Exercises

$\gamma = \beta \cdot 10 / (1 + 10\beta)$. For concreteness, let $\beta = 0.1$: $\gamma = 0.1 \cdot 10 / (1 + 1) = 0.5$.

$(\sum_{l=1}^{16} \gamma)^2 = (16 \cdot 0.5)^2 = 64$.

$\text{SINR} = P_t \cdot 64 / (\sigma^2 \cdot 16 \cdot 0.5) = P_t \cdot 64 / (8 \sigma^2) = 8 \cdot \text{SNR}$. With $\text{SNR} = 10$ dB, $\text{SINR} = 8 \cdot 10 = 80 = 19$ dB.

Number of subcarriers in coherence bandwidth: $B_c / \Delta f = 200/15 \approx 13$. Number of OFDM symbols in coherence time: $T_c \cdot \Delta f = 1 \times 10^{-3} \times 15 \times 10^3 = 15$. Coherence interval: $\tau_c = 13 \times 15 = 195$ samples.

$\tau_p = 20$ (one pilot per user). Pilot fraction: $20/195 \approx 10.3\%$. Pre-log factor: $(195 - 20)/195 = 175/195 \approx 0.897$.

About 10% of resources are consumed by pilots. With pilot reuse ($\tau_p = 10$), the fraction drops to 5.1% but pilot contamination appears.

Transmit: $100 \times 0.2 / 0.4 = 50$ W. Hardware: $100 \times (0.5 + 1.0) = 150$ W. Total (excluding CPU): $200$ W.

$50 / 200 = 25\%$. Hardware power is 3x the transmit power, dominating the total consumption. This motivates low-power AP design.

With MRC ($v_{lk} = \hat{h}_{lk}^*$), $\mathbb{E}[\hat{h}_{lk}^* h_{lk}] = \gamma_{lk}$. Numerator: ${P_t}_{k} (\sum_l \gamma_{lk})^2$.

$\text{Var}[\hat{h}_{lk}^* h_{lk}] = \mathbb{E}[|\hat{h}_{lk}|^2 |h_{lk}|^2] - \gamma_{lk}^2 = \gamma_{lk} \beta_{lk}$. The last equality uses $\mathbb{E}[|\hat{h}_{lk}|^4] = 2\gamma_{lk}^2$ and the independence of $\hat{h}_{lk}$ and $\tilde{h}_{lk}$.

For $j \neq k$ with orthogonal pilots: $\mathbb{E}[|\hat{h}_{lk}^* h_{lj}|^2] = \gamma_{lk} \beta_{lj}$.

$\text{SINR}_k = \frac{{P_t}_{k} (\sum_l \gamma_{lk})^2}{\sum_{j=1}^{K} {P_t}_{j} \sum_l \gamma_{lk} \beta_{lj} + \sigma^2 \sum_l \gamma_{lk}}$$where the interference-plus-noise denominator includes both beamforming uncertainty (from the$j = k$ term) and inter-user interference.

$\text{num} = {P_t}_{k} (\sum_l \gamma_{lk})^2$. By the law of large numbers, $\frac{1}{L} \sum_l \gamma_{lk} \to \mathbb{E}[\gamma_{lk}] \triangleq \bar{\gamma}_k$. So $\text{num} \approx {P_t}_{k} L^2 \bar{\gamma}_k^2$.

Each term $\sum_l \gamma_{lk} \beta_{lj}$ scales as $L \cdot \mathbb{E}[\gamma_{lk} \beta_{lj}]$. The noise term $\sigma^2 \sum_l \gamma_{lk}$ also scales as $L$. Total denominator $\approx L \cdot C$ for some constant $C$ independent of $L$.

$\text{SINR}_k \approx {P_t}_{k} L^2 \bar{\gamma}_k^2 / (L \cdot C) = ({P_t}_{k} \bar{\gamma}_k^2 / C) \cdot L$. This is $O(L)$, growing linearly in the number of APs. With pilot contamination, the coherent interference term also scales as $L^2$, creating a ceiling — the SINR becomes $O(1)$.

APs on a $6 \times 6$ grid with spacing $500/6 \approx 83$ m. Maximum distance to nearest AP: $\sqrt{2} \times 83/2 \approx 59$ m.

The user at a corner of the Voronoi cell is equidistant from 4 APs at distance $\sim 59$ m. Path loss: $\beta = (59 + 10)^{-3.8} = 69^{-3.8} \approx 5.1 \times 10^{-7}$. With 4 dominant APs: coherent gain $\approx (4\gamma)^2$ where $\gamma \approx 0.8 \beta$ at 10 dB pilot SNR. SINR $\approx 16 \times 0.64 \beta^{2} \cdot 10 / (10 \times 36 \times 0.8\beta^{2}) \approx 14$ dB.

Served by nearest BS at distance 59 m. Dominant interferer at same distance. SIR $\approx 1 = 0$ dB (no array gain with $N = 1$). With noise: SINR $< 0$ dB.

Cell-free: $\text{SINR}_{\text{worst}} \approx 14$ dB $\Rightarrow R \approx 4.8$ bits/s/Hz. Small cells: $\text{SINR}_{\text{worst}} \approx 0$ dB $\Rightarrow R \approx 1.0$ bits/s/Hz. Cell-free provides $\sim 5\times$ higher rate for the worst user.

$\text{SINR}_k = \frac{p_k (\sum_l \gamma_{lk})^2}{\sum_{j \neq k} p_j \sum_l \gamma_{lk} \beta_{lj} + \sigma^2 \sum_l \gamma_{lk}}$. Define $a_k = (\sum_l \gamma_{lk})^2$, $b_{kj} = \sum_l \gamma_{lk} \beta_{lj}$, $c_k = \sigma^2 \sum_l \gamma_{lk}$.

$p_k a_k \geq \gamma^* (\sum_{j \neq k} p_j b_{kj} + c_k)$, or equivalently $p_k a_k - \gamma^* \sum_{j \neq k} p_j b_{kj} \geq \gamma^* c_k$.

For fixed $\gamma^*$, find $\{p_k\}$ such that $p_k a_k - \gamma^* \sum_{j \neq k} p_j b_{kj} \geq \gamma^* c_k$ for all $k$, and $0 \leq p_k \leq P_{\max}$. This is a system of linear inequalities in $\{p_k\}$, i.e., a linear feasibility problem. Bisection over $\gamma^*$ finds the max-min SINR.

$\alpha_q = \frac{\pi\sqrt{3}}{2} \cdot 2^{-12} = \frac{2.72}{4096} \approx 6.64 \times 10^{-4}$.

SINR = $10^{20/10} = 100$. $\text{SINR}_q = 100 / (1 + 6.64 \times 10^{-4} \times 100) = 100 / 1.0664 \approx 93.8 = 19.7$ dB.

$\text{SE} = \log_2(1 + 100) = 6.66$ bits/s/Hz. $\text{SE}_q = \log_2(1 + 93.8) = 6.57$ bits/s/Hz. Loss: $0.09$ bits/s/Hz ($\sim 1.4\%$). At $b = 6$, fronthaul quantization is negligible for SINR $\leq 20$ dB.

$\{\lambda : f(\lambda) / g(\lambda) \geq \alpha\} = \{\lambda : f(\lambda) - \alpha g(\lambda) \geq 0\}$. Since $f$ is concave and $g$ is affine, $h(\lambda) = f(\lambda) - \alpha g(\lambda)$ is concave. The superlevel set of a concave function $\{x : h(x) \geq 0\}$ is convex.

Since all superlevel sets of $\text{EE}$ are convex, $\text{EE}$ is quasi-concave. A continuous quasi-concave function on a connected set has a unique maximum (or a plateau at the maximum).

At the optimum, $\frac{d}{d\lambda} \text{EE} = 0$: $\frac{f'(\lambda^*) g(\lambda^*) - f(\lambda^*) g'(\lambda^*)}{g(\lambda^*)^2} = 0$, giving $f'(\lambda^*) / g'(\lambda^*) = f(\lambda^*) / g(\lambda^*)$. This means the marginal efficiency equals the average efficiency at the optimum.

With shared pilots, $\gamma_{lk}^{\text{cont}} = \frac{{P_t}_{k} \tau_p \beta_{lk}^{2}}{({P_t}_{k} \beta_{lk} + {P_t}_{j} \beta_{lj}) \tau_p + \sigma^2}$ and $\mathbb{E}[\hat{h}_{lk}^* h_{lj}] = \frac{{P_t}_{j} \tau_p \beta_{lk} \beta_{lj}}{({P_t}_{k} \beta_{lk} + {P_t}_{j} \beta_{lj}) \tau_p + \sigma^2} \triangleq \xi_{lkj}$.

The interference from user $j$ contains the coherent term $p_j |\sum_l \xi_{lkj}|^2$, which scales as $L^2$ by the same law-of-large-numbers argument.

As $L \to \infty$: $\text{SINR}_k \to \frac{p_k (\bar{\gamma}_k^{\text{cont}})^2}{p_j (\bar{\xi}_{kj})^2} = \frac{p_k}{p_j} \cdot \frac{(\bar{\gamma}_k^{\text{cont}})^2}{(\bar{\xi}_{kj})^2}$ which is a finite constant independent of $L$. This is the pilot contamination ceiling.

When $p_k = p_j$ and $\beta_{lk} = \beta_{lj}$ for all $l$ (identical spatial profiles): $\bar{\gamma}_k^{\text{cont}} = \bar{\xi}_{kj}$ and the ceiling is $\text{SINR}_k \to 1 = 0$ dB.

The RZF precoder is $\mathbf{W} = \hat{\mathbf{H}} (\hat{\mathbf{H}}^H \hat{\mathbf{H}} + \mathbf{\Lambda})^{-1} \mathbf{D}$ where $\mathbf{D} = \text{diag}(\sqrt{\eta_1}, \ldots, \sqrt{\eta_K})$ controls per-user power and $\mathbf{\Lambda} = \text{blkdiag}(\lambda_1 \mathbf{I}_N, \ldots, \lambda_L \mathbf{I}_N)$ with per-AP regularization parameters.

$\text{SINR}_k = \frac{|\mathbb{E}[\mathbf{H}_{k}^{H} \mathbf{w}_k]|^2}{\sum_{j \neq k} \mathbb{E}[|\mathbf{H}_{k}^{H} \mathbf{w}_j|^2] + \text{Var}[\mathbf{H}_{k}^{H} \mathbf{w}_k] + \sigma^2}$.

The per-AP constraint is $\sum_k \eta_k \|\hat{\mathbf{H}}_{lk}^H (\hat{\mathbf{H}}^H \hat{\mathbf{H}} + \mathbf{\Lambda})^{-1} \mathbf{e}_k\|^2 \leq {P_t}_{l}$. This determines $\mathbf{\Lambda}$ and $\mathbf{D}$ jointly through a fixed-point iteration.

$2 b \cdot 20 \times 10^6 \leq 100 \times 10^6 \Rightarrow b \leq 2.5$. So $b = 2$ bits. $\alpha_q = \frac{\pi\sqrt{3}}{2} \cdot 2^{-4} = 0.170$. SINR ceiling: $1/\alpha_q = 5.88 = 7.7$ dB.

$2b \cdot 20 \times 10^6 \leq 10^9 \Rightarrow b \leq 25$. In practice $b = 12$ is sufficient. $\alpha_q = \frac{\pi\sqrt{3}}{2} \cdot 2^{-24} = 1.62 \times 10^{-7}$. SINR ceiling: $1/\alpha_q = 6.16 \times 10^6 = 67.9$ dB.

With 100 Mbps fronthaul, the system is severely fronthaul-limited (7.7 dB ceiling). With 1 Gbps, the ceiling is negligible (68 dB). For useful cell-free operation, each AP needs at least $\sim 250$ Mbps fronthaul ($b \geq 6$) to avoid significant SE loss.

$R_1 = \log_2(1 + p_1 \gamma_1) = \log_2(1 + p_1)$ and $R_2 = \log_2(1 + p_2 \gamma_2) = \log_2(1 + 0.1 p_2)$ where $p_1 + p_2 = 1$.

$p_1 = p_2 = 0.5$: $R_1 = \log_2(1.5) = 0.585$, $R_2 = \log_2(1.05) = 0.070$.

Maximize $\log R_1 + \log R_2$. KKT: $\frac{\gamma_1}{R_1 (1 + p_1 \gamma_1) \ln 2} = \frac{\gamma_2}{R_2 (1 + p_2 \gamma_2) \ln 2}$. Numerically: $p_1 \approx 0.32$, $p_2 \approx 0.68$. $R_1 = \log_2(1.32) = 0.40$, $R_2 = \log_2(1.068) = 0.095$.

Set $R_1 = R_2$: $\log_2(1 + p_1) = \log_2(1 + 0.1 p_2)$. Solution: $p_1 \approx 0.065$, $p_2 \approx 0.935$. $R_1 = R_2 = \log_2(1.065) = 0.091$. Max-min sacrifices user 1 heavily.

Power to weak user: equal = 0.50, PF = 0.68, max-min = 0.935. Proportional fairness gives more to the weak user than equal power but less than max-min.

$\min_{\{a_l, \eta_{lk}\}} \sum_l a_l (P_{\text{AP}} + P_{\text{fh}} + \sum_k \eta_{lk} \gamma_{lk} / \zeta)$ subject to $\text{SINR}_k(\{a_l \eta_{lk}\}) \geq \gamma_{\min}$ for all $k$, $\sum_k \eta_{lk} \gamma_{lk} \leq P_t \cdot a_l$ for all $l$, $a_l \in \{0, 1\}$.

1. Start with all APs active. Compute max-min SINR.
2. For each AP $l$, compute the SINR reduction if AP $l$ is deactivated.
3. Deactivate the AP with the smallest SINR impact, provided $\text{SINR}_k \geq \gamma_{\min}$ still holds.
4. Repeat until no more APs can be deactivated.

The greedy algorithm requires $O(L^2)$ SINR evaluations in the worst case (deactivating $L$ APs, each requiring checking $K$ constraints). Each SINR evaluation is $O(L K)$, giving $O(L^3 K)$ total — tractable for moderate $L$.

With LSFD-optimized weights, the Level 1 SINR scales as $c_1 \cdot L$ where $c_1 = {P_t}_{k} \bar{\gamma}_k^2 / (\sum_j {P_t}_{j} \overline{\gamma_k \beta_{j}} + \sigma^2 \bar{\gamma}_k)$.

Centralized MMSE suppresses inter-user interference. As $L \to \infty$, interference is fully canceled: $\text{SINR}_k \to {P_t}_{k} L \bar{\gamma}_k / \sigma^2 = c_4 \cdot L$ where $c_4 = {P_t}_{k} \bar{\gamma}_k / \sigma^2$.

The ratio $c_4 / c_1 = (\sum_j {P_t}_{j} \overline{\gamma_k \beta_{j}} + \sigma^2 \bar{\gamma}_k) / (\sigma^2 \bar{\gamma}_k) = 1 + \sum_j ({P_t}_{j} / \sigma^2) \overline{\gamma_k \beta_{j}} / \bar{\gamma}_k$. At high SNR, this ratio is large (Level 4 suppresses strong interference that Level 1 cannot). At low SNR, the ratio approaches 1 (noise dominates interference in both levels).

At low SNR, the absolute SE gap is small (both levels are noise-limited). But the relative improvement $\text{SE}_4 / \text{SE}_1$ is larger because Level 4 achieves a better noise-vs-signal tradeoff through MMSE combining. At high SNR, both levels are interference-limited and the relative gap is determined by the interference rejection capability — Level 4 still wins but the SE is already high.

Transmit: $80 \times 0.2/0.4 = 40$ W. Hardware: $80 \times 1.5 = 120$ W. CPU: 10 W. Total: 170 W. Throughput: $20 \times 10^6 \times 10 \times 3.0 = 600$ Mbps. EE: $600 / 170 = 3.53$ Mbits/J.

Transmit: $40 \times 0.2/0.4 = 20$ W. Hardware: $40 \times 1.5 = 60$ W. CPU: 10 W. Total: 90 W. Throughput: $20 \times 10^6 \times 10 \times 2.5 = 500$ Mbps. EE: $500 / 90 = 5.56$ Mbits/J.

$L = 40$ is 57% more energy-efficient despite 17% lower throughput. Doubling the APs from 40 to 80 adds 89% more power but only 20% more throughput. The optimal EE operating point is at lower AP density than the optimal SE point.

The average rate is dominated by cell-center users who have strong channels, masking the experience of cell-edge users who suffer from weak signal and strong interference. In small cells, the top 10% of users may have 50x higher rate than the bottom 10%, but the average looks reasonable. The 95%-likely rate captures the experience of the worst-served 5% of users — exactly the population that cell-free targets. Since cell-free provides its largest gains at the cell edge, the average rate understates the improvement (showing perhaps 2-3x gain) while the 95%-likely rate reveals the true 5-10x gain where it matters most.

APs: PPP with density $\lambda_{\text{AP}}$. Users: PPP with density $\lambda_u$. User-centric: each user is served by the $M$ nearest APs. Path loss: $\beta(d) = d^{-\alpha}$. Channel: Rayleigh fading.

With MRC, the signal power from the $M$ nearest APs is $S = (\sum_{m=1}^{M} \gamma_m)^2$ where $\gamma_m = \beta(d_m) \cdot \text{SNR}_{p} / (1 + \beta(d_m) \text{SNR}_{p})$. The distances $d_1, \ldots, d_M$ to the $M$ nearest APs follow the order statistics of the PPP, with $d_m^2 \sim \text{Gamma}(m, \pi \lambda_{\text{AP}})$.

$\Pr[\text{SINR} \geq \gamma] = \mathbb{E}_{d_1, \ldots, d_M}[\Pr[S / (I + N) \geq \gamma \mid d_1, \ldots, d_M]]$. The interference $I$ comes from non-serving APs transmitting to other users. The Laplace transform $\mathcal{L}_I(s) = \exp(-2\pi\lambda_{\text{AP}} \int_{d_M}^{\infty} \frac{s r^{-\alpha}}{1 + s r^{-\alpha}} r \, dr)$ captures the aggregate interference. The exact coverage probability requires numerical integration over the distance distribution, but the key scaling insight is that $\Pr[\text{SINR} \geq \gamma]$ increases with $\lambda_{\text{AP}} / \lambda_u$ and saturates when $M$ is large enough to capture most of the signal energy.