Exercises

Adjacent 8-PSK points differ in angle by $2\pi/8 = \pi/4$, so $$\Delta_0^2 = 2(1 - \cos(\pi/4)) = 2 - \sqrt{2} \approx 0.586.$$

Adjacent points in a parity-$k$ subset differ by $2 \cdot \pi/4 = \pi/2$, so $$\Delta_1^2 = 2(1 - \cos(\pi/2)) = 2.$$

Adjacent points in a deeper subset differ by $\pi$ (antipodal), so $$\Delta_2^2 = 2(1 - \cos(\pi)) = 4.$$ All three match the theorem. $\blacksquare$

Adjacent 16-QAM points differ by $(\pm 1, 0)$ or $(0, \pm 1)$ in the unit-spacing convention (actual coordinates $\pm 1, \pm 3$ with inter-point distance $2$). Rescaling to unit inter-point distance: $\Delta_0^2 = 1$. But in Ungerboeck's normalization $\Delta_0^2 = 4$ — this is just a scaling of $2\times$. Using the $\pm 1, \pm 3$ coordinates: $\Delta_0^2 = 4$ (nearest neighbour distance is $2$).

Split by parity of $a + b$. Inside either subset, adjacent points differ by $(\pm 2, \pm 2)$, giving $\Delta_1^2 = 4 + 4 = 8$.

Split each level-1 coset by parity of $a$. Inside a level-2 coset, adjacent points differ by $(\pm 4, 0)$ or $(0, \pm 4)$, giving $\Delta_2^2 = 16$.

Split each level-2 coset into antipodal pairs. Inside a pair, points differ by $(\pm 4, \pm 4)$, giving $\Delta_3^2 = 32$. The full chain $4, 8, 16, 32$ matches the lattice-index-2 doubling. $\blacksquare$

Any index-2 sublattice $\Lambda_{i+1} \subset \Lambda_i$ has $|\Lambda_i / \Lambda_{i+1}| = 2$ and can be written as $\Lambda_{i+1} = \{\mathbf{v} \in \Lambda_i : \mathbf{a}_i \cdot \mathbf{v} \in 2\mathbb{Z}\}$ for some lattice functional $\mathbf{a}_i$. The nearest-neighbour vector of $\Lambda_{i+1}$ has squared length $\geq 2 \cdot (\text{squared length of nearest neighbour of } \Lambda_i)$ because going from $\Lambda_i$ to $\Lambda_{i+1}$ removes the odd-parity cosets, leaving only the even-parity lattice — equivalently, a $\sqrt{2}$-rescaled rotation of $\Lambda_i$.

In the Ungerboeck partition, the split is always chosen as an index-2 sublattice (maximizing the intra-subset MSED at each step). The rescaling is exactly by $\sqrt{2}$ for 2D lattices (this is the key lemma), so squared distances double exactly: $\Delta_{i+1}^2 = 2 \Delta_i^2$.

Remark: For non-lattice 2D constellations (like 8-PSK at level 0) the first doubling is not exact — it is $\Delta_1^2 / \Delta_0^2 = 2 + \sqrt{2} \approx 3.4$ (super-doubling) because 8-PSK points are not on a lattice. From level 1 onward, the rotated QPSK sublattice is a 2D lattice, and subsequent steps double exactly. $\blacksquare$

$g_0(D) = 1 + D + D^2$, $g_1(D) = 1 + D^2 = (1 + D)^2$. We compute $\gcd(g_0, g_1)$. Since $g_1(D) = (1+D)^2$ has a double root at $D = 1$, and $g_0(1) = 1 + 1 + 1 = 1 \neq 0$ in $\mathbb{F}_2$, the polynomials share no factor of $(1+D)$. A direct polynomial division confirms $\gcd = 1$. Hence $(7,5)$ is non-catastrophic.

Its free Hamming distance, from standard tables or by enumerating error events of length $\leq 7$, is $d_{\rm free}^{(H)} = 5$.

$g_0(D) = D + D^2 = D(1 + D)$, $g_1(D) = D^2$. Then $\gcd(g_0, g_1) = D$, which is a factor of both. Hence the code is catastrophic.

Intuitively: the information bit shifted into the register at time 0 produces $(01, 01)$ at time 0 and $(11, 11)$... wait, actually this encoder has the issue that $D$ is a factor of both outputs, meaning a single input-zero symbol at time 0 (after which the shift register was in state $0$) produces zero output no matter what comes next — which looks harmless until you realize a constant-all-one input followed by all-zero input produces a finite Hamming-weight output from an infinite Hamming-weight input. $\blacksquare$

The parallel-transition bound is $\Delta_2^2 = 4$. For the 4-state code this was the minimum; for the 8-state code, a longer error event becomes competitive.

The rate-$1/2$ convolutional code in the 8-state TCM has $d_{\rm free}^{(H)} = 4$ (improved from $d_{\rm free}^{(H)} = 3$ of the 4-state code). Hence the coset-sequence bound gives $$d_{\rm free}^{(H)} \cdot \Delta_1^2 = 4 \cdot 2 = 8 > 4 = \Delta_2^2.$$ So the parallel-transition bound is still the active one — but wait, Ungerboeck's table lists $4.586$, not $4$.

The value $d_{\rm free}^2 = 4.586$ arises because the 8-state code has no parallel transitions at its inner trellis — its dominant error event is a length-3 coset-sequence event with total squared Euclidean distance $\Delta_0^2 + \Delta_1^2 + \Delta_1^2 = (2 - \sqrt{2}) + 2 + 2 = 6 - \sqrt{2} \approx 4.586$. The "level-0" distance enters because the two competing paths at one time step differ within the same level-1 coset but at different points on the 8-PSK circle at angular separation $\pi/4$.

Take-away: the simple lower bound of TLower Bound on Free Euclidean Distance via Partition Levels is not always tight — refined analysis via the transfer-function bound is needed when parallel transitions are eliminated by the design.

$\gamma_c = 4/2 = 2$, so $\gamma_c^{\rm dB} = 10 \log_{10} 2 \approx 3.01$ dB.

The union-bound form of the BER is $$P_e^{\rm TCM} \approx N_{\rm free} \, Q\!\left(\sqrt{d_{\rm free}^2 E_s / (2 N_0)}\right).$$ At a fixed target $P_e^{\rm TCM} = 10^{-6}$, we have $$Q\!\left(\sqrt{d_{\rm free}^2 E_s^{\rm TCM} / (2 N_0)}\right) = 10^{-6} / N_{\rm free} = 2.5 \cdot 10^{-7}.$$ For uncoded, $Q(\cdot) = 10^{-6}/2 = 5 \cdot 10^{-7}$.

Using $Q^{-1}(2.5 \cdot 10^{-7}) \approx 5.02$ and $Q^{-1}(5 \cdot 10^{-7}) \approx 4.89$: $\sqrt{d_{\rm free}^2 E_s^{\rm TCM}/(2 N_0)} = 5.02$ and $\sqrt{d_{\rm uncoded}^2 E_s^{\rm unc}/(2 N_0)} = 4.89$.

Let $r_{\rm TCM} = E_s^{\rm TCM}/N_0$ and $r_{\rm unc} = E_s^{\rm unc}/N_0$. Then: $4 \cdot r_{\rm TCM} / 2 = 25.2$, so $r_{\rm TCM} = 12.6$. $2 \cdot r_{\rm unc} / 2 = 23.9$, so $r_{\rm unc} = 23.9$.

Shift = $10 \log_{10}(23.9/12.6) = 10 \log_{10} 1.897 \approx 2.78$ dB. That is, the 4-state 8-PSK TCM saves about $2.78$ dB at BER $10^{-6}$ relative to uncoded QPSK — about $0.23$ dB short of the asymptotic $3.01$ dB, due to the extra multiplicity $N_{\rm free} = 4 > 2 = N_{\rm unc}$. $\blacksquare$

For $y_k = 0.6 - 0.4j$:

• $D_0 = \{+1, -1\}$: nearest is $+1$; $\|y - 1\|^2 = 0.16 + 0.16 = 0.32$.
• $D_1 = \{e^{j\pi/4}, e^{j 5\pi/4}\} \approx \{0.707 + 0.707 j, -0.707 - 0.707 j\}$: distance to the first is $(0.6 - 0.707)^2 + (-0.4 - 0.707)^2 = 0.011 + 1.226 = 1.237$; to the second is $(0.6 + 0.707)^2 + (-0.4 + 0.707)^2 = 1.709 + 0.094 = 1.803$. Nearest: $e^{j\pi/4}$; $\lambda = 1.237$.
• $D_2 = \{+j, -j\}$: distance to $+j$ is $0.36 + 1.96 = 2.32$; to $-j$ is $0.36 + 0.36 = 0.72$. Nearest: $-j$; $\lambda = 0.72$.
• $D_3 = \{e^{j 3\pi/4}, e^{j 7\pi/4}\} \approx \{-0.707 + 0.707 j, 0.707 - 0.707 j\}$: to the second is $(0.6 - 0.707)^2 + (-0.4 + 0.707)^2 = 0.011 + 0.094 = 0.105$. Nearest: $e^{j 7\pi/4}$; $\lambda = 0.105$.

So $\lambda = (0.32, 1.237, 0.72, 0.105)$.

Using the table from EOne ACS Step of Viterbi on the 4-State 8-PSK TCM:

• $s' = 0$: from $s = 0$ via $D_0$ (cand $1.0 + 0.32 = 1.32$) or $s = 1$ via $D_2$ (cand $0.3 + 0.72 = 1.02$). $\Lambda_{k+1}(0) = 1.02$.
• $s' = 1$: from $s = 2$ via $D_1$ (cand $0.7 + 1.237 = 1.937$) or $s = 3$ via $D_3$ (cand $0.5 + 0.105 = 0.605$). $\Lambda_{k+1}(1) = 0.605$.
• $s' = 2$: from $s = 0$ via $D_2$ (cand $1.0 + 0.72 = 1.72$) or $s = 1$ via $D_0$ (cand $0.3 + 0.32 = 0.62$). $\Lambda_{k+1}(2) = 0.62$.
• $s' = 3$: from $s = 2$ via $D_3$ (cand $0.7 + 0.105 = 0.805$) or $s = 3$ via $D_1$ (cand $0.5 + 1.237 = 1.737$). $\Lambda_{k+1}(3) = 0.805$.

$\Lambda_{k+1} = (1.02,\; 0.605,\; 0.62,\; 0.805)$. State $1$ is now the most likely end-state. $\blacksquare$

Per symbol:

• Distance evaluations: $2^{\tilde{m}+1} \cdot 2^{m - \tilde{m}} = 2^{m+1}$.
• ACS adds: $N_s \cdot 2^{\tilde{m}}$.
• ACS compares: $N_s \cdot (2^{\tilde{m}} - 1)$.

• Distance evaluations: $2^{2+1} \cdot 2^1 = 8 \cdot 2 = 16$.
• ACS adds: $16 \cdot 4 = 64$.
• ACS compares: $16 \cdot 3 = 48$.

Total: about $128$ arithmetic operations per symbol — well within the reach of 1990s DSP chips at 10 Msps, and a trivial fraction of today's FPGA cycles. $\blacksquare$

$\gamma_c = 10/2 = 5$, so $\gamma_c^{\rm dB} = 10 \log_{10} 5 \approx 6.99$ dB. This would be an exceptional coding gain — more typical of 4D TCM designs (V.34) than of plain 2D schemes. $\blacksquare$

Label outgoing edges from state 0 with points $e^{j 0}$ and $e^{j 3\pi/4}$ (angular separation $3\pi/4$, not in the same level-1 coset): this violates (R2). If we further arrange that all length-2 error events involve pairs at angular separation $\geq 3\pi/8$ on both diverging and re-merging edges, the length-2 event has squared distance $$\|e^{j 0} - e^{j 3\pi/4}\|^2 + \|e^{j\pi} - e^{j 7\pi/4}\|^2 = 2(1 - \cos(3\pi/4)) + 2(1 - \cos(5\pi/4)) \approx 3.41 + 3.41 = 6.83,$$ which is larger than $\Delta_2^2 = 4$. So $d_{\rm free}^2 = 4$ is still attained from parallel transitions (R1 still holds).

Two reasons:

1. When $\tilde{m} < m$, parallel transitions dominate for 4-state codes anyway, so violating (R2) does not help. (R2) only becomes active for larger-$\tilde{m}$ designs where length-2 events can be the bottleneck.
2. For rotationally invariant designs (important for V.32/V.34 for carrier phase ambiguity resolution), (R2) is essentially forced by the symmetry group of the constellation. Violating (R2) makes the code non-rotationally-invariant and unusable for phase-ambiguous carrier recovery.

Conclusion: (R2) is not logically necessary, but it is engineeringly required once we go beyond 4-state codes and need rotational invariance. $\blacksquare$

Parallel-transition bound: $\Delta_{\tilde{m}+1}^2 = \Delta_3^2 = 32$. Hamming-distance bound: $d_{\rm free}^{(H)} \cdot \Delta_1^2 = 4 \cdot 8 = 32$. Lower bound: $d_{\rm free}^2 \geq \min(32, 32) = 32$. Balanced design.

16-QAM in $\pm 1, \pm 3$ coordinates has $E_s^{\rm 16QAM} = 10$. Rescale so $E_s = 1$: divide distances by 10. $d_{\rm free}^2 / E_s = 32/10 = 3.2$.

Unit-energy 8-PSK has $d_{\rm uncoded}^2 = 2 - \sqrt{2} \approx 0.586$.

$\gamma_c = 3.2 / 0.586 \approx 5.46$, so $\gamma_c^{\rm dB} = 10 \log_{10} 5.46 \approx 7.37$ dB. This would be a very strong 2D TCM, comparable to the best 64- or 128-state codes in Ungerboeck's Table II. $\blacksquare$

State $s = (u_{k-1}, u_{k-2})$. On input $u_k$: Output $= (g_0 \cdot u, g_1 \cdot u)$ with $g_0(D) = 1 + D^2$ and $g_1(D) = D$: $y_0 = u_k + u_{k-2}$ (mod 2), $y_1 = u_{k-1}$.

(Notation: (output, next-state).)

An error event is an input sequence starting with $u_k = 1$ (diverge), with total input Hamming weight $w$, whose first remerge to state 00 occurs at time $k + L$ (length $L$ event). Output Hamming weight $d_H$ equals the sum of output Hamming weights along the event.

• Length 1: impossible (can't diverge and remerge in one step from state 00 with $u_k = 1$).

• Length 2: $u = (1, 0)$. Outputs: $(10, 00)$ — weight 1. Final state: 00. But wait, after $(1, 0)$ starting from 00, the state path is $00 \to 10 \to 00$. Outputs: $10, 00$ = total weight $1$. Actually, let me re-examine: after input $1$, state goes $00 \to 10$ with output $10$ (weight 1). After input $0$, state goes $10 \to 00$ with output $00$ (weight 0). Event length 2, $d_H = 1$.

  Hmm — but $d_{\rm free}^{(H)}$ should be $\geq 2$ for any non-trivial code. Let me recheck. Actually for $(5, 2)$: $g_0 = 101$, $g_1 = 010$. Output of input $u_k$: $y_{k,0} = u_k \oplus u_{k-2}$, $y_{k,1} = u_{k-1}$. For input $(1, 0, 0, \ldots)$ from state 00: outputs at $k=0,1,2$: $(1 \oplus 0, 0) = (1,0)$; $(0 \oplus 0, 1) = (0,1)$; $(0 \oplus 1, 0) = (1, 0)$; $(0 \oplus 0, 0) = (0,0)$. Total weight $= 3$ over the error event that returns to state 00.

  So $d_{\rm free}^{(H)} \leq 3$. And this length-3 event is the minimum-weight event. Direct tables confirm $d_{\rm free}^{(H)} = 3$. $\blacksquare$

The $d_{\rm free}^{(H)} = 3$ result plugs into the TCM bound: $d_{\rm free}^{(H)} \cdot \Delta_1^2 = 3 \cdot 2 = 6$ for unit-energy 8-PSK, which is larger than $\Delta_2^2 = 4$. So parallel transitions bound $d_{\rm free}^2$ at 4 — the 4-state 8-PSK TCM is parallel-transition-limited, which is why the 4-state code has only $3$ dB of gain: the weak link is the partition, not the convolutional code.

TCM transmits $m$ information bits per symbol via a $2^{m+1}$-point constellation at rate $1/T_s$ symbols/s. Bandwidth consumed: $1/T_s$ Hz (actually $\sim 1/T_s$ after excess-BW factor). Same as uncoded $2^m$-point at rate $1/T_s$.

A rate-$1/2$ binary code takes $m$ information bits and produces $2m$ coded bits; each pair of coded bits maps to one QPSK symbol, so the scheme uses $m$ QPSK symbols per $m$ information bits — bit rate = $1/T_s$ bits/s. But each QPSK symbol carries 2 bits, so without the rate-$1/2$ code we would get $2/T_s$ bits/s. To get the same $2/T_s$ bit rate with the coded scheme, we need to double the symbol rate — i.e., double the bandwidth.

TCM avoids this by expanding the constellation (cheap — no extra bandwidth) instead of the symbol rate (expensive). That's the whole slogan. $\blacksquare$

As in Exercise 2.3, an index-2 sublattice of a 2D lattice has nearest-neighbour squared distance exactly $2\times$ the parent. This is because the determinant of the lattice generator matrix squares as the index, and the nearest-neighbour distance is proportional to $\sqrt{|\det|}$. Hence exact doubling.

For an $n$-dimensional lattice, an index-$k$ sublattice has $(\det)^{1/n}$-scaled nearest-neighbour distance — for $k = 2$, $(2)^{1/n}$. Hence $\Delta_{i+1}^2 / \Delta_i^2 = 2^{2/n}$. In 2D: $2$. In 4D: $\sqrt{2}$. In 8D: $2^{1/4} \approx 1.19$. For V.34's 4D TCM this gives $\sqrt{2} \approx 1.41$ per level — less than the 2D ratio but still useful, and 4D gives more partition levels per bit of code rate, compensating.

$M$-PSK is not a lattice: it has curvature, and the chord $2 \sin(\pi/M)$ between adjacent points is only approximately $2\pi/M$ at small $\pi/M$. The level-1 split gains more than a factor $2$ at level 0 (super-doubling from $2 - \sqrt{2}$ to $2$), but subsequent splits do double exactly (they are happening on a sublattice-like subset, since level-1 of 8-PSK is rotated QPSK, a lattice). Hence the "exact doubling" fails only at level 0 of $M$-PSK partitions and holds exactly from level 1 onward. $\blacksquare$

$C = 3000 \log_2(1 + 3162) \approx 3000 \cdot 11.63 \approx 34.9$ kbit/s.

V.34 achieves $R = 33.6$ kbit/s at $\text{SNR}_{\rm actual} = 3162$ (35 dB). To achieve $R = 33.6$ kbit/s at Shannon-optimal SNR: $33.6 = 3 \log_2(1 + \text{SNR}^*) \;\Rightarrow\; 1 + \text{SNR}^* = 2^{11.2} \approx 2353 \;\Rightarrow\; \text{SNR}^* \approx 2352 \approx 33.7$ dB. Gap: $35 - 33.7 \approx 1.3$ dB.

Without TCM and shaping, uncoded 16-QAM at 4 bit/symbol would sit about 9 dB worse than Shannon (the coded-modulation loss). V.34 recovers about 5 dB via TCM and 1 dB via shell-mapping shaping, leaving approximately $$9 - 5 - 1 = 3 \;\text{dB unrecovered},$$ which is close to the actual 1.3 dB gap once we also account for the $\sim 1$ dB extra benefit of the 4D (vs. 2D) trellis and some optimization of symbol rate. The "headline" answer: TCM contributes $\sim 5/6 = 83\%$ of the coding-plus-shaping gain; shaping contributes $\sim 17\%$. $\blacksquare$

$\text{latency} = D / R_s \leq 3\,\text{ms} = 3 \times 10^{-3}\,\text{s}$. Hence $R_s \geq 60 / (3 \times 10^{-3}) = 2 \times 10^4\,\text{symbols/s} = 20$ kbaud.

At baud rates below $20$ kbaud (say, a high-frequency HF radio at $3$ kbaud), the Viterbi traceback latency exceeds 3 ms. At V.34's $3.2$ kbaud the $D = 60$ latency would be $60/3200 \approx 19$ ms — well above our 3 ms budget. This is why V.34 uses a modest 16-state TCM ($\nu = 4$) with only $D \approx 20$–$25$, yielding $\sim 6$–$8$ ms decoder latency — acceptable for data, unacceptable for voice echoes (hence the separate voice-cancellation filter in V.34's echo-canceller section). $\blacksquare$

Points: $1, j, -1, -j$. Label: $00, 01, 11, 10$. 1-bit-flip pairs:

• Flip bit 0: $(00, 01) = (1, j)$, dist $\sqrt{2}$; $(11, 10) = (-1, -j)$, dist $\sqrt{2}$. Min: $\sqrt{2}$.
• Flip bit 1: $(00, 10) = (1, -j)$, dist $\sqrt{2}$; $(01, 11) = (j, -1)$, dist $\sqrt{2}$. Min: $\sqrt{2}$.

Gray min-flip distance = $\sqrt{2}$ (adjacent QPSK points).

Points: $1, j, -1, -j$. Label: $00, 10, 11, 01$. 1-bit-flip pairs:

• Flip bit 0: $(00, 01) = (1, -j)$, dist $\sqrt{2}$; $(10, 11) = (j, -1)$, dist $\sqrt{2}$. Min: $\sqrt{2}$.
• Flip bit 1: $(00, 10) = (1, j)$, dist $\sqrt{2}$; $(01, 11) = (-j, -1)$, dist $\sqrt{2}$. Actually in Ungerboeck the high-order bit selects the antipodal pair: flip bit 1 at fixed bit 0 → flip from $\{1, j\}$ coset to $\{-1, -j\}$ coset.
• Flip bit 0 (the low-order bit, within coset): $(00, 01) = (1, -j)$, dist $\sqrt{2}$; $(10, 11) = (j, -1)$, dist $\sqrt{2}$. Min: $\sqrt{2}$.
• Flip bit 1 (the high-order bit, between cosets): $(00, 10) = (1, j)$, dist $\sqrt{2}$; $(01, 11) = (-j, -1)$, dist $\sqrt{2}$.

Min-flip distance = $\sqrt{2}$ for both labelings. In QPSK the difference is invisible because all non-identical pairs have distance $\sqrt{2}$ or $2$. The difference appears in larger constellations like 8-PSK, where Gray's 1-bit-flip pairs are adjacent (distance $\Delta_0$) but Ungerboeck's 1-bit-flip pairs at the high-order bit are antipodal (distance $\Delta_2$). This is the essential difference — and it is what makes Ungerboeck's labeling optimal for trellis codes and Gray's labeling optimal for BICM. $\blacksquare$

• Ungerboeck labeling maximizes the distance between points whose high-order bits differ — useful when high-order bits are coded.
• Gray labeling minimizes the Hamming distance between geometric nearest-neighbours — useful when bits are interleaved and each bit sees an independent binary channel (BICM).

A standard rate-$2/3$ $\nu = 4$ code from Biglieri's tables: $G(D) = \begin{pmatrix} 1 + D^2 + D^4 & D^2 & 1 + D^4 \\ D & 1 + D^2 & 1 + D + D^4 \end{pmatrix}$, with $d_{\rm free}^{(H)} = 5$. (Verify non-catastrophic via full rank of $G$ over $\mathbb{F}_2(D)$.)

Parallel-transition term: $\Delta_{\tilde{m}+1}^2 = \Delta_3^2 = 32$. Hamming-distance term: $d_{\rm free}^{(H)} \cdot \Delta_1^2 = 5 \cdot 8 = 40$. Bound: $d_{\rm free}^2 \geq \min(32, 40) = 32$. Parallel-transition limited (no gain from having $d_{\rm free}^{(H)} > 4$).

In unit-energy 16-QAM: $d_{\rm free}^2 / E_s = 32/10 = 3.2$. Unit-energy 8-PSK: $d_{\rm uncoded}^2 = 2 - \sqrt{2} \approx 0.586$. $\gamma_c = 3.2 / 0.586 \approx 5.46$; $\gamma_c^{\rm dB} \approx 7.4$ dB.

To break parallel transitions, one would need $\tilde{m} = 3$ (3 coded bits, so the coset level is 4 — but 16-QAM only has 4 levels, so this pushes to uncoded bits = 0 and all bits are coded). At $\tilde{m} = 3$ and $\nu = 4$: rate-$3/4$ code with $d_{\rm free}^{(H)} \geq 3$ gives Hamming-term $\geq 24$; parallel-transition term = $\Delta_4^2 = \infty$ (singleton). So $d_{\rm free}^2 \geq 24$; in unit energy $24/10 = 2.4$; gain $= 2.4/0.586 = 4.1 \;\Rightarrow\; 6.1$ dB. Actually worse than the $\tilde{m} = 2$ design — the Hamming term is now the bottleneck. This illustrates why 16-QAM TCM designs with $\tilde{m} = 2$ and 16–64 states are the sweet spot. $\blacksquare$

$d_{\rm free}^{(H)}$ is a property of the binary convolutional code: it measures the minimum number of time steps at which two distinct code paths disagree. In the TCM trellis, each disagreeing time step contributes at least $\Delta_1^2$ to the squared Euclidean distance between the two paths (because the two cosets are in different level-1 subsets, and the level-1 MSED is $\Delta_1^2$). Hence $$d_{\rm free}^2 \;\geq\; d_{\rm free}^{(H)} \cdot \Delta_1^2$$ for longer error events. The Hamming distance is the combinatorial enabler of the geometric distance bound.

So $d_{\rm free}^{(H)}$ and $\Delta_1^2$ are the two levers the designer has: the code provides the Hamming distance; the partition provides the per-position distance. $\blacksquare$

• Gray: flipping any bit moves to a geometric nearest-neighbor (distance $\Delta_0$). Hamming distance 1 corresponds to Euclidean distance $\Delta_0$ always.
• Ungerboeck: flipping the highest-order bit moves to a farthest neighbor (distance $\Delta_1$ at level 1). Hamming distance 1 on high-order bit corresponds to Euclidean distance $\Delta_1 > \Delta_0$.