Asked by Mary Jon
on 18 Sep 2013

(v(i,j))^(k+1)= (1-ω)(v(i,j))^(k)-ω[(W1(i,j)(v(i+1,j))^(k)+ W2(i,j)(v(i-1,j))^(k+1)+ W3(i,j)(v(i,j+1)^(k)+ W4(i,j)(v(i,j-1))^(k+1)(/(W i,j)]

v matrix k iteration W may by matrix (have i &j) w constant

*No products are associated with this question.*

Answer by John Petersen
on 18 Sep 2013

The trick is to update v(i-1,j) and v(i,j-1) without updating v(i+1,j) and v(i,j+1) before going on the v(i,j). After that do:

v(i,j)= (1-ω)*v(i,j) - ω*[(W1(i,j)*v(i+1,j) + W2(i,j)*v(i-1,j) + W3(i,j)*v(i,j+1) + W4(i,j)*v(i,j-1))/W(i,j)]

Mary Jon
on 18 Sep 2013

where is the (k) number of iteration?? I am have two matrices new matrix with iteratuon (k+1) and old matrix with iteration(k)

This equation consist of two part, one of it i take it from initial matrix (k) and second part with (k+1)

Image Analyst
on 18 Sep 2013

convolution, conv2(), and imfilter() do this "trick" inherently and efficiently. With your code, as you move over to the next j, your v(i,j) will now be the new v(i, j-1) and will be affected. So it will have some kind of recursive/hysteresis effect or trailing blur or something that's not right. imfilter() and conv2() do not suffer from this. If she insists on using two nested for loops (a much less efficient approach especially for larger windows) then she'll have to at least assign the result (right hand side of the equation) to a new output variable, like vout(i,j), not the original input variable.

conv2() and imfilter are more efficient because they only need to read in and discard a few of the pixels on the edge, not the whole array. For example with a 7x7 window, you have 49 elements/pixels in the window but as you slide over one pixel you only need to read in 7 new pixels, not all 49.

Mary Jon
on 21 Sep 2013

I do conv2(), and imfilter() ,but no result getting yet ,where is wrong?

Answer by Image Analyst
on 18 Sep 2013

Edited by Image Analyst
on 18 Sep 2013

Looks like unsharp masking, so you can do this

kernel = [0, W1, 0; W3, 0, W4; 0, W2, 0]; filtered = conv2(vk, kernel, 'same'); vk = (1-omega)*vk + filtered./W;

Show 5 older comments

Image Analyst
on 19 Sep 2013

What are the values of W1, W2, etc.? They're probably arrays, what are they?

Image Analyst
on 20 Sep 2013

Just tell me what it's supposed to do. It looks like you might be able to use convolution but if W1, etc. are not constants, then maybe not. I really need to know what the intent of that code is. For example if it's to do unsharp masking then you should use a different output variable if you're going to do it in a loop so that you avoid hysteresis. I suspect that you transcribed the equation incorrectly but I can't say exactly how to fix it unless I know what you want to accomplish. John's code will probably work in that it will do what you tried to do, but I'm not sure that is really what you want to do because I don't feel that you described in words properly what you want to do.

Mary Jon
on 20 Sep 2013

when using this "trick"

Z = zeros(size(W1)); kernel = [Z, W1, Z; W3, Z, W4; Z, W2, Z]; filtered = conv2(vk, kernel, 'same'); vk = (1-w)*vk + filtered./W(i,j);

get this error

Undefined function or variable "vk".

Error in ==> SOR at 69 filtered = conv2(vk, kernel, 'same');

Related Content

MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi

Learn moreOpportunities for recent engineering grads.

Apply Today
## 0 Comments