Fundamentals of Digital Image and Video Processing - Week 7 Solutions

Hi Coursera people,

After hours of struggle, I had found the solution to question number 7 of week 7! There is absolutely no material available online regarding calculation of frequency response of the CLS filter! Finally, the answer sprung up from the discussions forum of the course itself :) I must thank my fellow Courserans for helping me out with the code. Here goes the question number 7 of week 7.

In this problem, you will implement the Constrained Least Squares (CLS) filter and examine its performance when the regularization parameter is set at different values. You will be provided with the original image and a set of MATLAB files. Follow the instructions below to finish this problem. (1) Download the original image and the MATLAB code from here. Place the original image and all the provided MATLAB files in the same directory. (2) The file "wrapper.m" is the entry or the "main" code. It loads the original image, applies a motion blur to it, and degrades the image by adding noise. The 17th line in "wrapper.m" sets the value of the regularization parameter "alpha". (3) The MATLAB file "cls_restoration.m" has an incomplete implementation of the CLS filter. You need to un-comment line 24 in "cls_restoration.m" and complete the implementation of the CLS filter. (4) After you complete the implementation of the CLS filter, you should run "wrapper.m" with different values of alpha. Specifically, we ask you to try the following values of alpha: {0.0001, 0.001, 0.01, 0.1, 1, 10, 100}. For each value of alpha, we ask you to compute the Improvement in SNR (ISNR). Note that the computation of ISNR involves there images: the original image, the blurred and noisy image, and the restored image. After you obtain the ISNR values, enter in the box below the largest ISNR value. Enter the number with at least two decimal points.

Here is my attempt at the code :

wrapper.m

clear all
close all

%% Simulate 1-D blur and noise
image_original = im2double(imread('C:\Image and Video processing\Week 7\downloaded codes\Cameraman256.bmp', 'bmp'));
[H, W] = size(image_original);
blur_impulse = fspecial('motion', 7, 0);
image_blurred = imfilter(image_original, blur_impulse, 'conv', 'circular');
noise_power = 1e-4;
randn('seed', 1);
noise = sqrt(noise_power) * randn(H, W);
image_noisy = image_blurred + noise;

figure; imshow(image_original, 'border', 'tight');
figure; imshow(image_blurred, 'border', 'tight');
figure; imshow(image_noisy, 'border', 'tight');

%% CLS restoration
alpha = 0.0001;  % you should try different values of alpha
image_cls_restored = cls_restoration(image_noisy, blur_impulse, alpha);
figure; imshow(image_cls_restored, 'border', 'tight');

%% computation of ISNR

e1=image_original-image_noisy;
e2=image_original-image_cls_restored;
E1=mean2(e1.*e1);
E2=mean2(e2.*e2);
result=10*log(E1/E2)/log(10)


cls_restoration.m

function image_restored = cls_restoration(image_noisy, psf, alpha)

%% find proper dimension for frequency-domain processing
[image_height, image_width] = size(image_noisy);
[psf_height, psf_width] = size(psf);
dim = max([image_width, image_height, psf_width, psf_height]);
dim = next2pow(dim);

%% frequency-domain representation of degradation
psf = padarray(psf, [dim - psf_height, dim - psf_width], 'post');
psf = circshift(psf, [-(psf_height - 1) / 2, -(psf_width - 1) / 2]);
H = fft2(psf, dim, dim);

%% frequency-domain representation of Laplace operator
Laplace = [0, -0.25, 0; -0.25, 1, -0.25; 0, -0.25, 0];
Laplace = padarray(Laplace, [dim - 3, dim - 3], 'post');
Laplace = circshift(Laplace, [-1, -1]);
C = fft2(Laplace, dim, dim);

%% Frequency response of the CLS filter
% Refer to the lecture for frequency response of CLS filter
% Complete the implementation of the CLS filter by uncommenting the
% following line and adding appropriate content

R = conj(H)./(abs((H.*H))+(alpha*abs((C.*C))));

%% CLS filtering
Y = fft2(image_noisy, dim, dim);
image_restored_frequency = R .* Y;
image_restored = ifft2(image_restored_frequency);
image_restored = image_restored(1 : image_height, 1 : image_width);


next2pow.m

function result = next2pow(input)
if input <= 0
    fprintf('Error: input must be positive!\n');
    result = -1;
else
    index = 0;
    while 2 ^ index < input
        index = index + 1;
    end
    result = 2 ^ index;
end


And don't get freaked if you get negative ISNR values as the result; it is absolutely normal. Here are the observations for the various values of alpha.
0.0001    -5.9191
0.001      -1.5292
0.01        3.4933
0.1          4.3048
1             2.1471
10           0.4866

-Cheers,
Vijay.

Fundamentals of Digital Image and Video Processing - Week 6 Solutions

Hi Coursera people,

This week's lectures were very exasperating with regards to their length! Yes, I understand that there is a time limit of 12 weeks to complete the material; but it is equally important for followers to understand what is happening right? Anyways, with what I could manage to understand, I've cooked up this week's solution. Its very surprising to learn that a few lines of code will solve the purpose...

Here is question number 6 of week 6.

This problems pertains to inverse filtering. You should review the corresponding slides in the video lectures to refresh your memory before attempting this problem. To help you understand how inverse filter is implemented and applied, we have provided you with a MATLAB script here. Download the script and the original image, and open the script using MATLAB. Once you open the script, you will see on Line 8 the statement "T = 1e-1". This defines the threshold value used in the inverse filter. The script simulates the blur due to motion and applies inverse filtering for its removal. We encourage you to try different values of the threshold and see how it affects the performance of the inverse filter. We ask you to enter the ISNR value below when the threshold is set to 0.5. Make sure you enter the number with at least 2 decimal points.

Here's the code that was given as a part of the question. I've added the code that gives the solution at the end.

% inverse filter with thresholding

clear all
close all
clc

% specify the threshold T
T = 0.5;

%% read in the original, sharp and noise-free image
original = im2double(rgb2gray((imread('C:\Image and Video processing\Week 6\original_cameraman.jpg'))));
[H, W] = size(original);

%% generate the blurred and noise-corrupted image for experiment
motion_kernel = ones(1, 9) / 9;  % 1-D motion blur
motion_freq = fft2(motion_kernel, 1024, 1024);  % frequency response of motion blur
original_freq = fft2(original, 1024, 1024);
blurred_freq = original_freq .* motion_freq;  % spectrum of blurred image
blurred = ifft2(blurred_freq);
blurred = blurred(1 : H, 1 : W);
blurred(blurred < 0) = 0;
blurred(blurred > 1) = 1;
noisy = imnoise(blurred, 'gaussian', 0, 1e-4);


%% Restoration from blurred and noise-corrupted image
% generate restoration filter in the frequency domain
inverse_freq = zeros(size(motion_freq));
inverse_freq(abs(motion_freq) < T) = 0;
inverse_freq(abs(motion_freq) >= T) = 1 ./ motion_freq(abs(motion_freq) >= T);
% spectrum of blurred and noisy-corrupted image (the input to restoration)
noisy_freq = fft2(noisy, 1024, 1024);
% restoration
restored_freq = noisy_freq .* inverse_freq;
restored = ifft2(restored_freq);
restored = restored(1 : H, 1 : W);
restored(restored < 0) = 0;
restored(restored > 1) = 1;

%% analysis of result
noisy_psnr = 10 * log10(1 / (norm(original - noisy, 'fro') ^ 2 / H / W));
restored_psnr = 10 * log10(1 / (norm(original - restored, 'fro') ^ 2 / H / W));


%% visualization
figure; imshow(original, 'border', 'tight');
figure; imshow(blurred, 'border', 'tight');
figure; imshow(noisy, 'border', 'tight');
figure; imshow(restored, 'border', 'tight');
figure; plot(abs(fftshift(motion_freq(1, :)))); title('spectrum of motion blur'); xlim([0 1024]);
figure; plot(abs(fftshift(inverse_freq(1, :)))); title('spectrum of inverse filter'); xlim([0 1024]);

%% Calculation of ISNR

e1=original-noisy;
e2=original-restored;
E1=mean2(e1.*e1);
E2=mean2(e2.*e2);
result=10*log(E1/E2)/log(10)




P.S. : The answer is varying with regards to the second decimal as this code is run several times on the same machine. Don't ask me why! I'm as clueless as you are :p
Just type "2.85" in the answer area, and you get a full 3 points!

-Cheers,
Vijay.