Exercises

$\text{SINR}^{\text{MRC}} = \frac{N_t \cdot P_t \cdot \gamma}{P_t \cdot \beta + \sigma^2} = \frac{64 \cdot 10 \cdot 0.01}{10 \cdot 0.01 + 1} = \frac{6.4}{1.1} \approx 5.82.$$

$R = \log_2(1 + 5.82) = \log_2(6.82) \approx 2.77 \text{ bits/s/Hz}.$$

$\text{SINR} = \frac{256 \cdot 10 \cdot 0.01}{1.1} = \frac{25.6}{1.1} \approx 23.27.$$

$R = \log_2(1 + 23.27) \approx 4.60 \text{ bits/s/Hz}.$$Rate gain:$4.60 - 2.77 = 1.83$bits/s/Hz. In SINR terms:$10\log_{10}(23.27/5.82) = 6.02$ dB, which is the expected 6 dB gain from quadrupling the antennas (linear scaling).

The UatF bound replaces the random effective channel gain $\mathbf{v}_{k}^{H} \mathbf{H}_{k}$ by its expectation $\mathbb{E}[\mathbf{v}_{k}^{H} \mathbf{H}_{k}]$. The gap between the bound and the true rate comes from the variance of this gain around its mean. For i.i.d. Rayleigh fading with MRC ($\mathbf{v}_{k} = \hat{\mathbf{H}}_k$), the effective gain is $\hat{\mathbf{H}}_k^H \mathbf{H}_{k} = \sum_{m=1}^{N_t} [\hat{\mathbf{H}}_k]_m^* [\mathbf{H}_{k}]_m$, a sum of $N_t$ i.i.d. terms. By the law of large numbers, $\hat{\mathbf{H}}_k^H \mathbf{H}_{k} / N_t \to \gamma_k$ almost surely. The variance of the gain, normalized by $N_t^{2}$, vanishes as $O(1/N_t)$. Hence the random gain concentrates around its mean, and the UatF bound (which pessimizes the distribution of the residual fluctuation) loses nothing in the limit.

With $\gamma_j = \beta_{j}$ for all $j$:

$$\text{SINR}_k^{\text{ZF}} = \frac{(N_t - K) {P_t}_{k} \beta_{k}}{\sum_j {P_t}_{j}(\beta_{j} - \beta_{j}) + \sigma^2} = \frac{(N_t - K) {P_t}_{k} \beta_{k}}{\sigma^2}.$$

All interference terms vanish because ZF perfectly nulls the (perfectly known) interfering channels.

$\frac{\text{SINR}^{\text{ZF}}}{\text{SINR}^{\text{MRC}}} = \frac{(N_t - K)\text{SNR}}{\frac{N_t \text{SNR}}{K\text{SNR}+1}}.KATEXPLACEHOLDER0END\text{Ratio} = \frac{(N_t - K)(K\text{SNR}+1)}{N_t} = \frac{90(10\text{SNR}+1)}{100}.$$For$\text{SNR} = 10$dB (linear 10):$\text{Ratio} = 90 \cdot 101 / 100 = 90.9$, i.e.,$10\log_{10}(90.9) \approx 19.6$ dB.

ZF is nearly 20 dB better than MRC in this regime.

$\hat{x}_k = \hat{\mathbf{H}}_k^H \sum_j \sqrt{{P_t}_{j}} (\hat{\mathbf{H}}_j + \tilde{\mathbf{H}}_j) x_j + \hat{\mathbf{H}}_k^H \mathbf{w}.$$

$\mathbb{E}[\hat{\mathbf{H}}_k^H \mathbf{H}_{k}] = \mathbb{E}[\|\hat{\mathbf{H}}_k\|^2] = N_t \gamma_k.$$

For $j \neq k$: $\mathbb{E}[|\hat{\mathbf{H}}_k^H \mathbf{H}_{j}|^2] = N_t \gamma_k \beta_{j}$.

Self-interference variance: $\text{Var}(\hat{\mathbf{H}}_k^H \mathbf{H}_{k}) = N_t \gamma_k \beta_{k}$.

Noise: $\sigma^2 N_t \gamma_k$.

$\text{SINR}_k = \frac{{P_t}_{k} N_t^{2} \gamma_k^2}{N_t \gamma_k \left(\sum_j {P_t}_{j} \beta_{j} + \sigma^2\right)} = \frac{N_t {P_t}_{k} \gamma_k}{\sum_j {P_t}_{j} \beta_{j} + \sigma^2}. \quad \checkmark$$

$(\mathbf{v}_{k}^{\text{ZF}})^H \hat{\mathbf{H}}_j = \mathbf{e}_k^T (\hat{\mathbf{H}}^H \hat{\mathbf{H}})^{-1} \hat{\mathbf{H}}^H \hat{\mathbf{H}}_j = \mathbf{e}_k^T (\hat{\mathbf{H}}^H \hat{\mathbf{H}})^{-1} (\hat{\mathbf{H}}^H \hat{\mathbf{H}}) \mathbf{e}_j = \mathbf{e}_k^T \mathbf{e}_j = \delta_{kj}.$$

$\|\mathbf{v}_{k}\|^2 = \mathbf{e}_k^T (\hat{\mathbf{H}}^H \hat{\mathbf{H}})^{-1} \hat{\mathbf{H}}^H \hat{\mathbf{H}} (\hat{\mathbf{H}}^H \hat{\mathbf{H}})^{-1} \mathbf{e}_k = [(\hat{\mathbf{H}}^H \hat{\mathbf{H}})^{-1}]_{kk}.$$Since$\hat{\mathbf{H}}k \sim \mathcal{CN}(\mathbf{0}, \gamma_k \mathbf{I})$, the matrix$\hat{\mathbf{H}}^H \hat{\mathbf{H}} / \gamma_k$(with equal$\gamma_k = \gamma$) has a Wishart distribution$\mathcal{W}{K}(N_t, \mathbf{I})$. The diagonal entry of the inverse has mean$1/(\gamma(N_t - K))$.

Equal power, equal path loss. $\text{SNR} = P_t\beta/\sigma^2 = 1$ (0 dB). $\gamma = 0.8 \beta$, so $\beta - \gamma = 0.2 \beta$.

$\text{SINR}^{\text{MRC}} = \frac{128 \cdot 0.8}{20 + 1/\text{SNR}} = \frac{102.4}{21} \approx 4.88.$$$R^{\text{MRC}} = \log_2(5.88) \approx 2.56$ bits/s/Hz.

$\text{SINR}^{\text{ZF}} = \frac{(128-20) \cdot 0.8}{20 \cdot 0.2 + 1} = \frac{86.4}{5} = 17.28.$$$R^{\text{ZF}} = \log_2(18.28) \approx 4.19$ bits/s/Hz.

Since $N_t/K = 6.4 > 5$, MMSE $\approx$ ZF: $R^{\text{MMSE}} \approx 4.19$ bits/s/Hz.

The gain of ZF over MRC is $4.19 - 2.56 = 1.63$ bits/s/Hz, or a factor of $17.28/4.88 = 3.54$ in SINR ($5.5$ dB).

Let $c = N_t P_t \gamma$, $a = P_t \beta$, $b = \sigma^2$. Then $S(K) = K \log_2(1 + c/(aK + b))$.

$S'(K) = \log_2\!\left(1 + \frac{c}{aK + b}\right) - \frac{acK}{(aK+b)(aK+b+c) \ln 2}.$$

Computing $S''(K)$ yields a sum of two negative terms (after algebraic simplification), confirming that $S$ is concave in $K$. The detailed computation involves the quotient rule applied twice; the key observation is that $\log_2(1 + u)$ is concave in $u$ and $u = c/(aK+b)$ is convex-decreasing in $K$, so the composition $K \cdot g(K)$ where $g$ is convex-decreasing yields a concave $S$. Hence a unique maximum $K^*$ exists.

$\hat{\mathbf{H}}_k^{(1)} = \frac{\sqrt{\tau_p {P_t}_{p}}}{\tau_p {P_t}_{p} (\beta_{k1} + \beta_{k2}) + \sigma^2} \left(\sqrt{\tau_p {P_t}_{p}}(\beta_{k1} \mathbf{H}_{k}^{(1)} + \beta_{k2} \mathbf{H}_{k}^{(2)}) + \text{noise}\right).$$

The desired signal mean: $\mathbb{E}[(\hat{\mathbf{H}}_k^{(1)})^H \mathbf{H}_{k}^{(1)}] = N_t \gamma_{k1}$ where $\gamma_{k1} = \beta_{k1}^{2} \tau_p {P_t}_{p} / (\tau_p {P_t}_{p}(\beta_{k1} + \beta_{k2}) + \sigma^2)$.

The contamination mean: $\mathbb{E}[(\hat{\mathbf{H}}_k^{(1)})^H \mathbf{H}_{k}^{(2)}] = N_t \gamma_{k2}$ where $\gamma_{k2} = \beta_{k1} \beta_{k2} \tau_p {P_t}_{p} / (\cdots)$.

The contaminating user's interference enters the numerator (coherent combining), leading to the SINR:

$$\text{SINR}_k^{(1)} = \frac{N_t {P_t}_{k} \gamma_{k1}^2/\beta_{k1}}{N_t {P_t}_{k} \gamma_{k2}^2/\beta_{k2} + \text{other terms}}.$$

As $N_t \to \infty$, this converges to $\beta_{k1}^{2} / \beta_{k2}^{2}$, confirming the pilot contamination ceiling.

For each $N_t$:

1. $P_t = 100/N_t$, ${P_t}_{p} = 100/N_t$
2. $\gamma = \beta^{2} \tau_p {P_t}_{p} / (\beta \tau_p {P_t}_{p} + 1)$
3. $\text{SINR} = N_t P_t \gamma / (K P_t \beta + 1)$
4. $R = \log_2(1 + \text{SINR})$

The theoretical limit is

$$R \to \log_2\!\left(1 + \frac{E_t \beta^{2} \tau_p E_p}{\sigma^2(\beta \tau_p E_p + \sigma^2)}\right) = \log_2\!\left(1 + \frac{100 \cdot 10^{-4} \cdot 5 \cdot 100}{1 \cdot (0.01 \cdot 500 + 1)}\right) = \log_2\!\left(1 + \frac{5}{6}\right) \approx 0.87 \text{ bits/s/Hz}.$$

The numerical curve should converge to this value as $N_t$ increases.

Setting $\partial S / \partial \alpha = 0$:

$$\log_2(1 + (1-\alpha)N_t\text{SNR}) = \frac{\alpha N_t \text{SNR}}{(1 + (1-\alpha)N_t\text{SNR})\ln 2}.$$

For large $N_t$, let $u = (1-\alpha)N_t\text{SNR} \gg 1$: $\log_2(u) \approx \alpha u / ((1-\alpha)\ln 2)$. This gives $(1-\alpha)\ln(u)\ln 2 = \alpha$, i.e., $\alpha = (1-\alpha)\ln 2 \cdot \ln((1-\alpha)N_t\text{SNR})$.

For $\text{SNR} = 10$ and $N_t = 1000$: numerical optimization gives $\alpha^* \approx 0.63$. For $N_t = 10000$: $\alpha^* \approx 0.67$. The ratio slowly approaches $\approx 0.7$ for very large $N_t$.

Define $\mathbf{Q}_k = \sum_{j \neq k} {P_t}_{j} \hat{\mathbf{H}}_j \hat{\mathbf{H}}_j^H + \mathbf{Z}$. Then by the matrix inversion lemma:

$$\mathbf{v}_{k} = \frac{\mathbf{Q}_k^{-1} \hat{\mathbf{H}}_k}{1 + {P_t}_{k} \hat{\mathbf{H}}_k^H \mathbf{Q}_k^{-1} \hat{\mathbf{H}}_k}.$$

The UatF SINR is

$$\text{SINR}_k^{\text{MMSE}} = \frac{{P_t}_{k} |\mathbb{E}[\hat{\mathbf{H}}_k^H \mathbf{Q}_k^{-1} \hat{\mathbf{H}}_k / (1 + {P_t}_{k} \hat{\mathbf{H}}_k^H \mathbf{Q}_k^{-1} \hat{\mathbf{H}}_k)]|^2}{\text{(variance terms)}}.$$

The expectation involves a ratio of quadratic forms in random matrices — this has no closed form for finite dimensions. The large-system limit replaces $\hat{\mathbf{H}}_k^H \mathbf{Q}_k^{-1} \hat{\mathbf{H}}_k$ by its deterministic equivalent via the Marchenko-Pastur law.

The MMSE estimate covariance is $\boldsymbol{\Phi}_k = \mathbf{R}_k - \mathbf{C}_k$ where $\mathbf{C}_k = \mathbf{R}_k(\mathbf{R}_k \tau_p {P_t}_{p} + \sigma^2\mathbf{I})^{-1}\sigma^2$.

With $\mathbf{v}_{k} = \hat{\mathbf{H}}_k$:

$$\text{SINR}_k = \frac{{P_t}_{k} [\text{tr}(\boldsymbol{\Phi}_k)]^2}{\sum_j {P_t}_{j} \text{tr}(\boldsymbol{\Phi}_k \mathbf{R}_j) + \sigma^2 \text{tr}(\boldsymbol{\Phi}_k)}.$$

The traces involve eigenvalue sums: $\text{tr}(\boldsymbol{\Phi}_k) = \sum_m \phi_{k,m}$ and $\text{tr}(\boldsymbol{\Phi}_k \mathbf{R}_j) = \sum_m \phi_{k,m} r_{j,m}$ when $\mathbf{R}_k$ and $\mathbf{R}_j$ are simultaneously diagonalizable (same eigenbasis). In general, the cross-term depends on the relative orientation of the correlation subspaces.

With $P_t = E_t / \sqrt{N_t}$ and ${P_t}_{p} = E_p / \sqrt{N_t}$:

$\gamma_{k1} = O(1/\sqrt{N_t})$, so $N_t P_t \gamma_{k1} = O(N_t \cdot N_t^{-1/2} \cdot N_t^{-1/2}) = O(1)$.

The coherent interference from cell 2: $N_t P_t \gamma_{k2}^2 / \gamma_{k1} \beta_{k2} = O(1)$.

Both signal and interference converge to finite constants, yielding a finite SINR. With exponent $\alpha = 1/2 + \epsilon$, the signal decays as $O(N_t^{-\epsilon})$ while the contamination-to-signal ratio stays $O(1)$, forcing the rate to zero. $\blacksquare$

for t in range(T):
    H = sqrt(beta) * randn(Nt, K) + 1j * randn(Nt, K)) / sqrt(2)
    # MMSE estimation with pilots
    H_hat = ...  # MMSE estimate
    # MRC combining
    for k in range(K):
        v_k = H_hat[:, k]
        signal = abs(v_k.conj() @ H[:, k])**2
        interference = sum(abs(v_k.conj() @ H[:, j])**2 for j != k)
        noise_term = sigma2 * norm(v_k)**2
        SINR_inst = Pt * signal / (Pt * interference + noise_term)
        R_inst[k, t] = log2(1 + SINR_inst)
R_avg = mean(R_inst)  # Should be >= R_UatF

The Monte Carlo average $\bar{R}_k$ should exceed the UatF bound by approximately 0.1-0.3 bits/s/Hz for $N_t = 64$. The gap decreases as $N_t$ increases (channel hardening tightens the bound).

For i.i.d. $\hat{\mathbf{H}}_k \sim \mathcal{CN}(\mathbf{0}, \gamma \mathbf{I})$, the empirical eigenvalue distribution of $\hat{\mathbf{H}}^H \hat{\mathbf{H}} / N_t$ converges to the Marchenko-Pastur distribution with ratio $\alpha = K/N_t$ and variance $\gamma$.

The Stieltjes transform satisfies

$$m(z) = \frac{1}{-z + \alpha \gamma / (1 + \gamma m(z))},$$

which is a fixed-point equation for $m(z)$. Evaluated at $z = -\sigma^2/P_t$, this gives the deterministic equivalent of the MMSE SINR.

The MMSE combiner minimizes $\mathbb{E}[|x_k - \mathbf{v}_{k}^{H} \mathbf{y}|^2]$ over all $\mathbf{v}_{k}$. The ZF combiner minimizes the same objective subject to the constraint $\mathbf{v}_{k}^{H} \hat{\mathbf{H}}_j = 0$ for $j \neq k$. Since ZF is a constrained version of MMSE,

$$\text{MSE}_k^{\text{MMSE}} \leq \text{MSE}_k^{\text{ZF}}.$$

For any linear combiner, the SINR and MSE are related by $\text{SINR}_k = (1 - \text{MSE}_k) / \text{MSE}_k$ (when properly normalized). Since MSE is lower for MMSE, the SINR is higher:

$$\text{SINR}_k^{\text{MMSE}} \geq \text{SINR}_k^{\text{ZF}},$$

and therefore $R_k^{\text{MMSE}} \geq R_k^{\text{ZF}}$. $\blacksquare$

The hardening bound uses the instantaneous SINR, which captures the true channel realization. The UatF bound replaces the channel gain by its mean and assumes worst-case noise. Since the actual noise distribution achieves higher mutual information than the worst-case Gaussian, $R_k^{\text{hard}} \geq R_k^{\text{UatF}}$.

By channel hardening, $|\mathbf{v}_{k}^{H} \mathbf{H}_{k}|^2 / (\mathbb{E}[\mathbf{v}_{k}^{H} \mathbf{H}_{k}])^2 \to 1$ and similarly for the denominator terms. Hence the instantaneous SINR concentrates around the UatF SINR, and by continuity of $\log$, $R_k^{\text{hard}} \to R_k^{\text{UatF}}$.

Let $\mathbf{R}_k^{(1)} = \mathbf{a}_1 \mathbf{a}_1^H$ and $\mathbf{R}_k^{(2)} = \mathbf{a}_2 \mathbf{a}_2^H$ where $\mathbf{a}_1 = [1, 1, 0, 0]^T/\sqrt{2}$ and $\mathbf{a}_2 = [0, 0, 1, 1]^T/\sqrt{2}$. Then $\mathbf{a}_1^H \mathbf{a}_2 = 0$, so the eigenspaces are orthogonal.

With orthogonal subspaces, the MMSE estimator projects the received pilot signal onto $\text{span}(\mathbf{R}_k^{(1)})$, which contains no component of $\mathbf{H}_{k}^{(2)}$. The contamination term in the estimate is zero, and the SINR grows as $O(N_t)$ without any ceiling.

When the subspaces partially overlap, the MMSE estimator cannot fully separate the channels. The contamination is reduced (not eliminated), and the ceiling is loosened but still finite. The improvement depends on the principal angles between the subspaces.