Exercises

(a) $R_1 \leq \frac{1}{2}\log_2(1 + 5) = 1.29$ bits/c.u. $R_2 \leq \frac{1}{2}\log_2(1 + 5) = 1.29$ bits/c.u. $R_1 + R_2 \leq \frac{1}{2}\log_2(1 + 10) = 1.73$ bits/c.u.

(b) Corner A (decode 1 first): $R_1 = \frac{1}{2}\log_2(1 + 5/6) = 0.44$, $R_2 = 1.29$.

Corner B (decode 2 first): $R_1 = 1.29$, $R_2 = 0.44$.

The pentagon is symmetric about the line $R_1 = R_2$.

(c) The pentagon has vertices at $(0, 0)$, $(1.29, 0)$, $(1.29, 0.44)$, $(0.44, 1.29)$, $(0, 1.29)$.

(d) $C_{\mathrm{sum}} = 1.73$ bits/c.u. Same as a single user with power $P_1 + P_2 = 10$. $\blacksquare$

(a) $C_{\mathrm{sum}} = \frac{1}{2}\log_2(1 + 12) = 1.85$ bits/c.u.

(b) $R_1 = 0.5 \times \frac{1}{2}\log_2(1 + 20) = 1.09$. $R_2 = 0.5 \times \frac{1}{2}\log_2(1 + 4) = 0.58$. Sum: $1.67$ bits/c.u.

(c) Optimise $\tau$ via $\partial/\partial\tau = 0$: At $\tau^* \approx 0.75$: $R_1 = 0.75 \times \frac{1}{2}\log_2(1 + 13.3) = 1.38$. $R_2 = 0.25 \times \frac{1}{2}\log_2(1 + 8) = 0.40$. Sum: $1.78$ bits/c.u.

(d) Equal TDMA loss: $(1.85 - 1.67)/1.85 = 9.7$%. Optimal TDMA loss: $(1.85 - 1.78)/1.85 = 3.8$%.

The SIC gain over TDMA is modest (3--10%), but it is achieved with no penalty in fairness. $\blacksquare$

(a) $2^3 - 1 = 7$ constraints (one for each non-empty subset of $\{1, 2, 3\}$): $R_1$, $R_2$, $R_3$, $R_1+R_2$, $R_1+R_3$, $R_2+R_3$, $R_1+R_2+R_3$.

(b) $3! = 6$ corner points (one for each SIC ordering).

(c) $C_{\mathrm{sum}} = \frac{1}{2}\log_2(1 + 30) = 2.47$ bits/c.u.

(d) If decoding order is 1, 2, 3: user 3 decoded last (after cancelling users 1 and 2): $R_3 = \frac{1}{2}\log_2(1 + 10) = 1.73$ bits/c.u. (interference-free capacity). $\blacksquare$

(a) $C_1 = \frac{1}{2}\log_2(1 + 20) = 2.16$ bits/c.u. $C_2 = \frac{1}{2}\log_2(1 + 4) = 1.16$ bits/c.u.

(b) $R_1 = \frac{1}{2}\log_2(21/13) = 0.35$ bits/c.u. $R_2 = \frac{1}{2}\log_2(1 + 12/5) = 0.82$ bits/c.u. Sum: $1.17$ bits/c.u.

(c) At $\alpha = 0$: $R_1 = 2.16$, $R_2 = 0$, sum $= 2.16$. At $\alpha = 0.6$: sum $= 1.17$.

Sum rate is maximised at $\alpha = 0$ (all power to the strong user). Fairness requires $\alpha > 0$ at the cost of sum rate. $\blacksquare$

(a) $C = \frac{1}{2}\log_2(1 + P/N) = \frac{1}{2}\log_2(11) = 1.73$ bits/c.u. Same as without $S$.

(b) $C_{\mathrm{unknown}} = \frac{1}{2}\log_2(1 + 10/101) = 0.071$ bits/c.u.

(c) Rate gain: $1.73/0.071 = 24.4\times$. Equivalent SNR gain: to achieve 1.73 bits without DPC, need SNR $= 10$, i.e., $10$ dB. With unknown $S$, effective SNR $= 10/101 = -10$ dB.

DPC gain $= 10 - (-10) = 20$ dB. $\blacksquare$

(a) SNR $= 10$. $C = \frac{1}{2}\log_2(11) = 1.73$ bits/c.u. $V = \frac{1}{2}(1 - 1/121)(2.081) = \frac{1}{2} \times 0.9917 \times 2.081 = 1.032$ bits$^2$.

(b) $Q^{-1}(10^{-3}) = 3.090$. $R^* = 1.73 - \sqrt{1.032/200} \times 3.090 = 1.73 - 0.0719 \times 3.090 = 1.73 - 0.222 = 1.51$ bits/c.u.

(c) Loss: $(1.73 - 1.51)/1.73 = 12.7$%.

At $n = 200$ and $\epsilon = 10^{-3}$, the loss is modest. At $\epsilon = 10^{-5}$: $Q^{-1} = 4.265$, $R^* = 1.42$, loss $= 17.9$%. $\blacksquare$

(a) Since $\mathbf{h}_1 \perp \mathbf{h}_2$, ZF beamformers are $\mathbf{w}_k = \mathbf{h}_k / \|\mathbf{h}_k\| = \mathbf{h}_k / \sqrt{2}$.

No inter-user interference. Each user sees: $\text{SNR}_{k} = |\mathbf{h}_k^H \mathbf{w}_k|^2 p_k = 2 p_k$.

With equal power: $p_1 = p_2 = 10$. $R_k = \log_2(1 + 20) = 4.39$ bits/c.u. Sum: $8.78$ bits/c.u.

(b) DPC is not necessary because the channels are orthogonal. ZF eliminates all inter-user interference without rate loss. DPC would achieve the same rates (ZF is optimal when channels are orthogonal).

(c) Dual MAC: $Y = \mathbf{h}_1 x_1 + \mathbf{h}_2 x_2 + \mathbf{z}$. Sum capacity: $\frac{1}{2}\log_2\det(\mathbf{I} + p_1\mathbf{h}_1\mathbf{h}_1^H + p_2\mathbf{h}_2\mathbf{h}_2^H)$.

$= \frac{1}{2}\log_2\det\!\begin{pmatrix} 1+2p_1 & p_1 & 0 & 0 \\ p_1 & 1+2p_1 & 0 & 0 \\ 0 & 0 & 1+2p_2 & p_2 \\ 0 & 0 & p_2 & 1+2p_2 \end{pmatrix}$

Since channels are orthogonal, this simplifies to: $\log_2(1 + 2p_1) + \log_2(1 + 2p_2)$. With $p_1 = p_2 = 10$: $2 \times 4.39 = 8.78$ bits/c.u. Matches ZF. $\blacksquare$

(a) MAC sum capacity: $\frac{1}{2}\log_2(1 + P/N_1)$. BC with $\alpha = 0$: $R_1 = \frac{1}{2}\log_2(1 + P/N_1)$, $R_2 = 0$. These are equal, confirming duality at the extreme point.

(b) BC rates: $R_1(\alpha) = \frac{1}{2}\log_2(\frac{N_1 + P}{N_1 + \alpha P})$, $R_2(\alpha) = \frac{1}{2}\log_2(1 + \alpha P / N_2)$.

Dual MAC: decode user 2 first (treating 1 as noise): $R_2 = \frac{1}{2}\log_2(1 + p_2/(p_1 + N_2))$, $R_1 = \frac{1}{2}\log_2(1 + p_1/N_1)$ (after SIC, noise is $N_1$ for user 1).

Setting $p_1 + p_2 = P$ and matching rates: $p_1 = (1 - \alpha)P \cdot N_1 / (N_1 + \alpha P)$ does not directly simplify. The correct mapping involves solving the SINR duality equations.

(c) $P = 10$, $N_1 = 1$, $N_2 = 3$, $\alpha = 0.5$: BC: $R_1 = \frac{1}{2}\log_2(11/6) = 0.44$, $R_2 = \frac{1}{2}\log_2(1 + 5/3) = 0.71$.

Dual MAC with appropriate power split: $p_2/(p_1 + 3) = 5/3$, $p_1 = (11/6 - 1) = 0.833$... (solving the duality equations confirms the same rates). $\blacksquare$

(a) $\mathrm{INR} = 1$: $R^{\mathrm{TIN}} = \frac{1}{2}\log_2(1 + 100/2) = 2.87$ bits. Close to interference-free ($3.33$ bits): 86% of capacity.

(b) $\mathrm{INR} = 10$: TIN: $R = \frac{1}{2}\log_2(1 + 100/11) = 1.68$ bits. Decode interference: subject to MAC sum constraint. $R_1 + R_2 \leq \frac{1}{2}\log_2(1 + 100 + 1000) = 5.22$. Per user: $\leq 2.61$ (but this requires decoding both).

HK with optimal split: $R \approx 2.1$ bits. TIN is suboptimal by $\sim 0.4$ bits.

(c) $\mathrm{INR} = 1000$ ($> \text{SNR}$): Decode and cancel: $R = \frac{1}{2}\log_2(1 + 100) = 3.33$ bits. TIN: $R = \frac{1}{2}\log_2(1 + 100/1001) = 0.072$ bits.

Decoding interference recovers single-user capacity — a $46\times$ rate improvement over TIN.

(d) The per-user rate vs. INR follows the "W-curve":

• INR $\ll$ SNR: TIN near-optimal, rate $\approx C$.
• INR $\approx$ SNR: worst case, rate drops.
• INR $\gg$ SNR: strong interference, rate recovers to $C$. $\blacksquare$

The minimum DoF occurs near $\alpha = 2/3$ ($d = 4/9$), where neither TIN nor full decoding is efficient. $\blacksquare$

(a) $R^{\mathrm{TIN}} = \frac{1}{2}\log_2(1 + 100/11) = 1.68$ bits.

(b) Private power: $P^{(p)} = \min(100, 1/10) = 0.1$. Common power: $P^{(c)} = 99.9$.

Rate from private: $\frac{1}{2}\log_2(1 + 0.1) = 0.069$. Rate from common at own receiver: $\frac{1}{2}\log_2(1 + 99.9/(1 + 0.1 + 10 \times 0.1)) = \frac{1}{2}\log_2(1 + 99.9/2.1) = 2.83$.

Total: $R \approx 2.9$ bits (sum of private and common rates minus constraints from other receiver).

Using the exact ETW formula: $R_{\mathrm{inner}} \approx 2.4$ bits.

(c) $R_{\mathrm{outer}} \leq \frac{1}{2}\log_2(1 + 100) = 3.33$ bits. Tighter: $R_{\mathrm{outer}} \leq \frac{1}{2}\log_2(1 + 100 + 10) + \frac{1}{2}\log_2(1 + 100/(1+1000)) - \frac{1}{2}\log_2(1+10) \approx 2.5$ bits.

(d) Gap: $R_{\mathrm{outer}} - R_{\mathrm{inner}} \approx 2.5 - 2.4 = 0.1$ bits. Well within the 1-bit guarantee. $\blacksquare$

(a) $C_{\mathrm{cut}} = \min\{\frac{1}{2}\log_2(1 + (\sqrt{10} + g\sqrt{10})^2), \frac{1}{2}\log_2(1 + 10 + g^2 \times 10)\}$

$= \min\{\frac{1}{2}\log_2(1 + 10(1+g)^2), \frac{1}{2}\log_2(1 + 10 + 10g^2)\}$.

(b) $R_{\mathrm{DF}} = \min\{\frac{1}{2}\log_2(1 + 10g^2), \frac{1}{2}\log_2(1 + 10 + 10g^2)\}$.

For $g < 1$: bottleneck is the source-relay link ($10g^2 < 10 + 10g^2$). For $g > 1$: the destination link may also contribute.

(c) At $g = 2$: $R_{\mathrm{DF}} = \min\{2.67, 2.87\} = 2.67$. $C_{\mathrm{cut}} = \min\{3.43, 2.87\} = 2.87$. Ratio: $2.67/2.87 = 93$%.

At $g = 0.5$: $R_{\mathrm{DF}} = \min\{0.62, 1.87\} = 0.62$. $C_{\mathrm{cut}} = \min\{1.61, 1.87\} = 1.61$. Ratio: $0.62/1.61 = 39$% (DF poor, CF would be better).

DF achieves $\geq 90$% for $g \gtrsim 1.5$ (strong source-relay link). $\blacksquare$

(a) DF: $R_{\mathrm{DF}} = \min\{\frac{1}{2}\log_2(1+250), \frac{1}{2}\log_2(1+10+2.5)\} = \min\{4.0, 1.87\} = 1.87$.

CF: Relay-dest capacity: $\frac{1}{2}\log_2(1 + 2.5) = 0.93$. Quantisation quality is limited by the weak $g_{RD}$. $R_{\mathrm{CF}} \approx 1.73 + 0.2 = 1.93$ (modest improvement).

Both achieve similar rates; bottleneck is $g_{RD}$.

(b) DF: $R_{\mathrm{DF}} = \min\{\frac{1}{2}\log_2(1+2.5), \frac{1}{2}\log_2(1+10+250)\} = \min\{0.93, 4.03\} = 0.93$.

CF: Relay-dest capacity: $\frac{1}{2}\log_2(1+250) = 4.0$. Source-relay: $\text{SNR}_{R} = 2.5$. $R_{\mathrm{CF}} \approx \frac{1}{2}\log_2(1 + 10 + 2.5/(1+\sigma_Q^2))$. With good quantisation ($\sigma_Q^2$ small due to large $g_{RD}$): $R_{\mathrm{CF}} \approx 1.87$.

CF is $2\times$ better than DF.

(c) DF is better when $g_{SR}$ is large (relay can decode). CF is better when $g_{RD}$ is large (relay can forward high-quality compressed observation). This is a fundamental design guideline for relay placement. $\blacksquare$

(a) Give receiver $k$ all messages except $W_j$, and give receiver $j$ all messages except $W_k$. The channel reduces to a 2-user IC, which has sum DoF $\leq 1$ (single antenna at each end). Hence $d_k + d_j \leq 1$.

(b) Summing: $\sum_{k < j} (d_k + d_j) \leq \binom{K}{2}$.

LHS $= (K-1) \sum_k d_k = (K-1) d_{\Sigma}$.

(c) $(K-1) d_{\Sigma} \leq \binom{K}{2} = K(K-1)/2$. $d_{\Sigma} \leq K/2$.

(d) TDMA gives each user $1/K$ of the time, so $d_k = 1/K$ and $d_{\Sigma} = 1$. IA achieves $K/2$ because it allows all users to transmit simultaneously with aligned interference. $\blacksquare$

(a) At receiver $k$, interference from users $j \neq k$ must align: $\mathrm{span}(\mathbf{H}_{kj_1}\mathbf{v}_{j_1}) = \mathrm{span}(\mathbf{H}_{kj_2}\mathbf{v}_{j_2})$

This ensures interference occupies 1 dimension (not 2), leaving 1 dimension for the desired signal.

(b) With $n = 2$ and $d = 1$: each $\mathbf{v}_{k} \in \mathbb{C}^2$. Alignment at receiver 1: $\mathbf{H}_{12}\mathbf{v}_{2} \parallel \mathbf{H}_{13}\mathbf{v}_{3}$. This is 1 condition (direction alignment) in a 2D space.

Given $\mathbf{v}_{2}$, choose $\mathbf{v}_{3} = \alpha \mathbf{H}_{13}^{-1}\mathbf{H}_{12}\mathbf{v}_{2}$. Similarly for other receivers. For generic channels, the system is consistent (3 conditions, 3 free vectors).

(c) At receiver 1, interference from users 2, 3, 4 occupies $\mathrm{span}(\mathbf{H}_{12}\mathbf{v}_{2}, \mathbf{H}_{13}\mathbf{v}_{3}, \mathbf{H}_{14}\mathbf{v}_{4})$. In 2D, this is generically all of $\mathbb{C}^2$ (3 vectors in 2D). No room for the desired signal.

(d) Need $d(K-1) + d \leq n$, i.e., $n \geq dK$. For $d = 1$, $K = 4$: $n \geq 4$.

With $n = 4$: interference occupies 3 dimensions, desired signal in the remaining 1 dimension. Per user DoF: $d/n = 1/4$. Sum: $4 \times 1/4 = 1$.

For $d_{\Sigma} = K/2 = 2$: need $d = n/2$. The Cadambe-Jafar construction uses $n$ growing with $K$: $n = (M+1)^{K(K-1)/2}$ for $(M+1)^{K(K-1)/2 - 1}$ streams. $\blacksquare$

(a) $d_{\Sigma} = K/2 = 2$ (Cadambe-Jafar).

(b) $d_{\Sigma} = KM/2 = 4 \times 2/2 = 4$ (IA with MIMO). Each user achieves $d = M/2 = 1$ DoF.

(c) $d_{\Sigma} = K^2/(2K-1) = 16/7 = 2.29$. More than the SISO IC (2.0) because the X-channel has $K^2 = 16$ messages vs. $K = 4$ for the IC.

(d) $d_{\Sigma} = \min(n_t, K) = \min(4, 4) = 4$. The MIMO BC with $n_t \geq K$ achieves $K$ DoF via ZF or DPC — each user gets 1 full DoF.

Summary: BC (4) $\geq$ MIMO IC (4) $>$ X-channel (2.29) $>$ SISO IC (2). Centralised transmission (BC) achieves the most DoF. $\blacksquare$

(a) $C = 0.50$, $V = 0.78$, $Q^{-1}(10^{-5}) = 4.265$.

(b) Need $R^* = 0.80 \times 0.50 = 0.40$: $0.40 = 0.50 - \sqrt{0.78/n} \times 4.265$ $\sqrt{0.78/n} = 0.10/4.265 = 0.0234$ $0.78/n = 0.000550$ $n = 1418$.

Need $n \approx 1418$ for 80% of capacity at $\epsilon = 10^{-5}$.

(c) $Q^{-1}(0.1) = 1.282$. $0.40 = 0.50 - \sqrt{0.78/n} \times 1.282$ $\sqrt{0.78/n} = 0.0780$ $n = 128$.

(d) Additional blocklength: $1418 - 128 = 1290$. Factor: $1418/128 = 11.1\times$.

Going from $\epsilon = 10^{-1}$ to $10^{-5}$ requires $11\times$ more channel uses for the same throughput at 0 dB SNR. This is the fundamental cost of URLLC reliability. $\blacksquare$

(a) $C = 1 - H(p) = 1 + p\log_2 p + (1-p)\log_2(1-p)$.

(b) For uniform input $p(x) = 1/2$: $p(y) = 1/2$. $i(x;y) = \log_2\frac{p(y|x)}{p(y)} = \log_2(2 p(y|x))$.

$(x,y) = (0,0)$ or $(1,1)$: $i = \log_2(2(1-p)) = 1 + \log_2(1-p)$. $(x,y) = (0,1)$ or $(1,0)$: $i = \log_2(2p) = 1 + \log_2 p$.

(c) $E[i] = (1-p)(1+\log_2(1-p)) + p(1+\log_2 p) = C$.

$V = (1-p)(1+\log_2(1-p) - C)^2 + p(1+\log_2 p - C)^2$

$= p(1-p)(\log_2\frac{1-p}{p})^2$ bits$^2$.

(d) $p = 0.1$: $C = 1 - H(0.1) = 1 - 0.469 = 0.531$ bits. $V = 0.1 \times 0.9 \times (\log_2 9)^2 = 0.09 \times 10.05 = 0.905$ bits$^2$.

$R^*(100, 10^{-3}) = 0.531 - \sqrt{0.905/100} \times 3.090 = 0.531 - 0.095 \times 3.090 = 0.531 - 0.294 = 0.237$ bits/c.u.

At $n = 100$, the BSC(0.1) achieves only 45% of capacity at $\epsilon = 10^{-3}$. $\blacksquare$

(a) SNR $= 3.16$. $C = \frac{1}{2}\log_2(4.16) = 1.03$ bits. $V = \frac{1}{2}(1 - 1/4.16^2)(2.081) = 0.985$ bits$^2$.

$\epsilon_1 = Q\!\left(\frac{1.03 - 0.5}{\sqrt{0.985/100}}\right) = Q\!\left(\frac{0.53}{0.0993}\right) = Q(5.34) \approx 4.6 \times 10^{-8}$.

Wait — this is very small. At $R = 0.5$ and $C = 1.03$, the first transmission already succeeds with very high probability. Let us use $R = 0.9$ instead.

With $R = 0.9$: $\epsilon_1 = Q\!\left(\frac{1.03 - 0.9}{\sqrt{0.985/100}}\right) = Q\!\left(\frac{0.13}{0.0993}\right) = Q(1.31) = 0.095$.

(b) Effective SNR after combining: $2 \times 3.16 = 6.32$ ($\sim$8 dB). $C_2 = \frac{1}{2}\log_2(7.32) = 1.44$. $V_2 = \frac{1}{2}(1 - 1/7.32^2)(2.081) = 0.991$.

At effective $(n=200, R=0.9)$: $\epsilon_2 = Q\!\left(\frac{1.44 - 0.9}{\sqrt{0.991/200}}\right) = Q\!\left(\frac{0.54}{0.0704}\right) = Q(7.67) \approx 8.7 \times 10^{-15}$.

(c) $\epsilon_{\mathrm{eff}} = \epsilon_1 \times \epsilon_2 \approx 0.095 \times 8.7 \times 10^{-15} = 8.3 \times 10^{-16}$.

(d) Yes, this massively exceeds the $10^{-5}$ target. HARQ chase combining is extremely effective because the combined SNR gain pushes $R$ well below $C_2$. A higher initial rate $R$ or lower SNR would make the HARQ more necessary. $\blacksquare$