# What is the duration of the spreading sequence

I have a question regarding the spreading sequence duration using Walsh matrix. I've read a paper related to that, they consider the signal $$S = R(S) + I(S)$$, where $$R(S), I(S)$$ are the real and imaginary parts of the signal $$S$$ resulted from the modulation $$QAM$$.

what is done, briefly, a Walsh-Hadmard matrix is built with dimension $$4$$x$$4$$, then the real are imaginary parts are multiplied by the spreading code taken from the Walsh matrix and then transmitted into receiver. So the resulted signal before transmitting is $$S = W_1R(S) + W_2I(S)$$ where $$W_1, W_2$$ are two different spreading code taken from the Walsh matrix.

So, without talking about why that process was done, My question is related into the signal after spreading itself:

Will the duration of transmitted signal "after spreading" is similar to duration of signal $$S$$ before spreading?I think yes, because the duration of chip of spreading sequence is shorter by $$n$$ times of duration of signal $$S$$, it's similar to this question here.. If that right, what's the downside of using spreading sequence? I mean when using it, what's the disadvantages we face compared with any other technique. Finally, does using spreading sequence as above process mean that we are using CDMA? ..

thank you

• You might be missing $j$ in various places. – Dilip Sarwate Jan 26 '19 at 14:43

When you multiply the real and imaginary parts by $$\mathbf{W}_1$$ and $$\mathbf{W}_2$$, respectively, the resulting signal is no longer $$S$$. It will be a matrix.

1- Yes, the signal after spreading has the same duration.

2- One of the disadvantages of Walsh-Hadamard codes is that they don't have clear auto-correlation peak, which makes synchronization at the receiver side challenging. Also, they must have a dimension of $$2^k$$, i.e., you cannot have arbitrary dimension to support an arbitrary number of users. In your example, to make the system efficient, you need 4 users, otherwise, one user will occupy 4 times the minimum required bandwidth, which isn't spectrally efficient.

3- Yes. This is called direct sequence (DS) CDMA.

• Thank you so much ... that's clear now. but I have a question regarding your comment, why did you say that I need 4 users? .. Why not needing just 2 users which is 2 power 1. – New_student Jan 27 '19 at 6:24
• @New_student It's not necessary, but since you are using $4\times 4$ Walsh-Hadamard matrix, this means you have $4$ orthogonal rows/columns. Since the spreading factor is $4$, this means that each signal will occupy $4$ times the minimum bandwidth. In $4$ times the minimum bandwidth, you could transmit $4$ signals using FDMA, for example. Spreading $2$ signals over $4$ times the minimum bandwidth is possible, but it's not the most efficient way. – BlackMath Jan 27 '19 at 15:13

I totally agree with the above comments, but I'd like just to add a small note:

CDMA is a special case of spread sequence, I mean using spread sequence doesn't mean that you are using CDMA. Spread sequence is using the sequence to spread your signal. for example if your signal is 1, then you can use the sequence 0101001 in order to spread the signal 1 on your sequence 101001 using the known process. So the transmitted signal is the spread code instead of the signal itself. Now, when using many users with orthogonal sequences codes.. . that mean you are working with CDMA.

• CDMA is a spread spectrum multiple access technique. – BlackMath Jan 27 '19 at 18:22