Exercises

Peak power: $p_{k,64} = 1$. Threshold: $\exp(-|m-64|/20) \geq 0.1$.

$-|m-64|/20 \geq \ln(0.1) = -2.303$, so $|m-64| \leq 20 \cdot 2.303 \approx 46.05$. Therefore $m \in [18, 110]$, yielding $|\mathcal{V}_k| = 110 - 18 + 1 = 93$ elements.

$|\mathcal{V}_k|/N_t = 93/128 \approx 0.73$. The user illuminates $73$ percent of the aperture — on the higher end of what visibility-region measurements typically show. $\blacksquare$

$\Delta_k(0.3) = -5.23$ dB, $\Delta_k(0.5) = -3.01$ dB, $\Delta_k(0.7) = -1.55$ dB.

The loss exceeds $3$ dB precisely at $V_k/N_t \leq 10^{-0.3} = 0.501$. Measurement campaigns typically find $V_k/N_t \in [0.4, 0.6]$, placing most practical systems near or below the $3$ dB threshold. $\blacksquare$

Centralized: $N^2 K = (400)^2 \cdot 40 = 6.4 \times 10^6$ FLOPs. Consensus: $N_{\text{iter}} \cdot 100 \cdot 16 \cdot 40 = 64000 N_{\text{iter}}$.

$64000 N_{\text{iter}} = 6.4 \times 10^6$, so $N_{\text{iter}} = 100$. Equivalent cost at $N_{\text{iter}} = 100$ iterations — but consensus rarely needs more than $10$-$20$ iterations in practice, so it is typically $5$-$10\times$ cheaper. $\blacksquare$

Required $\xi = 30 - (-100) = 130$ dB. Already provided: $40 + 32 + 34 = 106$ dB. Remaining: $130 - 106 = 24$ dB from spatial nulling.

Spatial null from $N_t - K_{DL}$ free DoF gives approximately $10\log_{10}(N_t - K_{DL})$ dB. Setting this $\geq 24$ gives $N_t - K_{DL} \geq 10^{2.4} \approx 251$. With $N_t = 64$, this is impossible: the free DoF is at most $63$, giving at most $18$ dB of spatial cancellation. The conclusion is that a $64$-antenna FD system cannot close against a $-95$ dBm noise floor with $5$ dB margin using only the four-stage cascade; either more antennas or better analog/digital cancellation is needed. $\blacksquare$

$\lambda = 3 \times 10^8 / 6 \times 10^9 = 0.05$ m.

$L_F(5) = \sqrt{0.05 \cdot 5} = 0.5$ m. $L_F(50) = \sqrt{0.05 \cdot 50} \approx 1.58$ m. $L_F(500) = \sqrt{0.05 \cdot 500} = 5$ m.

At $d = 5$ m: $L = L_F = 0.5$ m — borderline near field. At $d = 50$ m: $L = 0.5 < L_F = 1.58$ — transitional. At $d = 500$ m: $L \ll L_F = 5$ — far field. The $(L/L_F)^4$ holographic gain is relevant only for the first case ($d = 5$ m), which is characteristic of indoor or short-range outdoor hotspots. $\blacksquare$

$\lambda = 0.0107$ m. Free-space loss over 50 m at 28 GHz: $\beta_{1} = 20\log_{10}(4\pi \cdot 50 / 0.0107) = 95.4$ dB. Same for $\beta_{2}$.

Combined: $\beta_{1} + \beta_{2} = 190.8$ dB (ignoring coherent gain and element gains).

$N_{\text{RIS}}^2 = 160000$, i.e. $52.0$ dB of coherent combining. Element gains: $2 \cdot G_{\text{element}} = 4$ dB. Tx array gain: $10\log_{10}(64) = 18$ dB. Net effective loss: $190.8 - 52.0 - 4 - 18 = 116.8$ dB.

$P_{\text{Rx}} = 30 - 116.8 = -86.8$ dBm. Compared to a $-95$ dBm noise floor, this gives $\sim 8$ dB of SINR — usable for a low-rate control channel but marginal for high-throughput data. To reach acceptable data rates, $N_{\text{RIS}}$ would need to grow or the geometry must shorten. $\blacksquare$

User 1: $\log_2(1/0.5) = 1.00$ bits/s/Hz. User 2: $\log_2(1/0.3) \approx 1.74$ bits/s/Hz.

Total loss: $\approx 2.74$ bits/s/Hz at high SNR. Not catastrophic on its own but a meaningful tax that a VR-aware combiner would avoid. Over a $100$ MHz channel, $\approx 274$ Mbps of throughput. $\blacksquare$

Centralized: $\sim N^2 K = N_{\text{AP}}^2 M^2 K$. Consensus: $N_{\text{iter}} N_{\text{AP}} M^2 K$.

Equating: $N_{\text{iter}}^\star N_{\text{AP}} M^2 K = N_{\text{AP}}^2 M^2 K$, so $N_{\text{iter}}^\star = N_{\text{AP}}$.

Consensus remains cheaper as long as the iteration count stays below the number of APs. In practice with $N_{\text{iter}} \approx 5$-$20$, this is always satisfied for $N_{\text{AP}} \gtrsim 20$. The risk is that performance-driven $N_{\text{iter}}$ might need to scale with $N_{\text{AP}}$ on poorly connected graphs, eliminating the advantage. $\blacksquare$

$(L/L_F)^4 = L^4 / L_F^4 = L^4 / (\lambda\,d)^2 = L^4/(\lambda^{2} d^2)$.

Doubling $L$: DoF scales by $2^4 = 16\times$. This is the headline holographic result.

Doubling $f_0$ halves $\lambda$, so $\lambda^{2}$ shrinks by $4$, giving $4\times$ more DoF.

Halving $d$: $d^2$ shrinks by $4$, giving $4\times$ more DoF.

The aperture size is by far the most effective knob; the power-of-4 dependence of DoF on $L$ is why "very large surface" proposals dominate the holographic MIMO literature. $\blacksquare$

Capex: $10 \cdot \$1000 = \$10000$. Opex (fronthaul): $10 \cdot \$200 \cdot 60 = \$120000$. Total: $\$130000$.

Capex: $2 \cdot \$1000 + 10 \cdot \$50 = \$2500$. Opex: $2 \cdot \$200 \cdot 60 = \$24000$. Total: $\$26500$.

Option A covers $100$ percent of the service area. Option B covers: the two active APs' intrinsic coverage, plus RIS-assisted coverage for $0.6 \cdot 0.3 = 18$ percent of the area (the blocked-LOS portion that RIS can restore). Under uniform distribution, Option B covers roughly $(2/10) + 0.6 \cdot 0.3 = 0.38$ of the area relative to Option A — $38$ percent of the capacity.

Option A: $\$130000 / 1.0 = \$130000$ per unit capacity. Option B: $\$26500 / 0.38 \approx \$69700$ per unit capacity. Option B wins on cost per capacity by nearly $2\times$, but provides only $38$ percent of the absolute capacity — which matters if the service-area demand exceeds that. The operator's decision hinges on whether the capacity ceiling or the cost per bit is the binding constraint. $\blacksquare$

$\sigma_\phi^2 = \int S_\phi(f)\,df = 100 \times 10^6 \cdot S_{\phi,0}$ (linear), where $S_{\phi,0}$ is linear per Hz.

The residual SI floor due to phase noise relative to the full SI power is $\sigma_\phi^2$ in small-angle approximation. Cancellation depth = $-10\log_{10}(\sigma_\phi^2)$ dB.

For $40$ dB cancellation: $-10\log_{10}(100 \times 10^6 \cdot S_{\phi,0}) = 40$, so $100 \times 10^6 \cdot S_{\phi,0} = 10^{-4}$, thus $S_{\phi,0} = 10^{-12}$ per Hz $= -120$ dBc/Hz.

Typical COTS LOs sit at $-110$ dBc/Hz at $100$ kHz, giving only $\approx 30$ dB cancellation. Achieving $40$ dB requires a $-120$ dBc/Hz oscillator, which is feasible but substantially more expensive. This is why laboratory prototypes use OCXOs or external reference locks. $\blacksquare$

After $N_{\text{iter}}$ iterations the residual SNR gap is $(1 - \lambda_2/2)^{N_{\text{iter}}} \leq \epsilon$. Taking logs: $N_{\text{iter}} \cdot \ln(1-\lambda_2/2) \leq \ln\epsilon$, so $N_{\text{iter}} \approx 2\ln(1/\epsilon)/\lambda_2$.

With $\lambda_2 \sim N_{\text{AP}}^{-1/2}$, $N_{\text{iter}} \sim \sqrt{N_{\text{AP}}} \ln(1/\epsilon)$.

Total cost becomes $N_{\text{iter}} N_{\text{AP}} M^2 K \sim N_{\text{AP}}^{3/2} M^2 K$. Super-linear in $N_{\text{AP}}$ but sub-quadratic — better than centralized ($N_{\text{AP}}^2$) but not the promised $\mathcal{O}(N_{\text{AP}})$ linear scaling. This is the core reason the ultra-dense scaling problem remains open: the graph topology penalty prevents fixed-iteration counts from being enough. $\blacksquare$

For each element, the phase quantization error $\delta$ is uniform on $[-\pi/2^b, \pi/2^b]$. The expected per-element amplitude contribution is $\mathbb{E}[e^{j\delta}] = \operatorname{sinc}(\pi/2^b)$. Coherent combining preserves the amplitude factor; power picks up the square: $\operatorname{sinc}^2(\pi/2^b)$.

$b = 1$: $\operatorname{sinc}(\pi/2) = 2/\pi$; loss $= 20\log_{10}(2/\pi) \approx -3.92$ dB. $b = 2$: $\operatorname{sinc}(\pi/4) \approx 0.9003$; loss $\approx -0.91$ dB. $b = 3$: $\operatorname{sinc}(\pi/8) \approx 0.9745$; loss $\approx -0.22$ dB. $b = 4$: $\operatorname{sinc}(\pi/16) \approx 0.9936$; loss $\approx -0.056$ dB.

A 1-bit RIS loses $\approx 4$ dB of coherent gain relative to a continuous-phase ideal — significant but not catastrophic. A 2-bit RIS loses less than $1$ dB. Beyond 3 bits, further resolution is wasted. This is why practical RIS hardware uses 1-2 bit phase resolution: the cost-benefit curve flattens very quickly. $\blacksquare$

Monostatic ISAC requires the transmitter and receiver to be on the same array — exactly the FD configuration. The same SI cancellation infrastructure serves both purposes.

FD radios run the Tx continuously. Monostatic sensing needs continuous Tx for Doppler estimation. Half-duplex sensing introduces ambiguities that FD does not.

The SI channel in FD massive MIMO is precisely the sensing channel when the target is absent. Cancellation and sensing share the same estimator.

Amortizing the high cost of FD hardware (calibration, high-quality LOs, ADC dynamic range) across two use cases (data doubling + sensing) halves the per-use-case cost. Commercial FD may ship first because of the ISAC business case, not the SE doubling. $\blacksquare$

(a) Result: Publish and validate a two-cluster visibility-region channel model for a $128$-element XL-MIMO array at $3.5$ GHz that reproduces the Lund and Eurecom measured VR statistics within $1$ dB of measured rate loss under MRC combining.

(b) Tools: Expectation-maximization for cluster membership, Kolmogorov-Smirnov goodness-of-fit tests, QuaDRiGa-compatible link-level simulation, Python/MATLAB scientific stack.

(c) Target venue: Workshop paper at IEEE GLOBECOM 2025 with fitted model and validation plots; extended journal version at IEEE Transactions on Wireless Communications.

(d) Risk and fallback: Main risk: the two-cluster model might not fit the measurements well enough. Fallback: extend to a mixture model with variable cluster count and use information criteria (BIC) to select; this is a less clean result but still publishable. Secondary risk: data access restrictions. Fallback: synthesize reference data using a full-wave simulator and publish a methodology paper as a backup artifact. $\blacksquare$

Centralized: $N_{\text{AP}}^2 M^2 K \sim N_{\text{AP}}^2$. Distributed MRC: $N_{\text{AP}} M^2 K \sim N_{\text{AP}}$. Consensus (5 iter): $5 N_{\text{AP}} M^2 K \sim 5 N_{\text{AP}}$.

Centralized: slope $2$. Distributed MRC: slope $1$, intercept at $M^2 K = 640$. Consensus: slope $1$, intercept at $5 \times 640 = 3200$.

Consensus = Centralized when $5 N_{\text{AP}} M^2 K = N_{\text{AP}}^2 M^2 K$, i.e. $N_{\text{AP}} = 5$. For $N_{\text{AP}} > 5$, consensus beats centralized on cost. The actual engineering crossover is typically higher because centralized can use Woodbury identities that shave constants. $\blacksquare$