Exercises

(a) $q = \sqrt{3 \times 3} = \sqrt{9} = 3$.

(b) $\text{SIR} = 3^4/6 = 81/6 = 13.5$.

(c) $\text{SIR}_{\text{dB}} = 10\log_{10}(13.5) = 11.3$ dB. $\blacksquare$

(a) $\text{SIR}_{\min} = 10^{18/10} = 63.1$.

(b) $N_{\min} = (6 \times 63.1)^{2/3.5}/3 = (378.6)^{0.571}/3 = 29.69/3 = 9.90$.

(c) Valid sizes: $N = 9$ ($3^2 + 0$) and $N = 12$ ($4 + 4 + 4$). Since $9.90 > 9$, the smallest valid $N = 12$. $\blacksquare$

(a) First tier: $D_1/R = \sqrt{3 \times 7} = \sqrt{21} \approx 4.58$. All 6 at approximately this distance.

Second tier: $D_2/R = \sqrt{3} \cdot D_1/R = \sqrt{63} \approx 7.94$. 6 second-tier interferers at this distance.

(b) $\text{SIR}^{-1} = \sum_{i=1}^{6}(D_{1}/R)^{-4} + \sum_{i=1}^{6}(D_{2}/R)^{-4}$

$= 6 \times 21^{-2} + 6 \times 63^{-2}$ $= 6/441 + 6/3969$ $= 0.01361 + 0.001512 = 0.01512$

$\text{SIR} = 1/0.01512 = 66.1$ (18.2 dB).

(c) First-tier only: $\text{SIR} = 21^2/6 = 73.5$ (18.7 dB).

Second-tier interference reduces SIR by 0.5 dB — a relatively small correction, justifying the first-tier approximation for most purposes. $\blacksquare$

(a) The hexagonal lattice has basis vectors $\mathbf{a}_1 = (\sqrt{3}R, 0)$ and $\mathbf{a}_2 = (\sqrt{3}R/2, 3R/2)$.

The co-channel cell at $(i, j)$ is at position $\mathbf{d} = i\mathbf{a}_1 + j\mathbf{a}_2$.

(b) $|\mathbf{d}|^2 = |i\mathbf{a}_1 + j\mathbf{a}_2|^2$ $= 3R^2 i^2 + 2 \cdot i j \cdot \mathbf{a}_1 \cdot \mathbf{a}_2 + 3R^2 j^2 (3/4 + 1/4)$

With $\mathbf{a}_1 \cdot \mathbf{a}_2 = 3R^2/2$:

$D^2 = 3R^2(i^2 + ij + j^2) = 3R^2 N$.

Hence $D = R\sqrt{3N}$.

(c) $N = 1$ (1,0), $N = 3$ (1,1), $N = 4$ (2,0), $N = 7$ (2,1), $N = 9$ (3,0), $N = 12$ (2,2), $N = 13$ (3,1), $N = 16$ (4,0), $N = 19$ (3,2), $N = 21$ (4,1). $\blacksquare$

(a) $\mathbb{E}[r_0] = \frac{1}{2\sqrt{\lambda}} = \frac{1}{2\sqrt{5}} = 0.224$ km $= 224$ m.

(b) $\Pr[r_0 > 0.2] = \exp(-\pi \times 5 \times 0.04) = \exp(-0.628) = 0.534$.

53.4% chance of no BS within 200 m.

(c) $\mathbb{E}[N(B(0, 0.5))] = \lambda \pi r^2 = 5 \times \pi \times 0.25 = 3.93$ BSs. $\blacksquare$

(a) $\int_{\tau^{-1/2}}^{\infty}\frac{du}{1+u^2} = \frac{\pi}{2} - \arctan(\tau^{-1/2}) = \arctan(\sqrt{\tau})$

(using $\pi/2 - \arctan(1/x) = \arctan(x)$ for $x > 0$).

Therefore $\rho(\tau, 4) = \sqrt{\tau}\arctan(\sqrt{\tau})$.

(b) $p_c(\tau) = \frac{1}{1 + \rho(\tau, 4)} = \frac{1}{1 + \sqrt{\tau}\arctan(\sqrt{\tau})}$.

(c) As $\tau \to 0$: $\sqrt{\tau}\arctan(\sqrt{\tau}) \to 0$, so $p_c \to 1$. As $\tau \to \infty$: $\sqrt{\tau}\arctan(\sqrt{\tau}) \to \pi\sqrt{\tau}/2 \to \infty$, so $p_c \to 0$.

(d) $p_c(\tau^{\star}) = 0.5$ requires $\sqrt{\tau^{\star}}\arctan(\sqrt{\tau^{\star}}) = 1$.

Numerical solution: $\tau^{\star} \approx 0.61$ ($-2.1$ dB). Half the users achieve SINR above $-2.1$ dB. $\blacksquare$

(a) $\Pr[\text{SINR} > \tau \mid r] = \mathbb{E}[\exp(-\tau r^{\alpha}(I_r + \sigma^2/P))]$ $= \mathcal{L}_{I_r}(\tau r^{\alpha}/P) \cdot \exp(-\tau r^{\alpha}\sigma^2/P)$

The noise term $\exp(-\tau r^{\alpha}\sigma^2/P)$ depends on $r$ and does not cancel with the PPP averaging.

(b) $p_c(\tau) = \int_0^{\infty} 2\pi\lambda r \exp\!\left(-\pi\lambda r^2(1+\rho) - \frac{\tau\sigma^2}{P}r^4\right)dr$

For small noise ($\sigma^2 \ll P\lambda^2$), expanding to first order and completing the Gaussian integral:

$p_c(\tau) \approx \frac{1}{1+\rho} \exp\!\left(-\frac{\tau\sigma^2}{P} \cdot\frac{\Gamma(1+2/\alpha)}{(\pi\lambda(1+\rho))^{2/\alpha}}\right)$

(c) The exponential noise penalty $\exp(-c/\lambda^{2/\alpha})$ vanishes as $\lambda \to \infty$ but penalises coverage when $\lambda$ is small.

For $\alpha = 4$: penalty $\sim \exp(-c/\sqrt{\lambda})$. Doubling $\lambda$ reduces the exponent by factor $\sqrt{2}$, improving coverage. In the noise-limited regime, densification does improve coverage. $\blacksquare$

(a) $P_2/P_1 = 10^{(20-46)/10} = 10^{-2.6} = 0.00251$.

$(P_2/P_1)^{2/4} = 0.00251^{0.5} = 0.0501$.

$A_2 = \frac{50 \times 0.0501}{1 + 50 \times 0.0501} = \frac{2.505}{3.505} = 0.715$.

71.5% of users associate with femto cells (due to high density). $A_1 = 0.285$.

(b) Macro cell area $\approx 1/\lambda_1 = 1$ km$^2$. Expected femtos: $\lambda_2/\lambda_1 = 50$.

Each macro cell overlaps with about 50 femto cells. $\blacksquare$

(a) $B = 10^{0.6} = 3.981$. $(B P_2/P_1)^{0.5} = (3.981 \times 0.00251)^{0.5} = 0.00999^{0.5} = 0.0999$.

$A_2 = \frac{50 \times 0.0999}{1 + 50 \times 0.0999} = \frac{5.0}{6.0} = 0.833$.

CRE increases femto association from 71.5% to 83.3%.

(b) $B^{1/\alpha} = 3.981^{1/4} = 1.413$.

The effective femto coverage radius increases by 41%.

(c) During ABS ($\beta$ fraction): SINR $\approx$ femto signal only $\Rightarrow$ high SINR (assume 15 dB, $\eta = 4.5$ bits/s/Hz).

During non-ABS ($1-\beta$): SINR $= -10$ dB ($\eta \approx 0.12$ bits/s/Hz).

Average: $\bar{\eta} = \beta \times 4.5 + (1-\beta) \times 0.12$.

With $\beta = 0.25$: $\bar{\eta} = 1.125 + 0.09 = 1.22$ bits/s/Hz. Without ABS ($\beta = 0$): $\bar{\eta} = 0.12$ bits/s/Hz.

ABS provides 10$\times$ rate improvement. $\blacksquare$

(a) $\mathcal{A}_A = 10 \times 1.49 = 14.9$ bits/s/Hz/km$^2$. $\mathcal{A}_B = 40 \times 1.05 = 42.0$ bits/s/Hz/km$^2$.

(b) Network B has 2.8$\times$ higher ASE despite the lower per-link rate, because the 4$\times$ higher density more than compensates for the $C(3)/C(4) = 0.70$ per-link penalty.

(c) Per-user rate $= C(\alpha)/K$: $R_A = 1.49/20 = 0.075$ bits/s/Hz (per user per slot). $R_B = 1.05/20 = 0.053$ bits/s/Hz.

Network A gives higher per-user rate despite lower ASE, because it has fewer users competing per cell. $\blacksquare$

(a) When the average ISD $\approx 1/\sqrt{\lambda} < d_0$, the serving BS and the strongest interferers are all within LOS range. The system transitions from the NLOS regime (high $\alpha_H$, good SINR) to the LOS regime (low $\alpha_L$, poor SINR). The per-link rate decreases with density, partially cancelling the linear density gain.

(b) In the ultra-dense limit, all relevant BSs are at distances $\ll d_0$, so $\ell(r) = r^{-\alpha_L}$. The PPP coverage formula gives $p_c(\tau) = 1/(1 + \rho(\tau, \alpha_L))$. With $\alpha_L = 2$: $\rho(\tau, 2) = \tau/(1 - 2/\alpha_L)$ diverges as $\alpha_L \to 2$, severely degrading coverage.

(c) For $\alpha_L = 2$: the integral $\int_{\tau^{-1}}^{\infty}\frac{du}{1+u} = \ln(1+\tau)/\tau$ gives $\rho(\tau,2) \sim \pi\tau/\sin(\pi) \to \infty$.

More precisely, $C(2)$ diverges to 0 as $\alpha_L \to 2$ from above. The ASE saturates at a finite limit determined by the LOS propagation quality, showing that ultra-densification has fundamental limits. $\blacksquare$

(a) $v = 30/3.6 = 8.33$ m/s. $\mu_{\text{HO}} = 2 \times 8.33/(\pi \times 300) = 0.0177$ HO/s $= 63.6$ HO/hour.

(b) $d = 8.33 \times 0.32 = 2.67$ m.

(c) $d/R = 2.67/300 = 0.89\%$. The UE moves less than 1% of the cell radius during TTT — this is a very conservative setting. The TTT could be reduced to 160 ms or even 80 ms for faster handover response. $\blacksquare$

(a) $p_{\text{pp}}(0) = Q(0) = 0.500$ (50%). $p_{\text{pp}}(2) = Q(2/6) = Q(0.333) \approx 0.370$ (37%). $p_{\text{pp}}(4) = Q(4/6) = Q(0.667) \approx 0.252$ (25%). $p_{\text{pp}}(6) = Q(1.0) \approx 0.159$ (16%).

(b) Additional distance for the UE to travel $\propto 10^{H_{\text{hys}}/10\alpha}$: $H = 0$: no delay. $H = 2$: $\sim$0.5 m extra. $H = 4$: $\sim$1.0 m extra. $H = 6$: $\sim$1.5 m extra. (at $v = 10$ m/s: 0--150 ms additional delay.)

(c) None of the listed values achieves $p_{\text{pp}} < 10\%$ with hysteresis alone. Need $H_{\text{hys}}/\sigma > 1.28$: $H_{\text{hys}} > 7.7$ dB. In practice, combining $H_{\text{hys}} = 4$ dB with $T_{\text{TTT}} = 160$ ms reduces $p_{\text{pp}}$ below 10%. $\blacksquare$

(a) $J(H, T) = w_1 p_{\text{fail}}(H, T, v) + w_2 p_{\text{pp}}(H, T, v) + w_3 \bar{d}_{\text{HO}}(H, T)$

where $w_1 \gg w_2 > w_3$ (failures are most critical).

(b) $\min_{H, T} J(H, T)$ subject to: $p_{\text{fail}} \leq 2\%$, $p_{\text{pp}} \leq 10\%$, $\bar{d}_{\text{HO}} \leq 500$ ms.

(c) Ping-pong constraint: $p_{\text{pp}} \leq 0.1$ defines a curve $T \geq f_1(H)$ (higher $T$ or $H$ reduces pp).

Failure constraint: $p_{\text{fail}} \leq 0.02$ defines $T \leq f_2(H)$ (lower $T$ and $H$ allows timely HO).

Feasibility requires $f_1(H) \leq f_2(H)$, which holds only for a range of $H$ values — too little hysteresis requires very long TTT (eventually exceeding the failure boundary).

(d) In practice, MRO is a SON function that:

1. Monitors HO failure and pp rates per cell pair.
2. Adjusts $H$ and $T$ using gradient-free optimisation (e.g., tabular Q-learning or Bayesian optimisation).
3. Adapts per-cell-pair to local conditions.
4. Operates on a slow timescale (hours). $\blacksquare$

(a) 2 of 6 interferers fall within the 120-degree sector.

(b) FBR $= 25$ dB: $G_{\text{back}}/G_{\text{main}} = 10^{-2.5} = 0.00316$.

$N_I^{\text{eff}} = 2 + 4 \times 0.00316 = 2.013$.

(c) $G_3 = 6/2.013 = 2.98$ (4.75 dB).

Very close to the ideal gain of 3 due to the excellent 25 dB FBR. $\blacksquare$

BW/sector: 20/7 = 2.86 MHz. SIR: $21^2/6 = 73.5$ (18.7 dB). FoM: $1 \times 2.86 \times \log_2(74.5) = 2.86 \times 6.22 = 17.8$.

BW/sector: 20/4 = 5 MHz. SIR: $2.5 \times 12^2/6 = 60$ (17.8 dB). FoM: $3 \times 5 \times \log_2(61) = 15 \times 5.93 = 89.0$.

BW/sector: 20/1 = 20 MHz. SIR: $2.5 \times 3^2/6 = 3.75$ (5.7 dB). FoM: $3 \times 20 \times \log_2(4.75) = 60 \times 2.25 = 134.9$.

Option (c) has 7.6$\times$ higher FoM than (a) and 1.5$\times$ higher than (b). This is why modern networks use $N = 1$ with tri-sector and ICIC: the bandwidth gain dominates despite the SIR penalty. $\blacksquare$

(a) With 60-degree sectors: 1 of 6 first-tier interferers is in the main beam.

(b) FBR $= 20$ dB ($10^{-2} = 0.01$). $N_I^{\text{eff}} = 1 + 5 \times 0.01 = 1.05$. $G_6 = 6/1.05 = 5.71$ (7.57 dB).

(c) With $S$ sectors: $N_I^{\text{eff}} \approx 6/S + (6 - 6/S) \times 10^{-\text{FBR}/10}$.

As $S$ increases, the residual back-lobe interference approaches $6 \times 10^{-\text{FBR}/10}$, setting a floor on $N_I^{\text{eff}}$. For FBR $= 20$ dB: floor $= 6 \times 0.01 = 0.06$, so $G_{\max} = 6/0.06 = 100$. But the practical gain is limited by sector-to-sector isolation, pilot overhead, and antenna coupling.

(d) With $N = 1$: each sector has full 20 MHz. SIR per sector: $G_6 \times 3^2/6 = 5.71 \times 1.5 = 8.57$ (9.3 dB). Site capacity $\propto 6 \times \log_2(1 + 8.57) = 6 \times 3.26 = 19.5$ vs. $1 \times \log_2(1+1.5) = 1.32$ for omni.

Total gain: $19.5/1.32 = 14.8\times$. $\blacksquare$

(a) SIR: $2.5 \times 9^2/6 = 33.75$ (15.3 dB). Rate per sector: $\log_2(1+33.75) = 5.12$ bits/s/Hz. Effective density: $\lambda_M \times 3 = 6$ sectors/km$^2$. ASE: $6 \times 5.12 = 30.7$ bits/s/Hz/km$^2$.

(b) Effective total BS density: $6 + 20 = 26$/km$^2$. PPP ASE: $\mathcal{A} \approx 26 \times 1.49 = 38.7$ bits/s/Hz/km$^2$.

(c) The hexagonal model overestimates SIR for the regular macro deployment (actual SIR is worse due to irregular placement). The PPP model is more appropriate for the HetNet where small cell locations are truly random. The PPP per-link rate of 1.49 bits/s/Hz is lower than the hexagonal 5.12 bits/s/Hz because the PPP accounts for the full interference landscape.

(d) Initial: $3 \times 30 = 90$ users per macro site. Per-user rate: $3 \times 5.12/90 = 0.171$ bits/s/Hz.

HetNet: macro sectors serve $6 \times 30 = 180$ users/km$^2$. Small cells serve $20 \times 5 = 100$ users/km$^2$. Total: 280 users/km$^2$. Per-user rate: $38.7/280 = 0.138$ bits/s/Hz.

Despite higher ASE, per-user rate decreases because of more total users. The benefit is serving 3.1$\times$ more users per km$^2$. $\blacksquare$