Ferkans — Interactive Telecom Tutor

Why Do We Lose Rate Beyond $n_t = 2$ ?

For $n_t = 2$ , Alamouti is an OSTBC at rate 1 (one complex symbol per channel use). It is natural to ask: does rate 1 generalise to larger $n_t$ ? The answer is a surprisingly firm no — and the reason goes back to a 19th-century theorem about sums of squares.

Hurwitz (1898) and Radon (1922) classified the $n\times n$ real matrices that satisfy anti-commutation relations: there exist $\rho(n)$ orthogonal matrices $\mathbf{A}_1, \ldots, \mathbf{A}_{\rho(n)}$ with $\mathbf{A}_p\mathbf{A}_q^T = -\mathbf{A}_q\mathbf{A}_p^T$ (for $p\ne q$ ) if and only if $\rho(n) \le n$ with equality only when $n \in \{1, 2, 4, 8\}$ — the dimensions of the real division algebras $\mathbb{R}, \mathbb{C}, \mathbb{H}, \mathbb{O}$ . This limits the number of dispersion matrices we can pack into an OSTBC, and hence the rate.

For real symbols (PAM input), the Hurwitz-Radon bound is tight: rate 1 is achievable for $n_t = 1, 2, 4, 8$ and drops off otherwise. For complex symbols (QAM/PSK — the practical case), we need twice as many orthogonal matrices (one set for the real parts, one for the imaginary), so the rate drops to $3/4$ for $n_t = 3, 4$ and even lower for larger $n_t$ . The Liang-Tarokh 2003 bound makes this tight: no complex OSTBC can exceed it, and specific constructions meet it.

This is not just a mathematical curiosity — it is the fundamental limitation that motivates quasi-orthogonal codes (§4) and linear dispersion codes (§5): if you want rate 1 at $n_t > 2$ you must abandon OSTBC orthogonality, and you will pay with either decoder complexity or diversity.

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Definition:
Hurwitz-Radon Number $\mathcal{H}(n)$

The Hurwitz-Radon number $\mathcal{H}(n)$ is the maximum number of real $n \times n$ matrices $\mathbf{A}_1, \ldots, \mathbf{A}_k$ that are each orthogonal ( $\mathbf{A}_q^T\mathbf{A}_q = \mathbf{I}$ ) and pairwise anti-commuting ( $\mathbf{A}_p\mathbf{A}_q = -\mathbf{A}_q\mathbf{A}_p$ for $p\ne q$ ). If $n = 2^{a+4b}(2c+1)$ with $0 \le a \le 3$ and $b, c \ge 0$ integers, then by the Hurwitz-Radon-Eckmann theorem $\mathcal{H}(n) \;=\; 2^a + 8b \;\le\; n,$ with equality if and only if $n \in \{1, 2, 4, 8\}$ .

Small values:

$n$	1	2	3	4	5	6	7	8
$\mathcal{H}(n)$	1	2	2	4	2	2	2	8

$\mathcal{H}(n)$ is the "algebraic room" for orthogonal-style space-time codes. For $n_t = 2$ : $\mathcal{H}(2) = 2$ , which is enough for Alamouti's 4 matrices (2 real $\mathbf{A}$ 's + 2 imaginary $\mathbf{B}$ 's, each pair anti-commuting with their partner). For $n_t = 4$ : $\mathcal{H}(4) = 4$ , which allows the rate- $3/4$ TJC construction but not rate-1 (rate-1 would need 8 matrices, exceeding $\mathcal{H}(4) = 4$ for each of the real/imaginary partitions).

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Review: The Hurwitz-Radon Sum-of-Squares Identity

The classical Hurwitz-Radon theorem gives the maximum $k$ such that there exists a bilinear sum-of-squares identity $\bigl(x_1^2 + \cdots + x_n^2\bigr)\bigl(y_1^2 + \cdots + y_k^2\bigr) \;=\; z_1^2 + \cdots + z_n^2,$ where the $z_i$ are bilinear in $(\mathbf{x}, \mathbf{y})$ . The answer: such an identity exists if and only if $k \le \mathcal{H}(n)$ . The classical $(n, k, n) = (1, 1, 1), (2, 2, 2), (4, 4, 4), (8, 8, 8)$ cases are the only ones where $n = k$ — they correspond to the real norms on $\mathbb{R}, \mathbb{C}, \mathbb{H}$ (quaternions), and $\mathbb{O}$ (octonions). Hurwitz proved in 1898 that no similar identity with $n = k = 16$ exists, confirming that $\mathbb{O}$ is the end of the line for division algebras over $\mathbb{R}$ .

This is not a historical footnote — it is the reason Alamouti gives rate 1 and no larger OSTBC can. Alamouti is the $(n, k, n) = (2, 2, 2)$ identity in disguise: the codeword matrix's orthogonality is the multiplicative norm on $\mathbb{C}$ . There is a corresponding rate-1 quaternionic code for $n_t = 4$ with real input (using the $(4, 4, 4)$ identity), but for the practically relevant complex input, the rate drops to $3/4$ .

Theorem: Liang-Tarokh Rate Upper Bound for Complex OSTBCs

For any complex OSTBC with $n_t$ transmit antennas, the maximum rate (in complex symbols per channel use) is $R_{\max}(n_t) \;=\; \frac{\lfloor n_t/2 \rfloor + 1}{2 \lfloor n_t/2 \rfloor}.$ Specifically:

$n_t$	2	3	4	5	6	7	8
$R_{\max}$	1	$3/4$	$3/4$	$2/3$	$2/3$	$5/8$	$5/8$

In particular, $R_{\max} = 1$ only for $n_t \le 2$ , $R_{\max} = 3/4$ for $n_t \in \{3, 4\}$ , and $R_{\max}$ decreases monotonically for larger $n_t$ , tending to $1/2$ as $n_t \to \infty$ .

The proof chains two algebraic observations. First, an OSTBC encodes $K$ complex symbols via $2K$ real dispersion matrices — $K$ matrices $\mathbf{A}_q$ for the real parts and $K$ matrices $\mathbf{B}_q$ for the imaginary parts. Second, these $2K$ matrices must satisfy mutually anti-commuting relations over an $n_t$ -dimensional space. The Hurwitz- Radon-Eckmann theorem bounds the maximum number of such matrices, which then bounds $K$ , which bounds $R = K/T$ . The specific formula falls out after optimising over $T$ .

Show Hint

Count the anti-commuting dispersion matrices needed for complex input: $2K$ matrices satisfying Hurwitz-Radon-Eckmann.

Apply the Hurwitz-Radon-Eckmann theorem to bound the maximum number of such matrices over $\mathbb{R}^{n_t}$ .

Optimise over $T$ to get the tightest bound on $K/T$ .

Proof

Reduction to a Hurwitz-Radon-Eckmann counting problem

Fix an OSTBC with $K$ complex symbols in $T$ channel uses and dispersion matrices $\{\mathbf{A}_q, \mathbf{B}_q\}_{q=1}^K$ . The OSTBC conditions $\mathbf{A}_p\mathbf{A}_q^H + \mathbf{A}_q \mathbf{A}_p^H = 2\delta_{pq}\mathbf{I}$ etc. say that the real matrices $\{\mathbf{A}_q, j\mathbf{B}_q\}$ (treating each as a real $2n_t \times 2T$ matrix via the standard complex-to-real embedding) satisfy the Hurwitz-Radon-Eckmann anti-commutation relations over the ambient space $\mathbb{R}^{2 n_t}$ acting on a column space $\mathbb{R}^{2T}$ .

Apply Hurwitz-Radon-Eckmann

The Hurwitz-Radon-Eckmann theorem bounds the number of such matrices. Via the Tarokh-Jafarkhani-Calderbank / Liang refinement for the complex case, the specific bound is that the number $2K$ of matrices is at most $2T \cdot \lfloor n_t/2 \rfloor / (\lfloor n_t/2 \rfloor + \lceil n_t/2 \rceil)$ — equivalently, $K \;\le\; T \cdot \frac{\lfloor n_t/2 \rfloor + 1}{2\lfloor n_t/2 \rfloor}.$ See Liang 2003, Thm. 1, for the detailed combinatorial argument (which uses the Radon-Eckmann-Hurwitz bound on mutually anti- commuting orthogonal matrices in representations of Clifford algebras).

Rate formula

Dividing both sides by $T$ : $R \;=\; \frac{K}{T} \;\le\; \frac{\lfloor n_t/2 \rfloor + 1}{2\lfloor n_t/2 \rfloor}.$ This is the Liang-Tarokh upper bound. Specialising: $n_t = 2: R_{\max} = 2/2 = 1$ ; $n_t = 4: R_{\max} = 3/4$ ; $n_t = 8: R_{\max} = 5/8$ ; and so on. The bound is achievable — Tarokh- Jafarkhani-Calderbank 1999 constructed OSTBCs meeting it for $n_t \le 8$ , and Liang 2003 showed achievability for all $n_t$ . $\blacksquare$

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Example: Rate Cost of Full Diversity at $n_t = 4$

Compare three space-time schemes at $n_t = 4, n_r = 4$ : (i) V-BLAST at rate 4 symbols per channel use with receive-ZF-SIC, (ii) TJC rate- $3/4$ OSTBC, (iii) a hypothetical rate-1 OSTBC. Which is achievable? What does the Liang-Tarokh bound say about (iii)?

Solution

V-BLAST at rate 4

V-BLAST splits the data into $n_t = 4$ independent streams, one per antenna. At $n_r = 4$ the receiver has enough degrees of freedom to separate them (zero-forcing or SIC). Rate = 4 symbols/channel use, but diversity order per layer = $n_r - n_t + 1 = 1$ (for ZF) or at most $n_r$ for the last-decoded SIC layer. This is the multiplexing extreme.

TJC rate-$3/4$ OSTBC

The TJC code of Ex. $3/4$ $3/4$ 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 carries 3 symbols in 4 channel uses, rate $3/4$ , and achieves diversity $n_t n_r = 16$ with linear decoding. At moderate SNR (say, $\text{SNR} = 10$ dB) this 16-fold diversity buys a much lower error probability than V-BLAST's single-diversity layers — OSTBC wins in the low-SNR, high- reliability regime.

Rate-1 OSTBC at $n_t = 4$ is impossible

By Thm. TLiang-Tarokh Rate Upper Bound for Complex OSTBCs, $R_{\max}(4) = 3/4$ . A rate-1 complex OSTBC for $n_t = 4$ does not exist: no matter how long the block length $T$ and no matter how cleverly you choose the dispersion matrices, the Hurwitz-Radon-Eckmann algebra cannot accommodate $2K = 2T$ anti-commuting matrices in $\mathbb{R}^{2 n_t = 8}$ unless $T \le K \cdot (3/4)$ .

Any rate-1 linear code for $n_t = 4$ must therefore break orthogonality. Jafarkhani's quasi-orthogonal code (§4) is the classic solution: rate 1, diversity 2 (not 4), and a pair-wise decoder that is no longer fully scalar.

Design-space summary

At $n_t = 4, n_r = 4$ , the Pareto front in (rate, diversity) plane is:

Scheme	Rate	Diversity	Decoder
V-BLAST	4	1 (per layer, ZF)	Linear ZF
TJC OSTBC	$3/4$	16	Linear MF
Jafarkhani QOSTBC	1	2	Pair-wise ML

No rate-1 full-diversity linear code exists at $n_t = 4$ . This is the exact content of the Liang-Tarokh bound. Chapter 13's Golden codes and Perfect codes solve the problem by being non-linear (cyclic division algebra codes) and decoding with a lattice decoder — paying $O(M^2)$ lattice search complexity for rate 2, diversity 4 simultaneously.

Complex OSTBC Rate Ceiling vs $n_t$

The Liang-Tarokh rate upper bound $R_{\max}(n_t) = (\lfloor n_t/2\rfloor + 1)/(2\lfloor n_t/2\rfloor)$ for complex OSTBCs (blue), with the rates actually achieved by the Tarokh-Jafarkhani-Calderbank 1999 construction marked (red dots). Toggle the real curve to see the rate-1 cases for $n_t \le 8$ that hold only when the information symbols are real-valued (PAM constellations).

The message: for the practical complex case (QAM / PSK input), rate 1 is available only for $n_t = 2$ . At $n_t = 4$ you drop to $3/4$ ; at $n_t = 8$ to $5/8$ ; and the ceiling asymptotes to $1/2$ as $n_t \to \infty$ . This is the reason full-diversity full-rate codes for $n_t > 2$ require the non-orthogonal constructions of §4–5 and of Chapter 13.

Parameters

Include real OSTBC (rate 1 for

n_t \le 8

)

Historical Note: Wolniansky et al. V-BLAST (1998): The Opposite Extreme

1998

While Alamouti was inventing the diversity-optimal rate-1 code, a team at Bell Labs (Peter Wolniansky, Gerard Foschini, Glen Golden, and Robert Valenzuela) published V-BLAST at the 1998 URSI ISSSE conference. Where Alamouti spends all of the $n_t$ transmit antennas on diversity, V-BLAST spends them on multiplexing: $n_t$ independent data streams, one per antenna, at a total rate of $n_t$ symbols/channel use. The receiver uses successive interference cancellation (SIC) to separate the streams — a nulling-and-cancelling architecture that in hindsight is MMSE-SIC.

Bell Labs built a lab prototype in 1998 at 8 Tx × 12 Rx antennas, achieving 20-40 bits/s/Hz in a $1/30$ -s coherence channel — what was then an astronomical spectral efficiency. The V-BLAST paper is the historical demonstration that the channel capacity (not just the diversity) of a MIMO link could be approached with practical receiver architectures. The tradeoff between Alamouti (all diversity) and V-BLAST (all multiplexing) became the subject of the Zheng-Tse 2003 DMT framework of Chapter 12 — which shows that the two extremes are endpoints of a continuum, and that neither is simultaneously optimal except on the end points.

V-BLAST uses no space-time code per se: it is the absence of coding across antennas, paired with smart sequential detection. It is the canonical counterpoint to every STBC in this chapter.

🔧Engineering Note

OSTBC Decoder Complexity in Practice

The scalar-decoupled decoder of an OSTBC has complexity $O(n_t n_r T)$ per block for matched-filter computation plus $O(M K)$ for $K$ scalar slicer decisions on an $M$ -ary constellation — a total of $O(n_t n_r T + M K)$ operations per codeword. By contrast, the ML decoder of a general linear STBC has complexity $O(M^K)$ (exhaustive search over $K$ symbols), which for $K = 3, M = 64$ is $64^3 = 262\,144$ metric evaluations per codeword.

This exponential-vs-polynomial gap is the pragmatic reason that OSTBC codes were rapidly adopted in early MIMO standards (WCDMA, 802.16, LTE-TM2), while more general constructions (Perfect codes, LDCs with non-orthogonal dispersion matrices) remained in the research literature until cheaper sphere-decoder implementations emerged circa 2005-2010.

Practical Constraints

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Matched-filter decoder requires only $O(n_t n_r T)$ complex multiplications per codeword
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Scalar slicer complexity is $O(M K)$ — flat in constellation size
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No sphere decoding, no lattice basis reduction — all front-end linear algebra

Common Mistake: Rate and Diversity Cannot Both Be Maximised — But OSTBCs Make It Look Like They Can

Mistake:

Reading "OSTBCs achieve full diversity $n_t n_r$ " as the claim that OSTBCs are Pareto-optimal in (rate, diversity) plane, and therefore that abandoning OSTBC orthogonality is a strictly dominated choice.

Correction:

OSTBCs are Pareto-optimal among rate- $R_{\max}(n_t)$ linear codes — within the rate they offer, they are diversity-optimal. But the Zheng-Tse DMT of Chapter 12 shows that the Pareto front of the MIMO channel is much richer: for any multiplexing gain $r \in [0, \min(n_t, n_r)]$ , the diversity-optimal code achieves $d^*(r) = (n_t - r)(n_r - r)$ . OSTBCs lie at the single point $(r, d) = (R_{\max}, n_t n_r)$ on this curve — the upper-left corner. Every other point on the DMT curve is unreachable by OSTBCs because their rate is stuck at $R_{\max}$ .

To fill in the rest of the curve you need non-orthogonal codes: QOSTBCs give a different point (rate 1, diversity 2 for $n_t = 4$ ), LDCs give a family of points interpolating between extremes, and Chapter 13's CDA codes achieve the entire DMT curve simultaneously. The OSTBC "no-compromise" pitch is true only if your operating point happens to coincide with the OSTBC rate.

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Quick Check

A design team wants a rate-1 full-diversity linear space-time code for $n_t = 4$ Tx antennas. Which of the following is correct?

No such code exists — the Liang-Tarokh bound caps complex-OSTBC rate at $3/4$ for $n_t = 4$

The code exists and is the TJC rate- $3/4$ OSTBC — just pad 1 dummy symbol to reach rate 1

Use V-BLAST — it is rate 4 at $n_t = 4$ , well above rate 1

Use the Jafarkhani QOSTBC — rate 1 with full diversity $n_t n_r$

Correction:

No such code exists — the Liang-Tarokh bound caps complex-OSTBC rate at

3/4

for

n_t = 4

Exactly right. Thm. TLiang-Tarokh Rate Upper Bound for Complex OSTBCs gives $R_{\max}(4) = 3/4$ for complex OSTBCs; no clever construction will beat it. Rate-1 at $n_t = 4$ with full diversity requires non-linear codes (cyclic division algebras, Ch. 13) or non-orthogonal codes at the cost of diversity (Jafarkhani's QOSTBC, §4, has rate 1 but diversity only 2).

Hurwitz-Radon Number

The maximum number $\mathcal{H}(n)$ of pairwise anti-commuting orthogonal $n\times n$ real matrices. Equals $2^a + 8b$ where $n = 2^{a+4b}(2c+1)$ with $0 \le a \le 3$ . Bounds the number of dispersion matrices usable in an OSTBC and, through the Liang-Tarokh bound, bounds the rate of complex OSTBCs.

Rate Constraints for Complex OSTBCs

Why Do We Lose Rate Beyond nt=2n_t = 2nt​=2?

Definition: Hurwitz-Radon Number H(n)\mathcal{H}(n)H(n)