Ferkans — Interactive Telecom Tutor

ch18-ex01

Easy

For a MISO BC with $n_t = 2$ antennas and $K = 50$ users with i.i.d. Rayleigh fading ( $\mathbf{h}_k \sim \mathcal{CN}(\mathbf{0}, \mathbf{I}_2)$ ), estimate the multiuser diversity gain at $\text{SNR} = 20$ dB. Compare the sum rate with $K = 2$ users (no scheduling) and $K = 50$ users (optimal scheduling).

Show Hint

The multiuser diversity gain is approximately $\log K$ for the effective SNR per stream.

With $K = 2$ : no scheduling choice, sum rate $\approx n_t \log(1 + P/n_t)$ .

With $K = 50$ : effective SNR enhanced by $\log 50 \approx 3.9$ .

Solution

$K = 2$ (no scheduling)

Sum rate $\approx 2 \log(1 + 100/2) = 2\log(51) \approx 11.3$ bits/use.

$K = 50$ (with scheduling)

Effective per-stream SNR $\approx (P/n_t) \cdot \log K = 50 \times 3.9 = 195$ . Sum rate $\approx 2 \log(1 + 195) = 2\log(196) \approx 15.3$ bits/use. Scheduling gain: $\approx 4$ bits (35% improvement).

ch18-ex02

Easy

In the ISAC framework, a system allocates 70% of its power to the communication signal and 30% to the deterministic sensing waveform. If the total SNR is 20 dB and the communication channel has a single antenna ( $n_t = 1$ ), what is the communication rate?

Show Hint

The communication rate depends only on the random signal power.

$R_{c} = \frac{1}{2}\log(1 + 0.7 \cdot \text{SNR})$ .

Solution

Communication rate

$R_{c} = \frac{1}{2}\log(1 + 0.7 \times 100) = \frac{1}{2}\log(71) \approx 3.08$ bits/use. Compared to the rate without sensing ( $\frac{1}{2}\log(101) \approx 3.33$ bits), the sensing allocation costs about 0.25 bits — a modest penalty for 30% sensing power.

ch18-ex03

Medium

For the MISO ISAC system with $n_t = 2$ , show that when the communication user and the sensing target are at the same angle ( $\phi_c = \phi_s$ ), the ISAC tradeoff vanishes — both functions can be optimized simultaneously.

Show Hint

When both are at the same angle, the optimal communication beam and the optimal sensing beam coincide.

The communication signal (beamformed toward the user) also illuminates the target optimally.

All power can be random (communication) and still achieve the best sensing.

Solution

Same-angle channels

With $\phi_c = \phi_s$ : $\mathbf{h}_c \propto \mathbf{a}(\phi_s)$ . The optimal communication beamformer $\mathbf{K}_r = P \cdot \mathbf{h}_c \mathbf{h}_c^H / \|\mathbf{h}_c\|^2$ points toward both the user and the target.

No tradeoff

The FIM is $\frac{1}{\sigma^2_{s}} \dot{\mathbf{H}}_s^H \mathbf{K}_r \dot{\mathbf{H}}_s$ . Since $\mathbf{K}_r$ concentrates all power in the direction of $\mathbf{a}(\phi_s)$ , this is the maximum FIM achievable with power $P$ — same as the sensing-only solution. Meanwhile, the communication rate is maximized because all power goes to $\mathbf{K}_r$ . Both are simultaneously optimal: no tradeoff.

Conclusion

The ISAC tradeoff exists only when the communication and sensing channels require different spatial resources. When they are aligned, "sensing is free" is literally true.

ch18-ex04

Medium

For a 5-user IC with the following interference topology — Tx 1 interferes with Rx 2 and Rx 5; Tx 2 interferes with Rx 3; Tx 3 interferes with Rx 4; Tx 4 interferes with Rx 5; Tx 5 interferes with Rx 1 — find the topological DoF. Compare with the IA DoF.

Show Hint

Draw the interference graph and find its complement.

Find the maximum independent set in the complement graph.

IA DoF for 5 users: $5/2 = 2.5$ .

Solution

Interference graph

Edges: $(1,2), (1,5), (2,3), (3,4), (4,5), (5,1)$ . This is not fully connected — some pairs do not interfere.

Complement graph

Non-interfering pairs: $(1,3), (1,4), (2,4), (2,5), (3,5)$ . (And their reverses for the undirected version.)

Maximum independent set

In $\bar{\mathcal{G}}$ : $\{1, 3, 4\}$ ? Check: $(1,3) \in \bar{\mathcal{G}}$ (no interference), $(1,4) \in \bar{\mathcal{G}}$ , $(3,4) \notin \bar{\mathcal{G}}$ (they interfere!). So $\{1, 3, 4\}$ is not independent. Try $\{1, 3\}$ : independent (no mutual interference). Try $\{2, 4\}$ : $(2,4) \in \bar{\mathcal{G}}$ , independent. Try $\{2, 5\}$ : $(2,5) \in \bar{\mathcal{G}}$ , independent. Try $\{1, 3, 5\}$ ? $(3,5) \in \bar{\mathcal{G}}$ , $(1,3) \in \bar{\mathcal{G}}$ , $(1,5) \notin \bar{\mathcal{G}}$ (they interfere). Not independent. Try $\{2, 4, 1\}$ ? $(2,4) \in \bar{\mathcal{G}}$ , $(1,4) \in \bar{\mathcal{G}}$ , $(1,2) \notin \bar{\mathcal{G}}$ . Not independent. Maximum independent set size: $\beta = 2$ . Topological DoF = 2.

Comparison

Topological DoF = 2. IA DoF = 5/2 = 2.5. Full CSIT provides 25% more DoF than topological CSIT for this topology.

ch18-ex05

Medium

Derive the Pareto boundary of the ISAC rate-CRB tradeoff for a scalar ( $n_t = 1$ ) ISAC system where the communication channel has gain $h_c$ and the sensing channel has gain $h_s$ . Express the CRB as a function of the communication rate.

Show Hint

With $n_t = 1$ : $x = x_d + x_r$ where $x_d$ is deterministic (power $P_d$ ) and $x_r \sim \mathcal{N}(0, P_r)$ .

Power constraint: $P_d + P_r \leq P$ .

Rate: $R_{c} = \frac{1}{2}\log(1 + h_c^2 P_r / \sigma^2_{c})$ .

CRB: $\text{CRB} = \sigma^2_{s} / (|\dot{h}_s|^2 (P_d + P_r))$ where $\dot{h}_s = dh_s/d\theta$ .

Solution

Express $P_r$ in terms of rate

$R_{c} = \frac{1}{2}\log(1 + h_c^2 P_r / \sigma^2_{c})$ gives $P_r = \sigma^2_{c} (2^{2R_{c}} - 1) / h_c^2$ .

CRB as function of rate

$P_d = P - P_r = P - \sigma^2_{c}(2^{2R_{c}}-1)/h_c^2$ . Total signal power for sensing: $P_d + P_r = P$ (all power contributes to sensing). But wait — the FIM for sensing depends on whether the signal is known or random. For the deterministic part: FIM $\propto |P_d| / \sigma^2_{s}$ . For the random part: FIM $\propto P_r / \sigma^2_{s}$ (averaged contribution). Total FIM $= |\dot{h}_s|^2 (P_d + P_r) / \sigma^2_{s} = |\dot{h}_s|^2 P / \sigma^2_{s}$ .

Surprising result for scalar case

In the scalar case ( $n_t = 1$ ), the total power $P$ always contributes to sensing regardless of the split — because there is only one spatial dimension. The CRB is $\sigma^2_{s} / (|\dot{h}_s|^2 P)$ , independent of $R_{c}$ ! This means there is no tradeoff in the scalar case — sensing performance is determined entirely by total power, not by the split.

The tradeoff appears only with $n_t \geq 2$ , where the spatial allocation (beamforming direction) creates competition between communication and sensing.

ch18-ex06

Hard

For the index coding problem corresponding to a 4-user TIM instance with cycle topology (edges $(1,2), (2,3), (3,4), (4,1)$ ), find the optimal index coding rate. Show that it equals the topological DoF computed via the complement graph.

Show Hint

The complement of the 4-cycle is... also a 4-structure (two disjoint edges).

The index coding capacity is $1/\chi_f(\bar{\mathcal{G}})$ per user.

For the complement of a 4-cycle: the fractional chromatic number is 2.

Solution

Complement graph

4-cycle edges: $(1,2), (2,3), (3,4), (4,1)$ . Complement: $(1,3), (2,4)$ — two disjoint edges (a perfect matching).

Independence number

$\beta(\bar{\mathcal{G}}) = 2$ : $\{1,3\}$ or $\{2,4\}$ are maximum independent sets.

Index coding rate

The optimal index coding broadcasts at rate $K/\beta = 4/2 = 2$ symbols per channel use. Each user achieves DoF = 1/2. Total DoF = 2, matching the TIM computation.

Optimal code

The index code transmits $W_1 \oplus W_3$ and $W_2 \oplus W_4$ in two uses. Rx 1 has side info $W_3$ (no interference from Tx 3), so it decodes $W_1 = (W_1 \oplus W_3) \oplus W_3$ . Rx 3 has side info $W_1$ , decodes $W_3$ similarly. Rx 2 has side info $W_4$ , decodes $W_2$ . Rx 4 has side info $W_2$ , decodes $W_4$ . This achieves 4 messages in 2 uses: DoF = 2.

ch18-ex07

Hard

For the MISO ISAC system with $n_t = 4$ antennas, a ULA with half-wavelength spacing, a communication user at $\phi_c = 0°$ and a target at $\phi_s = 45°$ , design the Pareto-optimal beamforming strategy that splits power between communication and sensing. Parameterize by the fraction $\alpha$ of power allocated to the deterministic sensing beam.

Show Hint

The steering vectors are $\mathbf{a}(\phi) = [1, e^{j\pi\sin\phi}, e^{j2\pi\sin\phi}, e^{j3\pi\sin\phi}]^T / 2$ .

Deterministic beam: $\mathbf{x}_d = \sqrt{\alpha P} \cdot \mathbf{a}(\phi_s) / \|\mathbf{a}(\phi_s)\|$ .

Communication covariance: $\mathbf{K}_r = (1-\alpha)P \cdot \mathbf{a}(\phi_c)\mathbf{a}(\phi_c)^H / \|\mathbf{a}(\phi_c)\|^2$ (MRT to comm user).

Solution

Steering vectors

$\mathbf{a}(0°) = [1, 1, 1, 1]^T / 2$ (broadside). $\mathbf{a}(45°) = [1, e^{j\pi/\sqrt{2}}, e^{j2\pi/\sqrt{2}}, e^{j3\pi/\sqrt{2}}]^T / 2$ . Inner product: $|\mathbf{a}(0°)^H \mathbf{a}(45°)|^2 / (4 \cdot 4) = |\sum_m e^{jm\pi/\sqrt{2}}|^2 / 16$ .

Communication rate

$R_{c} = \log(1 + (1-\alpha)P \cdot |\mathbf{h}_c^H \mathbf{a}(0°)|^2 / (\sigma^2 \|\mathbf{a}(0°)\|^2))$ . With $\mathbf{h}_c = \mathbf{a}(0°)$ : $R_{c} = \log(1 + (1-\alpha)P n_t / \sigma^2)$ .

Sensing FIM

The FIM depends on $\mathbf{R}_x = \alpha P \mathbf{a}(45°)\mathbf{a}(45°)^H + (1-\alpha)P \mathbf{a}(0°)\mathbf{a}(0°)^H$ . The FIM for the angle parameter $\phi_s$ is proportional to $\dot{\mathbf{a}}(45°)^H \mathbf{R}_x \dot{\mathbf{a}}(45°)$ . The deterministic component ( $\alpha P$ in the target direction) contributes more effectively to the FIM than the communication component.

Pareto boundary

As $\alpha$ increases from 0 to 1: $R_{c}$ decreases linearly (in dB), while the CRB decreases (sensing improves). The Pareto boundary is the curve $(R_{c}(\alpha), \text{CRB}(\alpha))$ for $\alpha \in [0, 1]$ .

ch18-ex08

Medium

Show that for a fully connected interference graph ( $K$ users, every Tx interferes with every Rx), the topological DoF equals 1, regardless of $K$ . What does this imply about the value of topological CSIT in dense networks?

Show Hint

Fully connected graph: every pair interferes.

Complement graph: no edges — independence number = 1.

Solution

Complement graph

If every pair interferes, the complement graph has no edges. Every vertex is isolated.

Independence number

$\beta(\bar{\mathcal{G}}) = 1$ (a single vertex is the maximum independent set).

Topological DoF

$d_{\Sigma}^{\text{top}} = 1$ . Only one user can transmit at a time — TDMA.

Implication

In fully connected (dense) interference graphs, topological CSIT provides no advantage over no CSIT. The topology is uninformative. Full CSIT is needed (IA achieves $K/2$ ). This highlights that TIM is most useful in sparse interference graphs where many pairs do not interfere.

ch18-ex09

Medium

For a 3-user MISO BC with $n_t = 2$ antennas and $K = 3$ users at angles $0°, 60°, 120°$ , explain why the base station cannot serve all 3 users simultaneously. How many users should be scheduled per slot, and what is the optimal scheduling strategy?

Show Hint

With $n_t = 2$ , at most 2 spatial streams can be multiplexed.

The optimal strategy selects the best pair of users per slot.

Three possible pairs: $\{1,2\}, \{1,3\}, \{2,3\}$ . The pair with the most orthogonal channels gives the highest sum rate.

Solution

Spatial constraint

With $n_t = 2$ , the rank of the total channel matrix is at most 2. Serving 3 users simultaneously would require rank 3, which is impossible.

Scheduling

Schedule 2 users per slot. Three options:

$\{1,2\}$ : angular separation $60°$
$\{1,3\}$ : angular separation $120°$ (most orthogonal for ULA)
$\{2,3\}$ : angular separation $60°$ The pair $\{1,3\}$ has the largest angular separation and thus the most orthogonal channels, giving the highest ZF sum rate.

Round-robin fairness

For fairness, alternate between all three pairs. On average, each user is served 2/3 of the time, achieving effective DoF = 2/3 per user.

ch18-ex10

Challenge

(Research-level) Extend the Liu-Caire ISAC framework to a $K$ -user MIMO BC setting where the base station must simultaneously communicate with $K$ users and sense $M$ targets. Formulate the multi-user multi-target rate-CRB tradeoff and discuss the challenges compared to the single-user single-target case.

Show Hint

The communication rate region is the MIMO BC DPC region with covariance $\{\mathbf{K}_{r,k}\}$ .

Each target has its own FIM depending on the total signal covariance $\mathbf{R}_x$ .

The tradeoff is now multi-dimensional: $K$ rates + $M$ CRB constraints.

Solution

Multi-user communication

The communication rate region is the DPC region with random signal covariances $\{\mathbf{K}_{r,k}\}$ for $K$ users and the deterministic sensing component $\mathbf{x}_d$ .

Multi-target sensing

The FIM for target $m$ depends on $\mathbf{R}_x = \mathbf{x}_d \mathbf{x}_d^H + \sum_k \mathbf{K}_{r,k}$ . Each target may require different spatial emphasis (beam directions).

Challenges

The optimization is over $K$ communication covariances + the sensing waveform.
Multiple targets may require contradictory beam directions.
The Pareto boundary is $(K+M)$ -dimensional, making it hard to visualize and optimize.
The MAC-BC duality for the communication part still applies, but the sensing constraints break the clean duality structure.

Exercises

ch18-ex01

$K = 2$ (no scheduling)

$K = 50$ (with scheduling)

ch18-ex02

Communication rate

ch18-ex03

Same-angle channels

No tradeoff

Conclusion

ch18-ex04

Interference graph

Complement graph

Maximum independent set

Comparison

ch18-ex05

Express $P_r$ in terms of rate

CRB as function of rate

Surprising result for scalar case

ch18-ex06

Complement graph

Independence number

Index coding rate

Optimal code

ch18-ex07

Steering vectors

Communication rate

Sensing FIM

Pareto boundary

ch18-ex08

Complement graph

Independence number

Topological DoF

Implication

ch18-ex09

Spatial constraint

Scheduling

Round-robin fairness

ch18-ex10

Multi-user communication

Multi-target sensing

Challenges