Exercises

(a) Each user's slot duration is $T_k = T/K = 2$ ms. Burst power: $P_{\text{burst}} = P \cdot K = 200 \times 5 = 1000$ mW = 1 W.

(b) $\text{SNR}_{\text{burst}} = P_{\text{burst}} |h_k|^2 / (W\sigma^2) = 1 / (5 \times 10^6 \times 10^{-9}) = 200$

$R_k = \frac{1}{K} W \log_2(1 + K \cdot \text{SNR}) = \frac{1}{5} \times 5 \times 10^6 \times \log_2(1 + 200)$ $= 10^6 \times 7.65 = 7.65$ Mbps.

(c) TDMA sum rate: $5 \times 7.65 = 38.3$ Mbps.

MAC sum capacity: $C_{\text{sum}} = W\log_2(1 + 5 \times 200/5) = 5 \times 10^6 \times \log_2(201) = 38.3$ Mbps.

For equal channel gains and equal power, TDMA achieves the sum capacity! The penalty arises only with unequal channels. $\blacksquare$

(a) Per-subcarrier SNR: assume $p_{k,n} = P/(K \cdot N_{\text{sc}}/K) = P/N_{\text{sc}}$ and $\text{SNR}_{0} = P/(N_{\text{sc}} \Delta f \sigma^2) = 10$ (for concreteness).

$R_k = 32 \times \Delta f \times \mathbb{E}[\log_2(1 + 10|h_{k,n}|^2)]$

For Rayleigh fading: $\mathbb{E}[\log_2(1 + 10X)]$ where $X \sim \text{Exp}(1)$ $\approx 2.86$ bits/s/Hz (by numerical integration).

$R_k \approx 32 \times 156.25 \times 10^3 \times 2.86 = 14.3$ Mbps.

(b) Each subcarrier goes to $\arg\max_k |h_{k,n}|^2$. By symmetry, each user gets $128/4 = 32$ subcarriers on average, but with better channels:

$\mathbb{E}[\max_k |h_{k,n}|^2] = H_4 = 1 + 1/2 + 1/3 + 1/4 = 2.083$

The effective per-subcarrier rate (with the best user): $\mathbb{E}[\log_2(1 + 10 \times 2.083)] \approx \log_2(21.83) = 4.45$ bits/s/Hz.

Per-user throughput: $32 \times 156.25 \times 10^3 \times 4.45 = 22.3$ Mbps.

Multi-user diversity gain: $22.3/14.3 = 1.56\times$ (56% improvement). $\blacksquare$

With $\alpha_k = 1/K$ and $\text{SNR}_{k} = \gamma$:

$$R_{\text{sum}}^{\text{orth}} = K \cdot \frac{1}{K} \log_2\!\left(1 + \frac{\gamma}{1/K}\right) = \log_2(1 + K\gamma)$$

The MAC sum capacity is:

$$C_{\text{sum}} = \log_2\!\left(1 + \sum_{k=1}^{K} \gamma\right) = \log_2(1 + K\gamma)$$

These are identical, confirming that orthogonal access with equal allocation is sum-rate optimal when all users are symmetric. $\blacksquare$

$\text{SNR}_{1} = P_1|h_1|^2/(W\sigma^2) = 10/1 = 10$

$\text{SNR}_{2} = P_2|h_2|^2/(W\sigma^2) = 1/1 = 1$

Sum capacity: $C_{\text{sum}} = \log_2(1 + 10 + 1) = \log_2(12) = 3.585$ bits/s/Hz.

Corner A (decode user 2 first): $R_1 = \log_2(1 + 10) = 3.459$, $R_2 = \log_2(12) - \log_2(11) = 0.126$.

Corner B (decode user 1 first): $R_2 = \log_2(1 + 1) = 1.0$, $R_1 = \log_2(12) - \log_2(2) = 2.585$.

$R_{\text{FDMA}}(\alpha) = \alpha\log_2(1 + 10/\alpha) + (1-\alpha)\log_2(1 + 1/(1-\alpha))$

Taking the derivative and setting to zero (numerically): $\alpha^{\star} \approx 0.77$.

$R_{\text{FDMA}} \approx 0.77\log_2(1 + 12.99) + 0.23\log_2(1 + 4.35) = 0.77 \times 3.81 + 0.23 \times 2.42 = 2.93 + 0.56 = 3.49$.

Loss: $(3.585 - 3.49)/3.585 = 2.6\%$. $\blacksquare$

(a) Decode user 2 first: $R_2 = \log_2(1 + 0.7 \times 1 / (0.3 \times 20 + 1)) = \log_2(1 + 0.7/7) = \log_2(1.1) = 0.137$ bits/s/Hz.

Decode user 1 (interference-free): $R_1 = \log_2(1 + 0.3 \times 20 / 1) = \log_2(7) = 2.807$ bits/s/Hz.

(b) $R_1 + R_2 = 2.807 + 0.137 = 2.944$.

$\log_2(1 + 0.3 \times 20 + 0.7 \times 1) = \log_2(1 + 6 + 0.7) = \log_2(7.7) = 2.944$. $\checkmark$ $\blacksquare$

MAC sum rate: $C_{\text{sum}} \approx \log_2(\text{SNR}_{1} + \text{SNR}_{2}) \approx \log_2(\text{SNR}_{1})$ for $\text{SNR}_{1} \gg \text{SNR}_{2}$.

TDMA with equal sharing ($\alpha = 1/2$): $R_{\text{TDMA}} = \frac{1}{2}\log_2(1 + 2\text{SNR}_{1}) + \frac{1}{2}\log_2(1 + 2\text{SNR}_{2})$ $\approx \frac{1}{2}\log_2(2\text{SNR}_{1}) + \frac{1}{2}\log_2(2\text{SNR}_{2})$ $= \frac{1}{2}\log_2(4\text{SNR}_{1}\text{SNR}_{2})$

$\Delta R = C_{\text{sum}} - R_{\text{TDMA}} \approx \log_2(\text{SNR}_{1}) - \frac{1}{2}\log_2(4\text{SNR}_{1}\text{SNR}_{2})$ $= \log_2(\text{SNR}_{1}) - 1 - \frac{1}{2}\log_2(\text{SNR}_{1}) - \frac{1}{2}\log_2(\text{SNR}_{2})$ $= \frac{1}{2}\log_2(\text{SNR}_{1}/\text{SNR}_{2}) - 1$ $= \frac{1}{2}\log_2(\gamma) - 1$

For $\gamma = 100$ (20 dB): $\Delta R \approx 3.32 - 1 = 2.32$ bits/s/Hz.

The NOMA gain grows logarithmically with the channel gain asymmetry, which is why NOMA is most beneficial in near-far scenarios. $\blacksquare$

With $P_k|h_k|^2 = P$ for all $k$, the MAC constraints are:

$$\sum_{k \in \mathcal{S}} R_k \leq \log_2(1 + |\mathcal{S}| \cdot P/\sigma^2) \quad \forall\, \mathcal{S}$$

The sum rate $C_{\text{sum}} = \log_2(1 + KP/\sigma^2)$ is achievable by any SIC order.

Equal TDMA: $R_k = \frac{1}{K}\log_2(1 + KP/\sigma^2)$ for all $k$. Sum: $\log_2(1 + KP/\sigma^2) = C_{\text{sum}}$. Matches.

Consider the rate tuple where user 1 gets rate $R_1 = \log_2(1 + P/\sigma^2)$ (decoded last, seeing all others as noise already cancelled) and all others share the remaining sum rate.

With TDMA: $R_1 \leq \frac{1}{K}\log_2(1 + KP/\sigma^2)$ or $R_1 \leq \log_2(1 + KP/\sigma^2)$ if user 1 gets the entire frame. But giving the full frame to user 1 leaves $R_j = 0$ for $j \neq 1$.

With NOMA: $R_1 = \log_2(1 + P/\sigma^2)$ while $\sum_{j \neq 1} R_j = \log_2(1 + KP/\sigma^2) - \log_2(1 + P/\sigma^2) > 0$.

This asymmetric rate tuple is inside the MAC region but outside the orthogonal region, proving the strict inclusion. $\blacksquare$

(a) $\mathbf{c}_1^T\mathbf{c}_2 = \frac{1}{8}(1\cdot1 + 1\cdot(-1) + 1\cdot1 + 1\cdot(-1) + \cdots) = \frac{1}{8}(1-1+1-1+1-1+1-1) = 0$. $\checkmark$

(b) $z_1 = \mathbf{c}_1^T(\mathbf{c}_1 b_1 + \mathbf{c}_2 b_2) = b_1 + 0 = +1$.

$z_2 = \mathbf{c}_2^T(\mathbf{c}_1 b_1 + \mathbf{c}_2 b_2) = 0 + b_2 = -1$.

Perfect separation with zero MAI.

(c) With a 1-chip delay, the received code for user 2 becomes a cyclic shift: $\mathbf{c}_2' = \frac{1}{\sqrt{8}}[-1,1,-1,1,-1,1,-1,1]^T$.

$\mathbf{c}_1^T\mathbf{c}_2' = \frac{1}{8}(-1+1-1+1-1+1-1+1) = 0$.

In this case, orthogonality is preserved. But for general Walsh code pairs, a chip delay can introduce non-zero cross-correlation of up to $1/\sqrt{N}$, destroying orthogonality. $\blacksquare$

(a) $\mathbb{E}[\mathbf{c}_i^T\mathbf{c}_j] = \sum_{n=1}^{N} \mathbb{E}[c_{i,n}c_{j,n}] = 0$ (independent codes).

$\text{Var}[\mathbf{c}_i^T\mathbf{c}_j] = \sum_{n=1}^{N} \mathbb{E}[c_{i,n}^2 c_{j,n}^2] = N \cdot \frac{1}{N^2} = \frac{1}{N} = \frac{1}{32}$.

(b) MAI power from each interferer: $\mathbb{E}[|\mathbf{c}_k^T\mathbf{c}_j|^2] = 1/N$.

Total MAI variance: $(K-1)/N = 15/32 = 0.469$.

Noise variance: $\sigma^2 = 1/(2 E_b/N_0) = 1/(2 \times 10) = 0.05$.

Total interference + noise: $0.469 + 0.05 = 0.519$.

SINR: $1/0.519 = 1.926$ (2.85 dB).

BER $\approx Q(\sqrt{2 \times 1.926}) = Q(1.963) \approx 0.025$.

The MAI dominates the noise, illustrating the MAI-limited regime of CDMA. $\blacksquare$

The CDMA system with random codes is equivalent to a $K \times N$ MIMO MAC. The sum capacity per chip is:

$$C = \frac{1}{N}\log_2\det\!\left(\mathbf{I}_K + \frac{\text{SNR}}{N}\mathbf{C}\mathbf{C}^H\right)$$

In the limit $K, N \to \infty$ with $K/N = \beta$, by the Marcenko-Pastur law:

$$C \to \int \log_2(1 + \text{SNR}\lambda)\,d F_{\beta}(\lambda)$$

where $F_\beta$ is the Marcenko-Pastur distribution. For the sum capacity with equal power and optimal multiuser detection:

$$C_{\text{sum}} = \beta\log_2\!\left(1 + \text{SNR} - \frac{\beta}{4}\mathcal{F}(\text{SNR}, \beta)\right) + \text{terms}$$

At low $\beta$ (few users per chip): $C_{\text{sum}} \approx \beta\log_2(1 + \text{SNR})$.

The simple formula $\log_2(1 + \beta \cdot \text{SNR})$ applies to the single-user bound (all users decoded jointly). $\blacksquare$

(a) After despreading, the MAI variance from each near user is $P \cdot d_{\text{near}}^{-\alpha} / N$. The signal power for user 1 is $P \cdot d_{\text{far}}^{-\alpha}$.

$$\text{SIR}_1 = \frac{P \cdot d_{\text{far}}^{-\alpha}} {(K-1) \cdot P \cdot d_{\text{near}}^{-\alpha} / N} = \frac{N}{K-1} \cdot \left(\frac{d_{\text{near}}}{d_{\text{far}}}\right)^{\alpha}$$

(b) $\text{SIR}_1 = \frac{64}{15} \cdot (1/10)^4 = 4.267 \times 10^{-4} = -33.7$ dB.

The far user is completely overwhelmed by the near users. Without power control, communication is impossible.

(c) To equalise received power: $P_{\text{near}} d_{\text{near}}^{-\alpha} = P_{\text{far}} d_{\text{far}}^{-\alpha}$.

$$\frac{P_{\text{near}}}{P_{\text{far}}} = \left(\frac{d_{\text{near}}}{d_{\text{far}}}\right)^{\alpha} = (1/10)^4 = 10^{-4} = -40 \;\text{dB}$$

Near users must reduce power by 40 dB (a factor of 10,000). $\blacksquare$

(a) $G = K\lambda = 50 \times 0.005 = 0.25$.

(b) $S = Ge^{-2G} = 0.25 \times e^{-0.5} = 0.25 \times 0.6065 = 0.1516$.

Packet success probability: $P_s = e^{-2G} = 0.6065$. Average transmissions: $1/P_s = 1.649$. Average retransmissions: $0.649$ per packet.

(c) Peak at $G = 0.5$: $K^{\star} = 0.5/0.005 = 100$ users.

$S_{\max} = 0.5e^{-1} = 0.184$. $\blacksquare$

Given $n$ users transmit in a slot, a tagged user captures if its power exceeds $\gamma_0$ times the sum of the other $n-1$ users' powers. For i.i.d. $\text{Exp}(\bar{P})$:

$$P_{\text{capture}}(n) = \Pr\!\left[P_1 > \gamma_0 \sum_{j=2}^{n} P_j\right] = \frac{1}{(1 + \gamma_0)^{n-1}}$$

By symmetry, any of the $n$ users can capture, so the probability of a successful transmission in the slot is $n \cdot P_{\text{capture}}(n) / 1 = n/(1+\gamma_0)^{n-1}$, but at most one succeeds.

$S = \sum_{n=1}^{K} \binom{K}{n} p^n(1-p)^{K-n} \cdot \frac{n}{(1+\gamma_0)^{n-1}}$$For$\gamma_0 \to \infty$(no capture): only$n = 1$terms survive, recovering standard slotted ALOHA$S = Kp(1-p)^{K-1} \to Ge^{-G}$. For$\gamma_0 = 0$(perfect capture): the strongest user always wins,$S = 1 - (1-p)^K$, approaching 1 as$G$grows. Capture significantly improves throughput beyond$1/e$.$\blacksquare$

(a) In a slot with $n$ backlogged users:

• Expected new arrivals: $\lambda$.
• Expected departures: probability that exactly one backlogged user transmits and no new arrival collides, approximately $np_r(1-p_r)^{n-1}e^{-\lambda}$ for large $n$.

$D(n) = \lambda - np_r(1-p_r)^{n-1}e^{-\lambda}$

(b) $D(n) < 0$ requires $\lambda < np_r(1-p_r)^{n-1}e^{-\lambda}$.

The throughput $g(n) = np_r(1-p_r)^{n-1}$ is maximised at $p_r = 1/n$, giving $g_{\max}(n) = (1 - 1/n)^{n-1} \to 1/e$.

(c) As $n \to \infty$: $\max_{p_r} g(n) \to 1/e$.

For $\lambda < e^{-\lambda}/e \approx 1/e$ (at small $\lambda$): the drift is negative for large $n$, ensuring stability.

For $\lambda > 1/e$: $D(n) > 0$ for all sufficiently large $n$, meaning the backlog grows without bound.

Therefore $\lambda_{\max} = 1/e \approx 0.368$, matching the throughput analysis. $\blacksquare$

(a) Pilot length: $\tau_p = K = 16$ symbols. Overhead: $16/196 = 8.2\%$.

(b) Data symbols: $196 - 16 = 180$. UL data: $180/4 = 45$ symbols. DL data: $180 \times 3/4 = 135$ symbols.

(c) DL efficiency: $135/196 = 68.9\%$. Effective rate: $5 \times 0.689 = 3.44$ bits/s/Hz.

Per-user throughput: if bandwidth is 100 MHz, $3.44 \times 10^8 / 16 = 21.5$ Mbps per user. $\blacksquare$

$\eta_{\text{TDD}} = (200 - 10)/(200) = 0.95$ (constant).

For $M = 64$: $R = \log_2(1 + 64) = 6.02$, sum $= 10 \times 6.02 \times 0.95 = 57.2$. For $M = 128$: $R = \log_2(1 + 128) = 7.01$, sum $= 66.6$. For $M = 256$: $R = \log_2(1 + 256) = 8.00$, sum $= 76.0$.

$\eta_{\text{FDD}}(M) = (200 - 2M)/(200)$ (pilots + feedback).

$M = 64$: $\eta = (200-128)/200 = 0.36$, sum $= 10 \times 6.02 \times 0.36 = 21.7$. $M = 100$: $\eta = 0$. FDD becomes infeasible. $M = 128, 256$: negative $\eta$ — impossible.

TDD provides $57.2/21.7 = 2.6\times$ higher efficiency at $M = 64$ and is the only viable option for $M > 100$. $\blacksquare$

$\max_{\tau_u, \tau_d \geq 0} \; \alpha \tau_d r + (1-\alpha)\tau_u rKATEXPLACEHOLDER0END\text{s.t. } \tau_u + \tau_d = T_c - KKATEXPLACEHOLDER1END\max_{\tau_u} \; r[\alpha(T_c - K - \tau_u) + (1-\alpha)\tau_u]KATEXPLACEHOLDER2END= r[\alpha(T_c - K) + (1 - 2\alpha)\tau_u]$$

For $\alpha > 1/2$ (DL-heavy): $\tau_u^{\star} = 0$, $\tau_d^{\star} = T_c - K$. All data symbols go to DL.

For $\alpha < 1/2$ (UL-heavy): $\tau_d^{\star} = 0$, $\tau_u^{\star} = T_c - K$. All to UL.

For $\alpha = 1/2$: any split is optimal.

In practice, the UL and DL rates differ (due to power asymmetry), leading to an interior optimum. With DL rate $r_d$ and UL rate $r_u$:

$\tau_d^{\star} / \tau_u^{\star} = \alpha r_d / ((1-\alpha)r_u)$

matching the traffic weight ratio to the rate ratio. $\blacksquare$

(a) Active devices per ms: $K \times (100/10^6) / 10^{-3} \approx 1$ device. But the BS does not know which device is active.

To allow any device to access, we need $K = 10{,}000$ pilot sequences. Each pilot has length $\tau_p \geq K = 10{,}000$. With 1 MHz bandwidth and 1 ms coherence: $T_c = 1000$ symbols. $\tau_p = 10{,}000 \gg T_c$: infeasible.

(b) Phase 1 (grant-free): Each active device picks a random pilot from a pool of $L = 64$ non-orthogonal pilots and transmits its 100-bit packet. Expected active devices per slot: $\sim 1$. Collision probability with $L = 64$ pilots: $\approx 1/L = 1.6\%$.

Phase 2 (scheduled): Collided devices are identified (via NACK) and scheduled in subsequent slots with orthogonal pilots.

Expected throughput: $\sim 0.984 \times 100$ kbps $= 98.4$ kbps in phase 1, with $\sim 1.6$ kbps retransmitted in phase 2.

(c) Pure slotted ALOHA: $S = 1/e \approx 36.8\%$ of the channel capacity. With 1 active device per ms on average, slotted ALOHA works well (success $\approx e^{-1/1} = 36.8\%$ with $G = 1$; but since $G \approx 0.001$ here, success $\approx 99.9\%$).

Pure OFDMA: infeasible (pilot overhead).

Hybrid scheme: near-100% success with $< 2$ ms average latency, combining the low overhead of random access with the reliability of scheduling. $\blacksquare$