Question
For the two-sender, two-receiver example, give an example of two CDMA codes containing 1 and #1 values that do not allow the two receivers to extract the original transmitted bits from the two CDMA senders.
Answer
The following is the example of two CDMA (Code division multiple access) codes that do not allow to extract the original data bits by the two receivers that are transferred by the two CDMA senders is as follows:
Consider a two sender and two receiver CDMA system.
Consider the CDMA code sent by the Sender1 is
(1, 1, 1, -1, 1, -1, -1, -1) and the CDMA code sent by the Sender2 is
(-1, -1, 1, 1, 1, 1, 1, -1) and assume that the senders are using 8-bit CDMA code.
Let us denote the data sent by two senders as 
=1 , 
=1 .
Sender 1 encoding process:
Sender 1’s 8-bit CDMA code is (1, 1, 1, -1, 1, -1, -1, -1).
|
Data bits ( |
||||||||
|
CDMA Code( |
1 |
1 |
1 |
-1 |
1 |
-1 |
-1 |
-1 |
|
Output ( |
1 |
1 |
1 |
-1 |
1 |
-1 |
-1 |
-1 |
)
)
Sender 2 encoding process:
Sender 2’s 8-bit CDMA code is (-1, -1, 1, 1, 1, 1, 1, -1).
|
Data bits ( |
||||||||
|
CDMA Code ( |
-1 |
-1 |
1 |
1 |
1 |
1 |
1 |
-1 |
|
Output ( |
-1 |
-1 |
1 |
1 |
1 |
1 |
1 |
-1 |
)
)
Calculate the output generated by the senders as follows:
|
|
1 |
1 |
1 |
-1 |
1 |
-1 |
-1 |
-1 |
|
|
-1 |
-1 |
1 |
1 |
1 |
1 |
1 |
-1 |
|
z*= |
0 |
0 |
2 |
0 |
2 |
0 |
0 |
-2 |
The following the data bits recovering process at receiver end:
Receiver 1 decoding process:
|
Bits in received signal Z* |
0 |
0 |
2 |
0 |
2 |
0 |
0 |
-2 |
|
CDMA Code |
1 |
1 |
1 |
-1 |
1 |
-1 |
-1 |
-1 |
Calculate the data received by the receiver:
Receiver 2 decoding process:
|
Bits in received signal Z* |
0 |
0 |
2 |
0 |
2 |
0 |
0 |
-2 |
|
CDMA Code |
-1 |
-1 |
1 |
1 |
1 |
1 |
1 |
-1 |
Calculate the data received by the receiver:
Therefore, the above example illustrates the case in which two CDMA codes that do not allow to extract the original data bits by the two receivers that are transferred by the two CDMA senders