Aufgaben:Exercise 5.3Z: Realization of a PN Sequence

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Two possible realizations for PN generators

The diagram shows two possible generators for generating PN sequences in unipolar representation:   $u_ν ∈ \{0, 1\}$.

• The upper generator with the coefficients

$$ g_0 = 1 \hspace{0.05cm}, \hspace{0.2cm}g_1 = 0 \hspace{0.05cm}, \hspace{0.2cm}g_2 = 1 \hspace{0.05cm}, \hspace{0.2cm}g_3 = 1 \hspace{0.05cm}$$

is denoted by the octal identifier   $(g_3, g_2, g_1, g_0)_{\rm octal} = (15)$. 

• Accordingly, the octal identifier of the second PN generator is equal to  $(17)$.

• One speaks of an M-sequence if for the period length of the sequence   $〈u_ν〉$  holds:

$$P = 2^G – 1.$$

Here   $G$  denotes the degree of the shift register, which is equal to the number of memory cells.

Notes:

• The exercise belongs to the chapter  Spreading Sequences for CDMA.
• However, reference is also made to the chapter  Generation of Discrete Random Variables  in the book "Theory of Stochastic Signals".
• We would also like to draw your attention to the learning video  Explanation of PN generators using an example. 

Questions

1

What is the degree  $G$  of the two PN generators considered here?

$G \ = \ $

2

Give the period length  $P$  of the PN generator with the octal identifier  $(15)$  an.

$P\ = \ $

3

Which of the following statements are true for each M-sequence?

	The number of zeros and ones is the same.
	In each period there is one more one than zeros.
	The maximum number of consecutive ones is  $G$.
	The sequence  $1 0 1 0 1 0$ ...   is not possible.

4

Specify the period length  $P$  of the PN generator with the octal identifier $(17)$  an.

$P\ = \ $

5

Which PN generator produces an M-sequence?

	The generator with the octal identifier  $(15)$.
	The generator with the octal identifier  $(17)$.

Solution

Solution

(1)  The degree  $\underline{G = 3}$  is equal to the number of memory cells of the shift register.

(2)  From the given sequence the period length  $\underline{P = 7}$  can be read.  Because of  $P = 2^G –1$  it is an M-sequence.

(3)  Solutions 2, 3 and 4 are correct:

• The maximum number of consecutive ones is  $G$  (whenever there is a one in all  $G$  memory cells).
• On the other hand, it is not possible that all memory cells are filled with zeros  (otherwise only zeros would be generated).
• Therefore, there is always one more one than zeros.
• The period length of the last sequence is  $P = 2$.  For an M-sequence  $P = 2^G –1$.  For no value of  $G$  is  $P = 2$  possible.

(4)  If all memory cells are occupied with ones, the generator with the octal identifier  $(17)$  returns a  $1$ again:

$$u_{\nu} \big [ u_{\nu-1} + u_{\nu-2} + u_{\nu-3} \big ] \,\,{\rm mod} \,\,2 =1 \hspace{0.05cm}.$$

• Since this does not change the memory allocation, all further binary values generated will also be  $1$  each   ⇒   $\underline{P = 1}$.

(5)  Answer 1 is correct:

• One speaks of an M-sequence only if  $P = 2^G –1$  holds.
• Here, "M" stands for "maximum".