Exercises

The multiplexing gain is $d = \min(N_t, K) = \min(100, 5) = 5$. The system is user-limited.

With $\sigma^2 = 1/N_t = 0.01$, per-user SNR: $\text{SNR}_k = P\beta_k/\sigma^2 = 1 \cdot 1 / 0.01 = 100$. Asymptotic sum rate: $$C \approx 5\log_2(1 + N_t \cdot \text{SNR}_k) = 5\log_2(1 + 10000) \approx 5 \times 13.3 = 66.5 \text{ bps/Hz}.$$

With $N_t = 200$, $\sigma^2 = 1/200 = 0.005$, SNR $= 200$: $C \approx 5\log_2(1 + 200 \times 200) = 5\log_2(40001) \approx 5 \times 15.3 = 76.5$ bps/Hz. Gain: $76.5 - 66.5 = 10$ bps/Hz $= 5\log_2(2) \times 2 = 10$. Consistent with $\Delta C = K\log_2(\text{doubling factor}) = 5\log_2(2) = 5$ — wait, doubling $N_t$ increases SNR by $2^2 = 4\times$ (since $\sigma^2 = 1/N_t$ also changes), adding $5\log_2(4) = 10$ bps/Hz. $\blacksquare$

For $h_i \sim \mathcal{CN}(0, \beta)$, we have $|h_i|^2 \sim \beta\chi^2_2/2$ (an exponential with mean $\beta$). Hence $\mathbb{E}[|h_i|^2] = \beta$ and $\text{Var}[|h_i|^2] = \mathbb{E}[|h_i|^4] - \beta^2 = 2\beta^2 - \beta^2 = \beta^2$ (using $\mathbb{E}[|h|^4] = 2\beta^2$ for $h \sim \mathcal{CN}(0,\beta)$).

$\mathbb{E}[\|\mathbf{h}\|^2] = \sum_{i=1}^{N_t}\mathbb{E}[|h_i|^2] = N_t\beta$. $\text{Var}[\|\mathbf{h}\|^2] = \sum_{i=1}^{N_t}\text{Var}[|h_i|^2] = N_t\beta^2$. Dividing by $N_t$: $\mathbb{E}[\|\mathbf{h}\|^2/N_t] = \beta$, $\text{Var}[\|\mathbf{h}\|^2/N_t] = \beta^2/N_t$.

$\text{CV} = \sigma / \mu = \sqrt{\text{Var}[\|\mathbf{h}\|^2]} / \mathbb{E}[\|\mathbf{h}\|^2] = \sqrt{N_t\beta^2} / (N_t\beta) = \sqrt{N_t}\beta/(N_t\beta) = 1/\sqrt{N_t}$. Thus the relative fluctuation decreases as $1/\sqrt{N_t}$. At $N_t = 100$, the standard deviation is 10% of the mean; at $N_t = 10000$, it is 1%. $\blacksquare$

$|\mathbf{h}_k^H\mathbf{h}_j|^2 = \sum_{i=1}^{N_t}\sum_{\ell=1}^{N_t} h_{k,i}^* h_{j,i} h_{j,\ell}^* h_{k,\ell}$. Taking expectation and using independence of $\mathbf{h}_k$ and $\mathbf{h}_j$, then independence of different components: $\mathbb{E}[h_{k,i}^* h_{j,i} h_{j,\ell}^* h_{k,\ell}] = \mathbb{E}[h_{k,i}^* h_{k,\ell}]\mathbb{E}[h_{j,i} h_{j,\ell}^*]$.

For $i \neq \ell$: $\mathbb{E}[h_{k,i}^*h_{k,\ell}] = 0$ (independent, zero mean). For $i = \ell$: $\mathbb{E}[|h_{k,i}|^2] = \beta_k$ and $\mathbb{E}[|h_{j,i}|^2] = \beta_j$. Therefore: $\mathbb{E}[|\mathbf{h}_k^H\mathbf{h}_j|^2] = N_t\beta_k\beta_j$. Dividing by $N_t^2$: $\mathbb{E}[|\mathbf{h}_k^H\mathbf{h}_j/N_t|^2] = \beta_k\beta_j/N_t \to 0$. $\blacksquare$

Let $\mathbf{v}_{k}$ satisfy $\mathbf{H}^{H}\mathbf{v}_{k} = \mathbf{e}_k$, i.e., $\mathbf{H}^H\mathbf{v}_k = \mathbf{e}_k$. We want to minimize $\|\mathbf{v}_k\|^2$ subject to this linear equality.

The minimum-norm solution to $\mathbf{A}\mathbf{x} = \mathbf{b}$ (with full row rank $\mathbf{A}$) is $\mathbf{x} = \mathbf{A}^H(\mathbf{A}\mathbf{A}^H)^{-1}\mathbf{b}$. Here $\mathbf{A} = \mathbf{H}^{H}$ and $\mathbf{b} = \mathbf{e}_k$: $\mathbf{v}_{k} = \mathbf{H}(\mathbf{H}^{H}\mathbf{H})^{-1}\mathbf{e}_k$.

Stacking all $K$ combining vectors as columns: $\mathbf{W}_{\text{ZF}} = [\mathbf{v}_{1}, \ldots, \mathbf{v}_{K}] = \mathbf{H}(\mathbf{H}^{H}\mathbf{H})^{-1}[\mathbf{e}_1, \ldots, \mathbf{e}_K] = \mathbf{H}(\mathbf{H}^{H}\mathbf{H})^{-1}.$ $\blacksquare$

$\beta_k = d_k^{-3}$: $\beta_1 = 50^{-3} = 8.0 \times 10^{-6}$, $\beta_2 = 60^{-3} = 4.6 \times 10^{-6}$, $\beta_3 = 70^{-3} = 2.9 \times 10^{-6}$, $\beta_4 = 80^{-3} = 2.0 \times 10^{-6}$, $\beta_5 = 90^{-3} = 1.4 \times 10^{-6}$, $\beta_6 = 100^{-3} = 1.0 \times 10^{-6}$, $\beta_7 = 120^{-3} = 0.58 \times 10^{-6}$, $\beta_8 = 150^{-3} = 0.30 \times 10^{-6}$. Mean: $\bar{\beta} = 2.6 \times 10^{-6}$.

$\tau_p = K = 8$. Fraction: $\eta = 8/200 = 4\%$. Negligible.

Using $P/\sigma^2 = P_t/N_0 B$ where, with typical values, the per-antenna SNR is $N_t P \beta_k / \sigma^2$. Heterogeneous: sum of 8 individual log terms. Homogeneous: $8\log_2(1 + N_t P \bar{\beta}/\sigma^2)$. Due to Jensen's inequality ($\log_2$ is concave), the heterogeneous sum rate is lower than the homogeneous case when the geometric mean $<$ arithmetic mean, which is always true unless all $\beta_k$ are equal. Power control (Chapter 5) addresses this by boosting weaker users. $\blacksquare$

Using block inversion, the $(k,k)$ entry of $(\mathbf{H}^H\mathbf{H})^{-1}$ is $[(\mathbf{H}^H\mathbf{H})^{-1}]_{kk} = 1/\|\mathbf{P}_{-k}^\perp \mathbf{h}_k\|^2$, where $\mathbf{P}_{-k}^\perp = \mathbf{I} - \mathbf{H}_{-k}(\mathbf{H}_{-k}^H\mathbf{H}_{-k})^{-1}\mathbf{H}_{-k}^H$ is the projection onto the null space of $\mathbf{H}_{-k}$ (all columns of $\mathbf{H}$ except $\mathbf{h}_k$).

Since $\mathbf{P}_{-k}^\perp$ is an orthogonal projection (idempotent Hermitian), $\|\mathbf{P}_{-k}^\perp \mathbf{h}_k\|^2 \leq \|\mathbf{h}_k\|^2$. Therefore $[(\mathbf{H}^H\mathbf{H})^{-1}]_{kk} = 1/\|\mathbf{P}_{-k}^\perp\mathbf{h}_k\|^2 \geq 1/\|\mathbf{h}_k\|^2$.

$\text{SINR}_k^{\text{ZF}} = \frac{P_k}{\sigma^2[(\mathbf{H}^H\mathbf{H})^{-1}]_{kk}} \leq \frac{P_k}{\sigma^2/\|\mathbf{h}_k\|^2} = \text{SNR}_k^{\text{MF}}.$$Oops — we get the reverse inequality. ZF SINR$\leq$MF SNR because ZF pays a noise penalty for interference cancellation. The bound shows the noise enhancement factor.$\blacksquare$

$\text{tr}(\mathbf{R}_1) = N_t\beta_1$, $\text{tr}(\mathbf{R}_1^2) = N_t\beta_1^2$. $\zeta_1 = N_t\beta_1^2/(N_t\beta_1)^2 = 1/N_t = 0.01$.

The 5 nonzero eigenvalues are $\lambda = \beta_2 N_t/5$ each. $\text{tr}(\mathbf{R}_2) = 5\lambda = N_t\beta_2$. $\text{tr}(\mathbf{R}_2^2) = 5\lambda^2 = 5(\beta_2 N_t/5)^2 = \beta_2^2 N_t^2/5$. $\zeta_2 = (\beta_2^2 N_t^2/5)/(N_t\beta_2)^2 = 1/5$.

User 1 has $\zeta_1 = 1/100 = 0.01$ (strong hardening). User 2 has $\zeta_2 = 1/5 = 0.2$ (weaker hardening — only 5 effective antennas). The rank-5 user needs 20× more antennas to achieve the same hardening level. This motivates using spatially diverse antennas (uniform linear arrays in rich scattering) over co-located antennas. $\blacksquare$

With $\mathbf{v}_1 = \mathbf{h}_1$: $$\text{SINR}_1^{\text{MRC}} = \frac{P_1\|\mathbf{h}_1\|^4} {P_2|\mathbf{h}_1^H\mathbf{h}_2|^2 + \sigma^2\|\mathbf{h}_1\|^2}.$$

Replace random quantities with their expectations: $\|\mathbf{h}_1\|^2 \approx N_t\beta_1$, so $\|\mathbf{h}_1\|^4 \approx N_t^2\beta_1^2$. For the interference: $\mathbb{E}[|\mathbf{h}_1^H\mathbf{h}_2|^2] = N_t\beta_1\beta_2$ (from Exercise Eex-mimo-ch01-03). Therefore: $$\text{SINR}_1^{\text{MRC}} \approx \frac{P_1 N_t^2\beta_1^2}{P_2 N_t\beta_1\beta_2 + \sigma^2 N_t\beta_1} = \frac{N_t P_1\beta_1}{P_2\beta_2 + \sigma^2}.$$

As $N_t \to \infty$ with fixed $\sigma^2$: $\text{SINR}_1^{\text{MRC}} \to N_t P_1\beta_1 / (P_2\beta_2)$ — limited by residual interference from user 2, not noise. But since this diverges with $N_t$, there is no floor! At finite $N_t$, the interference $P_2|\mathbf{h}_1^H\mathbf{h}_2|^2$ has variance (not just mean) — the actual realized SINR fluctuates. The formula above is valid as a deterministic approximation. $\blacksquare$

$\tau_p^{\text{min}} = K = 10$. Fraction: $10/200 = 5\%$.

$f_D = v f_c / c = (30/3.6) \times 2 \times 10^9 / (3 \times 10^8) = 55.6$ Hz. $T_c \approx 1/(2f_D) = 9$ ms. $\tau_c = B_c T_c \times \text{symbol rate}$. With $15$ kHz subcarrier spacing and OFDM: symbol duration $\approx 71.4\ \mu$s. Symbols per $9$ ms: $\approx 126$ symbols. So $\tau_c \approx 126$ in this OFDM system.

$K \leq 0.05 \times 126 = 6.3$, so $K \leq 6$ users with at most 5% overhead. (With $\tau_c = 200$: $K \leq 10$ users.) $\blacksquare$

From Theorem TMRC Asymptotic Near-Optimality (with general path-losses): $$\text{SINR}_k^{\text{MRC}} \approx \frac{N_t P\beta_k^2}{P\beta_k\sum_{j\neq k}\beta_j + \sigma^2\beta_k} = \frac{N_t P\beta_k}{P\sum_{j\neq k}\beta_j + \sigma^2}.$$

For ZF with i.i.d. $\mathcal{CN}(0, \beta_k\mathbf{I})$, the matrix $\mathbf{H}^H\mathbf{H}/N_t \to \mathbf{B} = \text{diag}(\beta_1,\ldots,\beta_K)$ (favorable propagation). By continuous mapping: $[(\mathbf{H}^H\mathbf{H})^{-1}]_{kk} \approx 1/(N_t\beta_k)$. Therefore: $$\text{SINR}_k^{\text{ZF}} \approx \frac{P_k}{\sigma^2/(N_t\beta_k)} = \frac{N_t P_k\beta_k}{\sigma^2}.$$

Ratio: $\text{SINR}_k^{\text{ZF}} / \text{SINR}_k^{\text{MRC}} = (N_tP\beta_k/\sigma^2) / (N_tP\beta_k/(P\sum_{j\neq k}\beta_j + \sigma^2)) = 1 + P\sum_{j\neq k}\beta_j/\sigma^2$. Wait — this ratio is actually $\geq 1$, so ZF is always better than or equal to MRC at finite $N_t$! The claim that they converge should be interpreted in the high-$N_t$ regime where $\sigma^2 \to 0$. In that case: $\text{SINR}_k^{\text{MRC}} \approx N_t P\beta_k/(P\sum_{j\neq k}\beta_j)$ (interference limited) and $\text{SINR}_k^{\text{ZF}} \approx N_t P\beta_k/\sigma^2$ (noise limited). The ZF advantage vanishes only as $\sigma^2/P \to 0$ compared to interference. $\blacksquare$

import numpy as np
N_t, K, n_mc = 4, 2, 1000
diag_vals, offdiag_vals = [], []
for _ in range(n_mc):
    H = (np.random.randn(N_t, K) + 1j*np.random.randn(N_t, K)) / np.sqrt(2)
    G = H.conj().T @ H / N_t
    diag_vals.append(np.real(np.diag(G)))
    offdiag_vals.append(G[0, 1])

For $N_t = 4$: diagonal mean $\approx 1$, variance $\approx 0.25 = 1/4$. Off-diagonal mean $\approx 0$, mean-squared $\approx 0.25 = 1/4$. For $N_t = 100$: all variances $\approx 0.01 = 1/100$. The numerical results should match the theoretical predictions from Theorems TChannel Hardening for i.i.d. Rayleigh Fading and TFavorable Propagation for i.i.d. Rayleigh Fading. $\blacksquare$

Cell 1 BS receives during uplink training: $\mathbf{Y}_p = \mathbf{h}_k^{(1)}\boldsymbol{\phi}^T + \mathbf{h}_j^{(2)}\boldsymbol{\phi}^T + \mathbf{N}$ (both users transmit the same pilot $\boldsymbol{\phi}$). LS estimate: $\hat{\mathbf{h}}_k = \mathbf{h}_k^{(1)} + \mathbf{h}_j^{(2)} + \tilde{\mathbf{n}}$ where $\tilde{\mathbf{n}} = \mathbf{N}\boldsymbol{\phi}^*/\|\boldsymbol{\phi}\|^2 \to \mathbf{0}$ as pilot power $\to \infty$.

Even with infinite SNR, $\hat{\mathbf{h}}_k \to \mathbf{h}_k^{(1)} + \mathbf{h}_j^{(2)}$. The contamination term $\mathbf{h}_j^{(2)} \in \mathbb{C}^{N_t}$ is a random vector of length $N_t$ — it does not "average out" when $N_t$ grows. Increasing $N_t$ makes each component of $\mathbf{h}_j^{(2)}$ small relative to $N_t$, but the effect on beamforming is that the BS accidentally points a beam toward the contaminating user in the adjacent cell. This residual interference grows with $N_t$ (Marzetta, 2010). $\blacksquare$

(a) $N_t = 1$: $\zeta = 1$ (single antenna; no averaging; the channel is just a scalar random variable — no hardening).

(b) $N_t = 100$, i.i.d.: $\zeta = 1/N_t = 0.01$ — strong hardening. The gain concentrates within 10% of its mean.

(c) $N_t = 100$, rank-1: $\zeta = 1$ regardless of $N_t$ — no hardening. The array gain $\|\mathbf{h}\|^2 = N_t\beta|\tilde{h}|^2$ where $\tilde{h}$ is a single Rayleigh fading coefficient. Even with 100 antennas, the effective gain fluctuates like a single-antenna channel.

(d) $N_t = 100$, rank-10: $\zeta = \text{tr}(\mathbf{R}^2)/\text{tr}(\mathbf{R})^2$. With 10 equal eigenvalues $\lambda = \beta N_t/10$ each: $\zeta = 10\lambda^2/(10\lambda)^2 = 1/10 = 0.1$ — partial hardening. The system hardens like it has 10 effective antennas. $\blacksquare$

Single-user MRC SINR $\approx P N_t\beta/\sigma^2$. With $P = E_u/N_t$: SINR $= (E_u/N_t) \cdot N_t\beta/\sigma^2 = E_u\beta/\sigma^2$ — constant in $N_t$. So the achievable rate $R = \log_2(1 + E_u\beta/\sigma^2)$ is fixed while transmit power $P = E_u/N_t$ decreases.

Energy efficiency $= R/P = \log_2(1 + E_u\beta/\sigma^2) \cdot N_t/E_u$, which grows linearly with $N_t$. Adding more BS antennas allows the user to transmit with proportionally less power (maintaining the same rate), or equivalently, extending battery life by a factor of $N_t$.

In 5G NR, this is exploited via uplink power control: cell-edge users transmit at lower power when served by a massive MIMO BS (fewer retransmissions, longer battery life). $\blacksquare$

$C_{\text{sum}} = \log_2\det\!\left(\mathbf{I}_{N_t} + \frac{P}{\sigma^2}\mathbf{H}\mathbf{H}^H\right) = \log_2\det\!\left(\mathbf{I}_K + \frac{P}{\sigma^2}\mathbf{H}^H\mathbf{H}\right).$ Write $\frac{P}{\sigma^2}\mathbf{H}^H\mathbf{H} = \frac{PN_t}{\sigma^2} \cdot \frac{\mathbf{H}^H\mathbf{H}}{N_t}$: $$C_{\text{sum}} = \log_2\det\!\left(\mathbf{I}_K + \frac{PN_t}{\sigma^2} \cdot \frac{\mathbf{H}^H\mathbf{H}}{N_t}\right).$$

As $N_t \to \infty$ with fixed $K$, favorable propagation gives $\mathbf{H}^H\mathbf{H}/N_t \to \mathbf{I}_K$ a.s. By continuous mapping: $$C_{\text{sum}} \to K\log_2\!\left(1 + \frac{PN_t}{\sigma^2}\right) \approx K\log_2(N_t) + K\log_2(P/\sigma^2).$$ This is the same asymptote as derived in Theorem $tn{ntx}$" data-ref-type="theorem">TSum Rate Linear Scaling with $tn{ntx}$. The random matrix (Marchenko–Pastur) approach gives the full finite-$N_t$ correction terms, developed in Chapters 2 and 4. $\blacksquare$