Exercises

$P_r^{\text{RIS}}/P_r^{\text{direct}} = N_{\text{RIS}}^2 d_{\text{direct}}^2 / (d_1 d_2)^2 = 512^2 \cdot 3600 / 810000 \approx 1165$.

$10 \log_{10}(1165) \approx 30.7$ dB. The RIS link is about 30 dB above the direct link in this idealized model; the factor is smaller with realistic wavelength-aware constants, as discussed in Example EHow Many RIS Elements to Match a Direct Link?. $\blacksquare$

$P_{\text{DC}}^{\text{AF}} = 8 \cdot 0.2 + 256 \cdot 1.5 \times 10^{-4} = 1.6 + 0.0384 \approx 1.64$ W.

$P_{\text{DC}}^{\text{dig}} = 256 \cdot 0.2 = 51.2$ W.

$51.2 / 1.64 \approx 31.2\times$ lower power for the array-fed architecture. $\blacksquare$

$10 \log_{10}(\text{sinc}^2(\pi/2)) = 10 \log_{10}((2/\pi)^2) \approx -3.92$ dB. A 1-bit RIS pays about 3.9 dB compared with continuous phase.

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

$10 \log_{10}(\text{sinc}^2(\pi/8)) = 10 \log_{10}((0.9745)^2) \approx -0.22$ dB. Beyond 3 bits the penalty is negligible, which is why most prototypes stop at $b \in \{2, 3\}$. $\blacksquare$

$\text{rank}(\mathbf{H}_{\text{eff}}) \leq \min(N_r, N_a, N_{\text{RIS}}) = \min(32, 6, 4096) = 6$.

The effective channel supports at most 6 parallel streams. The 4096-tile surface does not contribute any extra spatial DoF — it only boosts the magnitude of the 6 non-zero singular values. $\blacksquare$

$\text{tr}(\mathbf{W}_{\text{ZF}}^H \mathbf{W}_{\text{ZF}}) = \text{tr}(\mathbf{\Lambda} (\mathbf{H}\mathbf{H}^H)^{-1}) = P_t$. With equal $p_k = p$, $p \cdot \text{tr}((\mathbf{H}\mathbf{H}^H )^{-1}) = P_t$, so $p = P_t / \text{tr}((\mathbf{H}\mathbf{H}^H )^{-1})$.

$\gamma_k = p / \sigma^2 = P_t / (\sigma^2\, \text{tr}((\mathbf{H}\mathbf{H}^H)^{-1}))$. All users see the same SINR under equal allocation.

$R = K \log_2\!\left(1 + \frac{P_t}{\sigma^2\, \text{tr}((\mathbf{H}\mathbf{H}^H)^{-1})}\right).$ The RIS phase profile $\boldsymbol{\phi}$ controls $\mathbf{H}$, so maximizing $R$ is equivalent to minimizing $\text{tr}((\mathbf{H}\mathbf{H}^H)^{-1})$ over $\boldsymbol{\phi}$. $\blacksquare$

Write $h_{\text{eff},k}^H \mathbf{v}_{j} = \sum_{m=1}^{N_{\text{RIS}}} r_{k,m}^* \phi_m g_m^H \mathbf{v}_{j} = (\text{const}_{k,j}) + (r_{k,n}^* g_n^H \mathbf{v}_{j}) e^{j\phi_n}$, where the constant lumps all other $m \neq n$. Define $\alpha_{k,j} =$ const and $\beta_{k,j} = r_{k,n}^* g_n^H \mathbf{v}_{j}$.

The signal power is $|\alpha_{k,k} + \beta_{k,k} e^{j\phi_n}|^2$ and the interference is $\sum_{j \neq k} |\alpha_{k,j} + \beta_{k,j} e^{j\phi_n}|^2$. Both are sinusoids in $\phi_n$ with DC plus a single harmonic, which is the claimed form.

Taking $dR/d\phi_n = 0$ yields a single equation of the form $\sum_k (\text{rational}) \cdot \sin(\phi_n + \delta_k) = 0$, a transcendental equation in $\phi_n$ solvable by a 1-D root finder. In practice, grid search over $[0, 2\pi)$ with 180 points plus a local refine is more than adequate and closed-form for the single-user case. $\blacksquare$

Let $\mathbf{v}_1 \in \mathbb{C}^{N_{\text{RIS}}}$ be the top right singular vector of $\mathbf{H}_{\text{RIS-Rx}}$, with $\sigma_1(\mathbf{H}_{\text{RIS-Rx}}) \sim \sigma_r \sqrt{N_{\text{RIS}}}$ for tall Gaussian matrices.

Pick $\phi_n = \arg(\mathbf{v}_1[n] \cdot \mathbf{e}_1[n])^*$, where $\mathbf{e}_1$ is the first (unit-norm) column of $\mathbf{G}_f$. This aligns the product so every summand in $\mathbf{H}_{\text{RIS-Rx}} \text{diag}(\boldsymbol{\phi}) \mathbf{e}_1$ is positive real.

$\|\mathbf{H}_{\text{RIS-Rx}} \text{diag}(\boldsymbol{\phi}) \mathbf{e}_1\| \geq \sum_n |[\mathbf{H}_{\text{RIS-Rx}}]_{1,n}| |\mathbf{e}_1[n]| \sim \sigma_r \sqrt{N_{\text{RIS}}} \cdot (N_{\text{RIS}} / \sqrt{N_{\text{RIS}} N_a}) = \sigma_r N_{\text{RIS}} / \sqrt{N_a}$ after applying a Cauchy-Schwarz bound. Multiplying by $\sqrt{N_a}$ from the $N_r$-dimensional row sum yields $\sigma_1(\mathbf{H}_{\text{eff}}) \geq c \sigma_r N_{\text{RIS}}$ for a constant $c$. $\blacksquare$

The cascaded channel $\mathbf{H}_{\text{eff}}$ has $N_r N_{\text{RIS}}$ complex unknowns (one per Rx antenna per RIS element after factoring the sum-of-rank-one decomposition). Each pilot phase pattern reveals $N_r$ complex measurements, so $T \geq N_{\text{RIS}}$ is necessary in the worst case.

Under an $s$-sparse angular representation (the reflected channels live in a subspace of dimension $s \ll N_{\text{RIS}}$), compressed sensing gives $T = \mathcal{O}(s \log(N_{\text{RIS}}/s))$ pilot phases with high probability of exact recovery.

The sparse saving is significant when $s \ll N_{\text{RIS}}$, which holds at mmWave because most user reflected channels have only a few dominant angular paths. This is the same compressed- sensing principle that drives JSDM (Chapter 7) and XL-MIMO estimation (Chapter 18). $\blacksquare$

$N_{\text{eq}} \approx 8 \cdot (1 - 0.5/8)(1 + 6/8) = 8 \cdot 0.9375 \cdot 1.75 \approx 13.1$.

$N_{\text{eq}} \approx 8 \cdot 0.9375 \cdot (1 + 12/8) = 8 \cdot 0.9375 \cdot 2.5 \approx 18.75$.

64x larger RIS (case b vs case a) buys only a 43% increase in $N_{\text{eq}}$. The gain is logarithmic in $N_{\text{RIS}}$, so doubling the RIS size once buys $\sim$10% more equivalent chains. Beyond a few hundred tiles, the return is small — practical designs plateau around $N_{\text{RIS}} \in [256, 2048]$. $\blacksquare$

$G^{\text{passive}} = N_{\text{RIS}}^2 d_{\text{direct}}^2 / (d_1 d_2)^2 = 1024^2 \cdot 2500 / 360000 \approx 7281$, i.e. $\approx 38.6$ dB.

Let $G_a \approx N_a = 16$ and $G_{\text{RIS}} = N_{\text{RIS}} = 1024$. Using the array-fed formula (ignoring wavelength factors) yields $G^{\text{AF}} \approx G_a G_{\text{RIS}} d_{\text{direct}}^2 / d_2^2 = 16 \cdot 1024 \cdot 2500/400 \approx 10^5$, i.e. $\approx 50$ dB.

The array-fed RIS outperforms the passive RIS by about 11.4 dB in this geometry, and the gap grows as $d_1$ increases. The active feed eliminates the double-fading penalty at only a small DC-power cost. $\blacksquare$

Write $\mathbf{H}_{\text{RIS-Rx}} = \mathbf{U}_r \mathbf{\Sigma}_r \mathbf{V}_r^H$ and $\mathbf{G}_f = \mathbf{U}_g \mathbf{\Sigma}_g \mathbf{V}_g^H$. The dominant right singular vector of $\mathbf{H}_{\text{RIS-Rx}}$ is $\mathbf{v}_{r,1}$, and the dominant left singular vector of $\mathbf{G}_f$ is $\mathbf{u}_{g,1}$.

Tightness requires $\text{diag}(\boldsymbol{\phi}) \mathbf{u}_{g,1} = \mathbf{v}_{r,1}$, i.e. $\phi_n = \arg(\mathbf{v}_{r,1}[n] / \mathbf{u}_{g,1}[n])$ for all $n$. Since $|e^{j\phi_n}| = 1$, this is feasible exactly when $|\mathbf{v}_{r,1}[n]| = |\mathbf{u}_{g,1}[n]|$ for all $n$ — the two vectors must have the same amplitude profile. Otherwise the bound is only approximately achievable.

In the near-field reactive approximation of $\mathbf{G}_f$, the left singular vectors are Fourier-like and have roughly uniform amplitudes across RIS elements. For i.i.d. Gaussian $\mathbf{H}_{\text{RIS-Rx}}$, the right singular vector concentrates as $\mathbf{v}_{r,1}[n] \sim \mathcal{CN}(0, 1/N)$, also approximately uniform in amplitude. The alignment is therefore approximate but tight up to $o(1)$ constants. $\blacksquare$

Each inner sub-step maximizes $R$ over either $\mathbf{W}$ (fixing $\boldsymbol{\phi}$) or a single $\phi_n$ (fixing everything else). Both are exact maximizations, so $R^{(t+1)} \geq R^{(t)}$. Combined with the upper bound from the Shannon cap, $R^{(t)}$ converges monotonically.

Construct a toy problem with $N_a = 2$, $N_{\text{RIS}} = 2$, $K = 2$, where the sum-rate landscape as a function of $(\phi_1, \phi_2)$ has two symmetric local maxima. Initialize $\boldsymbol{\phi}$ near the first: the alternating updates converge to a local maximum at the same height as the second but not higher. Both are stationary points; neither is dominated.

Global optimality is lost. In practice, 5–10 random restarts plus a warm-start from the previous coherence block empirically recover within 0.5 dB of the global optimum. $\blacksquare$

Want $N_{\text{eq}} \geq 0.9 \cdot 2048 = 1843$. However, the fully digital rate with $K = 8$ and MMSE saturates well before 2048 chains: the effective $N_{\text{eq,target}}$ for 90% rate is closer to $\sim 100$ (since rate is logarithmic in $N$). Re-interpret "90% of full rate" as $N_{\text{eq}} = 100$.

$100 = N_a (1 - 0.5/N_a)(1 + \log_2(2048)/N_a) = N_a (1 - 0.5/N_a)(1 + 11/N_a)$. Expanding: $100 \approx N_a + 11 - 5.5 = N_a + 5.5$, so $N_a \approx 94$. But this result is outside the validity regime ($N_a \ll N_{\text{RIS}}$ required).

The exercise illustrates that "match the digital rate" and "save power" are in tension: matching 90% requires many active elements, which erases the power saving. The sweet spot is around 70–80% rate at $N_a \in [8, 32]$, not 90% at $N_a = 94$. This is a realistic picture of the architecture's regime of advantage. $\blacksquare$

Single-user received amplitude is $S = \alpha + a_n e^{j\phi_n}$, where $\alpha = \sum_{m \neq n} a_m e^{j\phi_m}$ is a complex constant. $|S|^2 = |\alpha|^2 + |a_n|^2 + 2\Re(\bar\alpha a_n e^{j\phi_n})$.

The only $\phi_n$-dependent term is $2\Re(\bar\alpha a_n e^{j\phi_n})$. This is maximized when $\bar\alpha a_n e^{j\phi_n} = |\bar\alpha a_n|$, i.e. $\phi_n = \arg(\alpha) - \arg(a_n) = \arg(\alpha / a_n)$.

The optimal $\phi_n$ co-phases the $n$-th rank-one term with the sum of all other terms — exactly what we would expect from a greedy constructive combining strategy. The per-element update is $\mathcal{O}(1)$ complex operations per iteration. $\blacksquare$

$\text{rank}(\mathbf{H}_{\text{eff}}) \leq N_a = 4 < K = 8$. At most 4 users can be served simultaneously.

(a) Round-robin: Schedule users $\{1,2,3,4\}$ in slot 1 and $\{5,6,7,8\}$ in slot 2, alternating across the 4 slots. Each user is served in half the slots, and the long-term rate is $\frac{1}{2} \mathbb{E}[R^{\text{inst}}(\text{user } k)]$ per user. (b) Max-sum-rate (greedy): At each slot, pick the 4 users whose joint effective channel $\mathbf{H}(\boldsymbol{\phi})$ is best-conditioned, e.g., those with largest $\log_2 \det(\mathbf{I} + \mathbf{H} \mathbf{H}^H \text{SNR})$.

Round-robin gives equal per-user time shares (fairness). Max-rate achieves higher total throughput but is unfair to users with poor channels. A weighted-proportional-fair scheduler interpolates between the two: maximize $\sum_k w_k R_k$ with $w_k = 1/\bar R_k$ where $\bar R_k$ is the running average rate. This is the standard 5G NR MU-MIMO scheduling rule, adapted to the array-fed RIS DoF constraint. $\blacksquare$

Let $\mathbf{H}_{\text{eff}} = \mathbf{U} \mathbf{\Sigma} \mathbf{V}^H$ with $\mathbf{\Sigma} = \text{diag}(\sigma_1, \ldots, \sigma_{N_a}, 0, \ldots, 0)$. Substitute $\mathbf{x} = \mathbf{V} \tilde{\mathbf{s}}$ so the channel decomposes into $N_a$ parallel SISO links with gains $\sigma_k^2$.

With sum-power budget $P_t$, the optimal allocation is $p_k^\star = (\mu - \sigma^2/\sigma_k^2)^+$ with $\mu$ chosen so $\sum_k p_k^\star = P_t$. Modes with small $\sigma_k$ receive no power.

The key difference is that the $\sigma_k$ are functions of $\boldsymbol{\phi}$, not fixed. A joint optimization over $\{p_k, \boldsymbol{\phi}\}$ shapes the singular values before waterfilling. In the RIS-aided case the singular values are comparable (from Theorem $N_{ ext{RIS}}$" data-ref-type="theorem">TDominant Singular Values Scale with $N_{ ext{RIS}}$), so all $N_a$ modes typically receive power. Conventional MIMO with random $\mathbf{H}$ has broadly spread singular values and waterfilling often shuts off weaker modes. $\blacksquare$

$N_a = 16$ (one per user, minimum for ZF). Pick $N_{\text{RIS}} = 4096$ tiles ($64 \times 64$ UPA at $\lambda/2$ spacing, total aperture $0.16 \times 0.16$ m).

$G_a \approx 16$ (array gain of feed), $G_{\text{RIS}} \approx 4096$. Path loss: $(\lambda/(4\pi r))^2 = (0.005/(12.57 \cdot 25))^2 = (1.59 \times 10^{-5})^2 \approx 2.5 \times 10^{-10}$. Combined: $G_a G_{\text{RIS}} \cdot 2.5 \times 10^{-10} = 65536 \cdot 2.5 \times 10^{-10} \approx 1.6 \times 10^{-5}$.

For 10 dB SNR per user at $\sigma^2 = -84$ dBm (4 GHz BW at 60 GHz), the required received power is $-74$ dBm $= 4 \times 10^{-11}$ W. So $P_t = 4 \times 10^{-11} / 1.6 \times 10^{-5} = 2.5 \times 10^{-6}$ W $= -26$ dBm. This is extraordinarily low — a few microwatts suffice at the cell edge thanks to the enormous $G_{\text{RIS}}$.

The link-budget calculation assumes the user sits on the boresight of the RIS beam. At 25 m from a 0.16 m aperture, the diffraction-limited beamwidth is $\lambda/A \approx 30$ mrad, i.e. 0.75 m at 25 m. Serving 16 randomly placed users requires spatial division — hence the $N_a = 16$ active chains doing multiuser precoding. The link budget per user holds when the precoder puts all of each user's power into the narrow RIS-focused beam. $\blacksquare$

A small rotation of the feed introduces a phase gradient across the RIS: $[\mathbf{G}_f(\epsilon)]_{n,a} = [\mathbf{G}_f(0)]_{n,a} e^{j 2\pi p_n \sin(\epsilon)/\lambda}$, where $p_n$ is the $n$-th RIS element position.

The phase gradient is linear in position, so it can be cancelled by subtracting a matching linear ramp from $\boldsymbol{\phi}$. If the RIS has $b$-bit phase resolution, the residual error is at most $\pi/2^b$ per element — a random dephasing of the cascaded channel.

At $\epsilon = 1^\circ = 0.0175$ rad and $\lambda = 1.07$ cm, the phase gradient across a $0.3$ m aperture is $2\pi \cdot 0.3 \cdot 0.0175 / 0.0107 \approx 30.8$ rad, so the tilt wraps around roughly 5 times. With 2-bit phase quantization the residual per-element error is $\pi/4 = 0.785$ rad. The SNR loss is $\sim |\mathbb{E}[e^{j\text{err}}]|^2 = \text{sinc}^2(\pi/4) \approx 0.81$, i.e. $\sim 0.9$ dB. Lower than one might fear, thanks to the RIS phase flexibility. $\blacksquare$

An AF relay with gain $G$ amplifies received signal plus noise: $y_{\text{out}} = G(\mathbf{H}_{1} \mathbf{x} + \mathbf{w}_1)$. At the destination, $y_r = G \mathbf{H}_{2} \mathbf{H}_{1} \mathbf{x} + G \mathbf{H}_{2} \mathbf{w}_1 + \mathbf{w}_2$. The relay adds amplified noise power $G^2 |\mathbf{H}_{2}|^2 \sigma^2$.

The passive RIS does not amplify, so the only noise is at the user. The active feed is on-site and injects its own signal without hop noise. The effective SNR is higher than AF relay by roughly the AF noise figure of $3$–$6$ dB.

When $d_1$ is large and the relay has its own transmit budget, it can boost signal above the AF noise by allocating more power. If the relay budget is much larger than the array-fed RIS active-feed budget (e.g., 10x), the amplification advantage wins. In typical equal-power comparisons at mmWave, the array- fed RIS wins by 3–5 dB because of the absence of hop noise. $\blacksquare$

With optimal input covariance $\mathbf{Q}$, capacity is $C = \max_{\mathbf{Q} \succeq 0, \text{tr}(\mathbf{Q})\leq P_t} \log_2\det(\mathbf{I}_{N_r} + \frac{1}{\sigma^2} \mathbf{H}_{\text{eff}} \mathbf{Q} \mathbf{H}_{\text{eff}}^H)$. With $\mathbf{H}_{\text{eff}}$ of rank $N_a$ and waterfilling, $C = \sum_{k=1}^{N_a} \log_2(1 + p_k \sigma_k^2 / \sigma^2)$.

From Theorem $N_{ ext{RIS}}$" data-ref-type="theorem">TDominant Singular Values Scale with $N_{ ext{RIS}}$, $\sigma_k \asymp \sigma_r N_{\text{RIS}}$ for $k \leq N_a$. So $\sigma_k^2 \asymp N_{\text{RIS}}^2$ and $C \asymp N_a \log_2(P_t N_{\text{RIS}}^2 / (N_a \sigma^2))$. Capacity grows like $N_a \log_2(N_{\text{RIS}}^2)$ — linearly in $\log N_{\text{RIS}}$ but with slope only $N_a$, not $N$.

A fully digital $N$-element array achieves $C_{\text{dig}} \sim \min(N_r, N) \log_2(P_t N / (\min(N_r, N) \sigma^2))$, which grows both in the coefficient (DoF) and in the argument. At $N = N_{\text{RIS}}$, the DoF is $\min(N_r, N_{\text{RIS}})$, potentially much larger than $N_a$.

The array-fed RIS achieves lower asymptotic capacity by a factor of $\min(N_r, N_{\text{RIS}}) / N_a$ in the DoF coefficient. It does not approach the fully digital array as $N_{\text{RIS}} \to \infty$ — the gap is a constant fraction, not vanishing. In exchange, the DC-power savings are roughly a factor $N_{\text{RIS}}/N_a$. The engineering tradeoff is precise: capacity/N DoF vs power/N chains. $\blacksquare$