Exercises

$(\mathcal{F}_s F)(f, t) = \iint \delta(\tau)\delta(\nu)\,e^{-j 2\pi (\nu t - f \tau)}\,d\tau\,d\nu = e^{-j 2\pi (0 \cdot t - f \cdot 0)} = 1$.

The SFT of a point impulse at $(0, 0)$ is the constant function 1 on the TF plane — a DC signal. Dually, by inverting, the SFT of a constant function is a point impulse at the DD origin.

The ISFFT is defined as a linear combination of the entries of its input matrix, with fixed coefficients $e^{j 2\pi(k n / N - m \ell / M)} / \sqrt{MN}$.

Any linear combination of entries preserves linearity: $\text{ISFFT}(\alpha A + \beta B) = \alpha\,\text{ISFFT}(A) + \beta\,\text{ISFFT}(B)$. $\blacksquare$

$X_{TF}[n, m] = \frac{1}{4}\sum_{\ell, k} 1 \cdot e^{j 2\pi(k n / 4 - m \ell / 4)} = \frac{1}{4}\,\sum_{\ell} e^{-j 2\pi m \ell / 4}\,\sum_k e^{j 2\pi k n / 4}$.

Each inner sum is $4\,\mathbf{1}_{n = 0}$ and $4\,\mathbf{1}_{m = 0}$.

$X_{TF}[n, m] = \frac{1}{4} \cdot 16\,\mathbf{1}_{n = 0}\mathbf{1}_{m = 0} = 4\,\mathbf{1}_{n = 0}\mathbf{1}_{m = 0}$. The all-ones DD grid produces a single TF impulse at the origin. This is the SFT-duality of Exercise 1.

The SFT is a 2D Fourier transform composed with a coordinate flip $(\nu, \tau) \to (\nu, -\tau)$, which is a unitary involution on $L^2(\mathbb{R}^2)$.

The ordinary 2D Fourier transform preserves inner products (Plancherel). A unitary involution preserves inner products. The composition is therefore unitary. $\blacksquare$

$\Delta\tau = 1/W = 100$ ns, $\Delta\nu = 1/T = 500$ Hz.

Number of delay taps: $\lceil 5000/100\rceil = 50$.

Number of Doppler taps (one-sided): $\lceil 300/500\rceil = 1$. Counting negative Doppler: 3 taps ($-1, 0, +1$) in Doppler index.

Support $\leq 50 \times 3 = 150$ cells. Grid total: $600 \times 16 = 9600$. Fraction: $150/9600 \approx 1.6\%$. The channel occupies less than 2% of the grid — the sparsity we expected.

$X_{TF}[n, m] = \frac{1}{4}\,e^{j 2\pi(2 n / 4 - m \cdot 1 / 4)} = \frac{1}{4}\,e^{j \pi n}\,e^{-j \pi m / 2}$. Explicit values: $X_{TF}[0, 0] = 1/4$, $X_{TF}[0, 1] = -j/4$, etc.

$\text{SFFT}(X_{TF})[\ell', k'] = \frac{1}{4}\sum_{n, m} X_{TF}[n, m]\,e^{-j 2\pi(k' n / 4 - m \ell' / 4)} = \frac{1}{16}\sum_{n, m}e^{j 2\pi((2 - k') n / 4 - (\ell' - 1) m / 4)}$.

The double sum is $16\,\mathbf{1}_{k' = 2}\mathbf{1}_{\ell' = 1}$. So $\text{SFFT}(X_{TF})[\ell', k'] = \mathbf{1}_{(\ell', k') = (1, 2)}$. The original grid is recovered. $\blacksquare$

$Y_{TF}[n, m] = \frac{1}{\sqrt{MN}}\sum_{\ell, k} X_{DD}[(\ell - \ell_0) \bmod M, (k - k_0) \bmod N]\,e^{j 2\pi (k n/N - m \ell / M)}$.

$Y_{TF}[n, m] = \frac{1}{\sqrt{MN}}\sum_{\ell', k'} X_{DD}[\ell', k']\,e^{j 2\pi ((k' + k_0) n / N - m(\ell' + \ell_0) / M)}$ $\quad = e^{j 2\pi (k_0 n / N - m \ell_0 / M)}\,X_{TF}[n, m]$. $\blacksquare$

The ISFFT is a unitary map (Theorem TParseval's Theorem for the DD Grid). Applied to a Gaussian vector, it produces a Gaussian vector with same covariance (identity).

Each entry $X_{TF}[n, m]$ is a linear combination of i.i.d. $\mathcal{CN}$ with unit total variance: $\mathcal{CN}(0, 1)$.

$\mathbb{E}[X_{TF}[n, m]\overline{X_{TF}[n', m']}] = \delta_{n, n'}\delta_{m, m'}$ — pairwise uncorrelated. Hence $X_{TF}$ is an i.i.d. $\mathcal{CN}(0, 1)$ field.

The peak-to-average power ratio of an i.i.d. $\mathcal{CN}$ field is $O(\log(MN))$, matching OFDM exactly. OTFS does not degrade (or improve) PAPR relative to OFDM in the Gaussian-input case. Real QAM constellations yield slightly better PAPR; see Chapter 20 for pulse shaping that further reduces PAPR.

The ISFFT: $X_{TF} = \mathbf{F}_M^H\,X_{DD}\,\mathbf{F}_N^T / \sqrt{MN}$ (column DFT by $\mathbf{F}_M^H$, then row DFT by $\mathbf{F}_N$).

$\text{vec}(X_{TF}) = (\mathbf{F}_N \otimes \mathbf{F}_M^H)\,\text{vec}(X_{DD})$.

$\mathbf{F}_N$ and $\mathbf{F}_M^H$ are unitary, so their Kronecker product is unitary. The ISFFT is therefore unitary as a linear map on $\mathbb{C}^{MN}$. $\blacksquare$

$\phi_{0, 0}(t) = \sum_{n = 0}^{N-1}\sum_{m = 0}^{M-1} e^{0}\,g_{tx}(t - n T_s)\,e^{j 2\pi m \Delta f (t - n T_s)} = \sum_n g_{tx}(t - n T_s)\,\sum_m e^{j 2\pi m \Delta f(t - n T_s)}$.

$\sum_{m=0}^{M-1} e^{j 2\pi m \Delta f \tau} = \text{Dirichlet kernel}$, which is a sharp peak at $\tau = 0$ with width $\sim 1/(M \Delta f) = 1/W$.

$\phi_{0, 0}(t)$ is a train of $N$ sharp peaks, one per OFDM symbol, each of width $1/W$. In other words, the "DD cell at the origin" corresponds to a time-domain signal concentrated at the start of each symbol. In the Fourier sense, its spectrum is approximately flat over the bandwidth.

$MN = (W/\Delta f)(T/T_s) = (WT_s)(T/T_s) = TW$. Since $\Delta f T_s = 1$ in OFDM, $MN = TW$ exactly.

The DD signal space on $[0, T] \times [0, W]$ (DD rectangle of continuous area $TW$) has $TW$ degrees of freedom on the DD grid — one QAM symbol per cell.

OTFS and OFDM have the same capacity on the same TF rectangle — both transmit $TW$ QAM symbols per frame. The difference is not in total capacity but in robustness: OTFS achieves the capacity uniformly across DD cells, while OFDM's per-cell capacity degrades when a subcarrier fades. At the same SNR and same channel, both give the same expected throughput; OTFS gives the same per-channel throughput more reliably.

Consider a symplectic matrix $S = \begin{pmatrix}a & b\\ c & d\end{pmatrix}$ with $ad - bc = 1$. The corresponding rotation acts on $(\tau, \nu)$ by $(\tau, \nu) \mapsto (a\tau + b\nu, c\tau + d\nu)$.

The phase $\nu t - f \tau$ under simultaneous rotation of $(\tau, \nu)$ and $(f, t)$ by $S$ (resp. $(S^T)^{-1}$) is preserved — this is the defining invariance of the symplectic form.

The SFT intertwines symplectic rotations (delay-Doppler rotations corresponding to chirp modulation of the signal) with unitary transformations on $L^2(\mathbb{R})$. This is why linear canonical transforms (LCTs, which are representations of symplectic rotations) act on the DD plane in a geometrically clean way — and why chirp-OTFS variants are mathematically natural extensions.

$\Delta f = W/M = 10^7/128 = 78.125$ kHz.

$T_s = 1/\Delta f = 12.8\,\mu\text{s}$.

$T = N T_s = 16 \cdot 12.8\,\mu\text{s} = 204.8\,\mu\text{s}$.

$\Delta\tau = 1/W = 100$ ns. $\Delta\nu = 1/T = 4.88$ kHz.

$\tau_{\max}/\Delta\tau = 30$: fits in $M = 128$ (20% utilization). $f_D/\Delta\nu = 0.08$: less than one Doppler bin. Paths with Doppler differences $\ll \Delta\nu$ are not resolvable — this is an underspread regime where Doppler is not even visible on the grid. The OTFS advantages over OFDM are small here; OFDM would work fine.

For ordinary 2D DFT, $\hat{Y}[n, m] = \hat{X}[n, m]\,e^{-j 2\pi(k_0 n / N + m \ell_0 / M)}$ (both signs negative). The SFT: $X_{TF}[n, m] \to X_{TF}[n, m]\,e^{j 2\pi(k_0 n / N - m \ell_0 / M)}$ (mixed signs).

The Heisenberg-Weyl group is defined by non-commuting delay and Doppler operators with a specific commutation phase. The SFT mixed-sign pattern is what makes the 2D DFT intertwine these non-commuting operators correctly. Ordinary 2D DFT would treat delay and Doppler as if they commuted.

Under the SFT, a DD-domain translation maps to a TF-domain modulation, and the TF channel "sees" the DD translation as a 2D convolution. Under the ordinary DFT, this correspondence breaks — the TF input-output relation would have residual phase errors that accumulate nonlinearly. This is why the SFT, not the ordinary DFT, is used in OTFS.

Doppler resolution: $\Delta\nu = 1/T = 1/(N T_s)$. Larger $N$ gives finer $\Delta\nu$ but longer $T$ (more latency). Delay resolution: $\Delta\tau = 1/W = T_s/M$. Larger $M$ gives finer $\Delta\tau$ — unrelated to $N$.

For LEO ($f_D \sim 30$ kHz): require $\Delta\nu \ll f_D$, e.g., $\Delta\nu \approx 500$ Hz, giving $T = 2$ ms and $N = 2\,\text{ms}/T_s$. At numerology-1 ($T_s = 33\,\mu\text{s}$), $N \approx 60$ — three times larger than typical terrestrial OTFS frames.

A non-uniform DD grid — e.g., finer Doppler sampling near $\nu = 0$, coarser at high Doppler — could be designed but loses the unitarity and FFT-friendliness of the uniform grid. Efficient implementation requires non-uniform FFT techniques. This is an active research direction in LEO-OTFS; the uniform-grid OTFS of this chapter is the baseline, not the final word.