I'll briefly reply here. if you need additional explanations, please email me. I'm not familiar with the concept circular bi-linear interpolation. Regular bi-linear interpolation is basically about calculating value lying between nearest neighbors (interpolation ) values, via weighted average based on their distances to the desired value position (bi-linear). See First Order Hold as well.
In case of LBP, having a central point, and angle, and a radius, usually results in a point "falling" between pixels. This demands calculation of this value via interpolation. bi-linear is the easiest one. Bi-cubic for example is another, more expensive option.
Now again- in LBP image, each pixel value is calculated via sum of neighboring pixels (of when speaking of the whole image via it's shifted versions). As the shift is not always an integer value it demands interpolation. that's about all. See matlab imresize function, or any bi-linear based image zoom method for visualization.
Hope I've made things a bit more clear.
Hi Nikolay ..
I am very much new to LBP.
I have successfully calculated LBP.
Can you tell me how to calculate circular bi-linear interpolation. I am confused about radius of circular LBP and how circular neighbourhoods are calculated using bilinear interpolation.
Pls help me. I read Ojala's paper , but confused about uniform LBP and circular neighbourhood.
You may solve your problem as 3 independent examples, because there is no coupling among the equations. In that case, you build a body of your function FUN for residuals in the form
res = fi(x,c) - yi;
where yi is the value of required fi.
Of course, you may to solve your problem also in one go, if you prepare the body of function FUN in the following way:
res = [f1(x,c(1:2)) - y1
f2(x,c(3:4)) - y2
f3(x,c(5:6)) - y3];
This approach seems to be better, because you do not need to prepare 3 functions FUN.
If you have any issue with you problem, don't hesitate to contact me again or send me just data of your problem.