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The GNSS integer ambiguities: estimation and validation

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oundary of the aperture pull-in region Ω0,E could be chosen as:<br />

y = ϱ · 1<br />

arg min<br />

2 z∈Zn \{0} z2Qâ with ϱ =<br />

1<br />

2<br />

µ<br />

min<br />

z∈Zn \{0} zQâ<br />

(5.59)<br />

since then y 2 Qâ = µ2 . Again the aperture parameter for OIA <strong>estimation</strong> is approximated<br />

with equation (5.58), so that y is also an element of the boundary of Ω0,OIA. It<br />

is important to note that this approach will not work in the case the ellipsoidal pull-in<br />

regions overlap, since then y /∈ S0. In that case ϱ in equation (5.59) should be set equal<br />

to one, so that y is on the boundary of S0.<br />

Figure 5.18 also shows the OIA aperture parameter as function of the fail rate as approximated<br />

using EIA. Instead of choosing y as in equation (5.59), one could also choose<br />

y = ϱ· 1<br />

2u with u the adjacent <strong>integer</strong> with the largest distance to the true <strong>integer</strong> a = 0.<br />

With the first option, the approximated aperture parameter will be too large, with the<br />

second option the approximated aperture parameter will be too small. For this example,<br />

approximation with IAB <strong>estimation</strong> works better, but still not good.<br />

5.7.2 Determination of the aperture parameter using simulations<br />

As explained at the beginning of this section, computation of the aperture parameter µ<br />

for a fixed fail rate is not possible for most IA estimators. An alternative is to determine<br />

µ numerically. <strong>The</strong>refore, simulated data are required. <strong>The</strong> procedure will be described<br />

here for the OIA estimator, but a similar approach can be followed for other IA estimators.<br />

<strong>The</strong> procedure is as follows:<br />

1. Generate N samples of normally distributed float <strong>ambiguities</strong>:<br />

âi ∼ N(0, Qâ), i = 1, . . . , N<br />

2. Determine the ILS solution:<br />

ǎi, ˇɛi = âi − ǎi, ri = fˇɛ(ˇɛi)<br />

fâ(ˇɛi)<br />

3. Choose the fail rate: Pf = β ≤ Pf,LS<br />

<strong>The</strong> fail rate as function of µ is given by: Pf (µ) = Nf<br />

N<br />

with Nf = N<br />

<br />

1 if ri ≤ µ ∧ ǎi = 0<br />

ω(ri, ǎi) <strong>and</strong> ω(ri, ǎi,) =<br />

0 otherwise<br />

4. Choose:<br />

i=1<br />

µ1 = (min(ri) − 10 −16 ) since this results in: Nf = 0 ⇒ Pf (µ1) = 0<br />

µ2 = max(ri) since this results in: Pf (µ2) = Pf,LS<br />

5. Use a root finding method to find µ ∈ [µ1, µ2] so that Pf (µ) − β = 0.<br />

This will give the solution since Pf (µ1) − β < 0 <strong>and</strong> Pf (µ2) − β > 0, <strong>and</strong> the fail<br />

rate is monotonically increasing for increasing µ, as will be shown below.<br />

118 Integer Aperture <strong>estimation</strong>

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