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EE 5351 Digital Video Coding Project #4 1001112760 Sai Surya Chintala

1 a)
Code: - clc; % Clears screen clear all; % Clears all variables from memory close all; % Closes all running figures and plots datafile='C:\Users\Venkata\Desktop\images\sena.img'; % File location of image image=fopen(datafile); % Open image as file senaimage=fread(image); % Read image from location specified senaimage2=imrotate(senaimage,270); % Rotates images by 270' fclose(image); % Closes image hist(senaimage2,1000); % Makes histogram of the rotated image

Result:

1 b)

Code: clc; %Clears screen clear all; % Clears all variables close all; % Closes all exiting graphs and plots datafile='C:\Users\Venkata\Desktop\images\sena.img'; % File location image= fopen(datafile); % Opens image file senaimage = fread(image); % Reads image to image bp= input('Enter no of bits/pixel:'); % Taking input for no of bits/pixel n = 2^bp; m = 256/n;
A(1) = 0 ;
B(1) =A(1)+ (m-1); for i=2:n % for loop A(i)=B(i-1)+1; B(i)=A(i)+ (m-1); end senaimage2 = uint8(senaimage); for i = 1:n % for loop for j=1:65536 if(senaimage2(j) >= A(i) && senaimage2(j) <= B(i)) % Checking condition senaimage3(j) = (A(i)+B(i)+1)/2; end end end senaimage4 = reshape(senaimage,256,256); % Reshaping image senaimage5 = reshape(senaimage3,256,256); % Rehsaping image
MSE = sum(sum((senaimage4-senaimage5).^2))/(256*256) % Calculating Mean Square error imshow(uint8(senaimage5')); % Shows the ouput image file fclose(image); %Closes file

For 1 bit/pixel:
MSE=893.9784

For 2 bit/pixel:
MSE=459.6786

For 3 bit/pixel:
MSE=94.1136

1 c)
Code:
clc; clear all; close all; datafile='C:\Users\Venkata\Desktop\images\sena.img'; sena=fopen(datafile); senaimage=fread(sena); bp=input('Enter No. of Bits/pixel:'); n = 2^bp; sq = 256/n; sq1=16/n; sq2= sq1; sqindex(1) = sq1/2; for i=2:n sqindex(i)= sqindex(i-1)+ sq1; end senaimage2 = uint8(senaimage); senaimage3 = reshape(senaimage,256,256);
[count,x]=imhist(uint8(senaimage3'));
resl=256*256; array = [resl*0.0625;resl*0.125; resl*0.1875; resl*0.25; resl*0.3125; resl*0.375; resl*0.4375; resl*0.50; resl*0.5625; resl*0.625; resl*0.6875; resl*0.75; resl*0.8125 ;resl*0.875 ;resl*0.9375]; m=0; bp=1; for j=1:15 while m<array(j) m=m+count(bp); bp=bp+1; end mid(j)=bp; end L(1) = 0 ;
H(1) = mid(sq1)-1; if(n>2) for i=2:n-1 sq2 = sq2+ sq1; L(i)=H(i-1)+1; H(i)=mid(sq2); end L(n)=H(n-1)+1;
H(n)=255;
else
L(2)=H(1)+1;
H(2)=255; end for i = 1:n for j=1:65536 if(senaimage2(j)>=L(i) && senaimage2(j)<=H(i)) senaimage3(j) = mid(sqindex(i)); end end end senaimage4 = reshape(senaimage,256,256); senaimage5= reshape(senaimage3,256,256);
MSE = sum(sum((senaimage4-senaimage5).^2))/(256*256) imshow(uint8(senaimage5')); fclose(sena);

Results:

For 1 bit/pixel
MSE=442.1536

For 2 bit/pixel MSE=151.8293

For 3 bit/pixel:
MSE=56.8859

Conclusion: 1. On observing the first histogram, we conclude that it is a Laplacian distribution. 2. As the bit/pixel increases the image quality increases, the darkness in picture decreases and the mean square error also decreases. 3. The mean square error in case “1B” is much higher than case “1C”. 4. The quantization in the last case is much better, as non-uniform quantization concentrates more on the higher probabilities of the image than the uniform quantization in case “1B”. The quantization in the last case proves out to be much better.