Free-Space Path Loss
Signal power lost simply by spreading through free space, growing with both distance and frequency.
Formula
Variables
Example
At 100 km, 1090 MHz: FSPL ≈ 20×2 + 20×3.04 + 32.44 ≈ 133 dB
How it works
Free-space path loss is the reduction in received signal power caused only by the wavefront spreading out as it travels through unobstructed space. The signal energy is distributed over an ever-larger spherical surface, so the power density falls with the square of distance; in dB this becomes the 20 log10(d) term. It also increases with frequency because, for a fixed antenna size and separation, higher-frequency waves have shorter wavelengths and the same physical aperture captures a smaller fraction of the radiated power, which is why 20 log10(f) appears in the formula.
When to use it
Use this when estimating link budgets for line-of-sight radio paths such as aircraft communications, GPS, radar, or satellite links, especially when obstacles, rain, and atmospheric effects are being ignored. It is a first-order baseline for checking whether a transmitter power and antenna setup can support a connection over a given distance and frequency.
Worked example, step by step
Given: d = 50 km, f = 2450 MHz FSPL_dB = 20 log10(d) + 20 log10(f) + 32.44 FSPL_dB = 20 log10(50) + 20 log10(2450) + 32.44 20 log10(50) = 20 × 1.699 = 33.98 dB 20 log10(2450) = 20 × 3.389 = 67.78 dB FSPL_dB = 33.98 + 67.78 + 32.44 FSPL_dB = 134.20 dB Final answer: FSPL ≈ 134.2 dB
Common mistakes to avoid
- Using metres and hertz directly instead of kilometres and megahertz, which changes the constant
- Treating free-space path loss as total received loss even when antennas, obstructions, or atmospheric effects matter
- Forgetting that doubling distance adds about 6 dB of loss, while increasing frequency also raises the loss
Did You Know?
Path loss quadruples (−6 dB) each time distance doubles, which is why long-range ADS-B and radar links need high transmit power and sensitive receivers.
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