Exercises

$\mathbf{v} = \mathbf{h}/\|\mathbf{h}\| = [1, j, -1, -j]^T / 2$.

$\|\mathbf{h}\|^2 = 4$, so $\text{SNR} = 100 \times 4 = 400 = 26$ dB.

$\|\mathbf{v}_{k}^{\text{MRT}}\| = \|\mathbf{h}_k/\|\mathbf{h}_k\|\| = \|\mathbf{h}_k\|/\|\mathbf{h}_k\| = 1$ whenever $\|\mathbf{h}_k\| > 0$. $\blacksquare$

$\mathbf{H}\mathbf{W}^{\text{ZF}} = \mathbf{H}\mathbf{H}^{H}(\mathbf{H}\mathbf{H}^{H})^{-1} = \mathbf{I}_{K}$ since any invertible matrix times its inverse is the identity. $\blacksquare$

The matrix $\mathbf{H}\mathbf{H}^{H} \in \mathbb{C}^{K \times K}$ has rank at most $\min(N_t, K)$. For invertibility, we need $\text{rank}(\mathbf{H}\mathbf{H}^{H}) = K$, which requires $N_t \geq K$.

The matrix $\mathbf{H}\mathbf{H}^{H}$ is rank-deficient and not invertible. There are not enough spatial dimensions to null all inter-user interference. In this case, one must either reduce the number of simultaneously served users (user scheduling) or accept nonzero interference (MRT/RZF). $\blacksquare$

The effective channel gain for user $k$ under ZF is $g_k^{\text{ZF}} = 1/[(\mathbf{H}\mathbf{H}^{H})^{-1}]_{kk}$. Since $\mathbf{H}\mathbf{H}^{H} \sim \mathcal{W}_{K}(N_t, \mathbf{I})$ for i.i.d. Rayleigh, the diagonal of the inverse satisfies $\mathbb{E}[(\mathbf{H}\mathbf{H}^{H})^{-1}_{kk}] = 1/(N_t - K)$.

By Jensen's inequality, $\mathbb{E}[g_k^{\text{ZF}}] \geq 1/\mathbb{E}[(\mathbf{H}\mathbf{H}^{H})^{-1}_{kk}] = N_t - K$. More precisely, $g_k^{\text{ZF}} \sim \chi^2_{2(N_t - K + 1)}/(2)$ (scaled chi-squared), so $\mathbb{E}[g_k^{\text{ZF}}] = N_t - K + 1$. The "$N_t - K$" approximation is tight for large systems.

For MRT, $g_k^{\text{MRT}} = \|\mathbf{h}_k\|^2$ with $\mathbb{E}[\|\mathbf{h}_k\|^2] = N_t$. The penalty is $N_t - (N_t - K) = K$ degrees of freedom. $\blacksquare$

The key identity is: for conformable matrices $\mathbf{A}$ and $\mathbf{B}$,

$$(\mathbf{A}\mathbf{B} + \alpha\mathbf{I})^{-1}\mathbf{A} = \mathbf{A}(\mathbf{B}\mathbf{A} + \alpha\mathbf{I})^{-1}.$$

Setting $\mathbf{A} = \mathbf{H}^{H}$, $\mathbf{B} = \mathbf{H}$:

$$(\mathbf{H}^{H}\mathbf{H} + \alpha\mathbf{I})^{-1}\mathbf{H}^{H} = \mathbf{H}^{H}(\mathbf{H}\mathbf{H}^{H} + \alpha\mathbf{I})^{-1}.$$

Multiplying both sides on the right by $\mathbf{e}_k$ gives $\mathbf{h}_k$ on the left and the $k$-th column on the right. $\blacksquare$

At 0 dB: $\alpha = 4$ is large, so RZF behaves like MRT — noise dominates and we should maximise signal power. At 10 dB: moderate regularization balances signal, interference, and noise. At 20 dB: $\alpha \approx 0$, so RZF approaches ZF — interference dominates and should be nulled. $\blacksquare$

The MMSE receiver is $\mathbf{g}^{\star} = \mathbf{R}_y^{-1}\mathbb{E}[\mathbf{y}s_k^*]$ where $\mathbf{R}_y = \mathbf{h}_k\mathbf{h}_k^H + \sigma^2\mathbf{I}$. Since $\mathbb{E}[\mathbf{y}s_k^*] = \mathbf{h}_k$, we get $\mathbf{g}^{\star} = (\mathbf{h}_k\mathbf{h}_k^H + \sigma^2\mathbf{I})^{-1}\mathbf{h}_k$. By the matrix inversion lemma, this is proportional to $\mathbf{h}_k$, confirming MRT/MRC optimality for the single-user case. $\blacksquare$

By the resolvent identity and trace lemma, the per-user SINR under RZF converges almost surely to

$$\gamma^{\text{RZF}} = \frac{P_t}{K} \cdot \frac{(1 - \beta + \beta m(\alpha)\alpha)^2}{m(\alpha)(1 + \alpha m(\alpha)) + \sigma^2 m(\alpha)}$$

where $m(\alpha) = N_t^{-1}\text{tr}(\mathbf{H}\mathbf{H}^{H}/N_t + \alpha\mathbf{I})^{-1}$ is the Stieltjes transform evaluated at $-\alpha$.

Setting $\alpha = K\sigma^2/P_t$ and solving the Marchenko--Pastur fixed-point equation yields

$$\gamma^{\text{RZF}} \to \rho(N_t - K) = \frac{P_t}{K\sigma^2}(N_t - K)$$

giving the stated per-user rate $R = \log_2(1 + \rho(N_t - K))$. $\blacksquare$

$\mathbf{h}_1^H \mathbf{h}_2 = 0$, so the channels are orthogonal. ZF does not need to null any interference (it is already zero), so $\mathbf{v}_{k}^{\text{ZF}} = \mathbf{v}_{k}^{\text{MRT}}$ and no power penalty occurs.

$\|\mathbf{h}_k\|^2 = 2$ for both users. With equal power $p_k = P_t/2$:

$R_k = \log_2(1 + (5)(2)/1) = \log_2(11) = 3.46$ bits/s/Hz.

Sum rate $= 6.92$ bits/s/Hz for both ZF and DPC.

The gap is exactly zero. When channels are orthogonal, linear precoding is capacity-achieving: no DPC is needed. This is the favorable propagation condition of massive MIMO. $\blacksquare$

$\alpha^{\star} = K\sigma^2/P_t \to \infty$ as $P_t/\sigma^2 \to 0$. Then $(\mathbf{H}\mathbf{H}^{H} + \alpha\mathbf{I})^{-1} \approx (1/\alpha)\mathbf{I}$ and $\mathbf{W}^{\text{RZF}} \approx (1/\alpha)\mathbf{H}^{H} =$ MRT up to scaling.

$\alpha^{\star} \to 0$ as $P_t/\sigma^2 \to \infty$. Then $(\mathbf{H}\mathbf{H}^{H} + \alpha\mathbf{I})^{-1} \to (\mathbf{H}\mathbf{H}^{H})^{-1}$ and $\mathbf{W}^{\text{RZF}} \to \mathbf{H}^{H}(\mathbf{H}\mathbf{H}^{H})^{-1} = \mathbf{W}^{\text{ZF}}$. The sum rates converge accordingly. $\blacksquare$

$\rho = 10 / 16 = 0.625$ (linear scale, since $P_t/\sigma^2 = 10$).

$R_k = \log_2(1 + 0.625 \times 48) = \log_2(31) = 4.95$ bits/s/Hz.

$R_{\text{sum}} = 16 \times 4.95 = 79.2$ bits/s/Hz. This is the large-system approximation; Monte Carlo gives $\approx 78.5$ bits/s/Hz. $\blacksquare$

The MAC sum capacity with sum power $P_t$ and equal allocation is

$$C_{\text{MAC}} = \log_2\det\!\left(\mathbf{I}_{N_t} + \frac{P_t}{K\sigma^2} \mathbf{H}^{H}\mathbf{H}\right).$$

By the DPC achievability and the chain rule for mutual information, the BC sum capacity also equals

$$C_{\text{BC}} = \log_2\det\!\left(\mathbf{I}_{N_t} + \frac{1}{\sigma^2} \sum_k \mathbf{S}_k\right)$$

optimised over $\mathbf{S}_k \succeq 0$ with $\sum_k \text{tr}(\mathbf{S}_k) \leq P_t$.

Both expressions have the same form: $\log_2\det(\mathbf{I} + (1/\sigma^2)\mathbf{Q})$ where $\text{tr}(\mathbf{Q}) \leq P_t$. The optimal $\mathbf{Q}$ is the same in both cases (water-filling over the eigenvalues of $\mathbf{H}^{H}\mathbf{H}$), so $C_{\text{MAC}} = C_{\text{BC}}$. $\blacksquare$

For large $N_t$ with $\|\mathbf{h}_k\|^2 \approx N_t$:

$$\mathbf{H}\mathbf{H}^{H} \approx N_t \begin{bmatrix} 1 & r e^{j\phi} \\ r e^{-j\phi} & 1 \end{bmatrix}$$

for some phase $\phi$.

$(\mathbf{H}\mathbf{H}^{H})^{-1} \approx \frac{1}{N_t(1 - r^2)} \begin{bmatrix} 1 & -r e^{j\phi} \\ -r e^{-j\phi} & 1 \end{bmatrix}$$The diagonal entries are$1/(N_t(1-r^2))$, so the effective gain$g_k^{\text{ZF}} \approx N_t(1-r^2)$, which decreases to zero as$r \to 1$(fully correlated channels).$\blacksquare$

Compute $\mathbf{H}\mathbf{H}^{H} = \begin{bmatrix} 6 & 4 \\ 4 & 6 \end{bmatrix}$, $(\mathbf{H}\mathbf{H}^{H})^{-1} = \frac{1}{20}\begin{bmatrix} 6 & -4 \\ -4 & 6 \end{bmatrix}$.

$\mathbf{W}^{\text{ZF}} = \mathbf{H}^{H}(\mathbf{H}\mathbf{H}^{H})^{-1} = \frac{1}{20}\begin{bmatrix} 6 & -4 \\ 4 & -2 \\ -2 & 4 \\ -4 & 6 \end{bmatrix}$.

With $p_1 = p_2 = P_t/2 = 2$ (assuming $P_t = 4$), the per-antenna power of the normalised precoder varies across antennas. The maximum per-antenna power exceeds $P_{\max}/N_t = 1$ for some configurations, requiring the iterative PAPC algorithm or SOCP solution.

The PAPC-constrained precoder redistributes power more uniformly, resulting in a small sum-rate reduction. The exact numerical values depend on the power allocation, but the rate loss is typically 2--5% for this moderate system size. $\blacksquare$

With favorable propagation ($N_t \gg K$), the SINR approximates $\text{SINR}_k \approx (P_t/K) \cdot N_t/\sigma^2 = (1/8) \times 128 = 16$.

Per-user rate: $R_k = \log_2(1 + 16) = 4.09$ bits/s/Hz. Sum rate: $R_{\text{sum}} = 8 \times 4.09 = 32.7$ bits/s/Hz. $\blacksquare$

The Gram matrix $\mathbf{H}\mathbf{H}^{H}$ costs $O(N_tK^{2}) = O(64 \times 256) = O(16384)$. The Cholesky factorisation costs $O(K^{3}) = O(4096)$. The back-multiplication $\mathbf{H}^{H} \times (\mathbf{H}\mathbf{H}^{H})^{-1}$ costs $O(N_tK^{2}) = O(16384)$. Total: $\sim 37000$ complex multiply-adds per subcarrier. $\blacksquare$

The received signal at user $k$ is $y_k = \mathbf{h}_k^H \sum_j \sqrt{p_j} \mathbf{v}_{j} s_j + w_k = \hat{\mathbf{h}}_k^H \sum_j \sqrt{p_j} \mathbf{v}_{j} s_j - \tilde{\mathbf{h}}_k^H \sum_j \sqrt{p_j} \mathbf{v}_{j} s_j + w_k$.

The last two terms form an effective noise with variance $\sigma^2 + \sigma_e^2 \sum_j p_j \|\mathbf{v}_{j}\|^2 \approx \sigma^2 + \sigma_e^2 P_t$.

Replacing $\sigma^2$ with $\sigma^2_{\text{eff}} = \sigma^2 + \sigma_e^2 P_t$ in the optimal $\alpha$ formula:

$$\alpha_{\text{eff}} = \frac{K(\sigma^2 + \sigma_e^2 P_t)}{P_t} = \frac{K\sigma^2}{P_t} + K\sigma_e^2.$$

The estimation error "raises the noise floor" by $\sigma_e^2 P_t$, pushing RZF away from ZF and toward MRT — which makes sense since imperfect CSI degrades interference cancellation. $\blacksquare$