Exercises

$\mathbf{X}_A = \begin{pmatrix} 1+j & -(1+j) \\ 1-j & 1-j \end{pmatrix} = \begin{pmatrix} 1+j & -1-j \\ 1-j & 1-j \end{pmatrix}.$ Wait, recheck: $-s_2^* = -(1-j)^* = -(1+j) = -1-j$. And $s_1^* = 1-j$. So $\mathbf{X}_A = \begin{pmatrix} 1+j & -1-j \\ 1-j & 1-j \end{pmatrix}$.

$(1,1)$ entry: $(1+j)(1-j) + (-1-j)(-1+j) = 2 + 2 = 4$. $(2,2)$ entry: $(1-j)(1+j) + (1-j)(1+j) = 2 + 2 = 4$. $(1,2)$ entry: $(1+j)(1+j) + (-1-j)(1+j)= (1+j)^2 - (1+j)^2 = 0$. $(2,1)$ entry: 0 by Hermitian symmetry. Hence $\mathbf{X}_A \mathbf{X}_A^H = 4\mathbf{I}_2$, and $|s_1|^2 + |s_2|^2 = 2 + 2 = 4$. $\blacksquare$

$\mathbb{E}[\text{SNR}_{\mathrm{eff}}^{\mathrm{Alamouti}}] = (\text{SNR}/2) \cdot \mathbb{E}[\sum_{i,j}|h_{i,j}|^2] = (\text{SNR}/2) \cdot n_t n_r = (10)/2 \cdot 4 = 20 \approx 13$ dB.

$\mathbb{E}[\text{SNR}_{\mathrm{eff}}^{\mathrm{MRC}}] = \text{SNR} \cdot \mathbb{E}[\sum |h_k|^2] = 10 \cdot 4 = 40 \approx 16$ dB.

MRC is 3 dB better in expected effective SNR — the power- splitting penalty. The diversity order (BER slope at high SNR) is the same, $d = 4$, for both. Alamouti $2\times 2$ and $1\times 4$ MRC are therefore equivalent up to this 3 dB shift.

Because $\mathbf{X}_A$ is linear in $(s_1, s_2)$, the difference $\boldsymbol{\Delta} = \mathbf{X}_A(s_1 - \hat s_1, s_2 - \hat s_2)$ is itself an Alamouti codeword with input $(\delta_1, \delta_2) = (s_1 - \hat s_1, s_2 - \hat s_2)$.

Thus $\boldsymbol{\Delta}\boldsymbol{\Delta}^H = (|\delta_1|^2 + |\delta_2|^2)\mathbf{I}_2$. If $(s_1, s_2) \ne (\hat s_1, \hat s_2)$ then at least one $\delta_q \ne 0$, so the coefficient is positive. Therefore $\boldsymbol{\Delta}\boldsymbol{\Delta}^H \succ 0$, which implies $\boldsymbol{\Delta}$ has rank $n_t = 2$.

By the rank criterion (Ch. 10 Thm. [?ch10:thm-rank-determinant]), the Alamouti code's diversity is $n_r \cdot 2 = 2 n_r$, the maximum possible for $n_t = 2$. This is a code-level (not distribution-level) statement: full diversity holds for every codeword pair, not just almost-every.

$y_2^* = -h_1^* s_2 + h_2^* s_1 + w_2^*$. Stack: $\tilde{\mathbf{y}} = \begin{pmatrix} y_1 \\ y_2^* \end{pmatrix} = \begin{pmatrix} h_1 & h_2 \\ h_2^* & -h_1^* \end{pmatrix} \begin{pmatrix} s_1 \\ s_2 \end{pmatrix} + \begin{pmatrix} w_1 \\ w_2^* \end{pmatrix}.$

$\tilde{\mathbf{H}}^H \tilde{\mathbf{H}} = (|h_1|^2 + |h_2|^2)\mathbf{I}_2$, confirming orthogonality.

$\tilde{\mathbf{H}}^H \tilde{\mathbf{y}}$ gives: $\tilde y_1 = h_1^* y_1 + h_2 y_2^*$ (for $s_1$), $\tilde y_2 = h_2^* y_1 - h_1 y_2^*$ (for $s_2$). Each is a scalar Gaussian observation of the corresponding $s_q$ scaled by $(|h_1|^2 + |h_2|^2)$, with i.i.d. $\mathcal{CN}(0, (|h_1|^2+|h_2|^2)\sigma^2)$ noise.

The ML decoder is two scalar slicers: $\hat s_q = \arg\min_{s\in\mathcal{X}} |s - \tilde y_q / (|h_1|^2 + |h_2|^2)|^2$ for $q = 1, 2$. For QPSK, this reduces to sign-thresholding the real and imaginary parts of $\tilde y_q$.

$\tilde{\mathbf{w}} = \tilde{\mathbf{H}}^H \tilde{\mathbf{w}}$ where $\tilde{\mathbf{w}} = (w_1, w_2^*)^T$ with i.i.d. $\mathcal{CN}(0, \sigma^2)$ entries.

$\mathrm{Cov}(\tilde{\mathbf{w}}) = \tilde{\mathbf{H}}^H \cdot \sigma^2 \mathbf{I}_2 \cdot \tilde{\mathbf{H}} = \sigma^2 \tilde{\mathbf{H}}^H \tilde{\mathbf{H}} = \sigma^2(|h_1|^2 + |h_2|^2)\mathbf{I}_2$. Off-diagonal zero, so the two components are uncorrelated.

For circularly symmetric complex Gaussian random vectors, uncorrelated implies independent. Since $\tilde{\mathbf{w}}$ is circularly Gaussian (its two entries are i.i.d. $\mathcal{CN}$), and $\tilde{\mathbf{H}}^H$ is a unitary transformation (up to scalar), the post-MF noise $\tilde{\mathbf{w}}$ is also circularly Gaussian and its uncorrelated components are independent. $\blacksquare$

$|(\mathbf{X}_A)_{1,1}|^2 = |s_1|^2$ (antenna 1, time 1). $|(\mathbf{X}_A)_{1,2}|^2 = |s_2|^2$ (antenna 1, time 2, since $|-s_2^*|^2 = |s_2|^2$). $|(\mathbf{X}_A)_{2,1}|^2 = |s_2|^2$ (antenna 2, time 1). $|(\mathbf{X}_A)_{2,2}|^2 = |s_1|^2$ (antenna 2, time 2). Total over 4 slots: $2|s_1|^2 + 2|s_2|^2 = 2(|s_1|^2 + |s_2|^2)$.

Averaging over the 2 time slots and 2 antennas per slot: per channel use, the 2 antennas together radiate $|s_1|^2 + |s_2|^2$ on average. For the constraint $E_s = \mathbb{E}[|s_1|^2 + |s_2|^2]$ this matches total per-slot energy $E_s$. Per antenna per channel use: $E_s/2$.

A $1\times n_r$ SIMO link radiates $E_s$ from its single antenna, while each of Alamouti's two antennas radiates only $E_s/2$. The MRC receiver at the SIMO link thus collects $\text{SNR} \cdot n_r$ in expected SNR, while Alamouti (with $n_t = 2, n_r$ fixed) collects $(\text{SNR}/2) \cdot 2 n_r = \text{SNR} n_r$ — the same expected effective SNR at the same total transmit energy but twice the diversity order (2 branches on transmit, $n_r$ branches on receive). That is why Alamouti beats SIMO: same array gain, more diversity.

By Ex. Eex-ch11-03, $\boldsymbol{\Delta}\boldsymbol{\Delta}^H = (|\delta_1|^2 + |\delta_2|^2)\mathbf{I}_2$. So $\|\boldsymbol{\Delta}\|_F^2 = \mathrm{tr}(\boldsymbol{\Delta} \boldsymbol{\Delta}^H) = 2(|\delta_1|^2 + |\delta_2|^2)$.

The minimum of $|\delta_1|^2 + |\delta_2|^2$ over distinct codeword pairs $(s_1, s_2) \ne (\hat s_1, \hat s_2)$ occurs when exactly one symbol changes by $\delta_{\min}$ (the QPSK minimum distance), so $\min_{\boldsymbol{\Delta}}\|\boldsymbol{\Delta}\|_F^2 = 2 \delta_{\min}^2$.

The coding gain is proportional to $\det(\boldsymbol{\Delta} \boldsymbol{\Delta}^H)^{1/n_t} = (|\delta_1|^2 + |\delta_2|^2)$. At the minimum, this is $\delta_{\min}^2$ — the same as QPSK's minimum Euclidean distance squared. Alamouti's gain is purely the diversity order jump from 1 (QPSK SISO) to $2 n_r$ (Alamouti MIMO); there is no improvement in per-symbol coding gain.

Row 1 squared-modulus sum: $|s_1|^2 + |s_2|^2 + |s_3|^2 + |s_4|^2 + |s_1|^2 + |s_2|^2 + |s_3|^2 + |s_4|^2 = 2(|s_1|^2 + |s_2|^2 + |s_3|^2 + |s_4|^2)$. Similar for rows 2 and 3.

Row 1 × Row 2$^H$ in the first 4 columns: $s_1 \overline{s_2} + (-s_2) \overline{s_1} + (-s_3) \overline{s_4} + (-s_4)(-\overline{s_3}) = \overline{s_2} s_1 - s_2 \overline{s_1} - s_3 \overline{s_4} + s_4 \overline{s_3} = 2j\,\mathrm{Im}(\overline{s_2} s_1) + 2j\, \mathrm{Im}(s_4 \overline{s_3})$. The second block of 4 columns contributes $+ s_1^* \overline{s_2^*} + \cdots = -$ (first block), so they cancel. Off-diagonals vanish.

For $n_t = 3$, the Liang-Tarokh bound gives $R_{\max}(3) = 3/4$. The above code achieves only $1/2$ — it is not rate-optimal. The rate-$3/4$ OSTBC for $n_t = 3$ requires a more complicated construction that Liang 2003 presents. The $T = 8$ block length here is longer than strictly necessary — using it we get a simpler code at the cost of rate efficiency. The production OSTBC for $n_t = 3, 4$ is the $T = 4$ rate-$3/4$ code of Ex. $3/4$ OSTBC for $n_t = 4$: Tarokh-Jafarkhani-Calderbank 1999" data-ref-type="example">ERate-$3/4$ OSTBC for $n_t = 4$: Tarokh-Jafarkhani-Calderbank 1999.

Note: for $n_t = 3$ the bound gives $R_{\max} = 1$ but this is the real case; for complex OSTBC at $n_t = 3$, the bound is $R_{\max} = 3/4$ (Tirkkonen-Hottinen 2002). The formula above is the coarse upper bound; the precise complex-case formula has slight corrections for odd $n_t$.

As $n_t \to \infty$: $R_{\max}(n_t) = (\lfloor n_t/2 \rfloor + 1)/(2 \lfloor n_t/2 \rfloor) = 1/2 + 1/(2\lfloor n_t/2 \rfloor) \to 1/2$. The complex-OSTBC rate asymptotes to $1/2$ — half the maximum possible Shannon-capacity rate $\min(n_t, n_r)$. This is the fundamental orthogonality penalty for large $n_t$.

$s_1 s_4^* = (1+j)(-1+j)/2 \cdot \mathrm{(conjugation)}$... Let me redo with the given inputs: $s_1 = (1+j)/\sqrt 2, s_4 = (-1+j)/\sqrt 2$; $s_1 s_4^* = (1+j)(-1-j)/2 = (-1-j-j-j^2)/2 = (-1-2j+1)/2 = -j$. $s_2 = (-1-j)/\sqrt 2, s_3 = (1-j)/\sqrt 2$; $s_2 s_3^* = (-1-j)(1+j)/2 = -(1+j)^2/2 = -(2j)/2 = -j$. $s_1 s_4^* - s_2 s_3^* = -j - (-j) = 0$.

$\mathrm{Re}(0) = 0$, so $\mathbf{E} = 0$. The QOSTBC Gramian for this specific input is block-diagonal scalar: $\mathbf{X}_Q \mathbf{X}_Q^H = 4 \mathbf{I}_4$ — as if full-rate full-diversity. This is the "lucky" input discussed in Ex. EJafarkhani QOSTBC for QPSK: The 4-Symbol Codeword: for a generic QPSK input the off-block is not zero and the ML decoder has to work pair-wise.

In the unrotated Jafarkhani QOSTBC, an error event that changes $(s_1, s_3)$ but leaves $(s_2, s_4)$ fixed produces $\boldsymbol{\Delta}$ of rank at most 2: the column space is spanned by 2 vectors.

Pre-rotate: let $(s_1', s_3') = e^{j\theta}(s_1, s_3)$ for an irrational $\theta$ (say $\pi/4$ or $\arctan(2)$). Now the "only $(s_1, s_3)$ change" error pattern becomes a rotation that mixes $(s_1, s_3)$ with all four underlying coordinates after expanding the complex rotation in real / imaginary parts.

After rotation, the worst-case error matrix $\boldsymbol{\Delta}$ has minimum rank 4 — because the rotation distributes each rotated pair's error over all 4 row dimensions of $\mathbf{X}_Q$. Sharma- Papadias 2003 proves this rigorously for specific $\theta$. Therefore the diversity becomes $4 n_r$, full MIMO diversity.

The decoder still requires pair-wise ML search ($O(M^2)$ per pair), but now over the rotated constellation. This is the rotated QOSTBC — rate 1, full diversity, pair-wise ML.

$s_1 = \alpha_1 + j\beta_1, s_2 = \alpha_2 + j\beta_2$. $-s_2^* = -\alpha_2 + j\beta_2$. $s_1^* = \alpha_1 - j\beta_1$.

$\mathbf{A}_1$ (coefficient of $\alpha_1$): $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. $\mathbf{B}_1$ (coefficient of $\beta_1$): $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ (so that $j\beta_1 \mathbf{B}_1$ contributes $j\beta_1$ at $(1,1)$ and $-j\beta_1$ at $(2,2)$, matching $s_1 = \alpha_1 + j\beta_1$ in $(1,1)$ and $s_1^* = \alpha_1 - j\beta_1$ in $(2,2)$). $\mathbf{A}_2$ (coefficient of $\alpha_2$): $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. $\mathbf{B}_2$ (coefficient of $\beta_2$): $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$.

$\mathbf{X}_A = \alpha_1 \mathbf{A}_1 + j\beta_1 \mathbf{B}_1 + \alpha_2 \mathbf{A}_2 + j\beta_2 \mathbf{B}_2 = \begin{pmatrix} \alpha_1 + j\beta_1 & -\alpha_2 + j\beta_2 \\ \alpha_2 + j\beta_2 & \alpha_1 - j\beta_1 \end{pmatrix} = \begin{pmatrix} s_1 & -s_2^* \\ s_2 & s_1^* \end{pmatrix}.$ $\blacksquare$

Verify Hurwitz-Radon: $\mathbf{A}_1\mathbf{A}_2^T = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ and $\mathbf{A}_2\mathbf{A}_1^T = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, so $\mathbf{A}_1 \mathbf{A}_2^T + \mathbf{A}_2 \mathbf{A}_1^T = 0$ — the anti-commutation relation is satisfied.

$(n_t, T) = (2, 2)$: $Q = 2 \cdot 2 \cdot 2 = 8$. $(4, 4)$: $Q = 2 \cdot 4 \cdot 4 = 32$. $(4, 8)$: $Q = 2 \cdot 4 \cdot 8 = 64$.

$Q = 2K$, so $K = Q/2$ and $R = K/T = Q/(2T)$. $(2,2)$: $R = 8/4 = 2$ symbols/cu — matches the capacity-achieving rate $\min(n_t, n_r) = 2$ for $n_r \ge 2$. $(4,4)$: $R = 32/8 = 4$ — matches $\min(n_t, n_r) = 4$. $(4,8)$: $R = 64/16 = 4$ — same as $(4,4)$ but twice the block length (lower complexity per symbol).

In each case, the capacity-achieving LDC requires $Q = 2 n_t T$ dispersion matrices — the full dimension of the codeword space. Smaller $Q$ corresponds to lower rate or a structured sub-code (e.g., Alamouti is $Q = 4$, rate 1 at $(2,2)$, which is half the capacity-achieving dimension and delivers only half the capacity at high SNR).

By Thm. $2 n_r$ with Linear Decoding" data-ref-type="theorem">TAlamouti Achieves Full Diversity $2 n_r$ with Linear Decoding, the Alamouti receiver sees two parallel scalar channels, each with effective SNR $\text{SNR}/2 (|h_1|^2 + |h_2|^2)$. Two parallel independent Gaussian channels with the same effective SNR $\gamma_{\mathrm{eff}}$ carry a total rate $2 \log_2(1 + \gamma_{\mathrm{eff}})$ bits per vector use. The vector use is 2 channel uses, so the per-channel-use rate is $\log_2(1 + \gamma_{\mathrm{eff}}) = \log_2(1 + (\text{SNR}/2)(|h_1|^2 + |h_2|^2))$, averaged over channel realisations.

$C_{\mathrm{erg}} = \mathbb{E}[\log_2 \det(\mathbf{I}_1 + (\text{SNR}/2)\mathbf{H}\mathbf{H}^{H})] = \mathbb{E}[\log_2(1 + (\text{SNR}/2)(|h_1|^2 + |h_2|^2))]$. Exactly equal to Alamouti's MI — Alamouti is capacity-achieving at $n_t = 2, n_r = 1$.

At $n_r > 1$, $\mathbf{H}\mathbf{H}^{H}$ has $\min(n_r, 2) = 2$ positive eigenvalues $\lambda_1, \lambda_2$. The capacity is $\mathbb{E} [\log_2(1 + (\text{SNR}/2)\lambda_1) + \log_2(1 + (\text{SNR}/2) \lambda_2)]$ — two waterfilling terms at high SNR scaling as $2 \log_2(\text{SNR})$. But Alamouti sees only one effective-SNR scalar channel (the sum $|h_1|^2 + \cdots$), scaling as $\log_2(\text{SNR})$ at high SNR. Hence Alamouti achieves capacity only at $n_r = 1$ (SIMO multiplexing gain = 1); at $n_r \ge 2$ it loses a factor of 2 in high-SNR multiplexing gain.

This is the fundamental multiplexing loss of Alamouti: it is capacity-optimal only when the receive dimension is 1.

The effective SNR is $\gamma = (\text{SNR}/2) \sum_{i=1}^{2} \sum_{j=1}^{2} |h_{i,j}|^2 \sim (\text{SNR}/2)\chi^2_{4}/2$ where $\chi^2_{4}$ is chi-squared with 4 d.o.f. (since each $|h|^2$ is exponential with mean 1, equivalently $\chi^2_2/2$). So $\gamma \sim (\text{SNR}/4)\chi^2_{4}$.

$P_e = \mathbb{E}_\gamma[2Q(\sqrt{\gamma})] = \int_0^\infty 2 Q(\sqrt{\gamma}) f_\gamma(\gamma) d\gamma$. Using standard Rayleigh-averaging, for $n_r = 2$ this evaluates to $P_e \approx (3/2)[1 - \mu]^2 [2 + \mu]/2$, with $\mu = \sqrt{\bar\gamma / (1 + \bar\gamma)}$ and $\bar\gamma = \text{SNR}/4$. At $\text{SNR} = 10$: $\bar\gamma = 2.5, \mu = \sqrt{2.5/3.5} = \sqrt{5/7} \approx 0.845$. Numerically $P_e \approx 2\times 10^{-3}$.

The approximation gives $P_e \approx 3/1600 \approx 1.9 \times 10^{-3}$ — within 5% of the exact value. The high-SNR approximation is excellent for $\text{SNR} \gtrsim 10$ dB.

An OSTBC has $Q = 2K$ real dispersion matrices, each a Hurwitz-Radon matrix (antipodal + orthogonal). These matrices act on the $n_t$-dimensional real row space. The Hurwitz-Radon-Eckmann bound says the maximum number of such anti-commuting orthogonal matrices is $\mathcal{H}(n_t)$.

For an OSTBC with $K$ complex symbols: need $2K \le \mathcal{H}(n_t)$. For rate 1, $K = T \cdot 1 = T$, so $2T \le \mathcal{H}(n_t)$. With the optimal block length $T \le n_t$ (codeword square/rectangular), we need $2 n_t \le \mathcal{H}(n_t)$, i.e., $\mathcal{H}(n_t) \ge 2 n_t$.

From the table in Def. $\mathcal{H}(n)$" data-ref-type="definition">DHurwitz-Radon Number $\mathcal{H}(n)$: $\mathcal{H}(1) = 1 < 2, \mathcal{H}(2) = 2 = 2 \cdot 1, \mathcal{H}(3) = 2 < 6, \mathcal{H}(4) = 4 < 8, \mathcal{H}(8) = 8 < 16$.

Only $n_t = 2$ has $\mathcal{H}(n_t) = 2 n_t$. For $n_t \ge 3$, $\mathcal{H}(n_t) < 2 n_t$, so rate 1 is algebraically impossible. Alamouti is therefore the unique full-rate full-diversity complex OSTBC (up to equivalence). $\blacksquare$

For $n_t = n_r = 2, \text{SNR} = 15$ dB $\approx 31.6$: $C_{\rm erg} = \mathbb{E}_\mathbf{H}[\log_2\det(\mathbf{I}_2 + (\text{SNR}/2)\mathbf{H}\mathbf{H}^{H})]$. For i.i.d. Rayleigh, numerical evaluation gives $C_{\rm erg} \approx 9.3$ bits/cu.

$I_{\rm Alamouti} = \mathbb{E}[\log_2(1 + (\text{SNR}/2)\|\mathbf{H}\|_F^2)]$. At $\text{SNR} = 15$ dB, $\mathbb{E}[\|\mathbf{H}\|_F^2] = n_t n_r = 4$, so $I_{\rm Alamouti} \approx \log_2(1 + 31.6 \cdot 2) = \log_2(64.2) \approx 6$ bits/cu (rough; the exact average is $\approx 5.5$).

Alamouti loses $\approx 3.8$ bits/cu — about 40% of the capacity — at $\text{SNR} = 15$ dB.

An optimised Hassibi-Hochwald rate-2 LDC achieves within $\lesssim 0.3$ bits/cu of the ergodic capacity at $\text{SNR} = 15$ dB — approximately 9.0 bits/cu — using a lattice decoder. This is a factor of $\sim 1.6\times$ the Alamouti rate at the same SNR, confirming that the orthogonality cost of Alamouti is real: full-rate full-diversity structures cannot coexist with linear scalar decoding, and LDCs give the designer the knob to trade.

For an OSTBC, $\boldsymbol{\Delta} = \sum_q (\delta_q^{Re} \mathbf{A}_q + j \delta_q^{Im} \mathbf{B}_q)$ where $\delta_q = s_q - \hat s_q$. The OSTBC property gives $\boldsymbol{\Delta}\boldsymbol{\Delta}^H = (\sum_q |\delta_q|^2) \mathbf{I}_{n_t} = d^2 \mathbf{I}_{n_t}$ with $d^2 = \sum_q |\delta_q|^2$.

All 4 eigenvalues of $\boldsymbol{\Delta}\boldsymbol{\Delta}^H$ are equal to $d^2$. The Chernoff bound becomes $P(\mathbf{X}\to\hat{\mathbf{X}}) \le \prod_{i=1}^4 (1 + (\text{SNR} d^2/4))^{-n_r} = (1 + (\text{SNR} d^2/4))^{-4 n_r}$.