A question about Loop Closure about licsbas HOT 6 CLOSED

The subtraction of the 2pi*n bias is to unbias the loop phase, as mentioned in the comment of the code.

Suppose that we have three complete 0 phases (phi1=phi2=phi3=0) and differential phases (phi12=phi23=phi13=0). The 2pin biases in each interferogram is completely unknown without the reference in the 1st and 2nd loop closure check (e.g., phi12 could be 2pi5). To get the reasonable loop phases taking into account the unknown biases, the 2pin bias estimated from the median must be subtracted.

You would understand why this unbias is necessary if you removed the lines for the unbias in the code.

from licsbas.

Hi, yumorishita. Thanks for your reply. But I'm still confused about this unbias operation.

As you said，suppose that we have three complete 0 phases SAR image (φ1=φ2=φ3=0), and it's differential phases should be (φ12=φ23=φ13=0), but why φ12 could be 5×2π without the reference. Do you mean it is still wrap phase? But we are dealing with unwrapping phase in loop close operation.

I still chouldn't understand where this bias comes from. Is it because my wrong understanding of interference and unwrap operation? I put my understanding of those operation below, please see which step is problematic.

Suppose we have three SAR image 1, 2, 3, the phases are φ1, φ2, φ3 separately. If the phase unwrapping operation is correct, Interferogram φ12，φ23，φ13 should be:

                φ12=φ1-φ2
                φ23=φ2-φ3
                φ13=φ1-φ3

Then:

               φ12 + φ23 = φ1-φ2 + φ2-φ3 =  φ13

Or:

               φ12 + φ23 - φ13 = 0

Example:

Suppose the phases of a pixel with same location for SAR image 1, 2, 3 are:

                 φ1 = n×2π + 0.3π
                 φ2 = (n+1)×2π - 0.3π
                 φ3 = (n+1)×2π + 0.3π

If the phase unwrapping operation is correct, Then:

                φ12 = φ1 - φ2  = -2π + 0.6π
                φ23 = φ2 - φ3  =  -0.6π
                φ13 = φ1 - φ3  = -2π

Then:

               φ12 + φ23 - φ13 = -2π + 0.6π + (-0.6π) - (-2π) = 0

According to my train of thought，I chouldn't understand where this bias comes from .

from licsbas.

I mean the phase is unwrapped. The each unwrapped phase has each unknown 2πn. The unwrapped phases for each interferogram are

φ12=φ1-φ2+2π*l
φ23=φ2-φ3+2π*m
φ13=φ1-φ3+2π*n

then,
φ12 + φ23 - φ13 = 2π(l+m-n)

from licsbas.

I think I understand where the bias comes from：
The unwrapping operation cann't generate a absolutely right unwrapped phase for every pixel, but with a same 2π*n bias for one interferogram. Is that right?

from licsbas.

Yes, you are right.

from licsbas.

OK，thanks for your reply. I will close this issue

from licsbas.