Implementation

See also [20]. See figure 31.4

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

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

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

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

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

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

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

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

The 'true' phase of the interferogram is

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

And corrected for the phase of the reference body

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

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

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

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

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

$$\phi' = -\frac{4\pi}{\lambda}[{B_{\parallel}}' - {B_{{\parallel0}}}' + \Delta r]$$ (12)

$$= -\frac{4\pi}{\lambda}[B'\sin(\theta-\alpha') - B'\sin(\theta_0-\alpha') + \Delta r]$$

$$= -\frac{4\pi}{\lambda}[B'\sin(\beta'+\delta\theta) - B'\sin\beta' + \Delta r]$$

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)

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

it follows from equation 31.13 that the 'flat earth' corrected phase equals

$$\phi' = -\frac{4\pi}{\lambda}[B'(\sin\beta' + \delta\theta\cos\beta') - B' \sin\beta' + \Delta r]$$ (14)

$$= -\frac{4\pi}{\lambda}[\delta\theta B'\cos\beta' + \Delta r]$$

$$= -\frac{4\pi}{\lambda}\delta\theta {B_{{\perp0}}}' - \frac{4\pi}{\lambda}\Delta r$$

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)

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

or

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

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

$$\phi_{\Delta r} = \phi' - \frac{{B_{{\perp0}}}'}{{B_{{\perp0}}}}\phi$$ (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 $1^\circ$.