Exercises

$y = (\mathbf{h}_d^H \mathbf{v}) s + (\mathbf{h}_2^H \boldsymbol{\Phi} \mathbf{H}_1 \mathbf{v}) s + w$.

$y = \big[\mathbf{h}_d^H + \mathbf{h}_2^H \boldsymbol{\Phi} \mathbf{H}_1\big] \mathbf{v}\, s + w$, so $\mathbf{h}_{\text{eff}}^H = \mathbf{h}_d^H + \mathbf{h}_2^H \boldsymbol{\Phi} \mathbf{H}_1$.

$\mathbf{h}_2^* = [2, 1 + j]^T$, so $\text{diag}(\mathbf{h}_2^*) = \begin{pmatrix} 2 & 0 \\ 0 & 1 + j \end{pmatrix}$.

$\mathbf{M} = \begin{pmatrix} 2 & 0 \\ 0 & 1 + j \end{pmatrix} \begin{pmatrix} 1 & j \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 2j \\ -(1+j) & 1 + j \end{pmatrix}$.

$|\boldsymbol{\phi}^T \mathbf{a}| = |\sum_n \phi_n a_n| \leq \sum_n |\phi_n||a_n| = \sum_n |a_n| = \|\mathbf{a}\|_1$.

Equality holds iff all $\phi_n a_n$ have the same complex argument. Writing $a_n = |a_n| e^{j\arg(a_n)}$, $\phi_n a_n = e^{j\theta_n} |a_n| e^{j\arg(a_n)}$ has argument $\theta_n + \arg(a_n)$. Setting this equal to the constant $0$ gives $\theta_n = -\arg(a_n)$, or $\phi_n^\star = e^{-j\arg(a_n)}$. $\blacksquare$

$\text{SNR}^\star / \mathbb{E}[\text{SNR}_{\text{rand}}] = N^2 / N = N = 512$.

$10 \log_{10}(512) = 27.1\text{ dB}$.

$f(d_1) = d_1(d_0 - d_1)$, $f'(d_1) = d_0 - 2d_1$, setting to zero yields $d_1 = d_0/2$ — a critical point inside the interval. $f''(d_1) = -2 < 0$, so it is a maximum, not a minimum.

$f(0) = f(d_0) = 0$. Since $f$ is continuous on a closed interval and the interior critical point is a maximum, the global minimum on the interval is at the endpoints. For received power ($\propto 1/f(d_1)^2$), the minimum of received power occurs at the maximum of $f$ — i.e., $d_1 = d_0/2$. So the RIS-path-worst placement is mid-link. $\blacksquare$

A BS with transmit power $P_t$ and gain $G_t$ produces power density $P_t G_t / (4\pi d_1^2)$ at the RIS, hence each element captures $P_t G_t A_e / (4\pi d_1^2)$ of incident energy.

Each element re-radiates its captured energy isotropically; at the UE at distance $d_2$, one element contributes field amplitude proportional to $\sqrt{A_e}/(d_2)$. Summing coherently over $N$ elements multiplies the amplitude by $N$.

Squared amplitude and including the receive aperture $\lambda^{2} G_r / (4\pi)$ (absorbed into the effective area), the total received power becomes $P_r = P_t G_t G_r \frac{N^2 A_e^2}{(4\pi)^2 d_1^2 d_2^2}$. $\blacksquare$

Write the sum as $Z = Z_c + Z_r$, where $Z_c$ is the coherent sum over $(1-\epsilon)N$ elements (giving $(1-\epsilon)N \alpha \beta$ in magnitude), and $Z_r$ is a random-phase sum over $\epsilon N$ elements with $\mathbb{E}|Z_r|^2 = \epsilon N \alpha^2 \beta^2$.

$\mathbb{E}|Z|^2 = |Z_c|^2 + \mathbb{E}|Z_r|^2$ (cross-term vanishes by independence and zero-mean $Z_r$) $= \big[(1-\epsilon)N\big]^2 \alpha^2\beta^2 + \epsilon N \alpha^2 \beta^2.$

For small $\epsilon$, the first term dominates and the SNR is roughly $(1-\epsilon)^2 N^2 \alpha^2 \beta^2$ — a penalty of $(1-\epsilon)^2$ on the coherent $N^2$ gain. A $10\%$ failure rate costs $(0.9)^2 = 0.81 \approx -0.9\text{ dB}$; a $50\%$ failure rate costs $4\times = -6\text{ dB}$. Much worse than the $(1-\epsilon)$ penalty one might naively expect. $\blacksquare$

For a panel at $d_1$, per-panel coherent power is proportional to $1/(d_1^2 (d_0 - d_1)^2)$, up to a common constant $K = P_t G_t G_r N^2 A_e^2 / (4\pi)^2$.

Both at $d_1 = 0.1\,d_0$: per-panel $1/(0.1\,d_0)^2 / (0.9\,d_0)^2 = 1/(0.0081 d_0^4) = 123.5/d_0^4$. Two panels: $2 \cdot 123.5 = 247/d_0^4$.

One at $0.1$, one at $0.9$: by symmetry, each gives $123.5/d_0^4$. Total: also $247/d_0^4$.

The expected powers are equal, but the second arrangement provides spatial diversity: if one panel's link is blocked (e.g., by a truck), the other still works. This is why multi-RIS deployment (Chapter 12) aims to place panels at both endpoints of the link. $\blacksquare$

Take $\phi_1 = 1$ and $\phi_2 = e^{j\pi/2} = j$. Both have modulus 1, so both are feasible.

$(\phi_1 + \phi_2)/2 = (1 + j)/2$, with modulus $|(1+j)/2| = \sqrt{2}/2 \approx 0.707 < 1$. Not feasible.

The feasible set is a circle in $\mathbb{C}$, which is not convex. Extending to $N$ elements, the feasible set is a product of circles — the $N$-dimensional complex torus — which is also not convex. $\blacksquare$

$\beta^2 = (0.1)^2 / (4\pi \cdot 30)^2 = 0.01 / 142\,000 \approx 7.0 \cdot 10^{-8}$.

With $\sigma_{\text{relay}}^2 = \sigma^2$: $N^2 > 1 / \beta^2 \approx 1.42 \cdot 10^7$, so $N \gtrsim 3\,770$.

For a symmetric 60-m RIS link at 3 GHz to beat a simple AF relay, one needs thousands of elements. This is steep but not unreasonable at mmWave where element sizes are much smaller: at 28 GHz, $\lambda = 1.07\text{ cm}$, and a $60\text{ cm} \times 60\text{ cm}$ panel holds about $N = 12\,544$ half-wavelength elements. $\blacksquare$

Writing $\text{SNR}^\star = N^2 \text{SNR}_{0}$ where $\text{SNR}_{0}$ is the single-element SNR, then $\log_2(1 + N^2 \text{SNR}_{0}) \approx \log_2(N^2 \text{SNR}_{0}) = 2\log_2 N + \log_2 \text{SNR}_{0}$ when $N^2 \text{SNR}_{0} \gg 1$.

The capacity gain over a single-element baseline (whose capacity is $\log_2(1 + \text{SNR}_{0})$) is thus $\approx 2 \log_2 N$ bits/s/Hz. Doubling $N$ adds two bits/s/Hz to the capacity — log gain, not linear.

The $N^2$ SNR advantage translates to a $2\log_2 N$ rate gain. This is significant but diminishes: going from $N = 256$ to $N = 1024$ gives $\Delta R = 2\log_2 4 = 4$ bits/s/Hz. Useful, but not the same drama as the linear-in-$N$ power gain. $\blacksquare$

The effective channel is $\mathbf{h}_{\text{eff}}^H = \mathbf{h}_2^H \boldsymbol{\Phi} \mathbf{H}_1 \in \mathbb{C}^{1 \times N_t}$. Under unit-norm beamforming, the optimal $\mathbf{v}^\star(\boldsymbol{\Phi}) = \mathbf{h}_{\text{eff}} / \|\mathbf{h}_{\text{eff}}\|$, giving SNR $= (P_t/\sigma^2) \|\mathbf{h}_{\text{eff}}\|^2$.

We need to maximize $\|\mathbf{h}_2^H \boldsymbol{\Phi} \mathbf{H}_1\|^2$ over unit-modulus $\boldsymbol{\phi}$. Using the diagonal-product identity: $\mathbf{h}_2^H \boldsymbol{\Phi} \mathbf{H}_1 = \boldsymbol{\phi}^T \mathbf{M}$, so we seek to maximize $\|\boldsymbol{\phi}^T \mathbf{M}\|^2$.

Unlike the $N_t = 1$ case, this is no longer a simple matched-filter problem: the objective is a quadratic form in $\boldsymbol{\phi}$, not a squared inner product. This is the core non-convex optimization problem addressed by SDR and manifold optimization in Chapter 6. A full closed-form solution does not exist in general — which is why the book has 17 more chapters. $\blacksquare$

False in the strict sense: each RIS element requires a bias voltage (for varactor or PIN-diode control) and the controller consumes power to compute and apply phase updates. However, the RF power budget is zero — the RIS does not amplify or generate signals. Typical idle consumption is $\sim 10\,\mu\text{W}$ per element, compared with hundreds of milliwatts per active-array element. So "passive" is a useful shorthand, but not a technical accuracy.

Each RIS update occupies $10\,\text{ms}$. In one coherence interval of $50\,\text{ms}$ there are at most 5 updates, so the controller can commit at most 5 new RIS configurations.

If the RIS controller were fast enough to reconfigure every channel sample ($50\,\text{ms} / 1\,\text{ms}$ = 50 samples), every sample gets a fresh RIS. With only 5 samples per coherence block served by 5 different configurations, the remaining 45 samples run on a stale configuration — the "effective $N$" is $N / 5 \approx 200$ if one splits elements across 5 configurations, or more conservatively $N$ is limited to what one configuration delivers. The update-limited architecture loses most of the potential $N^2$ gain.

RIS controller latency is a first-order design constraint. Chapter 18 studies this in detail; Chapter 4 covers CSI acquisition, which often dominates latency in the early deployments. $\blacksquare$

A well-argued answer identifies: (a) a RIS with $N$ elements at product path loss $d_1^2 d_2^2$ gains $N^2$ coherently but suffers the squared-distance penalty; (b) a small cell has a single path loss of $d_0^2$ but only single-element gain; (c) the RIS needs to estimate the cascaded channel, which scales as $\mathcal{O}(N)$ pilots in the best case (Chapter 4); (d) the RIS consumes $\sim\mu\text{W}$ per element while the small cell consumes $\sim\text{W}$ per radio chain; (e) RIS deployment per $\text{m}^2$ is currently comparable to LED panel manufacturing.

The RIS wins if: deployment is where power is unavailable; the link is blocked so the direct path doesn't exist anyway; $N \geq 256$ and the coherence time is long enough for CSI; the UE mobility is low. The small cell wins if: power is available and the operator already has backhaul; the UE is mobile; $N$ would need to be $> 10^4$ to match the small cell's throughput. This is precisely the analysis carried out in Chapter 17 on deployment optimization. $\blacksquare$