Exercises

$R_{\text{CPRI}} = 2 \cdot 12 \cdot 20 \times 10^6 \cdot 8 \cdot \frac{16}{15} = 4.096 \text{ Gbps}$$

This fits within a single 10G Ethernet link. For 100 MHz bandwidth, the rate would be $5 \times$ higher ($\approx 20.5$ Gbps), requiring a 25G link.

$R_{\text{sum}} \leq \min(50, \; 10 \cdot 5) = \min(50, 50) = 50 \text{ bits/s/Hz}$$The fronthaul is exactly sufficient. If$C_{\text{fh}} = 4$ bits/s/Hz, the sum rate would be limited to 40 bits/s/Hz.

$P_y = \beta \cdot P_t + \sigma^2$

$\sigma_q^2 = P_y / (2^{C_{\text{fh}}} - 1)$

$\text{SINR} = \frac{\beta \cdot P_t}{\sigma^2 + \sigma_q^2} = \frac{\beta \cdot P_t}{\sigma^2 + \frac{\beta \cdot P_t + \sigma^2}{2^{C_{\text{fh}}} - 1}}$$

$\text{SINR} = \frac{\beta \cdot P_t (2^{C_{\text{fh}}} - 1)}{\sigma^2 \cdot 2^{C_{\text{fh}}} + \beta \cdot P_t}$$As$C_{\text{fh}} \to \infty$,$\text{SINR} \to \beta P_t / \sigma^2$.

QF forwards $y_l = h_l x + w_l$ with variance $P_y = |h_l|^2 P_t + \sigma^2$. Required rate: $C_{\text{fh},l}^{\text{QF}} = \log_2(1 + P_y / \sigma_q^2)$ bits/symbol.

EF forwards $\hat{x}_l = h_l^* y_l / (|h_l|^2 + \sigma^2/P_t)$ with variance $P_{\hat{x}} = |h_l|^2 P_t / (|h_l|^2 + \sigma^2/P_t)$. Since $P_{\hat{x}} \leq P_y$, EF requires $C_{\text{fh},l}^{\text{EF}} \leq C_{\text{fh},l}^{\text{QF}}$.

The ratio $P_y / P_{\hat{x}} = 1 + \sigma^2/(|h_l|^2 P_t) = 1 + 1/\text{SNR}_{l}$. At high SNR, the gain vanishes; at low SNR, EF can save up to $\log_2(1 + 1/\text{SNR}_{l})$ bits/symbol per AP. With $K = 1$ and single-antenna APs, both forward a scalar, so the dimension reduction benefit of EF does not apply.

Require $\sigma_q^2 \leq 0.1 \sigma^2$.

$C_{\text{fh}} \geq N_t \log_2\!\left(\frac{P_y}{\sigma_q^2}\right) \geq N_t \log_2\!\left(\frac{P_y}{0.1 \sigma^2}\right)$$

With $P_y = K P_t \beta + \sigma^2$: $$C_{\text{fh}} \geq N_t \log_2\!\left(\frac{K P_t \beta + \sigma^2}{0.1 \sigma^2}\right) = N_t \log_2\!\left(10(K \text{SNR} + 1)\right)$$ For $\text{SNR} = 10$ dB and $K = 4$: $C_{\text{fh}} \geq N_t \cdot \log_2(410) \approx 8.7 N_t$ bits/symbol.

$R = \log_2\!\left(1 + \frac{|h_1 v_1 + h_2 v_2|^2}{|h_1|^2 \sigma_{e,1}^2 + |h_2|^2 \sigma_{e,2}^2 + \sigma^2}\right)$$

$\log_2(1 + |v_l|^2/\sigma_{e,l}^2) \leq C_{\text{fh},l}$, so $\sigma_{e,l}^2 \geq |v_l|^2 / (2^{C_{\text{fh},l}} - 1)$. Equality at the minimum: $\sigma_{e,l}^2 = |v_l|^2 / (2^{C_{\text{fh},l}} - 1)$.

The optimal compression noise is proportional to $|v_l|^2$, the precoded signal power. APs with more transmit power need proportionally higher fronthaul resolution.

$\text{Var}(y_1) = |h|^2 P_t + \sigma^2$. The correlation between $y_1$ and $y_2$ is $\text{Cov}(y_1, y_2) = |h|^2 P_t$ (from the common signal $x$). $$\text{Var}(y_1 | y_2) = \text{Var}(y_1) - \frac{|\text{Cov}(y_1, y_2)|^2}{\text{Var}(y_2)} = \sigma^2 - \frac{(|h|^2 P_t)^2}{|h|^2 P_t + \sigma^2}$$

$\text{Var}(y_1 | y_2) = \frac{\sigma^2(|h|^2 P_t + \sigma^2) - |h|^4 P_t^{2} / (|h|^2 P_t + \sigma^2)}{1}$$After algebra:$\text{Var}(y_1 | y_2) = \sigma^2 \cdot \frac{\sigma^2}{|h|^2 P_t + \sigma^2}$.

Independent: $R_1^{\text{ind}} = \log_2(\text{Var}(y_1) / D)$. Wyner-Ziv: $R_1^{\text{WZ}} = \log_2(\text{Var}(y_1|y_2) / D)$. The rate saving is $R_1^{\text{ind}} - R_1^{\text{WZ}} = \log_2(\text{Var}(y_1)/\text{Var}(y_1|y_2)) = \log_2(1 + \text{SNR})$. At high SNR, the saving grows logarithmically. $\blacksquare$

AP $l$ transmits $\hat{\mathbf{x}}_l = \mathbf{x}_l + \mathbf{e}_l$. Total power: $P_{\text{total}} = \text{tr}(\mathbf{R}_x) + \text{tr}(\mathbf{R}_e) \leq P$. Useful signal power: $P_{\text{useful}} = \text{tr}(\mathbf{R}_x) = P - \text{tr}(\mathbf{R}_e)$.

For isotropic quantization: $\sigma_e^2 = \text{tr}(\mathbf{R}_x)/(N_t(2^{C_{\text{fh}}/N_t} - 1))$. So $\text{tr}(\mathbf{R}_e) = N_t \sigma_e^2 = P_{\text{useful}} / (2^{C_{\text{fh}}/N_t} - 1)$.

$P_{\text{useful}} + P_{\text{useful}}/(2^{C_{\text{fh}}/N_t} - 1) = P$, so $P_{\text{useful}} \cdot 2^{C_{\text{fh}}/N_t} / (2^{C_{\text{fh}}/N_t} - 1) = P$, giving $P_{\text{useful}} = P \cdot (1 - 2^{-C_{\text{fh}}/N_t})$. $\blacksquare$

$\max_{\{C_{\text{fh},l}\}} \min_k R_k \quad \text{s.t.} \quad \sum_{l=1}^{L} C_{\text{fh},l} = C_{\text{total}}, \quad C_{\text{fh},l} \geq 0$$where$R_k$depends on the quantization noise at each AP serving user$k$.

For max-min fairness with EF, the optimal strategy equalizes the per-user quantization resolution: $r_l = C_{\text{fh},l} / |\mathcal{C}_l|$. Setting $r_l = r^*$ for all $l$: $C_{\text{fh},l} = r^* |\mathcal{C}_l|$. From the total budget: $r^* \sum_l |\mathcal{C}_l| = C_{\text{total}}$, so $r^* = C_{\text{total}} / \sum_l |\mathcal{C}_l|$.

$C_{\text{fh},l}^* = \frac{|\mathcal{C}_l|}{\sum_{l'} |\mathcal{C}_{l'}|} \cdot C_{\text{total}}$$ This allocates fronthaul proportional to cluster size, matching the load balancing result from Section 14.4.

• RU: RF front-end, DAC/ADC, FFT/IFFT, CP addition/removal.
• DU: Channel estimation, MIMO precoding/combining, scheduling, MAC layer.
• CU: PDCP, RRC, SDAP, mobility management.

• RU $\leftrightarrow$ DU: Open Fronthaul (O-RAN 7.2x, frequency-domain I/Q)
• DU $\leftrightarrow$ CU: F1 interface (midhaul, transport blocks)
• CU $\leftrightarrow$ core: NG interface (backhaul, IP packets)

$R_{7.2x} = 2 \cdot 12 \cdot 100 \times 10^6 \cdot 32 \cdot \frac{16}{15} \approx 81.9 \text{ Gbps}$$

$R_{7.1} = 2 \cdot 12 \cdot 100 \times 10^6 \cdot 4 \cdot \frac{16}{15} \approx 10.2 \text{ Gbps}$$

Option 7.1 requires $8\times$ less fronthaul ($32/4 = 8$) by performing beamforming at the RU. The tradeoff: the DU loses control over per-antenna signals, limiting centralized MIMO processing.

For AP $l$: $\text{tr}(\mathbf{R}_{y,l}) = N_t(\beta_{l1} + \beta_{l2} + \sigma^2)$. $\sigma_{q,l}^2 = (\beta_{l1} + \beta_{l2} + 1)/(2^{10/4} - 1) = (\beta_{l1} + \beta_{l2} + 1)/4.66$.

EF forwards $K = 2$ dimensions per AP. Bits per dimension: $10/2 = 5$ bits. The MMSE estimate variance is lower than $P_y$, so $\sigma_{q,l}^{\text{EF}} = P_{\hat{x},l}/(2^5 - 1) = P_{\hat{x},l}/31$.

QF: $\sigma_q^2 \sim P_y/4.66$ (2.5 bits/dimension). EF: $\sigma_q^2 \sim P_{\hat{x}}/31$ (5 bits/dimension). The per-dimension resolution is $2\times$ higher for EF because it forwards half the dimensions. Combined with the lower source variance ($P_{\hat{x}} < P_y$), EF achieves significantly higher rates.

Power loss fraction: $2^{-C_{\text{fh}}/N_t} = 2^{-4} = 1/16 = 6.25\%$. Effective power: $P_{\text{eff}} = P(1 - 1/16) = 15P/16$.

Loss in dB: $-10\log_{10}(15/16) = -10\log_{10}(0.9375) \approx 0.28$ dB. This is well below 1 dB, confirming that 4 bits per antenna dimension provides adequate fronthaul resolution for the downlink.

Define $\mathcal{C}_l$ as the set of users served by AP $l$. The EF fronthaul rate is $R_{\text{fh},l} = |\mathcal{C}_l| \cdot r$ where $r$ is the per-user resolution. The constraint is $|\mathcal{C}_l| \cdot r \leq C_{\text{fh},l}$.

Fiber APs: can serve up to $C_{\text{fh}}/r = 100/r$ users. Wireless APs: can serve up to $10/r$ users. At $r = 5$ bits/user: fiber APs serve up to 20 users, wireless APs serve up to 2 users. Strategy: assign each user to the AP with the best $\beta_{lk} \cdot \min(C_{\text{fh},l}/|\mathcal{C}_l|, r_{\max})$ metric, which balances channel quality and available fronthaul resolution.

(a) Nearest-AP ignores fronthaul: wireless APs get overloaded. (b) Best-channel ignores fronthaul: same issue. (c) Fronthaul-aware: limits wireless AP clusters to 2 users while fiber APs handle the rest, avoiding the fronthaul bottleneck. Expected sum rate gain: 20--40% over (a) and (b) in heterogeneous scenarios.

Total fronthaul: $C_{\text{total}}(r) = \sum_l |\mathcal{C}_l| \cdot r = r \sum_l |\mathcal{C}_l|$. This is linear in $r$.

From Theorem 14.4: the SINR for user $k$ has a quantization term $\propto \sigma_q^2 = P_{\hat{x}} \cdot 2^{-2r}$ in the denominator. As $r$ increases, $\sigma_q^2$ decreases and SINR increases.

$R_k(r) = \log_2(1 + \text{SINR}_k(r))$. Since $\text{SINR}_k$ is an increasing concave function of $2^{2r}$ (diminishing returns from additional resolution), and $\log_2(1 + \cdot)$ is concave, the composition $R_k(r)$ is concave in $r$.

The Pareto frontier plots $R_{\text{sum}}(r)$ vs. $C_{\text{total}}(r) = r \sum_l |\mathcal{C}_l|$. Since $R_{\text{sum}}$ is concave in $r$ and $C_{\text{total}}$ is linear, the Pareto frontier is a concave curve. This means small increases in fronthaul yield large rate gains at low resolution, with diminishing returns at high resolution.

Useful capacity: $25 \times 0.9 = 22.5$ Gbps.

$22.5 \times 10^9 = 2 \cdot b \cdot 100 \times 10^6 \cdot 8$, so $b = 22.5 \times 10^9 / (1.6 \times 10^9) \approx 14$ bits.

A 25G link supports approximately 14-bit I/Q resolution per port, more than sufficient for most deployments (12 bits is standard).

Each AP forwards $N_t = 4$ complex dimensions. Per AP: $2 \cdot 12 \cdot 4 = 96$ bits per subcarrier. Total: $16 \cdot 96 = 1536$ bits per subcarrier.

Each AP forwards $K = 8$ scalar combining outputs. Per AP: $2 \cdot 12 \cdot 8 = 192$ bits per subcarrier. Wait --- this is higher because $K > N_t$. Actually, each AP only forwards for users in its cluster. With cluster size $|\mathcal{C}_l| \leq N_t = 4$: Per AP: $2 \cdot 12 \cdot 4 = 96$ bits (same dimension as Level 4).

When $K \leq N_t$, Level 1 and Level 4 forward comparable dimensions. The real fronthaul saving comes from the reduced coordination overhead (Level 1 does not need to share channel estimates across APs).

Available for propagation: $100 - 30 = 70$ $\mu$s. Maximum distance: $70/5 = 14$ km.

A single DU at the center of a 10 km $\times$ 10 km area has maximum distance to corners: $\sqrt{50} \approx 7.1$ km $< 14$ km. One centrally placed DU can serve the entire area. For a 20 km $\times$ 20 km area, the maximum corner distance is $\approx 14.1$ km, marginally exceeding the budget. Two DUs would be needed.

The latency constraint limits the cell-free cluster radius. Dense urban deployments (1--5 km) are comfortably within budget. Rural deployments may require multiple DUs, creating a hybrid centralized-distributed architecture.