RF probe isotropy, averaging and uncertainty
Practical uncertainty budgets and spatial averaging methods for defensible RF exposure measurements.
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ContinueRepeated readings can reduce uncertainty in a mean without removing a calibration contribution shared by every reading. Spatial averaging can represent a defined region without making its highest point disappear. To use either operation correctly, first decide what result you are estimating and which information the averaging preserves.
Use RMS field quantities and calibration relationships. State the quantity, bands, time definition, source state, location or region, geometry and averaging rule. A point maximum, an arithmetic field average and a quadratic field average are different measurands.
Convert contributors before combining them
Standard uncertainty is expressed on a standard-deviation basis. An expanded calibration uncertainty U with stated factor k gives u = U/k. A justified uniform distribution over a bound ± a gives u = a/square root of 3; a specification bound alone does not establish that distribution.
In the simple model where independent contributors already refer to the same result and units, combine uc = square root of (∑ᵢ uᵢ²). For a general model y = f(x₁,…,xₘ), small-uncertainty propagation also uses sensitivities cᵢ = ∂ f/∂ xᵢ and covariance: uc²(y) = ∑ᵢ cᵢ²u²(xᵢ) + 2∑_(i < j)cᵢcⱼ cov(xᵢ,xⱼ). Do not combine unrelated percentages or ignore a shared influence. A calibration term can correlate many readings; it does not become independent because those readings occupy separate rows. Strong nonlinearity or larger uncertainty may require a method beyond first-order propagation.
Report U = k uc with the factor and basis stated. The choice k = 2 is not a universal guarantee of exactly 95% coverage; distribution and available statistical information matter. The JCGM uncertainty guide, sections 4.2, 5.2 and 6–7, develops repetition, correlation and reporting.
Standardise a complete illustrative budget
Worked example 1
Combine an RF measurement uncertainty budget
Problem
Measured result at a fixed point under a defined operating state:
- E = 18.0 V/m (RMS)
Use illustrative inputs: calibration expanded uncertainty 8% with k = 2, rectangular bounds ±5% for isotropy, ±3% for linearity and ±4% for positioning, and Type A standard uncertainty of repeatability 2.0%. Assume these contributions are independent, refer to the same field result and do not overlap.
Find the combined standard uncertainty, expanded uncertainty with k = 2, and a defensible report of the measured field.
Show full solution
Convert the uncertainty inputs
Method
Express each input as a standard uncertainty before combination.Reason
Calibration coverage factors and rectangular bounds must be put on the same standard-uncertainty basis.Working
Answer
Assume the following uncertainty inputs (illustrative; replace with your instrument’s certificate/specs and field estimates):
- Calibration expanded uncertainty: U_cal = 8% (k = 2) → u_cal = 4.0%
- Isotropy spec: ±5% (rectangular) → uᵢₛₒ = 5/square root of 3 = 2.9%
- Linearity spec: ±3% (rectangular) → uₗᵢₙ = 3/square root of 3 = 1.7%
- Positioning due to local gradient: ±4% (rectangular) → uₚₒₛ = 4/square root of 3 = 2.3%
- Repeatability (Type A): uᵣₑₚ = 2.0%
Combine and expand the budget
Method
Use the root-sum-square of the independent contributors, then apply the coverage factor.Reason
Independent standard uncertainties combine quadratically, and the reporting factor gives the expanded uncertainty.Working
Answer
Combine:
u_c = sqrt(4.0^2 + 5^2/3 + 3^2/3 + 4^2/3 + 2.0^2)% = sqrt(36.6667)% = 6.06% U = 2 * u_c = 12.1% (k=2)Report and interpret the uncertainty
Working
Answer
Report (one defensible format):
- E = 18.0 V/m ± 12.1% (k = 2)
- Absolute: ±2.2 V/m (since 12.1% of 18.0 ≈ 2.2)
Interpretation: if your decision threshold is close, the next step is often to reduce dominant contributors (geometry/gradient control, selective attribution, verify operational state) rather than only increasing repeat count.
Know what repetition can reduce
For independent repeated observations of one stable measurand, estimate the mean E bar = ∑ Eᵢ/n and sample standard deviation s = square root of ((∑ᵢ(Eᵢ-E bar)²)/(n-1)). The standard uncertainty contribution of repeatability to that mean is u_(rep,mean) = s/square root of n, not s itself. Correlated observations, source drift or changed geometry need another model; taking samples faster does not automatically supply more independent information.
Type A and Type B describe the evaluation method—statistical observations or other information—not a universal random/systematic split. Known corrections should be applied as appropriate, with remaining uncertainty included. Repetition cannot remove an unknown offset common to the whole set.
Guided practice 2
Separate uncertainty in a mean from shared calibration
Problem
Four independent RMS field readings at the same fixed point under a stable source state are 10, 12, 14 and 12 V/m. The reported measurand is their mean. A separate calibration contribution has standard uncertainty 5% of that mean and is shared by all four readings. Assume it is independent of the repeatability contribution; neglect other terms for this learning example. Find the mean, the standard uncertainty of the mean from repeatability and the expanded combined uncertainty using the stipulated factor k = 2. Explain why the calibration term is not divided by square root of 4. Do not infer a coverage probability or real survey adequacy from this small illustrative data set.
Work through the model, then compare your reasoning
Hints
Hint 1
Check the model
The sample variance uses n minus one. Only the independent repeatability contribution receives the mean-reduction factor.Show solution step by step
Estimate the mean and repeatability contribution
Method
Use the observed deviations, then divide the sample standard deviation by the square root of n.Reason
The target is a mean of independent fixed-point observations, not one observation or a spatial average.Working
Answer
E bar = 48/4 = 12 V/m. Squared deviations sum to 4 + 0 + 4 + 0 = 8 V²/m². Thus s = square root of (8/3) = 1.63299… V/m and u_(rep,mean) = s/2 = 0.816497… V/m.
Retain the shared contribution
Method
Combine the standard uncertainties before expanding.Reason
A common calibration relationship shifts all the observations together and remains in their mean.Working
Answer
u_cal = 0.05(12) = 0.60 V/m, with no further division by two. Then uc = square root of (0.816497² + 0.60²) = 1.01325… V/m and U = 2uc = 2.02649… V/m. Report the stipulated calculation as about 12.0 V/m with expanded uncertainty 2.0 V/m, k = 2, and its assumptions. It is not a complete exposure budget or a claimed 95% interval.
Define spatial weights and the quantity being averaged
For normalised nonnegative spatial weights wᵢ with ∑ᵢ wᵢ = 1, arithmetic field average is ∑ᵢwᵢEᵢ, whereas quadratic field average is square root of ∑ᵢwᵢEᵢ². The latter preserves a mean-square field quantity. Equal point weights represent equal portions of the specified region only when the sampling plan justifies them; unequal represented areas need their defined weights.
Use the governing method’s region, location, sampling and metric. Do not replace its maximum or averaging rule with a lower number you prefer. Geometry, operator coupling, isotropy, source variation and incomplete coverage still affect the result.
Compare two calculations on an equal-weight grid
Worked example 3
Find the arithmetic spatial average
Problem
Define a 3×3 grid and measure RMS E (V/m) at 9 points:
[20, 22, 19, 18, 25, 21, 17, 23, 20]
Assume equal spatial weights. Find the arithmetic spatial average of E as a mathematical comparison, not a substitute for the framework’s required exposure average.
Show full solution
Select the grid averaging method
Method
Add the nine RMS field readings and divide by the point count.Reason
The arithmetic spatial average weights each grid reading equally.Working
Answer
∑ᵢ Eᵢ = 20 + 22 + 19 + 18 + 25 + 21 + 17 + 23 + 20 = 185 V/m.State the spatial average
Working
Answer
Sum = 185 → Eₗᵢₙ = 185/9 = 20.6 V/m.
Worked example 4
Find the RMS spatial average
Problem
Define a 3×3 grid and measure RMS E (V/m) at 9 points:
[20, 22, 19, 18, 25, 21, 17, 23, 20]
Assume equal spatial weights. Find the quadratic (RMS) spatial average across points. These samples illustrate the calculation only; a real survey must justify its region and sampling against the adopted method.
Show full solution
Select the grid averaging method
Method
Average the squared grid readings and take the square root.Reason
The quadratic spatial average combines the squared field readings rather than their amplitudes.Working
Answer
Sum of squares = 20² + 22² + … + 20² = 3853 V²/m², so mean square = 3853/9 = 428.1 V²/m²
State the spatial average
Working
Answer
E_(rms,spatial) = square root of 428.1 = 20.7 V/m, slightly above the arithmetic mean of 20.6 V/m because squaring weights the higher readings.
Separate physical variation from uncertainty in an estimate
Different fields across a region can be the signal being averaged. Their spread is not automatically the uncertainty of that average. However, a finite grid estimating a continuous area average can have sampling uncertainty from unresolved variation; positioning and temporal changes also need evaluation. Including a justified sampling contribution is not inherently double-counting. Avoid assigning the same effect twice, and explain which part of the model each contribution represents.
For weights wᵢ, uncertainty in an arithmetic mean includes ∑ᵢ∑ⱼwᵢwⱼcov(Eᵢ,Eⱼ). A shared calibration influence remains correlated across points. A quadratic average needs its own sensitivities; copying an uncertainty formula for an arithmetic mean may be inappropriate. Locate relevant high-field regions with the prescribed method, document constraints and keep the measured result separate from the adopted conformity decision rule.
Use area weights rather than treating every reading equally
Check your understanding 1
A fictional learning method requires the square root of the area-weighted mean of RMS field squared over a defined region. Three zones occupy 50%, 30% and 20% of that region, with uniform RMS fields 2, 6 and 10 V/m respectively under the same source state. Find the required quadratic average and the arithmetic area-weighted average. Explain why neither is the point maximum, and what would change if the zone readings were only sparse samples of varying fields rather than exact uniform-zone values.
Show answer
The required mean square is 0.50(2²) + 0.30(6²) + 0.20(10²) = 32.8 V²/m², giving square root of 32.8 = 5.72713… V/m. The arithmetic average is 0.50(2) + 0.30(6) + 0.20(10) = 4.8 V/m; using its square would give 23.04 rather than 32.8. The point maximum is 10 V/m, a different measurand. Sparse samples would require a justified model of how well they represent each zone, with uncertainty from unsampled spatial variation as appropriate. The supplied exact uniform-zone model removes that sampling question for this exercise only.
Check whether a shared influence cancels in a comparison
A common calibration term remains in an absolute field estimate or an average, but its effect on another measurand depends on that measurand’s model. Write the relationship before carrying the term forward. Correlation is not a reason to discard calibration uncertainty generally.
Check your understanding 2
Two indicated RMS fields are R₁ = 10.0 V/m and R₂ = 12.0 V/m in separate stable source conditions, using the same frequency, probe and range. The supplied model is E₁ = cR₁ and E₂ = cR₂, with one shared multiplicative calibration factor c whose relative standard uncertainty is 5%. There is no additive offset, drift or response change. Each indication has independent 1% relative standard uncertainty from repeatability, independent of c. Find the ratio r = E₂/E₁ and its relative standard uncertainty using first-order propagation. Explain whether this makes the absolute fields free of calibration uncertainty.
Show answer
Under this specific model, r = cR₂/cR₁ = R₂/R₁ = 1.20. The shared factor cancels algebraically. The independent repeatability terms remain, so u(r)/r = square root of ((0.01)² + (0.01)²) = 0.0141421…, which is about 1.4% relative standard uncertainty. The absolute standard uncertainty is u(r) ≈ 0.0170, a dimensionless ratio uncertainty.
Each absolute field still depends on c and retains its calibration contribution. Cancellation would not follow automatically if different factors, an additive offset, changed frequency response or drift entered the two measurements. Check that your explanation names the common factor, the independent terms and the distinction between a ratio and an absolute field result. This ratio calculation does not establish exposure compliance.
Review the measurement decisions
Check your understanding 3
What is the first thing you should write down before building an uncertainty budget?
Show answer
The measurand definition: quantity, frequency scope, detector/averaging, spatial definition, and scenario geometry.
Check your understanding 4
Why does repeating the same measurement not remove a shared calibration uncertainty?
Show answer
All repeats share the same uncertain calibration relationship, so their errors are correlated. Averaging more readings may reduce independent fluctuations but cannot remove that shared contribution. Type A or Type B labels alone do not decide whether a term reduces.
Check your understanding 5
Why is “spatial variability” not automatically an uncertainty term?
Show answer
Physical variation may be the quantity represented by a defined spatial average, rather than an error in it. A finite sampling plan can still introduce uncertainty about that average; evaluate unresolved variation, positioning and time effects in their actual roles, without counting one effect twice.
Check your understanding 6
Combine standard uncertainties. If u₁ = 3%, u₂ = 4%, u₃ = 2% are independent standard uncertainties, compute uc and expanded uncertainty U with k = 2.
Show answer
uc = square root of (3² + 4² + 2²) % = square root of 29 % ≈ 5.39%.
U = 2uc ≈ 10.8%.
What to keep
Define the target before averaging. Standardise and propagate uncertainty with its sensitivities and correlations; independent repetition reduces only the applicable contribution. Spatial weights and the required field metric determine the average, while incomplete sampling can still leave uncertainty in it.
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