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The Finite-Dimensional DD Signal Space

With the discrete Zak transform of Chapter 2 and the ISFFT/SFFT of this chapter, we can now assemble the full discrete delay-Doppler signal space. An OTFS frame of duration $T$ and bandwidth $W$ is characterized by an $M \times N$ DD grid, and every cell of this grid corresponds to a specific $(\tau, \nu)$ region in the continuous DD plane. This section works out the grid–continuum correspondence and discusses the practical constraints on $M$ , $N$ , $T$ , and $W$ .

Definition:
Discrete Delay-Doppler Grid

An OTFS frame has duration $T$ (seconds) and bandwidth $W$ (hertz). The time-frequency grid has $N$ time slots (OFDM symbols) and $M$ frequency slots (subcarriers), with $T = N T_s$ and $W = M \Delta f$ . The delay-Doppler grid has the same area: $M$ delay bins and $N$ Doppler bins. The grid resolutions are $\Delta \tau \;=\; \frac{1}{W}\;=\;\frac{1}{M \Delta f}, \qquad \Delta \nu \;=\; \frac{1}{T}\;=\;\frac{1}{N T_s}.$ A DD grid point $(\ell, k)$ corresponds to delay $\tau = \ell \Delta \tau$ and Doppler $\nu = k \Delta \nu$ . The extent of the grid is $[0, M\Delta\tau] \times [0, N\Delta\nu] = [0, 1/\Delta f] \times [0, 1/T_s]$ .

Theorem: Resolvability of DD Paths

Two physical paths with delays $\tau_1, \tau_2$ and Dopplers $\nu_1, \nu_2$ are resolvable on the discrete DD grid (i.e., they land in different grid cells) if and only if $|\tau_1 - \tau_2| \geq \Delta\tau = 1/W \quad \text{or} \quad |\nu_1 - \nu_2| \geq \Delta\nu = 1/T.$ Equivalently, two paths are unresolvable when they lie within the same DD grid cell of area $\Delta\tau \Delta\nu = 1/(TW)$ .

The DD grid resolution is the reciprocal of the observation window: more bandwidth gives finer delay resolution, longer frame gives finer Doppler resolution. This is the OTFS analog of the Rayleigh limit in radar: the $TW$ product determines the number of resolvable cells, and the grid density determines when paths blur together.

Proof

Grid cell interpretation

Each DD grid cell $(\ell, k)$ corresponds to the rectangle $\tau \in [\ell \Delta\tau, (\ell+1)\Delta\tau)$ and $\nu \in [k \Delta\nu, (k+1)\Delta\nu)$ .

Resolvable pair

If $|\tau_1 - \tau_2| \geq \Delta\tau$ or $|\nu_1 - \nu_2| \geq \Delta\nu$ , the two paths necessarily land in different cells.

Unresolvable pair

If both separations are below the grid resolution, the two paths occupy the same cell (the spreading function's 2D Dirac impulses are discretized into the same grid point), and the detector cannot distinguish them. $\blacksquare$

Discrete Zak Transform and Its Resolution

A finite time-domain signal of length $MN$ is mapped to the $M \times N$ DD grid via the discrete Zak transform. Observe how changing $M$ (while keeping $MN$ constant) reshapes the grid: fewer delay bins with more Doppler bins, or vice versa. The product $MN$ determines the total dimensionality of the signal space; the split between $M$ and $N$ sets the trade-off between delay and Doppler resolution.

Parameters

Signal type

Delay bins

M

16

Doppler bins

N

16

Random seed0

Example: Choosing OTFS Parameters for a Given Channel

A vehicular channel has $\tau_{\max} = 4\,\mu\text{s}$ , $f_D = 800$ Hz, at carrier $f_0 = 5.9$ GHz. The system operates with bandwidth $W = 20$ MHz and an OTFS frame duration $T = 4$ ms. (a) Choose $(M, N)$ such that both $\tau_{\max}$ and $f_D$ are within the grid. (b) Confirm that typical path separations are resolvable.

Solution

Delay-bin requirement

$\tau_{\max}$ must fit in the grid: $M \Delta\tau \geq \tau_{\max}$ . With $\Delta\tau = 1/W = 50$ ns, we need $M \geq \tau_{\max}/\Delta\tau = 4000/50 = 80$ . A convenient choice is $M = 512$ (a power of 2, close to 5G NR numerology-1).

Doppler-bin requirement

$f_D$ must fit within the grid (with room for a guard): $N \Delta\nu/2 \geq f_D$ . With $\Delta\nu = 1/T = 250$ Hz, we need $N \geq 2 f_D/\Delta\nu = 1600/250 \approx 6.4$ . A safe choice is $N = 16$ .

Resolvability

$\Delta\tau = 50$ ns: two paths with $|\tau_1 - \tau_2| \geq 50$ ns are resolved. Typical urban paths are $\sim 100$ ns apart, so resolvable. $\Delta\nu = 250$ Hz: two paths with $|\nu_1 - \nu_2| \geq 250$ Hz are resolved. For the given $f_D = 800$ Hz, up to $\lceil 2 f_D / \Delta\nu \rceil = 7$ distinct Doppler values are resolvable.

Full frame

$(M, N) = (512, 16)$ yields $MN = 8192$ QAM symbols per frame, fits $\tau_{\max}$ and $f_D$ with margin, and maps cleanly to 5G NR numerology-1 OFDM symbols. Two OTFS frames per subframe if the subframe is 1 ms, etc. This sizing is representative of realistic deployments.

⚠️Engineering Note

Aligning OTFS With 5G NR Numerology

For OTFS to be backward-compatible with 5G NR, the OFDM parameters must line up with standardized numerology:

Numerology 0: $\Delta f = 15$ kHz, $T_s = 66.7\,\mu\text{s}$ . Sub-6 GHz.
Numerology 1: $\Delta f = 30$ kHz, $T_s = 33.3\,\mu\text{s}$ . Sub-6 GHz / FR1 high-band.
Numerology 3: $\Delta f = 120$ kHz, $T_s = 8.33\,\mu\text{s}$ . FR2 (mmWave).

For a given numerology, $M$ is determined by bandwidth: $M = W/\Delta f$ . The frame duration $T = N T_s$ is then chosen based on how much Doppler resolution is needed. Typical choices: $N = 14$ for NR slot alignment, $N = 28$ or $N = 56$ for finer Doppler. Longer $N$ improves Doppler resolution but increases transmit-buffer size and processing latency.

Practical Constraints

•
NR numerology determines $\Delta f$ and $T_s$
•
One OTFS frame spans $N \geq 14$ NR symbols
•
Frame duration $T = N T_s$ is fixed by alignment; Doppler resolution $1/T$ follows

📋 Ref: 3GPP TS 38.211 §4.2

Guard Region Preview

With the grid fixed at $(M, N)$ and the channel with $\tau_{\max}, f_D$ known, the support of the channel's spreading function on the discrete DD grid is bounded by at most $l_{\max} = \lceil \tau_{\max}/\Delta\tau\rceil$ delay bins and $k_{\max} = \lceil f_D / \Delta\nu\rceil$ Doppler bins. This is the "channel footprint" — the region of the DD grid where channel taps actually appear.

For embedded pilot estimation (Chapter 7), we reserve a guard region of size $(2 l_{\max} + 1) \times (2 k_{\max} + 1)$ around the pilot to prevent data from corrupting the pilot response. In our running example ( $l_{\max} = 80, k_{\max} = 4$ ), the guard area is roughly $81 \times 9 = 729$ cells — less than 10% of the 8192-cell grid. This is the concrete meaning of "sparse channel = low pilot overhead."

Common Mistake: DD Grid Aliasing

Mistake:

Allowing $\tau_{\max} \geq M \Delta\tau$ (delay exceeds grid) or $2 f_D \geq N \Delta\nu$ (Doppler exceeds Nyquist on the Doppler axis). This causes aliasing: path taps wrap around to incorrect DD cells, and detection breaks.

Correction:

Always size $M$ and $N$ with margin: $M \geq 2\,\lceil \tau_{\max} W\rceil$ and $N \geq 2\,\lceil 2 f_D T\rceil$ . For safety, double these minimums when designing for unknown deployment scenarios (the required maximum delay and Doppler may be only approximately known at design time).

Theorem: Parseval's Theorem for the DD Grid

For any DD-grid matrix $X_{DD} \in \mathbb{C}^{M \times N}$ and its ISFFT image $X_{TF} \in \mathbb{C}^{N \times M}$ , $\sum_{\ell, k} |X_{DD}[\ell, k]|^2 \;=\; \sum_{n, m} |X_{TF}[n, m]|^2.$ The ISFFT/SFFT is unitary — it preserves the Frobenius norm exactly.

Unitary transforms preserve energy — no information loss, no amplification, no attenuation. In Shannon-theoretic terms, the mutual information $I(X_{DD}; Y_{TF})$ equals $I(X_{TF}; Y_{TF})$ if the transform is unitary. This means OTFS cannot achieve a higher total rate than the underlying OFDM; the gain is distributional — uniform channel quality across cells — rather than aggregate.

Proof

Frobenius norm preservation

Let $\mathbf{X} = X_{DD}$ and $\mathbf{X}' = X_{TF}$ . The ISFFT is the Kronecker product $F_N \otimes F_M^H$ applied to $\text{vec}(\mathbf{X})$ , which is unitary (product of unitaries). Hence $\|\text{vec}(\mathbf{X}')\|^2 = \|\text{vec}(\mathbf{X})\|^2$ .

Reinterpret as sums

$\|\text{vec}(\mathbf{X})\|^2 = \sum_{\ell, k} |X_{DD}[\ell, k]|^2$ and $\|\text{vec}(\mathbf{X}')\|^2 = \sum_{n, m} |X_{TF}[n, m]|^2$ . These are equal. $\blacksquare$

Delay-Doppler grid

The $M \times N$ discrete 2D grid with delay index $\ell \in \{0, \ldots, M-1\}$ and Doppler index $k \in \{0, \ldots, N-1\}$ . Grid resolutions are $\Delta\tau = 1/W$ and $\Delta\nu = 1/T$ . Total area = $M N = WT$ (time-bandwidth product). In OTFS, each grid cell carries one QAM data symbol, or a pilot, or a guard entry.

The Discrete Delay-Doppler Grid

The Finite-Dimensional DD Signal Space

Definition: Discrete Delay-Doppler Grid

Theorem: Resolvability of DD Paths

Grid cell interpretation

Resolvable pair

Unresolvable pair

Discrete Zak Transform and Its Resolution

Parameters

Example: Choosing OTFS Parameters for a Given Channel

Delay-bin requirement

Doppler-bin requirement

Resolvability

Full frame

Aligning OTFS With 5G NR Numerology

Guard Region Preview

Common Mistake: DD Grid Aliasing

Theorem: Parseval's Theorem for the DD Grid

Frobenius norm preservation

Reinterpret as sums

Delay-Doppler grid

Definition:
Discrete Delay-Doppler Grid