7060 Image Sensors & Processing / 4061 Image Processing Assignment 2 August 2018 (Total Marks: 12%) Overview This assignment is intended to give you some hands-on experience with manipulating image...

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7060 Image Sensors & Processing / 4061 Image Processing Assignment 2 August 2018 (Total Marks: 12%) Overview This assignment is intended to give you some hands-on experience with manipulating image data in MATLAB and the basic image manipulation concepts covered in the course during weeks 4-6. In this assignment work you will be required to modify several functions which implement spatial transformations and Fourier filtering. IMPORTANT: No routines from the image processing toolbox (such as histeq, medfilt2, imfilter) etc may be used in your part of the assignment solutions unless specifically advised. Using such routines will result in zero marks for that component of the assignment work. Assignment Submission Assignments are to be submitted as a standard ZIP archive file (please do NOT use other formats such as 7z, RAR, PKZIP) containing the key MATLAB functions and any test code you have written and a Word (MS Office) or PDF document summarising your results with comments on the performance of the processing steps employed (in 1A-1D) and written answers to questions (1E). • All submitted materials MUST be your own individual original work. • Marks will be deducted for late submission or incorrect submission format (please check myUni for the due date) As part of the submission, you are expected to write up the results of experimenting with your code for different settings and images. Do not simply submit your code with screen shots of the basic test script output. Source Materials Source code, test functions and example imagery for this assignment is located on the MyUni website (https://www.myuni.adelaide.edu.au). You are welcome to add your own example imagery for testing and reporting purposes. DO YOU NEED HELP? - The coding required to complete this assignment is not intended to be complicated but it does assume familiarity with MATLAB programming. If you are having problems getting your code working, then please ask questions during the tutorial session or come and see me during my consulting times. https://www.myuni.adelaide.edu.au/ Exercise 2A – ( 5.5% ) – Image Transformations and Resampling You have been assigned to the implementation of some image transformation functions including partly correcting for lens distortion. To do this however, you will firstly need to complete a resampling function and implement bi-linear and bi-cubic interpolation. STEP 1: - Resampling – modify the supplied function iminterp() such that it implements both bi-linear and bi-cubic interpolation. This function creates a resampled image based on the relationship between each output pixel and its corresponding position within the original image data. In other words, this function uses information about the inverse transformation from each output pixel back into the original image data to perform the resampling. The function takes four inputs, the image I, two arrays x and y and a string value (one of ‘nearest’,’bilinear’ or ‘bicubic’). The two arrays define each corresponding output pixel’s x and y position in terms of the original image data. For example if x(2,2)=1.6 and y(2,2)=3.1 then pixel (2,2) in the output image corresponds to location (1.6,3.1) in the input image. Currently the function only supports nearest neighbour interpolation which would mean that pixel (2,2) in the above example would be given the intensity of the original image data at location (2,3). Your task here is to complete this function and implement bi-linear and bi-cubic interpolation. 1.1: Bi-Linear Interpolation As described in lectures bi-linear interpolation is based on the 2x2 neighbourhood around the sample point ),( yx PP of interest. The bi-linear estimate from this patch is calculated by:              )1,1( ),1()1( )1,()1( ),()1)(1()( yxabI yxIba yxbIa yxIbapI ypbxpa pypx yx yx Here (x,y) is the nearest pixel to ),( yx PP rounded down and (a,b) are the rounding-off errors. In MATLAB use the floor() function to round down ),( yx PP . Remember MATLAB uses (row,column) indexing not (x,y) coordinates. It is strongly suggested you test your bi-linear code before implementing bi-cubic interpolation. 1.2: Bi-Cubic Interpolation In the case of bi-cubic interpolation, a 4x4 neighbourhood around the point of interest is used for the estimation (centred on element (2,2)). To do this we will start with the 1D bi-cubic interpolation expression presented in lectures as follows:                                2 1 0 1 32 1331 1452 0101 0020 1 2 1 )( x x x x tttth where 1x , 0x … 2x are the four samples around the point of interest and t is the fractional offset of interest between sample 0x and 1x . To use this expression for bi-cubic interpolation, we firstly calculate h(t) for each row of the 4x4 region of interest where t is the offset in the x-direction (same as a in the bi-linear expression earlier) and then use these four values to calculate h(t) where t is the offset in the y-direction. This is visually illustrated below where the blue represents the four interpolations in the x-direction and the red the final one in the y-direction. For points in the boundary of the image where there is no 4x4 neighbourhood simply use the nearest value. A test script iminterp_test.m has been supplied so you can check your calculations. An example output from this is shown on the next page. Remember that MATLAB natively handles array multiplication which significantly simplifies the calculation of h(t) above. Test your code on other examples and write up your observations in your short report. Did the bi- cubic interpolation improve the estimation in your examples? Above: Example output from iminterp_test.m which applies 2 x scaling and a small rotation to the input data. Note the differences in estimation between nearest, bilinear and bicubic interpolation. Input 20 40 60 80 20 40 60 Output - nearest 50 100 150 20 40 60 80 100 120 Output - bilinear 50 100 150 20 40 60 80 100 120 Output - bicubic 50 100 150 20 40 60 80 100 120 STEP 2: Simple Lens Distortion Correction – modify the function lensdistortion() such that it implements the simple form of lens distortion correction described below. In class the following was described as a simple approximation to lens distortion: 6 3 4 2 2 11 rkrkrkrd  where r is the radius (distance) of the pixel from the image centre in the undistorted image, dr is the radius of the pixel in the distorted image and 1k etc are constants determining the level of distortion. Unfortunately, inverting this expression to estimate r based on dr is non-trivial, however a simple approximate solution for undoing this distortion which estimates r based on dr is as follows: 6 3 4 2 2 11 ddd rkrkrkr  where 1k etc are again constants (albeit slightly different ones). Given a set of correction factors k1, k2 and k3, to implement this solution, you will need to: 1. Determine the row and col offset for each pixel in the output relative to the image centre and then estimate the distorted radius dr for each pixel from the row and col offset. Assume the image centre ),( cc colrow is the midpoint (ie. centre) of the image (rounded down). 2. Estimate r for each pixel based on the expression shown earlier and multiply each row and col offset by r to get the row and col offsets into the distorted image. 3. Offset these by the image centre ),( cc colrow to get them into normal image coordinates 4. Use these values and call iminterp() to construct the undistorted image (if you have step 1 working you may use iminterp with ‘bilinear’ or ‘bicubic’ otherwise simply use ‘nearest’) To keep things simple the output image will be the same size as the input image. If you choose to use (x,y) coordinates rather than (row,col) be careful not to mix them up when indexing arrays. Use the test script lens_test.m to check your solution (see example output on the next page). Experiment and write up your results. An example output generated using this code is shown on the next page. Note that the result will still contain small imperfections. Above: An example output from lens_test.m showing the results before and after distortion correction. Input 1 (pincushion) 100 200 300 400 100 200 300 Input 1 (fixed) 100 200 300 400 100 200 300 Input 2 (barrel) 100 200 300 400 100 200 300 Input 2 (fixed) 100 200 300 400 100 200 300 Exercise 2B – (4.5%) – De-Noising using a Butterworth and a Wiener Filter In another part of the company you work for, they are experiencing problems with excessive Gaussian noise in imagery coming out of a video capture system that is causing delays in their development work. You have been asked to help them out by implementing two de-noising functions for them that will be used to clean up the imagery. These involve a simple Butterworth filter and a Wiener de-noising filter both of which need to be implemented in the frequency domain. Above: An example noise-free and noisy image (see noisy_image.m). STEP 1: Butterworth filter – Firstly modify the function fft_filter() such that given and image I and a set of FFT filter coefficients B (where the coefficients are shifted such that the array centre represents zero frequency) the function returns the FFT filtered version of image I. That is: I2 = fft_filter(I,B); Use the functions fft2(), ifft2(), real(), fftshift(), ifftshift() and meshgrid() to achieve this. Once completed, modify the function butterworth() such that this function returns a set of FFT filter coefficients representing the Butterworth filter discussed in class. The function butterworth() is called as follows: B = butterworth(width, height, order, cutoff, x0,y0) Where width and height refer to the size of the filter, order and cutoff are the filter order and frequency cut-off. The values x0 and y0 are offsets of the Butterworth filter relative to zero frequency. If not supplied x0 and y0 both default to zero. The expression for the Butterworth filter weights (in the Fourier domain) is as follows: ???????(?,
Answered Same DaySep 05, 2020

Answer To: 7060 Image Sensors & Processing / 4061 Image Processing Assignment 2 August 2018 (Total Marks: 12%)...

Abr Writing answered on Sep 11 2020
150 Votes
A2 code/BBC_test_card_F.tif
A2 code/BBC_test_card_F_barrel.tif
A2 code/BBC_test_card_F_pincushion.tif
A2 code/butterworth.m
% butterworth_2d - a simple function for generating the weights array for a
% 2d butterwoth filter
%
% width,height - size of filter to construct
% n - filter order 1,2 etc
% cut - cutoff value
% x0,y0 - shift offset of filter away from centre (def. (0,0))
%
% B - resulting filter weights (assumes coefficients are fftshifted)
function B = butterworth(width,height,n,cut,x0,y0)
if (nargin<6); y0=0; end
if (nargin<5); x0=0;
end
if (nargin<4); error('butterworth function needs at least 4 arguments'); end
% -- insert your code below --
% HINT: meshgrid() might be useful here. Note that after FFT shift
% the zero frequency term (DC) is located at the image centre plus one
% row/column (eg. pixel (201,201) if the image was 400x400)
%B = rand(height,width); % <-- delete this dummy line
[x,y]=meshgrid(linspace(-width/2,width/2,width),linspace(-height/2,height/2,height));
B=1./(1+((sqrt((x-x0).^2+(y-y0).^2))./cut).^(2*n));
% -- insert your code above --
return
A2 code/fft_filter.m
function [Ifilt,B] = fft_filter(img,B)
if (size(img,1)~=size(B,1))||(size(img,2)~=size(B,2))
error('image and fft filter are of different sizes');
end
% -- (INSERT YOUR OWN CODE BELOW (DELETE THE DUMMY CODE) --
%Ifilt = rand(size(img,1),size(img,2)); B=Ifilt; % DELETE ME
F = (fft2((img)));
Ifilt=ifft2((F));
%F = fft2(img);
% -- INSERT YOUR OWN CODE ABOVE --
return
end
% END OF FILE
A2 code/fft_test.m
k=0.01;
I=double(imread('pout.tif'))/255;
Inoisy = imnoise(I,'gaussian',0,k);
% butterworth test
B = butterworth(size(Inoisy,2),size(Inoisy,1),1,32);
Ib = fft_filter( Inoisy , B );
% wiener test
[Iw,Fw] = wiener_denoise_filter(I,k,Inoisy);
figure
colormap(gray);
subplot(2,4,1);
imagesc(I); title('I original');
subplot(2,4,5);
imagesc(Inoisy); title('I noisy');
subplot(2,4,2);
imagesc( log(abs(fftshift(fft2(I)))) ); title('FFT original');
subplot(2,4,6);
imagesc( log(abs(fftshift(fft2(Inoisy)))) ); title('FFT noisy');
subplot(2,4,3);
imagesc( log(abs(B)) ); title('Butterworth filter (n=1,fcut=32)');
subplot(2,4,4);
imagesc(Ib); title('Butterworth filtered (n=1,fcut=32)');
subplot(2,4,7);
imagesc( log(abs(fftshift(Fw))) ); title('FFT Wiener filter');
subplot(2,4,8);
imagesc(Iw); title('Wiener noise filtered k=0.01');
drawnow;
A2 code/iminterp.m
function Ifixed = iminterp(I,xd,yd,interpmethod)
% Inputs:
% I - original image (either NxM or NxMx3
% xd,yd - arrays defining mapping of new values BACK to input image
% coordinates s.t. Ifixed(i,j) <-- I(yd(i,j),xd(i,j)). Note xd and yd
% may be a different size to I.
% interpmethod - interpolation method ('nearest','bilinear','bicubic')
% Outputs:
% Ifixed - reconstructed image based on xd,yd mappings and interpolation
% the output is NxM or NxMx3 depending on the input image.
%
% NOTES: The value of each pixel is based on examining xd and yd to
% determine which location in the original image data relates to this
% pixel. As the reference point in the original image is likely to be
% a non-integer value, the stated interpolation method is used to
% estimate the value bsed on the local neighbouring values.
%
% convert uint8 imagery to doubles in range 0..1
if (isa(I,'uint8'))
I=double(I)/255;
end
if (nargin<4)
interpmethod='bilinear';
end
% size of output image
outrows=size(xd,1);
outcols=size(yd,2);
% size of input image I
inrows=size(I,1);
incols=size(I,2);
layers=size(I,3);
%
Ifixed=zeros(outrows,outcols,layers);
for lay=1:layers % for each colour layer
for row=1:outrows
for col=1:outcols
switch( interpmethod)
case 'nearest'
% use nearest pixel in I to required location
xd_rc=round(xd(row,col));
yd_rc=round(yd(row,col));
% if inside image then...
if (xd_rc>0)&&(xd_rc<=incols)&&(yd_rc>0)&&(yd_rc<=inrows)
Ifixed(row,col,lay)=I(yd_rc,xd_rc,lay);
end

case 'bilinear'
% bi-linear interpolation
% Note: Use floor() not round()
% ---- INSERT YOUR 'BILINEAR' SOLUTION BELOW ----
% use nearest pixel in I to required location
xd_rc_b=floor(xd(row,col));
yd_rc_b=floor(yd(row,col));
a=xd(row,col)-xd_rc_b;
b=yd(row,col)-yd_rc_b;
% if inside image then...
if (xd_rc_b>0)&&(xd_rc_b<=incols-1)&&(yd_rc_b>0)&&(yd_rc_b<=inrows-1)
Ifixed(row,col,lay)=(1-a)*(1-b)*I(yd_rc_b,xd_rc_b,lay)+(1-a)*(b)*I(yd_rc_b+1,xd_rc_b,lay)+(a)*(1-b)*I(yd_rc_b,xd_rc_b+1,lay)+(a)*(b)*I(yd_rc_b+1,xd_rc_b+1,lay);

end
%Ifixed(row,col,lay)= rand(1,1); % DUMMY CODE - REMOVE ME

% ---- INSERT YOUR 'BILINEAR' SOLUTION ABOVE ----

case 'bicubic'
% bicubic (see assignment sheet for details)
% Use NEAREST for locations where there is no 4x4 neighbourhood
% Fill in and Use the function bicubic() to simplify your code
% You may need to ensure the bicubic estimate lies within 0..1
% ---- INSERT YOUR 'BICUBIC' SOLUTION BELOW ----
xd_rc_b=floor(xd(row,col));
yd_rc_b=floor(yd(row,col));
ta=xd(row,col)-xd_rc_b;
tb=yd(row,col)-yd_rc_b;
% if inside image then...
if (xd_rc_b>1)&&(xd_rc_b<=incols-2)&&(yd_rc_b>1)&&(yd_rc_b<=inrows-2)
A=I(yd_rc_b-1:yd_rc_b+2,xd_rc_b-1:xd_rc_b+2,lay);
Ifixed(row,col,lay)=bicubic(ta,tb,A);
else
if (xd_rc_b>0)&&(xd_rc_b<=incols)&&(yd_rc_b>0)&&(yd_rc_b<=inrows)
Ifixed(row,col,lay)=I(yd_rc_b,xd_rc_b,lay);
end
end
%Ifixed(row,col,lay)= rand(1,1); % DUMMY CODE - REMOVE ME

% ---- INSERT...
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