RF wave propagation and boundary conditions
How reflections, multipath, and boundaries create hotspots and measurement variability in real RF exposure assessments.
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ContinueA low reading near a wall can be a cancellation point, while a nearby position has a much larger electric field. Learn to predict that variation, distinguish an electric-field-derived equivalent quantity from actual power flow, and design a search that does not repeatedly sample the same minimum.
Build on RMS fields and traveling-wave power. We first use one incident wave and one coherent reflection in air; real rooms can contain several paths and need a broader assessment.
Characterise the source and the boundaries
Record frequency, antenna geometry, operating state, accessible positions and the objects that can change the paths. A free-space far-field screen is
It estimates the incident wave under its stated conditions. It does not establish the total field next to a reflector. The relevant boundary effects include:
- Reflection and transmission: a boundary can send energy back and pass some through, with frequency-, material- and angle-dependent amplitudes and phases.
- Absorption: lossy material converts some energy to heat; moisture and material properties can change the result.
- Diffraction and scattering: edges and irregular objects redistribute fields, including into regions with no direct line of sight.
- Guiding and re-radiation: corridors, enclosures, rails and cables can channel energy or become secondary radiators.
Changing a door, vehicle or piece of metal staging changes this geometry. A result for one configuration is not automatically a result for another.
Add fields before interpreting the reading
At normal incidence, the electric-field reflection coefficient at a boundary between media of intrinsic impedances η₁ and η₂ is
It is an amplitude ratio with a phase, not a power fraction. For the following model let ρ = |Γ|, with 0 ≤ ρ ≤ 1, and let Eᵢ be the incident RMS electric-field magnitude. The waves have the same frequency and linear polarisation, travel in opposite directions, and have a stable relative phase. The surrounding air is treated as lossless, with η₀ = 377 Ω.
At a position where the electric contributions add in phase,
Where they oppose,
The general local RMS result, with electric relative phase φ, is
Adding only the separate E² contributions leaves out the interference term. For incoherent contributions, the cross term can average away under justified measurement conditions. For this coherent pair it generally remains and varies with position; vanishing at a particular phase does not make the pair incoherent or justify dropping the term at other positions.
The magnetic field reinforces at different positions
For each traveling wave separately, H = E/η₀. Reversing propagation reverses the magnetic-field direction relative to the electric field, so the total fields instead obey
At the electric maximum, the magnetic magnitude is Eᵢ(1-ρ)/η₀; at the electric minimum, it is Eᵢ(1 + ρ)/η₀. An electric minimum is therefore not a magnetic minimum. For a perfect conductor at normal incidence, the tangential electric field cancels at the surface; the electric maximum is displaced from it.
These field relations follow the normal-incidence boundary model in MIT’s electromagnetics notes, section 9.1.
Keep the equivalent quantity separate from net flow
The electric-field-derived plane-wave-equivalent quantity is
The time-averaged net Poynting flux along the incident direction for this counter-propagating pair is instead
It is the incident traveling-wave power minus the reflected traveling-wave power. At ρ = 1, the net average flow is zero, while Eₘₐₓ = 2Eᵢ and S_(eq,max) = 4Eᵢ²/η₀. Zero net flow does not mean zero local fields, stored energy or coupling to an introduced object.
The factor of four concerns this one-reflection model. It is not a universal ceiling for a multi-path environment or a compliance criterion. Use the adopted exposure framework’s valid field or dosimetric method and its averaging definitions.
Matching the local RMS magnitude ratio E/H = η₀ at one point does not by itself identify a single traveling wave. Average Poynting flux also depends on field directions and the relative temporal phase of the electric and magnetic oscillations. The two-wave equations must still be applied to the characterised incident and reflected contributions.
Bound exposure near a reflecting metal wall
Worked example 1
Bound exposure near a reflecting metal wall
Problem
An independently characterised incident plane wave has Eᵢ = 12 V/m RMS in air. One reflected wave has the same electric-field magnitude. Calculate the incident power density, maximum local electric field and its plane-wave-equivalent quantity. Explain why the last quantity is not the standing wave’s net power flow. Use η₀ = 377 Ω.
Show full solution
Identify the traveling-wave quantity
Method
Convert the incident RMS field.Reason
The incident contribution is one traveling plane wave; the resultant will not be.Working
Answer
S_inc = 12²/377 ≈ 0.382 W/m².Add amplitudes at the electric maximum
Method
Use ρ = 1 before squaring the resultant.Reason
Coherent electric contributions reinforce at an antinode.Working
Answer
Eₘₐₓ = 12(1 + 1) = 24 V/m,S_(eq,max) = 24²/377 ≈ 1.53 W/m².Check what the result measures
Working
Answer
The incident density is 0.382 W/m² and the electric maximum is 24 V/m. Its equivalent quantity is four times the incident density. Equal oppositely traveling powers give zero net average flow, however. Scan for local fields and assess them with a method valid for this geometry; neither number alone establishes exposure compliance.
Guided practice 2
Compare a field maximum with power flow
Problem
A 2.40 GHz incident wave has Eᵢ = 10.0 V/m RMS and one coherent, oppositely traveling reflection with ρ = 0.600. Use the model above, η₀ = 377 Ω and c = 3.00 × 10⁸ m/s.
Find the maximum and minimum electric fields, the maximum electric-field-derived equivalent quantity, the net average flux along the incident direction, and the spacing of neighbouring electric maxima. Explain why the two power-density numbers differ and whether an electric minimum establishes a low magnetic field.
Calculate the quantities, then compare their meanings
Hints
Hint 1
Choose what to add
The coherent fields add as amplitudes. The counter-propagating traveling-wave powers enter net flow with opposite signs.
Hint 2
Track relative phase over distance
Moving by Δ z changes the waves’ relative phase by 2kΔ z, where k = 2π/λ. A full relative-phase cycle brings the next electric maximum.
Show solution step by step
Find the two electric extremes
Method
Use the reflected-to-incident amplitude ratio.Reason
Partial reflection leaves a nonzero minimum.Working
Answer
The reflected RMS magnitude is 6.00 V/m. Adding 10.0 + 6.00 gives a maximum of 16.0 V/m; subtracting 10.0-6.00 gives a minimum of 4.00 V/m. All are RMS magnitudes.
Calculate and name each power quantity
Method
Square the maximum for the equivalent quantity; subtract traveling-wave powers for net flow.Reason
A resultant with reflected waves need not obey the single-traveling-wave relation.Working
Answer
S_(eq,max) = 16.0²/377 ≈ 0.679 W/m².⟨Sz⟩ = (10.0²(1-0.600²))/377,⟨Sz⟩ ≈ 0.170 W/m².At the electric minimum, H = 16.0/377 A/m, about 0.0424 A/m, its maximum in this model. A low E reading alone does not establish a low H reading.
Locate the repeating pattern
Method
Convert frequency to wavelength, then use the relative-phase cycle.Reason
The waves gain opposite phase shifts as position changes.Working
Answer
The wavelength is 0.125 m. Neighbouring electric maxima are λ/2 = 6.25 cm apart; a maximum and its neighbouring minimum are λ/4 = 3.125 cm apart. For self-review, name the two different power quantities and explain why the magnetic maximum coincides with the electric minimum.
Test a local field interpretation
Check your understanding 1
In the same lossless-air two-wave model, Eᵢ = 10.0 V/m RMS and ρ = 0.600. The electric contributions have relative phase φ = 90° at the observation point; the waves remain coherent, with the same linear polarisation and opposite propagation directions. Use η₀ = 377 Ω.
Calculate the local RMS electric and magnetic magnitudes, the electric-field-derived equivalent quantity and net average flux along the incident direction. A colleague finds E/H = 377 Ω at this point and concludes that E²/377 must therefore be the actual net flux. Does the local magnitude ratio establish that conclusion?
Show answer
At φ = 90°, cos φ = 0. Thus E = 10 square root of (1 + 0.600²) ≈ 11.7 V/m and H = (10/377) square root of (1 + 0.600²) ≈ 0.0309 A/m. Their RMS magnitude ratio is indeed 377 Ω, even though both traveling waves remain present.
The local equivalent quantity is E²/377 = 136/377 ≈ 0.361 W/m². Net average flux is instead (100-36)/377 ≈ 0.170 W/m², the difference between the oppositely traveling powers. The magnitude ratio alone has not established the single-wave phase relation: the total electric and magnetic oscillations need not be in phase.
Check that your explanation keeps the supplied coherent two-wave model, the local RMS magnitudes and the actual average power flow distinct. None of these illustrative values is an exposure-compliance determination.
Search for variation rather than a convenient low point
Along the propagation axis in this two-wave model, electric maxima repeat every λ/2. A grid with exactly that pitch can repeatedly sample a minimum. Finer sampling and offset scans help expose this failure, but a simple wavelength rule does not certify a survey of a complex room: several directions, near fields, apertures and instrument response can change the relevant scale.
Check your understanding 2 Challenge a repeated low reading
In an illustrative 3.00 GHz two-wave field, Eᵢ = 10.0 V/m RMS and ρ = 0.800. The x-coordinate runs along the propagation axis and x = 0 is an electric minimum. An E-only scan records 2.00 V/m at x = 0, 5.00 and 10.0 cm.
Use c = 3.00 × 10⁸ m/s to find the wavelength, the model’s electric maximum and the position of the first positive-x electric maximum. Is the repeated low reading enough to characterise this region? Propose a follow-up search and identify what must be recorded before an exposure comparison.
Show answer
The wavelength is 10.0 cm. Electric minima repeat every 5.00 cm, so every measured position can be a minimum. The maximum is 18.0 V/m, first reached at x = 2.50 cm; another occurs at 7.50 cm.
Scan between the original positions and use offset passes, rather than treating repeatability at those positions as coverage of the region. In a real survey, search relevant heights and accessible locations, consider H where the adopted method requires it, and record source operation, probe response, geometry, configurations, averaging and uncertainty. The illustrative maximum predicts this model’s field; it is not a compliance result or a universal maximum for a room.
For self-review, explain the repeating phase, identify the missed positions and distinguish a field search from the subsequent criterion-matched averaging and comparison.
Make the survey reproducible
- Define the accessible region and credible source operating states, including maintenance access.
- Inventory reflectors, apertures, rails, cables and movable objects; specify the configurations to test.
- Select instruments and a field or dosimetric method valid for the regime. Do not infer net flux from E alone in a standing wave.
- Search the spatial variation over relevant positions and heights. Record where maxima occur and how the search could detect them; minimise probe, cable and operator perturbation.
- Apply the adopted framework’s metric-specific averaging, source-combination and uncertainty rules. A maximum-search reading is not automatically its spatially averaged result.
- Select controls, retain the tested geometry and state, and document which changes require reassessment.
ITU-T K.91 provides assessment and monitoring guidance. Choose the edition and method applicable to the installation; this lesson does not prescribe one universal grid or exposure limit.
Check the replacement model
Check your understanding 3
A reflection doubles the electric-field magnitude at an antinode. How does E²/377 change, and does it establish actual net power flow there?
Show answer
It increases by a factor of four. In this standing wave it is an electric-field-derived plane-wave-equivalent quantity, not the actual net power flow. Net flow depends on the E/H relation and phase; equal counter-propagating waves give zero average net flux.
Check your understanding 4
How should you search for standing-wave hotspots near reflectors?
Show answer
Use spatial scanning or multiple-point sampling over accessible positions and relevant heights, with offsets that can reveal variation between points. Document the maxima, configurations and search method, then apply the governing averaging and uncertainty rules.
Check your understanding 5
Why should you record doors, vehicles and temporary metalwork when reporting a survey?
Show answer
They can change propagation paths, reflections and the field distribution. A report must identify the configuration to be reproducible and show what changes may invalidate its conclusion.
Check your understanding 6
A single indoor reading is low. Give two different reasons it does not establish a low level throughout the accessible region and operating period.
Show answer
Spatially, the point may be a cancellation minimum with larger fields nearby. Temporally, a snapshot may miss higher traffic, duty or beam states. Search both relevant spatial variation and source operating conditions; repeating only the same point does not resolve both gaps.
Keep secondary propagation estimates in their scope
For a good conductor in the skin-effect regime,
Here σ is conductivity and μ permeability. Smaller skin depth can reduce penetration through bulk metal, but seams, vents and cable penetrations can dominate an assembly’s leakage. Skin depth alone does not establish installed shielding performance.
For a line-of-sight path with distances d₁ and d₂ from an obstruction to its endpoints, a first-Fresnel-zone screen is
It describes a propagation-clearance scale, not an exposure limit or a stand-off distance for a probe. Obstructions can change the field even when geometric line of sight remains.
Match the conclusion to the characterised scenario
Add coherent fields with their phase before interpreting E. Keep local field strength, electric-field-derived equivalent density and net average power flow distinct. Search for variation, record configurations and apply the adopted method’s averaging and uncertainty rules before making an exposure conclusion.
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Syllabus and review details
No official syllabus alignment is listed for this lesson.
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