Exercises

$C = \mathbb{E}_{\mathbf{H}}[\log_2\det(\mathbf{I}_{n_r} + (\text{SNR}/n_t)\mathbf{H}\mathbf{H}^{H})]$ bits/channel use, where $\mathbf{H}\in\mathbb{C}^{n_r\times n_t}$ has i.i.d. $\mathcal{CN}(0,1)$ entries.

The SVD $\mathbf{H} = \mathbf{U}\boldsymbol{\Sigma}\mathbf{V}^H$ with $\sigma_i \ge 0$ diagonalises the log-det: $\log\det(\cdot) = \sum_{i=1}^{\min(n_t,n_r)}\log_2(1+(\text{SNR}/n_t) \sigma_i^2)$. At large SNR, each log is $\log_2(\text{SNR}/n_t) + \log_2\sigma_i^2 + o(1)$; summing over $\min(n_t, n_r)$ non-zero singular values gives $C = \min(n_t,n_r)\log_2\text{SNR} + O(1)$. $\blacksquare$

$P_{\mathrm{out}}(R) = \Pr[\log_2\det(\mathbf{I} + (\text{SNR}/n_t) \mathbf{H}\mathbf{H}^{H}) < R]$, and $C_\epsilon = \sup\{R : P_{\mathrm{out}}(R) \le \epsilon\}$, the $\epsilon$-quantile of the random capacity.

On a block-fading channel, a codeword sits on a single random $\mathbf{H}$. For any rate $R > 0$ the event $\{C(\mathbf{H}) < R\}$ has strictly positive probability, and on this event no code achieves low error. Ergodic capacity averages over fades assuming the code spans many — not the case here. The operational rate is therefore the largest $R$ such that the bad event $P_{\mathrm{out}}(R)$ is tolerable, i.e., $C_\epsilon$. $\blacksquare$

With $\mathbf{Y} = \mathbf{H}\mathbf{X} + \mathbf{W}$, $\mathbf{W} \sim \mathcal{CN}(0, \sigma^2\mathbf{I})$, the ML decoder prefers $\hat{\mathbf{X}}$ to $\mathbf{X}$ iff $\|\mathbf{Y} - \mathbf{H}\hat{\mathbf{X}}\|_F^2 \le \|\mathbf{Y} - \mathbf{H}\mathbf{X}\|_F^2$. Compute the difference $D = \|\mathbf{Y} - \mathbf{H}\hat{\mathbf{X}}\|_F^2 - \|\mathbf{Y} - \mathbf{H}\mathbf{X}\|_F^2$.

Substituting $\mathbf{Y} = \mathbf{H}\mathbf{X} + \mathbf{W}$: $D = \|\mathbf{H}\boldsymbol{\Delta} + \mathbf{W}\|_F^2 - \|\mathbf{W}\|_F^2 = \|\mathbf{H}\boldsymbol{\Delta}\|_F^2 + 2\Re\mathrm{tr}(\boldsymbol{\Delta}^H\mathbf{H}^{H}\mathbf{W})$, with $\boldsymbol{\Delta} = \hat{\mathbf{X}} - \mathbf{X}$ (flipped sign). Given $\mathbf{H}, \mathbf{X}$, the first term is deterministic and the second is Gaussian, zero-mean, with variance $2\sigma^2\|\mathbf{H}\boldsymbol{\Delta}\|_F^2$.

$P(\mathbf{X} \to \hat{\mathbf{X}} \mid \mathbf{H}) = \Pr[D \le 0] = \Pr[Z \ge \|\mathbf{H}\boldsymbol{\Delta}\|_F^2 / \sqrt{2\sigma^2\|\mathbf{H}\boldsymbol{\Delta}\|_F^2}]$ with $Z$ standard normal, which simplifies to $Q(\|\mathbf{H} \boldsymbol{\Delta}\|_F/\sqrt{2\sigma^2}) = Q(\sqrt{\text{SNR} \|\mathbf{H}\boldsymbol{\Delta}\|_F^2/(2 n_t)})$ after plugging $\text{SNR} = P/\sigma^2$ and the per-antenna power normalisation $P \to P/n_t$. Note: the sign of $\boldsymbol{\Delta}$ doesn't matter inside the Frobenius norm. $\blacksquare$

From Exercise Eex-ch10-03, $P(\mathbf{X}\to\hat{\mathbf{X}} \mid\mathbf{H}) \le \tfrac{1}{2}\exp(-(\text{SNR}/4n_t) \|\mathbf{H}\boldsymbol{\Delta}\|_F^2)$.

$\|\mathbf{H}\boldsymbol{\Delta}\|_F^2 = \sum_{j=1}^{n_r} \mathbf{h}_j^H \mathbf{A} \mathbf{h}_j$ with $\mathbf{A} = \boldsymbol{\Delta}\boldsymbol{\Delta}^H$, $\mathbf{h}_j \sim \mathcal{CN}(0,\mathbf{I}_{n_t})$ i.i.d.

Diagonalise $\mathbf{A} = \mathbf{V}\mathbf{\Lambda}\mathbf{V}^H$; with $\tilde{\mathbf{h}}_j = \mathbf{V}^H\mathbf{h}_j \sim \mathcal{CN}(0,\mathbf{I})$, $\mathbf{h}_j^H\mathbf{A}\mathbf{h}_j = \sum_{i=1}^r \lambda_i|\tilde h_{ji}|^2$, a sum of $r$ i.i.d. exponential random variables weighted by $\lambda_i$. Each $|\tilde h_{ji}|^2 \sim \mathrm{Exp}(1)$ with MGF $(1 - t)^{-1}$ for $t < 1$. Independence across $j$ and $i$: $\mathbb{E}[e^{-(\text{SNR}/4n_t)\|\mathbf{H}\boldsymbol{\Delta}\|_F^2}] = \prod_i (1 + (\text{SNR}/4n_t)\lambda_i)^{-n_r}$.

Taking expectation of the conditional Chernoff bound over $\mathbf{H}$: $P(\mathbf{X}\to\hat{\mathbf{X}}) \le \tfrac{1}{2} \prod_i(1+(\text{SNR}/4n_t)\lambda_i)^{-n_r}$; dropping the $1/2$ gives the stated form. $\blacksquare$

The diversity order of a space-time code on an i.i.d. Rayleigh MIMO channel is $d = r_{\min} \cdot n_r$, where $r_{\min}$ is the minimum rank of $\boldsymbol{\Delta} = \mathbf{X} - \hat{\mathbf{X}}$ over all codeword pairs. Maximum diversity $n_t n_r$ requires $r_{\min} = n_t$ (full rank for all pairs).

For $r_{\min} = 3, n_r = 2$: $d = 3\cdot 2 = 6$. Maximum possible is $n_t n_r = 4\cdot 2 = 8$. The code is rank- deficient and loses $2$ units of diversity — $2$ steps of slope on the log-log PEP curve. $\blacksquare$

$\gamma_c = 4^{1/2} = 2$. The second code has $\gamma_c' = 16^{1/2} = 4$ — twice as large.

Doubling $\det_{\min}$ shifts the BER curve left by $3 n_r/n_t = 3\cdot 2/2 = 3$ dB. Since $\det_{\min}' = 4\det_{\min}$ (two doublings), the shift is $6$ dB — a substantial gain at the same rate and diversity. $\blacksquare$

$\det\boldsymbol{\Delta} = e_1 \cdot e_1^* - (-e_2^*)\cdot e_2 = |e_1|^2 + |e_2|^2$, a real non-negative number.

$\det(\boldsymbol{\Delta}\boldsymbol{\Delta}^H) = (\det\boldsymbol{\Delta}) (\det\boldsymbol{\Delta})^* = (|e_1|^2 + |e_2|^2)^2$. This is strictly positive whenever $(e_1, e_2) \ne (0,0)$, so rank is $2$. Moreover, the value depends only on $|e_1|^2 + |e_2|^2 = \|\boldsymbol{\Delta}\|_F^2/2$ — the squared Euclidean distance between symbol pairs, not the specific symbols — so the Alamouti code is geometry-invariant under symbol permutations: a clean mathematical feature. $\blacksquare$

At fixed $R = 6$, as $\text{SNR} \to \infty$, the ratio $R/\log_2 \text{SNR} \to 0$ — the code operates at zero multiplexing gain ($r = 0$).

By Theorem $P_{\mathrm{out}}(R)$ (DMT Preview)" data-ref-type="theorem">THigh-SNR Scaling of $P_{\mathrm{out}}(R)$ (DMT Preview), $d^\star(0) = n_t n_r = 3\cdot 3 = 9$. The outage probability decays as $\text{SNR}^{-9}$ at high SNR — slope $-9$ in log-log.

This is the maximum MIMO diversity; a $3\times 3$ space-time code with full rank achieves the matching BER slope. The outage curve is the ceiling that space-time codes chase, and the rank criterion of §4 is the algebraic condition for a code to hit it. $\blacksquare$

$\boldsymbol{\Delta} = (s-\hat s)(1,1)^T$, a rank-$1$ matrix.

$d = 1\cdot n_r = n_r$, half of the Alamouti $2n_r$. At high SNR, the spatial-repetition BER slope is $\text{SNR}^{-n_r}$, whereas Alamouti's is $\text{SNR}^{-2n_r}$.

Spatial repetition wastes exactly one order of diversity by collapsing the two transmit antennas to a single effective beamformer. The fix is to transmit linearly independent functions on the two antennas — Alamouti's design. $\blacksquare$

$\eta = (\log_2 4 \cdot 3)/4 = 6/4 = 1.5$ bits/channel use per transmit stream. Per spatial dimension: $\eta/n_t = 0.5$ bits. The rate per complex symbol is $3/4$, matching Tarokh- Jafarkhani-Calderbank's maximum-rate theorem for complex OSTBC with $n_t > 2$.

Complex OSTBCs encode $k$ complex symbols into an $n_t\times T$ matrix with orthogonal rows. Tarokh-Jafarkhani-Calderbank 1999 proved that the maximum rate $k/T$ for complex symbols is $1$ only for $n_t = 2$ (Alamouti), $3/4$ for $n_t = 3, 4$, and smaller for $n_t > 4$. The proof uses representation theory of Clifford algebras — see Ch. 11. $\blacksquare$

The nonzero eigenvalues of $\boldsymbol{\Delta}\boldsymbol{\Delta}^H$ (a $n_t\times n_t$ PSD matrix, full rank assumed) are $\lambda_1, \ldots, \lambda_{n_t}$. AM-GM: $\prod_i \lambda_i \le ((\sum_i \lambda_i)/n_t)^{n_t}$.

$\sum_i \lambda_i = \mathrm{tr}(\boldsymbol{\Delta}\boldsymbol{\Delta}^H) = \|\boldsymbol{\Delta}\|_F^2$.

$\det(\boldsymbol{\Delta}\boldsymbol{\Delta}^H) = \prod_i\lambda_i \le (\|\boldsymbol{\Delta}\|_F^2/n_t)^{n_t}$. Equality iff all $\lambda_i$ equal, i.e., $\boldsymbol{\Delta}\boldsymbol{\Delta}^H = \lambda\mathbf{I}$ for some $\lambda > 0$.

Designer's corollary: for a fixed transmit power budget (which upper-bounds $\|\boldsymbol{\Delta}\|_F^2$ on worst pairs), maximising $\det_{\min}$ is equivalent to forcing the eigenvalues of $\boldsymbol{\Delta}\boldsymbol{\Delta}^H$ to be as equal as possible — the AM-GM-optimal structure that Alamouti achieves via orthogonality. $\blacksquare$

$\log P_e = -n_t n_r \log(\text{SNR}/4n_t) - n_r \log\det_{\min} + \text{const}$. At target $P_e$: $n_t n_r \log\text{SNR} = -n_r \log\det_{\min} + C$. So $\log\text{SNR} = -(1/n_t)\log\det_{\min} + C/(n_t n_r)$.

$\Delta\log\text{SNR} = -(1/n_t)\Delta\log\det_{\min}$. For $\det_{\min}\to 2\det_{\min}$ (doubling), $\Delta\log\det_{\min} = \log 2 = \ln 2 / \ln e$.

$\Delta\text{SNR (dB)} = 10\Delta\log_{10}\text{SNR} = (10/\ln 10)\Delta\ln\text{SNR} = -(10/n_t\ln 10)\ln 2$. With $10\ln 2/\ln 10 = 3.010$ dB, $\Delta\text{SNR} = -3/n_t$ dB — a left shift of $3/n_t$ dB per doubling of $\det_{\min}$. Multiplying by the receive-antenna factor $n_r$ (since the PEP exponent is $n_r$ times each eigenvalue contribution) gives $\Delta\text{SNR} = 3 n_r/n_t$ dB total improvement per doubling. $\blacksquare$

$P_e \approx c_a \cdot 100^{-2} = 10^{-4} c_a$. With a typical $c_a \sim 1$, $P_e \approx 10^{-4}$.

$P_e \approx c_b \cdot 100^{-4}/4^2 = c_b \cdot 10^{-8}/16 \approx 6\cdot 10^{-10} c_b$. With $c_b\sim 10$ (more codeword pairs), $P_e \sim 6\cdot 10^{-9}$.

$P_e \approx c_c \cdot 100^{-4}/16^2 = c_c \cdot 10^{-8}/256 \approx 4\cdot 10^{-11} c_c$. With $c_c\sim 10$, $P_e\sim 4\cdot 10^{-10}$, about $16\times$ better than code (b) (confirming the dB-shift calculation: $16 = 2^4$, so $4$ doublings of $\det_{\min}$, or $3n_r/n_t \cdot 4 = 6$ dB of horizontal shift).

At high SNR the rank is decisive (code (a) is catastrophically worse); within full-rank, $\det_{\min}$ delivers the coding gain. Both observations match the rank and determinant criteria. $\blacksquare$

By Exercise Eex-ch10-11, $\det(\boldsymbol{\Delta} \boldsymbol{\Delta}^H) \le (\|\boldsymbol{\Delta}\|_F^2/n_t)^{n_t}$ for any full-rank pair. The minimum is achieved on the pair with smallest determinant; among these, the Frobenius norm may or may not be minimal. If the minimum-$\det$ pair is also the minimum- $d_E$ pair (often true for "symmetric" codes), then $\det_{\min} \le (d_E^2/n_t)^{n_t}$.

Equality holds iff the minimum-$d_E$ error matrix has equal eigenvalues, $\boldsymbol{\Delta}\boldsymbol{\Delta}^H = (d_E^2/n_t) \mathbf{I}$. Alamouti achieves this on every one-symbol-error pair (Example EAlamouti Code: Verify Full Rank Over All Symbol Pairs), so it is isoperimetrically optimal among $2\times 2$ STCs at the symbol-distance level.

This is the coding analogue of the isoperimetric inequality: among codewords with a fixed Euclidean-distance budget, the best coding gain is achieved by codes with maximally spread-out eigenvalues (equal $\lambda_i$). Codes that cluster eigenvalues waste distance-budget on low-diversity modes. $\blacksquare$

Alamouti-QPSK: $R = 2 < 3.9 \approx C_{0.1}$. The code operates well below the $10\%$-outage capacity, so the outage probability at rate $2$ is much smaller than $10\%$.

Alamouti is full-rank ($r = 2$), so its diversity order is $r \cdot n_r = 4 = n_t n_r$, matching the outage exponent $d^\star(0) = 4$. At high SNR, both $P_e$ (code error) and $P_{\mathrm{out}}(R = 2)$ decay as $\text{SNR}^{-4}$ — the code is diversity-optimal.

Operating below $C_\epsilon$ means the outage event is rare; operating with matching diversity exponent means the BER slope tracks the outage slope. At $\text{SNR} = 10$ dB and rate $R = 2$, Alamouti with QPSK and ML decoding (essentially trivial for Alamouti — linear) has BER $\sim 10^{-3}$, and each additional $3$ dB reduces this by a factor of $2^4 = 16$. The code is approaching the channel ceiling. $\blacksquare$

$P_e^{\rm GC} \lesssim K \text{SNR}^{-4}/\delta(M)^2$. As $M$ grows (higher rate), $K$ and $\delta(M)$ are roughly constant — the Golden Code's coding gain does not vanish with rate. This is the non-vanishing determinant property.

$P_e^{\rm naive} \lesssim K \text{SNR}^{-4}/(1/M)^2 = K M^2 \text{SNR}^{-4}$. As $M\to\infty$, the constant explodes quadratically in rate: each doubling of $M$ loses $6$ dB of coding gain at any fixed $\text{SNR}$.

At rate $R = 2\log_2 M$ the Golden Code has fixed coding gain; the naive code loses coding gain as $M\to\infty$. Forward reference: the Elia-Kumar-Pawar-Kumar-Caire 2006 CommIT contribution generalises this to arbitrary $(n_t, n_r)$ — CDA codes with non-vanishing determinant that achieve the entire DMT curve (Ch. 13). The rank-determinant framework of this chapter is the entry point; Ch. 13 is the full development. $\blacksquare$

Any code of rate $R$ cannot decode reliably when the channel supports conditional capacity $C(\mathbf{H}) < R$. Therefore, the probability of (block) error is at least the outage probability: $P_e \ge P_{\mathrm{out}}(R)$.

At fixed $R$ and $\text{SNR} \to \infty$, the multiplexing gain $r = R/\log_2\text{SNR} \to 0$. By Theorem $P_{\mathrm{out}}(R)$ (DMT Preview)" data-ref-type="theorem">THigh-SNR Scaling of $P_{\mathrm{out}}(R)$ (DMT Preview), $P_{\mathrm{out}}(R) \doteq \text{SNR}^{-d^\star(0)} = \text{SNR}^{-n_t n_r}$.

$P_e \ge P_{\mathrm{out}}(R) \doteq \text{SNR}^{-n_t n_r}$, so $\log P_e \ge -n_t n_r \log\text{SNR} + o(\log\text{SNR})$, and $d = -\lim\log P_e/\log\text{SNR} \le n_t n_r$. No code can outperform this ceiling. The rank criterion's "maximum diversity $n_t n_r$" is therefore not merely the ceiling for full-rank codes, but the ceiling among all codes. $\blacksquare$

$\mathbf{X}(s_1, s_2) = \begin{pmatrix} \alpha_1+j\beta_1 & -(\alpha_2-j\beta_2) \\ \alpha_2+j\beta_2 & \alpha_1-j\beta_1 \end{pmatrix}$.

Grouping: $\mathbf{X} = \alpha_1 \mathbf{A}_1 + j\beta_1 \mathbf{B}_1 + \alpha_2 \mathbf{A}_2 + j\beta_2 \mathbf{B}_2$ with $\mathbf{A}_1 = \begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}$, $\mathbf{B}_1 = \begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}$, $\mathbf{A}_2 = \begin{pmatrix}0 & -1\\1 & 0\end{pmatrix}$, $\mathbf{B}_2 = \begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}$. So $Q = 2$ complex symbols (or $4$ real symbols), and the LDC form exposes Alamouti as a linear combination of four fixed matrices.

The LDC framework is the most general linear STC structure: every linear code fits this template. Hassibi-Hochwald 2002 then ask "what $\{\mathbf{A}_q, \mathbf{B}_q\}$ maximise the coding gain $\det_{\min}$ subject to full rank?" — a joint optimisation problem solved via gradient methods and capacity-optimised designs. The rank-determinant framework of this chapter is the theoretical target; LDC is the implementation vehicle. $\blacksquare$

Alamouti-QPSK: $R = 2 < 3.9 = C_{0.1}$, safely operating below the $10\%$-outage capacity. Alamouti-16QAM: $R = 4 > 3.9$, above $C_{0.1}$ — outage exceeds $10\%$ at $10$ dB.

QPSK's outage at $R = 2$ is much smaller than $10\%$ (probably $\sim 10^{-2}$ at $10$ dB), so the codeword error is dominated by the PEP bound on non-outage realisations — where Alamouti's full-diversity PEP is rapidly decaying. 16-QAM at rate $4$ is operating near or slightly above $C_{0.1}$; outage events dominate the error floor, so the BER will plateau even as $\text{SNR}\to\infty$ (until the rate drops below the growing outage capacity).

The designer chooses the constellation based on the operational SNR and outage tolerance. Ch. 11 generalises this to full orthogonal STBCs; Ch. 13 introduces rate-optimal codes that approach $C_\epsilon$ via the CDA / Golden Code constructions. $\blacksquare$

Code A (Alamouti QPSK, rate $2$): $P_e \sim c_A \text{SNR}^{-4} / 16^2 = c_A \text{SNR}^{-4}/256$. Code B (V-BLAST QPSK, rate $4$, rank-$1$ one-symbol-error matrices): $P_e \sim c_B \text{SNR}^{-2}$. Constants $c_A, c_B$ depend on constellation, decoder.

At high SNR, the $\text{SNR}^{-4}$ slope of Code A decays faster; Code A eventually wins. The crossover is where $c_A/256 = c_B/\text{SNR}^{2}$, i.e., $\text{SNR}^{2} = 256 c_B/c_A$. For $c_A = c_B$, $\text{SNR} = 16 = 12$ dB. Above $12$ dB, Code A dominates; below, Code B is competitive (if it matters — both codes have significant BER at this SNR).

This exercise is a preview of the Zheng-Tse DMT (Ch. 12): Code A trades rate ($2$ vs $4$) for diversity ($4$ vs $2$); Code B does the opposite. On the DMT curve, Code A sits near $(r=0, d=4)$ and Code B near $(r=2, d=2)$. A DMT-optimal code would operate on the tradeoff line connecting them (specifically at $(r=1, d=1)$ or $(r=2, d=0)$ endpoints, or any interior point). Ch. 12 formalises and solves this. $\blacksquare$