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θ s ( g 1 ) ,s ( g 2 ) ( τ )
N− 1
=
s ( g 1 , 0 ,t + τ ) s ( g 2 , 0 ,t )
t =0
ı u 1 ( t + τ ) +2 ı u 0 ( t + τ ) ı −v 1 ( t ) +2 ı −v 0 ( t ) +
=2
t even
ı u 1 ( t + τ )
2 ı u 0 ( t + τ ) ı −v 1 ( t )
2 ı −v 0 ( t )
2
t odd
=2( θ u 1 ,v 1 ( τ )+2 θ u 1 ,v 0 ( τ )+2 θ u 0 ,v 1 ( τ )+4 θ u 0 ,v 0 ( τ )) +
2( θ u 1 ,v 1 ( τ )
2 θ u 1 ,v 0 ( τ )
2 θ u 0 ,v 1 ( τ )+4 θ u 0 ,v 0 ( τ ))
=4( θ u 1 ,v 1 ( τ )+4 θ u 0 ,v 0 ( τ ))
Since θ u 0 ,v 0 ( τ ) is orthogonal to θ u 1 ,v 1 ( τ ) [1], we bound the magnitude of the
number θ s ( g 1 ) ,s ( g 2 ) ( τ ) in the above expression as
θ s ( g 1 ) ,s ( g 2 ) ( τ )
4
|
Γ (1)(1 + 4 ı )
|
16 . 49 N/ 2
11 . 66 N.
(7)
Now, if τ
1(mod2),weget
θ s ( g 1 ) ,s ( g 2 ) ( τ )
N− 1
=
s ( g 1 , 0 ,t + τ ) s ( g 2 , 0 ,t )
t =0
ı u 1 ( t + τ ) +2 ı u 0 ( t + τ ) ı −v 1 ( t )
2 ı −v 0 ( t ) +
2 ı
t even
=
ı u 1 ( t + τ )
2 ı u 0 ( t + τ ) ı −v 1 ( t ) +2 ı −v 0 ( t )
2 ı
t odd
2 ı ( θ u 1 ,v 1 ( τ )
2 θ u 1 ,v 0 ( τ )+2 θ u 0 ,v 1 ( τ )
4 θ u 0 ,v 0 ( τ )) +
=
2 ı ( θ u 1 ,v 1 ( τ )+2 θ u 1 ,v 0 ( τ )
2 θ u 0 ,v 1 ( τ )
4 θ u 0 ,v 0 ( τ ))
=8 ı ( θ u 1 ,v 0 ( τ )
θ u 0 ,v 1 ( τ )) .
Since we do not have any orthogonality relation between θ u 1 ,v 0 ( τ )and θ u 0 ,v 1 ( τ ),
we assume the worst case and bound the magnitude of θ s ( g 1 ) ,s ( g 2 ) ( τ )intheabove
expression as
θ s ( g 1 ) ,s ( g 2 ) ( τ )
8
|
Γ (1)(1 + 1)
|
16 N/ 2
11 . 31 N.
(8)
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