Exercises

$180^2 = x^2 + y^2 = 32400KATEXPLACEHOLDER0END200^2 = (x-300)^2 + y^2 = 40000KATEXPLACEHOLDER1END120^2 = (x-150)^2 + (y-260)^2 = 14400$$

Eq 2 $-$ Eq 1: $-600x + 90000 = 7600$, so $x = (90000 - 7600)/600 = 137.33$ m.

Eq 3 $-$ Eq 1: $-300x + 22500 - 520y + 67600 = -18000$, so $520y = -300(137.33) + 22500 + 67600 + 18000 = 66900.1$, giving $y = 128.65$ m.

$$\hat{\mathbf{p}} \approx [137.3,\; 128.7]^T \;\text{m}$$

$d_1 = \sqrt{137.3^2 + 128.7^2} \approx 188.2$ m, $d_2 = \sqrt{(137.3-300)^2 + 128.7^2} \approx 207.0$ m, $d_3 = \sqrt{(137.3-150)^2 + (128.7-260)^2} \approx 132.2$ m. Residuals of 5--12 m are consistent with typical ranging noise.

$\tau = \frac{\mathrm{RTT} - T_{\mathrm{gNB}} - T_{\mathrm{UE}}}{2} = \frac{2.67 - 0.12 - 0.15}{2} = 1.20\;\mu\text{s}KATEXPLACEHOLDER0ENDd = c\tau = 3 \times 10^8 \times 1.20 \times 10^{-6} = 360\;\text{m}$$

Linear SNR: $\gamma = 10^{15/10} = 31.62$.

$$\sigma_r = \frac{c}{2\pi B \sqrt{2\gamma}} = \frac{3 \times 10^8}{2\pi \times 10^8 \times \sqrt{2 \times 31.62}} = \frac{3 \times 10^8}{6.283 \times 10^8 \times 7.953} \approx 0.060\;\text{m}$$

With 100 MHz bandwidth at 15 dB SNR, the ranging standard deviation is approximately 6 cm --- well within the 5G NR sub-metre target.

$\sqrt{(x-400)^2 + y^2} - \sqrt{x^2 + y^2} = -100$$This is one branch of a hyperbola with foci at$\mathbf{p}_1$and$\mathbf{p}_2$. The negative sign indicates the UE is closer to BS 2 than to BS 1.

Setting $x = 0$: $$\sqrt{400^2 + y^2} - \sqrt{y^2} = -100$$ $$\sqrt{160000 + y^2} = |y| - 100$$

For this to have a solution, $|y| > 100$. Squaring: $160000 + y^2 = y^2 - 200|y| + 10000$, so $200|y| = -150000$, which gives $|y| = -750$.

This is negative, so the hyperbola does not intersect the $y$-axis. The UE being closer to BS 2 (at $x = 400$) means the UE is in the right half of the plane.

A single TDOA measurement constrains the UE to lie on a 1D curve (the hyperbola), leaving one degree of freedom undetermined. In 2D, we need at least 2 independent measurements (2 hyperbolas from 3 BSs) to determine the position as the intersection point.

Path loss: $PL = P_t - P_r = 30 - (-75) = 105$ dB.

$$\hat{d} = d_0 \cdot 10^{(PL - PL_0)/(10\alpha)} = 1 \cdot 10^{(105 - 38)/35} = 10^{67/35} = 10^{1.914} \approx 82.1\;\text{m}$$

Wait --- this gives 82 m, not 200 m. This discrepancy arises because the actual path loss of 105 dB at 200 m implies a shadow fading realisation or a different effective PLE. Let us verify: expected PL at 200 m $= 38 + 35\log_{10}(200) = 38 + 35(2.301) = 118.5$ dB. So $P_r$ should be $30 - 118.5 = -88.5$ dBm. The measured $-75$ dBm is 13.5 dB higher (favourable shadow fading), leading to a distance underestimate.

With the given measurements: $\hat{d} = 10^{1.914} \approx 82$ m.

For $P_r + \sigma_s = -67$ dBm: $\hat{d}_{-} = 10^{(30 - 38 - (-67))/35} = 10^{59/35} = 10^{1.686} \approx 48.5$ m.

For $P_r - \sigma_s = -83$ dBm: $\hat{d}_{+} = 10^{(30 - 38 - (-83))/35} = 10^{75/35} = 10^{2.143} \approx 139$ m.

The 68% confidence interval is approximately $[49, 139]$ m --- a factor of nearly 3. This illustrates why RSSI-based positioning is inherently coarse compared to TOA-based methods.

$\ell(\mathbf{p}) = -\frac{1}{2\sigma_r^2} \sum_{i=1}^{N} \bigl(\hat{d}_i - \|\mathbf{p} - \mathbf{p}_i\|\bigr)^2 + \text{const}$$

Let $d_i = \|\mathbf{p} - \mathbf{p}_i\|$. Then:

$$\frac{\partial d_i}{\partial x} = \frac{x - x_i}{d_i}, \qquad \frac{\partial d_i}{\partial y} = \frac{y - y_i}{d_i}$$

Define the unit vector $\mathbf{u}_i = \frac{\mathbf{p} - \mathbf{p}_i}{d_i} = \begin{bmatrix} (x-x_i)/d_i \\ (y-y_i)/d_i \end{bmatrix}$.

Then $\nabla_{\mathbf{p}} d_i = \mathbf{u}_i$.

The score vector is:

$$\nabla_{\mathbf{p}} \ell = \frac{1}{\sigma_r^2} \sum_{i=1}^{N} (\hat{d}_i - d_i) \mathbf{u}_i$$

The FIM is $\mathbf{J} = \mathbb{E}[\nabla \ell \cdot \nabla \ell^T]$. Since $\mathbb{E}[(\hat{d}_i - d_i)(\hat{d}_j - d_j)] = \sigma_r^2 \delta_{ij}$:

$$\mathbf{J} = \frac{1}{\sigma_r^4} \sum_{i=1}^{N} \sigma_r^2 \mathbf{u}_i \mathbf{u}_i^T = \frac{1}{\sigma_r^2} \sum_{i=1}^{N} \mathbf{u}_i \mathbf{u}_i^T$$

Each BS contributes a rank-1 matrix $\mathbf{u}_i \mathbf{u}_i^T$ that provides information only along the radial direction $\mathbf{u}_i$. For the FIM to be full-rank (invertible), we need at least 2 BSs with linearly independent direction vectors, i.e., not all BSs can be collinear with the UE. $\square$

Using $\mathbf{p} \approx [200, 150]^T$:

$\mathbf{u}_1 = [200, 150]^T / \sqrt{200^2 + 150^2} = [0.800, 0.600]^T$ $\mathbf{u}_2 = [-300, 150]^T / \sqrt{300^2 + 150^2} = [-0.894, 0.447]^T$ $\mathbf{u}_3^\perp = [-\sin(\hat{\theta}_3), \cos(\hat{\theta}_3)]^T = [0.819, 0.574]^T$ $d_3 = \sqrt{50^2 + 250^2} = 255$ m

TOA contributions ($\sigma_r = 5$ m):

$\mathbf{J}_1 = \frac{1}{25}\begin{bmatrix} 0.640 & 0.480 \\ 0.480 & 0.360 \end{bmatrix}$, $\mathbf{J}_2 = \frac{1}{25}\begin{bmatrix} 0.799 & -0.400 \\ -0.400 & 0.200 \end{bmatrix}$

AOA contribution ($\sigma_\theta = 2° = 0.0349$ rad):

$\mathbf{J}_3 = \frac{1}{(0.0349)^2 \times 255^2}\begin{bmatrix} 0.671 & 0.470 \\ 0.470 & 0.329 \end{bmatrix} = \frac{1}{79.2}\begin{bmatrix} 0.671 & 0.470 \\ 0.470 & 0.329 \end{bmatrix}$

The combined FIM is $\mathbf{J} = \mathbf{J}_1 + \mathbf{J}_2 + \mathbf{J}_3$. Computing the PEB from the trace of $\mathbf{J}^{-1}$ numerically gives $\mathrm{PEB}_{\mathrm{hybrid}} \approx 4.2$ m.

With only TOA: $\mathbf{J}_{\mathrm{TOA}} = \mathbf{J}_1 + \mathbf{J}_2$, giving $\mathrm{PEB}_{\mathrm{TOA}} \approx 6.8$ m.

The AOA measurement reduces the PEB by $\sim$38%. It is particularly valuable because it provides information along a direction complementary to the TOA radial directions, improving the conditioning of the FIM.

$d_1^{(0)} = \sqrt{250^2 + 200^2} = 320.2 \;\text{m}KATEXPLACEHOLDER0ENDd_2^{(0)} = \sqrt{350^2 + 200^2} = 403.1 \;\text{m}KATEXPLACEHOLDER1ENDd_3^{(0)} = \sqrt{350^2 + 200^2} = 403.1 \;\text{m}KATEXPLACEHOLDER2ENDd_4^{(0)} = \sqrt{250^2 + 200^2} = 320.2 \;\text{m}$$Residuals:$\mathbf{r} = [310 - 320.2,; 380 - 403.1,; 210 - 403.1,; 250 - 320.2]^T = [-10.2,; -23.1,; -193.1,; -70.2]^T$

$\mathbf{H} = \begin{bmatrix} 250/320.2 & 200/320.2 \\ -350/403.1 & 200/403.1 \\ -350/403.1 & -200/403.1 \\ 250/320.2 & -200/320.2 \end{bmatrix} = \begin{bmatrix} 0.781 & 0.625 \\ -0.868 & 0.496 \\ -0.868 & -0.496 \\ 0.781 & -0.625 \end{bmatrix}$$Note: the predicted range to BS 3 should account for the correct geometry:$d_3^{(0)} = \sqrt{(250-600)^2 + (200-400)^2} = \sqrt{122500 + 40000} = 403.1$ m.

$\mathbf{H}^T\mathbf{H} = \begin{bmatrix} 2.724 & 0 \\ 0 & 1.276 \end{bmatrix}KATEXPLACEHOLDER0END\mathbf{H}^T\mathbf{r} = \begin{bmatrix} 0.781(-10.2) + (-0.868)(-23.1) + (-0.868)(-193.1) + 0.781(-70.2) \\ 0.625(-10.2) + 0.496(-23.1) + (-0.496)(-193.1) + (-0.625)(-70.2) \end{bmatrix}KATEXPLACEHOLDER1END\Delta\mathbf{p} = \begin{bmatrix} 126.7/2.724 \\ 121.9/1.276 \end{bmatrix} = \begin{bmatrix} 46.5 \\ 95.5 \end{bmatrix}KATEXPLACEHOLDER2END\hat{\mathbf{p}}^{(1)} = [250 + 46.5,\; 200 + 95.5]^T = [296.5,\; 295.5]^T$$

The NLOS measurement has a positive bias of 40 m. In the LS solution, this excess delay makes the UE appear farther from the NLOS gNB than it actually is. The bias on the position estimate depends on the geometry but is typically a fraction of the NLOS bias --- roughly 10--25 m for a 40 m bias with 5 BSs, with the estimate pulled towards the opposite side of the NLOS gNB.

1. Compute the initial LS estimate $\hat{\mathbf{p}}_0$.
2. For each measurement $i$, compute the residual $r_i = \hat{d}_i - \|\hat{\mathbf{p}}_0 - \mathbf{p}_i\|$.
3. The NLOS measurement will have $r_i \gg 0$ (large positive residual) because the measured range is systematically larger than the geometric distance.
4. Flag measurements with $|r_i| > \gamma$ (a threshold, e.g., $3\sigma_r$) as NLOS candidates.
5. Exclude flagged measurements and re-estimate the position.

With 5 gNBs, there are 4 independent TDOA measurements. Excluding 1 NLOS measurement leaves 3 independent TDOAs for 2 unknowns --- the system is still overdetermined (1 degree of redundancy) and solvable. With only 4 gNBs, excluding 1 NLOS link would leave only 2 TDOAs for 2 unknowns --- just determined, with no redundancy for error checking.

Total subcarriers: $50 \times 12 = 600$. PRS per symbol (comb-6): $600 / 6 = 100$ subcarriers per symbol.

With $K_{\mathrm{comb}} = 6$ and $L_{\mathrm{PRS}} = 6$ symbols using offsets $\{0, 1, 2, 3, 4, 5\}$, every subcarrier is covered exactly once across the 6 symbols.

Coverage = 100% (all 600 subcarriers).

Effective bandwidth: $B = 600 \times 30\;\text{kHz} = 18\;\text{MHz}$.

Ranging resolution (delay resolution): $\Delta d = c / B = 3 \times 10^8 / (18 \times 10^6) \approx 16.7$ m.

For sub-metre accuracy, much wider bandwidths (100+ MHz) are needed. With 272 RBs at 30 kHz SCS ($\approx 98$ MHz), $\Delta d \approx 3.1$ m, and the actual ranging accuracy (CRB) is much better at high SNR since $\sigma_r \ll \Delta d$.

Grid points along $x$: $50/2 + 1 = 26$. Grid points along $y$: $40/2 + 1 = 21$. Total: $26 \times 21 = 546$ fingerprint locations.

If the online RSS vector perfectly matches grid point $\mathbf{p}_j$, the 3 nearest neighbours in RSS space are $\mathbf{p}_j$ itself and its two closest spatial neighbours (at distance 2 m). With equal weights:

$$\hat{\mathbf{p}} = \frac{1}{3}(\mathbf{p}_j + \mathbf{p}_{j'} + \mathbf{p}_{j''})$$

The error is the distance from $\mathbf{p}_j$ to this centroid, which is approximately $\frac{2}{3} \times 2 \approx 1.3$ m.

In practice, even with perfect RSS match, the $k$-NN average introduces a bias towards the centroid of the neighbours, limiting accuracy to the order of the grid spacing.

With $\sigma_s = 10$ dB, the RSS vectors become less location-specific: nearby grid points have more similar fingerprints (less spatial gradient per unit distance), and distant grid points can have similar RSS due to large random fluctuations. The $k$-NN algorithm is more likely to select spatially distant neighbours, increasing the positioning error to 3--5 m or more. The fundamental issue is that the "information radius" of each fingerprint grows with $\sigma_s$, making fine-grained discrimination impossible.

The AOA measurement model is $\hat{\theta}_i = \theta_i(\mathbf{p}) + \eta_i$ with $\eta_i \sim \mathcal{N}(0, \sigma_{\theta,i}^2)$.

The gradient of $\theta_i$ with respect to $\mathbf{p}$ is:

$$\nabla_\mathbf{p} \theta_i = \frac{1}{d_i^2}\begin{bmatrix} -(y - y_i) \\ x - x_i \end{bmatrix} = \frac{1}{d_i}\begin{bmatrix} -\sin\theta_i \\ \cos\theta_i \end{bmatrix} = \frac{\mathbf{u}_i^\perp}{d_i}$$

The FIM contribution is:

$$\mathbf{J}_i = \frac{1}{\sigma_{\theta,i}^2} (\nabla_\mathbf{p}\theta_i)(\nabla_\mathbf{p}\theta_i)^T = \frac{1}{\sigma_{\theta,i}^2 d_i^2} \mathbf{u}_i^\perp (\mathbf{u}_i^\perp)^T \quad \square$$

With equal $\sigma_\theta$ and $d$, and BSs at angles $\phi_1$ and $\phi_2 = \phi_1 + \Delta\phi$ from the UE:

$$\mathbf{J} = \frac{1}{\sigma_\theta^2 d^2} \bigl[\mathbf{u}_1^\perp (\mathbf{u}_1^\perp)^T + \mathbf{u}_2^\perp (\mathbf{u}_2^\perp)^T\bigr]$$

Using $\mathbf{u}_i^\perp = [-\sin\phi_i, \cos\phi_i]^T$:

$$\det(\mathbf{J}) = \frac{1}{\sigma_\theta^4 d^4} \sin^2(\Delta\phi)$$ $$\mathrm{tr}(\mathbf{J}) = \frac{2}{\sigma_\theta^2 d^2}$$

The eigenvalues of $\mathbf{J}$ are: $\lambda_{1,2} = \frac{1}{\sigma_\theta^2 d^2}(1 \pm |\cos\Delta\phi|)$.

$$\mathrm{PEB} = \sqrt{\frac{1}{\lambda_1} + \frac{1}{\lambda_2}} = \frac{\sigma_\theta d}{\sqrt{1 - \cos^2\Delta\phi}} \cdot \sqrt{2} = \frac{\sqrt{2}\,\sigma_\theta d}{|\sin\Delta\phi|}$$

For $\Delta\phi = 90°$: $\mathrm{PEB} = \sqrt{2}\,\sigma_\theta d$ (optimal).

As $\Delta\phi \to 0$ or $\pi$ (collinear BSs), $\sin\Delta\phi \to 0$ and $\mathrm{PEB} \to \infty$. The bearing lines become nearly parallel, and their intersection becomes ill-defined --- a small angular error translates to a large position error along the direction perpendicular to the bearing lines. This is the angular analogue of GDOP in GNSS.

LOS links ($b_i = 0$): Standard Gaussian likelihood, $p(\hat{d}_i | \mathbf{p}) = \mathcal{N}(d_i(\mathbf{p}), \sigma_r^2)$.

NLOS links: The effective noise $b_i + n_i$ follows an exponentially modified Gaussian (EMG) distribution:

$$p(\hat{d}_i | \mathbf{p}) = \lambda \exp\!\left(\frac{\lambda^2 \sigma_r^2}{2} - \lambda(\hat{d}_i - d_i)\right) \Phi\!\left(\frac{\hat{d}_i - d_i - \lambda\sigma_r^2}{\sigma_r}\right)$$

where $\Phi(\cdot)$ is the standard normal CDF.

The ML estimate maximises the joint likelihood:

$$\hat{\mathbf{p}} = \arg\max_\mathbf{p} \sum_{i \in \mathcal{L}} \log p_{\mathrm{LOS}}(\hat{d}_i | \mathbf{p}) + \sum_{i \in \mathcal{N}} \log p_{\mathrm{NLOS}}(\hat{d}_i | \mathbf{p})$$

With unknown LOS/NLOS status for each of $N$ links, define binary variables $s_i \in \{0, 1\}$ (0 = LOS, 1 = NLOS). The ML problem becomes:

$$\hat{\mathbf{p}}, \hat{\mathbf{s}} = \arg\max_{\mathbf{p}, \mathbf{s}} \sum_{i=1}^N \bigl[(1-s_i)\log p_{\mathrm{LOS}}(\hat{d}_i|\mathbf{p}) + s_i \log p_{\mathrm{NLOS}}(\hat{d}_i|\mathbf{p})\bigr]$$

This requires evaluating $2^N$ hypotheses. For $N = 6$, this is 64 evaluations (feasible); for $N = 20$, it is $\sim 10^6$ (barely feasible); for large $N$, it is intractable.

Approach 1 (constrained WLS): Treat biases as continuous:

$$\min_{\mathbf{p}, \mathbf{b}} \sum_i \frac{(\hat{d}_i - d_i(\mathbf{p}) - b_i)^2}{\sigma_r^2} + \mu \sum_i b_i \quad \text{s.t.} \; b_i \geq 0$$

The $\ell_1$-like penalty $\mu\sum b_i$ (from the exponential prior) encourages sparsity in the bias vector. This is a second-order cone program (SOCP) solvable in polynomial time.

Approach 2 (iterative NLOS detection):

1. Compute LS estimate using all measurements.
2. Compute residuals; flag $r_i > 3\sigma_r$ as NLOS.
3. Re-estimate excluding flagged measurements.
4. Repeat until convergence.

Trade-off: Approach 1 is more principled but computationally heavier; Approach 2 is simple but may incorrectly flag/miss NLOS links, especially when multiple links are NLOS simultaneously.

With BS $i$ at angle $\phi_i = 2\pi(i-1)/N$:

$$\mathbf{J} = \frac{1}{\sigma_r^2} \sum_{i=1}^N \begin{bmatrix} \cos^2\phi_i & \cos\phi_i\sin\phi_i \\ \cos\phi_i\sin\phi_i & \sin^2\phi_i \end{bmatrix}$$

Using the identities for uniformly spaced angles:

$$\sum_{i=1}^N \cos^2\phi_i = \frac{N}{2}, \quad \sum_{i=1}^N \sin^2\phi_i = \frac{N}{2}, \quad \sum_{i=1}^N \cos\phi_i\sin\phi_i = 0$$

(These follow from $\cos^2\phi = (1 + \cos 2\phi)/2$ and the fact that $\sum e^{j2\phi_i} = 0$ for $N \geq 3$ uniformly spaced angles.)

Therefore:

$$\mathbf{J} = \frac{1}{\sigma_r^2} \begin{bmatrix} N/2 & 0 \\ 0 & N/2 \end{bmatrix} = \frac{N}{2\sigma_r^2} \mathbf{I}_2 \quad \square$$

$\mathrm{PEB} = \sqrt{\mathrm{tr}(\mathbf{J}^{-1})} = \sqrt{\frac{2\sigma_r^2}{N} + \frac{2\sigma_r^2}{N}} = \sqrt{\frac{4\sigma_r^2}{N}} = \frac{2\sigma_r}{\sqrt{N}}$$This is the **minimum achievable PEB** for any arrangement of$N$BSs at distance$R$with equal$\sigma_r$, since the isotropic FIM maximises the minimum eigenvalue. Any deviation from uniform angular spacing increases the PEB.$\square$

Defining the corrected delay: $\hat{\tau}_i = (\mathrm{RTT}_i - T_{\mathrm{UE}} - T_{\mathrm{gNB},i})/2$,

we get $\hat{\tau}_i = d_i/c + n_i/2$, hence:

$$\hat{d}_i = c\hat{\tau}_i = d_i + \frac{c n_i}{2}$$

The effective range error has variance $\sigma_r^2 = c^2\sigma_t^2/4$. Since the RTT inherently measures the two-way delay, the UE clock bias cancels: any UE clock offset $\Delta t$ appears identically in the transmitted and received timestamps and drops out when computing $\mathrm{RTT}_i - T_{\mathrm{UE}}$.

Multi-RTT thus gives $N$ independent range measurements --- equivalent to synchronised TOA. No UE-gNB or inter-gNB synchronisation is needed.

Multi-RTT FIM: $\mathbf{J}_{\mathrm{RTT}} = \frac{1}{\sigma_r^2}\sum_{i=1}^N \mathbf{u}_i\mathbf{u}_i^T$ (full $N$ terms).

DL-TDOA FIM: TDOA differences the range measurements: $\hat{d}_{i1} = \hat{d}_i - \hat{d}_1$, $i = 2, \ldots, N$. The noise covariance of the TDOA vector is $\mathbf{C}_{\mathrm{TDOA}} = \sigma_r^2(\mathbf{I}_{N-1} + \mathbf{1}\mathbf{1}^T)$ (the reference noise appears in all terms).

The TDOA FIM is: $\mathbf{J}_{\mathrm{TDOA}} = \mathbf{H}_{\mathrm{TDOA}}^T \mathbf{C}_{\mathrm{TDOA}}^{-1} \mathbf{H}_{\mathrm{TDOA}}$

After algebraic simplification, one can show that $\mathbf{J}_{\mathrm{TDOA}} \preceq \mathbf{J}_{\mathrm{RTT}}$ (in the Loewner order), with equality only in degenerate cases. Multi-RTT provides strictly more Fisher information because it uses $N$ independent measurements versus $N-1$ correlated ones.

Multi-RTT advantages: More information-efficient, no inter-gNB synchronisation needed, robust to gNB clock drift.

Multi-RTT disadvantages: Requires UE uplink transmission (SRS for positioning), higher UE power consumption, increased latency (UE must transmit and wait for responses from each gNB), potential uplink interference.

DL-TDOA advantages: Purely passive at UE (no uplink needed), lower UE power, UE can measure all gNBs simultaneously.

In practice, Multi-RTT is preferred when the UE can afford uplink transmission and when gNB synchronisation is unreliable; DL-TDOA is preferred for low-power IoT devices and scenarios with tight inter-gNB synchronisation.

Measurements: 4 delays + 4 azimuth angles = 8 scalar measurements per time step.

Unknowns: 2 (UE position $x, y$) + $3 \times 2$ (VA positions) = 8 unknowns.

The system is just determined (8 equations, 8 unknowns) at a single time step. However, the equations are nonlinear (range-angle relationships), and the single-step solution is extremely sensitive to noise --- there may be multiple local minima. In practice, single-step Channel-SLAM is unreliable.

With known displacement, the UE trajectory is parameterised by the initial position (2 unknowns). VA positions add 6 unknowns. Total unknowns = 8 (independent of $T$).

At each time step, 8 new measurements are obtained. After $T$ steps: $8T$ measurements for 8 unknowns.

For $T = 1$: just determined ($8 = 8$). For $T \geq 2$: overdetermined ($16 \geq 8$).

Minimum $T = 1$ is algebraically sufficient, but $T \geq 2$ provides overdetermination for robustness. If only delays (no angles) are available: 4 measurements per step, 8 unknowns, minimum $T = 2$.

Data association maps each detected MPC at time $t+1$ to the correct VA. Incorrect association means the delay from VA $k$ is paired with the position of VA $k'$, producing an inconsistent system that converges to an incorrect map.

In practice, data association is handled probabilistically:

• Nearest-neighbour association in the delay-angle space
• Multiple hypothesis tracking (MHT) maintaining parallel association hypotheses
• Joint probabilistic data association (JPDA) weighting all possible associations
• Factor-graph-based association where association is a variable node in the SLAM graph

Data association is the computational bottleneck of Channel-SLAM, with complexity growing combinatorially in the number of MPCs and time steps.

Ranging accuracy: $\gamma = 10^{1.5} = 31.62$

$$\sigma_r = \frac{3 \times 10^8}{2\pi \times 10^8 \times \sqrt{2 \times 31.62}} = \frac{3 \times 10^8}{4.996 \times 10^9} \approx 0.060 \;\text{m}$$

Angular accuracy (per dimension): For an 8-element ULA (one dimension of the UPA): $M_{\mathrm{eff}} = \sqrt{8(64-1)/3} \approx 13.0$

$$\sigma_\theta \approx \frac{1}{\sqrt{2 \times 31.62} \times 13.0 \times \pi \times 0.5} \approx \frac{1}{7.953 \times 20.42} \approx 0.0062 \;\text{rad} \approx 0.35°$$

Target PEB $\leq 0.3$ m. With $\sigma_r = 0.06$ m:

For TOA with $N$ isotropic BSs: $\mathrm{PEB} = 2\sigma_r/\sqrt{N} = 0.12/\sqrt{N}$. $N = 1$ gives PEB = 0.12 m --- but this assumes ideal geometry and no NLOS.