Question 5 of 12: EDM Standard Deviation by Variance Propagation — why

Question 5: EDM Standard Deviation by Variance Propagation — why $\sigma=\pm(a+b\,\text{ppm})$ (8 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The phase-difference range equation above, with independent uncertainties in the phase measurement ($\sigma_\phi$), the modulation frequency ($\sigma_f$), the refractive index ($\sigma_n$) and the zero constant ($\sigma_K$).

Find. Show, via the law of variance propagation, that $\sigma_D$ splits into a distance-independent part $a$ and a distance-proportional part $b\cdot D$.

Approach. Write $D$ as a function of the independent variables, apply first-order variance propagation $\sigma_D^2=\sum(\partial D/\partial x_i)^2\sigma_{x_i}^2$, and group the resulting terms by whether their coefficient contains $D$ or not.

1. Identify the two error groups by inspection. Let $\Lambda=\dfrac{c}{2nf}$ (the unit length of a half-wavelength). Then $D=(N+\phi/2\pi)\,\Lambda + K$. Two of the variables ($\phi$, $K$) enter with coefficients that do not contain $D$; two ($f$, $n$) enter through $\Lambda$, which multiplies the whole count $(N+\phi/2\pi)$ — i.e. their effect is proportional to $D$.
2. Distance-independent terms (constant $a$). The fractional-phase and zero-error partials are $$\frac{\partial D}{\partial \phi}=\frac{\Lambda}{2\pi},\qquad \frac{\partial D}{\partial K}=1,$$ neither of which contains $D$. Their combined contribution is a constant: $$\sigma_a^2=\left(\frac{\Lambda}{2\pi}\right)^2\sigma_\phi^2+\sigma_K^2 \;\equiv\; a^2 .$$ Physically: the phase can be resolved only to a fixed fraction of a wavelength, and the additive (zero/prism) constant is fixed — both give the same absolute error whether the line is 10 m or 10 km long.
3. Distance-proportional terms (scale $b$). Since $D\propto \Lambda \propto 1/(nf)$, a fractional error in $n$ or $f$ produces a fractional error in $D$: $$\frac{\partial D}{\partial f}=-\frac{D}{f},\qquad \frac{\partial D}{\partial n}=-\frac{D}{n}.$$ Hence $$\sigma_b^2=D^2\!\left[\left(\frac{\sigma_f}{f}\right)^2+\left(\frac{\sigma_n}{n}\right)^2\right]\equiv (b\,D)^2,$$ so $b=\sqrt{(\sigma_f/f)^2+(\sigma_n/n)^2}$ is a dimensionless constant — a parts-per-million scale error from the frequency standard and the atmospheric (refractive-index) model.
4. Combine. Adding the independent groups gives the rigorous propagation result $$\boxed{\;\sigma_D=\pm\sqrt{a^2+(b\,D)^2}\;}$$ Because the two groups are of comparable size only near a cross-over distance, manufacturers quote the simpler, slightly conservative linear envelope $\sigma_D=\pm(a+b\,D)$, which is why the specification reads $\sigma=\pm(a\,[\text{mm}]+b\,[\text{ppm}])$: $a$ in millimetres (the additive constant) and $b$ in ppm (the scale term, $b\cdot D$ giving mm when $D$ is in km).
5. Numerical illustration. For a typical $a=2$ mm, $b=2$ ppm at $D=1.5$ km, the scale term is $b\,D = 2\times1.5 = 3$ mm, so $$\sigma_D=\pm(2+3)=\boxed{\pm5\ \text{mm}}\ \text{(linear)},\qquad \sqrt{2^2+3^2}=\pm3.6\ \text{mm}\ \text{(RSS)}.$$ Short lines are dominated by $a$; long lines by $b\,D$.