Ferkans — Interactive Telecom Tutor

The Main Theorem

The central theoretical result of OTFS — and the main reason the waveform is a 6G candidate — is the full delay-Doppler diversity theorem. It asserts that OTFS with ML detection achieves diversity order equal to the number of resolvable delay-Doppler paths $P$ . This is true independent of the frame size $MN$ and independent of the constellation $\mathcal{X}$ .

The point is that OTFS's diversity is not bought via multiple antennas or channel coding — it comes "for free" from the DD-domain signaling structure. This is a qualitative advance over OFDM, which can only achieve diversity $> 1$ via outer coding or MIMO.

Theorem: Full Delay-Doppler Diversity of OTFS

For an OTFS system with $M$ delay bins, $N$ Doppler bins, and a $P$ -path channel with distinct integer-Doppler indices $\{(\ell_i, k_i)\}_{i=1}^P$ and independent Rayleigh path gains:

The ML detector achieves uncoded BER $\mathrm{BER}_{\text{OTFS-ML}}(\mathrm{SNR}) \;\sim\; C_{\mathcal{X}}\,\mathrm{SNR}^{-P},$ at high SNR, where $C_{\mathcal{X}}$ depends on the QAM constellation but is independent of $\mathrm{SNR}$ and (asymptotically) of $MN$ .

In words: OTFS has diversity order exactly $P$ .

Each data symbol, after ISFFT, is spread across all $MN$ TF cells. Every TF cell is subjected to all $P$ paths (the channel is a dense $MN$ -dimensional multiply in TF). When we return to the DD grid via SFFT, each received cell contains a weighted sum of all $P$ path contributions — a $P$ -order diversity sum. ML detection uses this full sum; diversity $P$ is the natural outcome.

Alternatively: the OTFS transmit signal "probes" all $P$ paths simultaneously, and the receiver has $P$ independent observations to combine. Classical diversity combining theory then gives diversity $P$ .

Proof

Worst-case error vector

Consider all possible error pairs $(\mathbf{x}, \hat{\mathbf{x}})$ with $\Delta\mathbf{x} = \mathbf{x} - \hat{\mathbf{x}} \neq 0$ . By Theorem TPEP Scaling Under Rayleigh Path Gains, the diversity order is $d = \min_{\Delta\mathbf{x} \neq 0} \text{rank}(\mathbf{A}(\Delta\mathbf{x}))$ .

Rank bound

$\mathbf{A}(\Delta\mathbf{x})$ is a $P \times P$ positive semidefinite matrix built from the error vector. Its rank is at most $P$ and at least $\min(P, \text{support}(\Delta\mathbf{x}))$ .

Generic full rank

For any non-zero $\Delta\mathbf{x}$ with $\text{support} \geq 1$ , and for distinct $(\ell_i, k_i)$ path indices, the columns of $\mathbf{A}$ are linearly independent. Hence rank is exactly $P$ .

Diversity

$\min_{\Delta\mathbf{x}} \text{rank}(\mathbf{A}) = P$ . By the PEP scaling theorem, the BER slope is $P$ . $\blacksquare$

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Key Takeaway

OTFS delivers diversity $P$ — the maximum physically achievable. No matter the constellation size, frame size, or specific path geometry (as long as path indices are distinct), ML-detected OTFS achieves diversity order $P$ . This is the information-theoretic justification of OTFS's claim to outperform OFDM under mobility. Chapter 8's detectors (MP, LCD + IDD) realize this bound in practice.

🎓CommIT Contribution(2019)

Diversity Theorem for OTFS

G. D. Surabhi, R. Madhava Augustine, A. Chockalingam, G. Caire — IEEE Trans. Wireless Communications, vol. 18, no. 6

Surabhi, Augustine, Chockalingam, and Caire established the full-diversity theorem for OTFS in their 2019 paper. The main contribution is a rigorous proof that the minimum-rank $\mathbf{A}(\Delta\mathbf{x})$ over all error pairs equals $P$ for integer-Doppler paths, and hence the uncoded BER diversity is exactly $P$ . Prior work had demonstrated this via simulation; the 2019 paper provided the algebraic proof.

The paper also generalizes to fractional Doppler: the diversity becomes $\min(P, k_{\max})$ when Doppler is fractionally offset, with the additional factor accounting for inter-Doppler leakage. This generalization is reviewed in Chapter 10.

As a CommIT contribution, this work is one of the foundational performance-theoretic results of the OTFS literature and is referenced throughout the diversity-and-detection analysis.

commitdiversityotfs-performance

Example: BER at SNR = 25 dB for $P = 4$ vs OFDM

An OTFS receiver with $P = 4$ paths and ML detection operates at $\mathrm{SNR} = 25$ dB. An equivalent OFDM receiver (diversity 1) operates at the same SNR. Compare uncoded BERs.

Solution

OFDM BER

Diversity 1: $\mathrm{BER}_{\text{OFDM}} \sim 1/(4\,\mathrm{SNR}) = 1/(4 \cdot 316) = 8 \times 10^{-4}$ .

OTFS BER

Diversity $P = 4$ : $\mathrm{BER}_{\text{OTFS}} \sim C/\mathrm{SNR}^4$ . Using the constant $C \sim 1$ from simulations: $\mathrm{BER}_{\text{OTFS}} \approx 1/316^4 \approx 10^{-10}$ .

Gap

At 25 dB, OTFS's uncoded BER is $\sim 6$ orders of magnitude lower than OFDM's. This is the "full diversity gain" headline number.

Caveats

(1) This assumes ML detection. MP or LCD loses 1-2 dB. (2) At lower SNR, the gap is smaller (not diversity-dominated). (3) Channel coding applied to both closes much of the gap — the uncoded comparison is the extreme case. Typical coded comparison: OTFS wins 1-3 dB at BER $10^{-5}$ .

Uncoded BER: OTFS vs OFDM at Varying $P$

For OFDM (diversity 1) and OTFS (diversity $P$ ), plot uncoded BER at a fixed SNR as a function of $P$ . OFDM's BER is flat in $P$ (diversity-limited at 1). OTFS's BER drops exponentially as $P$ grows. At $P = 2$ , OTFS's BER is roughly the OFDM BER; at $P = 10$ , OTFS's BER is millions of times smaller.

Parameters

SNR (dB)20

Max

P

8

The Resolvability Caveat

The $P$ in the diversity theorem is the number of resolvable DD paths — i.e., paths that occupy distinct DD grid cells. If two paths have the same $(\ell_i, k_i)$ (falling in the same grid cell), they merge into a single effective path with summed amplitude, and the diversity is reduced to the count of distinct cells.

For realistic channels, this is rare — delay and Doppler values are generically distinct. But two edge cases exist:

Same delay, same Doppler: two reflectors coincident in DD space. Physically unusual.
Coarse grid resolution: if $\Delta\tau$ or $\Delta\nu$ are large, multiple physical paths may land in the same bin. To preserve full diversity, make the DD grid fine enough to resolve all paths.

In deployment, use $\Delta\tau < \tau_i - \tau_j$ for distinct paths (typically achievable with $W \geq 10$ MHz at urban delay spreads).

Theorem: Coding Gain Beyond Diversity

In addition to the diversity gain of $P$ , OTFS has a coding gain determined by the constant $C_{\mathcal{X}}$ in $\mathrm{BER} = C_{\mathcal{X}}\,\mathrm{SNR}^{-P}$ . For QPSK, $C_{\mathcal{X}} \approx P!/\prod_i |h_i|^2_{\text{avg}}$ ; for higher-order constellations, $C_{\mathcal{X}}$ grows (more closely-spaced constellation points are harder to distinguish).

The ratio of OTFS BER to OFDM BER at a given SNR is approximately $\frac{\mathrm{BER}_{\text{OTFS}}}{\mathrm{BER}_{\text{OFDM}}} \;=\; \frac{C_{\mathcal{X}, P}}{4}\,\mathrm{SNR}^{1 - P}.$ At large SNR, OTFS wins by $\mathrm{SNR}^{P - 1}$ ; the coding-gain ratio $C_{\mathcal{X}, P}/4$ is a modest constant offset (order 1) determined by the DD-channel geometry.

"Diversity gain" is the slope advantage: $P$ -fold SNR decay. "Coding gain" is the offset advantage: at which SNR the curves cross. OTFS has both — the diversity gain is the dominant factor at high SNR, but the coding gain matters at low-to-moderate SNR.

Proof

Union bound tight at dominant pair

BER $\sim \mathrm{PEP}$ at the dominant error pair. At full rank, $\mathrm{PEP} \sim (\text{const})/\mathrm{SNR}^P$ .

Constant derivation

The constant involves the determinant $\det(\mathbf{A}(\Delta\mathbf{x}^*))$ at the worst error $\Delta\mathbf{x}^*$ . For QPSK and independent Rayleigh paths: $C_{\mathcal{X}, P} = P!/(4^P \prod_i |h_i|^2_{\text{avg}})$ .

Evaluate for equal-power paths

When all $|h_i|^2_{\text{avg}} = 1/P$ : $C_{\mathcal{X}, P} = P!\,P^P/4^P$ . For $P = 4$ : $24 \cdot 256/256 = 24$ . For $P = 8$ : $8! \cdot 8^8/4^8 \approx 2.6 \times 10^7$ .

OFDM reference

$\mathrm{BER}_{\text{OFDM}} \sim 1/(4\,\mathrm{SNR})$ . Ratio: $C_{P}/(4^{P-1})\,\mathrm{SNR}^{1-P}$ . For $P = 4$ , SNR = 20 dB: ratio $= 24/64 \cdot 10^{-3} = 4 \times 10^{-4}$ . $\blacksquare$

Full Delay-Doppler Diversity Theorem