Fresnel Zone Calculator & Clearance

What is the Fresnel Zone?

The Fresnel zone is an ellipsoidal region around the direct line-of-sight (LOS) path between two antennas. Radio waves do not travel only in a straight line — they also diffract around objects, and energy from slightly curved paths arrives at the receiver. These indirect paths can either reinforce or cancel the direct signal depending on their path length difference.

The first Fresnel zone (n=1) defines the primary energy corridor. Any obstacle that intrudes into this zone causes signal diffraction loss. Industry practice requires at least 60% of the first Fresnel zone radius to be clear of obstructions — this threshold provides essentially free-space performance. Obstructions deeper than 60% clearance cause increasing signal loss via knife-edge diffraction.

Earth curvature also contributes to effective obstacle height over long paths. The earth bulge formula accounts for this, assuming a standard refractivity gradient with k-factor = 4/3.

Why Does It Matter?

In practice, many "line-of-sight" links fail to achieve expected performance because the Fresnel zone is partially obstructed by trees, hills, or buildings — even when a visual LOS exists. Diffraction loss from a 50% Fresnel intrusion can add 6 dB or more of attenuation, potentially causing link failure. Key engineering uses:

Determining the minimum antenna mast height to achieve 60% Fresnel clearance
Quantifying diffraction loss from a specific obstacle to include in the link budget
Comparing different frequency bands — lower frequencies have larger Fresnel zones requiring more clearance
Evaluating earth bulge effect on long paths where the curvature of the earth is significant

Quick Fresnel Calculator

Frequency (MHz)

Total distance (km)

Obstacle from Tx (km)

Tx antenna height (m)

Rx antenna height (m)

Obstacle height (m)

Formulas Used by LinkBudgetPro

\[ r_n = \sqrt{\dfrac{n \cdot \lambda \cdot d_1 \cdot d_2}{d_1 + d_2}} \]

Zone \(n\) radius at obstacle · \(\lambda = c/f\) (wavelength) · \(d_1, d_2\) = distances from Tx/Rx to obstacle in metres

\[ h_e = \dfrac{d_1 \cdot d_2}{12.75 \cdot k} \quad [\text{m}] \]

Earth bulge · \(d_1, d_2\) in km · \(k = 4/3\) standard atmosphere

\[ v = h \sqrt{\dfrac{2(d_1+d_2)}{\lambda \cdot d_1 \cdot d_2}}, \qquad L_{\text{diff}} = 6.9 + 20\log_{10}\!\left(\sqrt{(v-0.1)^2+1} + v - 0.1\right) \]

Knife-edge diffraction · \(h\) = obstacle height above LOS (positive = above) · Loss = 0 if \(v \leq -0.78\)

Clearance target: 60% of first Fresnel zone radius (free-space performance threshold). LOS height at obstacle = Tx_height + (Rx_height − Tx_height) × d⊂1;/d_total. Effective obstacle height = obstacle_height + earth_bulge.

Parameter Explanation

Parameter	Symbol	Unit	Description
Frequency	f	MHz	RF center frequency — lower frequency means larger Fresnel zone
Distance Tx to obstacle	d₁	km / m	Path length from transmitter to the obstacle point
Distance obstacle to Rx	d₂	km / m	Path length from obstacle to receiver
Wavelength	λ	m	c / f — determines Fresnel zone size
Fresnel zone radius	r₁	m	Maximum radius of first Fresnel zone at the obstacle location
Earth bulge	h_e	m	Effective height added to obstacle due to earth curvature
k-factor	k	—	Atmospheric refractivity gradient; 4/3 = standard, 1 = flat earth
Diffraction parameter	v	—	Normalized intrusion depth; positive values cause significant loss
Clearance target	—	%	60% of first Fresnel zone radius required for LOS performance

Worked Example

A 900 MHz link spans 5 km total. An obstacle sits 2 km from the transmitter at a height that intrudes 5 m into the first Fresnel zone. Find the Fresnel radius and diffraction loss:

f = 900 MHz → λ = 0.333 m
d₁ = 2000 m, d₂ = 3000 m
r₁ = sqrt(1 × 0.333 × 2000 × 3000 / 5000)
r₁ = sqrt(0.333 × 1200) = sqrt(400) = 20.0 m
60% clearance required: 0.6 × 20.0 = 12.0 m above LOS
Obstacle intrudes 5 m above LOS → v = 5 × sqrt(2×5000 / (0.333×2000×3000))
v = 5 × sqrt(10000 / 1998000) = 5 × 0.0707 = 0.354
Loss = 6.9 + 20·log₁₀(sqrt((0.354−0.1)²+1) + 0.354−0.1)
= 6.9 + 20·log₁₀(sqrt(0.0645+1) + 0.254) = 6.9 + 20·log₁₀(1.0317 + 0.254)
= 6.9 + 20·log₁₀(1.2857) = 6.9 + 20×0.1092
Diffraction Loss ≈ 9.1 dB

When Should You Use It?

Tower-to-tower backhaul links — verify antenna height achieves 60% first Fresnel clearance over terrain
Links over forested terrain — estimate additional loss from tree canopy intrusion
Long rural links (>10 km) — earth bulge becomes significant and must be added to obstacle height
Urban rooftop deployments — check for building intrusion into the Fresnel ellipse
Frequency selection — lower bands (e.g., 900 MHz) have larger Fresnel zones than 5 GHz, requiring taller masts

Related Calculations

Free Space Path Loss (FSPL) — baseline path loss without obstacle effects
Two-Ray Ground Reflection Model — propagation model for low-altitude links
Fade Margin Calculator — total link budget including Fresnel diffraction loss
EIRP Calculator — effective radiated power available to overcome path loss
RF Documentation Index — all RF engineering reference pages

Fresnel Zone & Clearance