DSSS features

"DSSS has the following features:

for generating spread-spectrum transmissions by [...]"

Is this supposed to make any sense? I cannot guess what the original poster wanted to mean. It would be nice if anyone with better knowledge on the subject corrected the statement. @SIG@

Bitrate raise vs. energy spreading

I am wondering why raising the bitrate "spreads the energy" on a wider frequency band. Does anyone have an idea?

Shannon's law gives us the maximum possible bit rate given a frequency band and a Signal to Noise ratio: MaxBitRate = Bandwidth * log2 (1 + SignalPower/NoisePower)

But I don't see that as an explanation since it allow us to calculate the MAX bitrate, not the bitrate itself.

Then there's Nyquist's law : the sampling rate must be at least twice the signal's max frequency. I don't see this as an explanation either.

One explanation I can imagine is that switching abruptly a signal from one frequency to another generates harmonics. If you switch very often, a greater part of the energy is used up by those harmonics. Does this even make sense? ;-)

Anyway, does anyone have any good explanation for this rule?

If you assume that you always use close to maximum bitrate at a given bandwidth (which makes sense, since otherwise you'd be wasting spectrum), you see that it's only by increasing bandwidth that you can increase the bitrate (or baud rate, really). Have a look at modem, Quadrature amplitude modulation and some of their linked pages for more info.

europrobe 08:50, 2004 Aug 1 (UTC)

Thank you for your answer, I appreciate your help. But IMHO, I feel that this does not really answer the question. True, Shannon's law tells us that with a wider bandwidth, you can get a higher channel capacity. So I agree with you that in order to get a higher bitrate, you need to use a larger bandwidth (or raise the S/N ratio exponentially). But what I do not understand is why raising the bitrate spreads the signal energy on a wider frequency band (as the DSSS article says). What is the physicle explanation for this? Let me summarize this:

A wider band allows for a greater bitrate.
But why does a greater bitrate cause a wider band?

Thanks again for you help.

--Ageron 10:32, 1 Aug 2004 (UTC)

Seems like this is more an issue of semantics. A wider band allows for a higher bitrate, but is not (to my knowledge), actually caused by it. –radiojon 00:39, 2004 Aug 2 (UTC)

Well, this is a chicken-and-egg problem. To send at a higher bitrate you need to allow the signal to use more bandwidth. A wider bandwidth does not "create" a higher bitrate. (Digital) frequency modulation, for example, uses two or more frequencies to send information. The more frequencies, the wider the band. Also, rapidly changing between the frequencies creates harmonics, which widen the band. Amplitude modulation uses changes in amplitude (as you no doubt are aware of), and rapid changes in amplitude also creates harmonics. Other modulations are just variations of these, so the same limitations apply. In other words, a high bitrate actually cause the signal to spead out due to harmonics and multiple frequencies being used.

europrobe 19:14, 2004 Aug 2 (UTC)

So the bottom line is that higher bitrates generate more harmonics, which spread the spectrum.

Thank you for your very clear answer!

--Ageron 09:15, 16 Aug 2004 (UTC)

How closely is this related to DS-CDMA? If it is similar could there be a mention of it, or even better a whole article for DS-CDMA.

Thanks

--Sparky132

The above explaination about why the signal spreads is incorrect. The spreading is an intentional operation, prior to transmission. Later, the receiver despreads the wideband transmitted signal and low pass filters to recovers the orignal narrowband information, with some residual noise.

The idea is that with the spread signal, it is very difficult to intercept the orignal narrowband without the transmitter's pseudorandom code, and it is extremely difficult to jam a wideband tranmission with narrowband noise because the despreading operation nearly eliminates the noise. Hence, a spread spectrum system has excellent antijam capability.

P

You haven't written anything that contradict what I wrote above. The method of spreading the signal in DSSS is by multiplying the bitstream with a higher-bitrate pseudorandom stream, thus increasing the bitrate of the transmission. The psudorandom operation is reversed in the receiver, which cause the noise to average out and the S/N to increase.

europrobe 20:35, 30 November 2005 (UTC)[reply]

Introduction / layout

I feel that the introduction to this article is not separated with the actual explanation of the DSSS. I think there should be a short introduction and then the explanation, separated with propoer headers. --153.109.5.120 14:35, 29 August 2006 (UTC)[reply]

Uses

I added Automatic meter reading (radio equipped water, electric, gas meters) as one of the common uses of DSSS. The entry was removed by User:Oli Filth. AMR is just as valid a "use" as cordless phones and CDMA cellular phones in my opinion... but I'll let other editors decide if it has noting to do with the uses of this RF technology and whether it should be included. Utilitysupplies 15:27, 26 April 2007 (UTC)[reply]