Formula
Free-space path loss in decibels is 20 log10 of the distance in metres, plus 20 log10 of the frequency in hertz, plus 20 log10 of 4π/c. c is 299,792,458 m/s. The engineering shortcut uses kilometres, megahertz, and 32.44. That constant is a rounding. This page’s answer is the exact expression. The shortcut is printed beside it.
FSPL = 20 log10(d) + 20 log10(f) + 20 log10(4π/c)
EIRP = Ptx + Gt
Pr = EIRP + Gr − FSPL − other
r = ½ √(λ D)
Fade margin is received power minus the sensitivity you type. The Fresnel radius is the first zone at mid-path, ½ √(λ D). Cable loss from the coax calculator belongs in the other-loss field if you want it in the budget. Antenna length is the dipole calculator, separate from this gain.
Worked examples
2.4 GHz, 1 km. Exact loss 100.052 dB. The 32.44 form is about 0.01 dB away.
20 dBm, 2 dBi on each end, no other loss, sensitivity −90 dBm. EIRP 22 dBm. Received power −76.05 dBm. Fade margin 13.95 dB.
Fresnel radius at the midpoint. 5.588 m.
Reference
Gains are in dBi, relative to an isotropic antenna. A dipole’s gain is not assumed.
Questions
What is the free-space loss at 2.4 GHz over 1 km?
100.052 dB, from 20 log10(d) + 20 log10(f) + 20 log10(4π/c), with d in metres, f in hertz, and c = 299,792,458 m/s.
Why does 32.44 disagree slightly?
20 log10(dkm) + 20 log10(fMHz) + 32.44 is a rounded shortcut. The constant that matches this c is about 32.45. The page shows both and treats the exact form as the answer.
What is the Fresnel radius here?
At mid-path, half of the square root of wavelength times distance. For this link that is 5.588 m. It is the first zone, not a clearance rule by itself.
Sources
- Friis free-space path loss, FSPL(dB) = 20 log10(4π d f / c), with c = 299,792,458 m/s.
- First Fresnel zone radius √(λ d1 d2 / (d1 + d2)). At mid-path that is ½ √(λ D).