Exercises

$\Delta\nu = (\sin(30°) - \sin(20°))/2 = (0.5 - 0.342)/2 = 0.079$.

$r_k \approx \lceil N_t \cdot \Delta\nu \rceil + 1 = \lceil 32 \times 0.158/2 \rceil + 1 = \lceil 2.53 \rceil + 1 = 4$. Verifying numerically by computing the covariance eigenvalues confirms $r_k = 4$ captures $>99\%$ of the trace.

$C_{\text{JSDM}} = 4 \times 4 + 4 \times 6 + 4 \times 5 = 16 + 24 + 20 = 60$.

$C_{\text{full}} = 12 \times 64 = 768$.

$\rho = 60/768 \approx 0.078$. JSDM requires only $7.8\%$ of the full-CSI feedback.

For a ULA, $\mathbf{h}_k = \int p_k(\theta)^{1/2} \mathbf{a}(\theta) \, dW(\theta)$ where $dW(\theta)$ is a white noise process and $p_k(\theta)$ is the angular power spectrum. The steering vector is $[\mathbf{a}(\theta)]_m = e^{j\pi m \sin\theta}$.

$[\mathbf{R}_k]_{m,n} = \int p_k(\theta) e^{j\pi(m-n)\sin\theta} \, d\theta$. This depends only on $m-n$, so $\mathbf{R}_k$ is Toeplitz. $\blacksquare$

The DFT matrix $\mathbf{F}$ is unitary: $\mathbf{F}^H \mathbf{F} = \mathbf{I}_{N_t}$. This means $\mathbf{f}_m^H \mathbf{f}_n = \delta_{mn}$ for any two columns.

$\mathbf{B}_g = \mathbf{F}(:, \mathcal{I}_g)$ and $\mathbf{B}_{g'} = \mathbf{F}(:, \mathcal{I}_{g'})$ with $\mathcal{I}_g \cap \mathcal{I}_{g'} = \emptyset$. Then $[\mathbf{B}_g^H \mathbf{B}_{g'}]_{ij} = \mathbf{f}_{\mathcal{I}_g(i)}^H \mathbf{f}_{\mathcal{I}_{g'}(j)} = 0$ since $\mathcal{I}_g(i) \neq \mathcal{I}_{g'}(j)$. Therefore $\mathbf{B}_g^H \mathbf{B}_{g'} = \mathbf{0}$. $\blacksquare$

The effective channel is $\tilde{\mathbf{H}}_g \in \mathbb{C}^{|\mathcal{S}_g| \times r_g}$. ZF precoding requires $\tilde{\mathbf{H}}_g$ to have full row rank, which needs $r_g \geq |\mathcal{S}_g|$. Here $4 \geq 2$, so ZF is feasible.

With ZF and no inter-group interference, user $k$'s SINR is $$\text{SINR}_k = \frac{\ntn{tx\_power}/K}{[(\tilde{\mathbf{H}}_g \tilde{\mathbf{H}}_g^H)^{-1}]_{kk} \cdot \sigma^2}$$ which grows with $\ntn{tx\_power}$ and benefits from the excess degrees of freedom ($r_g - |\mathcal{S}_g| = 2$ extra dimensions provide array gain).

$\tilde{\mathbf{R}}_k = \mathbb{E}[\mathbf{B}_g^H \mathbf{h}_k \mathbf{h}_k^H \mathbf{B}_g] = \mathbf{B}_g^H \mathbf{R}_k \mathbf{B}_g = (\mathbf{U}_g^{(r_g)})^H \mathbf{R}_k \, \mathbf{U}_g^{(r_g)} \in \mathbb{C}^{r_g \times r_g}$.

$\tilde{\mathbf{R}}_k$ is full rank if and only if $\mathcal{R}(\mathbf{R}_k) \cap \mathcal{R}(\mathbf{U}_g^{(r_g)})$ has dimension $r_g$, i.e., the column space of $\mathbf{R}_k$ includes the entire group eigenspace. This holds when user $k$'s eigenspace is contained within or well-aligned with the group eigenspace.

$\text{tr}(\mathbf{B}_2^H \mathbf{R}_k \mathbf{B}_2) = \sum_{n \in \mathcal{I}_2} \mathbf{f}_n^H \mathbf{R}_k \mathbf{f}_n = \sum_{n \in \mathcal{I}_2} \int p_k(\theta) |\mathbf{f}_n^H \mathbf{a}(\theta)|^2 d\theta$.

For $n \in \mathcal{I}_2$ and $\theta$ in the support of $p_k(\theta)$ (group 1's region), $|\nu_n - \nu(\theta)| \geq \Delta\nu_{\min}$. Therefore $|\mathbf{f}_n^H \mathbf{a}(\theta)|^2 \leq \frac{1}{N_t \sin^2(\pi \Delta\nu_{\min})}$.

$\text{tr}(\mathbf{B}_2^H \mathbf{R}_k \mathbf{B}_2) \leq |\mathcal{I}_2| \cdot \frac{1}{N_t \sin^2(\pi \Delta\nu_{\min})} \cdot \text{tr}(\mathbf{R}_k) = \frac{r_2}{N_t} \cdot \frac{\text{tr}(\mathbf{R}_k)}{\sin^2(\pi \Delta\nu_{\min})}.$$Normalizing by$\text{tr}(\mathbf{R}_k)$gives the stated bound.$\blacksquare$

Treating inter-group interference as noise, the sum rate within group $g$ is $$R_g \geq \log_2 \det\!\left(\mathbf{I} + \frac{P/K}{\sigma^2 + I_g} \tilde{\mathbf{H}}_g \tilde{\mathbf{H}}_g^H\right)$$ where $I_g = \sum_{g' \neq g} \text{tr}(\mathbf{B}_{g'}^H \mathbf{R}_k \mathbf{B}_{g'}) \cdot P/K$.

Summing over groups and bounding $I_g \leq I_{\max}$ for all $g$: $$R_{\text{sum}} \geq \sum_g \log_2 \det\!\left(\mathbf{I} + \frac{P/K}{\sigma^2} \tilde{\mathbf{H}}_g \tilde{\mathbf{H}}_g^H\right) - G \log_2(1 + I_{\max}/\sigma^2).$$ The first term is the interference-free sum rate; the second term is the penalty. $\blacksquare$

Users span $[-60°, 60°]$ in $\sin\theta$-space: $[-0.866, 0.866]$, total span $1.732$.

Each user has angular spread $15°$, so $r_k \approx \lceil 64 \times \Delta\sin/2 \rceil + 1$. At broadside ($\theta = 0°$), $\Delta\sin(15°) \approx 0.26$, giving $r_k \approx 9$. Near endfire, the $\sin$-domain spread is smaller.

With $G = 3$ groups: each group covers $40°$ in angle, with $4$ users per group. Group boundaries at $-20°, 20°$ ensure $\sim 5°$ guard bands. Each group has $r_g \approx 9$ and $|\mathcal{S}_g| = 4 < r_g$, so ZF is feasible within each group with $5$ excess DoF per group.

$G = 2$: fewer groups, more users per group ($6$ each), but wider angular regions ($r_g \approx 13$) — potentially more intra-group interference. $G = 4$: narrower groups ($r_g \approx 7$), but only $3$ users per group and tighter guard bands. Numerical optimization typically favors $G = 3$ or $G = 4$ for this configuration.

With orthogonal eigenspaces, all groups share the same pilot resources. The total pilot overhead is $\tau_p = \max_g r_g = 7$ symbols.

For comparison, without reuse: $\tau_p = \sum_g r_g = 5 + 7 + 6 + 4 = 22$ symbols.

$\|\mathbf{U}\mathbf{U}^H - \mathbf{V}\mathbf{V}^H\|_F^2 = \text{tr}(\mathbf{U}\mathbf{U}^H) + \text{tr}(\mathbf{V}\mathbf{V}^H) - 2\text{tr}(\mathbf{U}\mathbf{U}^H \mathbf{V}\mathbf{V}^H) = 2r - 2\|\mathbf{U}^H\mathbf{V}\|_F^2$.

Since $0 \leq \|\mathbf{U}^H\mathbf{V}\|_F^2 \leq r$ (the singular values of $\mathbf{U}^H\mathbf{V}$ are between 0 and 1), we get $0 \leq d_c^2 = r - \|\mathbf{U}^H\mathbf{V}\|_F^2 \leq r$. So $0 \leq d_c / \sqrt{r} \leq 1$. The standard normalization gives $d_c = \frac{1}{\sqrt{2}} \|\cdot\|_F \in [0, \sqrt{r}]$; with $r$-dimensional subspaces and the $1/\sqrt{2}$ factor, $d_c \in [0, \sqrt{r}]$. For $r = 1$, $d_c \in [0, 1]$. $\blacksquare$

$\tilde{\mathbf{h}}_k = \sum_{i=1}^{r_g} \alpha_{k,i} \mathbf{b}_i$ where $\mathbf{b}_i$ are the DFT beams and $|\alpha_{k,1}| \geq |\alpha_{k,2}| \geq \cdots$. The truncated version is $\hat{\mathbf{h}}_k = \sum_{i=1}^{L} \alpha_{k,i} \mathbf{b}_i$.

$\|\tilde{\mathbf{h}}_k - \hat{\mathbf{h}}_k\|^2 = \sum_{i=L+1}^{r_g} |\alpha_{k,i}|^2$. The fractional power loss is $1 - \frac{\sum_{i=1}^L |\alpha_{k,i}|^2}{\sum_{i=1}^{r_g} |\alpha_{k,i}|^2}$.

The SINR with truncated CSI is reduced by approximately $\frac{\sum_{i=1}^L |\alpha_{k,i}|^2}{\|\tilde{\mathbf{h}}_k\|^2}$, plus an interference term from the mismatched precoder. For $L = r_g - 1$ (dropping one beam), the loss is typically $< 0.5$ dB.

With JSDM-ZF, each user's SINR scales as $O(N_t)$ (the pre-beamformer provides $O(N_t)$ array gain, and inter-group interference is $O(1)$). Therefore $R_k = \log_2(1 + O(N_t)) = O(\log N_t)$.

Sum rate: $R_{\text{sum}} = K \cdot O(\log N_t) \to \infty$. Per user: $R_k = O(\log N_t) \to \infty$. Per antenna: $R_{\text{sum}} / N_t = K \cdot O(\log N_t) / N_t \to 0$.

Each additional antenna provides diminishing marginal returns (logarithmic growth per user), but the total system capacity grows without bound. This is the massive MIMO scaling regime.

Under the standard wideband channel model, the spatial covariance $\mathbf{R}_k$ depends on the angles of arrival, which are the same across all subcarriers. The path delays create frequency selectivity in $\mathbf{h}_{k,n}$ but not in $\mathbf{R}_k = \mathbb{E}[\mathbf{h}_{k,n} \mathbf{h}_{k,n}^H]$ (which is the same for all $n$).

Since $\mathbf{R}_k$ is the same for all $n$, the eigendecomposition and hence $\mathbf{B}_g$ are frequency-independent. A single pre-beamformer per group suffices for all subcarriers.

The effective channel $\tilde{\mathbf{h}}_{k,n} = \mathbf{B}_g^H \mathbf{h}_{k,n}$ varies with $n$ because the instantaneous channel $\mathbf{h}_{k,n}$ has frequency-selective fading. Therefore $\mathbf{P}_{g,n}$ must be computed independently for each subcarrier (or subband).

CSI feedback: $N \times \sum_g |\mathcal{S}_g| r_g$ complex scalars across all subcarriers, vs. $N \times K N_t$ for full CSI. The per-subcarrier reduction factor $r_g / N_t$ is the same as in the narrowband case.

$r_g = 8$ DFT beams cover a spatial frequency interval of $\Delta\nu = 8/128 = 0.0625$.

Near broadside, $\Delta\theta \approx \arcsin(2\Delta\nu) = \arcsin(0.125) \approx 7.2°$.

Forming $\mathbf{H}\mathbf{H}^{H}$: $O(K^{2} N_t) = O(16^2 \times 128) = O(32768)$. Inverting $\mathbf{H}\mathbf{H}^{H}$: $O(K^{3}) = O(4096)$. Matrix multiply $\mathbf{H}^{H} (\mathbf{H}\mathbf{H}^{H})^{-1}$: $O(N_t K^{2}) = O(32768)$. Total: $\sim 70$K FLOPs.

Pre-beamforming: $G \times O(r_g N_t |\mathcal{S}_g|) = 4 \times O(8 \times 128 \times 4) = O(16384)$. Per-group ZF: $G \times O(|\mathcal{S}_g|^2 r_g + |\mathcal{S}_g|^3) = 4 \times O(128 + 64) = O(768)$. Composite: $G \times O(N_t r_g |\mathcal{S}_g|) = O(16384)$. Total: $\sim 33$K FLOPs — about $2\times$ less than full ZF.

The complexity advantage of JSDM grows with $N_t$: the inner precoding cost is $O(r_g^2 |\mathcal{S}_g|)$ per group, independent of $N_t$. Only the pre-beamforming multiply scales with $N_t$.

The Davis-Kahan theorem states that $\sin\Theta(\hat{\mathbf{U}}_g, \mathbf{U}_g) \leq \frac{\|\boldsymbol{\Delta}\|}{\lambda_{r_g} - \lambda_{r_g+1}}$ where $\lambda_{r_g}$ and $\lambda_{r_g+1}$ are the $r_g$-th and $(r_g+1)$-th eigenvalues of $\mathbf{R}_k$. The eigengap $\gamma_g = \lambda_{r_g} - \lambda_{r_g+1}$ determines the sensitivity.

The additional inter-group interference due to the perturbed pre-beamformer is $\Delta I \leq O(\delta^2 / \gamma_g^2)$. The rate loss per user is $\Delta R \leq \log_2(1 + \Delta I / \sigma^2)$.

For $\Delta R < 1$ bit/s/Hz, we need $\Delta I < \sigma^2$, which requires $\delta < \gamma_g \sqrt{\sigma^2 / c}$ for a geometry-dependent constant $c$. In massive MIMO with well-separated eigenvalues ($\gamma_g$ large), the tolerance is generous.

The pre-beamformer that maximizes the average received signal energy for group $g$ is $$\mathbf{B}_g^* = \arg\max_{\mathbf{B} \in \mathbb{C}^{N_t \times r_g}, \mathbf{B}^H\mathbf{B} = \mathbf{I}} \text{tr}(\mathbf{B}^H \bar{\mathbf{R}}_g \mathbf{B}).$$

By the Ky Fan theorem (or equivalently, the Rayleigh-Ritz theorem for multiple vectors), the maximum is achieved when the columns of $\mathbf{B}_g^*$ are the eigenvectors corresponding to the $r_g$ largest eigenvalues of $\bar{\mathbf{R}}_g$. The maximum value is $\sum_{i=1}^{r_g} \lambda_i(\bar{\mathbf{R}}_g)$.

Therefore $\mathbf{B}_g = \mathbf{U}_g^{(r_g)}$ is the energy-optimal pre-beamformer under the rank-$r_g$ constraint. It acts as a spatial matched filter in the sense that it projects onto the subspace containing the most signal energy. $\blacksquare$