Hi,

I would like to analyze the radar backscatter change of agricultural fields between two acquisitions in parallel with the coherence between these two acquisition images. I have computed the VH and VV backscatter from IW GRD images and the coherence from IW SLC images using the TOPSAR Coreg Interferogram IW All Swaths graph. The coherence obtained has range and azimuth spacing: 2.33 x 13.924 m which I found surprising as I though that IW SLC images had 2.3 x 17.4 m pixel spacing. I wish to obtain the coherence image in same resolution and pixel spacing as the backscatter intensity images (10x10m). I tried to multilook (default 4:1) and apply terrain correction to resample to 10x10m but without success as resampling does not work for complex data.

Could somebody tell me if there’s a way to do it ?

Thanks a lot !

Pauline

Hi,

An IW SLC pixel size of 2.33 in slant range and 13.924m in azimuth is correct and compares well with the Product Definition values of 2.3m and 14.1m respectively.

To compare your coherence values (derived from SLC product) why not also calculate the radar backscatter from the SLC product as well? This will ensure the pixel spacing is exactly the same between the coherence and backscatter values.

Peter

Hi Peter,

Thanks for your answer.

I found several time in litterature the spacing 2.3 x 17.4 m, like in the ESA user guide for instance:

https://sentinel.esa.int/web/sentinel/user-guides/sentinel-1-sar/resolutions/level-1-single-look-complex

Do you know why it is different ?

Yes you’re right, it would be easier to recompute the backscatter from the SLC product but I was taken by the time and thought it would be easily possible.

Pauline

Dear People:

I have a question concerning this process too. It says that the images have a 2.33 m and 13.917 m in the metadata file, so… by not doing anything and applying

terrain correction then i get other values 13.92m and 3.69 m for Source GR pixel. I am confused about what is what.

If I apply a coherence window of lets say 7*2 do I have to multiply this 7 times 2.33 or 7*3.69 to know exactly which distance is on the ground??

Thank you very much