RF modulation, duty cycle and time averaging
How to handle time-varying RF sources correctly: duty cycle, peak vs RMS, sliding-window averaging, and common category errors.
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ContinueA time average answers a question about a specified quantity over a specified interval. Before calculating, identify what is being averaged, whether the “off” state is actually zero and whether the window can move. This lesson builds from a periodic on/off source to a changing exposure where those shortcuts need revision. The windows below are stipulated teaching examples; they do not prescribe a regulatory averaging time.
Next action: If your measurements use selective instruments, cross-check settings against Broadband vs selective meters.
At a glance
- Use this when: the source is time-varying (TDD, bursts, beam scheduling, radar pulses, intermittent uplink, power control) and you need a defensible time-averaged exposure metric.
- Core idea: most heating-relevant metrics are power-like (SAR, absorbed power density, incident power density). Power-like quantities average linearly in time; field amplitudes generally do not.
- Key deliverable: a statement of (a) averaging window and whether it is sliding, (b) time behavior assumptions (duty/occupancy), and (c) how the instrument/model implements averaging.
- Common trap: applying duty cycle to E instead of to E², S, or SAR.
Key equations (time averaging done correctly)
General time average
The time average is the area under a quantity’s time graph divided by the interval’s length. For example, integrating absorbed power density gives energy per area; integrating SAR gives absorbed energy per mass. The duration of a high state matters as well as its height.
For a quantity x(t) over a window starting at t₀ with duration T:
If the framework defines a time average over a window of length T, the compliance-relevant value is often the maximum over all window placements:
Duty cycle model (rectangular on/off)
For a power-like quantity y(t) that is yₒₙ during “on” and ~0 during “off”:
This applies directly to S bar, SAR bar, and P bar when the stated interval contains the corresponding on-time fraction. If the low state is not zero, include it: y bar = D yₒₙ + (1-D)y_low. A window containing exactly whole periods gives the cycle average, regardless of where it starts. A window with only part of a period need not contain the long-term high-state fraction. Place that window on the actual timeline.
If the window is longer than one period, separate the complete periods from the remaining interval. Each complete period contributes the same accumulated quantity. Only the remaining interval changes when the window moves, so place that remainder where its contribution is largest. For a rectangular two-state signal, that means capturing as much high-state time as the remainder permits. This argument uses periodic repetition; a non-repeating or changing source needs its actual time record.
Fields under duty cycling (far-field plane-wave intuition)
In a plane wave, S ∝ Eᵣₘₛ² and S ∝ Hᵣₘₛ². If S has duty cycle D, then RMS field averaged over the entire window scales as:
Duty factor multiplies power-like quantities; it square-roots into the full-interval RMS field for the stated on/off signal. RMS means the square root of the mean of the field squared, not the arithmetic mean of field amplitudes. If the reported RMS already includes the full interval, do not apply the duty factor again.
Average linear quantities before converting to decibels
Convert to linear units, average, then convert back. Never average dB values directly.
Metric selection guidance (what quantity do you average?)
Start from the quantity defined by your framework:
- SAR (whole-body or local),
- incident power density,
- absorbed/epithelial power density,
- induced-field metrics at lower frequencies,
- field-strength reference levels (E/H) in some bands.
Then implement averaging on the right underlying signal:
- If endpoint is power density S, average S(t) (not E(t)).
- If endpoint is SAR, average SAR(t) (or the validated power-like driver that maps to SAR).
- If you only have far-field E, use S ∝ E² and average E² (not E).
If the environment is near field, be cautious: plane-wave shortcuts can fail.
Intuition: why “peak vs limit” arguments often fail
Heating depends on energy deposition over time, and heat transfer smooths isolated spikes. Two waveforms can share the same peak but deposit very different energy.
Averaging is aligning the evaluated quantity with thermal time constants and repeatable measurement practice encoded in standards.
Average pulsed exposure and check field conversions
The pulse-cycle examples use a constant on state and zero between pulses. The moving-window example includes a nonzero low state. Use each prompt’s stated levels and interval rather than carrying the zero-off assumption into every calculation.
Average power density over a pulse cycle
Worked example 1
Average power density over a pulse cycle
Problem
Sₒₙ = 40 W/m² for tₒₙ = 5 ms every Tᵣₑₚ = 50 ms, with zero power density between pulses.
Find the duty cycle and average power density.
Show full solution
Identify the transmitting fraction
Method
Compare the on-time with the repetition period.Reason
The average power density depends on the fraction of each cycle that transmits.Working
Answer
D = (5 ms)/(50 ms) = 0.10.State the averaged power density
Working
Answer
D = 0.10, S bar = D Sₒₙ = 4.0 W/m².
Convert duty-averaged power density to RMS field
Worked example 2
Convert duty-averaged power density to RMS field
Problem
Use Sₒₙ = 40 W/m², D = 0.10, S bar = 4.0 W/m² and η₀ = 377 Ω in a plane-wave air field.
Find the on-state and averaged RMS electric fields in the far field and check the averaged result.
Show full solution
Find the on-state RMS field
Method
Convert the on-state power density using the plane-wave impedance.Reason
The field follows from the far-field power-density relation.Working
Answer
E_(rms,on) = square root of (η₀ Sₒₙ) ≈ square root of (377 × 40) ≈ 123 V/m.Apply the field duty scaling
Method
Scale the on-state RMS field by the square root of the duty cycle.Reason
Field squared is proportional to power density under the far-field assumption.Working
Answer
E_(rms,avg) = square root of D E_(rms,on) ≈ square root of 0.10 × 123 ≈ 39 V/m.Check the averaged field
Working
Answer
Check: E_(rms,avg) = square root of (377 S bar) ≈ 39 V/m.
Average SAR over a duty cycle
Worked example 3
Average SAR over a duty cycle
Problem
An on-state SAR of 3 W/kg persists for 25% of each averaging interval, with zero SAR otherwise:
Find the time-averaged SAR.
Show full solution
Select the on-state SAR and duty fraction
Method
Use the stated on-state SAR with the transmitting fraction.Reason
Time-averaged SAR weights the on-state absorption by its duration.Working
Answer
SARₒₙ = 3 W/kg with D = 0.25:
State the average SAR
Working
Answer
SAR bar = D SARₒₙ = 0.75 W/kg.
Choose the worst-case averaging window
Guided practice 4
Include the low state in a moving window
Problem
At a fixed point, incident power density is 20 W/m² for 10 s, then 2.0 W/m² for 110 s, repeating. For this exercise, evaluate every possible 60 s window. Find the maximum high-state time in a window, the maximum window-averaged power density and the equivalent full-window RMS field if travelling-plane-wave conditions hold. Use η₀ = 377 Ω.
Complete the decisions, then compare your working
Hints
Hint 1
choose the model
Draw the 120 s repeating timeline. Place a 60 s window to capture as much of the high state as possible, without inventing a second burst.Show solution step by step
Place the window
Method
Include a complete 10 s high-state burst.Reason
The 110 s low interval prevents a 60 s window from collecting more than 10 s of high-state time, even if it straddles a period boundary.Working
Answer
The maximum window contains 10 s at 20 W/m² and 50 s at 2.0 W/m².
Average both contributions
Method
Divide the accumulated energy per area by the window duration.Reason
The low state contributes exposure too; using only a zero-off duty model would omit it.Working
Answer
S bar ₘₐₓ = [(20)(10) + (2.0)(50)]/60 = 5.0 W/m². The equivalent RMS field is square root of 377(5.0) ≈ 43 V/m. The average lies between the two state values. The full-cycle average is 3.5 W/m² and does not replace the specified worst 60 s average.
Convert a peak-to-average ratio from decibels
Worked example 5
Convert a peak-to-average ratio from decibels
Problem
A waveform has a peak-to-average power ratio (PAPR) of 8 dB over a stated interval.
Find the linear peak-to-average power ratio and interpret a peak-hold reading.
Show full solution
Convert the power ratio
Method
Raise ten to the decibel value divided by ten.Reason
PAPR is a power ratio, so its decibel conversion uses the power-ratio rule.Working
Answer
Pₚₑₐₖ/P_avg = 10^(8/10) ≈ 6.3.Interpret the peak-hold reading
Working
Answer
The peak power is 6.3 times the interval’s average power. A peak-hold instrument with suitable bandwidth may capture that peak, but it does not establish the framework’s time-averaged exposure. Apply the required averaging and any separate short-duration or peak criterion.
“Averaging rules” checklist (what to write in a defensible report)
- Averaging window T (and whether sliding-window maximum was taken).
- Time behavior model: duty/occupancy/scheduling assumptions, beam states.
- Quantity averaged: S(t), SAR(t), or E²(t) (if converting far-field E to S).
- Detector mode and bandwidth: peak/RMS/average; measurement bandwidth; time constant.
- Treatment of variability: worst-case vs typical, and why that choice matches the task.
Common errors
- Applying duty cycle to E instead of E²/S/SAR.
- Comparing peak-hold readings to time-averaged reference levels.
- Averaging dB values directly.
- Using “whole day average” when the framework defines a shorter window.
- Ignoring worst-case window placement for non-stationary signals.
- Using plane-wave conversions in near-field conditions without justification.
- Using traffic/occupancy factors without stating whether the goal is typical characterization or worst-case compliance.
Further reading
- ICNIRP RF guidelines (2020) and IEEE C95.1 for their specific quantity, time and spatial averaging conditions. A time-average calculation alone does not check every applicable short-duration criterion.
- IEC 62232 and ITU RF measurement guidance (site-survey methodology for time-varying sources; detector modes and uncertainty).
- Instrument manuals for your meter/analyzer (how RMS/average/peak and time constants are defined).
Knowledge check
Check your understanding 1
If a guideline quantity is time-averaged power density, what should you average?
Show answer
Average the power density (or equivalently average E² in far-field plane-wave conditions). Do not average field amplitude directly unless the method explicitly defines it.
Check your understanding 2
Why is “peak vs limit” often a category error?
Show answer
Because many limits are defined as time averages over a specified window. A peak snapshot may be real but not directly comparable without applying the correct averaging window and detector interpretation.
Check your understanding 3
Why should a required arithmetic power average be calculated in linear units rather than by averaging dB values?
Show answer
The arithmetic mean of dB values gives a geometric mean of linear powers. Convert to linear power-like units, apply the required time weights, then convert back if needed to obtain the intended arithmetic average.
Exercises
Check your understanding 4
Sₒₙ = 60 W/m² for tₒₙ = 10 ms every Tᵣₑₚ = 100 ms, with zero between bursts. Average over complete periods and estimate S bar.
Show answer
D = tₒₙ/Tᵣₑₚ = 0.10. S bar ≈ D Sₒₙ = 6 W/m².
Check your understanding 5
At a fixed point in a justified travelling-wave field, power density is 8.0 W/m² for 40 s and 1.0 W/m² for 80 s, repeating without change. A report applies the duty factor 40/120 to 8.0 and gives 2.7 W/m². For this exercise, the required result is the maximum average over any 150 s window. The window is a supplied teaching condition, not a regulatory averaging time.
Find the greatest high-state time within a window, the requested maximum average and the full-cycle average as a separate comparison. Give one window placement that attains the maximum. Explain why the report’s calculation is insufficient.
Show answer
Every 150 s window contains a complete 120 s period plus 30 s. The complete period always contributes 40 s at the high level. The remainder can contribute at most another 30 s at that level, giving 70 s high and 80 s low in a worst window.
One placement starts at the beginning of a high state: 40 s high, 80 s low, then 30 s of the next high state. These durations sum to the required 150 s and attain the upper bound, so the maximum is not just a guessed placement.
The accumulated quantity per area is
Dividing by the window length gives
For a complete cycle instead, 40 s at the high level contributes 320 J/m² and 80 s at the low level contributes 80 J/m², giving 400 J/m² altogether. Thus
The report omits the nonzero low state and replaces the specified moving-window calculation with a long-term duty calculation. Its 2.7 W/m² therefore answers neither requested average. Adding the low-state contribution would repair the cycle average, but the 150 s maximum would still need the extra 30 s placed correctly.
The numerical checker verifies the three entries only. Compare your window placement and critique with the complete-cycle contribution, the attainable maximum of the remainder, and the distinction between cycle and window averages. No exposure criterion, instrument validation, uncertainty or other required short-duration condition has been evaluated here.
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Syllabus and review details
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