Independent Component Extraction and Cross-correlation Spectrum Analysis of Gravity Tide Signal
Dezhao Xing,Haiyan Quan

Table 1 Autocorrelation spectrum analysis of y 1 ( t ) , y 2 ( t ) and y 3 ( t )

$f 11 = 2.411 × 10 - 7$

$a 11 = 1.777 × 10 - 7$

$F 1 = f 11 + a 11 = 4.188 × 10 - 7$ $4.2004 × 10 - 7$
$F 2 = f 11 - a 11 = 6.34 × 10 - 8$ $6.3377 × 10 - 8$

$f 12 = 4.569 × 10 - 7$

$a 12 = 3.934 × 10 - 7$

$F 3 = f 12 + a 12 = 8.503 × 10 - 7$ $8.4725 × 10 - 7$
$F 4 = f 12 - a 12 = 6.35 × 10 - 8$ $6.3377 × 10 - 8$

$f 13 = 6.345 × 10 - 7$

$a 13 = 2.157 × 10 - 7$

$F 5 = f 13 + a 13 = 8.502 × 10 - 7$ $8.4725 × 10 - 7$
$F 6 = f 13 + a 13 = 4.188 × 10 - 7$ $4.2004 × 10 - 7$

$f 21 = 2.275 × 10 - 5$

$a 21 = 3.934 × 10 - 7$

$F 7 = f 21 + a 21 = 2.3143 × 10 - 5$ $2.3148 × 10 - 5$
$F 8 = f 21 - a 21 = 2.2357 × 10 - 5$ $2.2364 × 10 - 5$

$f 31 = 1.157 × 10 - 5$

$a 31 = 3.807 × 10 - 8$

$F 9 = f 31 + a 31 = 1.1608 × 10 - 5$ $1.1606 × 10 - 5$
$F 10 = f 31 - a 31 = 1.1532 × 10 - 5$ $1.1542 × 10 - 5$

$f 32 = 1.118 × 10 - 5$

$a 32 = 4.188 × 10 - 7$

$F 11 = f 32 + a 32 = 1.1599 × 10 - 5$ $1.1606 × 10 - 5$
$F 12 = f 32 - a 32 = 1.0761 × 10 - 5$ $1.0759 × 10 - 5$

$f 33 = 1.115 × 10 - 5$

$a 33 = 3.934 × 10 - 7$

$F 13 = f 33 + a 33 = 1.1543 × 10 - 5$ $1.1542 × 10 - 5$
$F 14 = f 33 - a 33 = 1.0757 × 10 - 5$ $1.0759 × 10 - 5$