Ww... € "ww": CE 
(12) 
are un- 
ing the 
um har- 
ar equa- 
olve the 
| residu- 
rent can 
he given 
ified as 
(13) 
os 0 and 
ed to be 
-. When 
(14) 
(15) 
his case, 
Jowever, 
be used, 
ce. The 
tion 4.4. 
solution 
(16) 
ue of the 
hould be 
he differ- 
ervations 
match is 
roperties 
atial do- 
requency 
1e weight 
Second, 
how is the mean square error from the residuals Val; Vb; Vel 
and Var, calculated? 
In order to answer the first question, variances and 
covariances between the Fourier coefficients should be ana- 
lyzed. Let the covariance matrix of the z and y coordinates 
be an identity matrix multiplied by a unit weight variance 
cà. From Eq. (1) we have the relationship between the 
coordinates and the coefficients as follows: 
ak " z(t) cos kt dt 
bl " e(t) sin kt dt 
al^]! n y(t) cos kt dt. | ' (m 
d, * J(t) sin kt dt 
One notable fact is that the off diagonal terms of the 
covariance matrix of the Fourier coefficients should all be 
zeros, because the functions used to calculate the coeffi- 
cients are orthogonal. Also according to the error propaga- 
tion law and Eq. (17), the diagonal terms (variances of the 
coefficients) can be calculated as follows 
2 
2 — 9 = Á [to zw 
Ca 7 Ca Jo. cos*ktdi — of, (18) 
Aa = a.=% Kum kd - o. 
It can, therefore, be concluded that the weight matrix 
is an identity matrix multiplied by 03. This conclusion is 
also appropriate to the case of open-line matching. 
The answer to the second question is that the mean- 
square error (MSE) is equal to the summation of the 
squared residuals in the frequency domain divided by 2. 
If the maximum harmonic is m, it can be expressed as 
1m 
MSE = 23 0 onn on). (19) 
Proof. 'The mean-square error is defined in the spatial do- 
main as 
2x 
)dt + f OE 
MSE= 2 ([ 2 
2: 9- Me -* 
By substituting Eq. (1) into above equation, the MSE can 
be expressed in the frequency domain as 
m 
2x 
MSE = x J SS (va, cos kt + vp, sin kt)’ dt+ 
2x Jo £^ 
2x; m 
| > (Ve, cos kt + va, sin kt)? a : 
0 k-1 
Because of the orthogonal property, the formula becomes 
1 = 2x 2 2 
MSE = — 2 ([ = 2 cos” ktdt + f v, Sin” ktdt+ 
27 € 0 
2x 
J e cos? ktdt + T vj, sin? ktdt) 4 
0 0 
Because 
2v 2v 
| cos” ktdt = | sin? ktdt — 7 , 
0 0 
we obtain 
1 m 
MSE = 3 2 (va, + tv t va). 
k= 
The theory and proof are also appropriate to the case of 
open-line matching, except there are no vy and vg, terms. 
4.4 First Approximations 
The first approximations may be crucial for solving 
a set of nonlinear equations. With poor approximations, 
the computation may converge to a wrong solution or even 
be divergent. It is, therefore, important to provide good 
approximations for the adjustment computations. 
In Eq. (12), let k — 1, then it seems possible to solve 
the approximations, So, 09 and Ato, from the four equa- 
lions. Unfortunately, the parameters 0 and At are depen- 
dent in each harmonic, so that 6, and At, solved from the 
first harmonic may be correct or incorrect with a difference 
of v. In order to assure the approximations are correct, the 
equations of the first two harmonics should be used. 
We firstly linearize Eq. (12) by letting 
ccl = Scosfcos At; cc2 = S cos 0 cos 2At; 
csl = Scos0sin Ât; cs2 = S cos sin 2At; 
scl = Ssinf cos At; sc2 = Ssin 0 cos2At; 
ssl = SsinÜsin At; ss2 — S sin 0 sin 2At. 
(20) 
Then they can be solved by using the following formulas: 
cel a £d a dl a^ 
esl] |&hà --« -d a b |. 
scl a d a b, a | 
ssl di —c b, —a, d. (21) 
cc2 da ba —ca —dı 7 a} 
cs2 = ba —as -—dı C2 b, 
sc2 C2 dj a2 b; C 
882 dy, —C2 b, —a, d 
According to Eq. (21), we define 
Cl = ccl—5s81 = Socos(60 + Ato); 
S1 = csl+scl = Spsin(fo + Ato); (22) 
C2 = cc2— 582 = 5Socos(6o + 2Ato); 
S2 = c82+ sc2 So sin(69 + 2Ato). 
Then the approximations can be calculated as 
Se - VO +5, 
6p = 2arctan(S1/C1) — arctan(S2/C2), (23) 
Ato — arctan(52/C2) — arctan(S1/C1). 
In fact, there are other combinations of Eq. (20) to solve the 
approximations. However, in practice, it is not necessary to 
elaborate the computation of the approximations. 
For open-line matching, although the system is lin- 
ear, the parameter At should be determined in advance. 
The observation equations of the first two harmonics are 
also required to solve the problem. According to Eq. (14), 
there are 2 unknowns and 2 equations, so that one can solve