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The First Design Objective: Full-Rank $\boldsymbol{\Delta}$

The high-SNR simplification of the PEP bound (Theorem TSpace-Time PEP Bound (Tarokh-Seshadri-Calderbank 1998 / Guey-Fitz-Bell-Kuo 1999)) produces two asymptotic numbers: the slope $r n_r$ (controlled by the rank $r$ of the error matrix) and the intercept $\prod_i \lambda_i$ (controlled by the determinant). At high SNR, slope dominates: a code with smaller slope is eventually worse than a code with larger slope, no matter how favourable the intercept.

The first question every space-time code designer asks is therefore: what is the minimum rank of $\boldsymbol{\Delta} = \mathbf{X} - \hat{\mathbf{X}}$ over all codeword pairs? This is the rank criterion, formalised by Tarokh, Seshadri, and Calderbank in 1998. The target is $r_{\min} = n_t$ — full rank — which yields the maximum possible diversity order $d = n_t n_r$ and matches the outage exponent $d^\star(0)$ of Theorem $P_{\mathrm{out}}(R)$ $P_{out} (R)$ (DMT Preview)" data-ref-type="theorem">THigh-SNR Scaling of $P_{\mathrm{out}}(R)$ (DMT Preview).

Only once this is achieved does the determinant criterion of §5 become relevant: within the full-rank class, the determinant tunes the coding gain. Without full rank, no amount of determinant optimisation can recover the lost diversity — the rank is a hard asymptotic ceiling.

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Definition:
Diversity Order of a Space-Time Code

The diversity order $d$ of a space-time code $\mathcal{C}$ on an i.i.d. Rayleigh MIMO channel is the high-SNR asymptotic slope of its codeword error probability on a log-log plot: $d \;\triangleq\; -\lim_{\text{SNR} \to \infty} \frac{\log P_e(\text{SNR})} {\log \text{SNR}}.$ Equivalently, $P_e(\text{SNR}) \doteq \text{SNR}^{-d}$ at high SNR.

The diversity order is a property of the code, not of the channel. The channel bounds it: no code on an $n_t \times n_r$ i.i.d. Rayleigh MIMO channel can achieve diversity greater than $n_t n_r$ (this is the $r = 0$ point of the Zheng-Tse DMT — see Ch. 12). The rank criterion of Theorem TRank Criterion (Tarokh-Seshadri-Calderbank 1998) tells us exactly which codes hit this ceiling.

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Theorem: Rank Criterion (Tarokh-Seshadri-Calderbank 1998)

For a space-time code $\mathcal{C} \subset \mathbb{C}^{n_t \times T}$ transmitted over an i.i.d. Rayleigh block-fading MIMO channel with $n_r$ receive antennas, CSIR, and ML decoding, the diversity order is $d \;=\; r_{\min} \cdot n_r, \qquad r_{\min} \;\triangleq\; \min_{\mathbf{X} \ne \hat{\mathbf{X}} \in \mathcal{C}} \mathrm{rank}(\boldsymbol{\Delta}), \quad \boldsymbol{\Delta} = \mathbf{X} - \hat{\mathbf{X}}.$ The maximum diversity order $d_{\max} = n_t n_r$ is achieved if and only if $\mathrm{rank}(\boldsymbol{\Delta}) = n_t \quad \text{for every pair} \;\; \mathbf{X} \ne \hat{\mathbf{X}} \in \mathcal{C}.$ Such a code is said to be full-rank or to satisfy the rank criterion.

The PEP upper bound decays as $\text{SNR}^{-r n_r}$ at high SNR (Theorem TSpace-Time PEP Bound (Tarokh-Seshadri-Calderbank 1998 / Guey-Fitz-Bell-Kuo 1999), final proof step). The union bound on codeword error is a finite sum of PEPs; at large SNR the slowest- decaying summand dominates, so $P_e \doteq \text{SNR}^{-r_{\min} n_r}$ .

Conversely, a lower bound on $P_e$ by the dominant PEP (one specific pair) is also $\text{SNR}^{-r_{\min} n_r}$ up to constants. Hence $d = r_{\min} n_r$ exactly at the level of exponents. The $n_t n_r$ ceiling is dictated by the dimension: $\boldsymbol{\Delta} \in \mathbb{C}^{n_t \times T}$ has rank $\le \min(n_t, T) = n_t$ (assuming $T \ge n_t$ ).

Show Hint

Upper-bound $P_e$ by the union bound over codeword pairs; the dominant term is the pair with smallest $r \cdot n_r$ -slope.

Lower-bound $P_e$ by the PEP of a single worst pair; the bound is $\text{SNR}^{-r_{\min} n_r}$ up to multiplicative constants.

Sandwich: $\text{SNR}^{-r_{\min} n_r}$ from above and below gives $d = r_{\min} n_r$ at the exponent level.

For the full-rank statement, observe that $\mathrm{rank}(\boldsymbol{\Delta}) \le \min(n_t, T)$ ; assuming $T \ge n_t$ , the ceiling is $n_t$ , achieved iff every pair has $\mathrm{rank} = n_t$ .

Proof

Upper bound on $P_e$ from PEP

Fix $\mathcal{C}$ with $|\mathcal{C}| = M$ . The union bound gives $P_e \;\le\; \frac{1}{M} \sum_{\mathbf{X} \in \mathcal{C}} \sum_{\hat{\mathbf{X}} \ne \mathbf{X}} P(\mathbf{X} \to \hat{\mathbf{X}}).$ By Theorem TSpace-Time PEP Bound (Tarokh-Seshadri-Calderbank 1998 / Guey-Fitz-Bell-Kuo 1999), each summand is bounded by $\prod_{i=1}^{r(\mathbf{X}, \hat{\mathbf{X}})} (1 + (\text{SNR}/4 n_t) \lambda_i)^{-n_r}$ , where $r(\mathbf{X}, \hat{\mathbf{X}}) = \mathrm{rank}(\mathbf{X} - \hat{\mathbf{X}})$ .

At high SNR, each term behaves as $(\text{SNR}/4 n_t)^{-r(\cdot,\cdot) n_r} \prod \lambda_i^{-n_r}$ . The finite sum is dominated by the term with the smallest exponent $r(\cdot,\cdot) n_r$ : $P_e \;\lesssim\; K \cdot \text{SNR}^{-r_{\min} n_r},$ where $K$ is a finite constant depending on $\mathcal{C}$ and $n_t, n_r$ .

Lower bound on $P_e$ from the worst pair

Let $(\mathbf{X}^\star, \hat{\mathbf{X}}^\star)$ be a codeword pair achieving $r(\mathbf{X}^\star, \hat{\mathbf{X}}^\star) = r_{\min}$ . Then $P_e \;\ge\; \tfrac{1}{M} P(\mathbf{X}^\star \to \hat{\mathbf{X}}^\star).$ A matching lower bound on the PEP $P(\mathbf{X}^\star \to \hat{\mathbf{X}}^\star) \gtrsim c \text{SNR}^{-r_{\min} n_r}$ at high SNR (obtainable from the exact PEP formula, which is $(1 + (\text{SNR}/4 n_t)\lambda_i)^{-n_r}$ up to an explicit pre-factor that scales as $\text{SNR}^{0}$ , and not the Chernoff bound; see Biglieri-Caire-Taricco 2000 §V.D for the lower-bound derivation) gives the matching exponent.

Sandwich: $c \text{SNR}^{-r_{\min} n_r} \le P_e \le K \text{SNR}^{-r_{\min} n_r}$ , so the high-SNR slope of $P_e$ is exactly $r_{\min} n_r$ .

Diversity order $d = r_{\min} n_r$

By Definition DDiversity Order of a Space-Time Code, the diversity order is $d = -\lim_{\text{SNR} \to \infty} \log P_e / \log \text{SNR} = r_{\min} n_r$ .

Maximum diversity $\Leftrightarrow$ full rank

The rank of $\boldsymbol{\Delta} \in \mathbb{C}^{n_t \times T}$ satisfies $\mathrm{rank}(\boldsymbol{\Delta}) \le \min(n_t, T)$ . Assuming $T \ge n_t$ (the natural setting), the ceiling is $\mathrm{rank}(\boldsymbol{\Delta}) \le n_t$ , so $r_{\min} \cdot n_r \le n_t n_r$ with equality iff every codeword pair gives $\mathrm{rank}(\boldsymbol{\Delta}) = n_t$ . Hence $d = n_t n_r$ iff every pair gives a full-rank error matrix. $\blacksquare$

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Key Takeaway

The designer's FIRST objective is full rank. Without $\mathrm{rank} (\boldsymbol{\Delta}) = n_t$ for every codeword pair, the code cannot reach the maximum diversity $n_t n_r$ — and the asymptotic PEP slope is strictly worse. No determinant optimisation, no power boost, no receiver cleverness can recover the lost diversity; rank deficiency is a permanent high-SNR handicap. Only after the code is full-rank does the determinant criterion of §5 become meaningful.

BER vs. SNR for Space-Time Codes of Different Rank

BER curves of three illustrative space-time codes on a $n_t \times n_r$ i.i.d. Rayleigh block-fading channel: (i) a rank- $1$ spatial-repetition code (diversity $n_r$ ), (ii) a rank- $2$ "half-diversity" code (diversity $2 n_r$ ), and (iii) a full-rank Alamouti-like code (diversity $n_t n_r$ ). The slopes are the diversity orders and illustrate the rank criterion at work.

Parameters

n_t

2

n_r

2

Example: Spatial Repetition: A Rank-Deficient Code (Diversity $n_r$ Only)

Consider the spatial repetition code with $n_t = 2$ : the same scalar QPSK symbol $s \in \{\pm 1 \pm j\}$ is transmitted simultaneously on both antennas for $T = 1$ channel use, i.e., $\mathbf{X}(s) = (s, s)^T$ . Compute the rank of $\boldsymbol{\Delta}$ for an arbitrary codeword pair $(s, \hat s)$ with $s \ne \hat s$ , determine the diversity order with $n_r$ receive antennas, and comment on the design implications.

Solution

Error matrix

$\boldsymbol{\Delta}(s, \hat s) = \mathbf{X}(s) - \mathbf{X}(\hat s) = (s - \hat s, s - \hat s)^T \in \mathbb{C}^{2 \times 1}$ . Every column is the same; $\mathrm{rank}(\boldsymbol{\Delta}) = 1$ .

Diversity order

By the rank criterion (Theorem TRank Criterion (Tarokh-Seshadri-Calderbank 1998)), the diversity order is $d = r_{\min} \cdot n_r = 1 \cdot n_r = n_r$ — NOT $n_t n_r = 2 n_r$ . The second transmit antenna contributes no diversity: it only contributes an array-gain factor of $2$ (a constant, not a slope).

Why rank is $1$

The code is essentially scalar: it sends the same symbol on both antennas and is therefore equivalent to a scalar code $s$ followed by a deterministic beamforming vector $(1,1)^T/\sqrt{2}$ . The error matrix is rank $1$ because $\boldsymbol{\Delta}$ is a scalar multiple of the beamformer. To achieve $\mathrm{rank} = 2$ , the two antennas must transmit different (linearly independent) functions of the information — Alamouti's design.

Designer takeaway

Spatial repetition is almost never used in practice: it wastes half the available diversity. The Alamouti scheme achieves $\mathrm{rank} = 2$ (full) with the same rate $1$ and the same number of antennas — it is strictly better. $\blacksquare$

Example: Alamouti Code: Verify Full Rank Over All Symbol Pairs

Recall the Alamouti codeword matrix $\mathbf{X}(s_1, s_2) \;=\; \begin{pmatrix} s_1 & -s_2^* \\ s_2 & s_1^* \end{pmatrix}, \quad s_1, s_2 \in \mathcal{X} \quad (\mathcal{X} \text{ QPSK}, M = 4).$ Show that $\mathrm{rank}(\boldsymbol{\Delta}) = 2$ for every distinct codeword pair — i.e., Alamouti satisfies the rank criterion.

Solution

Error matrix in closed form

For $(s_1, s_2) \ne (\hat s_1, \hat s_2)$ let $e_k = s_k - \hat s_k$ . Then $\boldsymbol{\Delta} = \begin{pmatrix} e_1 & -e_2^* \\ e_2 & e_1^* \end{pmatrix}$ .

Determinant

$\det(\boldsymbol{\Delta}) = e_1 e_1^* - (-e_2^*)e_2 = |e_1|^2 + |e_2|^2.$ This is strictly positive whenever $(e_1, e_2) \ne (0, 0)$ , i.e., whenever the two codewords differ. Therefore $\boldsymbol{\Delta}$ is always nonsingular, $\mathrm{rank} (\boldsymbol{\Delta}) = 2$ , and the rank criterion is satisfied.

Diversity order

By the rank criterion, Alamouti achieves $d = 2 \cdot n_r = n_t n_r$ — the full MIMO diversity. It is a full-rank code; in §5 we will see that it is also constant-determinant in a specific sense (all eigenvalue-pairs equal $|e_1|^2 + |e_2|^2$ ), which makes it a particularly clean example for the determinant criterion. $\blacksquare$

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⚠️Engineering Note

Rank Criterion in LTE Transmission Modes TM3 and TM4

LTE specifies eight transmission modes (TM1–TM8) in 3GPP TS 36.211; TM3 and TM4 are the open-loop MIMO modes that rely on the rank criterion for diversity:

TM3 (Open-Loop Spatial Multiplexing): Transmits up to $n_t = 4$ independent streams with large delay cyclic delay diversity (CDD). The CDD ensures that the effective error matrix is full rank over every codeword pair — i.e., TM3 satisfies the rank criterion without explicit space-time coding, at the cost of some rate loss.
TM4 (Closed-Loop Spatial Multiplexing): With CSIT feedback, uses codebook-based precoding (4-bit codebook for $n_t = 2$ , 6-bit for $n_t = 4$ ). The precoders are designed to be full-rank, so the effective channel is full-rank for every transmitted rank indicator (RI). The rank criterion is enforced by the codebook design.
Fallback to TM2 (SFBC/FSTD): Alamouti-based Space-Frequency Block Coding (SFBC, for $n_t = 2$ ) and Frequency-Switched Transmit Diversity (FSTD, for $n_t = 4$ ) are used when the channel is too correlated for rank- $\geq 2$ transmission. Both are full-rank by construction — SFBC is literally Alamouti on two OFDM subcarriers.

In 5G NR (3GPP TS 38.211), these are generalised to codebook-based precoding with rank indicator (RI) reporting, again with rank-preserving codebook entries. The "rank" in 5G terminology is precisely $\mathrm{rank}(\boldsymbol{\Delta})$ — the link from Tarokh 1998 theory to 2019 standards is direct.

Practical Constraints

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Cyclic delay diversity delay $\delta$ in TM3 must exceed the channel's delay spread to avoid subcarrier-level rank collapse
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Codebook size: LTE has 4-bit books for $n_t = 2$ (16 precoders) and 6-bit for $n_t = 4$ (64 precoders); 5G NR extends these
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SFBC requires flat fading across two adjacent subcarriers (typically true in LTE's 15 kHz spacing)

📋 Ref: 3GPP TS 36.211 §6.3.4; 3GPP TS 38.211 §7.3.1

Common Mistake: 'More Transmit Power' Does Not Fix Rank Deficiency

Mistake:

A prototype shows a rank-deficient code underperforming at high SNR; a well-meaning engineer proposes to boost the transmit power by $3$ dB to "close the gap". The new BER curve at the higher SNR still lies above the full-rank baseline — by exactly the same amount.

Correction:

At high SNR the error probability decays as $\text{SNR}^{-d}$ , where $d = r_{\min} n_r$ is the code's diversity order. A constant power boost multiplies every SNR by a constant, which shifts the curve horizontally by a constant dB amount but does not change the slope. A rank-deficient code is doomed to stay above a full-rank code at sufficiently high SNR, regardless of power. The only cure is to redesign the code for full rank — this is a structural change, not a tuning knob.

Key Takeaway

Rank is the ceiling. The diversity order is $r_{\min} \cdot n_r$ , bounded by $n_t n_r$ . A code that fails the rank criterion — even by a single codeword pair with rank-deficient $\boldsymbol{\Delta}$ — permanently loses that diversity. The second design objective (determinant) only ever bites after full rank has been secured. Codes in this chapter's forward pointers (Alamouti, OSTBCs, CDA codes) are engineered first for full rank and then for good determinant.

The Rank Criterion — Diversity Order

The First Design Objective: Full-Rank Δ\boldsymbol{\Delta}Δ

Definition: Diversity Order of a Space-Time Code

Theorem: Rank Criterion (Tarokh-Seshadri-Calderbank 1998)

Upper bound on $P_e$ from PEP

Lower bound on $P_e$ from the worst pair

Diversity order $d = r_{\min} n_r$

Maximum diversity $\Leftrightarrow$ full rank

Key Takeaway

BER vs. SNR for Space-Time Codes of Different Rank

Parameters

Example: Spatial Repetition: A Rank-Deficient Code (Diversity nrn_rnr​ Only)

Error matrix

Diversity order

Why rank is $1$

Designer takeaway

Example: Alamouti Code: Verify Full Rank Over All Symbol Pairs

Error matrix in closed form

Determinant

Diversity order

Rank Criterion in LTE Transmission Modes TM3 and TM4

Common Mistake: 'More Transmit Power' Does Not Fix Rank Deficiency

Key Takeaway

The First Design Objective: Full-Rank $\boldsymbol{\Delta}$

Definition:
Diversity Order of a Space-Time Code

Example: Spatial Repetition: A Rank-Deficient Code (Diversity $n_r$ Only)