The GNSS integer ambiguities: estimation and validation

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2-D example Figure 5.3 shows in black all float samples for which ā = ǎ for two different fail rates. 5.3 Ratio test, difference test and projector test In section 3.5 several integer validation tests as proposed in the past were described. It was explained that the test criteria cannot be derived using classical hypothesis tests. However, it can be shown that the ratio test (3.78), the difference test (3.86), and the projector test, on which the test statistics (3.87) and (3.89) are based, are all examples of Integer Aperture estimators, cf. (Teunissen 2003e; Verhagen and Teunissen 2004a). 5.3.1 Ratio test is an IA estimator The ratio test of (3.78) is a very popular validation test in practice. Instead of (3.78) the ’inverse’ of the ratio test is used here, i.e. accept ǎ if: â − ǎ 2 Qâ â − ǎ2 2 Qâ ≤ µ This guarantees that the critical value is bounded as 0 < µ ≤ 1, since by definition â − ǎ 2 Qâ ≤ â − ǎ2 2 Qâ . It can now be shown that integer estimation in combination with the ratio test is an IA estimator. The acceptance region is given as: ΩR = {x ∈ R n | x − ˇx 2 Qâ ≤ µx − ˇx2 2 Qâ , 0 < µ ≤ 1} (5.13) with ˇx and ˇx2 the best and second-best ILS estimator of x. Let Ωz,R = ΩR ∩ Sz, i.e. Ωz,R is the intersection of ΩR with the ILS pull-in region as defined in (3.16). Then all conditions of (5.1) are fulfilled, since: ⎧ ⎪⎨ Ω0,R = {x ∈ R ⎪⎩ n | x2 Qâ ≤ µx − z2Qâ , ∀z ∈ Zn \ {0}} Ωz,R = Ω0,R + z, ∀z ∈ Zn ΩR = z∈Zn Ωz,R (5.14) The proof was given in Teunissen (2003e) and Verhagen and Teunissen (2004a) and is 94 Integer Aperture estimation
as follows: Ωz,R = ΩR ∩ Sz = {x ∈ Sz | x − ˇx 2 Qâ ≤ µx − ˇx2 2 Qâ , 0 < µ ≤ 1} = {x ∈ Sz | x − ˇx 2 Qâ ≤ µx − u2Qâ , ∀u ∈ Zn \ {ˇx}, 0 < µ ≤ 1} = {x ∈ Sz | x − z 2 Qâ ≤ µx − u2 Qâ , ∀u ∈ Zn \ {z}, 0 < µ ≤ 1} = {x ∈ R n | x − z 2 Qâ ≤ µx − u2 Qâ , ∀u ∈ Zn \ {z}, 0 < µ ≤ 1} = {x ∈ R n | y 2 Qâ ≤ µy − v2 Qâ , ∀v ∈ Zn \ {0}, x = y + z, 0 < µ ≤ 1} = ΩR ∩ S0 + z = Ω0,R + z The first two equalities follow from the definition of the ratio test, and the third from ˆx− ˇx2 Qâ ≤ ˆx− ˇx2 2 Qˆx ≤ ˆx−u2Qâ , ∀u ∈ Zn \{ˇx}. The fourth equality follows since ˇx = z is equivalent to x ∈ Sz, and the fifth equality from the fact that 0 < µ ≤ 1. The last equalities follow from a change of variables and the definition of Ω0,R = ΩR ∩ S0. Finally note that Ωz,R = (ΩR ∩ Sz) = ΩR ∩ ( Sz) = ΩR z∈Z n z∈Z n This ends the proof of (5.14). z∈Z n It has thus been shown that the acceptance region of the ratio test consists of an infinite number of regions, each one is an integer translated copy of Ω0,R ⊂ S0. The acceptance region plays the role of the aperture space, and µ plays the role of an aperture parameter. The ratio test will be referred to as the RTIA estimator. It can now be shown how the aperture pull-in regions for the ratio test are ’constructed’, cf. (Verhagen and Teunissen 2004a). From the definition of Ω0 the following can be derived: Ω0,R : x 2 Qâ ≤ µx − z2 Qâ , ∀z ∈ Zn \ {0}, 0 < µ ≤ 1 1 − µ ⇐⇒ µ x + z2 1 Qâ ≤ ⇐⇒ x + µ 1 − µ z2 Qâ ≤ µ z2 Qâ , ∀z ∈ Zn \ {0} µ (1 − µ) 2 z2 Qâ , ∀z ∈ Zn \ {0} (5.15) This shows that the aperture pull-in region is equal to the intersection of all ellipsoids with centers − µ 1−µ z and size governed by √ µ zQâ 1−µ . In fact, the intersection region is only determined by the adjacent integers, since ∀u ∈ Z n that are not adjacent, (5.15) is always fulfilled ∀x ∈ S0. It follows namely from equations (5.13) and (5.14) that for all x ∈ S0 on the boundary of Ω0: x 2 Qâ = µx − c2Qâ , c = arg min z∈Zn \{0} x − z2Qâ So, c is the second-closest integer and must be adjacent. Therefore the following is always true: x 2 Qâ = µx − c2 Qâ ≤ µx − z2 Qâ , ∀z ∈ Zn \ {0, c} Ratio test, difference test and projector test 95
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The GNSS integer ambiguities: estim
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The GNSS integer ambiguities: estim
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Contents Preface v Summary vii Same
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4 Best Integer Equivariant estimati
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Preface It was in the beginning of
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• No attempt is made to fix the a
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Samenvatting (in Dutch) De GNSS geh
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gelegd door de keuze van de maximaa
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d, δ instrumental delay for code a
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Acronyms ADOP Ambiguity Dilution Of
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in practice, a user has to choose b
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GNSS observation model and quality
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Table 2.2: Signal and frequency pla
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2.2.2 Phase observations The phase
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the propagation delay does not depe
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2.2.9 The geometry-based observatio
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2.3.2 Single difference models If o
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and difficult to predict. Therefore
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The double difference observation v
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(appendix A.1): Qy = C sd pφ ⊗ D
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The quality of the solution can the
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is maximum. The degrees of freedom
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* â Figure 3.1: An ambiguity pull-
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2 1.5 1 0.5 0 −0.5 −1 −1.5
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estimator is given by: n Sz,B = x
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7 6 5 4 3 2 1 0 −1 −2 −3 −4
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Table 3.1: Overview of ambiguity re
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1 0.5 0 −2 −1 0 1 z 2 1 0 −1
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and has units of cycles. It was int
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chosen equal to half the number of
Page 70 and 71: Table 3.2: Two-dimensional example.
Page 72 and 73: The bias-affected success rate is a
Page 74 and 75: is close to one. This probability c
Page 76 and 77: 2 0 −2 0.5 0 −0.5 2 0 −2 −4
Page 78 and 79: error in pdf error in pdf error in
Page 80 and 81: 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 2 2.2
Page 82 and 83: acceptable. Moreover, the bounds wi
Page 84 and 85: Unfortunately, this relatively simp
Page 86 and 87: always larger than one. This shows
Page 88 and 89: where χ2 α(m−p, 0) is the criti
Page 90 and 91: Table 3.5: Overview of all test sta
Page 92 and 93: with the weight function wz(â) = 1
Page 94 and 95: IE IEU IU LU Figure 4.1: The set of
Page 96 and 97: 4.2 Approximation of the BIE estima
Page 98 and 99: 4.3 Comparison of the float, fixed,
Page 100 and 101: variane ratio 1.4 1.2 1 0.8 0.6 0.4
Page 102 and 103: ambiguity residual 0.5 0.4 0.3 0.2
Page 104 and 105: Table 4.1: Probabilities P1 = P (|
Page 106 and 107: Table 4.2: Probabilities Ps = P (ǎ
Page 108 and 109: Table 4.4: Probabilities that float
Page 110 and 111: probability probability 1 0.9 0.8 0
Page 113 and 114: Integer Aperture estimation 5 In se
Page 115 and 116: can be distinguished: â ∈ Ωa s
Page 117 and 118: and the hybrid distribution of ā i
Page 119: Figure 5.3: 2-D example for EIA est
Page 123 and 124: 3 2.5 2 1.5 1 0.5 0 −0.5 −1 −
Page 125 and 126: 5.3.3 Difference test is an IA esti
Page 127 and 128: Figure 5.9: 2-D example for DTIA es
Page 129 and 130: Figure 5.11: 2-D example for PTIA e
Page 131 and 132: Figure 5.12: 2-D example for IAB es
Page 133 and 134: 2-D example Figure 5.13 shows in bl
Page 135 and 136: µ 200 100 10 2 1 1 1.1 1.2 1.3 1.4
Page 137 and 138: 5.6 Optimal Integer Aperture estima
Page 139 and 140: The optimization problem of (5.49)
Page 141 and 142: Table 5.1: Comparison of IA estimat
Page 143 and 144: µ 1.5 1.45 1.4 1.35 1.3 1.25 1.2 1
Page 145 and 146: In order to show that the fail rate
Page 147 and 148: success rate percentage identical 0
Page 149 and 150: probability probability 1 0.9 0.8 0
Page 151 and 152: 5.8.2 OIA estimation versus RTIA an
Page 153 and 154: µ µ µ 0.7 0.6 0.5 0.4 0.3 0.2 0.
Page 155 and 156: 0.6 0.4 0.2 0 −0.2 −0.4 −0.6
Page 157 and 158: three vc-matrices which are scaled
Page 159 and 160: Table 5.3: Comparison of IAB and IA
Page 161 and 162: success rate if fixed 1 0.95 0.9 0.
Page 163: Table 5.4: Overview of IA estimator
Page 166 and 167: IE IA I Figure 6.1: The set of rela
Page 168 and 169: convex region symmetric with respec
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Mathematics and statistics A A.1 Kr
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The expectation, E{x}, and the disp
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Choose the tolerance ε. Determine
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Simulation and examples B Throughou
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Theory of BIE estimation C In chapt
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The first term on the right-hand si
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can be derived: L = {2h(x) T Q
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1. Choose fixed fail rate: Pf = β
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Bibliography Abidin HA (1993). Comp
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Jung J, Enge P, Pervan B (2000). Op
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Teunissen PJG (2003g). Towards a un
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Index adjacent, 33 ADOP, 42 alterna
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