Exercises

Signal power at user $k$: $|\mathbf{h}_{k,\text{eff}}^H \mathbf{v}_{k}|^2$. Interference: $\sum_{j \neq k} |\mathbf{h}_{k,\text{eff}}^H \mathbf{v}_{j}|^2$. Noise: $\sigma^2$.

$\text{SINR}_k = |\mathbf{h}_{k,\text{eff}}^H \mathbf{v}_{k}|^2 / (\sum_{j \neq k} |\mathbf{h}_{k,\text{eff}}^H \mathbf{v}_{j}|^2 + \sigma^2)$.

Take $\boldsymbol{\phi}^{(1)} = \mathbf{1}$, $\boldsymbol{\phi}^{(2)} = -\mathbf{1}$, both on the torus. Midpoint $\mathbf{0}$ has $|\phi_n| = 0$, not 1. Not convex.

The convex hull of the unit circle $\{z : |z| = 1\}$ is the closed unit disk $\{z : |z| \leq 1\}$. Hence the convex hull of the torus is $\{\boldsymbol{\phi} : |\phi_n| \leq 1 \forall n\}$. Relaxing to this hull gives a convex problem but loses the "passive" meaning; see Section 6.3 for when the relaxation is tight.

$f(\mathbf{W}^{(i+1)}, \boldsymbol{\Phi}^{(i)}) = \max_{\mathbf{W}} f(\cdot, \boldsymbol{\Phi}^{(i)}) \geq f(\mathbf{W}^{(i)}, \boldsymbol{\Phi}^{(i)})$. Similarly for the passive update.

Strict inequality holds whenever the current iterate is not the conditional optimum. Once an iterate is simultaneously the conditional optimum in both blocks (a stationary point), further updates do not change the iterate. In practice this convergence happens within $\sim 20$ iterations. $\blacksquare$

$|\mathbf{h}_{\text{eff}}^H \mathbf{v}|^2 \leq \|\mathbf{h}_{\text{eff}}\|^2 \|\mathbf{v}\|^2 \leq \|\mathbf{h}_{\text{eff}}\|^2 P_t$, with equality iff $\mathbf{v} \parallel \mathbf{h}_{\text{eff}}$.

$\mathbf{v}^\star = \sqrt{P_t} \cdot \mathbf{h}_{\text{eff}} / \|\mathbf{h}_{\text{eff}}\|$. Achieves SNR $= P_t \|\mathbf{h}_{\text{eff}}\|^2 / \sigma^2$. $\blacksquare$

Fixing all $\phi_m, m \neq n$: $|a + \phi_n b_n + c_n|^2$ where $c_n = \sum_{m \neq n} \phi_m b_m$ and $a = \mathbf{h}_d^H \mathbf{v}$, $b_n = [\mathbf{G}^H \mathbf{v}]_n$.

Triangle inequality is tight when $\phi_n b_n$ and $(a + c_n)$ have the same argument: $\phi_n^\star = e^{j\arg((a + c_n) / b_n)} = e^{j(\arg(a + c_n) - \arg(b_n))}$.

At high SNR, $|\mathbf{h}_{\text{eff}}|^2 \propto N^2$ (under coherent RIS alignment with matched-filter BS beamformer).

$R = \log_2(1 + \text{SNR}) \approx \log_2(P_t N^2 / \sigma^2) = \log_2(\text{SNR}_{0}) + 2 \log_2 N$. Each doubling of $N$ adds 2 bits/s/Hz.

$\kappa = 10\text{ dB}$ means the LoS component dominates. The effective channel is close to a rank-1 LoS, so the optimization landscape has few local minima.

In the high-K regime, AO converges in 3-5 iterations (nearly deterministic landscape, element-wise update nearly closed-form). In pure Rayleigh ($K = 0$), 20-30 iterations are typical. The rate of convergence scales with the spectral gap of the block Hessian, which is favorable under dominant LoS.

High-K deployments (mmWave, sub-THz, LoS scenarios) are algorithmically easier than rich-Rayleigh deployments. The RIS also gives more benefit there, because the $N^2$ coherent gain requires phase-aligned channels — achievable at high K. $\blacksquare$

$O(N_t^{3} K) = O(16^3 \cdot 4) = O(16384)$ flops per WMMSE step, with $\sim 10$ steps for convergence: $\sim 10^5$ flops.

$O(N^{6.5}) = 128^{6.5} \approx 2 \times 10^{13}$ flops — dominated entirely by the SDP interior-point method. $\sim 1$ second of wall-clock on a modern CPU.

$\sim 1$ s per outer iteration. With $10$-$20$ outer iterations, $\sim 10$-$20$ seconds total. For $N = 128$ this is already impractical for real-time; use element-wise or manifold instead for the passive step.

At high-SNR matched filter: $\log_2(1 + 10 \cdot 4) \approx 5.4$ bits/s/Hz.

$\log_2(1 + 10 \cdot 64) = \log_2(641) \approx 9.3$ bits/s/Hz. A $\sim 4$ bits/s/Hz improvement, or $\sim 70\%$.

AO under random channels typically achieves $80$-$95\%$ of the coherent upper bound after 10-20 iterations; expect $\sim 8.5$ bits/s/Hz. The gap is mostly due to the multi-user interference that does not exist here (single user); at $K = 1$ AO is nearly exact. $\blacksquare$

$\nabla_{\mathbf{W}} \sum_k \log_2(1 + \text{SINR}_k) - 2\mu \mathbf{W} = 0$.

$\nabla_{\mathbf{W}} [\sum_k w_k e_k] + 2\mu \mathbf{W} = 0$, $\nabla_{\mathbf{u}_k} e_k = 0$ (MMSE), $\partial/\partial w_k [w_k e_k - \log_2 w_k] = 0 \Rightarrow w_k = 1/(e_k \ln 2)$.

Plugging $w_k = 1/(e_k \ln 2)$ into the $\mathbf{W}$-stationarity: $\sum_k (1/(e_k \ln 2)) \nabla e_k + 2\mu \mathbf{W} = 0$. Using $\nabla \log_2(1 + \text{SINR}_k) = -1/(e_k \ln 2) \nabla e_k$ (an identity for the Gaussian channel), the WMMSE KKT reduces exactly to the sum-rate KKT. The two problems are equivalent. $\blacksquare$

The channel at block $t+1$ differs from block $t$ by a small perturbation (typically $\|\Delta \mathbf{h}\| / \|\mathbf{h}\| \lesssim 0.1$ for mobile at $\sim 100\,\text{ms}$ coherence).

AO from scratch takes $\sim 20$ iterations to converge. Warm-started from the previous iterate, only $\sim 3$-$5$ iterations are needed — a $4$-$7\times$ speedup. The precise number depends on how quickly the channel changes.

Warm-starting can get stuck in local optima: if the global optimum changes discontinuously (rare in practice but possible at coherence-block transitions across blockage events), warm-starting fails to track. Detect via sudden drop in objective; restart from random.

Water-filling allocates power based on per-user channel strengths. Weaker users get less power, stronger users more (in sum-rate maximization).

The RIS can reshape per-user strengths. A passive RIS that favors user 1 gives a stronger $\mathbf{h}_{1,\text{eff}}$, which justifies more power to user 1 under water-filling. The joint optimization over $(\mathbf{W}, \boldsymbol{\Phi})$ produces higher sum rate than optimizing power alone, because we have an extra control knob.

Under sum-rate maximization, the RIS amplifies the strong users (Matthew-effect: rich get richer). Under max-min fairness (Chapter 7), the RIS is steered toward the weakest user. The two objectives thus give genuinely different $\boldsymbol{\Phi}^\star$'s. $\blacksquare$

$R_{\text{AO}} \leq R^\star \leq R_{\text{SDR}}$, so the global optimum is in $[R_{\text{AO}}, R_{\text{SDR}}]$, an interval of width $R_{\text{SDR}} - R_{\text{AO}}$.

For $K \leq 4$, this interval is typically $< 1$ dB wide. AO's local optimum is within 1 dB of global. Empirically verified on random Rayleigh channels across many papers.

If the AO-to-SDR gap is small (< 1 dB), accept the AO result. If large, try more random initializations. If still large, the problem is genuinely hard and one may need a different algorithm (global branch-and-bound, semidefinite relaxation with randomization). $\blacksquare$

At iteration $i$: (a) run one AO step; (b) with probability $\epsilon_i$, add random perturbation $\delta \boldsymbol{\phi}$ to the current $\boldsymbol{\Phi}$ (e.g., rotate each $\phi_n$ by a small random angle); (c) accept the perturbation only if it doesn't decrease the objective by more than $T_i$ (the "temperature"); (d) decrease $T_i$ and $\epsilon_i$ over time.

Can escape shallow local optima that pure AO would miss. At convergence, closer to the global optimum (on average).

More iterations needed (perturbations are rejected often near convergence). Hyperparameters ($\epsilon_i, T_i$) are problem-dependent. In practice, multiple-start AO is simpler and often competitive.

For research, an interesting direction. For deployment, multiple-start AO plus warm-starting is the pragmatic winner. $\blacksquare$

MU-MIMO with $N_t = 8, K = 4$: degrees of freedom $= \min(N_t, K) = 4$. Sum rate $\sim 4 \log_2 \text{SNR}$.

The RIS reshapes the channel but does not create new spatial dimensions. DoF is still $\min(N_t, K) = 4$ — the RIS helps the SNR offset, not the pre-log factor. Sum rate $\sim 4 \log_2(\text{SNR}) + 4 \cdot \log_2 N^2 / K$ (very roughly).

RIS provides a power gain, not a multiplexing gain. In DoF-terms, RIS is a passive "SNR-booster" rather than a new spatial resource. $\blacksquare$