u vn

D u v D

Fig: a) Perspective plot of a Butterworth low pass filter transfer function

Fig : b) Filter displayed as an image

Fig : Filter radial Cross Sections on order 1 through 4

• Smooth transfer function, no sharp discontinuity, no clear cutoff

frequency. The vertical edges and sharp corners of Ideal low pass filter are non-realizable in the physical world. Although we can emulate these filter masks with a computer, side effects such as blurring and ringing become apparent.

BLPF does not have a sharp discontinuity that establishes a clear cutoff between passed and frequencies

H(u, v) = 0.5 (down 50% from its maximum value of 1) when D(u, v) = Do.

What is homomorphic filtering? Explain.

Homomorphic filtering is a generalized technique for signal and image processing, involving a nonlinear mapping to a different domain in which linear filter techniques are applied, followed by mapping back to the original domain.

The digital images are created from optical image that consist of two primary components:

– The lighting component – The reflectance component

The lighting component results from the lighting condition present

when the image is captured. – Can change as the lighting condition change.

The reflectance component results from the way the objects in the

image reflect light.

– Determined by the intrinsic properties of the object itself.

– Normally do not change.

In many applications, it is useful to enhance the reflectance component,

while reducing the contribution from the lighting component.

Homomorphic filtering is a frequency domain filtering process that

compresses the brightness (from the lighting condition) while

enhancing the contrast (from the reflectance properties of the object).

The homomorphic filtering process consists of five steps:

– A natural log transform (base e)

– The Fourier transform

– Filtering

– The inverse Fourier transform

– The inverse log function (exponential)

A simple image model

– f(x,y): the intensity is called the gray level for

monochrome image

– f(x, y) = i(x, y).r(x, y)

– 0 < i(x, y) < inf, the illumination

– 0< r(x, y) < 1, the reflectance

Fig : Homomorphic filtering approach for image enhancement

The illumination component – Slow spatial variations – Low frequency

The reflectance component – Vary abruptly, particularly at the junctions of dissimilar

objects – High frequency

, , ,

, ln , ln , ln ,

, ln , ln ,

( , ) ( , ) ( , )

, , , , ,

, , ,

, exp , exp , exp ,

f x y i x y r x y

z x y f x y i x y r x y

F z x y F i x y F r x y

Z u v F u v F u vri

S u v H u v F u v H u v F u vri

s x y i x y r x y

g x y s x y i x y r x y

Homomorphic filters – Affect low and high frequencies differently – Compress the low frequency dynamic range – Enhance the contrast in high frequency

Fig : Cross section of a circularly symmetric filter function. D(u,v) is the

distance from the origin of the centered transform

1

1H

L

2 2( ( , )/ )

0( , ) ( )[1 ]c D u v D

H u v eH L L

Explain with necessary diagrams how Histogram modeling techniques modify an image?

Histogram Useful to graphically represent the distribution of pixel values in a

histogram. The histogram of an image represents the relative frequency of

occurrence of the various grey levels in the image. Plots the number of pixels in the image (vertical axis) with a particular

brightness value (horizontal axis). Histogram modeling is the basis for numerous powerful spatial domain

processing techniques, especially for image enhancement. Histogram Processing

● Basic for numerous spatial domain processing techniques ● Used effectively for image enhancement ● Information inherent in histograms is also useful in image compression and segmentation

Histogram & Image Contrast

Dark Image

Components of histogram are concentrated on the low side of the gray scale.

Bright Image

Components of histogram are concentrated on the high side of the gray scale.

Low-contrast Image

Histogram is narrow and centred towards the middle of the gray scale.

High-contrast Image Histogram covers a broad range of the gray scale and the distribution of pixels is not too far from uniform, with very few vertical lines being much higher than others

We consider the gray values in the input image and output image as random variables in the interval [0, 1]. Let pin(r) and pout(s) denote the probability density of the Gray values in the input and output images. If pin(r) and T(r) are known, and r = T -1(s) satisfies condition 1, we can write (result from probability theory):

( ) ( )1( )

drp s p rout in ds r T s

One way to enhance the image is to design a transformation T(.) such that the gray values in the output is uniformly distributed in [0, 1], i.e. pout (s) = 1, 0≤s≤1 .

Histogram modeling techniques modify an image

Fig. Histogram modification

n

pv = f(u) = ( )xiu=0ix

1nnpu

ixf(u)= , n=2, 3,...1L-1x np x( )iu

=0ix

u v v'

Uniform

quantizer

f(u)

Approach of derivation

Step1: Equalize the levels of the original image

Step2: Specify the desired pdf and obtain the transformation function

Step3: Apply the inverse transformation function to the levels obtained

in step 1

Procedure Conclusion:

1. Obtain the transformation function T(r) by calculating the histogram equalization of the input image.

( ) ( )0

rs T r p w dwr

2. Obtain the transformation function G(z) by calculating histogram equalization of the desired density function.

( ) ( )0

zG z p t dt sz

3. Obtain the inversed transformation function G-1

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

4. Obtain the output image by applying the processed gray-level from the inversed transformation function to all the pixels in the input image.

Histogram specification is a trial-and-error process There are no rules for specifying histograms, and one must resort to

analysis on a case-by-case basis for any given enhancement task.