Numerical: λ/4 monopole and two-ray received power

Given: d = 5 km, f = 900 MHz, G_r = 2.55 dB, E₀ = 10⁻³ V/m at d₀ = 1 km, h_t = 50 m, h_r = 1.5 m.

a) Length and effective aperture of the antenna

Wavelength:

λ = c / f = (3 × 10⁸) / (900 × 10⁶) = 0.3333 m

Length of a quarter-wave monopole:

L = λ / 4 = 0.3333 / 4 = 0.0833 m  (8.33 cm)

Gain in linear form:

G = 10^(2.55/10) = 1.799

Effective aperture:

A_e = G λ² / (4π)
    = 1.799 × (0.3333)² / (4π)
    = 1.799 × 0.1111 / 12.566
    = 0.0159 m²

b) Received power using two-ray ground reflection model

First check that the far-distance approximation is valid:

20 h_t h_r / λ = 20 × 50 × 1.5 / 0.3333 = 4500 m
d = 5000 m > 4500 m  → approximation valid

For large d, the total received field is:

E_TOT(d) ≈ (2 E₀ d₀ / d) × (2π h_t h_r / (λ d))

Substitute:

2 E₀ d₀ / d        = 2 × 10⁻³ × 1000 / 5000 = 4 × 10⁻⁴ V/m
2π h_t h_r / (λ d) = 2π × 50 × 1.5 / (0.3333 × 5000)
                   = 471.24 / 1666.7 = 0.2827
E_TOT              = 4 × 10⁻⁴ × 0.2827
                   = 1.131 × 10⁻⁴ V/m = 113.1 µV/m

Received power (power density × effective aperture), with free-space impedance 120π Ω:

P_r = (|E|² / 120π) × A_e
    = ((1.131 × 10⁻⁴)² / 376.99) × 0.0159
    = (3.393 × 10⁻¹¹) × 0.0159
    = 5.40 × 10⁻¹³ W

In decibels:

P_r = 10 log(5.40 × 10⁻¹³) = −122.68 dBW = −92.68 dBm

Answer: L = 0.0833 m, A_e = 0.0159 m², E = 113.1 µV/m and P_r ≈ 5.40 × 10⁻¹³ W (−122.68 dBW, or −92.68 dBm).