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Numerical: two-ray vs free-space path loss at 800 MHz

Short answer

Two-ray model PL = 40 log(10000) − 20 log(30) − 20 log(2) = 124.44 dB. Free-space PL = 110.50 dB. Two-ray predicts ~13.9 dB more loss due to ground reflection causing 1/d⁴ power decay.

Determine the propagation path loss for signal at 800 MHz, with a transmitting antenna height of 30 m and a receiving antenna height of 2 m, over a distance of 10 km, using two-ray mobile point-to-point propagation model. How is it compared with that of free-space propagation path loss model?

Given: f = 800 MHz, h_t = 30 m, h_r = 2 m, d = 10 km. Antenna gains are taken as unity (0 dB) because they are not given.

Path loss by two-ray model

For large distances (d ≫ √(h_t h_r)), the two-ray ground reflection model gives a path loss that does not depend on frequency:

PL(dB) = 40 log d − (10 log G_t + 10 log G_r
                     + 20 log h_t + 20 log h_r)

With G_t = G_r = 1:

PL = 40 log(10000) − 20 log(30) − 20 log(2)
   = 160 − 29.54 − 6.02
   = 124.44 dB

Path loss by free-space model

λ  = c / f = 3 × 10⁸ / 800 × 10⁶ = 0.375 m
PL = 20 log(4π d / λ)
   = 20 log(4π × 10000 / 0.375)
   = 20 log(335103) = 110.50 dB

(Same using PL = 32.44 + 20 log f(MHz) + 20 log d(km) = 32.44 + 58.06 + 20 = 110.50 dB.)

Comparison

Difference = 124.44 − 110.50 = 13.93 dB
ModelPath lossFall-off with distance
Free space110.50 dB20 dB/decade (1/d²)
Two-ray124.44 dB40 dB/decade (1/d⁴)

Answer: Two-ray path loss = 124.44 dB, free-space path loss = 110.50 dB. The two-ray model predicts about 13.9 dB more loss, because at large distances the ground-reflected wave nearly cancels the direct wave, so power falls as 1/d⁴ instead of 1/d². The two-ray result is more realistic for a mobile link over flat ground.

✓ Answer checked by Claude Opus 5.5 (high effort)

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