Ferkans — Interactive Telecom Tutor

Numbers, Not Just Slopes

The diversity theorem gives an asymptotic slope. For deployment we need actual numbers: BER at SNR = 10 dB, BER at SNR = 25 dB, and so on. This section derives explicit closed-form BER expressions for OTFS under canonical channel models and validates them against Monte Carlo simulations. The result is a concrete performance calculator for OTFS links.

Theorem: Uncoded BER for QPSK-OTFS Over a $P$ -Path Rayleigh Channel

For QPSK input, ML detection, and a $P$ -path channel with equal-power Rayleigh paths ( $|h_i|^2_{\text{avg}} = 1/P$ ), the uncoded BER is $\mathrm{BER}_{\text{OTFS}}(\rho) \;=\; \left(\frac{1}{2}\right)^P \sum_{\ell = 0}^{P - 1} \binom{P - 1 + \ell}{\ell}\left(1 + \rho^{-1}\right)^{-\ell},$ where $\rho = \mathrm{SNR}$ . At high SNR this simplifies to $\mathrm{BER}_{\text{OTFS}}(\rho) \;\approx\; \binom{2P - 1}{P}\,(4\rho)^{-P}.$

The result is the standard maximal-ratio-combining (MRC) BER for $P$ -fold Rayleigh diversity. OTFS's DD-domain spreading acts as an inherent $P$ -fold combiner — every data symbol is received through $P$ independent fading realizations and combined coherently by the ML detector.

The binomial coefficient $\binom{2P-1}{P}$ is the "penalty" for having $P$ Rayleigh fades instead of a single AWGN channel. For $P = 4$ : $\binom{7}{4} = 35$ . For $P = 8$ : $\binom{15}{8} = 6435$ .

Proof

PEP for QPSK

$\mathrm{PEP}(\mathrm{SNR}, \mathbf{h}) = Q(\sqrt{2|\mathbf{H}\Delta\mathbf{x}|^2 \mathrm{SNR}})$ . For QPSK nearest-neighbor error, $|\Delta\mathbf{x}|^2 = 2$ .

Channel PDF

$|\mathbf{H}\Delta\mathbf{x}|^2/2 = \sum_i |h_i|^2$ (for a suitable error vector that spreads over $P$ cells). This is chi-squared with $2P$ DoF, mean 1.

Average over channel

$\mathrm{BER} = \int_0^\infty Q(\sqrt{2\rho x}) \cdot p_{\chi^2}(x; 2P)\,dx$ . This integral evaluates to the closed form above.

High-SNR asymptote

Using the expansion $Q(\sqrt{2\rho x}) \approx \exp(-\rho x)/\sqrt{4\pi\rho x}$ , the integral becomes a Gamma-function moment, yielding $\mathrm{BER} \approx \binom{2P - 1}{P}/(4\rho)^P$ . $\blacksquare$

,

Closed-Form BER vs Monte Carlo: $P$ -Path OTFS

Plot the closed-form BER formula alongside a Monte Carlo simulation for validation. The two curves should agree within a fraction of a dB for reasonable sample counts. Different $P$ values give different slopes. The asymptotic BER formula is the dashed straight line in log-log coordinates.

Parameters

P

4

Min SNR (dB)5

Max SNR (dB)30

MC trials2000

Theorem: QAM BER and Constellation Penalty

For $M$ -QAM with $M = 2^b$ (e.g., 16-QAM with $b = 4$ ), the uncoded BER in a $P$ -path OTFS channel has the same $\mathrm{SNR}^{-P}$ scaling but with a constellation-dependent multiplier: $\mathrm{BER}_{\text{QAM}}(\rho) \;\approx\; \frac{4(1 - 1/\sqrt{M})}{b}\,\binom{2P - 1}{P}\,(4\rho/(M - 1))^{-P}.$ At high $M$ , the BER is larger (more constellation points = more nearest-neighbor collisions), but the slope is still $P$ .

Higher-order QAM packs more bits per symbol but places constellation points closer together, increasing the nearest- neighbor error rate. The $1/(M - 1)$ factor in the SNR term represents this: effectively, at SNR $\rho$ , the QAM detector operates at "effective SNR" $\rho/(M-1)$ . For QPSK ( $M = 4$ ), factor is $1/3 \approx 1$ ; for 64-QAM ( $M = 64$ ), factor is $1/63$ , so 16 dB extra SNR needed for same BER.

OTFS preserves this scaling with its $P$ -fold diversity — a $P = 4$ system at 64-QAM with SNR = 25 dB still achieves $\mathrm{BER} \sim 10^{-5}$ , well above the OFDM baseline.

Example: Link Budget for 16-QAM OTFS

An OTFS link uses 16-QAM over a $P = 4$ -path channel. The target BER is $10^{-5}$ (uncoded). What SNR is required?

Solution

BER formula

$\mathrm{BER} = (4 \cdot 0.75/4)\,\binom{7}{4}\,(4 \rho/15)^{-4} = 0.75 \cdot 35 \cdot (15/(4\rho))^4 = 26.25 \cdot (15/(4\rho))^4$ .

Solve for $\rho$

$10^{-5} = 26.25 \cdot (15/(4\rho))^4$ $\Rightarrow (15/(4\rho))^4 = 3.8 \times 10^{-7}$ $\Rightarrow 15/(4\rho) = 0.025$ $\Rightarrow \rho = 150$ ( $\approx 22$ dB).

OFDM comparison

For OFDM with single-tap Rayleigh: 16-QAM at $10^{-5}$ needs $\rho \sim 10^4$ ( $40$ dB). OTFS saves 18 dB in this case — enormous margin.

Caveat: coded performance

Channel coding changes the comparison. With rate-1/2 LDPC, OFDM can reach $10^{-5}$ at around 15 dB and OTFS at around 13 dB — the gap shrinks to $\sim 2$ dB. The uncoded comparison is the extreme case.

OTFS BER for Multiple QAM Orders

Plot uncoded BER vs SNR for QPSK, 16-QAM, 64-QAM, 256-QAM in an OTFS system with $P = 4$ . All curves have diversity slope 4, but different coding-gain offsets. 256-QAM requires $\sim 18$ dB more than QPSK at equivalent BER. Observe where each curve crosses the BER $= 10^{-5}$ target line for link-budget design.

Parameters

P

4

Max SNR (dB)40

Theorem: Outage Probability in OTFS

The $\epsilon$ -outage capacity of an OTFS channel — the maximum rate achievable with failure probability $\epsilon$ — is $C_\epsilon \;=\; \log_2(1 + \rho \cdot F_{\chi^2, 2P}^{-1}(\epsilon)),$ where $F_{\chi^2, 2P}^{-1}(\epsilon)$ is the $\epsilon$ -quantile of the chi-squared distribution with $2P$ degrees of freedom. At small $\epsilon$ , $F^{-1}(\epsilon) \approx P \cdot \epsilon^{1/P}$ , and the outage capacity is approximately $\log_2(1 + \rho \cdot P \epsilon^{1/P})$ — much better than OFDM's $\log_2(1 + \rho \epsilon)$ .

Outage capacity is the rate guaranteed with high probability, not the average. Under single-tap Rayleigh (OFDM), the worst-case 10%-outage capacity is very small — deep fades are not averaged out. Under $P$ -fold diversity (OTFS), the chance of simultaneous deep fades in all paths is $\epsilon^P$ , making the outage probability much smaller. This is the operational advantage of OTFS for URLLC (ultra-reliable low-latency) traffic.

Proof

Effective SNR

For OTFS with ML combining, the effective SNR is $\rho \sum_i |h_i|^2$ . For equal-power Rayleigh, this is $\rho \cdot \chi^2_{2P}/P$ (after normalization).

Outage definition

$\Pr(\log_2(1 + \rho \chi^2_{2P}/P) < R) = \Pr(\chi^2_{2P}/P < (2^R - 1)/\rho) = F_{\chi^2, 2P}((2^R - 1)P/\rho)$ . Setting this to $\epsilon$ gives the $C_\epsilon$ formula.

Quantile approximation

For small $\epsilon$ , the chi-squared lower-tail has the approximation $F_{\chi^2, 2P}^{-1}(\epsilon) \approx P \epsilon^{1/P}$ . The outage capacity is therefore $\log_2(1 + \rho P \epsilon^{1/P})$ .

OFDM comparison

For $P = 1$ : OFDM's outage capacity is $\log_2(1 + \rho \epsilon)$ . For $P = 8, \epsilon = 10^{-3}$ : OTFS's outage is $\log_2(1 + 8\rho \cdot 10^{-3/8}) = \log_2(1 + 8\rho \cdot 0.42) = \log_2(1 + 3.4\rho)$ . OFDM: $\log_2(1 + 10^{-3}\rho)$ . At $\rho = 10^3$ : OTFS outage = $\log_2(3400) \approx 11.7$ bits; OFDM outage = $\log_2(2) = 1$ bit. OTFS 12× higher outage rate. $\blacksquare$

Key Takeaway

OTFS's outage advantage scales exponentially with $P$ . The probability of a deep fade in all $P$ paths is $\epsilon^P$ , giving the OTFS outage capacity the $\epsilon^{1/P}$ quantile scaling. For URLLC applications requiring $\epsilon = 10^{-5}$ outage at moderate SNR, OTFS with $P = 4$ is decisively better than OFDM. This is the quantitative argument behind OTFS's suitability for URLLC traffic in 6G.

🔧Engineering Note

Practical BER at Typical Deployment Parameters

Representative BER numbers at SNR = 20 dB for key deployment scenarios:

Pedestrian, 5 GHz, $P = 3$ : $\mathrm{BER}_{\text{OTFS}}\approx 10^{-6}$ , $\mathrm{BER}_{\text{OFDM}} \approx 10^{-3}$ . OTFS 3 orders better.
Urban vehicular, 3.5 GHz, $P = 6$ : $\mathrm{BER}_{\text{OTFS}}\approx 10^{-10}$ , $\mathrm{BER}_{\text{OFDM}} \approx 10^{-3}$ (with ICI).
HST, 3.5 GHz, $P = 4$ : $\mathrm{BER}_{\text{OTFS}}\approx 10^{-8}$ , $\mathrm{BER}_{\text{OFDM}} \approx 10^{-3}$ (floor due to ICI).
LEO sat, 10 GHz, $P = 2$ : $\mathrm{BER}_{\text{OTFS}}\approx 10^{-4}$ , OFDM: fails ( $\mathrm{BER} \approx 0.5$ , ICI dominant).

These are uncoded numbers; coded BER is 3-5 orders better, but the OTFS advantage ratio remains.

Practical Constraints

•
Numbers assume ML detection. MP/LCD add 1-2 dB
•
Monte Carlo simulation gives curves within 0.5 dB at $10^5$ trials
•
URLLC (10^-6 target) is where OTFS diversity pays most

BER Analysis: Asymptotic and Monte Carlo

Numbers, Not Just Slopes

Theorem: Uncoded BER for QPSK-OTFS Over a PPP-Path Rayleigh Channel

PEP for QPSK

Channel PDF

Average over channel

High-SNR asymptote

Closed-Form BER vs Monte Carlo: PPP-Path OTFS

Parameters

Theorem: QAM BER and Constellation Penalty

Example: Link Budget for 16-QAM OTFS

BER formula

Solve for $\rho$

OFDM comparison

Caveat: coded performance

OTFS BER for Multiple QAM Orders

Parameters

Theorem: Outage Probability in OTFS

Effective SNR

Outage definition

Quantile approximation

OFDM comparison

Key Takeaway

Practical BER at Typical Deployment Parameters

Theorem: Uncoded BER for QPSK-OTFS Over a $P$ -Path Rayleigh Channel

Closed-Form BER vs Monte Carlo: $P$ -Path OTFS