This is because the shift register can cycle through

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: j +2 - u j +1 h0 = 1 EECS 455 (Univ. of Michigan) - uj h =1  ?m−1   Fall 2012 - hm = 1 September 7, 2012 147 / 174 Lecture Notes 2 m-sequences 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 − 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 − 1 nonzero states before repeating. EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 148 / 174 Lecture Notes 2 m-sequences Properties of m-sequences 1 2 The sequence length is 2m − 1. [This is because the shift register can cycle through all nonzero states before repeating. The number of nonzero states is 2m−1 .] The number of ones in the sequence is 2m−1 and the number of zeros is 2m−1 − 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 0’s and 1’s. Not including the all zero vector decreases the number of 0’s by 1.] EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 149 / 174 Lecture Notes 2 m-sequences Properties of m-sequences 3 (Shift and Add) The sum of any two (distinct) shifts of a single m-sequence 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 + · · · + h n −1 a 2 + h n a 1 a n +j = h 1 a j +n −1 + · · · + h n −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 + · · · + h n −1 b 2 + h n b 1 b n +j = h 1 b j +n −1 + · · · + h n −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 m-sequences Shift and Add Property: Example 1 100010011010111100010011010111 110001001101011110001001101011 010011010111100010011010111100 EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 151 / 174 Lecture Notes 2 m-sequences Shift and Add Property: Example 2 100010011010111100010011010111 111000100110101111000100110101 011010111100010011010111100010 EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 152 / 174 Lecture Notes 2 m-sequences Shift and Add Property: Example 3 100010011010111100010011010111 111100010011010111100010011010 011110001001101011110001001101 EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 153 / 174 Lecture Notes 2 m-sequences 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 m-sequences Shift and Add Property: Example 5 100010011010111100010011010111 010111100010011010111100010011 110101111000100110101111000100 EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 155 / 174 Lecture Notes 2 m-sequences Shift and Add Property Fn EECS 455 (Univ. of Michigan) F n −1 F n −2 Fall 2012 F n −3 F n −4 September 7, 2012 156 / 174 Lecture Notes 2 m-sequences 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. ⊕ 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 m-sequences Equivalently if the two sequences are converted to ±1 by the mapping 0 → 1, 1 → −1 and then multiplied we get a third sequence. -1 × -1 1 +1 -1 -1 +1 +1 1 +1 -1 -1 EECS 455 (Univ. of Michigan) -1 +1 -1 +1 -1 -1 +1 -1 -1 -1 -1 1 Fall 2012 -1 -1 1 +1 +1 1 -1 +1 -1 +1 +1 1 -1 -1 1 September 7, 2012 -1 +1 -1 -1 +1 -1 158 / 174 Lecture Notes 2 m-sequences Properties of m-sequences 4 The (periodic) autocorrelation function of the ±1 sequence obtained by the transformation vj = (−1)uj is two valued. That is θv (l ) = N −1 i =0 vi vi +l = N , l = 0 mod (N ) −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 = (−1)ui (−1)ui +l = (−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 θv (l ) = N −1 i =0 EECS 455 (Univ. of Michigan) (−1)ui +k = 2m−1 (−1) + (2m−1...
View Full Document

Ask a homework question - tutors are online