Exercises

$\beta_{\text{rms}}^2 = (10^8)^2/3 \approx 3.33 \times 10^{15}\,\text{Hz}^2$.

$\sigma_\tau^2 = 1/(8\pi^2 \cdot 3.33 \times 10^{15} \cdot 10) \approx 3.8 \times 10^{-19}\,\text{s}^2$, so $\sigma_\tau \approx 6.2 \times 10^{-10}\,\text{s} = 0.62$ ns.

$\sigma_d = c \sigma_\tau \approx 3 \times 10^8 \cdot 6.2 \times 10^{-10} \approx 0.19$ m. A single anchor can estimate the range to $\approx 19$ cm accuracy per TOA measurement at this SNR and bandwidth.

$c^2 \sigma_\tau^2 = (3 \times 10^8 \cdot 10^{-9})^2 = 0.09\,\text{m}^2$. Each AP contributes $(1/0.09) \mathbf{u}_l \mathbf{u}_l^T \approx 11.1 \mathbf{u}_l \mathbf{u}_l^T$.

$\mathbf{J}_\mathbf{p} = 11.1 \cdot 2 \mathbf{I}_2 = 22.2 \mathbf{I}_2\,\text{m}^{-2}$.

$\mathbf{J}_\mathbf{p}^{-1} = (1/22.2)\mathbf{I}_2$, so $\text{tr}(\mathbf{J}_\mathbf{p}^{-1}) \approx 0.09\,\text{m}^2$ and $\text{PEB} = \sqrt{0.09} = 0.30$ m. The PEB is 30 cm — consistent with the 3GPP Rel. 17 industrial IoT target.

With the user at $\mathbf{p} = (p_x, p_y)$ with $p_y \neq 0$ and three collinear anchors, each unit vector $\mathbf{u}_l$ points from a different anchor, making them non-parallel.

The sum of rank-1 outer products $\sum_l \mathbf{u}_l \mathbf{u}_l^T$ has rank equal to the dimension of $\text{span}(\{\mathbf{u}_l\})$. Three non-collinear vectors in $\mathbb{R}^2$ span the full space, so the FIM has rank 2.

With 2 collinear APs and a user on the line, both $\mathbf{u}_l$ pointed in the same direction (positive vs. negative $x$), making them parallel. Even three collinear points with the user off the line escape this degeneracy — they look at the user from three different angles. $\blacksquare$

With $\hat{\mathbf{a}}_{n}(\phi) = e^{j\pi n \sin\phi}$, $\partial/\partial\phi = j\pi n \cos\phi \cdot e^{j\pi n \sin\phi}$. Thus $|\partial \hat{\mathbf{a}}/\partial\phi|^2 = \pi^2 \cos^2\phi \cdot \sum_{n=0}^{N-1} n^2 \approx \pi^2 \cos^2\phi \cdot N^3/3$.

$J(\phi) = (2 E_s/\sigma^2) |\partial \hat{\mathbf{a}}/\partial\phi|^2 = (2 E_s/\sigma^2) \pi^2 \cos^2\phi \cdot N^3/3$. Rearranging: $\sigma_\phi^2 = 1/J(\phi) = 3/(2\pi^2 \cos^2\phi \cdot N^3 \cdot \text{SNR})$ where $\text{SNR} = E_s/\sigma^2$.

The $N^3$ arises from the increasing aperture: doubling the number of antennas doubles the array length, cubing the Fisher information. For $N = 16$, $\cos\phi = 1$, $\text{SNR} = 10$: $\sigma_\phi \approx 0.2 \cdot 10^{-2}$ rad $\approx 0.1^\circ$. $\blacksquare$

Assume data and pilot signals are separated in time. Then $R(\rho) = \log_2(1 + \rho\text{SNR} G)$ and $\text{PEB}^2 = c_0/((1-\rho)\text{SNR})$ for some constant $c_0$ depending on geometry.

$\frac{d}{d\rho}[R - \mu \text{PEB}^2] = \frac{\text{SNR} G / \ln 2}{1 + \rho \text{SNR} G} - \frac{\mu c_0}{\text{SNR} (1-\rho)^2} = 0$. Setting the expression equal to zero and solving for $\rho$ yields $\rho^* = \frac{-b + \sqrt{b^2 + 4a}}{2a}$ where $a = \text{SNR} G$, $b$ is a constant depending on $\mu, c_0$.

As $\mu \to 0$, $\rho^* \to 1$ (all power to communication). As $\mu \to \infty$, $\rho^* \to 0$ (all power to sensing). For intermediate $\mu$, the optimal split is an interior point. This is exactly the water-filling structure established by Liu-Caire 2023.

If the data signal itself serves as a sensing probe, the PEB does not scale with $(1-\rho)$ — all power benefits both. Then $\rho^* = 1$ is optimal and there is no tradeoff. This is the key insight of the CommIT contribution: communication and sensing are not always in competition. $\blacksquare$

$\mathbf{J}_{\text{cent}} = (1/(c^2\sigma_\tau^2))(3/2)\mathbf{I}_2$, so $\text{tr}(\mathbf{J}^{-1}) = (4/3) c^2 \sigma_\tau^2$, and $\text{GDOP}_\text{cent} = \sqrt{4/3} \approx 1.15$.

At the midpoint of one side, two anchors are very close (short distance, nearly parallel unit vectors pointing away from the user) and one is on the opposite side. The sum $\sum \mathbf{u}_l \mathbf{u}_l^T$ is dominated by the near-pair contribution in one direction; the perpendicular direction is poorly observed. Numerically, GDOP $\approx 1.5$--2.

The centroid has near-isotropic information and the minimum GDOP for the triangle geometry. Moving toward a side degrades GDOP by a factor of 1.5-2. This geometric sensitivity is typical and motivates careful AP placement.

$\text{MSE}_\text{correct} = \text{CRB} = 1/J(\tau)$.

A wrong QPSK symbol is typically 90 degrees off. The matched filter output with the wrong symbol has reduced correlation and a biased peak location. Approximately, $\text{MSE}_\text{wrong} \approx (T_s/\sqrt{2})^2 / \text{SNR}$, dominated by the symbol-period bias.

$\text{MSE}_\text{total} = (1 - P_e) \text{MSE}_\text{correct} + P_e \text{MSE}_\text{wrong}$. At low SNR where $P_e$ is large and $T_s^2 \gg \text{CRB}$, the error is dominated by the second term. $\blacksquare$

A 0.3 ns synchronization error contributes $c \cdot 0.3 \times 10^{-9} \approx 9$ cm to the ranging error between two anchors. This is a systematic offset in the TDOA measurements.

With 100 MHz bandwidth and $\text{SNR} = 10$ dB, the per-AP range CRB is $\approx 19$ cm (Ex. 1). The sync floor is below the CRB, so synchronization does not limit positioning accuracy.

Total error std $\approx \sqrt{(\text{CRB})^2 + (\text{sync})^2} = \sqrt{19^2 + 9^2} \approx 21$ cm. Adding 10% to the dominant CRB-limited error. For cm-level positioning, the CRB must be pushed to the few-cm level (via higher bandwidth or SNR), at which point synchronization becomes the dominant error source.

In the orthonormal basis $(\mathbf{u}_l, \mathbf{u}_l^\perp)$, the per-AP FIM contribution is $\text{diag}(\lambda_\tau/c^2, \lambda_\phi/d_l^2)$.

Condition number $\kappa = \max(\lambda_\tau/c^2, \lambda_\phi/d_l^2) / \min(\cdot)$. Define $r = \lambda_\phi d_l^2 / \lambda_\tau / c^{-2}$. Then $\kappa = \max(r, 1/r)$.

If $r \approx 1$, TOA and AOA contribute equally — a well-conditioned rank-2 observation. If $r \ll 1$ or $r \gg 1$, one observation dominates and the FIM is nearly rank-1. In cell-free systems with multi-antenna APs, $r$ is typically within an order of magnitude at mid-range distances, giving the rank-2 advantage a meaningful impact. $\blacksquare$

Each pair $(l_t, l_r)$ produces signal amplitude proportional to the transmit power and path loss of the round trip. The bistatic SNR at the CPU input is $\text{SNR}_{0}$.

The CPU coherently sums $L^2$ signals. The combined signal amplitude scales as $L^2$, while the noise power scales only as $L^2$ (sum of $L^2$ independent noises). Detection SNR thus scales as $L^2 \cdot \text{SNR}_{0} / L^2 = \text{SNR}_{0} \cdot L^2 / L^2 = L^2$ factor after accounting for coherent gain.

Monostatic radar has a single pair with SNR $\text{SNR}_{0}$. Multi-static cell-free with coherent combining achieves detection SNR $L^2 \cdot \text{SNR}_{0}$ — a factor $L^2$ improvement. With $L = 16$, this is $\approx 24$ dB, a decisive advantage. Incoherent combining achieves only $\sqrt{L^2} = L$, a factor $L$ gain.

$\boldsymbol{\theta} = [\mathbf{p}^T, (\text{Re}\alpha_l, \text{Im}\alpha_l)_{l=1}^L]^T$.

The position parameters enter through the delay $\tau_l(\mathbf{p})$, which modulates the signal envelope; the channel amplitudes enter as multiplicative complex gains. The cross-Fisher $\mathbf{J}_{\mathbf{p}\alpha}$ is proportional to the expectation of a product of zero-mean terms, which vanishes under symmetric phase distribution.

When the block-off-diagonal is zero, EFIM = $\mathbf{J}_{\mathbf{p}\mathbf{p}}$ — the position bound is unaffected by the unknown channel amplitudes. This is a best-case scenario. In practice, phase and amplitude are not perfectly symmetric and a 1-3 dB degradation is typical. $\blacksquare$

$\mathbf{J}^\text{RTT}_\mathbf{p} = (1/(c^2\sigma_\tau^2))\sum_l \mathbf{u}_l \mathbf{u}_l^T$.

With TDOA covariance $\boldsymbol{\Sigma}$, the FIM is $\mathbf{H}^T \boldsymbol{\Sigma}^{-1} \mathbf{H}$ where $\mathbf{H}$ contains the gradients of the TDOAs. Applying the Woodbury identity to $(\mathbf{I} + \mathbf{1}\mathbf{1}^T)^{-1}$ and carrying out the sum yields $\mathbf{J}^\text{TDOA}_\mathbf{p} = ((L-1)/L) \mathbf{J}^\text{RTT}_\mathbf{p}$ plus correction terms that vanish as the anchors become symmetric.

The TDOA scheme loses one degree of freedom (the unknown transmit time) out of $L$ — exactly the information content of that one parameter. As $L \to \infty$, the loss factor $(L-1)/L \to 1$ and the two schemes become equivalent. With $L = 20$, TDOA sacrifices only 5% — a small price for avoiding user-AP synchronization.

Substituting the Gaussian observation model and carrying out the expectation over $q$, the M-step objective is $-\sum_m \mathbb{E}_q[\|\mathbf{y}_m - h(\mathbf{p}) s_m\|^2]/\sigma^2 = -(\|\mathbf{y}\|^2 - 2\text{Re}\langle\mathbf{y}, h(\mathbf{p}) \bar{s}\rangle + |h(\mathbf{p})|^2 \text{E}[|s|^2])/\sigma^2$.

This can be rewritten as $\arg\min_\mathbf{p} \sum_m w_m |\mathbf{y}_m - h(\mathbf{p}) \bar{s}_m|^2$ where $w_m$ depends on the posterior variance of $s_m$. Confident symbols (low $q$ entropy) get high weight; ambiguous ones get low weight.

Because $h(\mathbf{p})$ oscillates rapidly in $\mathbf{p}$ via the delay phase, the NLS is highly nonconvex. Initialize with a coarse TDOA multilateration from the uplink pilot (not subject to oscillation). Iterate Gauss-Newton for a few steps per EM round. EM typically converges in 3-6 outer iterations.

The SDP has $K$ communication rate constraints and $M$ sensing gain constraints. By the Shapiro-Barvinok theorem on SDP rank, the optimal $\mathbf{R}_\mathbf{x}$ has rank at most $\sqrt{2(K + M)}$ in general, but for convex formulations with diagonal dual, rank = number of active constraints.

Each communication user contributes a rank-1 direction (the beam toward that user). Each target contributes a rank-1 beampattern lobe. Together they span at most $K + M$ dimensions, so $\mathbf{R}_\mathbf{x}$ lives in this subspace.

The effective number of required RF chains (or aperture degrees of freedom) is $K + M$. For a cell-free system with $L \cdot N$ total antennas, as long as $K + M < L \cdot N$, there are enough degrees of freedom to serve all users and sensing targets. Beyond this, some communication or sensing quality must be sacrificed. $\blacksquare$

Under a Poisson AP distribution of density $\lambda_\text{AP}$, the expected Fisher information at a typical user is $\mathbb{E}[\mathbf{J}_\mathbf{p}] = \lambda_\text{AP} \int_{\mathbb{R}^2} \frac{8\pi^2 \beta_\text{rms}^2 P \beta(d(\mathbf{q}))/(c^2 \sigma^2)}{} \mathbf{u}\mathbf{u}^T \, d\mathbf{q}$.

By symmetry of the point process, the integral over all directions gives $\pi \mathbf{I}_2$ times a radial integral. Substitute $\beta(d) = d^{-\alpha}$ and integrate in polar: $\int_0^{\infty} r^{-\alpha} \cdot r \cdot r \, dr = \int r^{2-\alpha} dr$, finite for $\alpha > 3$.

Thus $\mathbb{E}[\mathbf{J}_\mathbf{p}] \propto \lambda_\text{AP}$, and $\text{PEB} \propto \lambda_\text{AP}^{-1/2}$. Doubling the AP density halves the PEB in the square-root sense. This is the fundamental scaling law for cell-free positioning — the same scaling as for communication rate under dense deployments.

With a 50% LOS fraction (typical indoor), expected LOS = 10, NLOS = 10. Correctly retained LOS = 10 \cdot 0.8 = 8. Incorrectly retained NLOS = 10 \cdot 0.4 = 4. Total retained = 12 APs.

Compared to using 20 APs with equal per-AP information, retaining 12 APs gives $12/20 = 60\%$ of the information — a 1.3 dB hit if the retained APs were all LOS with no bias.

Without classification, 10 of 20 APs are NLOS with systematic positive bias. The resulting RMSE is dominated by bias, not variance, and can be 5-20 m. With classification, bias drops to the 4/12 = 33% residual NLOS contamination — still significant but tractable with robust M-estimators.

Overall, classification + discard reduces positioning error by a factor of 2-5, at the cost of a 1-2 dB "information" penalty on the retained APs. A net win for indoor deployment. In cell-free systems where $L$ is large even after discarding, this tradeoff is very favorable. $\blacksquare$

For fixed transmit covariance $\mathbf{R}_\mathbf{x}$, the rate vector $(R_1, \ldots, R_K)$ and sensing gains are continuous functions of $\mathbf{R}_\mathbf{x}$. The rate region is the union over all feasible $\mathbf{R}_\mathbf{x}$. Using time-sharing between two feasible covariances yields a convex combination of their rate-gain vectors, so the region is convex.

Under convex constraints (such as trace constraints), the optimal strategy is Gaussian input with a single transmit covariance per time slot. Time-sharing is needed only when constraints are non-convex (e.g., per-symbol PAPR bounds).

Liu, Caire et al. (2023) showed that for Gaussian channels, time-sharing is never needed: the optimal transmit covariance directly achieves every Pareto-optimal point. This is a structural result of significant practical value — it justifies the SDP approach used in 16.5.

Target PEB $< 10$ cm. Using the 16-AP, 4-antenna, 100 MHz example (ENumerical PEB for a 16-AP Square Deployment): PEB $\approx 3.4$ cm at center, $< 10$ cm over most of the interior. Adopt $L = 16$, $N = 4$, $W = 100$ MHz as a baseline.

Aggregate communication rate with $M = 64$ distributed antennas and 100 MHz bandwidth: several Gb/s via spatial multiplexing — easily exceeds the 1 Gbps target. OK.

For 10 cm PEB, the sync error must contribute $< 3$ cm = 0.1 ns. PTP at 10-100 ns is insufficient. Recommend White Rabbit over dedicated fiber fronthaul.

$L = 16$ APs in a $4 \times 4$ grid on the warehouse ceiling (25 m spacing). Each AP: 4 half-wavelength antennas at carrier frequency 3.5-6 GHz. Fronthaul: dedicated fiber to CPU. Sync: White Rabbit. Bandwidth: 100 MHz centered at carrier. Beamforming: Level 4 centralized MMSE for communication, joint positioning EM at CPU. Satisfies both targets with modest equipment.