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When ( m, n )=(0 , 1), we have
s ( t
jd ))
+ s ( t +( i +1) d +1) s ( t + τ 1 +( j +1) d +1) , if τ 0 =0;
id ))(
s ( t + τ 1
C ( i, 0) , ( j, 1) ( τ )=
s ( t
id ) s ( t + τ 1 +( j +1) d +1)
+ s ( t +( i +1) d +1)(
s ( t + τ 1 +1
jd )) ,
if τ 0 =1
=
C s ( τ 1 + ad )+ C s ( τ 1
ad ) ,
if τ 0 =0;
C s ( τ 1 + bd +1)
C s ( τ 1
bd ) , if τ 0 =1 .
When ( m, n )=(1 , 0), we have
(
s ( t
jd )
+ s ( t +( i +1) d +1) s ( t + τ 1 +( j +1) d +1) , if τ 0 =0;
id )) s ( t + τ 1
C ( i, 1) , ( j, 0) ( τ )=
(
s ( t
id )) s ( t + τ 1 +( j +1) d +1)
+ s ( t +( i +1) d +1) s ( t + τ 1 +1
jd ) ,
if τ 0 =1
=
C s ( τ 1 + ad )+ C s ( τ 1
ad ) ,
if τ 0 =0;
C s ( τ 1 + bd +1)+ C s ( τ 1
bd ) , if τ 0 =1 .
T− 1
2
Case ii) d
|
: Similar to the Proof of Case i).
Remark : In Case i) of Lemma 1, it is easily checked that ad
=
ad mod T if
T− 1
2
0
i, j
f
1and( i, j )
=(0 , 0). The condition d
implies bd +1
=
bd
mod T for any 0
i, j
f
1. Hence,
{
s ( i,m ) ( t )
}
and
{
s ( j,n ) ( t )
}
are cyclically
distinct if ( i, m )
=( j, n ). Cyclic distinctness for Case ii) can also be checked
similarly. Therefore, the set
IS
in (2) contains 2 f cyclically distinct sequences
of period 2 T .
T
1
Theorem 2. Let 1 ≤ d ≤
.Theset IS in (2) is a (2 T, 2 f, 2 d, 2 ) M -ary
2
LCZ sequence set, where
2 T− 1
4 d
for d
T
1
2
f =
2 T− 3
4 d for d
T
1
|
.
2
T− 1
2
Proof. Consider the case that d
. Without loss of generality, we can assume
that i
j for C ( i,m ) , ( j,n ) ( τ ). By Lemma 1, we have
|
C ( i,m ) , ( j,n ) ( τ )
|≤
2
except for
( τ 0 1 )=(0 ,ad ) , (0 ,T
ad ) , (1 ,bd ) , (1 ,T
bd
1)
which are equivalent to
τ =2 ad, 2 T
2 ad, 2 bd +1 , 2 T
2 bd
1 ,
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