Exercises

$\hat{x}_1 = \mathbf{h}_1^H (\mathbf{h}_1 x_1 + \mathbf{h}_2 x_2 + \mathbf{w}) = \|\mathbf{h}_1\|^2 x_1 + \mathbf{h}_1^H \mathbf{h}_2 x_2 + \mathbf{h}_1^H \mathbf{w}$.

$\mathbb{E}[\hat{x}_1 | \mathbf{H}] = \|\mathbf{h}_1\|^2 \mathbb{E}[x_1] + \mathbf{h}_1^H \mathbf{h}_2 \mathbb{E}[x_2]$. If $x_k$ are zero-mean, then $\mathbb{E}[\hat{x}_1 | \mathbf{H}] = 0$, confirming MRC does not estimate $x_1$ directly but rather $\|\mathbf{h}_1\|^2 x_1$ (after normalization by $\|\mathbf{h}_1\|^2$, the estimate becomes unbiased).

$(\mathbf{G}^{\text{ZF}})^H \mathbf{H} = [(\mathbf{H}^{H}\mathbf{H})^{-1}]^H \mathbf{H}^{H} \mathbf{H} = (\mathbf{H}^{H}\mathbf{H})^{-1} \mathbf{H}^{H} \mathbf{H} = \mathbf{I}_{K}$. (We used the fact that $(\mathbf{H}^{H}\mathbf{H})^{-1}$ is Hermitian since $\mathbf{H}^{H}\mathbf{H}$ is Hermitian positive definite.) $\blacksquare$

$\mathbf{G}^{\text{MMSE}} = \mathbf{H}(\mathbf{H}^{H}\mathbf{H} + \alpha\mathbf{I})^{-1}$. For $\alpha \gg \|\mathbf{H}^{H}\mathbf{H}\|$: $(\mathbf{H}^{H}\mathbf{H} + \alpha\mathbf{I})^{-1} \approx \alpha^{-1}\mathbf{I} - \alpha^{-2}\mathbf{H}^{H}\mathbf{H} + \cdots$

$\mathbf{G}^{\text{MMSE}} \approx \mathbf{H} \cdot \alpha^{-1}\mathbf{I} = (P/\sigma^2)\mathbf{H}$. This is proportional to $\mathbf{H}$ (MRC), confirming the low-SNR limit. $\blacksquare$

$[(\mathbf{H}^{H}\mathbf{H})^{-1}]_{11} = 1/\|\mathbf{h}_1\|^2$. So $\text{SINR}_1^{\text{ZF}} = P/({\sigma^2}/{\|\mathbf{h}_1\|^2}) = P\|\mathbf{h}_1\|^2/\sigma^2$, which is the single-user matched-filter SNR. $\blacksquare$

$(P\mathbf{h}_2\mathbf{h}_2^H + \sigma^2\mathbf{I})^{-1} = \frac{1}{\sigma^2}\mathbf{I} - \frac{P/{\sigma^2}^{2}}{1 + P\|\mathbf{h}_2\|^2/\sigma^2} \mathbf{h}_2\mathbf{h}_2^H$.

$\text{SINR}_1 = \frac{P}{\sigma^2}\|\mathbf{h}_1\|^2 - \frac{P^2|\mathbf{h}_1^H\mathbf{h}_2|^2/{\sigma^2}^{2}}{1 + P\|\mathbf{h}_2\|^2/\sigma^2}$

$= \frac{P\|\mathbf{h}_1\|^2}{\sigma^2} \left(1 - \frac{P|\mathbf{h}_1^H\mathbf{h}_2|^2/(\sigma^2\|\mathbf{h}_1\|^2)}{1 + P\|\mathbf{h}_2\|^2/\sigma^2}\right)$.

The first factor is the single-user SNR; the second factor is a penalty from the interferer, which vanishes when $\mathbf{h}_1 \perp \mathbf{h}_2$ (favorable propagation) or when $\sigma^2/P \to \infty$. $\blacksquare$

$\text{SINR}_1 = P\mathbf{h}_1^H(P\mathbf{h}_2\mathbf{h}_2^H + \sigma^2\mathbf{I})^{-1}\mathbf{h}_1$. After cancelling user 1: $\text{SINR}_2 = P\|\mathbf{h}_2\|^2/\sigma^2$.

$\det(\mathbf{I} + \frac{P}{\sigma^2}\mathbf{H}\mathbf{H}^{H}) = \det(\mathbf{I} + \frac{P}{\sigma^2}(\mathbf{h}_1\mathbf{h}_1^H + \mathbf{h}_2\mathbf{h}_2^H))$.

Apply iteratively: $= (1 + \frac{P\|\mathbf{h}_2\|^2}{\sigma^2})(1 + P\mathbf{h}_1^H(P\mathbf{h}_2\mathbf{h}_2^H + \sigma^2\mathbf{I})^{-1}\mathbf{h}_1)$ $= (1 + \text{SINR}_2)(1 + \text{SINR}_1)$.

$\log_2\det(\cdot) = \log_2(1 + \text{SINR}_1) + \log_2(1 + \text{SINR}_2)$. $\blacksquare$

$R_{\text{sum}}^{\text{ZF}} \approx K \log_2(1 + (N_t - K)\text{SNR}\beta) = 8 \log_2(1 + 56 \times 1 \times 1) = 8 \log_2(57) \approx 46.7$ bits/s/Hz.

$R_{\text{sum}}^{\text{MMSE}} \approx K \log_2(1 + N_t \text{SNR}\beta) = 8 \log_2(1 + 64) = 8 \log_2(65) \approx 48.1$ bits/s/Hz.

The gap is approximately $48.1 - 46.7 = 1.4$ bits/s/Hz, or about 3% — small because $N_t/K = 8$ is a comfortable ratio. At $N_t/K = 2$, the gap would be much larger. $\blacksquare$

$(\mathbf{H}^{H}\mathbf{H} + \alpha\mathbf{I})^{-1} \approx \mathbf{D}^{-1}$ where $\mathbf{D} = \text{diag}(\|\mathbf{h}_1\|^2 + \alpha, \ldots, \|\mathbf{h}_{K}\|^2 + \alpha)$.

$\mathbf{G} \approx \mathbf{H}\mathbf{D}^{-1}$, so $\mathbf{g}_k \approx \frac{\mathbf{h}_k}{\|\mathbf{h}_k\|^2 + \alpha}$. This is a normalized MRC vector — MRC with a per-user SNR-dependent scaling factor. $\blacksquare$

The total MSE decomposes (approximately, assuming the two error terms are uncorrelated for large $N_t$) as $\text{MSE}_L \approx \text{MMSE} + \text{tr}(\mathbf{E}_L \mathbf{R}_y \mathbf{E}_L^H)$ where $\mathbf{E}_L = \hat{\mathbf{A}}_L^{-1} - \mathbf{A}^{-1}$.

$\|\mathbf{E}_L\| \leq \|\mathbf{D}^{-1}\| \cdot \frac{\rho^L}{1-\rho}$ where $\rho = \rho(\mathbf{D}^{-1}\mathbf{E})$. Therefore the excess MSE is $\mathcal{O}(\rho^{2L})$, decaying exponentially with $L$. $\blacksquare$

$[\mathbf{R}_{ry}]_{mm} = \mathbb{E}[\text{sign}(\Re\{y_m\}) \cdot \Re\{y_m\}] + j \mathbb{E}[\text{sign}(\Im\{y_m\}) \cdot \Im\{y_m\}]$. Since $\Re\{y_m\}$ and $\Im\{y_m\}$ each have variance $\frac{1}{2}(PK\beta + \sigma^2)$, using the sign-correlation identity: each term equals $\sqrt{\frac{2}{\pi}} \cdot \sqrt{\frac{1}{2}(PK\beta + \sigma^2)}$.

$\mathbf{Q} = \mathbf{R}_{ry}\mathbf{R}_y^{-1} = \sqrt{\frac{2}{\pi(PK\beta + \sigma^2)}}\mathbf{I}$. $\blacksquare$

For any permutation $\pi$ of $\{1, \ldots, K\}$: $I(\mathbf{x}; \mathbf{y}) = \sum_{k=1}^{K} I(x_{\pi(k)}; \mathbf{y} | x_{\pi(1)}, \ldots, x_{\pi(k-1)})$.

The $k$-th term is the rate achieved by decoding user $\pi(k)$ after perfectly cancelling users $\pi(1), \ldots, \pi(k-1)$, which is $\log_2(1 + \text{SINR}_{\pi(k)}^{\text{SIC}})$.

Since the left side $I(\mathbf{x}; \mathbf{y}) = \log_2\det(\mathbf{I} + \frac{1}{\sigma^2}\mathbf{H}\mathbf{P}\mathbf{H}^{H})$ does not depend on $\pi$, the sum of rates is the same for all orderings. $\blacksquare$

$\frac{1}{N_t}\|\mathbf{h}_k\|^2 = \frac{1}{N_t}\mathbf{g}_k^H\mathbf{R}_k\mathbf{g}_k \to \frac{1}{N_t}\text{tr}(\mathbf{R}_k)$. $\frac{1}{N_t}|\mathbf{h}_k^H\mathbf{h}_j|^2 \to \frac{1}{N_t^{2}}\text{tr}(\mathbf{R}_k\mathbf{R}_j)$ for $k \neq j$.

Dividing numerator and denominator by $N_t^{2}$: $\text{SINR}_k \to \frac{P[\frac{1}{N_t}\text{tr}(\mathbf{R}_k)]^2} {\sum_{j \neq k} P \frac{1}{N_t^{2}}\text{tr}(\mathbf{R}_k\mathbf{R}_j) + \frac{\sigma^2}{N_t} \frac{1}{N_t}\text{tr}(\mathbf{R}_k)}$.

Multiplying through by $N_t^{2}$ gives the stated result. $\blacksquare$

$I(\mathbf{x}; \mathbf{r}) = h(\mathbf{r}) - h(\mathbf{r}|\mathbf{x})$. Using $\mathbf{r} = \mathbf{Q}\mathbf{H}\mathbf{x} + \mathbf{Q}\mathbf{w} + \boldsymbol{\eta}$: $h(\mathbf{r}|\mathbf{x}) = h(\mathbf{Q}\mathbf{w} + \boldsymbol{\eta}|\mathbf{x})$.

Since $\boldsymbol{\eta}$ given $\mathbf{x}$ is not Gaussian, $h(\mathbf{Q}\mathbf{w} + \boldsymbol{\eta}|\mathbf{x}) \leq h(\mathbf{Q}\mathbf{w}) + \text{correction}$. The correction depends on the kurtosis of $\boldsymbol{\eta}$.

The non-Gaussian nature of $\boldsymbol{\eta}$ means the effective noise has lower entropy than a Gaussian with the same covariance, yielding a higher mutual information than the Bussgang-LMMSE bound. The improvement is typically 0.1–0.5 dB for 1-bit quantization. $\blacksquare$

H = (np.random.randn(n_mc, Nt, K) + 1j*np.random.randn(n_mc, Nt, K)) / np.sqrt(2)
x = (np.random.choice([-1,1], (n_mc, K, n_sym))
     + 1j*np.random.choice([-1,1], (n_mc, K, n_sym))) / np.sqrt(2)

For each SNR: compute the combining matrices in batch, detect, slice to QPSK, and count bit errors. For MMSE-SIC: sort users, apply MMSE at each stage, cancel, repeat.

MMSE-SIC should show the lowest BER, followed by MMSE, ZF, and MRC. At high SNR all four converge due to favorable propagation.

$\mathbf{G}^{\text{MMSE}} = \mathbf{H}(\mathbf{H}^{H}\mathbf{H} + 0 \cdot \mathbf{I})^{-1} = \mathbf{H}(\mathbf{H}^{H}\mathbf{H})^{-1} = \mathbf{G}^{\text{ZF}}$.

In the noiseless case, there is no noise to balance against interference, so the MMSE receiver puts all its effort into interference suppression — exactly what ZF does. $\blacksquare$

$\text{SINR}_1 = 2P\|\mathbf{h}_1\|^2/(P\|\mathbf{h}_2\|^2 + \sigma^2) = 2P/(P + \sigma^2)$. After cancelling user 1: $\text{SINR}_2 = P/\sigma^2$. Sum rate $= \log_2(1 + \frac{2P}{P+\sigma^2}) + \log_2(1 + \frac{P}{\sigma^2})$.

With orthogonal channels: $\text{SINR}_2 = P/(2P + \sigma^2)$, then $\text{SINR}_1 = 2P/\sigma^2$. Sum rate $= \log_2(1 + \frac{P}{2P+\sigma^2}) + \log_2(1 + \frac{2P}{\sigma^2})$.

Both sum to $\log_2(\frac{(2P + P + \sigma^2)(P + \sigma^2)}{\sigma^2(P + \sigma^2)}) = \log_2(1 + 2P/\sigma^2) + \log_2(1 + P/\sigma^2)$. (For orthogonal channels, the sum simplifies since there's no cross-interference.) $\blacksquare$

$[\mathbf{D}]_{kk} = \|\mathbf{h}_k\|^2 + \alpha$, $[\mathbf{E}]_{kj} = \mathbf{h}_k^H\mathbf{h}_j$ for $k \neq j$. $[\mathbf{D}^{-1}\mathbf{E}]_{kj} = \frac{\mathbf{h}_k^H\mathbf{h}_j}{\|\mathbf{h}_k\|^2 + \alpha}$.

$\rho \leq \max_k \frac{\sum_{j \neq k} |\mathbf{h}_k^H\mathbf{h}_j|}{\|\mathbf{h}_k\|^2 + \alpha}$. Convergence requires this to be $< 1$, i.e., the diagonal of $\mathbf{A}$ must dominate the off-diagonal — diagonal dominance.

For i.i.d. Rayleigh: numerator $\sim (K-1)\sqrt{N_t}$, denominator $\sim N_t$. Convergence holds when $N_t/K$ is sufficiently large. $\blacksquare$

$C_{\text{sum}} = 16 \log_2(1 + 128 \times 10 / 16) = 16 \log_2(81) \approx 101.2$ bits/s/Hz.

MRC: $16 \log_2(1 + 128 \times 10 / (15 \times 10 + 1)) \approx 16 \log_2(1 + 8.48) \approx 50.4$ bits/s/Hz ($\approx 50\%$). ZF: $16 \log_2(1 + (128-16) \times 10 / 16) = 16 \log_2(71) \approx 98.1$ bits/s/Hz ($\approx 97\%$). MMSE: approximately $99\%$ of sum capacity at this operating point.

ZF and MMSE are near-capacity at $N_t/K = 8$; MRC captures only about half of the sum capacity due to residual interference. $\blacksquare$