block processing to find local orientation of each block

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Gunjan 2016-1-7

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https://ww2.mathworks.cn/matlabcentral/answers/263003-block-processing-to-find-local-orientation-of-each-block

移动：DGM 2023-12-21

I want to implement an algorithm in MATLAB on a Fingerprint image in order to enhance it. The algorithm starts with normalization then orientation image estimation followed by three more steps. I have completed the normalization part but stuck at the orientation step for which following are the steps: 1) Divide the image into 16x16 blocks. 2) Compute the gradient at each pixel. 3) Estimate the local orientation of each block centered at pixel(i,j) using the below

Here delX and delY are gradients at each pixel. Vx(i,j) and Vy(i,j) are orientations.

    function [imgOrient] = orientation(imgNorm)
    Grad =@(block_struct)gradient(block_struct.data)
    imgBlock=blockproc(ingNorm, [16 16], 'Grad)

Please let me know if this is correct way to use blockproc. Now I want to ask that how to save both the both the components (x as well as y ) returned by gradient function and use them in above formula. Any help would be appreciated.

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Leon 2023-12-21

移动：DGM 2023-12-21

Hello, I would like to know if you are using the original image or the normalized image for estimating the image direction field

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Guillaume 2016-1-7

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https://ww2.mathworks.cn/matlabcentral/answers/263003-block-processing-to-find-local-orientation-of-each-block#answer_205424

编辑：Guillaume 2016-1-7

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Your code at present is just equivalent to

imgBlock = gradient(imgNorm);

Not really what you want.

What you want to do is pass to blockproc a function that returns a single value for each block, the orientation. You can't do it with an anonymous function (well, you could but that would be very unwieldy), so use a normal function:

function orientation = CalculateBlockOrientation(block_struct)
   %block_struct: block structure generated by blockproc
   %orientation: the orientation of the block
   [delx, dely] = gradient(block_struct.data);
   Vx = sum(sum(2*delx.*dely));
   Vy = ... %I'm sure you can work it out
   orientation = 0.5*atan(Vy/Vx);
end

And in your main function

blockorientation = blockproc(imgNorm, [16, 16], @CalculateBlockOrientation);

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Guillaume 2016-1-26

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I'm afraid I don't understand anymore what you are attempting to do. Within a block, θ is scalar, and hence your Φx and Φy are scalar. There is nothing to average. You could of course, filter the whole θ image but that would be done after the call to blockproc, so:

blockorientation = blockproc(imgNorm, [16, 16], @CalculateBlockOrientation); %CalculateBlockOrientation is unchanged
phi_x = cos(2*blockorientation);
phi_y = sin(2*blockorientation);
w = ones(5) / 25;
phi_x_prime = conv2(phi_x, w, 'same');
phi_y_prime = conv2(phi_y, w, 'same');

My answer was based on my understanding that the filter was to be done inside the blockproc and assumed that Φx and Φy was the thing that was summed in your original question to obtain νx and νy. The |sum(sum(| is used to convert the result of the convolution (a matrix) into a scalar, as you did originally for ν. That is what I understood:

original:

ν = Σ Σ Φ

new:

&Phi;' = &Sigma; &Sigma; (W &otimes; &Phi;) %convolution
&nu;' = &Sigma; &Sigma; &Phi;'

therefore:

&nu' = Σ Σ conv2(W, &Phi, 'same');

Guillaume 2016-1-27

在 MATLAB Online 中打开

Right, so the filtering is done on the orientation and is done outside of blockproc. This corresponds to the 1st part of my answer:

blockorientation = blockproc(imgNorm, [16, 16], @CalculateBlockOrientation); %CalculateBlockOrientation is unchanged
phi_x = cos(2*blockorientation);
phi_y = sin(2*blockorientation);
w = ones(5) / 25;
phi_x_prime = conv2(phi_x, w, 'same');
phi_y_prime = conv2(phi_y, w, 'same');

ALI Salah 2021-12-18

编辑：ALI Salah 2021-12-18

hi Guillaume the function CalculateBlockOrientation will return the oriantation of each block in normimg but the question is how can i abtain new oriantation for each block based on

thetax(i,j) = cos(2*blockorientation);

thetay(i,j) = sin(2*blockorientation);

and theta=0.5*tan((thetax(i,j))'/(thetay(i,j))');

the last 3 equation below