I am reading an article that is about CDMA. In that article I encountered a term "spreading vetor". I searched some books such as Proakis and Gallager to see what it means. I even surfed internet for this statement, but I found some other terms such as "spreading signature" but not "Spreading vector". Can any one explain the meaning of "Spreading vector" in digital communication?
2 Answers
Here's an extremely simplified way to think about it. The 'spreading vector' you are talking about (made up from chips), is a bit vector of length N. Let's say that N is 8. That means that the spreading factor is 8 and the process gain is \$10*log_{10}(8) = 9dB\$.
Let our code be C = [0,1,1,0,1,1,1,0]
To 'spread' our signal, we send one full code for every bit of our actual data. If our message bit is a 1 then we send C. If our message bit is a 0 then we send ~C = [1,0,0,1,0,0,0,1]
.
Since the chip rate in this example is 8x that of the message rate, the bandwidth is spread out in the frequency domain.
The codes used should ideally all be orthogonal to one another, however in practice PN codes give sufficient chip distance.
-
\$\begingroup\$ You are confusing chips with bits, which is ironically why chips are called chips, to prevent exactly that. Chips are emitted at the code rate, the sequence of chips makes up the code length (AKA vector) which is the same as the bit length. In your example there are 8 chips per bit. \$\endgroup\$ Commented Sep 19, 2014 at 14:37
CDMA is a generalized term for SSDS (Spread Spectrum Direct Sequence) which usually refers to PN (Pseudo Noise) generating codes. But it need not necessarily use PN, as long as the sequence spreads the information in the frequency domain. A vector, in general terms, is any finite length sequence of states. The detection process is one of correlation, which in a mathematically dense derivation refers to the inputs as vectors, thus the vector terminology. So that is not typically used but is accurate in context.
This term is exactly equivalent to "process gain".
-
\$\begingroup\$ I did not understand your response. Can you explain more? I did not even find relation between my question and your answer, so I will be thankful if you explain more \$\endgroup\$– CLAUDECommented Sep 18, 2014 at 16:48
-
1\$\begingroup\$ It's an accurate answer to my way of thinking. \$\endgroup\$– Andy akaCommented Sep 18, 2014 at 17:00
-
\$\begingroup\$ @Andyaka Can you explain more about his answer? \$\endgroup\$– CLAUDECommented Sep 18, 2014 at 17:07
-
1\$\begingroup\$ You need to read a few articles to get your head around the concept of spread spectrum. I suggest that is your 1st goal. It's not an easy subject so good luck. \$\endgroup\$– Andy akaCommented Sep 18, 2014 at 17:10