Wi-Fi, Bluetooth and indoor RF exposure

An indoor reading describes the signals, traffic and geometry present during that measurement. Reflections can move maxima across the room; packet traffic can change them over time. To interpret the result, decide whether you are estimating a time average, searching positions, or attributing a contribution to a source.

Treat the room as part of the measurement

Direct, reflected and scattered waves add as fields. Their relative phases can reinforce or cancel an electric field at a position, so a low reading can be a local minimum. Moving furniture, opening a door or changing where people stand can alter those paths. Multipath can be important without assuming that it dominates every room or every frequency.

For equal coherent counter-propagating waves, successive electric-field maxima are separated by λ/2 and a node lies λ/4 from a maximum. Real rooms have several path directions, so that one-dimensional spacing is an explanatory model, not a universal survey-grid specification.

Worked example 1

Relate Wi-Fi wavelength to changes across a room

Problem

Consider a 5 GHz signal in air and a reflective room with interference from multiple paths.

Find the wavelength and half-wavelength, and explain why moving 10 cm can change a reading in a reflective room.

Show full solution

1. Find the wavelength

   Method

   Divide the wave speed by the Wi-Fi frequency.

   Reason

   Wavelength sets the length scale of spatial interference patterns.

   Working

   Answer

   λ ≈ (3 × 10⁸)/(5 × 10⁹) = 0.06 m = 6 cm

2. Find the half-wavelength

   Method

   Halve the wavelength before comparing it with the movement distance.

   Reason

   The half-wavelength provides the stated scale for changes in the interference pattern.

   Working

   Answer

   Half-wavelength is:

   λ/2 ≈ 3 cm

3. Interpret the room sampling implication

   Working

   Answer

   The wavelength is 6 cm and the half-wavelength is 3 cm. A 10 cm move can cross several interference features, so a single reading may miss nearby maxima. The precise change depends on path directions and phases; movement does not guarantee a higher or lower result.

   Practical takeaway: if you want a room-level statement, use a grid and report a statistic (e.g., max + mean) instead of a single point.

Connect traffic to the time statistic

Access points transmit beacons and downlink traffic; clients transmit uplink traffic and acknowledgements. A client near the body can be important for local absorption even when the room survey is dominated by an access point. Record both source locations and activity instead of assuming all Wi-Fi comes from the router.

Wi-Fi and Bluetooth transmit in events. For a zero-off-state power-like quantity, a constant on level and complete periods give y bar = dyₒₙ. A true RMS field is related instead through Eᵣₘₛ² = E² bar. Do not multiply an on-state field by duty and then square it to obtain an average power.

Traffic, radio mode and link quality determine the time record. A band label or Bluetooth Classic/BLE label alone does not specify duty. Select detector response and dwell time that represent the required statistic; check sensitivity and overload as well as calibration.

At 2.4, 5.0 and 6.0 GHz, the wavelengths in air are about 12.5, 6.0 and 5.0 cm respectively. Shorter wavelength changes the spatial interference scale; transmission through walls also depends on their material and geometry. A band marketed as “6 GHz” is not a single frequency: use the actual channel and adopted framework to select the exposure criterion.

For the same payload, a higher usable data rate can shorten airtime, but retries and increased traffic can change that result. Bluetooth Classic and Low Energy also have different event and profile arrangements. Determine activity from the actual profile and traffic; “Low Energy” is not a measured duty fraction. These distinctions explain why comparing protocol or channel names cannot replace a time record.

Attribute a source only under stated conditions

A broadband instrument’s band, response and detector determine what contributes to its indication; not every instrument simply reports a sum of power densities. For uncorrelated contributions measured as matched true RMS electric fields, mean-square fields add. Coherent fields can have a cross term, so that power-sum assumption needs a basis.

A controlled source-on/source-off comparison also requires stable background sources, geometry and traffic. Switching an access point off can change client activity and cellular use. A change in the total then describes the changed system, not automatically the access point alone. Frequency-selective measurement can separate bands, although overlapping sources within one band still need attribution.

Guided practice 2

Interpret a controlled on/off comparison

About 8 min

Problem

At a fixed point, matched true RMS broadband electric-field measurements are 4.00 V/m with a source operating and 3.00 V/m with it off. For this illustrative comparison, independent checks establish unchanged background and geometry, uncorrelated contributions, linear unsaturated instrument response and identical time windows.

Find the added mean-square field in (V/m)², its effective RMS field in V/m, and its percentage of the on-state mean-square total. Explain why subtracting the two field amplitudes gives the wrong contribution. Do not convert the result to power density or SAR.

Make the decisions, then compare your working

Added mean-square electric field

(V/m)²Unit: (V/m)²

Effective RMS field of the added contribution

V/mUnit: V/m

Added contribution to the on-state mean-square total

%Unit: %

Check answer

Your subtraction model, percentage denominator and attribution conditions (self-review; paper working is also fine)

Hints

Hint 1

Subtract the additive quantities

The supplied conditions make Eₒₙ² = E_background² + E_added² applicable. Field amplitudes are not the additive quantities here.

Hint 2

Recover the requested field

Take the square root of the difference. For the percentage, compare that difference with 4.00², not with 4.00.

Show solution step by step

1. Extract the added mean square

   Method

   Square both true RMS totals before subtracting.

   Reason

   Uncorrelated contributions have zero averaged cross term.

   Working

   Answer

   The squared totals are 4.00² = 16.0 and 3.00² = 9.00, both in (V/m)². Their difference is 7.00 (V/m)². The amplitude difference 1.00 V/m would describe a different addition model.

2. Recover the effective RMS field

   Method

   Take the positive square root.

   Reason

   The effective RMS magnitude must reproduce the added mean square.

   Working

   Answer

   square root of 7.00 gives an effective RMS field of 2.65 V/m. Check the mean squares: 9.00 + 7.00 = 16.0, recovering the on-state total. No travelling-wave or tissue-absorption assumption has been supplied.

3. Express the fraction of the total

   Method

   Use the complete on-state mean square as the denominator.

   Reason

   This is a fraction of mean-square field, not a fraction of field amplitude.

   Working

   Answer

   100(7.00/16.0) = 43.75%, or 43.8% to three significant figures. The conclusion depends on the verified stability, independence and instrument conditions. Compare those yourself; the checker verifies the three numerical quantities only.

Test what the sampling and switching establish

Check your understanding 1 A regular grid with reassuring readings

In a simplified 5.00 GHz standing-wave model, electric-field nodes occur at x = λ/4 + nλ/2 for integer n. A survey samples only those positions along a line, obtaining a negligible electric field at every point. Use c = 3.00 × 10⁸ m/s.

Does adding more points at the same spacing establish negligible field throughout the line? Explain how to test the inference and why E²/377 would not represent the net power flow of this standing wave.

Your reasoning (self-review; paper working is also fine)

Show answer

The wavelength is 6.00 cm, so the survey starts at 1.50 cm and samples every 3.00 cm. Each chosen point is another node; more points with the same alignment continue missing the maxima between them. Shift the positions and investigate across the occupied region instead of treating one aligned grid as proof of negligible field.

The counter-propagating waves do not form one travelling plane wave. In an ideal equal-amplitude standing wave their time-averaged net energy flow is zero, while electric and magnetic fields still store energy and vary in space. Therefore a local electric-field square is not the net Poynting flux. Real-room assessment needs a method and metric suited to its geometry. Compare your response for the systematic sampling error and the failed travelling-wave assumption.

Check your understanding 2 The background changes during the test

Repeat the 4.00 V/m on / 3.00 V/m off comparison, but this time the access point’s shutdown makes clients switch to cellular data. The report calls the square-root difference “the access point’s field.”

What part of the earlier inference has failed, and what should the report state or measure next?

Your reasoning (self-review; paper working is also fine)

Show answer

The unchanged-background condition has failed. The two observations describe different combinations of transmitters, so their mean-square difference cannot isolate the access point’s contribution by the previous argument.

Report the measured totals with the two operating states and the attribution limitation. Restore a matched controlled comparison if possible, or use selective measurements and traffic records to distinguish contributions. Selectivity identifies frequency content; it does not automatically identify one transmitter among overlapping signals. Compare your answer for the failed condition and a way to obtain the missing evidence.

Make the room statement reproducible

Specify the occupied region, heights, positions and search procedure; active access points and clients; traffic states; instrument band and detector; dwell and averaging; and uncertainty. Distinguish a maximum among sampled points from a justified maximum over the accessible region. Match a framework comparison to its required spatial statistic rather than substituting a convenient room mean.

Review the indoor interpretation

Check your understanding 3

Why can moving 10 cm change an indoor reading a lot?

Your reasoning (self-review; paper working is also fine)

Show answer

Because standing-wave/multipath patterns can vary on the order of λ/2 (a few centimeters at Wi‑Fi bands). A 10 cm move can cross multiple maxima/minima.

Check your understanding 4

Why does burstiness matter for measurement interpretation?

Your reasoning (self-review; paper working is also fine)

Show answer

A meter’s detector mode and time constant determine whether you capture peaks, averages, or something in between. Bursty emissions can produce large peak-to-average differences.

Check your understanding 5

What’s a defensible way to claim something about “room level”?

Your reasoning (self-review; paper working is also fine)

Show answer

Define a spatial sampling plan (grid/points at an occupancy height), measure over a defined time window, and report a statistic (max/mean/percentile) with limitations/uncertainty.

Preserve the meaning of the statistic

A peak, a time average, an area average and a broadband total are different results. State the measured one and the conditions supporting it. A low or unstable Bluetooth indication may reflect instrument sensitivity or time response; it does not establish zero emission.