The number of zeros and ones in the original sequence

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Unformatted text preview: 1)(1) = −1. Fall 2012 September 7, 2012 159 / 174 Lecture Notes 2 m-sequences For example consider the sequence 100010011010111 from the first shift register. Consider transforming it to a ±1 sequence. Consider a cyclic shift of this sequence. Shift the sequence by 5 to the left (or 10 to the right). OriginalSequence ShiftedSequence Product −1 +1 −1 +1 +1 −1 EECS 455 (Univ. of Michigan) +1 −1 −1 +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 −1 +1 +1 September 7, 2012 −1 +1 −1 −1 −1 +1 160 / 174 Lecture Notes 2 m-sequences Now clearly the result of taking a sequence in the ±1 domain and multiplying by a shifted version of that sequence is yet another shifted version of the same sequence. The number of zeros and ones in the original sequence correspond to the number of ones and minus ones in the ±1 domain. When we add all the elements of this ±1 sequence all the ones will cancel all but one of the minus ones we end up with -1 for the autocorrelation. EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 161 / 174 Lecture Notes 2 m-sequences Properties of m-sequences 5 The number of distinct m-sequences is of length 2n − 1 is φ(2m − 1)/m where φ(x ) is Euler phi-function (also called totient function) φ(L) is the number of positive integers less than or equal to L that are relatively prime (greatest common divisor is 1) to L. m 2 3 4 5 6 7 8 9 10 EECS 455 (Univ. of Michigan) 2m − 1 3 7 15 31 63 127 255 511 1023 Number of Sequences 1 2 2 6 6 18 16 48 60 Fall 2012 September 7, 2012 162 / 174 Lecture Notes 2 m-sequences Fibonacci vs. Galois Implementation F n −1 Fn F n −4 EECS 455 (Univ. of Michigan) F n −2 Fn F n −1 Fall 2012 F n −3 F n −2 F n −4 F n −3 September 7, 2012 163 / 174 Lecture Notes 2 m-sequences Galois Implementation of maximal length sequences (m-sequences) + 00001 10100 01010 00101 10110 01011 10001 11100 01110 00111 10111 EECS 455 (Univ. of Michigan) 11111 11011 11001 11000 01100 00110 00011 10101 11110 01111 10011 Fall 2012 11101 11010 01101 10010 01001 10000 01000 00100 00010 00001 September 7, 2012 164 / 174 Lecture Notes 2 m-sequences Alternative Maximal length sequences (m-sequences) The alternative (Galois) implementation produces the same sequence as the pure feedback form if the feedback connections are reciprocal. This implementation avoids doing a large number of mod 2 additions that might have longer propagation delay of the previous (Fibonacci) implementation. EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 165 / 174 Lecture Notes 2 m-sequences Masking for different m-sequences EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 166 / 174 Lecture Notes 2 m-sequences Distinct m-sequences EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 167 / 174 Lecture Notes 2 m-sequences IS-95 Implementation of Long Code EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 168 / 174 Lecture Notes 2 m-sequences Waveform Autocorrelation Below we show a waveform consisting of a sequence of pulses of amplitude ±1 determined from an m-sequence and the continuous time autocorrelation function ˆ s(τ ) = ∞ −∞ s(t )s(t − τ )dt . It should be noted that the continuous time autocorrelation function defined above is the same as the discrete periodic correlation when the argument τ of the continuous correlation function is an integer multiple of the duration of a single bit. In between the autocorrelation function varies linearly. EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 169 / 174 Lecture Notes 2 m-sequences Waveform Autocorrelation s(t ) 1 6 - T = 15 t −1 EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 170 / 174 Lecture Notes 2 m-sequences Autocorrelation ˆ s(t ) 15 10 5 −15 −10 EECS 455 (Univ. of Michigan) −5 5 Fall 2012 10 15 τ September 7, 2012 171 / 174 Lecture Notes 2 m-sequences Problem with m-sequences The problem with m-sequences is that there are not enough distinct m-sequences to accommodate many users. Distinct means that the m-sequences are not just cyclic shifts of each other. This is solved by using Gold codes or Gold sequences (such as used in a GPS system). EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 172 / 174 Lecture Notes 2 m-sequences Table of m-sequences Below we give a table of m-sequences. The table lists the feedback connection for a Fibonacci style connection. As an example the entry for a 5 stage shift register of [3, 5] means that the 3rd and 5th stages are combined to enter the first stage. This can also be written as a binary vector, namely 100101 where the positions containing a 1 indicate a connection. The first bit is always a ’1’ indicating that the first stage has the feedback. The last bit is also always a ’1’. Sometimes this is converted to octal (100101 = [45]8 ). EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 173 / 174 Lecture Notes 2 m-sequences Table of m-sequences The following feedback connections generate sequences of maximal length. The reciprocal of a shift register also produces a maximum-length sequence. Sequence Length 3 7 7 15 15 31 31 31 31 31 31 Shift Register Size 2 3 3 4 4 5 5 5 5 5 5 Feedback = x n −1 + x n −2 = x n −2 + x n −3 = x n −1 + x n −3 = x n −3 + x n −4 = x n −1 + x n −4 = x n −3 + x n −5 = x n −2 + x n −5 = x n −1 + x n −2 + x n −3 + x n −5 = x n −2 + x n −3 + x n −4 + x n −5 = x n −1 + x n −3 + x n −4 + x n −5 = x n −1 + x n −2 + x n −4 + x n −5 xn xn xn xn xn xn xn xn xn xn xn Octal (111) 7 (1011) 13 (1101) 15 (10011) 23 (11001) 31 (100101) 45 (101001) 51 (111101) 75 (101111) 57 (110111) 67 (111011) 73 http://www.newwaveinstruments.com/resources/articles/m sequence linear feedback shift register lfsr.h EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 174 / 174...
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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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