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What Transform Do We Need?

Chapter 1 told us the channel lives naturally in the delay-Doppler plane. It did not tell us how to map an ordinary time-domain signal $x(t)$ into that plane. This chapter builds that map. The answer is the Zak transform.

A naive guess would be the 2D Fourier transform of $x$ , but $x(t)$ is a function of one variable, not two — we cannot simply 2D-Fourier-transform it. The Zak transform is, cleverly, a 1D transform of $x$ whose output is a function of two variables $(t, \nu)$ . It does this by folding $x$ onto itself at period $T_0$ and simultaneously collecting a Fourier series in that folding. The result lives on a 2D torus and has exactly the quasi-periodicity structure that the delay-Doppler plane needs.

The Zak transform was introduced by Joshua Zak in 1967 in the context of solid-state physics (Bloch functions in electron band theory). It was rediscovered in signal processing in the 1980s by Janssen, as the right tool for Gabor analysis. For OTFS, the Zak transform is how a continuous-time waveform meets the 2D DD grid.

Definition:
Zak Transform

Let $f: \mathbb{R} \to \mathbb{C}$ be a sufficiently regular function (e.g., in $L^2(\mathbb{R})$ ). For a parameter $T_0 > 0$ (the Zak period), the Zak transform of $f$ is $Z_f(t, \nu) \;=\; \sum_{k \in \mathbb{Z}} f(t - k\,T_0)\,e^{-j 2\pi k\,\nu\,T_0},$ where $t \in \mathbb{R}$ and $\nu \in \mathbb{R}$ are the time and Doppler variables on the delay-Doppler plane. The dual frequency period is $\nu_0 = 1/T_0$ .

Different authors use different sign conventions. Hadani–Rakib (2017) and Raviteja–Viterbo (2018) use $e^{-j 2\pi k\,\nu\,T_0}$ — we adopt that convention throughout this book to match the OTFS literature. Some mathematical references (e.g., Folland 1989) use the opposite sign, which flips the role of the Doppler axis.

, ,

The Zak Transform Is a Fourier Series in Disguise

Fix $t$ . The sequence $\{f(t - k T_0)\}_{k \in \mathbb{Z}}$ consists of samples of $f$ spaced by $T_0$ . The Zak transform in $\nu$ is exactly the Fourier series of this sequence. So $Z_f(t, \nu)$ takes a continuous function $f(t)$ and encodes, at each time $t$ , the Fourier series of the infinite samples $\{f(t - k T_0)\}$ .

The reason this is useful: the sparse DD channel acts by delay $(t \mapsto t - \tau_i)$ and Doppler $(f \mapsto f\,e^{j 2\pi \nu_i t})$ — exactly the two operations that the Zak transform naturally respects, as we now prove.

Theorem: Zak Transform of a Delay-and-Doppler-Shifted Signal

Let $\pi_{\tau_0, \nu_0}$ denote the delay-and-Doppler-shift operator: $(\pi_{\tau_0, \nu_0} f)(t) = f(t - \tau_0)\,e^{j 2\pi \nu_0 t}$ . Then $Z_{\pi_{\tau_0, \nu_0} f}(t, \nu) \;=\; e^{j 2\pi \nu_0 t}\,Z_f(t - \tau_0, \nu - \nu_0).$ That is: a delay and Doppler shift of $f$ becomes a translation in the DD plane of $Z_f$ , up to a phase factor.

This is the defining "covariance property" of the Zak transform — and the reason it is the correct signal space for the DD domain. A channel that delays and Doppler-shifts the transmit signal by $(\tau_0, \nu_0)$ simply translates the DD-domain picture by $(\tau_0, \nu_0)$ . The input-output relation becomes a 2D convolution, as promised in Chapter 1.

Proof

Substitute the operator

$(\pi_{\tau_0, \nu_0} f)(t) = f(t - \tau_0)\,e^{j 2\pi \nu_0 t}$ . Apply the Zak transform: $Z_{\pi_{\tau_0, \nu_0} f}(t, \nu) = \sum_k f(t - \tau_0 - k T_0)\,e^{j 2\pi \nu_0 (t - k T_0)}\,e^{-j 2\pi k \nu T_0}.$

Group the phase factors

Pull $e^{j 2\pi \nu_0 t}$ out (it does not depend on $k$ ), and combine the remaining exponentials: $= e^{j 2\pi \nu_0 t} \sum_k f(t - \tau_0 - k T_0)\,e^{-j 2\pi k (\nu - \nu_0) T_0}.$

Recognize the Zak transform

The sum is precisely $Z_f(t - \tau_0, \nu - \nu_0)$ . Hence $Z_{\pi_{\tau_0, \nu_0} f}(t, \nu) = e^{j 2\pi \nu_0 t}\,Z_f(t - \tau_0, \nu - \nu_0)$ . $\blacksquare$

Key Takeaway

The Zak transform is covariant under delay-Doppler translation. A time-domain signal that has been delayed by $\tau_0$ and Doppler-shifted by $\nu_0$ appears, in the DD domain, as the same DD picture shifted by $(\tau_0, \nu_0)$ . This single property is what makes the Zak transform the right signal space for the delay-Doppler channel. The rest of OTFS is, loosely, bookkeeping around this fact.

Example: Zak Transform of a Rectangular Pulse

Let $g(t) = \mathbf{1}_{[0, T_0)}(t)$ be the rectangular pulse of unit height supported on $[0, T_0)$ , i.e., the indicator of the fundamental interval. Compute $Z_g(t, \nu)$ and interpret the result.

Solution

Evaluate the sum

For $t \in [0, T_0)$ and any integer $k$ , $g(t - k T_0) = \mathbf{1}_{[0, T_0)}(t - k T_0)$ equals $1$ precisely when $k = 0$ (since $t \in [0, T_0)$ lies inside the support) and $0$ for all other $k$ .

Read off the Zak transform

Therefore $Z_g(t, \nu) = 1$ for $t \in [0, T_0)$ and all $\nu$ . Outside the fundamental interval, quasi-periodicity (Section 2) determines the values.

Interpret

The rectangular pulse of width $T_0$ produces a Zak transform that is constant over the fundamental rectangle $[0, T_0) \times [0, \nu_0)$ . This is the "prototype pulse" used in idealized OTFS — its flat Zak transform makes the DD grid perfectly orthonormal. In practice we use smoothed pulses (raised cosine, etc.), whose Zak transforms have slight ripple.

Zak Transform of a Rectangular Pulse of Variable Width

Plot $|Z_g(t, \nu)|$ for a rectangular pulse $g(t) = \mathbf{1}_{[0, W_0]}(t)$ of variable width $W_0$ on a Zak grid with period $T_0$ . When $W_0 = T_0$ the Zak transform is uniform and flat; when $W_0 \neq T_0$ it exhibits fringes that signal aliasing in the DD grid. Slide the pulse width through one period to see the clean case and the aliased cases.

Parameters

Pulse width

W_0 / T_0

1

Period

T_0

(

\mu

s)4

Example: Zak Transform of a Dirac Impulse

Compute $Z_\delta(t, \nu)$ where $\delta$ is the Dirac impulse at the origin.

Solution

Apply the definition

$Z_\delta(t, \nu) = \sum_k \delta(t - k T_0)\,e^{-j 2\pi k \nu T_0}$ . The delta $\delta(t - k T_0)$ vanishes except when $t = k T_0$ .

Impulse train in the DD plane

The Zak transform is a train of impulses along the time axis, at positions $t = k T_0$ : $Z_\delta(t, \nu) = \sum_k \delta(t - k T_0)\,e^{-j 2\pi k \nu T_0}.$

Interpret

At each $t = k T_0$ the Doppler factor is $e^{-j 2\pi k \nu T_0}$ — a phase that rotates around the Doppler axis at rate $k$ . A time-domain Dirac becomes an impulse train on the DD plane. Dually, the Zak transform of a constant signal is a Dirac at $\nu = 0$ (see exercises).

Building $Z_f(t, \nu)$ by Folding

The Zak transform is built by taking a time-domain signal

f(t)

, replicating it at integer multiples of

T_0

, and summing each copy with a phase rotation along the Doppler axis. The animation builds the first few terms of the sum and shows the resulting function on the 2D DD plane.

The Zak Transform is a Unitary Map

One can show that the Zak transform extends to a unitary isomorphism between $L^2(\mathbb{R})$ and $L^2(\mathbb{T}^2)$ , where $\mathbb{T}^2 = [0, T_0) \times [0, \nu_0)$ is the fundamental rectangle (the "quasi-periodic cell"). Concretely, $\int_{\mathbb{R}} |f(t)|^2\,dt \;=\; T_0 \int_0^{T_0}\!\!\int_0^{\nu_0} |Z_f(t, \nu)|^2\,d\nu\,dt.$ Unitarity is important because it means the Zak transform preserves signal energy (no free information loss). We will need unitarity in Chapter 6 when designing the OTFS transmitter.

⚠️Engineering Note

Choosing the Zak Period

In OTFS, the Zak period $T_0$ is tied to the OTFS frame parameters: $T_0 = T/N$ where $T$ is the frame duration and $N$ is the number of Doppler bins. Equivalently, $T_0$ is the OFDM symbol duration after Heisenberg embedding. Choosing $T_0$ determines the Doppler resolution $\Delta\nu = 1/T$ (since the DD plane has $N$ Doppler bins over $[0, \nu_0) = [0, 1/T_0)$ ).

Key rule: $T_0 \times f_{D,\max}$ must satisfy $T_0 f_{D,\max} < 1/2$ to avoid Doppler aliasing (the Zak-plane analog of Nyquist). For vehicular channels ( $f_D \approx 500$ Hz) this requires $T_0 < 1$ ms; LEO satellite channels ( $f_D \approx 30$ kHz) require $T_0 < 17\,\mu\text{s}$ .

Practical Constraints

•
$T_0\, f_{D,\max} < 1/2$ to avoid Doppler aliasing
•
Frame duration $T = N T_0$ determines Doppler resolution $\Delta\nu = 1/T$
•
Practical trade-off: smaller $T_0$ gives more Doppler slack but shorter frames

📋 Ref: Hadani et al. 2017, §II-B

Common Mistake: When the Zak Sum Converges

Mistake:

Treating the Zak transform as "always well-defined" — the infinite sum $\sum_k f(t - k T_0)\,e^{-j 2\pi k \nu T_0}$ can fail to converge for signals that do not decay, such as constant or slowly-varying signals.

Correction:

Strictly speaking, the Zak transform converges absolutely for $f \in L^2(\mathbb{R})$ in the $L^2$ -sense over the fundamental domain $\mathbb{T}^2$ . Pointwise convergence for $L^1$ signals is guaranteed almost everywhere. For rigor in OTFS, treat the Zak transform as a map $L^2(\mathbb{R}) \to L^2(\mathbb{T}^2)$ — the discrete version (Section 4) sidesteps these subtleties entirely.

Definition and Motivation

What Transform Do We Need?

Definition: Zak Transform

The Zak Transform Is a Fourier Series in Disguise

Theorem: Zak Transform of a Delay-and-Doppler-Shifted Signal

Substitute the operator

Group the phase factors

Recognize the Zak transform

Key Takeaway

Example: Zak Transform of a Rectangular Pulse

Evaluate the sum

Read off the Zak transform

Interpret

Zak Transform of a Rectangular Pulse of Variable Width

Parameters

Example: Zak Transform of a Dirac Impulse

Apply the definition

Impulse train in the DD plane

Interpret

Building Zf(t,ν)Z_f(t, \nu)Zf​(t,ν) by Folding

The Zak Transform is a Unitary Map

Choosing the Zak Period

Common Mistake: When the Zak Sum Converges

Definition:
Zak Transform

Building $Z_f(t, \nu)$ by Folding