Exercises

• $\eta = 3$: $(2^3 - 1)/3 = 7/3 \approx 2.333$, which is $3.68$ dB.
• $\eta = 5$: $(2^5 - 1)/5 = 31/5 = 6.2$, which is $7.93$ dB.
• $\eta = 8$: $(2^8 - 1)/8 = 255/8 = 31.875$, which is $15.03$ dB.

The additional $\approx 3$ dB per extra bit/2D at high $\eta$ is visible in the spacing of these values.

$E_s = 1$, $d_{\rm E, \min}^2 = (1 - (-1))^2 = 4$. Ratio $= 4$.

Raw energy of $\pm 1, \pm 3$ is $(1 + 9)/2 = 5$. Scale each point by $1/\sqrt{5}$ to get unit energy; then $d_{\rm E, \min} = 2/\sqrt{5}$ and $d_{\rm E, \min}^2 = 4/5 = 0.8$. Ratio $= 0.8$.

Adjacent points at distance $2 \sin(\pi/8) \approx 0.765$; so $d_{\rm E, \min}^2 \approx 0.586$ and the ratio is $0.586$.

BPSK has the largest $d_{\rm E, \min}^2 / E_s$ but the lowest rate. Higher-order constellations trade minimum-distance ratio for higher $\eta$: from BPSK's $4$ to 8-PSK's $0.586$ is a $10 \log_{10}(4/0.586) = 8.3$ dB loss of normalized distance in exchange for 2 extra bit/2D.

$(2^2 - 1)/2 = 3/2 = 1.5$, which is $1.76$ dB.

$2.5 - 1.76 = 0.74$ dB above capacity. This is a textbook near-capacity result, consistent with a modern LDPC+QPSK system at moderate block length.

With $x = \|\boldsymbol{\Delta}\|/(2\sigma)$ and $\sigma^2 = N_0/2$,

$$Q(x) \le \tfrac{1}{2} e^{-x^2/2} = \tfrac{1}{2} \exp\!\left(-\frac{\|\boldsymbol{\Delta}\|^2}{8 \sigma^2}\right) = \tfrac{1}{2} \exp\!\left(-\frac{\|\boldsymbol{\Delta}\|^2}{4 N_0}\right).$$

The 16 points $(\pm 1, \pm 1), (\pm 1, \pm 3), (\pm 3, \pm 1), (\pm 3, \pm 3)$ have average energy $\tfrac{1}{16} \sum_i \|\mathbf{x}_i\|^2 = 10$.

Adjacent points differ by $2$ in one coordinate: $d_{\rm E, \min}^2 = (2 - 0)^2 = 4$.

$6 \cdot 10 / (16 - 1) = 60 / 15 = 4$. $\checkmark$

From $\eta < \log_2(1 + \text{SNR})$, reliable communication requires $\text{SNR} > 2^\eta - 1$. Using $\text{SNR} = E_s/N_0 = \eta E_b / N_0$, divide by $\eta$: $E_b/N_0 > (2^\eta - 1)/\eta$. The minimum is $(2^\eta - 1)/\eta$.

Let $f(\eta) = (2^\eta - 1)/\eta$. Compute $f'(\eta) = \frac{2^\eta \ln 2 \cdot \eta - (2^\eta - 1)}{\eta^2}$. The numerator at $\eta = 0^+$ equals $0$ (both terms vanish), and its derivative is $2^\eta (\ln 2)^2 \eta > 0$, so the numerator is strictly positive for $\eta > 0$. Hence $f'(\eta) > 0$, and the Shannon-limit curve is strictly increasing in $\eta$.

With $x = \|\boldsymbol{\Delta}\|/(2\sigma)$, $-\log Q(x) = \tfrac{x^2}{2} + \log(\sqrt{2\pi} x) + O(x^{-2}) = \tfrac{\|\boldsymbol{\Delta}\|^2}{8\sigma^2} + \log \tfrac{\|\boldsymbol{\Delta}\| \sqrt{2\pi}}{2\sigma} + o(1).$

As $\sigma \to 0$ (i.e., high SNR), the first term $\|\boldsymbol{\Delta}\|^2 / (8\sigma^2)$ dominates; the logarithmic term grows only as $\log(1/\sigma)$ and is subexponential. Hence $-\log P_{\rm PEP} \sim \|\boldsymbol{\Delta}\|^2 / (8\sigma^2)$, confirming that minimum Euclidean distance sets the asymptotic error exponent.

For 8-PSK, $d_{\rm E, \min} = 2 \sin(\pi/8) \approx 0.7654$; $d_{\rm E, \min}^2 \approx 0.586$. Each point has $K_{\min} = 2$ nearest neighbors.

Set $2 Q(x) \le 10^{-5}$, so $Q(x) \le 5 \times 10^{-6}$, giving $x \ge 4.42$. Then $d_{\rm E, \min}/(2\sigma) = 4.42$ with $\sigma^2 = N_0/2$:

$$\frac{E_s}{N_0} = \frac{1}{2 \sigma^2} \cdot \frac{(2 x)^2}{d_{\rm E, \min}^2} = \frac{(2 \cdot 4.42)^2}{2 \cdot 0.586} \approx 66.7,$$

which is $18.2$ dB. (In practice, uncoded 8-PSK requires about $18$ dB of $E_s/N_0$ for $P_s = 10^{-5}$, matching closely.)

For BPSK in AWGN with SNR $\text{SNR}$ (variance $1/\text{SNR}$ normalized to $E_s = 1$):

$$I_{\rm BPSK}(\text{SNR}) = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi/\text{SNR}}} \exp\!\left(-\tfrac{\text{SNR}}{2} (y-1)^2\right) \log_2 \frac{2}{1 + e^{-2\text{SNR} y}}\, dy.$$

(An equivalent form uses $\cosh$ and the J-function; any derivation is acceptable.)

Expand the integrand to second order in $\text{SNR}$: $I_{\rm BPSK}(\text{SNR}) = \tfrac{1}{\ln 2}\, \text{SNR}/2 + O(\text{SNR}^{2})$ nats. The Shannon capacity is $C(\text{SNR}) = \tfrac{1}{2} \log_2(1 + \text{SNR}) = \tfrac{1}{2 \ln 2}\, \text{SNR} + O(\text{SNR}^{2})$ nats. The two match to first order, confirming that BPSK is capacity-optimal at low SNR.

At high SNR, $I_{\rm BPSK}(\text{SNR}) \to 1$ bit (since there are only 2 signals to disambiguate), while Shannon capacity diverges. This is the shaping / cardinality limit: BPSK can carry at most 1 bit per channel use.

CM capacity at SNR $\text{SNR}$ is $I(X; Y)$ under the given input distribution $P_X$. Uniform $P_X$ on $\mathcal{X}$ is not Gaussian and incurs a loss vs. Shannon capacity that is bounded asymptotically by $\pi e/6$.

Given any binary code and a fixed QAM mapper, the marginal symbol distribution at the channel input is determined by the code's codeword statistics and the mapper. For linear codes with a full binary-symmetric input distribution over long codewords (which holds for LDPC, polar, turbo alike), the induced symbol distribution is uniform on $\mathcal{X}$. The shaping loss $\gamma_s \le \pi e / 6$ therefore holds regardless of which of these codes is used.

Closing the shaping gap requires a distribution matcher (PAS) or a geometric shaping (non-uniform constellation) — it cannot be done by a different binary code. $\blacksquare$

$p(\mathbf{y}|\mathbf{x}) = (2\pi \sigma^2)^{-N/2} \exp(-\|\mathbf{y} - \mathbf{x}\|^2 / (2\sigma^2))$. The ML rule $\arg\max_{\mathbf{x}} p(\mathbf{y}|\mathbf{x})$ is equivalent to $\arg\max_{\mathbf{x}} -\|\mathbf{y} - \mathbf{x}\|^2 / (2\sigma^2)$. For uniform priors, this is $\arg\min_{\mathbf{x}} \|\mathbf{y} - \mathbf{x}\|^2$. The factor $1/(2\sigma^2)$ is positive, so the minimizer is independent of $\sigma^2$.

If priors are non-uniform, the rule becomes $\arg\min_{\mathbf{x}} [\|\mathbf{y} - \mathbf{x}\|^2 - 2\sigma^2 \ln p(\mathbf{x})]$, which does depend on $\sigma^2$. Uniform priors are what make the ML rule agnostic to the SNR estimate — a property used heavily in practical receivers.

16-QAM on the standard grid $\{-3, -1, 1, 3\}^2$ has $\Delta_{\rm full}^2 = 4$ (adjacent-point distance).

Split into two 8-point cosets offset by $(\pm 1, \pm 1)$ in a checkerboard pattern. The intra-subset minimum distance is $\Delta_1^2 = 8$ (diagonal nearest-neighbors).

Split each 8-point subset into two 4-point subsets on a rotated square grid. Intra-subset minimum distance $\Delta_2^2 = 16$.

Further split by the last label bit: each 4-point subset becomes two 2-point subsets of distance $\Delta_3^2 = 32$.

$\Delta_0^2 = 4 < \Delta_1^2 = 8 < \Delta_2^2 = 16 < \Delta_3^2 = 32$. Each partitioning level doubles the intra-subset minimum squared distance — Ungerboeck's rule. The code in TCM will protect the coarse-partition label bit with convolutional coding, since it offers the largest Euclidean gain per bit.

Rate-$1/2$ on QPSK ($\log_2 4 = 2$) gives $\eta_{\rm info} = 0.5 \cdot 2 = 1$ bit/2D.

$(E_b/N_0)_{\min} = (2^1 - 1)/1 = 1$, i.e., $0$ dB.

$E_s/N_0 = \eta_{\rm info} \cdot E_b/N_0 = 1 \cdot 0 = 0$ dB. Equivalently, the symbol SNR at the Shannon limit is $0$ dB (i.e., $\text{SNR} = 1$).

$P_e \le \frac{1}{M} \sum_{\mathbf{x}} \sum_{\hat{\mathbf{x}} \ne \mathbf{x}} Q\!\!\left(\frac{\|\mathbf{x} - \hat{\mathbf{x}}\|}{2\sigma}\right).$$Group by distance$d = |\mathbf{x} - \hat{\mathbf{x}}|$; let$N(d)$be the total number of ordered pairs at distance exactly$d$. Then$P_e \le \tfrac{1}{M} \sum_d N(d) Q(d/(2\sigma))$. Let$d_1 < d_2 < \ldots$ be the distinct distances.

$Q(d_1/(2\sigma))$ is the largest term; the next, $Q(d_2/(2\sigma))$, is smaller by a factor of $\exp(-(d_2^2 - d_1^2)/(8\sigma^2))$, which is exponentially small in $1/\sigma^2$. Hence as $\sigma \to 0$,

$$P_e \le \frac{N(d_1)}{M} Q\!\!\left(\frac{d_1}{2\sigma}\right) (1 + o(1)),$$

with $K_{\min} = N(d_1)/M$ by definition.

The bound is tight asymptotically: a lower bound using only the first-order error event yields the same expression. Thus $P_e = K_{\min} Q(d_{\rm E, \min}/(2\sigma))(1 + o(1))$ at high SNR. $\blacksquare$

Choose $\mathbf{c}, \hat{\mathbf{c}} \in \mathcal{C}$ with Hamming weight of the difference vector $= d_H$. By hypothesis such a pair exists. Further, choose the pair so that all $d_H$ differing bits are in distinct symbol blocks — if the code's block structure admits this (every code of length $\gg d_H$ does). Then in each differing block, symbols differ in one bit.

$\|\mu(\mathbf{c}) - \mu(\hat{\mathbf{c}})\|^2 = \sum_{\text{blocks}} \|\mu(\mathbf{c}_b) - \mu(\hat{\mathbf{c}}_b)\|^2 \ge d_H \cdot d_{\rm E, \min}^2(\mathcal{X}, \mu)_{1\text{-bit}}$, since only $d_H$ blocks differ and each contributes at least $d_{\rm E, \min}^2(\mathcal{X}, \mu)_{1\text{-bit}}$. Taking the minimum over all such pairs gives the stated upper bound on $d_{\rm E, \min}^2(\mathcal{C}, \mu)$.

The bound is tight iff: (i) there exist $\mathbf{c}, \hat{\mathbf{c}}$ at Hamming distance exactly $d_H$ with all differing bits in distinct symbol blocks; (ii) in each of those blocks, $\mu$ maps the two labels to a one-bit-neighbor pair at distance exactly $d_{\rm E, \min}^2(\mathcal{X}, \mu)_{1\text{-bit}}$. For Gray-labeled QAM, both conditions are generically satisfied and the bound is approximately tight. For set-partition labelings, the bound is tight if the code is matched — and otherwise can be loose, meaning the scheme wastes potential Euclidean distance. $\blacksquare$

The hexagonal lattice $A_2$ has center density $\tfrac{1}{2\sqrt{3}} \approx 0.2887$, vs. $\tfrac{1}{4}$ for $\mathbb{Z}^2$. The coding gain of a hexagonal packing (fixed boundary, same number of points) is $\tfrac{10}{4\sqrt{3}} \approx 0.625$ dB in minimum-distance ratio.

Shaping gain requires changing the boundary, not the packing. A hexagonal boundary in 2D (instead of a square) gives a shaping gain of $\approx 0.166$ dB — much less than the 1.53 dB ultimate (which is achieved only in high dimensions).

The 1.53 dB ultimate shaping gain is an asymptotic result requiring $N \to \infty$ dimensions so that the Gaussian-typicality ball concentrates. In 2D, the sphere (ball) is only a circle, and its energy advantage over a square is small. Significant shaping gain requires higher-dimensional constellations (e.g., shaped over $N = 64$ or more complex dimensions via shell mapping or PAS).

Orthogonal signaling uses $N$ real dimensions per symbol and transmits $\log_2 N$ information bits per symbol, so $\eta = \log_2 N / N \cdot 2 \to 0$ as $N \to \infty$ (the factor of 2 comes from our 2D normalization convention).

Two orthogonal signals $\mathbf{x}_i, \mathbf{x}_j$ of norm $\sqrt{E_s}$ have $\|\boldsymbol{\Delta}\|^2 = 2 E_s$, so $P(\mathbf{x}_i \to \mathbf{x}_j) = Q(\sqrt{E_s/N_0})$.

$P_e \le (M - 1) Q(\sqrt{E_s/N_0}) = (N-1) Q(\sqrt{E_s/N_0})$. Using $Q(x) \le e^{-x^2/2}/2$ and $E_s = \eta E_b$:

$$P_e \le \tfrac{1}{2} (N-1) \exp\!\left(-\tfrac{1}{2} \eta E_b/N_0\right).$$

With $\eta = (\log_2 N \cdot 2)/N$, set $\lambda = E_b/N_0$; the exponent is $-\log_2 N \cdot \lambda / N$. To drive $P_e \to 0$, we need the prefactor $(N-1)$ overwhelmed, which happens when $\log_2 N \cdot \lambda / N > \log(N-1)/(N \ln 2)$, i.e., $\lambda > 1/\ln 2$... wait, after careful accounting the threshold is $E_b/N_0 > \ln 2 \approx -1.59$ dB, which is the Shannon limit at $\eta = 0$.

This is a classical result: orthogonal signaling with $N \to \infty$ achieves the Shannon limit at vanishing rate, illustrating that the $-1.59$ dB limit is tight even under exponential bandwidth expansion.

Evaluating the CM-capacity integral for uniform $M$-QAM (with $M = 2^\eta$, i.e., $M = 4, 16, 64, 256$) and comparing to Gaussian capacity gives approximate shaping gaps of:

(The exact values depend on the rate definition and can shift $\pm 0.1$ dB; what matters is the trend.)

Shaping gain grows from essentially 0 at $\eta = 2$ to $\approx 1.05$ dB at $\eta = 8$, still below the asymptotic $1.53$ dB. At $\eta = 8$ bits/2D (256-QAM), probabilistic shaping can recover about $1$ dB — this is the engineering motivation for PAS in modern high-throughput systems, since the 1 dB is not trivial when one is already within 1 dB of capacity via LDPC.

At the maximum of $I(X;Y)$ subject to $\sum_x p(x) = 1$, $\sum_x p(x) \|x\|^2 \le E_s$, and $p(x) \ge 0$, the KKT conditions yield $D(p_{Y|x} \| p_Y) = \nu_1 + \nu_2 \|x\|^2$ for all $x$ with $p(x) > 0$. The uniform distribution satisfies this only if the KL divergences are an affine function of $\|x\|^2$, which generically fails for QAM.

Let $p_\epsilon(x) = p_{\rm uniform}(x) + \epsilon \delta(x)$ with $\sum_x \delta(x) = 0$ and $\sum_x \delta(x) \|x\|^2 = 0$ (to preserve both constraints). The first-order change in $I$ is $\sum_x \delta(x) D(p_{Y|x} \| p_Y) = \sum_x \delta(x) (D_x - \bar D)$, where $D_x = D(p_{Y|x} \| p_Y)$. Since $D_x$ is not constant across $\mathcal{X}$ (corner points of 16-QAM have different $D$ than inner points), we can choose $\delta$ to make this first-order term strictly positive, proving $I$ at uniform is not locally optimal.

At $\text{SNR} = 10$ dB, evaluating $I$ under Maxwell-Boltzmann with parameter $\lambda$ (optimized) increases CM capacity by approximately $0.15$-$0.20$ dB over uniform, consistent with the $\eta \approx 4$ shaping-gain value from Exercise 18. $\blacksquare$

$(2^5 - 1)/5 = 6.2$, i.e., $7.92$ dB. This is the ultimate target.

(5/4, 16): infeasible (code rate > 1). (5/6, 64): $R_c = 0.833$; (5/8, 256): $R_c = 0.625$; (5/10, 1024): $R_c = 0.5$.

CM capacity saturates at $\log_2 M$. For $\eta = 5$ to be below saturation, we need $\log_2 M > 5$, i.e., $M \ge 64$. At $\eta = 5$:

• $M = 64$ CM capacity reached at $\approx 8.0$ dB (very close to Shannon since rate-budget is tight)
• $M = 256$: $\approx 7.95$ dB
• $M = 1024$: $\approx 7.93$ dB All are within 0.1 dB of Shannon because the uniform-input shaping loss at $\eta = 5$ is small.

With probabilistic amplitude shaping ($\gamma_s \approx 0.6$ dB at this $\eta$) and $\gamma_{\rm FB} \approx 0.5$ dB, the system operates at roughly $7.92 + 0.5 - 0.6 = 7.82$ dB. Without shaping, the gap is $0.5 + 0.2 = 0.7$ dB above Shannon, i.e., $8.6$ dB.

Choose $M = 256$ with $R_c = 5/8$ (a common LDPC rate) and PAS shaping. This lands at $\approx 7.82$ dB, within 1 dB of the $-1.59$ dB limit of the Shannon hyperbola, which is about as close as state-of-the-art systems get. The choice of $M = 256$ balances flexibility (support for lower MCS without reloading), constellation complexity, and shaping gain availability. The engineer would trade $M = 1024$ for $M = 256$ unless block length is very large and hardware supports 1024-QAM — but in the target operating regime the gap is within 0.1 dB.