Ferkans — Interactive Telecom Tutor

ex01-sar-resolution-design

Easy

Design a stripmap SAR system for 1-meter cross-range resolution at $f_0 = 5$ GHz:

(a) What synthetic aperture length $L_{\text{sa}}$ is required at range $R_0 = 10$ km?

(b) If the platform velocity is $v_p = 100$ m/s, what is the required dwell time $T_{\text{sa}}$ ?

(c) What is the Doppler bandwidth $B_a$ ?

(d) What PRF is required to avoid azimuth ambiguities?

(e) What bandwidth $B$ is required for 1-meter range resolution?

Show Hint

Use $\Delta x = \lambda R_0 / (2 L_{\text{sa}})$ .

PRF must satisfy PRF $\geq B_a$ .

Solution

Synthetic aperture length

$\lambda = c/f_0 = 0.06$ m. $L_{\text{sa}} = \lambda R_0/(2\Delta x) = 0.06 \times 10{,}000 / (2 \times 1) = 300$ m.

Dwell time

$T_{\text{sa}} = L_{\text{sa}} / v_p = 300/100 = 3$ s.

Doppler bandwidth

$B_a = 2v_p L_{\text{sa}}/(\lambda R_0) = 2 \times 100 \times 300/(0.06 \times 10{,}000) = 100$ Hz.

PRF

PRF $\geq B_a = 100$ Hz. In practice, use PRF $= 200$ Hz for 2 $\times$ oversampling.

Bandwidth for range resolution

$B = c/(2\Delta R) = 3 \times 10^8/(2 \times 1) = 150$ MHz.

ex02-resolution-paradox

Easy

(a) Show that for stripmap SAR, the best cross-range resolution $\Delta x_{\min} = D/2$ is achieved when $L_{\text{sa}} = R_0\lambda/D$ .

(b) A SAR system uses an antenna of length $D = 10$ m. Compute $\Delta x_{\min}$ .

(c) The antenna is replaced with one of length $D = 2$ m. Compute the new $\Delta x_{\min}$ . Explain the paradox.

(d) For $D = 2$ m at $R_0 = 50$ km and $f_0 = 1.2$ GHz, compute $L_{\text{sa,max}}$ and $T_{\text{sa}}$ for $v_p = 200$ m/s.

Show Hint

Best resolution occurs when the target is illuminated for the maximum possible time.

Solution

Maximum aperture derivation

Beamwidth $\theta = \lambda/D$ , so $L_{\text{sa,max}} = R_0\theta = R_0\lambda/D$ . Substituting: $\Delta x_{\min} = \lambda R_0/(2 R_0\lambda/D) = D/2$ .

$D = 10$ m

$\Delta x_{\min} = 5$ m.

$D = 2$ m

$\Delta x_{\min} = 1$ m. The smaller antenna gives 5 $\times$ better resolution because its wider beam illuminates the target longer.

Numerical example

$\lambda = 0.25$ m. $L_{\text{sa,max}} = 50{,}000 \times 0.25/2 = 6{,}250$ m. $T_{\text{sa}} = 6{,}250/200 = 31.25$ s.

ex03-rda-implementation

Medium

Implement the range-Doppler algorithm for a simulated SAR dataset:

(a) Generate raw SAR data for 5 point targets at known $(x, R)$ positions. Use $f_0 = 10$ GHz, $B = 200$ MHz, $v_p = 100$ m/s, $T_{\text{sa}} = 2$ s, PRF $= 500$ Hz.

(b) Implement range compression via matched filtering. Plot the range-compressed data and identify the hyperbolic range migration.

(c) Implement RCMC using sinc interpolation in the range-Doppler domain.

(d) Implement azimuth compression. Plot the focused image and measure the $-3$ dB mainlobe widths in range and azimuth.

(e) Compare the measured resolution with the theoretical values.

Show Hint

Use range-dependent azimuth FM rate $K_a(R_0)$ .

Solution

Range compression

Apply matched filter in frequency domain for each azimuth sample: $S_{\text{rc}}(f, \eta) = S(f, \eta) \cdot H_r^*(f)$ where $H_r(f) = \text{rect}(f/B) \exp(j\pi f^2/K_r)$ .

RCMC

In the range-Doppler domain, shift each range profile by $\Delta R(f_\eta) = \lambda^{2} R_0 f_\eta^2/(8v_p^2)$ via sinc interpolation.

Azimuth compression

Multiply by $H_a(f_\eta) = \exp(j\pi f_\eta^2/K_a)$ where $K_a = -2v_p^2/(\lambda R_0)$ .

Resolution verification

Theoretical: $\Delta R = c/(2B) = 0.75$ m, $\Delta x = \lambda R_0/(2L_{\text{sa}}) = 0.15$ m. Measured values should match within 5%.

ex04-rcmc-analysis

Medium

For a SAR system with $f_0 = 5.3$ GHz, $B = 100$ MHz, $v_p = 200$ m/s:

(a) Compute the range migration $\Delta R_{\max}$ for targets at $R_0 = 10, 50, 100$ km, using a dwell time that achieves $\Delta x = D/2$ with $D = 10$ m.

(b) Express $\Delta R_{\max}$ in units of range cells. At what range does the migration exceed one range cell?

(c) Show that RCMC is unnecessary when $v_p^2 T_{\text{sa}}^2 / (8 R_0) < c/(4B)$ .

Show Hint

Maximum range migration is $\Delta R_{\max} = v_p^2 T_{\text{sa}}^2 / (8R_0)$ .

Solution

Dwell time

$T_{\text{sa}} = R_0\lambda/D / v_p$ . At $R_0 = 10$ km: $T_{\text{sa}} = 10{,}000 \times 0.0566/10/200 = 0.283$ s.

Range migration

$\Delta R_{\max} = v_p^2 T_{\text{sa}}^2/(8R_0)$ . At 10 km: $40{,}000 \times 0.08/(80{,}000) = 0.04$ m $= 0.027$ range cells. At 100 km: much smaller. RCMC unneeded at these parameters.

RCMC criterion

RCMC unneeded when $\Delta R_{\max} < \Delta R/2 = c/(4B) = 0.75$ m.

ex05-pga-implementation

Medium

Implement Phase Gradient Autofocus for a defocused SAR image:

(a) Generate focused SAR data for 10 point targets, then apply a phase error $\phi_e(\eta) = 3\sin(2\pi\eta/T) + 2\cos(4\pi\eta/T)$ .

(b) Implement the PGA algorithm.

(c) Run PGA for 10 iterations. Plot the estimated phase error at each iteration and the image entropy vs iteration number.

(d) Compare the final focused image with the original.

(e) Repeat with a random phase error. How many iterations are needed?

Show Hint

Window width should be $\sim$ 10 azimuth resolution cells.

Solution

Phase error application

Multiply range-compressed data by $\exp(j\phi_e(\eta))$ for each slow-time sample.

PGA iteration

Each iteration: circular shift, window, estimate gradient, integrate, correct. Entropy should decrease monotonically if convergent.

Convergence

Sinusoidal error: $\sim$ 5 iterations. Random error: $\sim$ 8--10 iterations.

ex06-minimum-entropy

Medium

Implement minimum-entropy autofocus for a distributed scene (uniform random reflectivity, no dominant scatterers):

(a) Apply a quadratic phase error with peak phase $= 5$ rad.

(b) Parameterize $\phi_e(\eta) = \sum_{p=2}^P c_p \eta^p$ and minimize image entropy via gradient descent.

(c) Compare with PGA applied to the same scene.

Show Hint

Start with $P = 2$ and increase.

Solution

Entropy gradient

$\partial H/\partial c_p$ involves the chain rule through the image formation and magnitude operations. Numerical gradient is acceptable.

Comparison

MEA should outperform PGA on distributed scenes where no dominant scatterer exists for phase gradient estimation.

ex07-isar-imaging

Medium

Simulate ISAR imaging of a simplified ship:

(a) Model the ship as 20 point scatterers on a $100 \times 20$ m hull.

(b) The ship rolls at $\omega = 0.1$ rad/s. Simulate radar returns at $f_0 = 10$ GHz, $B = 500$ MHz, PRF $= 1000$ Hz, $T = 5$ s.

(c) Apply range alignment and dominant-scatterer phase adjustment.

(d) Form the ISAR image. Identify structural features.

(e) Add translational acceleration and show defocusing. Apply autofocus.

Show Hint

Use the brightest scatterer as the reference.

Solution

Range alignment

Cross-correlate consecutive range profiles to estimate and correct bulk range shifts.

Phase adjustment

Extract phase history of the brightest scatterer and subtract from all range bins.

Image formation

FFT along slow time after TMC. Cross-range axis calibrated by $y = \lambda f_D/(2\omega)$ .

ex08-tomosar-elevation

Hard

Implement TomoSAR elevation inversion:

(a) Construct the elevation sensing matrix $\mathbf{A}_{\text{elev}}$ for $M = 15$ non-uniform baselines drawn from $[0, 300]$ m, with $\lambda = 3.1$ cm and $R_0 = 600$ km.

(b) Generate measurements from 3 scatterers at elevations $z = -10, 0, 15$ m. Add noise at SNR = 15 dB.

(c) Implement beamforming: $\hat{\gamma}_{\text{BF}} = \mathbf{A}^{H} \mathbf{y}$ .

(d) Implement ISTA for $\ell_1$ -regularized recovery.

(e) Compare the two reconstructions. At what SNR does ISTA fail to resolve the scatterers?

Show Hint

The Rayleigh resolution $\Delta z = 31$ m cannot resolve the 3 scatterers; ISTA can.

Solution

Sensing matrix

$[\mathbf{A}]_{m,n} = \exp(-j4\pi b_m z_n/(\lambda R_0))$ .

ISTA recovery

Step size $\alpha = 1/\sigma_{\max}^2(\mathbf{A})$ . Regularization $\lambda \sim 0.1 \max|\mathbf{A}^{H} \mathbf{y}|$ . 200 iterations typically sufficient.

SNR threshold

ISTA typically resolves 3 scatterers down to SNR $\sim 5$ dB. Below that, the weakest scatterer is lost.

ex09-cs-sar

Hard

Implement CS-SAR with sub-Nyquist azimuth sampling:

(a) Generate full SAR data ( $N = 512$ pulses) for $S = 15$ point targets on a $128 \times 128$ grid.

(b) Sub-sample to $M = 128$ randomly selected pulses (75% reduction).

(c) Implement FISTA with the Kronecker-structured forward operator.

(d) Compare with zero-padded FFT and Tikhonov regularization.

(e) Vary $M \in \{32, 64, 128, 256\}$ and plot RMSE vs $M$ . Identify the phase transition.

Show Hint

ISTA step size: $\alpha = 1/\sigma_{\max}^2(\mathbf{A})$ .

Solution

Kronecker efficiency

$\mathbf{A}\mathbf{c}$ computed via $\text{vec}(\mathbf{A}_{\text{rg}} \cdot \text{mat}(\mathbf{c}) \cdot \mathbf{A}_{\text{az}}^T)$ — never form the full matrix.

Phase transition

Exact recovery typically occurs around $M \approx 4S\log N \approx 4 \times 15 \times 9 \approx 540$ . With $M = 128$ and $S = 15$ , recovery is near the transition.

ex10-joint-autofocus-sparse

Challenge

Combine autofocus and sparse reconstruction:

(a) Formulate the joint problem: $\min_{\mathbf{c}, \phi_e} \frac{1}{2}\|\mathbf{y} - \mathbf{D}(\phi_e)\mathbf{A}\mathbf{c}\|_2^2 + \lambda\|\mathbf{c}\|_1.$

(b) Derive an alternating minimization algorithm.

(c) Implement for 10 point targets, SNR = 20 dB, random phase error.

(d) Compare: (i) ISTA without autofocus, (ii) PGA then ISTA, (iii) joint alternating minimization.

(e) Analyze convergence and discuss identifiability conditions.

Show Hint

Initialize $\phi_e = 0$ and $\mathbf{c} = \mathbf{A}^{H}\mathbf{y}$ .

Phase gradient: $\text{Im}\{d_m^* r_m\}$ where $r_m$ is the residual.

Solution

Alternating minimization

Step 1: Fix $\phi_e$ , run ISTA for $\mathbf{c}$ . Step 2: Fix $\mathbf{c}$ , gradient descent on $\phi_e$ . Repeat 10--20 outer iterations.

Comparison

Joint outperforms sequential PGA+ISTA by 2--4 dB in RMSE because the sparsity prior regularizes both image and phase.

Identifiability

The joint problem is identifiable when the scene has sufficient diversity (not all targets at the same range) and the phase error bandwidth is lower than the scene bandwidth.

ex11-polsar

Medium

Implement polarimetric SAR decomposition:

(a) Simulate a scene with 3 scattering mechanisms: surface (HH+VV dominant), double-bounce (HH-VV dominant), volume (HV dominant).

(b) Form the Pauli decomposition RGB image.

(c) Implement group-LASSO ( $\ell_{2,1}$ norm) for joint recovery of all 4 polarization channels, exploiting shared support.

(d) Compare with independent $\ell_1$ recovery per channel.

Show Hint

Group LASSO: $\|\mathbf{c}\|_{2,1} = \sum_q \sqrt{|c_q^{HH}|^2 + |c_q^{HV}|^2 + |c_q^{VH}|^2 + |c_q^{VV}|^2}$ .

Solution

Pauli decomposition

Red: $|S_{HH} - S_{VV}|/\sqrt{2}$ (double-bounce). Green: $|S_{HV}|$ (volume). Blue: $|S_{HH} + S_{VV}|/\sqrt{2}$ (surface).

Group LASSO advantage

Group LASSO enforces common support across polarizations, reducing false detections by $\sim$ 30% compared to independent recovery.

ex12-sar-vs-multi-static

Medium

Compare SAR with multi-static RF imaging for the same scene:

(a) Set up a 2D scene with 10 scatterers. Configure: (i) a SAR system with 256 pulses at 10 GHz, 500 MHz BW, and (ii) a multi-static system with $N_t = 4$ , $N_r = 8$ at the same frequency and bandwidth.

(b) Construct $\mathbf{A}$ for both systems. Compare the SVD spectra.

(c) Compute matched-filter images for both. Compare resolution and sidelobe structure.

(d) Compute LASSO images for both. Which system gives better reconstruction with the same number of measurements?

Show Hint

SAR has structured (Kronecker/Fourier) $\mathbf{A}$ ; multi-static is less structured.

Solution

Sensing matrix comparison

SAR: $\mathbf{A} = \mathbf{A}_{\text{az}} \otimes \mathbf{A}_{\text{rg}}$ (Fourier). Multi-static: $\mathbf{A}$ has rows indexed by (Tx, Rx, freq) with target-position-dependent phases.

SVD comparison

SAR has flat singular values (good conditioning). Multi-static may have faster decay depending on geometry.

Reconstruction comparison

SAR excels for range-azimuth imaging; multi-static provides better angular diversity for 3D scenes.

ex13-sar-modes

Easy

Compare stripmap, spotlight, and ScanSAR for a spaceborne system:

(a) An X-band satellite at 600 km altitude has a 5 m antenna. Compute the stripmap cross-range resolution.

(b) In spotlight mode with $2\times$ longer dwell, compute the improved resolution.

(c) In ScanSAR mode with 4 sub-swaths, compute the degraded resolution and the total swath width.

(d) For each mode, compute the DOF per second of imaging time.

Show Hint

Spotlight dwell $2\times$ means $L_{\text{sa}}$ doubles.

Solution

Stripmap

$\Delta x = D/2 = 2.5$ m.

Spotlight

$L_{\text{sa,spot}} = 2L_{\text{sa,strip}}$ , so $\Delta x_{\text{spot}} = D/4 = 1.25$ m.

ScanSAR

$\Delta x_{\text{scan}} = 4 \times D/2 = 10$ m. Swath width $= 4 \times$ single swath.

ex14-diff-tomosar

Hard

Implement differential TomoSAR:

(a) Simulate 20 passes over an urban scene with 2 buildings. Each building has scatterers at ground level and rooftop, with the ground scatterer subsiding at 5 mm/year.

(b) Construct the joint elevation-deformation sensing matrix.

(c) Recover elevation and deformation rates using $\ell_1$ regularization.

(d) Compare with separate elevation-only TomoSAR (ignoring deformation). Show that ignoring deformation biases the elevation estimates.

Show Hint

Joint model: $y_{m,t} = \sum_k \sigma_k \exp(-j4\pi b_m z_k/(\lambda R_0) - j4\pi d_k t/\lambda)$ .

Solution

Joint sensing matrix

Discretize $(z, d)$ on a 2D grid. Each measurement contributes a row to the joint matrix encoding both baseline and temporal phase.

Sparse recovery

ISTA on the 2D grid. The deformation resolution depends on the temporal baseline span.

ex15-isar-cross-range-scaling

Medium

Implement cross-range scaling for ISAR:

(a) Simulate ISAR data for a ship rotating at unknown $\omega$ .

(b) Form the unscaled range-Doppler image (cross-range in Hz).

(c) Estimate $\omega$ using: (i) two sub-aperture correlation, (ii) minimum-entropy search over $\omega$ .

(d) Scale the cross-range axis to meters. Compare with the true target geometry.

Show Hint

Search $\omega$ in a range consistent with typical ship roll rates (0.01--0.2 rad/s).

Solution

Sub-aperture method

Split data into two halves, form images, cross-correlate to estimate angular shift, divide by half-observation time.

Minimum entropy

For each candidate $\omega$ , scale cross-range and compute image entropy. Choose the $\omega$ that minimizes entropy.

Exercises

ex01-sar-resolution-design

Synthetic aperture length

Dwell time

Doppler bandwidth

PRF

Bandwidth for range resolution

ex02-resolution-paradox

Maximum aperture derivation

$D = 10$ m

$D = 2$ m

Numerical example

ex03-rda-implementation

Range compression

RCMC

Azimuth compression

Resolution verification

ex04-rcmc-analysis

Dwell time

Range migration

RCMC criterion

ex05-pga-implementation

Phase error application

PGA iteration

Convergence

ex06-minimum-entropy

Entropy gradient

Comparison

ex07-isar-imaging

Range alignment

Phase adjustment

Image formation

ex08-tomosar-elevation

Sensing matrix

ISTA recovery

SNR threshold

ex09-cs-sar

Kronecker efficiency

Phase transition

ex10-joint-autofocus-sparse

Alternating minimization

Comparison

Identifiability

ex11-polsar

Pauli decomposition

Group LASSO advantage

ex12-sar-vs-multi-static

Sensing matrix comparison

SVD comparison

Reconstruction comparison

ex13-sar-modes

Stripmap

Spotlight

ScanSAR

ex14-diff-tomosar

Joint sensing matrix

Sparse recovery

ex15-isar-cross-range-scaling

Sub-aperture method

Minimum entropy