Exercises

$\eta_{\text{FDD}} = 1 - (64 + 10)/300 = 1 - 74/300 = 0.753$. About 75.3% efficient.

$\eta_{\text{TDD}} = 1 - 10/300 = 1 - 0.033 = 0.967$. About 96.7% efficient. TDD is $\sim 28\%$ more efficient in this scenario.

$B_{\text{fb}} = 2 \times 4 \times 128 = 1024$ bits.

$16 \times 1024 = 16{,}384$ bits per coherence block.

$r_k \approx \lceil 32 \times 15/180 \rceil = \lceil 2.67 \rceil = 3$. Adding 1–2 guard bins for spectral leakage: $r_k \approx 4$–$5$.

$\epsilon = 2^{-B/15}$, so $-B/15 = \log_2(0.01) = -6.64$. Hence $B = 15 \times 6.64 \approx 100$ bits.

For $N_t = 16$ antennas and 1% quantization error, we need about 100 feedback bits ($\sim 6.6$ bits per antenna dimension). For $N_t = 64$, the same error would need $63 \times 6.64 \approx 418$ bits — the linear scaling with $N_t$.

With imperfect CSI, the ZF precoder is designed from $\hat{\mathbf{H}}_k$ (the quantized direction). The residual interference at user $k$ from user $j$'s stream is bounded by $|\mathbf{H}_{k}^{H} \hat{\mathbf{v}}_j|^2 \leq \|\mathbf{H}_{k}\|^2 \epsilon_k$ where $\epsilon_k = d_c^2(\bar{\mathbf{H}}_k, \hat{\mathbf{v}}_k)$.

The total interference at user $k$ from $K-1$ other users is at most $(K-1) \cdot (P_t/K) \cdot \|\mathbf{H}_{k}\|^2 \epsilon_k$.

The per-user rate loss is $\Delta R_k \leq \log_2(1 + \text{SNR}(K-1)\epsilon_k)$. Summing over $K$ users and taking expectation: $\Delta R_{\text{sum}} \leq K\log_2(1 + \text{SNR}(K-1)\epsilon(\mathcal{C}))$. $\blacksquare$

For $r_k = 8$:

• $N_t = 32$: $M = 32 \log_2(4) = 64 > 16$. No savings.
• $N_t = 64$: $M = 32 \log_2(8) = 96 > 32$. No savings.
• $N_t = 128$: $M = 32 \log_2(16) = 128 = 64$. Exactly $50\%$.
• $N_t = 256$: $M = 32 \log_2(32) = 160 < 128$. $62.5\%$ savings.

For $r_k = 8$, savings exceed $50\%$ when $N_t > 128$. For $r_k = 4$, savings exceed $50\%$ when $N_t > 32$. Compressed sensing provides large savings only when $N_t \gg r_k$.

Phase: $4 \times 3 = 12$ bits. Differential amplitude: $3 \times 1 = 3$ bits. Total per subband: $15$ bits.

$13 \times 15 = 195$ bits for subband feedback (rank 1).

$\gamma = 512 / 2048 = 1/4 = 0.25$.

$B_{\text{fb}} = 512 \times 4 = 2048$ bits.

$B_{\text{naive}} = 2048 \times 5 = 10{,}240$ bits. CsiNet achieves $5\times$ reduction in feedback bits.

Since $\mathbf{\Lambda}_g$ is diagonal, $[\mathbf{H}_{\text{eff},k}]_i \sim \mathcal{CN}(0, \lambda_i)$ are independent. The rate-distortion function decomposes: $R^*(D) = \sum_{i=1}^{r_g} R_i^*(D_i)$ subject to $\sum_i D_i = D$.

For a complex Gaussian source with variance $\lambda_i$: $R_i^*(D_i) = [\log_2(\lambda_i / D_i)]^+$ (bits per complex sample).

Minimizing $\sum_i R_i^*(D_i)$ subject to $\sum_i D_i \leq D$ via Lagrangian: $D_i^* = \min(\lambda_i, \nu)$ where $\nu$ satisfies $\sum_i \min(\lambda_i, \nu) = D$. This is reverse water-filling: components with $\lambda_i < \nu$ are described with distortion $\lambda_i$ (zero rate), others with distortion $\nu$. $\blacksquare$

DL pilots: $2 \times 10 = 20$ (both groups). UL pilots: $10$. $\eta = 1 - (20 + 10)/200 = 85\%$. Without JSDM: $\eta = 1 - (128+10)/200 = 31\%$.

With JSDM: $10 \times 2 \times 5 \times 10 = 1000$ bits total. Without: $10 \times 2 \times 5 \times 128 = 12{,}800$ bits total. Reduction: $12.8\times$.

Without JSDM: $\text{NMSE} = \sigma^2/(P_t \times 128)$. With JSDM: effective noise is projected to $r_g = 10$ dimensions, but pilot SNR is $P_t/\sigma^2$ per dimension (same power spread over fewer pilots). $\text{NMSE}_{\text{JSDM}} = \sigma^2/(P_t \times 10)$. The JSDM estimate has $128/10 = 12.8\times$ higher NMSE per dimension, but the channel is only $10$-dimensional. The total MSE is $10 \times \text{NMSE}_{\text{JSDM}}$ vs. $128 \times \text{NMSE}$, which are comparable.

$d_c(\mathbf{v}_{1}, \mathbf{v}_{2}) = \sqrt{1 - |\mathbf{v}_{1}^{H} \mathbf{v}_{2}|^2} \geq 0$ since $|\mathbf{v}_{1}^{H} \mathbf{v}_{2}| \leq 1$ by Cauchy–Schwarz. Symmetry: $|\mathbf{v}_{1}^{H} \mathbf{v}_{2}| = |\mathbf{v}_{2}^{H} \mathbf{v}_{1}|$.

$d_c = 0 \iff |\mathbf{v}_{1}^{H} \mathbf{v}_{2}| = 1 \iff \mathbf{v}_{1} = e^{j\phi} \mathbf{v}_{2}$ for some phase $\phi$, which means $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ span the same 1-dimensional subspace (the same point on $\mathcal{G}(1, N_t)$).

Let $\theta_{ij} = \arccos(|\mathbf{v}_{i}^{H} \mathbf{v}_{j}|)$ be the principal angle. Then $d_c(\mathbf{v}_{i}, \mathbf{v}_{j}) = \sin(\theta_{ij})$. The principal angles satisfy $\theta_{13} \leq \theta_{12} + \theta_{23}$. Since $\sin$ is subadditive on $[0, \pi/2]$: $\sin(\theta_{13}) \leq \sin(\theta_{12} + \theta_{23}) \leq \sin(\theta_{12}) + \sin(\theta_{23})$, hence $d_c(\mathbf{v}_{1}, \mathbf{v}_{3}) \leq d_c(\mathbf{v}_{1}, \mathbf{v}_{2}) + d_c(\mathbf{v}_{2}, \mathbf{v}_{3})$. $\blacksquare$

$\text{MSE} = \mathbb{E}[\|\mathbf{x} - \mathbf{BA}\mathbf{x}\|^2] = \text{tr}(\mathbf{R}_x) - 2\text{tr}(\mathbf{BA}\mathbf{R}_x) + \text{tr}(\mathbf{BA}\mathbf{R}_x\mathbf{A}^T\mathbf{B}^T)$.

The product $\mathbf{P} = \mathbf{BA}$ is a rank-$M$ projection-like operator. The MSE is minimized when $\mathbf{P}$ is the orthogonal projection onto the top-$M$ eigenspace of $\mathbf{R}_x$. By the Eckart–Young theorem, $\mathbf{P}^* = \mathbf{U}_M \mathbf{U}_M^T$, achieved by $\mathbf{A} = \mathbf{U}_M^T$, $\mathbf{B} = \mathbf{U}_M$.

$\text{MSE}^* = \sum_{i=M+1}^{N} \lambda_i$ — the sum of the discarded eigenvalues. This is exactly PCA truncation. CsiNet's nonlinear layers improve upon this baseline by capturing non-Gaussian structure in the channel distribution. $\blacksquare$

Group 1: $[15°, 25°]$, Group 2: $[45°, 55°]$, Group 3: $[-35°, -25°]$. Minimum gap: $45° - 25° = 20°$ between groups 1 and 2. Non-overlapping: yes.

$r_g \approx \lceil 64 \times 10/180 \rceil = \lceil 3.56 \rceil = 4$ for each group.

Group 1 spans spatial frequencies in $[\sin(15°)/(2d), \sin(25°)/(2d)]$ where $d = \lambda/2$. This maps to DFT indices $[64 \times 0.259/2, 64 \times 0.423/2] \approx [8.3, 13.5]$, so indices $\{8, 9, 10, 11, 12, 13\}$ (taking $r_1 = 6$ with guard bins). $\mathbf{B}_1 = [\mathbf{f}_8, \mathbf{f}_9, \ldots, \mathbf{f}_{13}]$ (DFT columns).

For a group-2 user, $\mathbf{H}_{2}$ has energy in DFT indices $\approx \{22, \ldots, 27\}$. $\|\mathbf{B}_1^H \mathbf{H}_{2}\|^2 / \|\mathbf{H}_{2}\|^2 \approx 0$ since the DFT columns $\{8, \ldots, 13\}$ and $\{22, \ldots, 27\}$ are orthogonal.

$\eta(r_g) = 1 - (r_g + K_g)/T_c$. This is linear and decreasing in $r_g$.

The effective channel energy is $\|\mathbf{H}_{\text{eff},k}\|^2 = \sum_{i=1}^{r_g} \lambda_i |g_i|^2$ where $|g_i|^2 \sim \text{Exp}(1)$. In expectation: $\mathbb{E}[\|\mathbf{H}_{\text{eff},k}\|^2] = \sum_{i=1}^{r_g} \lambda_i$. Including more eigenmodes (larger $r_g$) increases the beamforming gain, but the marginal gain $\lambda_{r_g+1}$ decreases (eigenvalues are ordered).

The rate is approximately $R(r_g) \approx (1 - (r_g+K_g)/T_c) \log_2(1 + \text{SNR} \sum_{i=1}^{r_g} \lambda_i / K_g)$. Setting $dR/dr_g = 0$: $$-\frac{1}{T_c}\log_2\!\left(1 + \text{SNR}\sum_{i=1}^{r_g}\lambda_i/K_g\right) + \eta(r_g)\frac{\text{SNR}\lambda_{r_g}/K_g}{(1+\text{SNR}\sum_{i=1}^{r_g}\lambda_i/K_g)\ln 2} = 0.$$ The optimal $r_g^*$ balances overhead loss (first term) against SNR gain (second term). It increases with $T_c$ (larger coherence blocks tolerate more overhead) and decreases as eigenvalues decay faster (diminishing returns from additional dimensions). $\blacksquare$

The autoencoder $(\mathcal{E}_\theta, \mathcal{D}_\phi)$ is trained to minimize $\mathbb{E}_{p_{\text{train}}}[\text{MSE}]$. By definition, this is the minimum achievable under $p_{\text{train}}$. Under $p_{\text{deploy}} \neq p_{\text{train}}$, the same encoder-decoder is suboptimal (it was not trained for this distribution), so $\text{NMSE}_{\text{mismatch}} \geq \text{NMSE}_{\text{matched}}$.

UMa channels have narrow angular spread ($\Delta\theta \sim 5°$–$15°$) and few clusters. UMi channels have wide angular spread ($\Delta\theta \sim 30°$–$60°$) and many clusters. A CsiNet trained on UMa learns to compress channels with $\sim 5$ significant angular bins. Deployed on UMi, it encounters channels with $\sim 20$+ significant bins — the encoder discards information that it was never trained to preserve. The NMSE can degrade by 10–20 dB.

Collect a small dataset $\{\mathbf{H}_{a}^{(n)}\}_{n=1}^{N'}$ from the deployment environment ($N' \ll N$, e.g., 100–1000 samples). Fine-tune only the decoder $\mathcal{D}_\phi$ (keeping the encoder fixed, since it runs on the UE and is harder to update). This "decoder-side adaptation" requires only BS-side computation and can be done online as the BS accumulates channel measurements. $\blacksquare$

For a ULA, $|\mathbf{a}(\theta)^H \mathbf{f}(\phi)|^2 = \frac{\sin^2(N_t\pi\Delta/2)}{N_t^{2}\sin^2(\pi\Delta/2)}$ where $\Delta = \sin\theta - 2\pi i / (2^B \cdot 2\pi/N_t)$.

The worst case is when the true angle falls exactly between two codewords: $\Delta_{\max} = \pi/(2^B)$. The gain is approximately $G \approx \text{sinc}^2(N_t/(2^{B+1}))$.

We need $\text{sinc}^2(N_t/(2^{B+1})) \geq 0.794$. $\text{sinc}(x) \geq 0.891$ requires $x \leq 0.36$, so $N_t/(2^{B+1}) \leq 0.36$, giving $B \geq \log_2(N_t/0.72) - 1 \approx \log_2(N_t) + 0.47$. For $N_t = 64$: $B \geq 6.47$, so $B = 7$ bits suffice.

JSDM: $2 \times 4 \times 12 = 96$ bits. Naive: $2 \times 4 \times 256 = 2048$ bits. Reduction factor: $2048/96 = 21.3\times$.

$\mathbf{H}_{k} = \alpha \mathbf{a}(\theta)$ where $\theta \sim \text{Unif}[-\pi/2, \pi/2]$. The channel direction $\bar{\mathbf{H}}_k = \mathbf{a}(\theta)$ traces a 1D curve on the complex unit sphere, parameterized by the spatial frequency $\psi = \pi\sin\theta$.

Since $\psi$ ranges over $[-\pi, \pi]$ and is uniformly distributed (for $\theta$ uniform on $[-\pi/2, \pi/2]$ and large $N_t$), the optimal codebook places $2^B$ points uniformly on this 1D arc. These are exactly the DFT vectors with spatial frequencies $2\pi i / 2^B$.

The maximum chordal distance to the nearest codeword is $d_c^{\max} = \sin(\pi/2^{B+1})$, achieved at the midpoints between consecutive codewords. No other placement of $2^B$ points on this arc achieves a smaller maximum distance — the DFT codebook is the optimal uniform quantizer. $\blacksquare$