Exercises

(a) $v_n = h_{r,n}^* g_n$: $v_1 = 1 \cdot 1 = 1$, $v_2 = 1 \cdot j = j$, $v_3 = 1 \cdot (-1) = -1$, $v_4 = 1 \cdot (-j) = -j$.

(b) Align each $e^{j\theta_n} v_n$ to be real and positive (same phase as $h_d = 0.5 > 0$):

$\theta_1^* = -\angle v_1 = 0$, $\theta_2^* = -\angle v_2 = -\pi/2$, $\theta_3^* = -\angle v_3 = -\pi$ (equivalently $\pi$), $\theta_4^* = -\angle v_4 = \pi/2$.

(c) With optimal phases: $e^{j\theta_n^*} v_n = |v_n| = 1$ for all $n$.

Effective channel: $h_d + \sum_n |v_n| = 0.5 + 4 \times 1 = 4.5$.

$\text{SNR} = (P/\sigma^2) \times |4.5|^2 = 10 \times 20.25 = 202.5$ $= 23.1$ dB.

(d) Without RIS: $\text{SNR}_{\text{direct}} = (P/\sigma^2) \times |h_d|^2 = 10 \times 0.25 = 2.5 = 4.0$ dB.

RIS gain: $23.1 - 4.0 = 19.1$ dB. The RIS with only 4 elements provides a dramatic improvement because the direct link is weak ($|h_d| = 0.5$) while the cascaded channels have unit magnitude. $\blacksquare$

(a) For Rayleigh fading with $\mathbb{E}[|h|^2] = 1$: $\mu_h = \mu_g = \sqrt{\pi/4} = 0.8862$.

(b) Optimised: $\text{SNR}_{\text{opt}} \approx (P/\sigma^2) N^2 \mu_h^2 \mu_g^2$. Random: $\text{SNR}_{\text{rand}} = (P/\sigma^2) N$ (each element contributes independently with power $\mathbb{E}[|v_n|^2] = 1$).

Gain: $N \mu_h^2 \mu_g^2 = 100 \times (\pi/4)^2 = 100 \times 0.617 = 61.7$. In dB: $10\log_{10}(61.7) = 17.9$ dB.

(c) $\text{SNR}_{\text{opt}} = 1 \times 100^2 \times (\pi/4)^2 = 6170 = 37.9$ dB.

(d) $20 \text{ dB} = 100$. Need $(P/\sigma^2) N^2 \mu_h^2 \mu_g^2 = 100$. $N^2 = 100 / (1 \times 0.617) = 162.1$. $N = \lceil\sqrt{162.1}\rceil = 13$.

Only 13 elements suffice for 20 dB SNR at 0 dB base SNR. $\blacksquare$

(a) $\mathrm{sinc}(\pi/2) = \sin(\pi/2)/(\pi/2) = 1/(pi/2) = 2/\pi = 0.6366$. Power loss: $(2/\pi)^2 = 0.4053$. In dB: $-10\log_{10}(0.4053) = 3.92$ dB.

(b) $\mathrm{sinc}(\pi/4) = \sin(\pi/4)/(\pi/4) = (\sqrt{2}/2)/0.7854 = 0.9003$. Power loss: $0.9003^2 = 0.8105$. In dB: $-10\log_{10}(0.8105) = 0.91$ dB.

(c) $\mathrm{sinc}(\pi/8) = \sin(\pi/8)/(\pi/8) = 0.3827/0.3927 = 0.9745$. Power loss: $0.9745^2 = 0.9497$. In dB: $-10\log_{10}(0.9497) = 0.22$ dB.

(d) Need $-10\log_{10}(\mathrm{sinc}^2(\pi/2^b)) < 0.1$ dB. $\mathrm{sinc}^2(\pi/2^b) > 10^{-0.01} = 0.9772$.

At $b = 3$: $0.9497 < 0.9772$ (fails). At $b = 4$: $\mathrm{sinc}(\pi/16) = \sin(0.1963)/0.1963 = 0.1951/0.1963 = 0.9936$. $0.9936^2 = 0.9872 > 0.9772$ (passes).

$b = 4$ bits suffice for $< 0.1$ dB loss. $\blacksquare$

(a) Overhead: $T/T_c = 65/200 = 32.5\%$. This leaves only $135$ symbols for data, a significant loss.

(b) Groups: $\lceil 64/4 \rceil = 16$. Pilot slots: $16 + 1 = 17$. Overhead: $17/200 = 8.5\%$.

(c) $T_{\text{CS}} = O(S \log(N/S)) = O(8 \log(64/8)) = O(8 \times 3) = O(24)$. Approximately 24 pilot slots, overhead $\approx 12\%$.

(d) Full DFT: highest accuracy (achieves CRLB) but largest overhead. Grouping: lower overhead but loses within-group spatial resolution; beamforming gains are limited by the group approximation. Compressed sensing: moderate overhead with near-full resolution, but requires angular sparsity and more complex algorithms (OMP/LASSO); performance degrades if the channel is not truly sparse. $\blacksquare$

(a) $\mathbf{G}^H \boldsymbol{\Theta} \mathbf{h}_r = \mathbf{G}^H \mathrm{diag}(\phi_1, \phi_2) [1, 1]^T = \mathbf{G}^H [\phi_1, \phi_2]^T = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} [\phi_1, \phi_2]^T = [\phi_1, \phi_1 + \phi_2]^T$.

$\mathbf{h}_{\mathrm{eff}} = [1, 0]^T + [\phi_1, \phi_1 + \phi_2]^T = [1 + \phi_1, \phi_1 + \phi_2]^T$.

(b) $\boldsymbol{\phi} = [1, 1]^T$: $\mathbf{h}_{\mathrm{eff}} = [2, 2]^T$. $\|\mathbf{h}_{\mathrm{eff}}\|^2 = 8$. SNR $= 10 \times 8 = 80 = 19.0$ dB.

(c) $\text{SNR} = 10 \|\mathbf{h}_{\mathrm{eff}}\|^2 = 10(|1 + \phi_1|^2 + |\phi_1 + \phi_2|^2)$.

Let $\phi_1 = e^{j\alpha}$, $\phi_2 = e^{j\beta}$. $f(\alpha, \beta) = |1 + e^{j\alpha}|^2 + |e^{j\alpha} + e^{j\beta}|^2 = 2(1 + \cos\alpha) + 2(1 + \cos(\alpha - \beta))$.

$\partial f / \partial \alpha = -2\sin\alpha - 2\sin(\alpha - \beta) = 0$ $\partial f / \partial \beta = 2\sin(\alpha - \beta) = 0$

From the second equation: $\alpha = \beta$. Substituting: $-2\sin\alpha - 0 = 0$, so $\alpha = 0$.

Optimal: $\phi_1^* = \phi_2^* = 1$, $\mathbf{h}_{\mathrm{eff}}^* = [2, 2]^T$. SNR $= 80 = 19.0$ dB.

(d) In this particular case, the all-ones configuration is already optimal because the channels are real-valued and positive. The gain over all-ones is 0 dB. This illustrates that when the channels are already aligned, phase optimisation provides no additional benefit. $\blacksquare$

(a) Define $\bar{\mathbf{q}} = [c, a_1, \ldots, a_N]^T$ and $\bar{\boldsymbol{\phi}} = [1, \phi_1, \ldots, \phi_N]^T$. Then $c + \mathbf{a}^T \boldsymbol{\phi} = \bar{\mathbf{q}}^T \bar{\boldsymbol{\phi}}$ and the objective is:

$$|\bar{\mathbf{q}}^T \bar{\boldsymbol{\phi}}|^2 = \bar{\boldsymbol{\phi}}^H \bar{\mathbf{q}}^* \bar{\mathbf{q}}^T \bar{\boldsymbol{\phi}} = \bar{\boldsymbol{\phi}}^H \mathbf{C} \bar{\boldsymbol{\phi}}$$

where $\mathbf{C} = \bar{\mathbf{q}}^* \bar{\mathbf{q}}^T \in \mathbb{C}^{(N+1) \times (N+1)}$.

(b) Let $\bar{\boldsymbol{\Phi}} = \bar{\boldsymbol{\phi}} \bar{\boldsymbol{\phi}}^H$. The objective becomes $\mathrm{tr}(\mathbf{C} \bar{\boldsymbol{\Phi}})$. The constraints $|\bar{\phi}_n| = 1$ translate to $\bar{\Phi}_{nn} = 1$ for $n = 0, 1, \ldots, N$. The constraint $\mathrm{rank}(\bar{\boldsymbol{\Phi}}) = 1$ is non-convex.

SDP relaxation: $$\max_{\bar{\boldsymbol{\Phi}} \succeq \mathbf{0}} \; \mathrm{tr}(\mathbf{C} \bar{\boldsymbol{\Phi}}) \quad \text{s.t.} \; \bar{\Phi}_{nn} = 1, \; n = 0, 1, \ldots, N$$

This is a convex SDP solvable by interior-point methods in $O(N^{3.5})$ time.

(c) The SDR is tight (rank-one optimal) when:

• $\mathbf{C}$ has rank one (always true here since $\mathbf{C} = \bar{\mathbf{q}}^* \bar{\mathbf{q}}^T$);
• The optimal dual variable satisfies certain complementarity conditions.

Wu and Zhang (2019) showed that for the single-user problem, the SDR achieves an approximation ratio of $\pi/4$, meaning the randomised solution is within a factor $\pi/4 \approx 0.785$ of the SDP upper bound.

(d) If $\bar{\boldsymbol{\Phi}}^*$ has rank $> 1$:

1. Generate $L$ random vectors $\bar{\boldsymbol{\phi}}_\ell \sim \mathcal{CN}(\mathbf{0}, \bar{\boldsymbol{\Phi}}^*)$.
2. Project: $\hat{\phi}_{\ell,n} = e^{j\angle \bar{\phi}_{\ell,n}}$.
3. Evaluate $|\bar{\mathbf{q}}^T \hat{\bar{\boldsymbol{\phi}}}_\ell|^2$ for each $\ell$.
4. Return the $\hat{\bar{\boldsymbol{\phi}}}_\ell$ with the largest objective.

Typically $L = 100$--$1000$ suffices. $\blacksquare$

(a) $f(\mathbf{w}, \boldsymbol{\phi}) = |\mathbf{h}_{\mathrm{eff}}^H(\boldsymbol{\phi}) \mathbf{w}|^2 / \sigma^2$ where $\mathbf{h}_{\mathrm{eff}}(\boldsymbol{\phi}) = \mathbf{h}_d + \mathbf{G}^H \mathrm{diag}(\boldsymbol{\phi}) \mathbf{h}_r$.

(b) $\mathbf{w}^{(i+1)} = \arg\max_{\|\mathbf{w}\|^2 \leq P} f(\mathbf{w}, \boldsymbol{\phi}^{(i)})$. Since $\mathbf{w}^{(i)}$ is in the feasible set and $\mathbf{w}^{(i+1)}$ is the maximiser: $f(\mathbf{w}^{(i+1)}, \boldsymbol{\phi}^{(i)}) \geq f(\mathbf{w}^{(i)}, \boldsymbol{\phi}^{(i)})$.

(c) $\boldsymbol{\phi}^{(i+1)} = \arg\max_{|\phi_n|=1} f(\mathbf{w}^{(i+1)}, \boldsymbol{\phi})$. Since $\boldsymbol{\phi}^{(i)}$ is feasible: $f(\mathbf{w}^{(i+1)}, \boldsymbol{\phi}^{(i+1)}) \geq f(\mathbf{w}^{(i+1)}, \boldsymbol{\phi}^{(i)})$.

(d) Combining (b) and (c): $\text{SNR}^{(i+1)} = f(\mathbf{w}^{(i+1)}, \boldsymbol{\phi}^{(i+1)}) \geq f(\mathbf{w}^{(i)}, \boldsymbol{\phi}^{(i)}) = \text{SNR}^{(i)}$.

The sequence is non-decreasing and bounded above (finite power, finite channel gains), so it converges by the monotone convergence theorem.

Why not global optimality: The unit-modulus constraint set $|\phi_n| = 1$ is non-convex. Each passive subproblem may converge to a local optimum rather than the global one. The final solution depends on initialisation. $\blacksquare$

(a) Let $q = c + \mathbf{a}^T \boldsymbol{\phi}$. Then $f = |q|^2 = q \bar{q}$. Using Wirtinger calculus ($\nabla_{\boldsymbol{\phi}^*} |q|^2 = q \cdot \overline{\partial q / \partial \boldsymbol{\phi}}$):

$\nabla_{\boldsymbol{\phi}^*} f = q \cdot \mathbf{a}^* = (c + \mathbf{a}^T \boldsymbol{\phi}) \mathbf{a}^*$

(b) The tangent space of $\mathcal{S}^1$ at $\phi_n$ is $T_{\phi_n}\mathcal{S}^1 = \{v \in \mathbb{C} : \mathrm{Re}(v \phi_n^*) = 0\}$ (purely imaginary in the rotated frame).

The Riemannian gradient is the tangent projection: $[\mathrm{grad}\, f]_n = [\nabla f]_n - \mathrm{Re}([\nabla f]_n \phi_n^*) \phi_n$

(c) After the gradient step $\tilde{\phi}_n = \phi_n + \alpha [\mathrm{grad}\, f]_n$, the retraction maps back to $\mathcal{S}^1$: $\phi_n^{(i+1)} = \tilde{\phi}_n / |\tilde{\phi}_n| = e^{j\angle \tilde{\phi}_n}$

(d) $q^{(0)} = 1 + (1+j) \cdot 1 + (1-j) \cdot 1 = 1 + 1 + j + 1 - j = 3$. $\nabla f = 3 \cdot [1-j, 1+j]^T = [3-3j, 3+3j]^T$.

Riemannian gradient at $\phi_1 = 1$: $\mathrm{Re}((3-3j) \cdot 1) = 3$. $[\mathrm{grad}]_1 = (3-3j) - 3 \cdot 1 = -3j$.

At $\phi_2 = 1$: $\mathrm{Re}((3+3j) \cdot 1) = 3$. $[\mathrm{grad}]_2 = (3+3j) - 3 \cdot 1 = 3j$.

Update: $\tilde{\phi}_1 = 1 + 0.5(-3j) = 1 - 1.5j$. $\phi_1^{(1)} = (1 - 1.5j)/|1 - 1.5j| = (1-1.5j)/1.803 = e^{-j\cdot 0.983}$.

$\tilde{\phi}_2 = 1 + 0.5(3j) = 1 + 1.5j$. $\phi_2^{(1)} = (1+1.5j)/1.803 = e^{j\cdot 0.983}$. $\blacksquare$

(a) $\mathrm{SINR}_k = \frac{|\mathbf{h}_{\mathrm{eff},k}^H(\boldsymbol{\phi}) \mathbf{w}_k|^2}{\sum_{j \neq k} |\mathbf{h}_{\mathrm{eff},k}^H(\boldsymbol{\phi}) \mathbf{w}_j|^2 + \sigma^2}$

where $\mathbf{h}_{\mathrm{eff},k}(\boldsymbol{\phi}) = \mathbf{h}_{d,k} + \mathbf{G}^H \mathrm{diag}(\boldsymbol{\phi}) \mathbf{h}_{r,k}$.

(b) $\max_{\{\mathbf{w}_k\}, \boldsymbol{\phi}} \sum_{k=1}^2 \log_2(1 + \mathrm{SINR}_k)$

s.t. $\sum_{k=1}^2 \|\mathbf{w}_k\|^2 \leq P$, $|\phi_n| = 1, \forall n$.

(c) The single-user case has no inter-user interference, so the active beamformer is simple MRT. With $K > 1$ users: (i) the active beamformers $\{\mathbf{w}_k\}$ must balance signal enhancement and interference suppression (e.g., MMSE or ZF beamforming); (ii) the RIS phases $\boldsymbol{\phi}$ affect all users' channels simultaneously — improving one user's channel may worsen another's; (iii) the SINR is a ratio of functions of $\boldsymbol{\phi}$, making the passive subproblem even more non-convex.

(d) Fix $\boldsymbol{\phi}$: compute effective channels $\{\mathbf{h}_{\mathrm{eff},k}\}$ and design $\{\mathbf{w}_k\}$ via MMSE beamforming or WMMSE (fractional programming).

Fix $\{\mathbf{w}_k\}$: optimise $\boldsymbol{\phi}$ to maximise the sum rate. This subproblem is more complex than the single-user case because the objective involves ratios. Approaches: successive convex approximation (SCA), penalty-based methods, or manifold optimisation with the sum-rate gradient. $\blacksquare$

(a) $\lambda = c/f_c = 3\times 10^8 / 28\times 10^9 = 10.71$ mm. $d = \lambda/2 = 5.36$ mm. $\tau_{\max} = (N-1)d\sin\theta/c = 255 \times 5.36\times 10^{-3} \times 0.5 / (3\times 10^8)$ $= 2.28$ ns.

(b) At element $n$ (relative to element 0), the extra delay is $\tau_n = nd\sin\theta/c$. The phase at frequency $f$ is $2\pi f \tau_n$. The phase optimised for $f_c$ is $2\pi f_c \tau_n$. The error at $f = f_c + B/2$:

$\Delta\psi_n = 2\pi (B/2) \tau_n = \pi B \cdot nd\sin\theta/c$.

At the last element ($n = 255$): $\Delta\psi_{255} = \pi \times 400\times 10^6 \times 2.28\times 10^{-9} = \pi \times 0.912 = 2.87$ rad $= 164°$.

(c) The effective array factor at the edge subcarrier has a linear phase error across elements. The normalised gain is:

$G_{\text{edge}} = \frac{1}{N}\left|\sum_{n=0}^{N-1} e^{j\Delta\psi_n}\right| = \frac{1}{N}\left|\frac{\sin(N\Delta\psi_1/2)}{\sin(\Delta\psi_1/2)}\right|$

$\Delta\psi_1 = \pi B d\sin\theta/c = \pi \times 400\times10^6 \times 5.36\times 10^{-3} \times 0.5 / 3\times 10^8 = 0.01124$ rad.

$N\Delta\psi_1/2 = 256 \times 0.01124/2 = 1.44$ rad. $G_{\text{edge}} = |\sin(1.44)/(256\sin(0.00562))| = |0.9915/(256 \times 0.00562)| = 0.9915/1.439 = 0.689$.

Power loss: $0.689^2 = 0.475 = 3.24$ dB.

(d) The 3 dB loss occurs approximately when the edge-subcarrier phase error across the full array reaches $\sim \pi$ radians. At the current $B/f_c = 1.4\%$, the loss is already $\sim$3 dB. More precisely, the beam squint loss exceeds 3 dB when $B/f_c > 2/(N\sin\theta) = 2/(256 \times 0.5) = 1.56\%$ for our parameters. With $N = 256$ and $\theta = 30°$, bandwidths above $\sim$1.5% of $f_c$ suffer significant squint. $\blacksquare$

The received signal power (without noise) is: $$|s|^2 = P\left|\sum_{n=1}^N h_{r,n}^* e^{j\theta_n} g_n\right|^2$$

where $\theta_n$ is random, uniformly distributed on $[0, 2\pi)$ and independent of the channels.

$\mathbb{E}[|s|^2] = P\sum_{n=1}^N \sum_{m=1}^N \mathbb{E}[h_{r,n}^* g_n e^{j\theta_n} \cdot h_{r,m} g_m^* e^{-j\theta_m}]$$**Self-terms** ($n = m$):$\mathbb{E}[|h_{r,n}|^2 |g_n|^2] = \mathbb{E}[|h_{r,n}|^2]\mathbb{E}[|g_n|^2]$(by independence of$h_r$and$g$). **Cross-terms** ($n \neq m$):$\mathbb{E}[e^{j(\theta_n - \theta_m)}] = \mathbb{E}[e^{j\theta_n}]\mathbb{E}[e^{-j\theta_m}] = 0 \cdot 0 = 0$since$\mathbb{E}[e^{j\theta}] = 0$for uniform$\theta$.

$\mathbb{E}[|s|^2] = P \sum_{n=1}^N \mathbb{E}[|h_{r,n}|^2]\mathbb{E}[|g_n|^2] = PN\sigma_h^2 \sigma_g^2KATEXPLACEHOLDER0END\mathbb{E}[\text{SNR}_{\text{random}}] = \frac{P}{\sigma^2} N \sigma_h^2 \sigma_g^2$$This is linear in$N$.

Under optimal phases: $\text{SNR}_{\text{opt}} \approx (P/\sigma^2) N^2 \mu_h^2 \mu_g^2$

The ratio: $\text{SNR}_{\text{opt}} / \mathbb{E}[\text{SNR}_{\text{random}}] = N \mu_h^2 \mu_g^2 / (\sigma_h^2 \sigma_g^2)$.

For Rayleigh fading: $\mu_h^2/\sigma_h^2 = \pi/4$. So the ratio is $N(\pi/4)^2 \approx 0.617 N$.

The phase optimisation converts the incoherent $N$-scaling (power adds) to coherent $N^2$-scaling (amplitudes add, then square). $\blacksquare$

(a) In the LOS-only case, $g_n = \sqrt{K_R/(K_R+1)} g_n^{\text{LOS}}$ where $g_n^{\text{LOS}} = e^{j2\pi(n-1)d\sin\theta_{\text{BS}}/\lambda}$ (array response). Similarly, $h_{r,n}^{\text{LOS}} = e^{j2\pi(n-1)d\sin\theta_{\text{user}}/\lambda}$.

The optimal phase: $\theta_n^* = -\angle(h_{r,n}^{*,\text{LOS}} g_n^{\text{LOS}}) = -2\pi(n-1)d(\sin\theta_{\text{BS}} - \sin\theta_{\text{user}})/\lambda$.

This depends only on the BS-RIS angle $\theta_{\text{BS}}$ and RIS-user angle $\theta_{\text{user}}$ — purely geometric, no fading randomness.

(b) For $K_R \to \infty$: all paths are LOS, $|h_{r,n}^{\text{LOS}}| = |g_n^{\text{LOS}}| = 1$. With optimal phases: $\text{SNR} = (P/\sigma^2) N^2$ (all terms contribute $+1$ to the coherent sum).

This is exactly $N^2$ scaling with coefficient 1.

(c) With LOS-only phases applied to the full Rician channel, the LOS components add coherently ($\propto N$) while the NLOS components add incoherently ($\propto \sqrt{N}$):

$\text{SNR} \approx \frac{P}{\sigma^2}\left(\frac{K_R}{K_R+1} N + \frac{1}{K_R+1} N\right)^2$

More precisely: $\text{SNR} \approx \frac{P}{\sigma^2}\left(\frac{K_R}{K_R+1} N^2 + \frac{1}{K_R+1} N\right)$

The dominant term for large $N$ is $K_R/(K_R+1) \cdot N^2$.

(d) The ratio of LOS-only to fully-optimised SNR: $\eta \approx K_R/(K_R + 1) + 1/((K_R+1)N)$.

For $\eta = 0.9$ and large $N$: $K_R/(K_R + 1) \geq 0.9 \Rightarrow K_R \geq 9$ (i.e., $\sim$9.5 dB).

Above $K_R = 9$ dB, the geometry-only phase design achieves $> 90\%$ of the fully optimised gain — no need for fast channel estimation. $\blacksquare$

(a) With no direct link: $\mathbf{h}_{\mathrm{eff}} = \mathbf{G}^H \boldsymbol{\Theta} \mathbf{h}_r$. Optimal $\mathbf{w} = \sqrt{P}\mathbf{h}_{\mathrm{eff}}/\|\mathbf{h}_{\mathrm{eff}}\|$ (MRT). SNR $= P\|\mathbf{h}_{\mathrm{eff}}\|^2/\sigma^2$. $C = \log_2(1 + P\|\mathbf{G}^H \boldsymbol{\Theta} \mathbf{h}_r\|^2/\sigma^2)$.

(b) Define $\mathbf{A} = \mathrm{diag}(\mathbf{h}_r^*)\mathbf{G} \in \mathbb{C}^{N \times M}$ (row $n$ is $h_{r,n}^* \mathbf{g}_n^T$). Then:

$\mathbf{G}^H \boldsymbol{\Theta} \mathbf{h}_r = \mathbf{A}^H \boldsymbol{\phi}$

$\|\mathbf{A}^H \boldsymbol{\phi}\|^2 = \boldsymbol{\phi}^H \mathbf{A}\mathbf{A}^H \boldsymbol{\phi}$

This is a Rayleigh quotient on the manifold $|\phi_n| = 1$. The unconstrained maximiser is $\boldsymbol{\phi} = \mathbf{u}_1$ (dominant eigenvector of $\mathbf{A}\mathbf{A}^H$). The unit-modulus constraint prevents exact achievement, but for large $N$ with i.i.d. channels, the projected solution is near-optimal.

(c) $\|\mathbf{A}^H \boldsymbol{\phi}\|^2 \leq \lambda_{\max}(\mathbf{A}\mathbf{A}^H)$.

For i.i.d. Rayleigh, $\mathbf{A}$ is $N \times M$ with i.i.d. $\mathcal{CN}(0,1)$ entries. By random matrix theory: $\lambda_{\max}(\mathbf{A}\mathbf{A}^H) \approx (\sqrt{N} + \sqrt{M})^2$.

With unit-modulus constraint (not eigenvector): the achievable value scales as $NM$ (coherent sum over $N$ elements, each contributing an $M$-dimensional vector).

Specifically: $\text{SNR} \sim (P/\sigma^2) N^2 M \mu^2$ (the $N^2$ from phase alignment, $M$ from the array gain).

Capacity: $C \approx \log_2(1 + (P/\sigma^2) N^2 M \mu^2)$.

(d) Massive MIMO (no RIS, $M$ BS antennas, direct link): $C_{\text{MIMO}} = \log_2(1 + PM/\sigma^2)$ (coherent array gain $M$).

RIS-assisted: $C_{\text{RIS}} \approx \log_2(1 + PN^2M\mu^2/\sigma^2)$.

The RIS provides an additional factor $N^2 \mu^2$ beyond the massive MIMO gain, but at the cost of double path loss (the $\mu^2$ factor includes the product path loss). The comparison depends critically on the relative path losses. $\blacksquare$

(a) $y = \alpha \mathbf{h}_r^H \boldsymbol{\Theta}_{\text{phase}} \mathbf{g} \sqrt{P} x + \alpha \mathbf{h}_r^H \boldsymbol{\Theta}_{\text{phase}} \mathbf{n}_a + n$

where $\boldsymbol{\Theta}_{\text{phase}} = \mathrm{diag}(e^{j\theta_1}, \ldots, e^{j\theta_N})$ and $\mathbf{n}_a \sim \mathcal{CN}(\mathbf{0}, \sigma_a^2 \mathbf{I}_N)$.

(b) Signal power: $\alpha^2 P (\sum_n |h_{r,n}||g_n|)^2$. Noise power: $\alpha^2 \sigma_a^2 \|\mathbf{h}_r\|^2 + \sigma^2$ $= \alpha^2 \sigma_a^2 \sum_n |h_{r,n}|^2 + \sigma^2$.

$\text{SNR} = \frac{\alpha^2 P (\sum_n |h_{r,n}||g_n|)^2}{\alpha^2 \sigma_a^2 \sum_n |h_{r,n}|^2 + \sigma^2}$