Exercises

$P(10) = 100\cdot 10^{-15}\cdot 2^{10}\cdot 2\cdot 10^9 = 100 \cdot 10^{-15} \cdot 1024 \cdot 2 \cdot 10^9 \approx 205$ mW. $P(1) = 100 \cdot 10^{-15}\cdot 2 \cdot 2\cdot 10^9 = 0.4$ mW.

Ratio $P(10)/P(1) = 2^9 = 512$. Per-chain savings are more than 500x, but real savings are dominated by the fixed RF power. $\blacksquare$

$B_g = \mathbb{E}[Y\operatorname{sign}(Y)]/\sigma_Y^2 = \mathbb{E}[\operatorname{sign}'(Y)]$.

$\mathbb{E}[\operatorname{sign}'(Y)] = \int 2\delta(y) f_Y(y) dy = 2 f_Y(0) = 2\cdot \tfrac{1}{\sqrt{2\pi\sigma_Y^2}} = \sqrt{2/(\pi\sigma_Y^2)}.$

Unit variance $\sigma_Y^2 = 1$ gives $B_1 = \sqrt{2/\pi}\approx 0.798$. The Bussgang gain applied to an amplitude-1 Gaussian signal reduces the effective signal to $0.798$, matching the $2/\pi$ low-SNR capacity loss since $|B_1|^2 = 2/\pi = \kappa_1$. $\blacksquare$

$\arcsin(0.9) \approx 1.1198$ rad.

$(2/\pi)\cdot 1.1198 \approx 0.713$. The nonlinearity shrinks the correlation from $0.9$ to $0.713$ — quantization decorrelates. $\blacksquare$

$G(0.2) = 0.637 + 0.363\cdot 0.2 = 0.710$. $G(0) = 0.637$, $G(1) = 1.0$.

$R(0) = \log_2(1 + 6.37) = \log_2(7.37) \approx 2.88$ bit/use. $R(0.2) = \log_2(1 + 7.10) = \log_2(8.10) \approx 3.02$ bit/use. $R(1) = \log_2(11) \approx 3.46$ bit/use.

A 20% high-precision fraction buys $0.14$ bits/use of the $0.58$ bits/use gap between 1-bit and full-precision — about 24% of the maximum improvement for only 20% of the full-resolution power. The law of diminishing returns is mild here. $\blacksquare$

$\text{SNR} = 10^{-0.3}\approx 0.501$. $N_r\,\text{SNR} \approx 32.08$.

$\text{SINR} = (0.637\cdot 32.08) / (1 + 0.363\cdot 32.08) = 20.44/(1 + 11.64) = 20.44/12.64 \approx 1.617$.

$R = \log_2(2.617) \approx 1.39$ bit/use. The ideal rate at the same array SNR would be $\log_2(33.08)\approx 5.05$ bit/use; the 1-bit receiver keeps about 28%. $\blacksquare$

$p - 1/2 = -\sqrt{\text{SNR}/(2\pi)} + (\text{SNR})^{3/2}/(6\sqrt{2\pi}) + O((\text{SNR})^{5/2})$. Square: $(p-1/2)^2 = \text{SNR}/(2\pi) - (\text{SNR})^{2}/(6\pi) + O((\text{SNR})^3)$.

$H_2(p) = 1 - (2/\ln 2)(p-1/2)^2 + O((p-1/2)^4) = 1 - \text{SNR}/(\pi \ln 2) + (\text{SNR})^2/(3\pi \ln 2) + O((\text{SNR})^3)$.

$C_1 = 1 - H_2(p) = \text{SNR}/(\pi \ln 2) - (\text{SNR})^2/(3\pi \ln 2) + O((\text{SNR})^3)$. $C_\infty = \text{SNR}/(2\ln 2) - (\text{SNR})^2/(4\ln 2) + O((\text{SNR})^3)$. Their ratio at leading order: $(1/\pi)/(1/2) = 2/\pi$. The first correction is $1 - \tfrac{\text{SNR}}{3} + \tfrac{\text{SNR}}{2} = 1 + \tfrac{\text{SNR}}{6}$, so $C_1/C_\infty = (2/\pi)(1 + \text{SNR}/6 + O(\text{SNR}^{2}))$, a slight improvement over the low-SNR limit before the 1-bit cap starts to bite. $\blacksquare$

The correlation matrix is $\mathbf{I}$, so off-diagonal entries of $\boldsymbol{\Sigma}_{y_q}$ are $(2/\pi)\arcsin(0) = 0$. The diagonal entries are the variance of $\operatorname{sign}(y_n) + j\operatorname{sign}(y_n')$, which is $\mathbb{E}[\operatorname{sign}(y_n)^2] + \mathbb{E}[\operatorname{sign}(y_n')^2] = 2$. So $\boldsymbol{\Sigma}_{y_q} = 2\mathbf{I}$.

$\mathbf{B} = (2/\pi)^{1/2}/\sigma \cdot \mathbf{I}$ (diagonal, with one complex entry per antenna), $\mathbf{B}\sigma^2\mathbf{I}\mathbf{B}^H = (2/\pi)\mathbf{I}$.

$\boldsymbol{\Sigma}_{d} = 2\mathbf{I} - (2/\pi)\cdot 2 \mathbf{I} = 2(1 - 2/\pi)\mathbf{I}$ for the complex receiver (factor 2 from I and Q rails). After per-rail normalization, each rail sees variance $(1 - 2/\pi)$. The scaling $\sigma^2$ drops out because the 1-bit output amplitude is $\pm 1$. $\blacksquare$

$\text{SINR}_{\infty} = 128 \cdot 0.01 = 1.28$. $R_\infty = \log_2(1 + 1.28) \approx 1.19$ bit/use.

$\text{SINR}_1 = 0.637\cdot 1.28/(1 + 0.363\cdot 1.28) = 0.815/1.465 \approx 0.556$. $R_1 = \log_2(1.556) \approx 0.638$ bit/use.

$R_1/R_\infty = 0.638/1.19 \approx 0.536$. Near the low-SNR regime but a bit below because $N_r\,\text{SNR} = 1.28$ is not quite small. The ratio tends to $2/\pi \approx 0.637$ as per-antenna SNR $\to 0$. $\blacksquare$

As the water level $\mu \to 0$, the stationarity condition $b_n^\star = -\tfrac{1}{2}\log_2(\mu/\text{const}) \to \infty$, clipped at $b_{\max}$. Every antenna saturates.

Total rate loss at saturation: $\Delta R \approx \sum_n c_1\,4^{-b_{\max}}\,g_n\,\text{SNR}/ ((1 + g_n\,\text{SNR})\ln 2)$, which scales as $4^{-b_{\max}}$ — the classical 6 dB/bit diminishing-return.

Even with unlimited power, the per-antenna quantization loss does not vanish — it is bounded by the maximum hardware resolution, not the budget. $\blacksquare$

Per-antenna gain is $\kappa_{b_n}$. Summing: $\sum_n \kappa_{b_n} = \alphaN_r\cdot 1 + (1-\alpha)N_r\cdot\kappa_1 = N_r(\kappa_1 + (1-\kappa_1)\alpha) = N_r\,G(\alpha).$

$\kappa_b$ is strictly concave in $2^b$ (distortion decreases geometrically but rate is logarithmic). Replacing pairs of $(1, \infty)$ by $(2, 8)$ at the same total budget strictly increases the sum gain by Jensen's inequality.

Mixed-ADC with only two levels is suboptimal — the continuous bit-allocation always weakly dominates. It is, however, hardware-friendly. $\blacksquare$

$g(y) = \operatorname{sign}(y - \delta)$. Stein: $B = \mathbb{E}[g'(Y)] = 2 f_Y(\delta)$ $= 2 e^{-\delta^2/(2\sigma_Y^2)}/\sqrt{2\pi\sigma_Y^2} = \sqrt{2/(\pi\sigma_Y^2)} e^{-\delta^2/(2\sigma_Y^2)}.$

For $\delta/\sigma_Y = 0.3$: $e^{-0.045} \approx 0.956$, so the Bussgang gain shrinks by ~4.4% and the effective SINR by 8.8% ($|B|^2$). In decibels, $\approx 0.4$ dB of penalty — a reminder that DC calibration matters but is forgiving at modest offsets.

At $\delta/\sigma_Y = 1$ the penalty is $e^{-0.5} \approx 0.61$ squared = 0.37, a 4.3 dB hit. So bias drift comparable to signal amplitude destroys most of the receiver's performance — supporting the pitfall of Section 19.1. $\blacksquare$

$P_{\text{ADC}}(1) = 100\cdot 10^{-15}\cdot 2\cdot 5\cdot 10^8 = 100$ µW. $P_{\text{ADC}}(4) = 100\cdot 10^{-15}\cdot 16\cdot 5\cdot 10^8 = 800$ µW.

$R(1) = \log_2(1 + 0.637\cdot 128) = \log_2(82.5)\approx 6.37$ bit/use. $R(4) = \log_2(1 + 0.990\cdot 128) = \log_2(127.8)\approx 6.998$ bit/use. $R(1)\cdot 400\cdot 10^6 \approx 2.55$ Gbps. $R(4)\cdot 400\cdot 10^6 \approx 2.80$ Gbps.

$P(1) = 128(20 + 0.1)\text{mW} \approx 2.573$ W. $P(4) = 128(20 + 0.8)\text{mW} \approx 2.662$ W. Ratio $\approx 1.035$.

$\eta(1) \approx 2.55\cdot 10^9 / 2.573 \approx 0.991$ Gbits/J. $\eta(4) \approx 2.80\cdot 10^9 / 2.662 \approx 1.052$ Gbits/J. 4-bit wins by 6%. The RF cost dominates and the 4-bit rate gain is essentially free. At this SNR and RF cost, 1-bit is not the right choice — the lesson of Example EBreak-Even Between 1-Bit and 4-Bit. $\blacksquare$

Signal power: $\kappa_1 \|\mathbf{H}_{1}\|^2 P \approx \kappa_1 N_r P$.

Using favorable propagation, $|\mathbf{H}_{1}^{H}\mathbf{H}_{j}|^2 / \|\mathbf{H}_{1}\|^2 \approx N_r$, so each interferer contributes $\kappa_1 P$ (not $\kappa_1N_r P$). Total interference $\approx \kappa_1 (K-1) P$.

Per-antenna distortion variance is $\rho_1 \sigma_y^2 = \rho_1(N_r KP + \sigma^2)/N_r$ (dominated by the total received power). After MRC by $\mathbf{H}_{1}/\|\mathbf{H}_{1}\|$ it contributes $\rho_1(KP + \sigma^2/N_r) \approx \rho_1 K P$ for large $N_r$.

$\text{SINR}_1 \approx \dfrac{\kappa_1\,N_r\,P} {\kappa_1(K-1)P + \rho_1 K P + \sigma^2} = \dfrac{\kappa_1\,N_r\,\rho} {\kappa_1(K-1)\rho + \rho_1 K\rho + 1}$.

As $N_r\to\infty$ with fixed $\rho$, the numerator grows linearly in $N_r$ while the denominator stays $O(1)$, so the SINR is unbounded in $N_r$. But in the fixed product $\rhoN_r \triangleq \gamma$ regime (common in low-SNR analyses), the SINR converges to $\kappa_1\gamma/(1 + \rho_1\gamma K/N_r + \ldots) \to \kappa_1\gamma$ as $N_r\to\infty$ and $\rho\to 0$. The 1-bit receiver does not suffer from a finite SINR ceiling under the combined limit — the array gain still pays — but the rate scales as $\log_2(N_r)$, same as infinite precision, with a constant $2/\pi$ penalty. $\blacksquare$

Any odd quantizer has $\mathbb{E}[\mathbf{y}_q] = 0$ when $\mathbf{y}$ is zero-mean symmetric (which holds for Gaussian). Since $\mathbf{B}\mathbb{E}[\mathbf{y}] = 0$ as well, $\mathbb{E}[\mathbf{d}] = 0$.

$\mathbb{E}[\mathbf{d}\mathbf{y}^H] = \mathbb{E}[\mathbf{y}_q\mathbf{y}^H] - \mathbf{B}\mathbb{E}[\mathbf{y}\mathbf{y}^H] = \mathbb{E}[\mathbf{y}_q\mathbf{y}^H] - \mathbb{E}[\mathbf{y}_q\mathbf{y}^H] = 0.$

The quantizer map is deterministic, so $\mathbf{d}$ is a deterministic nonlinear function of $\mathbf{y}$. Its marginal distribution is the push-forward of the Gaussian through that function, which has finite support (for 1-bit, $\mathbf{d}$ takes only $4^{N_r}$ values). The Gaussian assumption in rate analysis is a worst-case upper bound on mutual information loss (Gaussian-noise-worst-case), not an exact distribution. $\blacksquare$

Bisection on $\mu$ until $\sum_n 2^{b_n^\star}\approx 48$. Typical answer (round to two decimals): $\mathbf{b}^\star_{\text{cont}} \approx (2.98, 2.81, 2.64, 2.48, 2.31, 2.14, 1.98, 1.81, 1.64, 1.48, 1.31, 1.14, 0.98, 0.81, 0.64, 0.48)$.

Floor: $\mathbf{b} = (2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0)$, cost $6\cdot 4 + 6\cdot 2 + 4\cdot 1 = 40$. Leftover 8. Greedy adds one bit each to antennas 1, 2 (costs 4 each, total 8): $\mathbf{b}_\text{final} = (3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0)$, cost $2\cdot 8 + 4\cdot 4 + 6\cdot 2 = 44$. Add one bit to antenna 3 (cost 4, total 48). Final: $\mathbf{b}_\text{final} = (3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0)$.

Using $\kappa_1 = 0.637$, $\kappa_2 = 0.882$, $\kappa_3 = 0.965$ and $\kappa_0 = 0$: $R \approx \sum_n \log_2(1 + \kappa_{b_n} g_n) \approx 3\cdot\log_2(1.965) + 3\cdot\log_2(1.882\cdot g_n) + \ldots$ Numerically this comes to about $8.2$ bits/use.

The continuous relaxation gives $\approx 8.5$ bits/use, so the rounding loss is $\approx 0.3$ bits/use (~3-4%). The dropout of antennas 13-16 is correct for this budget — they contribute less marginal rate per bit than the added bit to antennas 1-3. $\blacksquare$

At $\rho = 0.1$, all $b_n^\star$ shift down by $\tfrac{1}{2} \log_2(10) \approx 1.66$ bits uniformly. New vector: $(1.32, 1.15, 0.98, 0.82, 0.65, 0.48, 0.32, 0.15, \ldots)$. Most antennas now sit below 1 bit, so the integer rounding sets them to 0 (off). Only the top 2-4 antennas cross 1 bit.

Floor: only 1 or 2 antennas have $b \geq 1$, total cost e.g. 4. Leftover 44 to spend. The greedy phase will walk up: give more bits to the top antenna first, then the second, reaching perhaps $\mathbf{b} = (5, 4, 3, 2, 1, 1, 0, 0, \ldots)$. Details depend on the cost-gain ratios.

At low SNR, the strongest antennas become more valuable relative to the rest — the water drops, and the relative prominence of the best antennas grows. So the optimum does not shift uniformly toward 1 bit; it concentrates more bits on fewer antennas, shutting off the weak ones. This is the opposite of what the pure Walden argument at low SNR suggests. The reconciliation: the Walden argument assumed all antennas have the same gain, while here they do not. $\blacksquare$

$N_r\,\text{SNR} = 256\cdot 0.063 \approx 16.1$.

$R_\infty = \log_2(17.1) \approx 4.09$ bit/use.

$\text{SINR}_1 = 0.637\cdot 16.1/(1 + 0.363\cdot 16.1) = 10.25/6.84 \approx 1.50$. $R_1 = \log_2(2.50) \approx 1.32$ bit/use.

$R_1/R_\infty \approx 0.32$. In decibels $10\log_{10}(0.32) \approx -4.9$ dB, about 3 dB worse than the low-SNR $-1.96$ dB prediction. The extra penalty comes from the fact that $N_r\,\text{SNR} > 1$ so we are not in the linear (low-SNR) regime — the quantization noise has started to cap the effective SINR. This shows that the $2/\pi$ loss is a best-case bound, not an always-achievable constant. $\blacksquare$

$f(b) \triangleq \log_2(1 + \kappa_b x)$ is concave in $b$ for fixed $x$ because $\kappa_b$ is concave in $2^b$ (by the Lloyd-Max table's decreasing marginal returns) and $\log_2(1+\cdot)$ is concave. Hence Jensen gives $\frac{1}{N}\sum_n f(b_n) \leq f(\bar b)$ for any allocation with the same average.

Restricting to the two-level set $\{1, \infty\}$ forces the allocation to be binary, which is suboptimal relative to allocating the same total budget continuously. The binary case gives exactly $G(\alpha) = \alpha\cdot \kappa_\infty + (1-\alpha)\kappa_1 = \kappa_1 + (1-\kappa_1)\alpha$, a linear function of $\alpha$.

The gap between the linear upper bound and Jensen's best continuous allocation is maximized at $\alpha = 1/2$ and equals $\kappa_2 - (\kappa_1 + 1)/2 \approx 0.882 - 0.819 = 0.063$. So mixed-ADC at the halfway point leaves about 7% of the per-antenna Bussgang-gain improvement on the table. In rate terms this is at most a few tenths of a bit/channel use. $\blacksquare$