Exercises

$\lambda = c/f = 3 \times 10^8 / 12 \times 10^9 = 0.025$ m.

$FSPL = 20\log_{10}\!\left(\frac{4\pi \times 36 \times 10^6}{0.025}\right) = 20\log_{10}(1.81 \times 10^{13}) = 205.2$ dB.

$P_r = P_t + G_t + G_r - FSPL = 43 + 30 + 40 - 205.2 = -92.2$ dBm. $\blacksquare$

$EIRP = P_t - L_{\text{cable}} + G_t = 40 - 2 + 18 = 56$ dBm.

New EIRP $= 40 - 2 + 12 = 50$ dBm. Change $= 50 - 56 = -6$ dB (6 dB reduction). $\blacksquare$

$FSPL_{\max} = P_t + G_t + G_r - P_{r,\min} = 14 + 2 + 2 - (-110) = 128$ dB.

$\lambda = 0.3456$ m. $128 = 20\log_{10}(4\pi d / \lambda)$ $\Rightarrow d = \frac{\lambda}{4\pi} \times 10^{128/20} = \frac{0.3456}{4\pi} \times 10^{6.4} = 0.02752 \times 2.512 \times 10^6 = 69.1$ km. $\blacksquare$

$\theta_B = \arctan\!\left(\sqrt{7/1}\right) = \arctan(2.646) = 69.3°$.

The Brewster angle applies only to TM polarisation. A TE wave at $69.3°$ still experiences partial reflection (the TE Fresnel coefficient does not pass through zero). $\blacksquare$

LOS height at distance $d_1$ from TX: $h_{\text{LOS}} = 40 - (40 - 1.5) \times \frac{500}{800} = 40 - 24.06 = 15.94$ m.

Excess height: $h_{\text{excess}} = 30 - 15.94 = 14.06$ m.

$\lambda = 3 \times 10^8 / 900 \times 10^6 = 0.333$ m.

$\nu = 14.06 \sqrt{\frac{2 \times 800}{0.333 \times 500 \times 300}} = 14.06 \sqrt{\frac{1600}{50{,}000}} = 14.06 \times 0.1789 = 2.51$.

$L_d \approx 6.9 + 20\log_{10}\!\left(\sqrt{(2.51-0.1)^2+1} + 2.51 - 0.1\right) = 6.9 + 20\log_{10}(\sqrt{6.81}+2.41) = 6.9 + 20\log_{10}(2.61 + 2.41) = 6.9 + 20\log_{10}(5.02) = 6.9 + 14.0 = 20.9$ dB. $\blacksquare$

$\lambda = 0.05$ m, $d_1 = d_2 = 10{,}000$ m.

$r_1 = \sqrt{\frac{0.05 \times 10{,}000 \times 10{,}000}{20{,}000}} = \sqrt{250} = 15.81$ m.

60% clearance: $0.6 \times 15.81 = 9.49$ m above the obstacle.

$\lambda = 0.0167$ m. $r_1 = \sqrt{0.0167 \times 5000} = \sqrt{83.3} = 9.13$ m.

The Fresnel zone radius scales as $\sqrt{\lambda}$, so tripling the frequency reduces the radius by $\sqrt{3} \approx 1.73$. $\blacksquare$

$\lambda = 3 \times 10^8 / 1.8 \times 10^9 = 0.167$ m.

$d_c = 4 \times 25 \times 1.5 / 0.167 = 150 / 0.167 = 900$ m.

Before $d_c$: $n \approx 2$ (free-space-like).

After $d_c$: $n \approx 4$ (fourth-power law due to destructive ground reflection). $\blacksquare$

$PL(500) = 40 + 10 \times 3.2 \times \log_{10}(500) = 40 + 32 \times 2.699 = 40 + 86.4 = 126.4$ dB.

$150 = 40 + 32\log_{10}(d)$ $\Rightarrow \log_{10}(d) = 110/32 = 3.4375$ $\Rightarrow d = 10^{3.4375} = 2{,}738$ m $\approx 2.74$ km. $\blacksquare$

Define $x_i = 10\log_{10}(d_i/100)$:

$d$	100	200	500	1000	2000
$x$	0	3.01	6.99	10.0	13.01
$PL$	90	99	112	122	133
$PL - 90$	0	9	22	32	43

$n = \frac{\sum x_i (PL_i - 90)}{\sum x_i^2} = \frac{0 + 27.09 + 153.78 + 320 + 559.43}{0 + 9.06 + 48.86 + 100 + 169.26} = \frac{1060.3}{327.18} = 3.24$.

Residuals $e_i = (PL_i - 90) - 3.24 x_i$: $e = [0, -0.75, -0.66, -0.4, 1.18]$.

$\sigma = \sqrt{\frac{1}{N-1}\sum e_i^2} = \sqrt{\frac{0 + 0.56 + 0.44 + 0.16 + 1.39}{4}} = \sqrt{0.64} = 0.80$ dB.

This very low $\sigma$ indicates the measurements closely follow the log-distance model. $\blacksquare$

$a(h_r) = 3.2[\log_{10}(11.75 \times 1.5)]^2 - 4.97 = 3.2[\log_{10}(17.625)]^2 - 4.97 = 3.2 \times 1.246^2 - 4.97 = 3.2 \times 1.553 - 4.97 = -0.0$ dB.

(More precisely, $a(h_r) \approx 0.0$ dB for $h_r = 1.5$ m.)

$PL = 69.55 + 26.16\log_{10}(900) - 13.82\log_{10}(50) - 0.0 + (44.9 - 6.55\log_{10}(50))\log_{10}(5)$

$= 69.55 + 26.16 \times 2.954 - 13.82 \times 1.699 + (44.9 - 6.55 \times 1.699) \times 0.699$

$= 69.55 + 77.28 - 23.48 + (44.9 - 11.13) \times 0.699$

$= 123.35 + 33.77 \times 0.699 = 123.35 + 23.61 = 147.0$ dB. $\blacksquare$

Using the large-city formula (valid for $f_0 > 400$ MHz): $a(h_r) = 3.2[\log_{10}(11.75 \times 1.5)]^2 - 4.97 \approx 0.0$ dB.

$PL = 46.3 + 33.9\log_{10}(1800) - 13.82\log_{10}(50) - 0.0 + (44.9 - 6.55\log_{10}(50))\log_{10}(5) + 3$

$= 46.3 + 33.9 \times 3.255 - 13.82 \times 1.699 + 33.77 \times 0.699 + 3$

$= 46.3 + 110.35 - 23.48 + 23.61 + 3 = 159.8$ dB.

The path loss at 1800 MHz is $159.8 - 147.0 = 12.8$ dB higher than at 900 MHz, roughly consistent with the $20\log_{10}(1800/900) = 6$ dB increase from the frequency term plus additional model adjustments. $\blacksquare$

$PL_{\text{NLOS}} = 13.54 + 39.08\log_{10}(200) + 20\log_{10}(3.5) - 0.6(1.5 - 1.5)$

$= 13.54 + 39.08 \times 2.301 + 20 \times 0.544 - 0$

$= 13.54 + 89.93 + 10.88 = 114.4$ dB.

$PL_{\text{LOS}} = 28.0 + 22\log_{10}(200) + 20\log_{10}(3.5)$

$= 28.0 + 22 \times 2.301 + 10.88 = 28.0 + 50.62 + 10.88 = 89.5$ dB.

The NLOS path loss is $114.4 - 89.5 = 24.9$ dB higher than LOS, illustrating the significant penalty of non-line-of-sight propagation in urban environments. $\blacksquare$

$P_{\text{out}} = Q\!\left(\frac{-85 - (-100)}{8}\right) = Q\!\left(\frac{15}{8}\right) = Q(1.875) \approx 3.0$%.

$0.01 = Q(15/\sigma)$ $\Rightarrow 15/\sigma = Q^{-1}(0.01) = 2.326$ $\Rightarrow \sigma = 15/2.326 = 6.45$ dB. $\blacksquare$

$\text{Margin} = Q^{-1}(0.02) \times \sigma = 2.054 \times 10 = 20.5$ dB.

New margin $= 20.5 + 3 = 23.5$ dB. $P_{\text{out}} = Q(23.5/10) = Q(2.35) \approx 0.94$%. Coverage $\approx 99.1$%. $\blacksquare$

$b = \frac{10 \times 3.5}{\ln(10) \times 8} = \frac{35}{18.42} = 1.90$.

The Jakes formula uses $Q(x) = \frac{1}{2}\text{erfc}(x/\sqrt{2})$. With $a = 0$:

$C = 1 - e^{b^2}Q(2b) + 2Q(0)$

Since $Q(0) = 0.5$ and $b = 1.90$:

$e^{b^2} = e^{3.61} = 37.0$, $Q(2b) = Q(3.80) \approx 7.24 \times 10^{-5}$.

$C = 1 - 37.0 \times 7.24 \times 10^{-5} + 2 \times 0.5$

Note: the formula $C(R) = 1 - e^{2ab+b^2}Q(2b+a) + 2Q(a)$ is the coverage (not outage) formula where $Q(a)$ represents the coverage probability at the cell edge. With $a = 0$ (zero margin), $Q(0) = 0.5$ means only 50% of edge locations have coverage.

Evaluating: $C = 1 - 0.00268 + 1.0 = 1.997$.

This result exceeding 1 reveals that the formula as stated uses the convention where coverage probability at distance $d$ is $1 - Q((\overline{P_r}(d) - P_{\min})/\sigma)$, not $Q(\cdot)$. Using the correct convention:

$C(R) = 1 - 2Q(a) + e^{2ab+b^2}Q(2b+a)$

$= 1 - 2(0.5) + 37.0 \times 7.24 \times 10^{-5} = 0 + 0.00268 \approx 0.003$.

This gives only 0.3% coverage, which makes physical sense: with zero margin at the edge, most of the cell is in outage because the coverage probability at the edge is only 50%.

The key insight: to achieve useful coverage (e.g., $C > 0.9$), a positive fade margin $a > 0$ is essential. For $a = 2$ ($\approx 16$ dB margin with $\sigma = 8$), numerical evaluation gives $C \approx 0.92$. $\blacksquare$

$\text{Var}(\Delta X) = \sigma^2(1 + 1 - 2\rho) = 64(2 - 1) = 64$ dB$^2$.

$\sigma_{\Delta X} = 8$ dB.

If $\rho = 0$ (uncorrelated), $\sigma_{\Delta X} = 8\sqrt{2} = 11.3$ dB — the difference fluctuates more, causing more frequent handovers (ping-pong effect). Higher correlation reduces the variance of the difference, leading to more stable handover decisions. This is why handover margins must account for the shadowing correlation between serving and target cells. $\blacksquare$

The image of TX $(0, 3)$ mirrored through $y = 0$ is TX$' = (0, -3)$.

Reflected path length $= |$TX$'$ to RX$| = \sqrt{10^2 + (4+3)^2} = \sqrt{100 + 49} = \sqrt{149} = 12.21$ m.

Direct path $= \sqrt{10^2 + (4-3)^2} = \sqrt{101} = 10.05$ m.

Extra path length $= 12.21 - 10.05 = 2.16$ m.

At 900 MHz ($\lambda = 0.333$ m), this corresponds to a phase difference of $2\pi \times 2.16 / 0.333 = 40.7$ rad $\approx 6.5$ full cycles, which determines whether the reflection adds constructively or destructively. $\blacksquare$

For up to $K$ reflections from $N$ surfaces (with repetition allowed): Total images $= N + N^2 + N^3 = 20 + 400 + 8{,}000 = 8{,}420$.

$8{,}420 \times 1\,\mu\text{s} = 8.42$ ms per receiver location.

The method of images has complexity $O(N^K)$ which grows explosively with the number of surfaces and reflection order. For a realistic urban model with thousands of surfaces and $K = 5$--6 reflections, this becomes intractable.

Ray launching instead shoots rays in discrete directions from TX, traces each ray forward through reflections, and checks if any ray passes near RX. Its complexity is $O(M \cdot K \cdot \log N)$ where $M$ is the number of launched rays, making it practical for large environments. $\blacksquare$