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Image Processing

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Introduction to Image Processing and Compression Using
Matlab/Octave
Table of content
Image Processing and Compression Using Matlab ........................................................................... 1
1.0 Basic image operation.............................................................................................................. 1
1.1 Image display ....................................................................................................................... 1
1.2: Color Image Display........................................................................................................... 2
1.3 Image Array Indexing .......................................................................................................... 3
1.4 Converting image array to image file .................................................................................. 8
1.5 Image Manipulation ............................................................................................................. 8
1.5.1 Resize image ..................................................................................................................... 9
1.6 Image Analysis Using Histogram ...................................................................................... 10
1.6.1 Generating feature vector for image using greyscale histogram .................................... 14
1.6.2 Comparing Image Descriptor ......................................................................................... 17
1.6.2 Colour Histogram as Image Descriptor .......................................................................... 18
1.6.2 Two dimensional Colour Histogram as Image Descriptor ............................................. 19
1.7 Image Database Using Cell Array ..................................................................................... 22
1.8 Edge Detection................................................................................................................... 24
2. Image Compression ................................................................................................................. 26
2.1 Forward 2D- DCT transform ............................................................................................. 26
2.2 Inverse 2D- DCT transform ............................................................................................... 30
2.3 DCT Transform Applied to 8x8 Image Sub-blocks .......................................................... 31

1.0 Basic image operation
1.1 Image display
Code
cd('C:\MATLABR11\work\lab1') a= imread('cameraman.tif');; % This is a grayscale image whos % view the image resolution and size

Output
Name
Size a 256x256

Bytes Class
65536 uint8

Note
Each pixel a(i,j) value represent the grayscale intensity value a(i,j) denote the pixel value at row i and column j a(1,1) is the pixel at the top left corner a(256,256) is the pixel at bottom right corner

2 figure(1), imshow(a); axis on

Figure 1 minval= min(min(a)) maxval= max(max(a))

% find the min value of the pixel in array a
% find max value of the pixel in array a

% convert matrix a from uint8 type to double so that b(i,j) is now in the range
[0,1]
b = im2double(a)
% convert b so that it has the same value as image a but in double format b = b * 255;

Result minval = maxval =

7
253

Code: Find image dimension
[nrow ncol] = size(b); % find the number of rows and columns of image array b

1.2: Color Image Display
% 1.2 color image display addpath('C:\MATLABR11\work\lab1\image'); imColor = imread('children.tif'); % read the color image figure(3),imshow(imColor) % display in figure 3

3

1.3 Image Array Indexing

Im = imread('cameraman.tif'); % This is a grascale image figure(1), imshow(Im); axis on pixval on; % move the pixel over the image and view the pixel value

The grayscale image stored in the file ‘kids.tif’ is represented by 2 dimensional array or matrix variable Im . Each element of the array Im(i,j) at location row i and column j
In Matlab the array index start with 1. The variable array f here store the grayscale value of an image having resolution MxN

4

Im(31,22) = 164
50

Grey level intensity at pixel
(31,22) is 164

100

For 8 bits pixel, gray intensity level range from 0 to 255

150

200

250
50

100

150

200

250

Colour Image
RGB colour image is stored in 3 dimensional array variable imC.
%color image display imC = imread('hand.jpg'); % read the color image imshow(imC) % display image pixel_val = imC(35,80,:); disp (pixel_val); ans(:,:,1) = 3 ans(:,:,2) = 121 ans(:,:,3) = 191

% red value
% green value
% blue value

5

From the image plot, the color of the pixel at location row=35 and column=80 imC(35,80) is represented by the 3 tupple number (14,126,192)
Red R =14 , Green G=126 and Blue = 192 imA = imread('redbox.bmp') ;% display in figure 3 imshow(imA) blue

green

Pixel value of the red color plane

» imA(:,:,1) ans =

Red colour plane imA(:,:,1) 10x10 array

255 255 255 255 255 255 255 255 255 255
255 255 255 255 255 255 255 255 255 255
255 255 255 255 255 255 255 255 255 255

6
255
255
255
255
255
255
255

255
255
255
255
255
255
255

255
255
255
255
255
255
255

255
255
255
255
255
255
255

255
255
255
255
255
255
255

255
255
255
255
255
255
255

255
255
255
255
255
255
255

255
255
255
255
255
255
255

255
255
255
255
255
255
255

255
255
255
255
255
255
255

» imA(:,:,2) ans =
0
0
0
0
0
0
0
0
0
0

0
0
0
0
0
0
0
0
0
0

0
0
0
0
0
0
0
0
0
0

0
0
0
0
0
0
0
0
0
0

0
0
0
0
0
0
0
0
0
0

0
0
0
0
0
0
0
0
0
0

0
0
0
0
0
0
0
0
0
0

0
0
0
0
0
0
0
0
0
0

0
0
0
0
0
0
0
0
0
0

0
0
0
0
0
0
0
0
0
0

0
0
0
0
0
0
0
0
0
0

0
0
0
0
0
0
0
0
0
0

0
0
0
0
0
0
0
0
0
0

0
0
0
0
0
0
0
0
0
0

0
0
0
0
0
0
0
0
0
0

0
0
0
0
0
0
0
0
0
0

0
0
0
0
0
0
0
0
0
0

0
0
0
0
0
0
0
0
0
0

» imA(:,:,3) ans =
0
0
0
0
0
0
0
0
0
0

0
0
0
0
0
0
0
0
0
0

Pixel A(1,1) has the value (255,0,0)
This implys that imA(1,1,1) = 255
% red intensity = 255 imA(1,1,2) = 0
% green intensity = 0 imA(1,1,3) = 0
% blue intensity = 0

and

7 imC = imread('hand.jpg'); imC_blue = imC(:,:,2); imshow(imC_blue); % read the color image

Image of the blue intensity of the color image

Color image can be converted to grayscale image imG = rgb2gray(imC) imshow(imG) ; % convert color image to grayscale

Exercise Q1
Display the red, green and blue component of the image children.tif. You should display 3 images one for each colour component.

8

1.4 Converting image array to image file
Use imwrite() function
% Convert a 3D array to a bmp image
% Create a red color square
A1(:,:,1) = ones(10,10) * 255 ; % set the red component to max value
A1(:,:,2) = ones(10,10) * 0 ;
A1(:,:,3) = ones(10,10) * 0 ;
% imwrite(A,filename,fmt) writes the image A to the file specified by filename in the format specified by fmt. imwrite(A1,'redbox.bmp','bmp'); 1.5 Image Manipulation
After storing the image in an array, the pixel value can be manipulated or changed.
The following example shows how a binary black and white image is transformed. The black pixel is converted to white and white pixel converted to block. In other word, the black (pixel value = 0) and white (pixel value = 1 ) pixels are inverted.
The invert operator ~ can be used. A=1 , ~A = 0
% invert the pixel value of the binary image imBW = imread('bwman.tif'); figure(1),imshow(imBW); imBW2 = ~imBW ; figure(2), imshow(imBW2);
% the for loop version
[nrow ncol] = size(imBW); for i=1:nrow for j=1:ncol imBW3(i,j) = ~imBW(i,j); % invert the pixel value end end figure(3), imshow(imBW3);

Output is a binary image that has been inverted
Figure 1

9

1.5.1 Resize image
When an image is resized into smaller image, the downsampling operation is used. When it is resized to a larger image, the interpolation operation is applied. imC = imread('hand.jpg'); imD = rgb2gray(imC);
[nrow ncol] = size(imD) disp(' Reduce image by half'); nrow2 = round(nrow/2); ncol2= round(ncol/2); imD2 = imresize(imD, [nrow2 , ncol2]); figure(1), imshow(imD); figure(2), imshow(imD2);

Exercise Q2
Down sample the image cameraman.tif four times to get a smaller image of dimension. Then resize the smaller image by upsampling (interpolation) to get back original resolution of 256x256.
Comment on the visual quality of image by comparing the original image and the image after undergoing down-sampling/interpolation. Write the code to compare the difference between the two image using mean square error E(a^, a) as given by the formula below. The original image is represented by the matrix a and the approximated image after downsampling/interpolation is a^ .

10

1.6 Image Analysis Using Histogram
The distribution of the pixel value in an image can be represented by the histogram. For the case of greyscale image, the histogram shows the distribution of the gray intensity of all the pixels in the image. From the histogram we can see that the majority of the gray value is from 90-150
I = imread('pout.tif'); imshow(I) figure, imhist(I)

The histogram consists of a set of bins on the x axis and the count of value that falls within the bin on the y axis.
1600
1400
1200
1000
800
600
400
200
0
0

Greyscale image

50

100

150

200

250

histogram of the image

Definition of Histogram
Given N data samples f1 , f2 , …fi….fN where each sample is a real scalar value
The data samples are grouped into m interval/bins . The m intervals are defined by the points b1 , b2 , ……bm+1 . The dynamic range of the data is usually uniformly spaced where the width of each bin is the same bw = bi+1 - bi
Each bin is defined by the bin index number, [lower, upper limit]
The following figure shows the division of the intervals for 5 bins.

b1

b2
Bin 1

b5

b6
Bin 5

The histogram H = [ h1 h2 …hj….hm ] where hj records the number of samples that falls within the interval [bj , bj+1 ]

11
The histogram H can be transformed to probability distribution by dividing hi with the total samples N.
The algorithm for the histogram function is very simple
Set all the bin count to 0, H = [ 0,0 , …..0] for i = 1 : N
% check if the sample falls into any particular bin if bj < f(i) > database(1).imageName
>> database(1).imageName ans = C:\MATLABR11\work\cbir\images2\110.jpg
>> im=database(1).imageName im = C:\MATLABR11\work\cbir\images2\110.jpg
>> imshow(im)
>> surf(database(1).Hist) % plot the 2D histogram
>>

0.05
0.04
0.03
0.02
0.01
0
15
15

10

10

5

5
0

Image 110.jpg

0

2D histogram (Saturation and Value Channel)

24
The following mat file contains the database that has been prepared using the code demo_createDatabase_all.m . To load the database into matlab workspace
>> cd('C:\MATLABR11\work\cbir')
>> load database_cbir.mat
View the first entry in the database
>> database(1).imageName imageName: 'C:\MATLABR11\work\cbir\images2\110.jpg' database(1).Hist: database(1).label

1.8 Edge Detection close all cd('C:\MATLABR11\work\lab1') I = imread('coins.png');
BW = edge(I, 'sobel'); imshow(I); figure, imshow(BW);

Edge detection result using sobel edge operator
Simple example image with sharp edges and no background clutter

coins.png

edge map (white colour denote edge pixel)

Image with some blurred edges

25
A suitable code can be used to detect the eyes and mouth. face_bush.jpg edge map
Image with large background clutter and non uniform texture

child1.jpg

edge map (white color denote edge pixel)

26

2. Image Compression
In order to exploit the spatial redundancies of the pixel value, the image is transformed to the frequency domain. In JPEG image compression, two dimensional DCT transform is used. In the experiment you will discover that some DCT coefficients are more important than the other because these coefficients have effect on the reconstructed image.
DCT coefficients matrix 10x10

Image I
10x10

Forward
2D- DCT transform J(1,1) J(1,2) ….J(10,10)
J(2,1) J(2,2) …..J(2,10)
.
.
.
J(10,1) ……….. J(10,10)

Inverse
2D- DCT transform Reconstructed image I’

J(1,1) represent the coefficient for the DC value (lowest frequency) of the image (average pixel value). If the DCT coefficient is J(m,n) then the larger value for m represents the larger vertical frequency component whereas the larger value for n represents the larger horizontal frequency component. The main aim of image compression is to reduce the image size as much as possible but at the same time tries to preserve the image quality of the reconstructed image. The reconstructed image
I’ is considered good if it appear similar to the original image I. There maybe some slight distortion to the pixel value in I’ and thus I’ is not exactly similar to a I. However since the distortion is not detected by the human eye, this is acceptable.

2.1 Forward 2D- DCT transform
The forward 2D-DCT transform represent the image in frequency domain and this information is stored in a matrix of DCT coefficients. In JPEG, the image is commonly divided into an 8x8 sub image blocks. The forward 2D-DCT transform is applied to the 8x8 image blocks of the image.
The following code creates a set of simple images that varies in spatial frequency. Observe the
DCT coefficients for the image with different vertical and horizontal frequency. A peak value at the bottom right corner of the DCT coefficients matrix indicate that the image is filled with fast changing pixel value on both the horizontal and vertical direction.
% The following code create the image with different spatial frequency
A1
A2
A3
A4
A5

=
=
=
=
=

ones(10,10); ones(5,10); ones(10,5); ones(2,10); ones(10,2);

M1
M2
M3
M4

=
=
=
=

A1 ; % dc value only
[A2 ; A2*0 ] % low vertical freq image
[A3 A3*0 ] % low horizontal freq image
[A4 ; A4*0 ; A4 ; A4*0 ; A4] ; % high vertical freq image

M5 =

[A5

A5*0

A5

A5*0

A5] ;% high vertical freq image

M6 = M4 & M5 ; % image with both vertical and horizontal freq

27 imwrite(M1,'M1.jpg','jpg'); imwrite(M2,'M2.jpg','jpg'); imwrite(M3,'M3.jpg','jpg'); imwrite(M4,'M4.jpg','jpg'); imwrite(M5,'M5.jpg','jpg'); imwrite(M6,'M6.jpg','jpg');
Im1
Im2
Im3
Im4
Im5
Im6

=
=
=
=
=
=

imread('M1.jpg'); imread('M2.jpg'); imread('M3.jpg'); imread('M4.jpg'); imread('M5.jpg'); imread('M6.jpg'); subplot(6,1,1), subplot(6,1,2), subplot(6,1,3), subplot(6,1,4), subplot(6,1,5), subplot(6,1,6), imshow imshow imshow imshow imshow imshow (Im1);
(Im2);
(Im3);
(Im4);
(Im5);
(Im6);

% Apply DCT Transform onto the Image and Observe the DCT coefficients
J1
J2
J3
J4
J5
J6

=
=
=
=
=
=

dct2(Im1); dct2(Im2); dct2(Im3); dct2(Im4); dct2(Im5); dct2(Im6); figure(2); subplot(6,1,1),imshow(abs(J1),[]), subplot(6,1,2),imshow(abs(J2),[]), subplot(6,1,3),imshow(abs(J3),[]), subplot(6,1,4),imshow(abs(J4),[]), subplot(6,1,5),imshow(abs(J5),[]), subplot(6,1,6),imshow(abs(J6),[]),

colormap(jet), colormap(jet), colormap(jet), colormap(jet), colormap(jet), colormap(jet), colorbar colorbar colorbar colorbar colorbar colorbar 28

Result
Image with different vertical and horizontal frequency
10x10 Images

10x10 DCT Coefficients Matrix
2000
1000
0

1000
500

Image with low vertical frequency 0

1000
500

Image with low horizontal frequency 0

1000

Image with higher vertical frequency 0

1000

0

Image with both high vertical and horizontal spatial frequency

1200
1000
800
600
400
200

29
Observe the distribution of the DCT coefficients stored in the variable J1, J2 and J6
J1 =

J2 =

1.0e+003 *

1.0e+003 *

Columns 1 through 7

Columns 1 through 7

2.5500
-0.0000
-0.0000
-0.0000
-0.0000
-0.0000
-0.0000
-0.0000
-0.0000
-0.0000

-0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000

-0.0000
0.0000
0.0000
0.0000
-0.0000
0.0000
0.0000
0.0000
-0.0000
-0.0000

-0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000

0.0000
-0.0000
-0.0000
-0.0000
-0.0000
0.0000
0.0000
0.0000
-0.0000
0.0000

-0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000

0.0000
0.0000
0.0000
-0.0000
0.0000
0.0000
0.0000
-0.0000
-0.0000
0.0000

1.2750
1.1473
-0.0044
-0.3941
-0.0032
0.2490
0.0010
-0.2045
-0.0032
0.1872

-0.0000
-0.0000
0.0000
0.0000
0.0000
-0.0000
-0.0000
0.0000
0.0000
-0.0000

-0.0000
-0.0000
0.0000
0.0000
-0.0000
-0.0000
-0.0000
0.0000
0.0000
0.0000

Columns 8 through 10

Columns 8 through 10

-0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000

-0.0000
-0.0000
0.0000
0.0000
0.0000
-0.0000
-0.0000
0.0000
0.0000
-0.0000

-0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000

-0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000

J6 =
1.0e+003 *
Columns 1 through 7
1.3238
-0.0002
0.0847
-0.0003
0.1882
-0.0006
-0.1373
-0.0007
-0.0284
0.0004

-0.0001
-0.0011
-0.0007
0.0003
-0.0013
-0.0002
0.0018
0.0015
-0.0020
-0.0054

0.0854
-0.0010
0.1405
-0.0005
0.3132
0.0001
-0.2265
0.0014
-0.0477
-0.0041

Columns 8 through 10
0.0001
-0.0026
-0.0020
0.0009
0.0017
0.0003
-0.0040
-0.0035
0.0023
0.0030

-0.0270
0.0010
-0.0446
-0.0014
-0.1015
0.0001
0.0760
0.0020
0.0127
-0.0010

0.0017
-0.0026
0.0017
-0.0007
0.0014
0.0000
-0.0011
0.0004
-0.0004
0.0006

-0.0000
0.0005
-0.0008
-0.0003
-0.0024
-0.0002
-0.0009
-0.0016
0.0027
0.0023

0.1875
-0.0005
0.3126
-0.0024
0.6900
0.0000
-0.5063
-0.0026
-0.1000
-0.0009

0.0002
-0.0006
0.0003
-0.0005
0.0002
-0.0002
-0.0002
0.0002
-0.0003
0.0002

-0.1359
-0.0039
-0.2293
0.0028
-0.5004
-0.0001
0.3615
-0.0035
0.0776
0.0062

-0.0000
-0.0000
-0.0000
-0.0000
0.0000
-0.0000
-0.0000
0.0000
0.0000
-0.0000

-0.0000
-0.0000
0.0000
0.0000
0.0000
-0.0000
-0.0000
0.0000
0.0000
-0.0000

-0.0000
-0.0000
0.0000
0.0000
0.0000
-0.0000
-0.0000
0.0000
0.0000
-0.0000

0.0000
-0.0000
0.0000
0.0000
0.0000
0.0000
-0.0000
0.0000
0.0000
-0.0000

-0.0000
-0.0000
0.0000
0.0000
0.0000
-0.0000
-0.0000
0.0000
0.0000
-0.0000

0.0000
-0.0000
0.0000
0.0000
0.0000
-0.0000
-0.0000
0.0000
-0.0000
-0.0000

30

2.2 Inverse 2D- DCT transform
The following code converts the DCT coefficients to zero if the magnitude is less than 10. The modified DCT coefficients are then used for reconstructing back the image. Compare the original image and the reconstructed image.

% Apply Inverse DCT figure(3); J6(abs(J6)

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