Exercises

The inter-user link supports $\alpha \cdot \frac{1}{2}\log(1 + |h_{12}|^2 P/\sigma^2) = \frac{1}{2} \cdot \frac{1}{2}\log(1 + 40) \approx 1.34$ bits/use per user.

Each user's Phase 1 rate to the destination: $\alpha \cdot \frac{1}{2}\log(1 + P/2) = \frac{1}{4}\log(6) \approx 0.65$ bits/use.

With equal power split $P_{1c} = P_{2c,1} = P/2 = 5$ and $\rho = 1$: $(1-\alpha)\frac{1}{2}\log(1 + (5 + 5 + 2\sqrt{25})/1) = \frac{1}{4}\log(21) \approx 1.10$ bits/use.

Per-user rate: $\min(1.34, 0.65 + 1.10) = \min(1.34, 1.75) = 1.34$ bits/use. Sum rate: $2 \times 1.34 = 2.68$ bits/use. The bottleneck is the inter-user link. Increasing $\alpha$ would help.

The outage event is $\{\log(1 + \text{SNR}|h|^2) < r\log\text{SNR}\}$, which simplifies to $\{|h|^2 < (\text{SNR}^{r} - 1)/\text{SNR}\}$.

For large $\text{SNR}$: $(\text{SNR}^{r} - 1)/\text{SNR} \approx \text{SNR}^{r-1}$. Since $|h|^2 \sim \text{Exp}(1)$: $P_{\text{out}} = 1 - e^{-\text{SNR}^{r-1}} \approx \text{SNR}^{r-1}$.

$d(r) = -\lim_{\text{SNR} \to \infty} \frac{\log \text{SNR}^{r-1}}{\log\text{SNR}} = -(r-1) = 1-r$.

$\sigma_q^2 = (10 + 1)/(2^6 - 1) = 11/63 \approx 0.175$.

With 2 APs and MRC, the effective capacity is: $\frac{1}{2}\log\det(\mathbf{I}_2 + \frac{P}{\sigma^2 + \sigma_q^2}\mathbf{1}\mathbf{1}^T) = \frac{1}{2}\log(1 + \frac{2P}{\sigma^2 + \sigma_q^2}) = \frac{1}{2}\log(1 + 2 \times 10/1.175) \approx 2.17$ bits/use.

Compare with ideal ($C_{\text{fh}} \to \infty$): $\frac{1}{2}\log(1 + 20) \approx 2.19$ bits/use. Loss: only 0.02 bits/use.

From Theorem 25.3, with $\gamma_{kl} = \beta$ and $\beta_{k'l} = \beta$:

$$\text{SINR}_k = \frac{p_k(L\beta)^2}{\sum_{k'=1}^K p_{k'} L\beta^2 + \sigma^2 L\beta} = \frac{p L^2\beta^2}{KpL\beta^2 + \sigma^2 L\beta} = \frac{pL\beta}{Kp\beta + \sigma^2}.$$

This confirms SINR $\propto L$, and the per-user rate $R_k = \log(1 + pL\beta/(Kp\beta + \sigma^2))$ grows as $\log L$ for large $L$.

Per-user fronthaul: $N_{\text{AP}} \times 2B \times b = 4 \times 2 \times 20\text{M} \times 5 = 800$ Mbps.

Compare with full cooperation ($N_{\text{AP}} = 64$): $64 \times 200\text{M} = 12.8$ Gbps. The user-centric approach reduces fronthaul by a factor of 16.

With dynamic DF, the effective channel to the destination has $K$ independent fading paths: the direct link and $K-1$ cooperative links. The outage probability at rate $R = r\log\text{SNR}$ requires all $K$ paths to be simultaneously weak:

$$P_{\text{out}} \le \Pr\!\left[\sum_{i=1}^K |h_i|^2 < \frac{\text{SNR}^{r}}{\text{SNR}}\right] = \Pr\!\left[\sum_{i=1}^K |h_i|^2 < \text{SNR}^{r-1}\right].$$

For $K$ independent exponential random variables, the CDF near zero satisfies:

$$\Pr\!\left[\sum_{i=1}^K |h_i|^2 < x\right] \approx \frac{x^K}{K!}$$

for small $x$. Therefore $P_{\text{out}} \approx \text{SNR}^{K(r-1)}/K!$ and $d(r) = K(1-r)$.

The dynamic DF protocol achieves this by adapting the listening duration at each relay. When relay $j$'s link is strong, it decodes quickly and cooperates for most of the block. The effective multiplexing gain is $r \le 1$ (full rate) because the protocol overhead vanishes in the high-SNR limit.

$\|\mathbf{g}_l\|^2 P = P \sum_k |g_{kl}|^2$:

• AP 1: $10(1 + 0.25) = 12.5$
• AP 2: $10(0.25 + 1) = 12.5$
• AP 3: $10(0.64 + 0.09) = 7.3$
• AP 4: $10(0.09 + 0.64) = 7.3$

For each AP: $\sigma_{q,l}^2 = (\|\mathbf{g}_l\|^2 P + 1)/(2^8 - 1)$.

• APs 1,2: $\sigma_q^2 = 13.5/255 \approx 0.053$
• APs 3,4: $\sigma_q^2 = 8.3/255 \approx 0.033$

Effective noise: $\mathbf{D}_q = \text{diag}(0.053, 0.053, 0.033, 0.033)$.

$R_{\text{CF}} = \log\det(\mathbf{I}_4 + P\,(\sigma^2\mathbf{I} + \mathbf{D}_q)^{-1}\mathbf{G}\mathbf{G}^H)$. Numerically computing this (or using the approximation $\mathbf{D}_q \approx 0$): the CF rate is within 0.1 bits/use of the ideal rate.

Ideal: $R_{\text{ideal}} = \log\det(\mathbf{I}_4 + 10\,\mathbf{G}\mathbf{G}^H)$.

$\gamma = \frac{\tau_p p \beta^2}{\tau_p p \beta + \sigma^2} = \frac{10 \times 0.1 \times 10^{-4}}{10 \times 0.1 \times 0.01 + 10^{-13}} = \frac{10^{-4}}{0.01} = 0.01 = \beta$.

Since $\tau_p p \beta \gg \sigma^2$, the channel estimation is nearly perfect.

$\text{SINR}_k = \frac{pL\gamma}{Kp\beta + \sigma^2} = \frac{0.1 \times 100 \times 0.01}{10 \times 0.1 \times 0.01 + 10^{-13}} = \frac{0.1}{0.01} = 10.$

$R_k = \log_2(1 + 10) \approx 3.46$ bits/use.

With $L = 200$: SINR $= 20$, $R_k \approx 4.39$ bits/use. With $L = 400$: SINR $= 40$, $R_k \approx 5.36$ bits/use.

Indeed $R_k \approx \log_2(pL\gamma/(Kp\beta)) = \log_2(L/K) = \log_2(L) - \log_2(K)$, confirming $\Theta(\log L)$ scaling.

The relay receives $Y_r = h_{12}X_1 + Z_r$ and transmits $X_r = \alpha Y_r$ where $\alpha = \sqrt{P_r/(|h_{12}|^2 P_1 + \sigma^2)}$ normalizes the power. The destination receives $Y_d^{(2)} = h_2 \alpha(h_{12}X_1 + Z_r) + Z_d^{(2)}$.

The SNR of the relay link is: $$\text{SNR}_{\text{relay}} = \frac{|h_2|^2 \alpha^2 |h_{12}|^2 P_1} {|h_2|^2 \alpha^2 \sigma^2 + \sigma^2} \le \min(|h_{12}|^2 P_1/\sigma^2,\; |h_2|^2 P_r/\sigma^2).$$

The total effective SNR is $\text{SNR}_{\text{eff}} = |h_1|^2 P/\sigma^2 + \text{SNR}_{\text{relay}}$.

At $r = 0$ (fixed rate), outage requires both the direct link and the relay link to be weak. Since the relay link depends on two independent channels ($h_{12}$ and $h_2$), the overall diversity is $d = 1 + 1 = 2$. However, the relay link SNR is bounded by the harmonic mean of $\text{SNR}|h_{12}|^2$ and $\text{SNR}|h_2|^2$, which gives a 3 dB coding gain loss compared to DF (where the relay decodes perfectly before re-encoding).

The 200-bit codeword is split into two 100-bit halves, each transmitted through an independent Rayleigh fading channel. The Viterbi decoder at the destination sees a virtual $2 \times 1$ space-time code. The diversity order is 2.

Non-cooperative (diversity 1): $P_e^{\text{nc}} \approx \frac{1}{4 d_{\text{free}} \cdot \text{SNR}}$ for a rate-$1/2$ convolutional code with free distance $d_{\text{free}}$.

Coded cooperation (diversity 2): $P_e^{\text{cc}} \approx \frac{c}{(d_{\text{free}} \cdot \text{SNR})^2}$.

At $\text{SNR} = 10$ dB (= 10 linear), with $d_{\text{free}} = 5$: $P_e^{\text{nc}} \approx 1/(4 \times 50) = 5 \times 10^{-3}$, $P_e^{\text{cc}} \approx c / 2500 \approx 4 \times 10^{-4}$ (about one order of magnitude improvement).

The CP computes beamformed signals $\mathbf{x} = \sum_k \mathbf{v}_k s_k$ where $\mathbf{v}_k \in \mathbb{C}^L$ is the beamforming vector for user $k$ and $s_k$ is the data symbol. Each AP $l$ must receive $x_l$ through the fronthaul.

The CP quantizes $x_l$ to $\hat{x}_l$ and sends the quantization index through the fronthaul. The fronthaul constraint is: $I(X_l; \hat{X}_l) \le C_{\text{fh}}$ for each $l$.

User $k$ receives $Y_k = \sum_l g_{kl}\hat{X}_l + \sum_l g_{kl}Q_l + Z_k$ where $Q_l = \hat{X}_l - X_l$ is the quantization noise. The achievable rates are: $R_k \le I(S_k; Y_k | S_{\bar{k}})$ subject to $I(X_l; \hat{X}_l) \le C_{\text{fh}}$ and power constraints $\mathbb{E}[\|X_l\|^2] \le P_l$.

This is a joint optimization over beamforming vectors, power allocation, and quantization distortion levels.

During pilot phase, user $k$ sends pilot of length $\tau_p$ at power $p$. AP $l$ receives: $\mathbf{y}_{\text{pilot},l} = \sqrt{\tau_p p}\, g_{kl} + \mathbf{w}_l$ where $\mathbf{w}_l$ includes noise and interference from pilot-sharing users. For orthogonal pilots (no contamination): $y_l = \sqrt{\tau_p p}\, g_{kl} + w_l$, $w_l \sim \mathcal{CN}(0, \sigma^2)$.

Since $g_{kl} \sim \mathcal{CN}(0, \beta_{kl})$ and $y_l | g_{kl} \sim \mathcal{CN}(\sqrt{\tau_p p}\, g_{kl}, \sigma^2)$:

$$\hat{g}_{kl} = \frac{\sqrt{\tau_p p}\,\beta_{kl}}{\tau_p p \beta_{kl} + \sigma^2} y_l.$$

$\gamma_{kl} = \mathbb{E}[|\hat{g}_{kl}|^2] = \frac{\tau_p p \beta_{kl}^2}{\tau_p p \beta_{kl} + \sigma^2}.$

By the MMSE property for jointly Gaussian variables: $\mathbb{E}[\hat{g}_{kl} \tilde{g}_{kl}^*] = 0$. The estimation error variance is $\beta_{kl} - \gamma_{kl} = \frac{\sigma^2\beta_{kl}}{\tau_p p \beta_{kl} + \sigma^2}$.

Consider a cut that places the users and APs in $\mathcal{T}$ on one side, and the remaining APs $\mathcal{T}^c$ and the CP on the other side. Information crosses this cut via: (1) the wireless channel from users to APs in $\mathcal{T}^c$, and (2) the fronthaul links from APs in $\mathcal{T}$ to the CP.

By the cut-set bound: $$R_{\text{sum}} \le I(\mathbf{X}; \mathbf{Y}_{\mathcal{T}^c} | \mathbf{Y}_{\mathcal{T}}) + \sum_{l \in \mathcal{T}} C_{\text{fh}}.$$ The first term is the mutual information through the wireless channel to the uncut APs (conditioned on the cut APs' observations), and the second is the total fronthaul capacity of the cut APs.

$\mathcal{T} = \emptyset$: $R_{\text{sum}} \le I(\mathbf{X}; \mathbf{Y})$ — the full cooperative MIMO capacity (no fronthaul constraint).

$\mathcal{T} = [L]$: $R_{\text{sum}} \le L \cdot C_{\text{fh}}$ — the sum rate cannot exceed the total fronthaul capacity (wireless channel not a bottleneck).

The tightest bound is the minimum over all $2^L$ possible cuts.

When users $k$ and $k'$ share a pilot, AP $l$ receives: $y_l = \sqrt{\tau_p p}(g_{kl} + g_{k'l}) + w_l$.

The MMSE estimate of $g_{kl}$ is: $$\hat{g}_{kl} = \frac{\sqrt{\tau_p p}\,\beta_{kl}}{\tau_p p(\beta_{kl} + \beta_{k'l}) + \sigma^2} y_l.$$

The key issue is that $\hat{g}_{kl}$ is correlated with $g_{k'l}$ (the contaminating user's channel). The interference term $\mathbb{E}[|\sum_l \hat{g}_{kl}^* g_{k'l}|^2]$ grows as $(\sum_l \gamma_{kl}\beta_{k'l}/\beta_{kl} \cdot \gamma_{kl})$, which scales as $L^2$ — the same rate as the desired signal.

Therefore SINR $\to \text{const}$ as $L \to \infty$. The ceiling is: $$\text{SINR}_{\infty} = \frac{(\sum_l \gamma_{kl})^2}{\sum_l \gamma_{kl}\beta_{k'l}\gamma_{kl}/\beta_{kl}} = \frac{(\sum_l \gamma_{kl})^2}{\sum_l \gamma_{kl}^2 \beta_{k'l}/\beta_{kl}}.$$

This ceiling depends on the spatial separation of the contaminating users: if $\beta_{k'l}/\beta_{kl} \ll 1$ for most APs (users are far apart), the ceiling is high. Pilot assignment must exploit this spatial structure.

Phase 1: $\mathbf{Y}^{(1)} = \mathbf{H}_{1} \mathbf{X}_1^{(1)} + \mathbf{H}_{2} \mathbf{X}_2^{(1)} + \mathbf{Z}^{(1)}$ where $\mathbf{H}_{k} \in \mathbb{C}^{2 \times 2}$. Inter-user channels: $\mathbf{Y}_{12} = \mathbf{H}_{12}\mathbf{X}_1^{(1)} + \mathbf{Z}_{12}$.

Phase 2: $\mathbf{Y}^{(2)} = [\mathbf{H}_{1}, \mathbf{H}_{2}]\begin{bmatrix}\mathbf{X}_1^{(2)} \\ \mathbf{X}_2^{(2)}\end{bmatrix} + \mathbf{Z}^{(2)}$ forming an effective $4 \times 2$ MIMO channel.

The achievable rate for user $k$ is: $$R_k \le \min\!\Big\{\alpha\log\det(\mathbf{I} + \mathbf{H}_{12}\mathbf{K}_1\mathbf{H}_{12}^H/\sigma^2),\; \alpha\log\det(\mathbf{I} + \mathbf{H}_{k}\mathbf{K}_k\mathbf{H}_{k}^{H}/\sigma^2) + (1-\alpha)\log\det(\mathbf{I} + \text{SNR}\,\tilde{\mathbf{H}}\tilde{\mathbf{K}}\tilde{\mathbf{H}}^H)\Big\}$$

where $\tilde{\mathbf{H}} = [\mathbf{H}_{1}, \mathbf{H}_{2}]$ is the effective $2 \times 4$ cooperative MIMO channel and $\tilde{\mathbf{K}}$ is the joint covariance matrix of the cooperative transmission.

The maximum multiplexing gain is $\min(N_{t,1} + N_{t,2}, N_r) = \min(4, 2) = 2$, limited by the destination's 2 antennas. Cooperation doubles the effective number of transmit antennas but does not increase the multiplexing gain beyond $N_r$. The benefit is in diversity: $d^*(r) = (N_{t,1} + N_{t,2} - r)(N_r - r)$.

For the single-relay Gaussian channel: $$C_{\text{cut-set}} = \min\!\Big\{\frac{1}{2}\log\!\Big(1 + \frac{(|h_1|^2 + |h_{12}|^2)P}{\sigma^2}\Big),\; C_{\text{fh}} + \frac{1}{2}\log\!\Big(1 + \frac{|h_1|^2 P}{\sigma^2}\Big)\Big\}.$$

With Gaussian quantization at distortion $D$: $$C_{\text{CF}} = \frac{1}{2}\log\!\Big(1 + \frac{|h_1|^2 P}{\sigma^2} + \frac{|h_{12}|^2 P}{\sigma^2 + D}\Big)$$ subject to $C_{\text{fh}} \ge \frac{1}{2}\log\!\Big(\frac{|h_{12}|^2 P + \sigma^2}{D}\Big)$.

Choosing $D = \sigma^2(|h_{12}|^2 P + \sigma^2)/(2^{2C_{\text{fh}}}(\sigma^2))$ and bounding the gap between CF and cut-set yields a constant independent of channel parameters. The gap is at most $\frac{1}{2}\log(2\pi e / 12) \approx 0.254$ bits, following the approach of El Gamal and Kim (2011, Theorem 16.4).

The per-user SINR is $\text{SINR}_k = f_k(\boldsymbol{\eta})/g_k(\boldsymbol{\eta})$ where both $f_k$ and $g_k$ are linear in $\boldsymbol{\eta}$. Therefore $R_k = \log(1 + f_k/g_k)$ is quasi-concave in $\boldsymbol{\eta}$ (a log of a ratio of affine functions). The minimum of quasi-concave functions is quasi-concave, so the problem is quasi-convex.

Fix a target rate $R_{\min}$. The feasibility problem $\{R_k(\boldsymbol{\eta}) \ge R_{\min},\; \forall k\}$ is equivalent to $\{\text{SINR}_k \ge 2^{R_{\min}} - 1,\; \forall k\}$, which is a system of linear inequalities in $\boldsymbol{\eta}$ (after rearranging the SINR expression). This is a linear program, solvable efficiently.

Bisect on $R_{\min}$: if feasible, increase $R_{\min}$; if infeasible, decrease. Converges to the optimal max-min rate in $O(\log(1/\epsilon))$ LP solves.

The CP computes $x_l = \sum_k v_{kl}s_k$ and quantizes: $\hat{x}_l = x_l + q_l$ with $q_l \sim \mathcal{CN}(0, D_l)$. User $k$ receives: $Y_k = \sum_l g_{kl}\hat{x}_l + Z_k = \sum_l g_{kl}(x_l + q_l) + Z_k = \mathbf{g}_k^T\mathbf{V}\mathbf{s} + \mathbf{g}_k^T\mathbf{q} + Z_k.$

$R_k = \log\!\Big(1 + \frac{|\mathbf{g}_k^T\mathbf{v}_k|^2} {\sum_{k' \ne k}|\mathbf{g}_k^T\mathbf{v}_{k'}|^2 + \sum_l |g_{kl}|^2 D_l + \sigma^2}\Big)$.

Subject to: $\sum_k |v_{kl}|^2 + D_l \le P_l$ (per-AP power), $\frac{1}{2}\log(\sum_k |v_{kl}|^2 / D_l + 1) \le C_{\text{fh}}$ (fronthaul).

The rate expression is non-convex because the SINR is a ratio of quadratic forms in $\mathbf{v}_k$. Using the weighted MMSE equivalence (Shi et al., 2011), the problem can be reformulated as a sequence of convex subproblems:

1. Fix $\{\mathbf{v}_k\}$, optimize $\{D_l\}$ (convex in $D_l$).
2. Fix $\{D_l\}$, optimize $\{\mathbf{v}_k\}$ (WMMSE is convex).
3. Iterate until convergence.

Each iteration increases the sum-rate, and the sequence converges to a stationary point of the non-convex problem.

The channel matrix $\mathbf{G} \in \mathbb{C}^{L \times K}$ has entries $g_{lk} = \sqrt{\beta_{lk}} h_{lk}$ with $h_{lk} \sim \mathcal{CN}(0,1)$ i.i.d. The MMSE receiver for user $k$ is: $\mathbf{w}_k = (\mathbf{G}\mathbf{P}\mathbf{G}^H + \sigma^2\mathbf{I}_L)^{-1}\mathbf{g}_k$.

By the resolvent identity and random matrix theory, as $L, K \to \infty$: $\text{SINR}_k \xrightarrow{\text{a.s.}} \bar{\text{SINR}}_k = p_k \mathbf{b}_k^T(\mathbf{B}\mathbf{P}\mathbf{T} + \sigma^2\mathbf{I})^{-1}\mathbf{b}_k$ where $\mathbf{b}_k = [\beta_{1k}, \ldots, \beta_{Lk}]^T$, $\mathbf{B} = \text{diag}(\mathbf{b}_1, \ldots, \mathbf{b}_K)$, and $\mathbf{T} = \text{diag}(t_1, \ldots, t_K)$ with $t_k$ determined by a system of fixed-point equations.

As $\alpha = K/L \to 0$, the MMSE receiver converges to the matched filter because inter-user interference vanishes. The SINR becomes $p_k \|\mathbf{b}_k\|^2 / \sigma^2 = p_k \sum_l \beta_{lk}^2 / \sigma^2$, which matches the MF result when $\gamma_{kl} = \beta_{kl}$ (perfect CSI).

1. Place 64 APs on an $8 \times 8$ grid with 125 m spacing.
2. Drop $K = 8$ users uniformly at random.
3. Compute $\beta_{kl}$ using the path-loss model (with 10 m minimum distance).
4. Generate i.i.d. Rayleigh small-scale fading.
5. Compute per-user rates for each scheme.
6. Repeat for 1000 random user drops to build CDFs.

Cell-free (MF): moderate median rate, high 5th-percentile rate. Cell-free (MMSE): highest median and 5th-percentile rate. Co-located: high median but low 5th-percentile (cell-edge users suffer). Small cells: lowest performance due to lack of cooperation.

The key metric is the 5th-percentile: cell-free should outperform co-located by 2-5x at the cell edge, validating the "no cell boundaries" claim.