hi mengdahl, i am new here,sorry to interrupt, i want to know where i can find the local incidence angle? thanks before
if your data was processed correctly, you find it under tiepointgrids (can also be used in the band maths):
http://forum.step.esa.int/uploads/default/original/2X/3/34d85adda31c7f070d245966ec7bc3aee37644db.png
hi Abraun, we meet again, thanks! fyi my LOS DinSAR show in meter units, how if i want to change in cm units?
calculate in meters and then multiply with 100
ohhaha, well forget for the easiest step tbh i used the ‘phase to displacement’ at my first try, the processing parameters window didn’t show anything, is there any alternative?
yes, you can apply this formula manually in the band maths:
wait, i’m sorry, i still confused between LOS and vertical displacement formula. to calculate LOS in band math is (Unw_Phase * wavelength in mm) / (4 * PI * cos(rad(incident_angle))) ? if i’m not wrong, that is vertical displacement formula right?
you are right, the cosine of the incidence angle is the LOS to vertical displacement factor.
so the LOS formula is unwrapped phase * wavelength/ 4*PI ? if the wavelength in cm, and then i want to calculate LOS to vertical displacement, should i use cm unit too?
I’m not sure, to be honest.
maybe i should do trial and error. well, thanks for your time to answer my questions, have a good day!
Since you say “not sure”, can you find an expert to describe the equation? Because (1) In optics the incident angle of a light line to a surface is between the “line of sight( LOS )” direction and the normal direction of the surface. If this is the same definition for the radio wave of InSAR, the ‘cos’ term in the equation zealandia_sarah shows above should be in the numerator to convert the LOSD to vertical displacement. (2) In my image processing by SNAP the result of vertical displacement looks too large( ~30cm ) on a region where there is no big geological event in the 12 days interval( S1A images ). Can you have some comments on this?
hello LiniC,

Incidence angle is the angle defined by the incident radar beam and the vertical (normal) to the intercepting surface as shown below.

In terms of vertical displacement, as it is given by Z.Lu and D.Dzurisin, 2014 is as follows:
Let’s say we have two pixels, m and n which corresponds to two targets on the ground, the SAR phase for these points can be defined as follows:
phase_m = (4pi/wavelength)*rm and phase_n = (4pi/wavelength)*rn (where rn and rm are the distance between the satellite and the points on the ground.)
A second SAR image is acquired for the same area (let’s assume that the target point n has moved up by an amount of h). Now the phase for the point n’ which has moved can be expressed as follows:
phase_n’ = (4pi/wavelength)(rn  hcos*theta) (where h is the vertical displacement of target n)
Hence, we form the interferogram by subtracting the phases of the two SAR images and the vertical displacement between the two points can be expressed as follows:
phase_n’ = (4pi / wavelength)(hcos*theta)
So, in order to find the vertical displacement, we need to get the cos
of the incidence angle
The radar measures LOSdisplacement. If one has only one observationgeometry that is not enough measurements to solve for NS/EW/updown components without extra assumptions (for example vertical movement only). If one can assume vertical movement only, then the conversion from LOS to vertical can be done simply using the incidenceangle.
You said “(let’s assume that the target point n has moved up by an amount of h)”, so “h” is the “vertical displacement”, right? By vector algebra “h” is the vertical component of vector of LOSD, what does “(_h_cos*theta)” in the equation “phase_n’ = (4pi / wavelength)(_h_cos*theta)” mean? it should be “h/cos(theta)”. So when converting the phase difference to the vertical displacement with the help of incident angle as defined above, the cos term should be in the numerator, not in denominator as someone told to zealandia_sarah above.
According to Z.Lu and D.Dzurisin, 2014, h*cos(theta) is used to convert LOS into vertical displacement. This equation is slightly different than the one posted here.
Why cos(theta) should be in the nominator?
Also, the big displacement value (30 cm) you see on your results is due to atmospheric errors. Additional fringes are created in your interferogram due to atmospheric delay and these fringes are interpreted as displacement. So, atmospheric corrections should be applied.
If “d” is LOSD and “h” is corresponding vertical displacement, then “h=dcos(theta)" which means “h” is “d”'s projection on a vertical line. So from the correct equation "(phaseDifference)=(4PId)/(waveLength)" ( from ESA’s report TM90, equation 2.6 ) the correct equation should be "h=(phaseDifference)(waveLength)cos(theta)/(4PI)”, here cos term is in numerator, “” sign means a positive “h” is an uplift. I can not find Lu’s paper online, but I think “hcostheta" or "hcos(theta)” has not a geometric meaning.
Thank you very much for pointing out that an atmospheric correction should be applied! Can SNAP do this? Can SNAP find the relative atmospheric data automatically ( just as it can find the relative orbit information and DEM automatically in processing )? If not, where can I find it and import it to SNAP when processing? Is there a tutorial material for doing atmospheric correction?
Atmospheric corrections is the tricky part of InSAR processing but very important. If SAR images were acquired under severe weather conditions, the tropospheric error (in the worst case scenario) can be up to 9cm in your measurements. This 9cm corresponds to a few fringes on your interferogram which must be removed.
SNAP does not perform atmospheric corrections for InSAR. For applying atpospheric corrections we need to have precise measurements of the composition of the troposphere (this is where the most signal delay occurs). The good news is that there is a python library (PyAPS) that perform atmospheric corrections.
http://earthdef.caltech.edu/projects/pyaps/wiki/Main
You can have a look at the website.
Hi liniC,
I think what @johngan said is correct, it should be h=d/cos(theta) when convert LOS displacement d to vertical displacement h, it means d is h’s projection on LOS direction. And I have confirmed it by levelling data used in my experiment.
Hope this helps,
Fei