Exercises

$\Delta r = \frac{c}{2B} = \frac{3 \times 10^8}{2 \times 75 \times 10^6} = 2.0\;\text{m}$

$\lambda = \frac{c}{f_0} = \frac{3 \times 10^8}{5.9 \times 10^9} = 50.85\;\text{mm}$

$\Delta v = \frac{\lambda}{2T} = \frac{50.85 \times 10^{-3}}{2 \times 20 \times 10^{-3}} = 1.271\;\text{m/s} \approx 4.58\;\text{km/h}$

$BT = 75 \times 10^6 \times 20 \times 10^{-3} = 1.5 \times 10^6$

$\Delta r \cdot \Delta v = 2.0 \times 1.271 = 2.542\;\text{m} \cdot \text{m/s}$

Alternatively, $\Delta r \cdot \Delta v = \frac{c\lambda}{4BT} = \frac{3 \times 10^8 \times 50.85 \times 10^{-3}}{4 \times 1.5 \times 10^6} = 2.542\;\text{m} \cdot \text{m/s}$.

$\chi(0, 0) = (1 - 0) \cdot \mathrm{sinc}(0) = 1 \cdot 1 \cdot T_p$ (when the amplitude is normalised to give energy $T_p$). More precisely, $\chi(0,0) = \int_0^{T_p} |s(t)|^2 dt = T_p$ for a unit-amplitude pulse.

$|\chi(\tau, 0)|^2 = (1 - |\tau|/T_p)^2 T_p^2$. Setting $(1 - |\tau|/T_p)^2 = 1/2$: $1 - |\tau|/T_p = 1/\sqrt{2}$, so $|\tau| = T_p(1 - 1/\sqrt{2}) \approx 0.293 T_p$. The full $-3$ dB width is $2 \times 0.293 T_p = 0.586 T_p$.

$|\chi(0, f_d)|^2 = T_p^2 \mathrm{sinc}^2(f_d T_p)$. The $-3$ dB width of $\mathrm{sinc}^2(x)$ is approximately $0.886/T_p$ in frequency, giving a Doppler resolution of $\Delta f_d \approx 0.886/T_p$.

The rectangular pulse has a time-bandwidth product $BT_p = 1$ (bandwidth $B \approx 1/T_p$), which is the minimum possible. This means improving range resolution (shorter pulse, larger $B$) degrades Doppler resolution (shorter observation time), and vice versa. Pulse compression waveforms (chirps, phase codes) achieve $BT_p \gg 1$, decoupling range and Doppler resolution.

$\mu = B/T_c = 10^9 / (50 \times 10^{-6}) = 2 \times 10^{13}\;\text{Hz/s} = 20\;\text{MHz/}\mu\text{s}$

$f_b = \mu \cdot \frac{2R}{c} = 2 \times 10^{13} \times \frac{2 \times 30}{3 \times 10^8} = 4 \times 10^6\;\text{Hz} = 4\;\text{MHz}$

$f_{b,\max} = f_s / 2 = 5\;\text{MHz}$

$R_{\max} = \frac{f_{b,\max} \cdot c}{2\mu} = \frac{5 \times 10^6 \times 3 \times 10^8}{2 \times 2 \times 10^{13}} = 37.5\;\text{m}$

FMCW: $\Delta r = c/(2B) = 3 \times 10^8 / (2 \times 10^9) = 0.15\;\text{m}$

OFDM (Ex. 1): $\Delta r = 2.0\;\text{m}$

The FMCW radar has $\sim$13 times better range resolution due to its much wider bandwidth (1 GHz vs 75 MHz).

$K = B / \Delta f = 100 \times 10^6 / (120 \times 10^3) \approx 833$ (in 5G NR, the actual value may differ due to guard bands, but this is the approximate count).

$\Delta r = c / (2B) = 3 \times 10^8 / (2 \times 10^8) = 1.5\;\text{m}$

$R_{\max} = c / (2\Delta f) = 3 \times 10^8 / (2 \times 120 \times 10^3) = 1250\;\text{m}$

$\lambda = c / f_0 = 3 \times 10^8 / (28 \times 10^9) = 10.71\;\text{mm}$

Total observation time: $T = M \cdot T_{\text{sym}} = 14 \times 8.93 \times 10^{-6} = 125.0\;\mu\text{s}$

$\Delta v = \lambda / (2T) = 10.71 \times 10^{-3} / (2 \times 125.0 \times 10^{-6}) = 42.86\;\text{m/s} \approx 154\;\text{km/h}$

$v_{\max} = \lambda \Delta f / 2 = 10.71 \times 10^{-3} \times 120 \times 10^3 / 2 = 642.9\;\text{m/s} \approx 2314\;\text{km/h}$

The velocity resolution from a single slot is quite coarse (42.86 m/s); coherent processing across multiple slots is needed for finer Doppler resolution in automotive applications.

After OFDM demodulation, the received signal on subcarrier $k$ of symbol $m$ is: $Y_{m,k} = \alpha X_{m,k} e^{-j2\pi k\Delta f \tau} e^{j2\pi m T_{\text{sym}} f_d} + N_{m,k}$

Dividing by $X_{m,k}$: $\tilde{Y}_{m,k} = \alpha e^{-j2\pi k\Delta f \tau} e^{j2\pi m T_{\text{sym}} f_d} + N_{m,k}/X_{m,k}$

The noise $\tilde{N}_{m,k} = N_{m,k}/X_{m,k}$ has variance $\sigma^2/|X_{m,k}|^2$. For constant-modulus constellations (e.g., PSK) where $|X_{m,k}| = 1$, the noise variance is unchanged: $\tilde{N}_{m,k}$ is still $\mathcal{CN}(0, \sigma^2)$. For QAM constellations with varying amplitudes, the noise becomes heteroscedastic but the sensing information is preserved.

The division requires $X_{m,k} \neq 0$ for all subcarriers used in radar processing. This is satisfied for all standard modulation schemes (QAM, PSK) but fails for null subcarriers (DC subcarrier, guard band subcarriers) where $X_{m,k} = 0$. These subcarriers must be excluded from the radar processing, which creates small gaps in the effective sensing bandwidth.

(i) For high-order QAM (e.g., 256-QAM), some symbols have very small amplitude, amplifying noise after division. (ii) Null subcarriers create spectral gaps that increase range sidelobes. (iii) If the transmitted symbols $X_{m,k}$ are not known to the radar processor (e.g., in a passive sensing scenario), the division cannot be performed.

Starting from the definition: $\chi(\tau, f_d) = \int_{-T_c/2}^{T_c/2} s(t) s^*(t-\tau) e^{j2\pi f_d t} dt$

Substituting $s(t) = e^{j\pi\mu t^2}$ and $s^*(t-\tau) = e^{-j\pi\mu(t-\tau)^2}$:

$s(t)s^*(t-\tau) = e^{j\pi\mu[t^2 - (t-\tau)^2]} = e^{j\pi\mu[2t\tau - \tau^2]} = e^{j2\pi\mu\tau t} \cdot e^{-j\pi\mu\tau^2}$

Therefore: $\chi(\tau, f_d) = e^{-j\pi\mu\tau^2} \int_{-T_c/2}^{T_c/2} e^{j2\pi(f_d + \mu\tau)t} dt$

The integral over the rectangular window of length $T_c$ (accounting for the overlap region of length $T_c - |\tau|$) gives:

$\chi(\tau, f_d) = (T_c - |\tau|) \cdot e^{-j\pi\mu\tau^2} \cdot \mathrm{sinc}\!\big((f_d + \mu\tau)(T_c - |\tau|)\big)$

Normalising by $T_c$ and taking the magnitude: $|\chi(\tau, f_d)| = (1 - |\tau|/T_c) \left|\mathrm{sinc}\!\big((f_d + \mu\tau)(T_c - |\tau|)\big)\right|$

The sinc function peaks when its argument is zero: $f_d + \mu\tau = 0$, i.e., $f_d = -\mu\tau$.

The $-3$ dB contour of $|\chi|^2$ forms an elongated ridge along the line $f_d = -\mu\tau$ in the $(\tau, f_d)$ plane. This ridge is tilted at an angle $\arctan(-\mu)$ with respect to the $\tau$ axis. The width of the ridge (perpendicular to it) is approximately $1/B$ in delay, while its length (along the ridge) extends to $\pm T_c/2$ in delay.

A target with Doppler shift $f_d$ shifts the peak of the matched filter output by $\Delta\tau = -f_d/\mu$ in delay. This causes an apparent range error of:

$\Delta R = \frac{c \Delta\tau}{2} = -\frac{c f_d}{2\mu} = -\frac{c f_d T_c}{2B}$

For example, with $B = 150$ MHz, $T_c = 50\;\mu$s, and $f_d = 10$ kHz: $\Delta R = -3 \times 10^8 \times 10^4 \times 50 \times 10^{-6} / (2 \times 150 \times 10^6) = -0.5$ m.

This range-Doppler coupling is the main disadvantage of the LFM chirp. It can be mitigated by transmitting up-chirps and down-chirps alternately or by Doppler compensation after velocity estimation.

$B = K\Delta f = 1024 \times 120 \times 10^3 = 122.88\;\text{MHz}$ $\Delta r = c/(2B) = 3 \times 10^8 / (2 \times 122.88 \times 10^6) = 1.221\;\text{m}$ Range separation: $|R_A - R_B| = 3$ m $> \Delta r = 1.221$ m. Yes, the targets are resolvable in range.

$\lambda = c/f_0 = 10.71$ mm. $T_{\text{sym}} \approx 1/\Delta f + T_{\text{CP}} \approx 8.33 + 0.59 = 8.93\;\mu$s. $T = M T_{\text{sym}} = 14 \times 8.93 \times 10^{-6} = 125.0\;\mu$s. $\Delta v = \lambda/(2T) = 10.71 \times 10^{-3} / (2 \times 125.0 \times 10^{-6}) = 42.86\;\text{m/s}$ Velocity separation: $|v_A - v_B| = 2$ m/s $< \Delta v = 42.86$ m/s. No, the targets cannot be resolved in velocity from one slot.

Need $\Delta v \leq 2$ m/s, so $T_{\text{total}} \geq \lambda/(2 \times 2) = 2.679$ ms. Each slot is 125.0 $\mu$s, so we need at least $\lceil 2679/125.0 \rceil = 22$ slots coherently processed. (With $M_{\text{total}} = 22 \times 14 = 308$ symbols, $\Delta v = 10.71 \times 10^{-3}/(2 \times 308 \times 8.93 \times 10^{-6}) = 1.95$ m/s $< 2$ m/s.)

Range bins: $p = 2R \cdot K\Delta f / c = 2R \cdot B / c$. $p_A = 2 \times 50 \times 122.88 \times 10^6 / (3 \times 10^8) = 40.96 \approx 41$ $p_B = 2 \times 53 \times 122.88 \times 10^6 / (3 \times 10^8) = 43.42 \approx 43$

Doppler bins (single slot): $q = 2v f_0 M T_{\text{sym}} / c = 2v M T_{\text{sym}} / \lambda$. $q_A = 2 \times 20 \times 14 \times 8.93 \times 10^{-6} / (10.71 \times 10^{-3}) = 0.467 \approx 0$ $q_B = 2 \times 18 \times 14 \times 8.93 \times 10^{-6} / (10.71 \times 10^{-3}) = 0.420 \approx 0$ Both targets fall in the same Doppler bin (bin 0), confirming they are not resolvable in velocity from a single slot.

The Fisher information for delay $\tau$ from an OFDM signal is: $J(\tau) = \frac{2}{\sigma^2} \sum_{k=0}^{K-1} |S_k|^2 (2\pi k\Delta f)^2$

Therefore: $\mathrm{CRB}(\tau) = \frac{1}{J(\tau)} = \frac{\sigma^2}{8\pi^2 \sum_k (k\Delta f)^2 |S_k|^2}$

Defining $\text{SNR}_{\text{eff}} = \sum_k |S_k|^2 / \sigma^2$ and the squared RMS bandwidth $\beta^2 = \sum_k (k\Delta f)^2 |S_k|^2 / \sum_k |S_k|^2$:

$\mathrm{CRB}(\tau) = \frac{1}{8\pi^2 \text{SNR}_{\text{eff}} \cdot \beta^2}$

Minimising the CRB is equivalent to maximising $\beta^2$ (the RMS bandwidth).

Let $p_k = |S_k|^2 \geq 0$. The problem is:

$\min_{\{p_k\}} \; \frac{1}{\sum_k (k\Delta f)^2 p_k}$

subject to: $\sum_k \log_2(1 + p_k |H_k|^2 / \sigma^2) \geq R_{\min}$ $\sum_k p_k \leq P$ $p_k \geq 0, \; \forall k$

The Lagrangian is: $\mathcal{L} = -\sum_k (k\Delta f)^2 p_k + \mu \bigl(R_{\min} - \sum_k \log_2(1 + p_k |H_k|^2/\sigma^2)\bigr) + \nu \bigl(\sum_k p_k - P\bigr)$

Setting $\partial\mathcal{L}/\partial p_k = 0$: $-(k\Delta f)^2 - \frac{\mu |H_k|^2/\sigma^2}{\ln 2 (1 + p_k |H_k|^2/\sigma^2)} + \nu = 0$

Solving: $p_k = \left[\frac{\mu |H_k|^2}{\sigma^2 \ln 2 (\nu - (k\Delta f)^2)} - \frac{\sigma^2}{|H_k|^2}\right]^+$

When $\mu = 0$ (no rate constraint): $p_k \propto (k\Delta f)^2$ — all power goes to edge subcarriers to maximise RMS bandwidth. When $\mu \to \infty$ (rate dominates): water-filling over $|H_k|^2$ — power goes to strong subcarriers. Intermediate $\mu$ values trade off between these extremes.

$|\chi(\tau, f_d)|^2 = \left|\int s(t) s^*(t-\tau) e^{j2\pi f_d t} dt\right|^2$

$= \int\int s(t) s^*(t-\tau) s^*(u) s(u-\tau) e^{j2\pi f_d (t-u)} dt \, du$

$\int |\chi(\tau, f_d)|^2 df_d = \int\int s(t) s^*(t-\tau) s^*(u) s(u-\tau) \underbrace{\int e^{j2\pi f_d(t-u)} df_d}_{\delta(t-u)} dt \, du$

$= \int |s(t)|^2 |s(t-\tau)|^2 dt$

$\iint |\chi(\tau, f_d)|^2 d\tau \, df_d = \int\int |s(t)|^2 |s(t-\tau)|^2 dt \, d\tau$

Substituting $w = t - \tau$ ($d\tau = -dw$):

$= \int |s(t)|^2 dt \int |s(w)|^2 dw = E_s \cdot E_s = E_s^{2}$

This completes the proof. $\square$

TX positions (in units of $\lambda/2$): $\{0, 4, 8, 12\}$ RX positions: $\{0, 1, 2, 3\}$

Virtual positions = $\{i \cdot 4 + j : i \in \{0,1,2,3\}, j \in \{0,1,2,3\}\}$ $= \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}$

All 16 positions are distinct, forming a filled ULA from 0 to 15 with spacing $\lambda/2$.

Yes, the virtual array is a filled ULA. This is because $d_T = N_R d_R = 4 \times \lambda/2 = 2\lambda$, which is the condition for a filled virtual array. Each group of 4 RX elements fills the gap between consecutive TX elements.

Virtual aperture: $L_v = 15 \times \lambda/2 = 7.5\lambda$. Angular resolution: $\Delta\theta \approx \lambda/L_v = 1/7.5 = 0.133$ rad $= 7.64°$.

For a 4-element receive-only phased array: $L_R = 3 \times \lambda/2 = 1.5\lambda$. $\Delta\theta_R = \lambda/L_R = 1/1.5 = 0.667$ rad $= 38.2°$.

Improvement factor: $38.2°/7.64° = 5 \approx N_T + 1$. The angular resolution improves by approximately $N_T = 4$ times.

With $d_T = 3\lambda/2$: TX positions: $\{0, 3, 6, 9\}$, RX positions: $\{0, 1, 2, 3\}$ Virtual positions: $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$ but with redundancies at positions $\{3, 6, 9\}$.

Checking: $0+3=3$, $3+0=3$ (redundant); $0+6=6$, $3+3=6$ (redundant); etc. Distinct positions: $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$ = 13 unique elements. Effective aperture: $12 \times \lambda/2 = 6\lambda$. We lose 3 virtual elements to redundancy but the array is still filled.

The MRT beamformer $\mathbf{w} = \sqrt{P}\,\mathbf{h}/\|\mathbf{h}\| = \sqrt{P}\,\mathbf{a}(\theta_c)/\sqrt{N_T}$ maximises the communication SNR:

$\text{SNR}_{c} = \frac{|\mathbf{h}^H\mathbf{w}|^2}{\sigma^2} = \frac{P|\sqrt{\gamma}\,\mathbf{a}^H(\theta_c)\mathbf{a}(\theta_c)/\sqrt{N_T}|^2}{\sigma^2} = \frac{P\gamma N_T}{\sigma^2}$

This is the maximum achievable SNR (Cauchy-Schwarz bound).

$P(\theta_s) = |\mathbf{a}^H(\theta_s)\mathbf{w}|^2 = \frac{P}{N_T}|\mathbf{a}^H(\theta_s)\mathbf{a}(\theta_c)|^2$

When $\theta_s = \theta_c$: $P(\theta_c) = \frac{P}{N_T}|\mathbf{a}^H(\theta_c)\mathbf{a}(\theta_c)|^2 = \frac{P}{N_T} N_T^2 = P N_T$.

For any unit-norm beamformer $\mathbf{w}$ ($\|\mathbf{w}\|^2 = P$): $P(\theta_s) = P|\mathbf{a}^H(\theta_s)\mathbf{w}/\sqrt{P}|^2 \leq P\|\mathbf{a}(\theta_s)\|^2 = PN_T$

by Cauchy-Schwarz, with equality iff $\mathbf{w} \propto \mathbf{a}(\theta_s)$. When $\theta_s = \theta_c$, the MRT beamformer $\mathbf{w} \propto \mathbf{a}(\theta_c) = \mathbf{a}(\theta_s)$ achieves this maximum.

Therefore, the same beamformer simultaneously maximises both the communication SNR and the sensing beampattern gain. The trade-off only arises when $\theta_c \neq \theta_s$, in which case $|\mathbf{a}^H(\theta_s)\mathbf{a}(\theta_c)| < N_T$.

Minimise $f(x) = \frac{1}{2}(x-z)^2 + \lambda|x|$ over $x \in \mathbb{R}$.

Case 1: $x > 0$. $f(x) = \frac{1}{2}(x-z)^2 + \lambda x$. $f'(x) = x - z + \lambda = 0 \Rightarrow x = z - \lambda$. Valid when $z > \lambda$, giving $x^* = z - \lambda > 0$.

Case 2: $x < 0$. $f(x) = \frac{1}{2}(x-z)^2 - \lambda x$. $f'(x) = x - z - \lambda = 0 \Rightarrow x = z + \lambda$. Valid when $z < -\lambda$, giving $x^* = z + \lambda < 0$.

Case 3: $x = 0$. $f(0) = z^2/2$, which is optimal when $|z| \leq \lambda$ (the subdifferential condition $0 \in z + \lambda\,\partial|x|$ is satisfied).

Combining: $x^* = \mathrm{sign}(z)\max(|z| - \lambda, 0) = \mathcal{S}_\lambda(z)$.

The soft-thresholding operator sets all values with $|z| \leq \lambda$ exactly to zero, and shrinks larger values by $\lambda$ towards zero. This creates a "dead zone" around the origin that promotes sparsity: small noisy components are killed, while large genuine signal components are preserved (with slight bias). The parameter $\lambda$ controls the aggressiveness of sparsification.

For $z = |z|e^{j\phi} \in \mathbb{C}$, write $x = re^{j\psi}$ and minimise: $f(r, \psi) = \frac{1}{2}|re^{j\psi} - |z|e^{j\phi}|^2 + \lambda r$

The phase minimiser is $\psi^* = \phi$ (align with $z$). The magnitude minimiser: $f(r) = \frac{1}{2}(r - |z|)^2 + \lambda r$, $f'(r) = r - |z| + \lambda = 0 \Rightarrow r^* = \max(|z| - \lambda, 0)$.

Therefore: $x^* = \max(|z| - \lambda, 0) \cdot e^{j\phi} = z \cdot \max(1 - \lambda/|z|, 0)$.

The objective combines weighted communication rate and sensing gain. Since $\mathbf{h} = \sqrt{\gamma_c}\,\mathbf{a}(\theta_c)$: $\mathbf{h}^H\mathbf{R}_{x}\mathbf{h} = \gamma_c \mathbf{a}^H(\theta_c)\mathbf{R}_{x}\mathbf{a}(\theta_c)$

The problem involves $\mathbf{R}_{x}$ only through its projections onto $\mathbf{a}(\theta_c)$ and $\mathbf{a}(\theta_s)$.

Since the objective depends on $\mathbf{R}_{x}$ only through $\mathbf{a}^H(\theta_c)\mathbf{R}_{x}\mathbf{a}(\theta_c)$ and $\mathbf{a}^H(\theta_s)\mathbf{R}_{x}\mathbf{a}(\theta_s)$, the optimal solution lies in the 2D subspace spanned by $\{\mathbf{a}(\theta_c), \mathbf{a}(\theta_s)\}$.

Decompose: $\mathbf{R}_{x} = p_c \mathbf{u}_c\mathbf{u}_c^H + p_s \mathbf{u}_s\mathbf{u}_s^H$ where $\mathbf{u}_c, \mathbf{u}_s$ are orthonormal vectors in $\mathrm{span}\{\mathbf{a}(\theta_c), \mathbf{a}(\theta_s)\}$ and $p_c + p_s = P$.

When $\theta_c = \theta_s$: $\mathbf{a}(\theta_c) = \mathbf{a}(\theta_s)$, so $\mathbf{R}_{x} = P\mathbf{a}(\theta_c)\mathbf{a}^H(\theta_c)/N_T$ (rank-1).

With $\mathbf{R}_{x} = p_c \mathbf{u}_c\mathbf{u}_c^H + p_s \mathbf{u}_s\mathbf{u}_s^H$, the problem reduces to a scalar optimisation over the power split $(p_c, p_s)$ with $p_c + p_s = P$:

$\max_{0 \leq p_c \leq P} (1-\rho)\log_2\!\left(1 + \frac{\gamma_c}{N_T\sigma_c^2}[p_c|\mathbf{a}^H(\theta_c)\mathbf{u}_c|^2 + (P-p_c)|\mathbf{a}^H(\theta_c)\mathbf{u}_s|^2]\right)$ $+ \rho[p_c|\mathbf{a}^H(\theta_s)\mathbf{u}_c|^2 + (P-p_c)|\mathbf{a}^H(\theta_s)\mathbf{u}_s|^2]$

The KKT condition $\partial/\partial p_c = 0$ yields: $(1-\rho)\frac{\gamma_c \Delta_c / (N_T\sigma_c^2)}{1 + \gamma_c(\ldots)/(N_T\sigma_c^2)\ln 2} + \rho \Delta_s = 0$ where $\Delta_c = |\mathbf{a}^H(\theta_c)\mathbf{u}_c|^2 - |\mathbf{a}^H(\theta_c)\mathbf{u}_s|^2$ and $\Delta_s = |\mathbf{a}^H(\theta_s)\mathbf{u}_c|^2 - |\mathbf{a}^H(\theta_s)\mathbf{u}_s|^2$. This is solved numerically via bisection.