Exercises

(a) $\mathbb{E}[\|\mathbf{h}_k\|^2] = M\beta_k = 128 \times 0.5 = 64$

$\text{Var}[\|\mathbf{h}_k\|^2] = M\beta_k^2 = 128 \times 0.25 = 32$

(b) For $Z = \|\mathbf{h}_k\|^2/M$: $\text{CV} = \sqrt{\text{Var}[Z]}/\mathbb{E}[Z] = \sqrt{0.25/128}/0.5 = (0.0442)/0.5 = 0.0884$

The channel gain fluctuates by only about 8.8% around its mean.

(c) $\Pr[|Z - 0.5| > 0.1] \leq \frac{\beta_k^2}{M\varepsilon^2} = \frac{0.25}{128 \times 0.01} = 0.195$

There is at most a 19.5% chance that $Z$ deviates from its mean by more than 0.1. $\blacksquare$

$\|\mathbf{h}_k\|^2 = \frac{\kappa}{\kappa+1}\|\bar{\mathbf{h}}_k\|^2 + \frac{1}{\kappa+1}\|\mathbf{g}_k\|^2 + \frac{2\sqrt{\kappa}}{(\kappa+1)}\text{Re}(\bar{\mathbf{h}}_k^H\mathbf{g}_k)$$

$\mathbb{E}\!\left[\frac{\|\mathbf{h}_k\|^2}{M}\right] = \frac{\kappa}{\kappa+1} + \frac{1}{\kappa+1} = 1KATEXPLACEHOLDER0END\text{Var}\!\left[\frac{\|\mathbf{h}_k\|^2}{M}\right] = \frac{1}{(\kappa+1)^2}\frac{1}{M} + \frac{4\kappa}{(\kappa+1)^2}\frac{1}{M} = \frac{1+4\kappa}{(\kappa+1)^2 M}$$This variance vanishes as$M \to \infty$, so the Rician channel also hardens. In fact, the LoS component hardens **faster** because the dominant term$\frac{\kappa}{\kappa+1}M$is deterministic. The Rician channel exhibits stronger hardening than Rayleigh for large$\kappa$.$\blacksquare$

(a) $\mathbb{E}[[\frac{1}{M}\mathbf{H}^{H}\mathbf{H}]_{kk}] = \beta_k = 1$

(b) $\mathbb{E}[[\frac{1}{M}\mathbf{H}^{H}\mathbf{H}]_{ij}] = 0$ for $i \neq j$

(c) Expected off-diagonal energy: $K(K-1) \times \frac{\beta_i\beta_j}{M} = 8 \times 7 \times \frac{1}{64} = 0.875$

Expected diagonal energy: $K \times (\beta_k^2 + \beta_k^2/M) \approx K\beta_k^2 = 8$

Ratio $\approx 0.875/8 = 0.109$.

The off-diagonal energy is about 11% of the diagonal, confirming that favourable propagation is already evident at $M/K = 8$. $\blacksquare$

$\frac{1}{M}\mathbf{h}_1^H\mathbf{h}_2 = \frac{1}{M}\mathbf{g}_1^H\mathbf{R}\mathbf{g}_2KATEXPLACEHOLDER0END\mathbb{E}\!\left[\frac{1}{M}\mathbf{g}_1^H\mathbf{R}\mathbf{g}_2\right] = 0KATEXPLACEHOLDER1END\text{Var}\!\left[\frac{1}{M}\mathbf{g}_1^H\mathbf{R}\mathbf{g}_2\right] = \frac{1}{M^2}\text{tr}(\mathbf{R}\mathbf{R}^{H}) = \frac{\|\mathbf{R}\|_F^2}{M^2}$$By the SLLN,$\frac{1}{M}\mathbf{h}_1^H\mathbf{h}_2 \to 0$ still holds. Favourable propagation is maintained for independent users even with identical correlation.

Favourable propagation fails when the correlation subspaces overlap significantly. If $\mathbf{R}$ has rank $r \ll M$, then channels are confined to an $r$-dimensional subspace and $\frac{1}{M}\mathbf{h}_1^H\mathbf{h}_2 \to 0$ but with variance $\frac{\|\mathbf{R}\|_F^2}{M^2}$ which may decrease slowly. The effective dimensionality, not just $M$, determines the degree of orthogonality. $\blacksquare$

$\frac{\mathbf{h}_i^H\mathbf{h}_j}{M} = \frac{1}{M}\mathbf{g}_i^H\mathbf{R}_{i}^{1/2}\mathbf{R}_{j}^{1/2}\mathbf{g}_jKATEXPLACEHOLDER0END\mathbb{E}\!\left[\left|\frac{\mathbf{h}_i^H\mathbf{h}_j}{M}\right|^2\right] = \frac{1}{M^2}\mathbb{E}\!\left[\mathbf{g}_i^H\mathbf{R}_{i}^{1/2}\mathbf{R}_{j}^{1/2}\mathbf{g}_j \mathbf{g}_j^H\mathbf{R}_{j}^{1/2}\mathbf{R}_{i}^{1/2}\mathbf{g}_i\right]KATEXPLACEHOLDER1END= \frac{1}{M^2}\text{tr}(\mathbf{R}_{i}^{1/2}\mathbf{R}_{j}\mathbf{R}_{i}^{1/2}) = \frac{\text{tr}(\mathbf{R}_{i}\mathbf{R}_{j})}{M^2}$$

This vanishes as $M \to \infty$ if and only if $\text{tr}(\mathbf{R}_{i}\mathbf{R}_{j}) = o(M^2)$.

• i.i.d. case: $\mathbf{R}_{k} = \beta_k\mathbf{I}$, so $\text{tr}(\mathbf{R}_{i}\mathbf{R}_{j}) = M\beta_i\beta_j = O(M)$. Vanishes.
• Non-overlapping supports: If $\mathbf{R}_{i}$ and $\mathbf{R}_{j}$ have orthogonal eigenvectors, $\text{tr}(\mathbf{R}_{i}\mathbf{R}_{j}) = 0$. Vanishes trivially.
• Identical rank-$r$ correlation: $\text{tr}(\mathbf{R}^{2}) \sim r\lambda_{\max}^2$. If $r = O(1)$, this is $O(1) \ll M^2$. Still vanishes.
• Identical full-rank: $\text{tr}(\mathbf{R}^{2}) = \|\mathbf{R}\|_F^2$. If eigenvalues are $O(1)$, $\text{tr}(\mathbf{R}^{2}) = O(M)$. Vanishes. $\blacksquare$

$\text{SINR}^{\text{MR}} = \frac{256}{15 + 1} = 16$

$R^{\text{MR}} = \log_2(17) \approx 4.09$ bits/s/Hz

$\text{SINR}^{\text{ZF}} = (256 - 16)/1 = 240$

$R^{\text{ZF}} = \log_2(241) \approx 7.91$ bits/s/Hz

The ZF-MR gap is $7.91 - 4.09 = 3.82$ bits/s/Hz. At $M/K = 16$, ZF substantially outperforms MR. For $M = 1024$ ($M/K = 64$), the MR SINR would be $1024/16 = 64$ ($R = 6.02$) while ZF gives $1008/1 = 1008$ ($R = 9.98$), and the gap narrows to 3.96 bits/s/Hz. The relative gap (as a fraction) decreases as $M$ grows. $\blacksquare$

$y_k = \sum_{j=1}^{K}\sqrt{\eta_j}\frac{\mathbf{h}_k^T\mathbf{h}_j^*}{\|\mathbf{h}_j\|}s_j + n_k$$The desired signal term ($j = k$):$\sqrt{\eta_k}\frac{|\mathbf{h}_k|^2}{|\mathbf{h}_k|}s_k = \sqrt{\eta_k}|\mathbf{h}_k|s_k$

By channel hardening: $\|\mathbf{h}_k\| \approx \sqrt{M\beta_k}$.

By favourable propagation: $|\mathbf{h}_k^T\mathbf{h}_j^*/\|\mathbf{h}_j\||^2 \approx \beta_k$ for $j \neq k$.

$$\text{SINR}_k^{\text{DL}} = \frac{\eta_k M\beta_k} {\sum_{j \neq k}\eta_j\beta_k + \sigma^2} \approx \frac{\eta_k M\beta_k}{\sum_{j\neq k}\eta_j\beta_j + \sigma^2}$$

using the UatF approach. $\blacksquare$

The ZF estimate for user $k$ uses the projection of $\mathbf{h}_k$ onto $\text{span}(\mathbf{h}_1,\ldots,\mathbf{h}_{k-1},\mathbf{h}_{k+1},\ldots,\mathbf{h}_K)^{\perp}$:

$$[(\mathbf{H}^{H}\mathbf{H})^{-1}]_{kk} = \frac{1}{\|\mathbf{P}_k^{\perp}\mathbf{h}_k\|^2}$$

where $\mathbf{P}_k^{\perp}$ projects onto the $(M-K+1)$-dimensional orthogonal complement.

Since $\mathbf{h}_k = \sqrt{\beta_k}\mathbf{g}_k$ with $\mathbf{g}_k \sim \mathcal{CN}(\mathbf{0}, \mathbf{I}_M)$ independent of the other channels:

$$\|\mathbf{P}_k^{\perp}\mathbf{h}_k\|^2 = \beta_k\|\mathbf{P}_k^{\perp}\mathbf{g}_k\|^2 \sim \beta_k \cdot \text{Gamma}(M-K+1, 1)$$

This is a chi-squared distribution with $2(M-K+1)$ real degrees of freedom, scaled by $\beta_k$.

The post-ZF SNR is: $\text{SNR}_{k} = p_k\|\mathbf{P}_k^{\perp}\mathbf{h}_k\|^2/\sigma^2$.

The outage probability at rate $R_0$ is: $$P_{\text{out}} = \Pr[\text{SNR}_{k} < 2^{R_0}-1] \propto \text{SNR}^{-(M-K+1)}$$

at high SNR. The diversity order is $M - K + 1$, which grows with $M$. For $M = 100, K = 10$, the diversity order is 91. $\blacksquare$

(a) $R^{\infty} = \log_2\!\left(1 + \frac{1}{\alpha^2}\right)$

For $\alpha = 0.1$: $R^{\infty} = \log_2(101) = 6.66$ bits/s/Hz. For $\alpha = 0.5$: $R^{\infty} = \log_2(5) = 2.32$ bits/s/Hz.

(b) $\log_2(1 + 1/\alpha^2) = 1 \Rightarrow 1/\alpha^2 = 1 \Rightarrow \alpha = 1$.

When $\alpha = 1$, the inter-cell and intra-cell path losses are equal (worst case), and the ceiling is just 1 bit/s/Hz.

(c) For $\alpha = 0.1$ with SNR = 0 dB:

With contamination: $R(M) = \log_2\!\left(1 + \frac{M}{0.01M + 1}\right)$

Without contamination: $R(M) = \log_2(1 + M)$

At $M = 100$: contaminated $= \log_2(1 + 100/2) = 5.67$, uncontaminated $= \log_2(101) = 6.66$. $\blacksquare$

(a) Reuse 1: $\tau_p = K$ pilots per cell. Reuse 3: $\tau_p = 3K$ pilots per cell (3 different pilot sets).

With $K = 10$: reuse 1 needs 10 pilots, reuse 3 needs 30 pilots.

(b) With reuse 3, only $\sim 2$ cells (at distance $2R$) share each pilot set.

$$R^{\infty} = \log_2\!\left(1 + \frac{1}{2 \times 0.01^2}\right) = \log_2(1 + 5000) \approx 12.3\;\text{bits/s/Hz}$$

vs. reuse 1 with 6 adjacent cells at $\beta = 0.05$:

$$R^{\infty}_{\text{reuse 1}} = \log_2\!\left(1 + \frac{1}{6 \times 0.0025}\right) = \log_2(67.7) \approx 6.08\;\text{bits/s/Hz}$$

(c) If $\tau_c = 200$ symbols, the pilot overhead is:

• Reuse 1: $10/200 = 5\%$
• Reuse 3: $30/200 = 15\%$

The effective rate is $R_{\text{eff}} = (1 - \tau_p/\tau_c)R$. At the ceiling: reuse 1 gives $0.95 \times 6.08 = 5.78$, reuse 3 gives $0.85 \times 12.3 = 10.5$ bits/s/Hz.

Reuse 3 nearly doubles the effective rate despite the increased overhead. $\blacksquare$

(a) $\beta_{\min} = 0.01$, $\eta = 200 \times 0.01 = 2$ mW.

$p_1 = 2/1 = 2$ mW, $p_2 = 2/0.1 = 20$ mW, $p_3 = 2/0.01 = 200$ mW.

(b) $p_1 = 3$ dBm, $p_2 = 13$ dBm, $p_3 = 23$ dBm.

(c) User 1 reduces from $P_{\max} = 200$ mW to 2 mW, a factor of 100 (20 dB reduction). $\blacksquare$

$\max_{\{p_k\}} \min_k \frac{p_k(M-K)\beta_k}{\sigma^2} \quad \text{s.t. } 0 \leq p_k \leq P_{\max}$$Since each SINR is independent, equalising them gives:$p_k\beta_k = c$for all$k$, i.e.,$p_k = c/\beta_k$. Setting$\max_k p_k = P_{\max}$:$c = P_{\max}\beta_{\min}$.

The optimal power allocation is identical to MR: $p_k^{\star} = P_{\max}\beta_{\min}/\beta_k$.

However, the achievable SINR is different:

• MR: $\text{SINR}^{\star} = P_{\max}\beta_{\min}M/((K-1)P_{\max}\beta_{\min}+\sigma^2)$
• ZF: $\text{SINR}^{\star} = P_{\max}\beta_{\min}(M-K)/\sigma^2$

ZF provides a higher equalised rate because it eliminates the inter-user interference term $(K-1)P_{\max}\beta_{\min}$ from the denominator. $\blacksquare$

Nearby APs: $4 \times 50^{-3.8} = 4 \times 2.19 \times 10^{-7} = 8.74 \times 10^{-7}$

Wait — let us normalise. Set $\beta_{nk} = (d_{nk}/d_0)^{-3.8}$ with $d_0 = 1$ m.

$4 \times 50^{-3.8} = 4 / 50^{3.8} = 4/3.01 \times 10^6 = 1.33 \times 10^{-6}$

This gets unwieldy. Let us use $d_0 = 50$ m as reference ($\beta = 1$ at 50 m):

$4 \times 1 + 12 \times (50/150)^{3.8} + 84 \times (50/400)^{3.8}$ $= 4 + 12 \times 0.333^{3.8} + 84 \times 0.125^{3.8}$ $= 4 + 12 \times 0.0148 + 84 \times 0.000366$ $= 4 + 0.178 + 0.031 = 4.21$

Co-located BS at 500 m with 100 antennas: $100 \times (50/500)^{3.8} = 100 \times 0.1^{3.8} = 100 \times 1.58 \times 10^{-4} = 0.0158$

Ratio: $4.21 / 0.0158 = 266$.

The cell-free architecture provides 266$\times$ (24 dB) more effective channel gain for this user. $\blacksquare$

After correlating with $\phi_k^*$:

$$\tilde{y}_{nk} = \sqrt{\tau_p p_p}\,g_{nk} + \sqrt{\tau_p p_p}\sum_{j \in \mathcal{P}_k \setminus \{k\}}g_{nj} + \tilde{w}_{nk}$$

where $\mathcal{P}_k$ is the set of users sharing pilot $k$ and $\tilde{w}_{nk} \sim \mathcal{CN}(0, \sigma^2)$.

$\hat{g}_{nk} = \frac{\sqrt{\tau_p p_p}\,\beta_{nk}} {\tau_p p_p\sum_{j \in \mathcal{P}_k}\beta_{nj} + \sigma^2} \tilde{y}_{nk}KATEXPLACEHOLDER0END\gamma_{nk} = \mathbb{E}[|\hat{g}_{nk}|^2] = \frac{\tau_p p_p\,\beta_{nk}^2} {\tau_p p_p\sum_{j \in \mathcal{P}_k}\beta_{nj} + \sigma^2}KATEXPLACEHOLDER1END\gamma_{nk} = \frac{\tau_p p_p\,\beta_{nk}^2} {\tau_p p_p\,\beta_{nk} + \sigma^2}$$$\blacksquare$

(a) $P_{\text{total}} = 5 + 6.4 + 10 = 21.4$ W.

Throughput $= 50 \times 20 \times 10^6 = 10^9$ bits/s = 1 Gbps.

$\text{EE} = 10^9/21.4 = 46.7$ Mbits/joule.

(b) $P_{\text{total}} = 5 + 12.8 + 10 = 27.8$ W.

Throughput $= 60 \times 20 \times 10^6 = 1.2$ Gbps.

$\text{EE} = 1.2 \times 10^9/27.8 = 43.2$ Mbits/joule.

The EE decreased by 7.5% despite the rate increase. This suggests $M = 64$ is closer to $M^{\star}$ than $M = 128$ for these parameters. $\blacksquare$

With $p = E_u/M^{\alpha}$:

$$\text{SINR}_k = \frac{(E_u/M^{\alpha})M\beta_k} {(K-1)(E_u/M^{\alpha})\beta_{\text{avg}} + \sigma^2} = \frac{E_u M^{1-\alpha}\beta_k} {(K-1)E_u M^{-\alpha}\beta_{\text{avg}} + \sigma^2}$$

As $M \to \infty$:

• Numerator grows as $M^{1-\alpha}$
• Denominator: first term decays as $M^{-\alpha}$, second is constant

So $\text{SINR} \sim M^{1-\alpha}/\sigma^2$.

For $p \propto 1/\sqrt{M}$: $\alpha = 1/2$.

$\text{SINR} \sim M^{1/2}$, so the rate grows as $\frac{1}{2}\log_2 M$.

The total transmit power per user is $p = E_u/\sqrt{M}$, which decreases with $M$. With $M = 100$, each user needs 10$\times$ less power than with a single antenna. $\blacksquare$

$\text{EE}(M) \geq t$ iff $R(M) \geq t P(M) = t(aM + b)$.

Define $\phi(M) = R(M) - t(aM + b)$.

$\phi(M)$ is concave in $M$ because $R(M)$ is concave and $t(aM+b)$ is affine.

The superlevel set $\{M : \phi(M) \geq 0\}$ of a concave function is a convex set (an interval in one dimension).

Therefore $\{M : \text{EE}(M) \geq t\}$ is convex for all $t$, which is the definition of quasi-concavity.

Since $R(M) \to \infty$ (logarithmically) and $P(M) \to \infty$ (linearly), and $R(M)/P(M) \to 0$ as $M \to \infty$, while $R(0)/P(0) = 0$, the quasi-concave EE function achieves a unique interior maximum $M^{\star} > 0$. $\blacksquare$

(a) With local MR at AP $n$ and UatF bound:

$$\text{SINR}_k = \frac{p_k\left(\sum_{n=1}^{N}\gamma_{nk}\right)^2} {\sum_{j=1}^{K}p_j\sum_{n=1}^{N}\gamma_{nk}\beta_{nj} + \sigma^2\sum_{n=1}^{N}\gamma_{nk} + \sum_{j \in \mathcal{P}_k \setminus \{k\}}p_j\left(\sum_{n=1}^{N}c_{nk}\sqrt{\tau_p p_p}\beta_{nj}\right)^2}$$

The last term is the coherent pilot contamination.

(b) In cell-free, the coherent contamination from user $j$ is:

$$\left(\sum_{n=1}^{N}c_{nk}\sqrt{\tau_p p_p}\beta_{nj}\right)^2$$

As $N \to \infty$, the APs near user $k$ (which dominate $\sum_n c_{nk}\beta_{nk}$) are far from user $j$, so $\beta_{nj} \approx 0$ for those APs.

The ratio of contamination to signal converges to zero:

$$\frac{\sum_n c_{nk}\beta_{nj}}{\sum_n c_{nk}\beta_{nk}} \to 0$$

(c) In co-located massive MIMO, all $M$ antennas see the same $\beta_{jk}^{(\ell)}$, so the contamination ratio is $\beta_{jk}^{(\ell)}/\beta_{\ell k}^{(\ell)}$, which is a fixed positive number independent of $M$.

Cell-free's distributed geometry is its fundamental advantage: it converts the spatially constant contamination of co-located systems into a spatially varying quantity that averages out. $\blacksquare$

For $M < M^{\star}_{\text{EE}}$: both sum rate and EE increase with $M$, so these points are dominated by $M^{\star}_{\text{EE}}$.

For $M > M^{\star}_{\text{EE}}$: sum rate increases but EE decreases. These points form the Pareto frontier — improving one objective requires sacrificing the other.

The frontier is the set:

$$\{(R(M), \text{EE}(M)) : M \in [M^{\star}_{\text{EE}}, M_{\max}]\}$$

At $M = M^{\star}_{\text{EE}}$: maximum EE, moderate sum rate. At $M = M_{\max}$: maximum sum rate, lower EE.

The $\alpha$-weighted problem $\max_M \alpha \cdot R(M) + (1-\alpha) \cdot \text{EE}(M)$ selects a point on the Pareto frontier that trades off rate for efficiency. For $\alpha = 0$: $M = M^{\star}_{\text{EE}}$. For $\alpha = 1$: $M = M_{\max}$. Intermediate $\alpha$ values select intermediate $M$ values. $\blacksquare$