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Free-Space Path Loss

Signal attenuation over distance in free space (d in km, f in MHz).

Formula

FSPL=20log⁡10d+20log⁡10f+32.44FSPL = 20\log_{10}d + 20\log_{10}f + 32.44

Variables

FSPLPath loss (dB)
dDistance (km)
fFrequency (MHz)

Example

d=10 km, f=2400 MHz: FSPL ≈ 120 dB

How it works

Free-space path loss quantifies how a radio signal weakens simply because it spreads out as it travels through unobstructed space. The formula comes from geometric spreading: the received power falls with the square of distance, and because higher-frequency waves have shorter wavelengths, they also suffer more loss for the same path length. The 32.44 constant converts the idealized Friis transmission equation into convenient engineering units with distance in kilometres and frequency in megahertz.

When to use it

Use this when estimating the theoretical loss on a clear line-of-sight radio link, such as satellite links, microwave backhaul, or long-range Wi‑Fi/radio planning. It is a baseline model for comparing antenna gains, transmit power, and link budget margins before adding real-world losses like buildings, foliage, rain, or atmospheric effects.

Worked example, step by step

Given: d = 2.5 km, f = 5800 MHz FSPL = 20\log10(d) + 20\log10(f) + 32.44 FSPL = 20\log10(2.5) + 20\log10(5800) + 32.44 20\log10(2.5) = 20(0.39794) = 7.96 dB 20\log10(5800) = 20(3.76343) = 75.27 dB FSPL = 7.96 + 75.27 + 32.44 FSPL = 115.67 dB Final answer: FSPL \u2248 115.7 dB

Common mistakes to avoid

  • Using frequency in GHz or Hz without converting to MHz first
  • Using distance in metres instead of kilometres, which changes the constant
  • Applying free-space path loss to obstructed or indoor environments without accounting for extra attenuation

Did You Know?

Path loss rises with the square of both distance and frequency, so higher 5G bands need denser towers.

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026