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See also [20]. See figure 31.4

Simple equations for topo-pair (no deformation, no atmosphere, no other errors, $ r_1 \parallel r_2$)

$\displaystyle \theta = \theta_0+\delta\theta$ (1)

$\displaystyle {B_{\parallel}}= r_1 - r_2$ (2)

$\displaystyle {B_{\perp}}= B\cos(\theta-\alpha) = B\cos(\alpha-\theta)$ (3)

$\displaystyle {B_{\parallel}}= B\sin(\theta-\alpha) = -B\sin(\alpha-\theta)$ (4)

The baseline components, for points on the reference ellipsoid ($ h=0$) are

$\displaystyle {\left[{B_{\perp}}\right]}_{h=0} = {B_{{\perp0}}}= B\cos(\theta_0-\alpha)$ (5)

$\displaystyle {\left[{B_{\parallel}}\right]}_{h=0} = {B_{{\parallel0}}}= B\sin(\theta_0-\alpha)$ (6)

The 'true' phase of the interferogram is

$\displaystyle \phi = -\frac{4\pi}{\lambda}{B_{\parallel}}$ (7)

And corrected for the phase of the reference body

$\displaystyle \phi = -\frac{4\pi}{\lambda}({B_{\parallel}}-{B_{{\parallel0}}})$ (8)

For the defo-pair (1,3), denoted with a prime, similar equations follow. Deformation in the line of sight (range), that occurred in between the acquisitions, is denoted by $ \Delta r$

$\displaystyle \Delta r = -\frac{\lambda}{4\pi} \phi_{\Delta r}$ (9)

A positive $ \Delta r$ implies deformation in the $ {B_{\parallel }}$ direction (away from the sensor, i.e., subsidence). The phase of this interferogram is

$\displaystyle \phi' = -\frac{4\pi}{\lambda}(r_1 - ( r_3 + \Delta r)) = -\frac{4\pi}{\lambda}({B_{\parallel}}' + \Delta r)$ (10)

Combining the expressions for the interferometric phase for the topo-pair (31.7) and defo-pair (31.10) yields:

$\displaystyle \phi' = \phi \frac{{B_{\parallel}}'}{{B_{\parallel}}} + \frac{4\pi}{\lambda}\Delta r$ (11)

The problem here is that the 'true' parallel baselines are unknown.

The (actually wrapped) phase of the deformation interferogram, corrected for reference phase, is defined as:

$\displaystyle \phi'$ $\displaystyle =$ $\displaystyle -\frac{4\pi}{\lambda}[{B_{\parallel}}' - {B_{{\parallel0}}}' + \Delta r]$ (12)
  $\displaystyle =$ $\displaystyle -\frac{4\pi}{\lambda}[B'\sin(\theta-\alpha')
- B'\sin(\theta_0-\alpha')
+ \Delta r]$  
  $\displaystyle =$ $\displaystyle -\frac{4\pi}{\lambda}[B'\sin(\beta'+\delta\theta) - B'\sin\beta' + \Delta r]$  

where $ \beta'=\theta_0-\alpha'$. Using the approximation for small $ \delta \theta $ (which is about $ 1^\circ $ or 0.0175 rad for terrain height differences of 5 km)

$\displaystyle \sin(\beta+\delta\theta) = \sin\beta \cos\delta\theta + \cos\beta \sin\delta\theta \approx \sin\beta + \delta\theta\cos\beta$ (13)

it follows from equation 31.13 that the 'flat earth' corrected phase equals
$\displaystyle \phi'$ $\displaystyle =$ $\displaystyle -\frac{4\pi}{\lambda}[B'(\sin\beta' + \delta\theta\cos\beta')
- B' \sin\beta' + \Delta r]$ (14)
  $\displaystyle =$ $\displaystyle -\frac{4\pi}{\lambda}[\delta\theta B'\cos\beta' + \Delta r]$  
  $\displaystyle =$ $\displaystyle -\frac{4\pi}{\lambda}\delta\theta {B_{{\perp0}}}' - \frac{4\pi}{\lambda}\Delta r$  

The corrected phase for the topo pair equals $ \phi=-\frac{4\pi}{\lambda}\delta\theta{B_{{\perp0}}}$ (using the same approximation), and combining this with 31.15 yields (for the 'flat earth' corrected phases)

$\displaystyle \phi' = \phi \frac{\delta\theta {B_{{\perp0}}}'} {\delta\theta {B...
... r = \phi \frac{{B_{{\perp0}}}'}{{B_{{\perp0}}}} - \frac{4\pi}{\lambda}\Delta r$ (15)


$\displaystyle \Delta r = -\frac{\lambda}{4\pi} [\phi' - \phi \frac{{B_{{\perp0}}}'}{{B_{{\perp0}}}}]$ (16)

or for the phase $ \phi_{\Delta r}$ caused by the deformation $ \Delta r$

$\displaystyle \phi_{\Delta r} = \phi' - \frac{{B_{{\perp0}}}'}{{B_{{\perp0}}}}\phi$ (17)

This important equation shows how to obtain offset vectors from 3 SLC images, i.e., by scaling the (reference phase corrected) unwrapped phase of the topo-pair by the ratio of the perpendicular baselines (to points on reference body), and subtracting this from the phase of the defo-pair. This can thus be performed without the 'true' values for $ \theta $ are required.

Figure 31.4: Geometric configuration for 3-pass differential insar. The orbits go 'into' the paper. All angles are defined counterclockwise. The terrain element $ P$ corresponding to the radar coordinate (l,p) is located at a height $ h$ above the ellipsoid. The perpendicular baseline required for this method is the one for points $ P$ located on the reference ellipsoid ($ h=0$). $ \delta \theta $, the change in $ \theta $ since $ P$ is on a height $ h$, due to a 5 km height difference, is approximately $ 1^\circ $.

next up previous contents
Next: Algorithm Up: DINSAR Previous: Output Description   Contents
Leijen 2009-04-14