Data sets from image

I am new to Matlab and I have a jpg image that I have loaded. the image is not shown as data points that are plotted... not sure if this makes sense.. you can display the image with imshow.

what i am wanting to do is take the fourier transform and multiply the image by functions of my choosing however i dont know how to go about doing this as the image dosnt seem to be displayed as "data points" per say??

Answer 1

[Jonas Reber]

does it help if I provide you the following code as an example...?

%%Two Dimensional DFT Example

% 1.Load the sample image HeadCT.jpg, calculate its 2D DFT and plot the
%   Fourier spectrum (increase its contrast if necessary) and phaseangles.

   img=double(imread('HeadCT.jpg'));
   figure, imshow(img,[]);

   IMG=fftshift(fft2(img));
   logspect=log(1+abs(IMG));
   theta=angle(IMG);

   figure
   subplot(1,2,1);
   imshow(theta,[]);title('Phase');
   subplot(1,2,2);
   imshow(logspect,[]);title('Spectrum');

% 2.Calculate the 2D Fourier transformation of the sample image using only
%   the 1D Fourier transform fft().

   IMG2=fftshift( fft(fft(img).').' );
   max(max(abs(IMG-IMG2)))

% 3.Keep the phase of the Fourier transform and replace the magnitude of 
%   each frequency sample by its square root. Take the inverse DFT.

   spect=sqrt(abs(IMG));
   Z=spect.*exp(i*theta);
   z=abs(ifft2(fftshift(Z)));

   figure, imshow(z,[]);

   % what do you see?
   % the operation enhanced the edges. the reason being, that the low
   % frequencies with lots of energy thus having an amplitude>1 are damped.
   % The high frequencies on the other hand generally have little energy,
   % i.e. amplitude<1, and are therefore enhanced by the sqrt.

% 4.Set the magnitude and phase of the high frequencies to zero and 
%   take the inverse DFT, and explain what you see.
  [M N]=size(img);
  u=0:M-1;
  v=0:N-1;
  %
  idx=find(u>M/2);
  u(idx)=u(idx)-M;
  idx=find(v>N/2);
  v(idx)=v(idx)-N;
  %
  [U,V]=meshgrid(v,u);
  D=fftshift(sqrt(U.^2+V.^2));
  mesh(D);
  %
  Z=IMG.*(D<75);
  z=abs(ifft2(fftshift(Z)));
  figure, imshow(z,[0 255]);
  % setting the high frequencies to 0 equals a low pass filter

[Jonas Reber]

this is actually for a grayscale image.

if you have rgb you should treat each color seperately.

[Brenden]

this does help thank you. however i am still unsure exactly how the image grid is organized.. for example i will use j for the 'yaxis' and i for the 'xaxis': how would i go about multiplying the image data by a function in terms of j and then a function in terms of i?

im not sure that i am making sence... is the image data a matrix that can be operated on by functions in terms of the axis-coulombs and and the axis- rows?

[Brenden]

yes my image is grayscale

[Jonas Reber]

exactly. if you have a grayscale image (IMG) of size MxN you can access the pixels IMG(y,x) (where y=1..N and x=1..M)

note. x and y are changed as indeces are handeled row x column.

M = 200; % the real width of the image

N = 100; % the real height of the image

IMG = zeros(N,M);

% put a white line to the half of the image,

% top left corner to center of left edge

IMG(1:end/2,1) = 1;

% put a gray square to the center of the image

IMG(end/2-10:end/2+10,end/2-5:end/2+5) = 0.5;

% put a horizontal line on the right side of the image

IMG(end/2,end/2:end) = 0.8;

imshow(IMG,[]);

it's basically just important to remember that x and y somehow seem to be changed :) as we are used to think in an x first way when we talk about images...

[Brenden]

This has been great help