# Signal spreading in spread spectrum technology (DSSS)

1. Apr 15, 2013

### mariano54

Hi, I've been trying to understand DSSS, but I'm not an engineer and have trouble with one point.

I get the fact that multiplying the user signal with the chip sequence is what spreads the signal. My question is, why is this?

If the data rate is, say, 1 bps, and the chip sequence 1000 bits long, then the final signal will have 1000 bps which requires 1000 Hz = 1 kHz.

From my perspective, this would mean that I have a signal that is oscillating at 1 kHz, but I know it's wrong and what it actually means is that the signal has a bandwidth of 1kHz (and is, therefore, spread). Why does increasing the frequency mean that the signal spans a whole range of frequencies instead of a single higher frequency?

2. Apr 15, 2013

### sophiecentaur

The chip sequence will have a very wide spectral bandwidth - a comb of frequencies, spaced by the chip sequence repeat frequency and extending to a maximum value which is determined by the rise and fall times of the chip waveform. When you multiply it by the user data, each spectral component of the chip sequence will be modulated by the user data, carrying its own version of the original data in its sidebands.

3. Apr 15, 2013

### mariano54

So the wide spectrum is basically encoded in the sequence? E.g., 111111 would be a low frequency sequence while 10101010 a high frequency one?

Do you mean the fact the sequence repeats itself in the time domain depending on the original data rate?

Do you mean the spectral density of the sequence?

4. Apr 16, 2013

### sophiecentaur

I think you may be jumping into this thing half way through.
First, you need to investigate the spectrum of a pseudorandom binary sequence (look it up and take time to read what you find). Then you need to consider what happens when you modulate a carrier with this sequence.
Then what I have already written my make sense.
That wiki article presents the info in a reasonable enough way. I don't know what it is that you don't know; you are the only one who knows that, so you can't rely on the right level of answers from me. Q&A is not a very efficient way of finding things out at this stage.

Read the following with care (and the wiki article again!), after having done what I suggested:- Find out what a pseudorandom sequence is, for a start.
The 'sequence' is fixed - as wiki says, and so is its spectrum. The spacing of the comb of frequencies in the spectrum is governed by the repeat rate of the sequence and the extent of the spectrum (total bandwidth) depends on the shape of the pulses. You then take your input (modulating) signal and multiply that wide range of components.

5. Apr 16, 2013

### mariano54

I'm sorry, what wiki article are you talking about? I can't see any link.

6. Apr 16, 2013

### sophiecentaur

Sorry - I'm losing it!
I though you'd quoted the link. Here it is.
I would recommend going to wiki as a first stop - even before PF - so that you can ask better informed questions and 'get there' faster.

7. Apr 16, 2013

### mariano54

Ok, thanks for your help.

I already read this (and many other) articles about DSSS. The only thing I didn't quite get was how the spectrum was actually spread by multiplying it with the PN sequence: now I get it, it's the PN sequence that has a large spectrum which get's carried to the baseband wave when the multiplication is carried out.

I'm not studying signal processing nor electrical engineering, only telecommunications related tech, so I'm in that middle ground in which I need to understand how things work without delving too much into the specifics.

Again, thanks a lot for your help.

8. Apr 16, 2013

### sophiecentaur

Exactly. And the clever thing is that all the components are at a very low level (no appreciable outgoing interference) and yet it can be demodulated at the receiver, multiply by an identical, synched PN sequence to produce the original input signal - information from each of the many components all adds up, back at the baseband frequencies. Because of the 'synchronous demodulation' only a small amount of power from any received interfering CW signal arrives at the baseband frequency and it turns up as a noiselike baseband signal.