24 Simple PN sequence generator block diagram.
Tc Tc
+
A B
Tc
C
Output PN
Sequence
Consider the initial loading approach first. If we assume the initial contents of the PN generator
are 100, this corresponds to a polynomial of x2 from the polynomial defined above. Next we will multiply
this initial contents polynomial by the PN code polynomial and maintain the variables with
degree less than m, in this case m 3.
(7.35)
(7.36)
(7.37)
Hence for this example, this corresponds to a delay or shift of 2 chips in the output PN sequence.
d(x) x2
d(x) [x2 # (1 x x3)]Degree 3
d(x) [x2 # p(x)]Degree 3
362 CHAPTER SEVEN
FIGURE 7.25 PN generator states for p(x) 1 x x3.
A B C
0 1 0 0
1 1 1 0
2 1 1 1
3 0 1 1
4 1 0 1
5 0 1 0
6 0 0 1
7 1 0 0
8 1 1 0
9 1 1 1
10 0 1 1
11 1 0 1
Chip-Time Index
FIGURE 7.26 Shifting the PN sequence output sequence.
a2 a1
+
PN Output
a0
X m0 X m1 X m2
+
Delayed
PN Output
a(x)
Mask
Now that was a rather simple example, let??™s consider initial contents equal to 001, which corresponds
to a polynomial of 1. Hence, we go through the same procedure as shown above to generate d(x).
(7.38)
(7.39)
(7.40)
In order to determine what offset this corresponds to, we need to perform the following step of
dividing by the PN code polynomial:
(7.
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