Exercises

$|\mathcal{M}_B| = 8^{64} = 2^{192} \approx 6.3 \times 10^{57}$.

$6 \times 10^{57}$ configurations at $10^9$ evaluations per second would require $\sim 2 \times 10^{41}$ years. Brute force is completely infeasible. Must use BCD or branch-and-bound.

$\eta_2 = \text{sinc}^2(\pi/4)$. $\sin(\pi/4)/(\pi/4) \approx 0.9003$. $\eta_2 \approx 0.811$. Loss $= -10 \log_{10}(0.811) \approx 0.91$ dB.

$\eta_4 = \text{sinc}^2(\pi/16)$. $\sin(\pi/16)/(\pi/16) \approx 0.9937$. $\eta_4 \approx 0.987$. Loss $\approx 0.056$ dB.

Going from 2-bit to 4-bit: loss reduction of $0.91 - 0.056 \approx 0.85$ dB. Only worth it if the extra 2 bits of control hardware are cheap. $\blacksquare$

For $\theta^\star \sim \text{Unif}(0, 2\pi)$, the projection error $\epsilon = \theta^\star - \mathcal{P}(\theta^\star)$ is uniform on $[-\Delta\theta/2, \Delta\theta/2]$.

For uniform $U(-a, a)$, variance $= (2a)^2/12 = a^2/3$. With $a = \Delta\theta/2$: variance $= \Delta\theta^2 / 12$.

Smaller grid step $\Delta\theta$ → smaller quantization variance → smaller SNR loss. Scales as $(1/L)^2 = 1/4^B$. This is the asymptotic form of $\eta_B$ for large $B$. $\blacksquare$

$\mathbb{E}[e^{j\epsilon}] = \frac{1}{\Delta\theta}\int_{-\Delta\theta/2}^{\Delta\theta/2} e^{j\epsilon} d\epsilon = \frac{2\sin(\Delta\theta/2)}{\Delta\theta} = \frac{\sin(\pi/2^B)}{\pi/2^B} = \text{sinc}(\pi/2^B)$ (using $\Delta\theta = 2\pi/2^B$).

At optimum $\theta_n^\star = -\arg a_n$: $a_n e^{j\theta_n^\star} = |a_n|$ (real positive). After quantization: $a_n e^{j\tilde\theta_n} = |a_n| e^{j\epsilon_n}$. Sum: $\sum_n |a_n| e^{j\epsilon_n}$, expected value $\text{sinc}(\pi/2^B) \sum_n |a_n|$.

$|\mathbb{E}[\sum]|^2 = \text{sinc}^2(\pi/2^B) \|\mathbf{a}\|_1^2 = \eta_B \cdot \|\mathbf{a}\|_1^2$. $\blacksquare$

$f(\phi_n) = A_{nn} |\phi_n|^2 + 2\Re(\phi_n^* \alpha_n) +$ const, where $\alpha_n$ collects the cross terms with all other $\phi_m$. $|\phi_n|^2 = 1$ is fixed; $f(\phi_n) = 2\Re(\phi_n^* \alpha_n) +$ const.

$\Re(\phi_n^* \alpha_n) = |\alpha_n| \cos(\arg \alpha_n - \theta_n)$ with $\theta_n \in \Theta_B$.

Best: $\theta_n^\star = \mathcal{P}_{\Theta_B}(\arg \alpha_n)$. Identical to the projection of the continuous-BCD update. Discrete BCD at each step uses already-quantized neighbors, which is the only difference from Section 8.2's approach. $\blacksquare$

$\eta_1 N^2 = 0.405 \cdot 768^2 = 0.405 \cdot 589\,824 = 2.39 \cdot 10^5$.

$\eta_3 N^2 = 0.949 \cdot 256^2 = 0.949 \cdot 65\,536 = 6.22 \cdot 10^4$.

1-bit wins by a factor of $3.85$, or $5.9$ dB. More elements at coarser resolution beats fewer elements at finer resolution under a fixed pin-count budget.

This argues for 1-bit prototypes with large $N$ when pin cost is the constraint. However, the 3-bit elegance might still win in scenarios with complex multi-user or ISAC objectives where sub-grid precision matters. $\blacksquare$

Projection errors $\epsilon_n$ are i.i.d. uniform on $[-\Delta\theta/2, \Delta\theta/2]$. For large $N$, the fluctuation of $\sum_n e^{j\epsilon_n}$ around its mean scales as $\sqrt{N}$, while the mean scales as $N \text{sinc}$.

Squared amplitude: $|\text{sum}|^2 \approx (N \text{sinc})^2 + O(N)$. Relative error: $O(1/N) \to 0$ as $N \to \infty$.

$\text{SNR}_{\text{disc}}/\text{SNR}_{\text{cont}} \to \eta_B = \text{sinc}^2(\pi/2^B)$ almost surely. Matches Theorem 8.1. $\blacksquare$

1. Start from discrete BCD output.
2. Define a "move": change one $\phi_n$ to any other grid level.
3. At each step, pick the best non-tabu move that improves (or least worsens) the objective.
4. Record this move in the tabu list (prevents immediate reversal).
5. Repeat for $T$ iterations; return the best seen.

BCD is pure greedy — can get stuck at local optima where single-element moves don't improve, but a sequence of moves would. Tabu search escapes by allowing temporarily worse moves, then finding better regions.

More compute per iteration; no monotonicity guarantee; needs tuning (tabu list length, iteration count). In practice, $\sim 10\%$ better solutions at $\sim 10\times$ cost. Only worth it for offline benchmarking. $\blacksquare$

Each user's SNR is scaled by $\eta_B$; rate loss $\approx -\log_2(\eta_B) = \log_2(\eta_B^{-1})$ per user.

$\Delta R_{\text{sum}} = K \log_2(\eta_B^{-1})$. Linear in $K$.

$\Delta R_{\text{mm}} = \log_2(\eta_B^{-1})$. Single-user rate loss; doesn't scale with $K$. $\blacksquare$

When the continuous-BCD iterates oscillate (e.g., near the boundary of different local basins), the "converged" continuous optimum is less stable, and the projection may miss a better discrete solution. Discrete BCD, by always staying in $\mathcal{M}_B$, tracks a single stable trajectory.

At $B = 1$ (only 2 levels), the grid is coarse. Small continuous-phase differences don't distinguish between the two levels — projection is essentially random at the boundary. Discrete BCD explicitly chooses the best option.

$N = 16, B = 1, K = 4$: discrete BCD typically beats projection by $\sim 0.3$-$0.6\text{ dB}$. The coarse grid and multi-user coupling both contribute. $\blacksquare$

$N \cdot B = 256 \cdot 3 = 768$ bits per RIS reconfiguration.

10 kHz → $10^4$ updates per second. Total: $768 \cdot 10^4 = 7.68 \cdot 10^6$ bps $= 7.68$ Mbps.

5-ms block carries 50 updates (at 10 kHz). Total bits per block: $50 \cdot 768 = 38.4$ kbits. Needs a dedicated 10-Mbps control link; typical RIS controllers offer this via Ethernet or a USB link. $\blacksquare$

In a coherence block $T_c$, channels decorrelate. CSI error grows with time within the block.

If CSI error $\sim \sigma_{\text{CSI}}$ is comparable to $\Delta\theta / 2 = \pi / 2^B$, the finest quantization level is below the noise floor. Higher $B$ buys nothing; the CSI limits the effective resolution.

High mobility: short $T_c$, large $\sigma_{\text{CSI}}$, small effective $B_{\text{eff}} = \log_2(\pi/\sigma_{\text{CSI}})$. At pedestrian speeds (1 m/s at 28 GHz): $\sigma_{\text{CSI}} \sim 0.1$ rad, $B_{\text{eff}} \sim 5$. At vehicular speeds (30 m/s): $\sigma_{\text{CSI}} \sim 1$ rad, $B_{\text{eff}} \sim 2$. Higher $B$ than $B_{\text{eff}}$ is wasted — use lower-$B$ hardware for high-mobility deployments. $\blacksquare$

$\Delta R_3 = \log_2(1/0.949) \approx 0.075$ bits/s/Hz per user.

With $\epsilon_{\text{CSI}}^2 \approx 0.05$ (5%), $\Delta R \approx \log_2(1/(1-0.05)) \approx 0.074$ bits/s/Hz. Comparable to $B = 3$ quantization. For $\epsilon_{\text{CSI}}^2 = 0.1$: $\Delta R \approx 0.15$ bits/s/Hz. Worse than $B = 3$.

CSI error and $B = 3$ quantization cause similar-magnitude losses in realistic deployments. Improving either in isolation has limited effect; both need attention. The coupling is studied in Ch. 4.5 (tau_p tradeoff). $\blacksquare$

$\sim 0.2$-$0.5$ dB (empirical; see Wu and Zhang 2020 Fig. 5).

At high SNR: $\Delta R \approx 0.3 / 3$ bits/s/Hz per user, so $\sim 0.1$-$0.2$ bits/s/Hz total. Small.

For $K = 2$ the gap is small enough that projection is usually preferred (simpler). At higher $K$ the gap grows; at $K = 8, B = 1$ it can be $> 1$ dB, then discrete BCD is worth the extra complexity. $\blacksquare$

In the cascaded channel $\mathbf{h}_2^H \boldsymbol{\Phi} \mathbf{H}_1$, elements with larger per-element amplitude (stronger channel) contribute more to the total — they deserve higher precision.

Rank elements by channel amplitude $|c_n|$. Top 25% get $B_1 = 4$ bits; middle 50% get $B_2 = 3$ bits; bottom 25% get $B_3 = 1$ bit. Total control budget: similar to uniform $B = 3$.

High-gain elements are tuned precisely, contributing the coherent peak. Low-gain elements are roughly tuned; their contribution is small anyway, so quantization loss is diluted. Expected: $\sim 0.5\text{ dB}$ improvement over uniform $B$ at same budget.

Requires per-element bit-depth control, which most hardware doesn't offer. Design would need to be baked into the RIS architecture (mixed 1-bit and 3-bit unit cells). An interesting research direction, not yet realized in practice. $\blacksquare$