# Is this a “valid” CDMA modulation?

I am banging my head against (specific aspects of) CDMA for a few days. At the basic level I am confused about the interrelation between implementations that are modeled as mixing and fully digital implementations. Example:

1. Mixing could use the the matrix $$[ 1, 1 ; 1, -1]$$ (appears in the example of https://en.wikipedia.org/wiki/Code-division_multiple_access#Example and https://en.wikipedia.org/wiki/Walsh_matrix). These matrices have elements -1 and 1 and flip signs and can be interpreted as mixing/multiplication. Makes sense

2. Other sources show these matrices as $$[ 0, 0 ; 0, 1]$$ (e.g. https://www.tutorialspoint.com/cdma/cdma_techniques.htm). My understanding is that these setups assume bit streams (as opposed to more general sample streams) and for each data "1" they would just send the normal PN sequence (e.g. $$[ 0, 1 ]$$) and for each data "0" they would send the 1's complement $$[ 1, 0]$$. Makes sense too.

Now I found a paper with yet another setup:

1. In the example station one has code $$01$$ and station two has code $$10$$. Station one sends $$111001$$ and station two sends $$100111$$. After CDMA code injection, station one transmits $$010101000001$$ and station two transmits $$100000101010$$. It should be noted that if these two transmissions are summed in the shared channel, the result is still a two-level waveform (=binary): $$110101101011$$! Demodulation is performed again by multiplying (=ANDing) the chip code and then OR-ing each bit. For example, station two would multiply/AND the received signal $$110101101011$$, and it with $$10$$ to get $$100000101010$$. OR-ing two adjacent bits gives the original sequence $$100111$$.

Questions:

• Is #3 a valid CDMA modulation/demodulation scheme?
• How does it relate to the setups #1 and #2 ?
• How do setup #1 and #2 relate? (I would expect multiplication in $$GF(2)$$ is AND and addition is XOR ... but 2. is different)
• How would setup #3 extend to code lengths $$\gt 2$$?
• Why would someone use setup #3 and call it "CDMA" (instead of setup #2)? Do I miss the relationship to #2?

One one hand #3 would provide a more logical match with #1 because AND is a multiplication in $$GF(2)$$. However, addition would correspond to XOR instead of OR and I don't know where OR comes from.

On the other hand, I cannot make setup #3 work with codes that resemble normal Walsh matrices (e.g. $$[ 1, 1 ; 0, 1]$$. All that works are codes like $$[ 1, 0, 0, 0 ; 0, 1, 0, 0 ; 0, 0, 1, 0; 0, 0, 0, 1]$$. Yes, they are orthogonal but somewhat miss the point of CDMA: code spreading. They implement more time division.

I think you are trying to mix orthogonality for $$GF(2)$$ and $$R^n$$. For the sequence $$010101000001$$, if station 2 had tried to demodulate it, he will have to take groups of 2 bits, multiple (AND) and add(XOR) them, to decode whether bit 0 or 1. Station 2 would have got all 0s because $$10$$ and $$01$$ is 'orthogonal' for station 1 and 2. So yes, #3 is indeed a valid CDMA scheme in $$GF(2)$$. I stress, it is a valid CDMA scheme only in $$GF(2)$$. It is not related to #1 and #2 (see below why).
Setups #1 and #2 are valid CDMA schemes in $$R^n$$ (real number vectors) because orthogonality is validated by $$x_1^Tx_2$$. For #2, you cannot validate orthogonality by dot product of vectors containing 0 and 1. Because for $$W_4$$, rows 3 and 4 are not orthogonal. So, $$0$$ has to be represented by $$-1$$. Even though logically we switch $$0$$ and $$1$$ as NOT operator while forming Walsh matrices, orthogonality is not valid in $$GF(2)$$.
$$[1,1;0,1]$$ would not correspond to CDMA because whether in $$GF(2)$$ or $$R^2$$, they are not orthogonal. $$[1,0,0,0;0,1,0,0;0,0,1,0;0,0,0,1]$$ is orthogonal in $$GF(2)$$ as well as in $$R^2$$ which is same as your multiplexing in time but it is still technically a CDMA scheme (if you transmit each bit at 4x higher rate).
• I believe that is not possible. I think it would work with either the above kind of diagonal schemes or something like this $[1,0,1,0;0,1,0,1]$, both orthogonal in GF(2) and R^4 but still kind of multiplexed because when one station transmits 0, other is 1. – jithin Apr 17 at 18:19