Exercises

$h(\tau, t) \;=\; 0.8\,e^{j\pi/3}\,e^{j 2\pi (200) t}\,\delta(\tau - 10^{-6}).$$

$h(\tau, \nu) \;=\; 0.8\,e^{j\pi/3}\,\delta(\tau - 10^{-6})\,\delta(\nu - 200).$$ One impulse in the plane, at the specified DD coordinates.

$v = 140$ km/h $= 38.9$ m/s. $f_D = (38.9 / 3 \times 10^8) \cdot 3.5 \times 10^9 = 453$ Hz. $T_c \sim 1/f_D \approx 2.2$ ms.

$f_D = (38.9 / 3 \times 10^8) \cdot 28 \times 10^9 = 3{,}627$ Hz. Coherence time shrinks by a factor of 8 to $\sim 0.28$ ms. This is the concrete reason mmWave 5G deployments struggle with mobility.

$H(f, t) = \int h(\tau, t)\,e^{-j 2\pi f \tau}\,d\tau = h_1\,e^{j 2\pi \nu_1 t}\,e^{-j 2\pi f \tau_1}$.

$h(\tau, \nu) = \int h(\tau, t)\,e^{-j 2\pi \nu t}\,dt = h_1\,\delta(\tau - \tau_1)\int e^{j 2\pi (\nu_1 - \nu) t}\,dt = h_1\,\delta(\tau - \tau_1)\,\delta(\nu - \nu_1)$.

From $H(f, t)$: $H(f, \nu) = h_1\,e^{-j 2\pi f \tau_1}\,\delta(\nu - \nu_1)$. From $h(\tau, \nu)$: $H(f, \nu) = h_1\,e^{-j 2\pi f \tau_1}\,\delta(\nu - \nu_1)$. Both routes agree, confirming the commuting square. $\blacksquare$

Assuming equal unit power per path, $$S_h(\tau, \nu) = \frac{1}{4}\left[\delta(\tau, \nu) + \delta(\tau - \tau_2, \nu) + \delta(\tau, \nu - \nu_3) + \delta(\tau - \tau_2, \nu - \nu_3)\right].$$

$P_\tau(\tau) = \int S_h(\tau, \nu)\,d\nu = \frac{1}{2}[\delta(\tau) + \delta(\tau - \tau_2)]$. $P_\nu(\nu) = \int S_h(\tau, \nu)\,d\tau = \frac{1}{2}[\delta(\nu) + \delta(\nu - \nu_3)]$.

The marginals are the same as those of a two-path channel with $(0, 0)$ and $(\tau_2, \nu_3)$, yet the scattering functions are different. The marginals lose the joint delay-Doppler structure; $S_h(\tau, \nu)$ carries strictly more information.

$P$ paths, each with one complex $h_i$ (2 real) + one real $\tau_i$

• one real $\nu_i$ = $4P$ real numbers. For $P = 10$: $40$ reals.

$MN$ complex samples $= 2MN$ reals. For the given parameters: $2 \cdot 1200 \cdot 14 = 33{,}600$ reals.

$33{,}600 / 40 = 840$. The TF representation is nearly three orders of magnitude larger — a direct measure of the compressibility gain offered by the DD representation.

$x_{\text{pass}}(t) = \Re\{s(t)\,e^{j 2\pi f_0 t}\}$, with $s(t)$ the complex baseband.

After reflection with delay $\tau_b$ and carrier Doppler $\nu_b$, the passband component is $y_{\text{pass}}(t) = h_b\,\Re\{s(t - \tau_b)\,e^{j 2\pi (f_0 + \nu_b)(t - \tau_b)}\}$.

Multiply by $e^{-j 2\pi f_0 t}$ and low-pass filter: $y_b(t) = h_b\,s(t - \tau_b)\,e^{j 2\pi \nu_b t}\,e^{-j 2\pi f_0 \tau_b}$. The $e^{j 2\pi \nu_b t}$ factor is the claimed Doppler modulation; $e^{-j 2\pi f_0 \tau_b}$ is a constant phase absorbed into $h_b$. $\blacksquare$

Powers sum to 1.0 already, so $p_i = |h_i|^2$.

$\bar{\tau} = 0.5 \cdot 0 + 0.3 \cdot 1 + 0.2 \cdot 2 = 0.7\,\mu\text{s}$.

$\mathbb{E}[\tau^2] = 0.3 \cdot 1 + 0.2 \cdot 4 = 1.1\,\mu\text{s}^2$, $\sigma_\tau^2 = 1.1 - 0.49 = 0.61$, $\sigma_\tau \approx 0.78\,\mu\text{s}$.

$\bar{\nu} = 0.3 \cdot 100 + 0.2 \cdot (-100) = 10$ Hz. $\mathbb{E}[\nu^2] = 0.3 \cdot 10^4 + 0.2 \cdot 10^4 = 5000$ Hz$^2$, $\sigma_\nu^2 = 5000 - 100 = 4900$, $\sigma_\nu = 70$ Hz.

$B_c \sim 1/\sigma_\tau \approx 1.28$ MHz. Over bandwidths narrower than this the channel is approximately flat.

To distinguish two paths with delay separation $\Delta \tau_i$ we need $\Delta \tau_i \geq 1/W$. To distinguish two Doppler separation $\Delta \nu_i$ we need $\Delta \nu_i \geq 1/T$.

Maximum delay must fit in the grid: $\tau_{\max} \leq M / W$, and maximum Doppler: $f_D \leq N / T / 2$ (Nyquist). Using $M = 1/(\Delta \tau \cdot f)$ and $N = T \cdot 2 f_D$, we get $MN \geq 2 \tau_{\max} f_D \cdot TW$.

An overspread channel ($\tau_{\max} f_D > 1$) requires $TW > 1$. At LEO parameters, $TW \sim 10^3$ is needed. An OTFS frame therefore cannot be arbitrarily short — frame duration and bandwidth must jointly exceed the channel's delay-Doppler area.

For any test function $\phi(\tau, \nu)$, let $X = \iint \phi(\tau, \nu)\,h(\tau, \nu)\,d\tau\,d\nu$. Clearly $\mathbb{E}[|X|^2] \geq 0$.

$\mathbb{E}[|X|^2] = \iint\iint \phi(\tau, \nu)\,\phi^*(\tau', \nu')\,\mathbb{E}[h(\tau, \nu)\,h^*(\tau', \nu')]\,d\tau\,d\nu\,d\tau'\,d\nu'.$$By WSSUS,$\mathbb{E}[h h^*] = S_h(\tau, \nu),\delta(\tau - \tau'),\delta(\nu - \nu')$. The integral collapses to$\iint |\phi(\tau, \nu)|^2,S_h(\tau, \nu),d\tau,d\nu \geq 0$.

Since this holds for arbitrary test $\phi$, and the integrand is a non-negative measure against $|\phi|^2$, $S_h(\tau, \nu) \geq 0$ pointwise (almost everywhere). $\blacksquare$

$\int_{-f_D}^{f_D} P_\nu(\nu)\,d\nu = \frac{1}{\pi}\int_{-1}^{1}(1-u^2)^{-1/2}\,du = 1$.

$\mathbb{E}[H(f, t) H^*(f, t + \Delta t)] = \int_{-f_D}^{f_D} P_\nu(\nu)\,e^{-j 2\pi \nu \Delta t}\,d\nu$. Using $\nu = f_D \cos\theta$: $= \frac{1}{\pi}\int_0^{\pi} e^{-j 2\pi f_D \Delta t \cos\theta}\,d\theta = J_0(2\pi f_D \Delta t)$, where $J_0$ is the zeroth Bessel function of the first kind.

The Jakes correlation is the signature of an isotropic ring of scatterers (a classical model for an outdoor mobile receiver surrounded by equidistant reflectors). The first zero of $J_0$ occurs at $2\pi f_D \Delta t \approx 2.4$, giving the Jakes coherence time $T_c \approx 0.4 / f_D$. This is the canonical "coherence time" cited in textbooks.

$T_c \approx 1/f_D$, $B_c \approx 1/\tau_{\max}$. The coherence area is $T_c B_c = 1/(\tau_{\max} f_D)$.

In a TF region of area $TW$, the number of coherence cells is $TW / (T_c B_c) = TW \cdot \tau_{\max} f_D$.

When $\tau_{\max} f_D = 1$, the coherence cell size equals the unit TF area — no "large" cells exist. For $\tau_{\max} f_D > 1$, the coherence cell is smaller than the resolvable TF unit, and OFDM cannot fit a symbol in one coherence cell. This is the information-theoretic version of the OFDM failure mode at high Doppler. OTFS, by signaling directly on the DD support, does not require coherence cells at all. $\blacksquare$

Uniform PDP: $P_\tau(\tau) = \mathbf{1}_{[0, \tau_{\max}]}(\tau)/\tau_{\max}$.

$\widehat{P_\tau}(\Delta f) = \int_0^{\tau_{\max}} \frac{1}{\tau_{\max}}\,e^{-j 2\pi \Delta f \tau}\,d\tau = \text{sinc}(\tau_{\max} \Delta f)\,e^{-j \pi \tau_{\max} \Delta f}$.

From Exercise 10, $\widehat{P_\nu}(-\Delta t) = J_0(2\pi f_D \Delta t)$.

$R_H(\Delta f, \Delta t) = \text{sinc}(\tau_{\max} \Delta f)\,e^{-j\pi \tau_{\max} \Delta f}\,J_0(2\pi f_D \Delta t)$. The product structure is a consequence of the WSSUS and separability assumptions; the frequency behavior is sinc-like (uniform PDP) while the time behavior is Bessel-like (isotropic scatterers).

$\mathcal{T}_{\tau_0} \mathcal{E}_{\nu_0}[x](t) = \mathcal{T}_{\tau_0}\{x(t)\,e^{j 2\pi \nu_0 t}\} = x(t - \tau_0)\,e^{j 2\pi \nu_0 (t - \tau_0)}$.

$\mathcal{E}_{\nu_0}\mathcal{T}_{\tau_0}[x](t) = x(t - \tau_0)\,e^{j 2\pi \nu_0 t}$.

$\text{LHS}/\text{RHS} = e^{-j 2\pi \nu_0 \tau_0}$, which matches the claim. $\blacksquare$

The single-path channel kernel $\pi_{\tau_i, \nu_i} = h_i\,\mathcal{T}_{\tau_i}\mathcal{E}_{\nu_i}$ does not commute with other $\pi_{\tau_j, \nu_j}$ except up to this phase factor. This non-commutativity is the Heisenberg–Weyl structure that underlies the Zak transform (Chapter 2) and the symplectic Fourier transform (Chapter 3). The "signal space" of OTFS is a projective representation of this group.

Treat the grid point as the sum of all path contributions with the corresponding $(\tau_i, \nu_i)$ close to $(\tau_0, \nu_0)$: $h(\tau_0, \nu_0) = \sum_{i \in \mathcal{I}_0} h_i$ where $\mathcal{I}_0$ is the set of paths within one DD resolution cell.

For $|\mathcal{I}_0|$ large and each $h_i \sim \mathcal{CN}(0, 1/P)$ i.i.d., the sum is approximately $\mathcal{CN}(0, \sigma^2)$ with $\sigma^2 = |\mathcal{I}_0|/P$.

$|X|^2$ for $X \sim \mathcal{CN}(0, \sigma^2)$ is exponentially distributed with mean $\sigma^2$. Thus the per-cell power in the dense regime is exponential — the DD-domain manifestation of Rayleigh fading.

When the number of physical paths is very large and the per-path gains are weak (classical Rayleigh channel), the DD representation is dense rather than sparse; the WSSUS picture applies directly. This is the limiting regime of the dense-multipath assumption in Telecom Ch. 6. For OTFS, the useful regime is the opposite extreme: few strong paths, each cleanly resolvable.

At the horizon the line-of-sight velocity equals the orbital velocity: $f_D = (v_{\text{orb}}/c) f_0 = (7600/3\times10^8)\cdot 10^{10} \approx 253$ kHz. At zenith $v_{\text{LOS}} = 0$; $f_D$ sweeps from $+253$ kHz through zero to $-253$ kHz over the pass.

Radial distance varies by $\sim 500$ km over the pass, giving $\tau_{\max} \approx 500\text{ km}/c = 1.67$ ms.

$\tau_{\max}\, f_D \approx (1.67 \times 10^{-3})(2.53 \times 10^5) \approx 420 \gg 1$. Deeply overspread.

OFDM with subcarrier spacing $\Delta f = 15$ kHz cannot tolerate a Doppler of 253 kHz — ICI spreads energy across 17 subcarriers, scrambling all data. Even with $\Delta f = 120$ kHz (NR numerology 3), the Doppler exceeds two subcarriers. OTFS, by contrast, resolves each path's Doppler into a specific $k$ bin on the DD grid; the channel remains sparse even in this extreme scenario. This is the core motivation of Chapter 18 (LEO satellite OTFS).

As $P \to \infty$, the DD density of support points per resolution cell grows without bound. Sparse recovery techniques (OMP, ISTA) no longer apply because there is no sparse representation to exploit. $h(\tau, \nu)$ is essentially a continuous function of the DD coordinates.

Under WSSUS, $K_h(\tau, \nu; \tau', \nu') = S_h(\tau, \nu)\,\delta(\tau-\tau')\,\delta(\nu-\nu')$. Restricted to a finite DD grid, the effective channel vector is $\mathbf{h}_{DD}\in \mathbb{C}^{MN}$ with diagonal covariance $\text{diag}(S_h[\ell, k])$.

Since $\mathbf{h}_{DD}$ has a known diagonal covariance, LMMSE detection is optimal for the linear Gaussian problem. The "number of paths" is replaced by the effective rank — the number of cells with significant $S_h$. The resulting detector is nearly identical to the LCD detector (low-complexity DD-domain, Chapter 8) under a slightly different statistical prior.

A fully principled treatment of OTFS in the dense-multipath limit is an active research problem. See Chapter 22 for a discussion of open questions in this area.