The GNSS integer ambiguities: estimation and validation

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