Unformatted text preview: j +2  u j +1 h0 = 1 EECS 455 (Univ. of Michigan)  uj h =1
?mâˆ’1
Fall 2012  hm = 1 September 7, 2012 147 / 174 Lecture Notes 2 msequences The period of the sequence is denoted by N .
If the length of the shift register is n the longest period possible for
the sequence is 2m âˆ’ 1.
This is because the shift register could be in one of 2m states.
However, the all zero state would only generate the all zero
sequence.
Thus a nonzero sequence could correspond to the shift register
going through all the 2m âˆ’ 1 nonzero states before repeating. EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 148 / 174 Lecture Notes 2 msequences Properties of msequences 1 2 The sequence length is 2m âˆ’ 1. [This is because the shift register
can cycle through all nonzero states before repeating. The
number of nonzero states is 2mâˆ’1 .]
The number of ones in the sequence is 2mâˆ’1 and the number of
zeros is 2mâˆ’1 âˆ’ 1. [This is because the states are all possible
length m binary vectors except the all zero vector. Including the all
zero vector would yield an equal number of 0â€™s and 1â€™s. Not
including the all zero vector decreases the number of 0â€™s by 1.] EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 149 / 174 Lecture Notes 2 msequences Properties of msequences
3 (Shift and Add) The sum of any two (distinct) shifts of a single
msequence is a different shift of the same sequence. To see this,
consider a sequence out of the shift register with initial state
a1 , ..., an . The sequence generated is
a n +1 = h 1 a n + Â· Â· Â· + h n âˆ’1 a 2 + h n a 1
a n +j = h 1 a j +n âˆ’1 + Â· Â· Â· + h n âˆ’1 a j +1 + h n a j . Consider another sequence started in a different initial state
b1 , ..., bn . The sequence generated is
b n +1 = h 1 b n + Â· Â· Â· + h n âˆ’1 b 2 + h n b 1
b n +j = h 1 b j +n âˆ’1 + Â· Â· Â· + h n âˆ’1 b j +1 + h n b j . The mod 2 sum of these two sequences is cj +n = aj +n + bj +n and
is exactly the sequence from the same shift register by starting in
state a1 + b1 , ..., an + bn where the addition is mod 2.
EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 150 / 174 Lecture Notes 2 msequences Shift and Add Property: Example 1 100010011010111100010011010111
110001001101011110001001101011
010011010111100010011010111100 EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 151 / 174 Lecture Notes 2 msequences Shift and Add Property: Example 2 100010011010111100010011010111
111000100110101111000100110101
011010111100010011010111100010 EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 152 / 174 Lecture Notes 2 msequences Shift and Add Property: Example 3 100010011010111100010011010111
111100010011010111100010011010
011110001001101011110001001101 EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 153 / 174 Lecture Notes 2 msequences Shift and Add Property: Example 4 100010011010111100010011010111
110001001101011110001001101011
111000100110101111000100110101
101011110001001101011110001001 EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 154 / 174 Lecture Notes 2 msequences Shift and Add Property: Example 5 100010011010111100010011010111
010111100010011010111100010011
110101111000100110101111000100 EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 155 / 174 Lecture Notes 2 msequences Shift and Add Property Fn EECS 455 (Univ. of Michigan) F n âˆ’1 F n âˆ’2 Fall 2012 F n âˆ’3 F n âˆ’4 September 7, 2012 156 / 174 Lecture Notes 2 msequences For example take the shift register of length 4 shown earlier that
produces a sequence of length 15.
One sequence out of the shift register is found by starting in state
0001. The sequence is 100010011010111 and then repeats. Another
sequence is found by starting in state 1011 and is 110101111000100
and then repeats. If these two sequences are added mod 2 we get a
100010011010111
third sequence. âŠ• 110101111000100 This sequence can be
010111100010011
generated from the same shift register by starting in state 1010. EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 157 / 174 Lecture Notes 2 msequences Equivalently if the two sequences are converted to Â±1 by the mapping
0 â†’ 1, 1 â†’ âˆ’1 and then multiplied we get a third sequence.
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1 158 / 174 Lecture Notes 2 msequences Properties of msequences
4 The (periodic) autocorrelation function of the Â±1 sequence
obtained by the transformation vj = (âˆ’1)uj is two valued. That is
Î¸v (l ) = N âˆ’1
i =0 vi vi +l = N , l = 0 mod (N )
âˆ’1, l = 0 mod (N ) This is obvious if l = 0 for then vi vi +l = 1 so the sum is N . If l = 0
then the result is obtained by realizing that
vi vi +l = (âˆ’1)ui (âˆ’1)ui +l = (âˆ’1)ui +ui +l .
Clearly the sum on the right matters only mod 2. Since l = 0,
ui + ui +l = ui +k for some integer k . (This is using the shift and add
property) . So for l = 0
Î¸v (l ) = N âˆ’1
i =0 EECS 455 (Univ. of Michigan) (âˆ’1)ui +k = 2mâˆ’1 (âˆ’1) + (2mâˆ’1 â...
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This note was uploaded on 02/12/2014 for the course EECS 455 taught by Professor Stark during the Fall '08 term at University of Michigan.
 Fall '08
 Stark
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