1 Introduction
AMC AWO.B.CATII.113 gives acceptable methods of demonstrating acceptable ILS and/or MLS tracking performance. This Appendix gives the mathematical derivation of these methods.
2 Numerical analysis
The maximum glide path or localiser deviation recorded during an xLS approach will vary from one approach to another and may be treated as a statistical variable. If it is assumed that the glideslope and localiser deviations recorded during an xLS approach have a normal distribution with mean zero, then it can be shown that the maximum deviations (ignoring the sign of the maximum value) during a certain approach interval follow a Rayleigh distribution of the form:
[Figure or form omitted from this preview — available in the Avioverse workspace library.]
Source transcription: P(x) = (x / λ₀²) × exp(−(1/2) × (x / λ₀)²)
where x is the maximum glideslope or localiser deviation and λ0 is the scale parameter of the Rayleigh Distribution function.
It follows that the probability of recording a maximum deviation less than some specified value xo is:
[Figure or form omitted from this preview — available in the Avioverse workspace library.]
Source transcription: P(x₀) = ∫ from 0 to x₀ P(x) dx = 1 − exp(−(1/2) × (x₀ / λ₀)²)
It can be shown that:
λ₀²= (1/2)∫_(0)^(x) x²P(x)dx
and, to a good approximation:
λ²= (1/(2n)) ∑_(i=1)^(n) (x_(i))²
where n is the number of approaches and xi the maximum deviation recorded on each approach.
If large numbers of approaches were made, λ0 could be calculated and used to find the probability that the maximum xLS deviation will not exceed the excess-deviation alert setting.
For example, if:
λ₀²= (1/(2n)) ∑_(i-1)^(n) (x_(i))²=625
and the excess-deviation alert setting is 75 µA, then:
x₀/λ₀=3∙0
and
P(x₀)=98.9 %
However, it is not economically practicable to make large numbers of approaches and the effects of small sample sizes should be considered. The usual method of doing so is to impose a confidence level (in this case, 90 %) on the results of the measured sample.
If values of λ² are calculated from a number of samples, sampling theory shows that they will be normally distributed with a mean value λ₀² and a standard deviation of λ₀²/(√n) where n is the number of approaches in each sample.
Parameter μ= ((λ²- λ₀²) √n)/λ₀² is normally distributed with a mean value 0 and a standard deviation 1.
The probability (or confidence level) that a value of µ is greater (or smaller) than a certain value is given by the probability distribution function of the normal distribution N (0,1):
[Figure or form omitted from this preview — available in the Avioverse workspace library.]
Source transcription: P(μ > −μ₁) = P(μ < μ₁) = τ = (1 / √(2π)) × ∫ from −∞ to μ₁ exp(−μ² / 2) dμ
Figure A1–1 shows numerical solutions of this integral, in percentages of the integral from – ∞ to ∞, representing one-sided exceedance probabilities (or confidence levels) τ for a range of µ1 values.
[Figure or form omitted from this preview — available in the Avioverse workspace library.]
FIGURE A1–1: Confidence level
From this Figure, it can be seen that for τ = 90 %, µ1 = 1.28.
Thus, there is a given level of confidence τ that: –
μ= ((λ-λ₀²)√n)/λ₀² > - μ₁
From which
λ₀²< λ²/(1- μ₁/(√n))
The value of λ² for the sample is, as shown earlier:
λ²= (1/(2n)) ∑_(i=1)^(n) (x_(i))²
Hence, the maximum value of λ0 can be calculated, followed by the minimum value of
α= x₀√((1- μ¹/(√n))/λ²)
where, as before, x0 is the excess-deviation alert setting.
The minimum probability of not exceeding the excess-deviation alert setting is found by using the probability equation:
P (α)=100 (1-e^(-α²/2))
3 Graphical analysis
As before, the distribution of the maximum deviation on an approach is assumed to be such that the probability that it is less than a value x0 is given by:
P (x₀)=1-e^(-(1/2)(x₀/λ₀)²)
From this equation, given that the required probability is 95 %, the value of x₀/λ₀ can be calculated as:
x₀/λ₀=2 ∙4477
The limiting deviations (x0) are the excess-deviation alert settings; 75 µA for the glide path and 25 µA for the localiser. Hence:
λ0 = 30.64 for the glide path λ0 = 10.21 for the localiser
As given earlier:
λ= √((1/(2n))∑_(i=1)^(n) (x_(i))²)
so that:
∑_(i=1)^(n) (x_(i))²=2nλ²
= 1 878 n for the glide path
= 209 n for the localiser
Thus, a 95 % success rate can be represented graphically as in Figure A1–2 showing Σxi² plotted against i:
[Figure or form omitted from this preview — available in the Avioverse workspace library.]
FIGURE A1–2: Examples of results of flight trials
If, now, a flight trials programme is carried out and the accuracy of the results needs to be checked against the 95 % success criterion, this can be achieved by plotting the value of Σxi², the sum of the squares of the maximum recorded deviations, against n, the number of runs as the trial progresses. If the results are better than required, the graph will cross the 95 % line as shown by line A above. If they are worse the results will appear as line B.
So far, the effect of sample size has not been considered. Its effect is to lower the 95 % success line.
For the sample:
λ= √(1/(2n)) ∑_(i=1)^(n) (x_(i))²
As shown earlier:
λ₀²< λ²/(1- μ₁/(√n))
which, in the limiting case becomes:
λ²= (λ²)₀ (1- μ₁/(√n))
Hence:
(λ²)₀ (1- μ₁/(√n))= (1/(2n)) ∑_(i=1)^(n) (x_(i))²
or
∑_(i=1)^(n) (x_(i))²=2n(λ²)₀ (1- μ₁/(√n))
| λ0 | = 30.64 for the glide path |
|---|
| λ0 | = 10.21 for the localiser |
| µ1 | = 1.28 for 90 % confidence level |
| ∑_(i=1)^(n) (x₁)² | = 1 878 n – 2 403 (√n) for the glide path |
| = 209 n – 267 (√n) for the localiser |
These expressions have been used to produce Figure 1 of AMC AWO.B.CATII.113.
4 Pass or fail method
Suppose the rate of failed approaches measured over a large number of approaches is r.
In a number of approaches T, the expected number of failures is n = rT.
In any given period of time, the number of failures occurring may be greater or less than n, and the small sample may not be typical.
If the failures are randomly distributed with respect to time, the probability p of observing F failures when the expected number is n is given by the various terms of the Poisson distribution, viz.:
| F | 0 | 1 | 2 | 3 | F |
|---|
| P | e⁻ⁿ | e⁻ⁿn | (e⁻ⁿn²)/(2!) | (e⁻ⁿn³)/(3!) | (e⁻ⁿn^(F))/(F!) |
This is a convenient form when the long-term average n is known and the probability of an occurrence of abnormally high or low numbers of failures over short periods is to be found. The problem here is the reverse of this. The observed number F is known and the value of n, which is consistent with it, is required.
In this case, n can have any value above zero and less than infinity. By considering all values of n from zero to some selected maximum N, the Poisson distribution can be used to find the probability of occurrence of each value of n. Summing all these probabilities gives the cumulative probability P that, for an observed value of F, the expected value is not in excess of N. Thus:
P= ∫_(0)^(N) (n^(F)/(F!))e⁻ⁿ dn
As F is a known whole number, then, for various values of F, the value of P may be determined as follows:
F=0, P= ∫_(0)^(N) e⁻ⁿ dn=1- e^(-N)
F=1, P= ∫_(0)^(N) ne⁻ⁿ dn=1-(N+1)e^(-N)
F=2,P= ∫_(0)^(N) (n²eⁿ)/(2!) dn=1- (N²+2N+2)e^(-N)/2
and generally for any value of F,
P=1-(N^(F)/(F!)+N^(F-1)/((F-1)!)+N^(F-2)/((F-2)!)---+N+1)e^(-N)
By evaluating the integral for various values of N, the variation of P with N is obtained. Then, for a given confidence level P, the value of N corresponding to the observed value F is obtained. Thus if the observed rate is F/T, then, for a selected confidence level, it is possible to determine the maximum value for the failure rate N/T.
[Figure or form omitted from this preview — available in the Avioverse workspace library.]
FIGURE A1–3: P, N and F Relationships
From Figure A1–3 it can be seen that for a failure rate r of 5 % and a 90 % confidence level, the required number of approaches T is:
| F | N | T |
|---|
| 0 | 2.30 | 46 |
| 1 | 3.9 | 78 |
| 2 | 5.3 | 106 |
| 3 | 6.65 | 133 |
| 4 | 8 | 160 |
| 5 | 9.2 | 184 |
For example, it is necessary to make 46 approaches without a failure, 78 if one failure occurs and so on as shown in Figure 2 of AMC AWO.B.CATII.113.
[Issue: CS-AWO/2]