Exercises

$d(\mathbf{s}, \mathbf{p}) \approx d(\mathbf{s}, \mathbf{p}_{0}) + \frac{(\mathbf{p}_{0} - \mathbf{s})^\mathsf{T}}{d(\mathbf{s}, \mathbf{p}_{0})}(\mathbf{p} - \mathbf{p}_{0}).$$Similarly for$d(\mathbf{p}, \mathbf{r})$.

The round-trip phase becomes $\kappa(d(\mathbf{s}, \mathbf{p}_{0}) + d(\mathbf{p}_{0}, \mathbf{r})) + (\boldsymbol{\kappa}_s + \boldsymbol{\kappa}_r)^\mathsf{T}(\mathbf{p} - \mathbf{p}_{0})$. The first term factors out as a known constant $\alpha$. The remaining integral over $\mathbf{p}$ is the Fourier transform $\tilde{c}(\kappa_{\mathbf{s},\mathbf{r}})$.

The unit vector from Tx toward target is $\hat{\mathbf{s}} = (\cos\phi, \sin\phi)$. Then $\boldsymbol{\kappa}_s = \kappa(\cos\phi, \sin\phi)$ and $\boldsymbol{\kappa}_r = \boldsymbol{\kappa}_s$.

$\kappa_{\mathbf{s},\mathbf{r}} = 2\kappa(\cos\phi, \sin\phi)$. As $\phi$ varies, this traces a circle of radius $2\kappa$ in k-space.

A signal with rectangular spectrum $[-W/2, W/2]$ has autocorrelation $R(\tau) = W\,\text{sinc}(W\tau)$, which has its first zero at $\tau = 1/W$. The two-way delay resolution is $\Delta\tau = 1/W$, giving $\Delta r = \text{c}/(2W)$.

The radial k-space extent is $\Delta\kappa = 4\piW/\text{c}$. The spatial resolution is $\Delta r = 2\pi/\Delta\kappa = \text{c}/(2W)$. Both approaches agree.

$\boldsymbol{\kappa}_s = \kappa(1, 0)$, $\boldsymbol{\kappa}_r = \kappa(0, 1)$. $\kappa_{\mathbf{s},\mathbf{r}} = \kappa(1, 1)$.

$\|\kappa_{\mathbf{s},\mathbf{r}}\| = \kappa\sqrt{2} \approx 1.41\kappa$. This is less than the monostatic $2\kappa$, so bistatic pairs sample lower spatial frequencies.

The monostatic case accesses $\|\kappa_{\mathbf{s},\mathbf{r}}\| = 2\kappa$ (outer ring). The 90-degree bistatic pair accesses $\|\kappa_{\mathbf{s},\mathbf{r}}\| = \sqrt{2}\kappa$ (intermediate radius). Points at $\|\boldsymbol{\kappa}\| > \sqrt{2}\kappa$ are inaccessible to this bistatic pair.

$m = 1: (i=1, j=1)$, $m = 2: (i=1, j=2)$, $m = 3: (i=2, j=1)$, $m = 4: (i=2, j=2)$.

$\mathbf{A} = \begin{bmatrix} \alpha_{11} e^{-j\boldsymbol{\kappa}_{11}^\mathsf{T}\mathbf{p}_{1}} & \alpha_{11} e^{-j\boldsymbol{\kappa}_{11}^\mathsf{T}\mathbf{p}_{2}} & \alpha_{11} e^{-j\boldsymbol{\kappa}_{11}^\mathsf{T}\mathbf{p}_{3}} & \alpha_{11} e^{-j\boldsymbol{\kappa}_{11}^\mathsf{T}\mathbf{p}_{4}} \\ \alpha_{12} e^{-j\boldsymbol{\kappa}_{12}^\mathsf{T}\mathbf{p}_{1}} & \cdots & \cdots & \cdots \\ \alpha_{21} e^{-j\boldsymbol{\kappa}_{21}^\mathsf{T}\mathbf{p}_{1}} & \cdots & \cdots & \cdots \\ \alpha_{22} e^{-j\boldsymbol{\kappa}_{22}^\mathsf{T}\mathbf{p}_{1}} & \cdots & \cdots & \alpha_{22} e^{-j\boldsymbol{\kappa}_{22}^\mathsf{T}\mathbf{p}_{4}} \end{bmatrix}$$where$\alpha_{ij}$contains the distance-dependent attenuation and phase, and$\boldsymbol{\kappa}{ij} = \boldsymbol{\kappa}{s,i} + \boldsymbol{\kappa}_{r,j}$. This is a partial DFT-like matrix with non-uniform frequency sampling.

Minimizing $\|\mathbf{y} - \mathbf{A}\mathbf{c}\|^2$ gives $\mathbf{A}^{H}\mathbf{A}\hat{\mathbf{c}} = \mathbf{A}^{H}\mathbf{y}$.

If $\mathbf{A}^{H}\mathbf{A} = \alpha\mathbf{I}$, then $\hat{\mathbf{c}} = (1/\alpha)\mathbf{A}^{H}\mathbf{y}$, which is BP up to a scalar. In practice $\mathbf{A}^{H}\mathbf{A} \neq \alpha\mathbf{I}$, so BP is only an approximation to the ML estimate.

$\tilde{c}(\kappa_x, \kappa_y) = e^{-\pi(\kappa_x^2 + \kappa_y^2)/(4\pi^2)} = e^{-(\kappa_x^2 + \kappa_y^2)/(4\pi)}$.

$\boldsymbol{\kappa}_s = \kappa(0, 1)$, $\boldsymbol{\kappa}_r = \kappa(-1, 0)$. $\kappa_{\mathbf{s},\mathbf{r}} = \kappa(-1, 1)$.

$x(\mathbf{s}, \mathbf{r}; f_0) = \alpha \cdot \tilde{c}(\kappa_{\mathbf{s},\mathbf{r}}) = \alpha \cdot e^{-\kappa^{2} \cdot 2/(4\pi)} = \alpha \cdot e^{-\kappa^{2}/(2\pi)}$.

$M \times N = 9$ pairs (3 monostatic + 6 bistatic).

$MNK = 9 \times 16 = 144$.

$MNK = 144 \ll Q = 16{,}384$. The problem is severely underdetermined — we have far fewer measurements than unknowns. This motivates regularization and sparsity-based reconstruction.

$(\mathcal{E}/N_0)_{\text{dB}} \geq 0 + 20\log_{10}(20) + 20\log_{10}(20) - 0 - 0 = 26 + 26 = 52$ dB.

$(\mathcal{E}/N_0)_{\text{dB}} \geq 0 + 52 - 30 - 30 = -8$ dB. With arrays, the required transmit energy is 60 dB lower — the arrays provide enormous gains that make imaging feasible at practical power levels.

$\Delta\theta \leq \lambda/(2L) = 0.03/(2 \times 16) = 9.375 \times 10^{-4}$ rad $\approx 0.054°$. Over $360°$, we need $N_{\text{angular}} = 2\pi/\Delta\theta \approx 6{,}702$ directions. This is impractically many for a static deployment.

With 6 directions over $360°$, $\Delta\theta = 60°$. Then $L_{\max} = \lambda/(2\Delta\theta) = 0.03/(2 \times \pi/3) = 0.0143$ m $= 1.4$ cm. This is tiny — in practice, the angular sampling is always far below Nyquist for room-scale scenes.

When angular sampling is below Nyquist, grating lobes appear as ghost images. Regularization (sparsity, total variation, learned priors) suppresses these artifacts by enforcing that the reconstructed image belongs to a restricted class of "reasonable" scenes. This is the main reason why classical backpropagation performs poorly and regularized methods are essential in RF imaging.

$[\hat{\mathbf{c}}^{\text{BP}}]_q = \sum_m [\mathbf{A}]^*_{m,q} \, y_m$. The $(m,q)$ entry of $\mathbf{A}$ is $\alpha_m \cdot e^{-j\boldsymbol{\kappa}_m^\mathsf{T}\mathbf{p}_{q}}$ where $\alpha_m$ absorbs attenuation and phase-to-center terms. Therefore $[\mathbf{A}]^*_{m,q} = \alpha_m^* \cdot e^{+j\boldsymbol{\kappa}_m^\mathsf{T}\mathbf{p}_{q}}$.

$[\hat{\mathbf{c}}^{\text{BP}}]_q = \sum_m \alpha_m^* \, y_m \, e^{+j\boldsymbol{\kappa}_m^\mathsf{T}\mathbf{p}_{q}}.$$This is a non-uniform inverse Fourier transform: we evaluate the weighted sum$\sum_m \tilde{y}_m , e^{+j\boldsymbol{\kappa}m^\mathsf{T}\mathbf{p}{q}}$where$\tilde{y}_m = \alpha_m^* y_m$ is the compensated measurement. This confirms that BP (View B) = non-uniform inverse FFT of calibrated k-space data (View A).

Full coverage: $\text{PSF}(x) = \sum_{n=-N}^{N} e^{jn\Delta\kappa x} = \text{diric}(x, 2N+1)$ (Dirichlet kernel).

Missing samples: $\text{PSF}_{\text{gap}}(x) = \text{PSF}_{\text{full}}(x) - \sum_{n \in \{-4,-3,3,4\}} e^{jn\Delta\kappa x}$. The subtracted terms create additional sidelobes that are periodic with spacing $2\pi/\Delta\kappa$.

The mainlobe width is determined by the total extent $2N\Delta\kappa$ (unchanged), so Rayleigh resolution is preserved. However, the sidelobes are elevated, which may mask weak scatterers near strong ones. The gap degrades dynamic range, not resolution per se.

The angular aperture is $\theta_{\max} \approx D/R = 1/50 = 0.02$ rad. $\Delta x = \lambda/(2\theta_{\max}) = 0.03/(2 \times 0.02) = 0.75$ m.

$\Delta x = \lambda/(2\sin(15°)) = (3 \times 10^8/(28 \times 10^9))/(2 \times 0.259) = 0.0107/(0.518) = 0.0207$ m $\approx 2.1$ cm.

$\nabla_{\mathbf{h}} [\|\mathbf{y} - \mathbf{A}\mathbf{h}\|^2 + \lambda\|\mathbf{h}\|^2] = -2\mathbf{A}^{H}(\mathbf{y} - \mathbf{A}\mathbf{h}) + 2\lambda\mathbf{h} = 0$. Solution: $\hat{\mathbf{c}} = (\mathbf{A}^{H}\mathbf{A} + \lambda\mathbf{I})^{-1}\mathbf{A}^{H}\mathbf{y}$.

The $\ell_1$ penalty makes the problem non-smooth. No closed-form exists; iterative algorithms (ISTA, FISTA, ADMM from Ch 04) are needed. If the scene is sparse ($\mathbf{c}$ has few nonzero entries), LASSO promotes sparsity and can super-resolve beyond the diffraction limit.

Place 4 nodes at corners of a 10 m square. Each node is both Tx and Rx, giving $M = N = 4$ (16 pairs: 4 monostatic + 12 bistatic). With $K = 8$ subcarriers, $MNK = 128$.

For each pair $(i,j)$ and subcarrier $k$, compute $\boldsymbol{\kappa}_{i,k}$ and $\hat{\boldsymbol{\kappa}}_{j,k}$ using the formulas from the chapter, then sum to get $\boldsymbol{\kappa}_{i,j,k}$.

The PSF is the sum of 128 complex exponentials evaluated on the grid. The $-3$ dB mainlobe widths should approximately match: range $\sim 0.75$ m, cross-range $\sim 2.1$ cm. Deviations arise from the irregular k-space sampling and finite number of samples.

With $L$ i.i.d. snapshots $\mathbf{y}_\ell = \mathbf{A}\mathbf{c} + \mathbf{w}_\ell$, the averaged observation is $\bar{\mathbf{y}} = \mathbf{A}\mathbf{c} + \bar{\mathbf{w}}$ where $\bar{\mathbf{w}} \sim \mathcal{CN}(\mathbf{0}, \sigma^2/L \cdot \mathbf{I})$.

The effective SNR improves by a factor of $L$ (or $10\log_{10}(L)$ dB). This is the standard coherent integration gain. In practice, $L$ corresponds to the number of pilot OFDM symbols per time slot, and typical values are $L = 16$-$64$ for 5G NR systems.

From DThe Sensing Matrix, the $q$-th column of $\mathbf{A}_{k}$ is $\mathbf{a}_{k,q} \otimes \mathbf{b}_{k,q}$. Collecting these: $\mathbf{A}_{k} = [\mathbf{a}_{k,1} \otimes \mathbf{b}_{k,1}, \ldots, \mathbf{a}_{k,Q} \otimes \mathbf{b}_{k,Q}]$.

Defining $\mathbf{A}_k = [\mathbf{a}_{k,1}, \ldots, \mathbf{a}_{k,Q}]$ and $\mathbf{B}_k = [\mathbf{b}_{k,1}, \ldots, \mathbf{b}_{k,Q}]$, we recognize $\mathbf{A}_{k} = \mathbf{A}_k \odot \mathbf{B}_k$ (Khatri-Rao product).

When the voxel grid is separable and the arrays have uniform geometry, $\mathbf{A}_k$ and $\mathbf{B}_k$ themselves have DFT-like structure, and $\mathbf{A}_{k}$ inherits a partial Kronecker structure. Ch 07 develops this in detail, showing how the full sensing matrix factors as a Kronecker product of frequency, Rx, and Tx components when the geometry is fully separable.