Hello Jihad,

could you explain more about the relationship between input1 and input2? Visually, it looks like input2 is just a smoothed version of input 1, i.e. i2=conv(i1,h). In this case, simply multiply FT1 with H (H is the FT of h) to obtain FT2.

Regards,

Michael

Hello Michael,

First of all, thank for your reply.

Actually yes it looks like a smothed version, but practically i didn use a filter to get input2 from input1. I have got these two images by projecting a number of vertices from a 3D object on a 2D plan, and since the two objects have different densities (number of vertices of the mesh), then number of projected vertices also is different. So to do what you proposed i normally have to know the filter or to know how to get it comparing these two images if there is a way of course, thing i don't know.

Hello Jihad,

For a better understanding, I try to split up the process. Please correct me, if I misunderstood something:

1. A number of points from a 3D object are projected onto a 2D plane (the image). The two images differ only in the number of points of the 3D objects, e.g. the object of image 1 is a subset of the points of the object of image2.

If the grid (pixel resolution) of projection plane would be infinitesimally small, then the difference between the images is only a masking out of some vertices, i.e. setting some values of the grid to zero.

I assume that the information on how this masking is done is not available at the stage where you are comparing the two FTs. Hence a random masking is our best guess.

2. The infinitesimal pixes are summed up to generate the pixels of the two images (subsampling). This transform is equal for image1 and image2, hence it does not affect the equality/unequality of FT1 and FT2.

3. The FT is calculated, which is linear and equal for both images.

So the only difference between image1 and image2 is the masking of the vertices before the filtering, subsampling and FT.

In my opinion, a direct comparison of the masked/unmasked image spectra is the best we can do, if we have no information about the masking process.

Probably further subsampling of the images (i.e. reduce number of pixels in the image by summarizing neighboring pixels) would reduce the "noise" introduced by the masking and might produce visually better results, however this information loss is disadvantageous if you want to further process the FTs.

Furthermore, I would double-check that the colormaps of the two images are equal. Probably the dissimilarty of the two FTs is only a plotting problem. Assure that the minimum values (dark blue) and maximum values (yellow) correspond to the same numbers in the two FTs. Matlab usually rescales the values to minimum/maximum of the images, which might cause the FTs to look different although they are actually similar. This is particularly important for the log-scaling which you seem to use.

Regards,

Michael

Hello Michael,

I apologize that i didn't reply early. I want to explain some other things clearly.

1. image1 is not a subset of the points of the object of image2 , only te form are slightly similar. So for example in a pixel (x1,y1) of image1 may contain n1 vertices and the pixel (x'1,y'1) in the same position in image2 may contain m1, and for (x2,y2) contains n2 and (x'2,y'2) contains m2...and so on..there relation is not lineair.

2. The grid is not infinitismalmy small for these two images max(image1) = 3.7 and max(image2) = 67.06.

3.Thus the TF1 and TF2 are different because i tested by commparing them like what is did in Fourier-Mellin Transform, by computing crosspowerspectrum and the peak is different than one. So it is not a problem of displaying.

4.I am thinking about what you proposed for subsumpling, I will learn more how to do it or if there is some spesific techniques.

Thank you so much

Best regards.

Could you post the Fourier transform images including the colorbar? Or could you upload them as matlab .fig files?

Ok here they are.

I was busy during the last week, I could only take a look at your data today. Sorry for that.

The spectra look different, simply because the values in image1 are significantly lower than the values in image2 (~2 in image 1 and ~40 in image2). Your plotting somehow scales the data before display.

You can easily avoid this by normalizing out the energy of the respective images before plotting. Then, you can use the values of the data as an absolute reference for comparison.

Regarding the the sidelobes, the difference between the two ffts is actually very small. The sidelobe level is ~30dB lower than the main lobe, which is a lot! Differences at such a small level are not surprising, since also your images are slightly different. My suggestion to overcome this issue is to limit the dynamic range of the fft-plot to, say, 15dB (however, you will loose quite a lot of information from the smaller sidelobes then).

So my recommendation is to rather compare the ffts of the two images in a linear scale, rather than in log (dB) scale. Very low sidelobes will have a small influence then.

I appended a small matlab example code for your problem. I hope this helps.