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Fall 2012 September 7, 2012 159 / 174 Lecture Notes 2 msequences 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 msequences 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 msequences Properties of msequences
5 The number of distinct msequences is of length 2n − 1 is
φ(2m − 1)/m where φ(x ) is Euler phifunction (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 msequences 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 msequences Galois Implementation of maximal length sequences
(msequences)
+ 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 msequences Alternative Maximal length sequences (msequences) 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 msequences Masking for different msequences EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 166 / 174 Lecture Notes 2 msequences Distinct msequences EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 167 / 174 Lecture Notes 2 msequences IS95 Implementation of Long Code EECS 455 (Univ. of Michigan) Fall 2012 September 7, 2012 168 / 174 Lecture Notes 2 msequences Waveform Autocorrelation Below we show a waveform consisting of a sequence of pulses of
amplitude ±1 determined from an msequence 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 msequences 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 msequences 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 msequences Problem with msequences The problem with msequences is that there are not enough
distinct msequences to accommodate many users.
Distinct means that the msequences 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 msequences Table of msequences
Below we give a table of msequences. 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 msequences Table of msequences
The following feedback connections generate sequences of maximal
length. The reciprocal of a shift register also produces a
maximumlength 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.
 Fall '08
 Stark
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