# Of michigan 1 5 2 n fall 2012 53 25 1 5 2 n september

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Unformatted text preview: 0 Fall 2012 September 7, 2012 128 / 174 Lecture Notes 2 m-sequences For F0 = 0, F1 = 1 we get the following relation 1 Fn = √ 5 √ 1+ 5 2 n 1 −√ 5 √ 1− 5 2 n Check this in Matlab EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 129 / 174 Lecture Notes 2 m-sequences Fibanacci sequences G n = G n −1 + G n −2 , G0 = 1 , G1 = 1 n Gn 0 1 1 1 2 2 3 3 4 5 5 8 6 13 7 21 8 34 9 55 n Gn 10 89 11 144 12 233 13 377 14 610 15 987 16 1597 17 2584 18 4181 19 6765 Gn = √ 5+1 √ 25 EECS 455 (Univ. of Michigan) √ 1+ 5 2 n + Fall 2012 √ 5−1 √ 25 √ 1− 5 2 n September 7, 2012 130 / 174 Lecture Notes 2 m-sequences Fibanacci sequences Hn = Hn − 1 + Hn − 2 , H0 = 1, H1 = 2 n Hn 0 1 1 2 2 3 3 5 4 8 5 13 6 21 7 34 8 55 9 89 n Hn 10 144 11 233 12 377 13 610 14 987 15 1597 16 2584 17 4181 18 6765 19 10926 Hn = √ 5+3 √ 25 EECS 455 (Univ. of Michigan) √ 1+ 5 2 n + Fall 2012 √ 5−3 √ 25 √ 1− 5 2 n September 7, 2012 131 / 174 Lecture Notes 2 m-sequences Fibanacci sequences Suppose F0 = 3 and F1 = 4. The the sequence is 3, 4, 7, 11, 18, 29, 47,... Find the general form of the Fibanacci sequence (ﬁnd A and B ) for this starting condition. A= B= Fn = A √ 1+ 5 2 n +B √ 1− 5 2 n F20 = EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 132 / 174 Lecture Notes 2 m-sequences Fibanacci sequences F n = F n −3 + F n −4 , F0 = 0, F1 = 0, F2 = 0, F3 = 1 n Fn 1 0 2 0 3 1 4 0 5 0 6 1 7 1 8 0 9 1 n Fn Fn 0 0 10 2 11 1 12 1 13 3 14 3 15 2 16 4 17 6 18 5 19 6 = (0.1593)(1.2207)n + (0.1187 + j 0.2045)(−0.2481 + j 1.0340)n +(0.1187 − j 0.2045)(−0.2481 − j 1.0340)n + (−0.3967)(−0.7245)n EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 133 / 174 Lecture Notes 2 m-sequences Fibanacci Sequences Fn EECS 455 (Univ. of Michigan) F n −1 F n −2 Fall 2012 F n −3 F n −4 September 7, 2012 134 / 174 Lecture Notes 2 m-sequences Fibanacci Sequences, mod 2 Fn = Fn−3 + Fn−4 mod (2), F0 = 0, F1 = 0, F2 = 0, F3 = 1 n Fn 0 0 1 0 2 0 3 1 4 0 5 0 6 1 7 1 8 0 9 1 n Fn 10 0 11 1 12 1 13 1 14 1 15 0 16 0 17 0 18 1 19 0 EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 135 / 174 Lecture Notes 2 m-sequences Maximal length sequences (m-sequences) - - - - - ?    0001 0101 1000 0100 1010 1101 0010 1001 1110 1111 1100 0111 0110 1011 0011 0001 The output of the shift register is the periodic sequence. 100010011010111 EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 136 / 174 Lecture Notes 2 m-sequences Maximal length sequences (m-sequences) 0 0 0 0 1 0 0 + 1 0 0 + EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 137 / 174 Lecture Notes 2 m-sequences Maximal length sequences (m-sequences) 0 1 0 0 0 0 0 + 0 0 1 + EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 138 / 174 Lecture Notes 2 m-sequences Maximal length sequences (m-sequences) 1 0 0 1 0 0 1 + 0 1 0 + EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 139 / 174 Lecture Notes 2 m-sequences Maximal length sequences (m-sequences) + EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 140 / 174 Lecture Notes 2 m-sequences Linear Shift Register Sequences Not all shift registers with linear feedback give a maximal length sequence. For example, the register shown below does not. + EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 141 / 174 Lecture Notes 2 m-sequences The state of the shift register is 0001 1000 0100 1010 0101 0010 0001 So the sequence is of length 6 (not 15). EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 142 / 174 Lecture Notes 2 m-sequences If the register is started in different states we get different sequences. 1 0 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 0 0 1 0 1 1 0 1 0 1 1 1 1 0 1 0 1 1 0 One is also a sequence of length 6. Another sequence is of length 3. With these three sequences we have exhausted all possible starting states except the all zero state. EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 143 / 174 Lecture Notes 2 m-sequences Not a maximal length sequences 1 0 0 + EECS 455 (Univ. of Michigan) 1 0 + Fall 2012 September 7, 2012 144 / 174 Lecture Notes 2 m-sequences Unique Cycles (for non-maximal length sequence) 1 11111 12 01111 10111 01011 00101 00010 00001 10000 01000 10100 11001 11100 11110 EECS 455 (Univ. of Michigan) 6 00111 00011 10001 11000 11100 01110 3 11011 01101 10110 1 00000 Fall 2012 2 10101 01010 3 01001 00100 10010 4 00110 10011 11001 01100 September 7, 2012 145 / 174 Lecture Notes 2 m-sequences Maximal length sequences (m-sequences) In general the feedback connection can be described by a polynomial h(x ). Let h(x ) = h0 x n + h1 x n−1 + · · · + hn−1 x + hn with hi ∈ {0, 1}. The taps of the shift register that are fed back correspond to hi = 1. We require hn = h0 = 1 (otherwise we could make the shift register shorter). The sequence at the output of the shift register can be expressed as fj +n = h1 fj +n−1 + · · · + hn−1 fj +1 + hn fj as can be seen below. EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 146 / 174 Lecture Notes 2 m-sequences Maximal length sequences (m-sequences) u j +m - u j +3 - u...
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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.

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