Lecture 8: SAR interferometry

Question 1

Phase

Answer 1

= “strength” of the backscattered signal
Depends on the physical and
electric properties of the surface

In SAR, the phase refers to the position of the wave within its cycle (oscillation) at the time the
radar signal is recorded.
→ The instrument measures “where” we are in the wave’s oscillation of the returning signal
The SAR phase is usually expressed in radian (0 → 2π)

The returing radar signal carries a phase that depends on the distance travelled
➢ This phase information = -π → +π (or -180° → +180°)
➢ A single phase image is useless, as it does not tell us how many cycles happened
However, comparing two phase images of the same area, acquired from with the same instrument, from the same orbit, can provide mm- to cm-scale changes
(providing that the backscattering properties of the ground
surface do not change)

Question 2

InSAR

Answer 2

InSAR = Interferometric Synthetic Aperture Radar
InSAR = Measurement of the phase difference of two images of the same area
➢ The phase is proportional to the distance (×2)

ϕ = 2π/𝜆 *2R = 4π/𝜆 *R
Equation = linear proportionality between the phase (ϕ), the range (R), at a given wavelength (𝜆)

➢ 1st image = reference / primary / master
➢ 2nd image = secondary / slave

➢ The interferometric phase (φ) is a double-difference observation in time (δt = j − i) and space (δx = x1 − x2)
➢ The double difference interferometric phase (φ) relates to three main and superimposed, contributions:
a) the spatial gradient in topography (φtopo),
b) the spatial gradient in surface displacements occurred between temporally separated acquisitions (φdefo) and
c) spatial variation in the two atmospheric conditions affecting the signal propagation at the two acquisitions (φatmo).
φ = φtopo + φdefo + φatmo + (φpath^j - φpath^i)
➢ Double difference of intrapixel path radar phases must cancel out (φpath^j ≈ φpath^i).

Question 3

Interferogram

Answer 3

Interferogram (cross-track, repeat pass) =
➢ Viewing geometry (flat earth phase, orbit error)
➢ Topography
➢ Ground surface deformation
➢ Atmospheric effects
➢ Noise

Answer 4

InSAR Workflow: Step-by-Step

Step 1. Co-registration
➢ Align two SAR images (master and slave) with cm-level precision using amplitude images.
➢ Result: Raw interferogram showing noisy phase differences (fringes represent displacement or topography).

Step 2. Phase Differencing
➢ Subtract phase of secondary from primary:
→ Shows how much the ground target moved between acquisitions.

3. Remove Effects of Viewing Geometry
   ➢ Correct for orbital effects and imaging geometry (e.g., satellite position, angle).
   ➢ Produces a flattened interferogram.
   ➢ You can already estimate a DEM (Digital Elevation Model) at this point.
4. Remove Topography
   ➢ Subtract synthetic interferogram (generated from DEM) from the flattened interferogram.
   ➢ Isolates signal due to ground deformation and atmospheric effects.
   Height/altitude of ambiguity = height difference needed to produce a change of 2𝛑 (= 1 fringe) in the interferogram
   ha = - (𝜆r sin𝜽) / (2b⊥)
   Where ha = altitude of ambiguity
   𝜆 = wavelength
   r = slant range
   𝜽 = look angle
   b⊥ = perpendicular baseline
   Smaller b⊥ → larger ambiguity → less height sensitivity
   Larger b⊥ → better topographic detail but more noise (slide 11 shows this clearly)
5. Spatial Filtering
   ➢ Smooth the result to reduce noise caused by:
   ➢ Low coherence (e.g., vegetation, water, shadows)
   ➢ Signal decorrelation
6. Phase Unwrapping
   ➢ Convert wrapped phase (0 to 2π) into continuous displacement field.
   ➢ Stacked unwrapped images allow accurate displacement time series.
   ➢ Important Concept: Height of Ambiguity

Question 5

Coherence image

Answer 5

= image expressing the similarity of the radar reflection in an image pair
➢ Effect of time lapse between two image acquisitions (= temporal baseline) on coherence

Question 6

DInSAR for ground deformation

Answer 6

A distance change between the satellite and the ground corresponds to:
∆ɸ = 4𝜋/𝜆 * ∆d
where:
∆ɸ = phase change (in radians),
𝜆 = radar wavelength (meters),
∆d = ground displacement along the
satellite’s line of sight (LOS) (meters).

If the phase difference is 2π (1 fringe),
then:
2π = 4𝜋/𝜆 * ∆d So, ∆d = 𝜆/2
e.g., Sentinel-1 has a wavelength of 5.6 cm → 1 fringe = 2.8 cm

Question 7

Ground deformation modelling

Answer 7

= comparison between a measured ground deformation and a modelled one, with the aim of explaining the source of the deformation.

Question 8

InSAR Time-Series: the Multidimensional Small Baseline Subset (MSBAS) technique

Answer 8

Advantages:
✓ Improved temporal resolution
✓ Improved signal to noise ratio
(average various sources of noise)
✓ Combined uninterrupted temporal coverage (> life time of single sensor)
✓ 2 or 3 components of ground deformation vector
+ Day/night acquisition
+ All-weather (! Still atmospheric effects)
+ Regional GD observation (vs. GNSS)
+ Possibility to measure long-term or
precursory deformation

Limitations
− No coherence → no measurement
− Deformation must be < 2𝛑 between 2 neighbouring pixels
− Polar orbits → no N-S horizontal
component in deformation time-series
− Ascending/Descending data not always available → prevents the calculation of vertical and horizontal components in time-series