Talk:Direct-sequence spread spectrum

message: "the sequence of chips produced by the transmitter is known a priori by the receiver"
Gold star for obfuscating the meaning. How about replacing the Latin 'a priori' with
the more common English phrase 'in advance', pipe to the 'a priori' article, or simply
omit it from the sentence altogether.  — Preceding unsigned comment added by 124.170.69.189 (talk) 08:39, 20 February 2012 (UTC)
replies: text: That section was so bad. I successively removed the incorrect wording, the wrong claims, the confusion between bits and symbols, such that I ended up with nothing worth keeping (because it's been said already at that point, and at least as concisively so). I hence ended up removing the whole section, see the edit history. Marcusmueller ettus (talk) 14:31, 30 July 2023 (UTC)
thread_title:  Features 

message: 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)
thread_title:  Bitrate raise vs. energy spreading 

message: 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)
thread_title: Introduction / layout

message: 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)
replies: text: My apologies; I originally misread, and thought that you had added this link under "See also", in which case it would have been inappropriate.  But under "Uses", it makes much more sense, so I've re-added the link. Oli Filth 21:39, 26 April 2007 (UTC)
thread_title: Uses

message: The article does mention, that this allows multiple transmitters to operate on the same frequency in the exact same time moment, but does not explain how this is possible.  —Preceding unsigned comment added by 195.148.99.241 (talk) 09:43, 9 October 2007 (UTC)
replies: text: The article does a fine job of describing this complex process. The editor placed a link to the Code division multiple access at just the right place. Reading that may enhance your understanding of the technology. It helped me. E_dog95'   Hi ' 07:17, 15 October 2007 (UTC)
replies: text: I have every confidence the explanation is clear to engineers, who are people who needn't look up DSSS in the encyclopedia.  For those who do have that need, I dimly recall seeing in Scientific American Decades ago a description of despreading using a multiply tapped delay line that made it clear to relatively ordinary people.  DSSS having become ever more impotant in subsequent decades, it would be a very good thing if a similarly simple explanation could be put in the encyclopedia.  Jim.henderson (talk) 06:58, 22 January 2008 (UTC)
thread_title: How does this allow multiple transmitters to operate on exact same frequency?

message: I've renamed this article from "Direct-Sequence Spread Spectrum" back to its orginal name of "Direct-sequence spread spectrum". The reason given for the previous rename on 1 January 2008 was "Title is an acronym and should be capitalized WP:MOS". In fact, the Manual of Style says "Initial capitals are not used in the full name of an item just because capitals are used in the abbreviation." (Wikipedia:MOS#Acronyms_and_abbreviations) --Dr Greg (talk) 17:44, 29 February 2008 (UTC)
thread_title:  Article renamed 

message: I added 900 MHZ in the phones section. I own a 900 MHZ that is indeed using DSSS. Further, I could not monitor it on  scanner. A full sweep on a frequency counter was done as well. Nothing.  Sprint cat. no. 43-5518  —Preceding unsigned comment added by 64.55.144.50 (talk) 00:42, 26 September 2008 (UTC)
thread_title:  Uses 

message: My professor drew smth like this:
And said: "We have our data in a narrow band. But instead of just sending that narrow band, we make the band wider by multiplying with a bit string. The volume under the graph is the same in the two because the amount of energy is the same.".
This article does not explain this. Have I misunderstood this?
And then: What is it being multiplied by the PN? Is it a series of bits (discrete) or is it a wave form (sort of analog)? It is easy to understand if it is a series of bits. But if so, then how can that affect the bandwidth? And if it is a series of bits, then how can my professor even talk about the initial narrow band? (bit strings don't have a bandwidth). 
188.58.21.168 (talk) 17:04, 14 March 2012 (UTC)
replies: text:  See. The problem is that many people that teach spread spectrum miss out various important and significant things, which leaves the student very confused. Direct sequence spread spectrum is usually the sending of modified digital bits. For example, 1 digital bit might represent a binary '1' or a binary '0'. Applying DSSS to 1 single message bit will involve constructing some OTHER pseudo-random sequence of bits, having a bit-rate that is usually much higher than the message bit rate. And the pseudorandom bit-rate is an integer multiple of the message bit-rate. For example, four pseudorandom bits may fully fit within one single message bit period (time). So when you XOR four pseudorandom bits with one single message bit, you will end up with four bits. And what those four resulting bits are will depend on whether the message bit is a binary 1 or a binary 0. Eg. pseudorandom bit pattern might be 1010, and message bit might be a '1'. So the resulting XOR (over 1 message bit time) will be 1010 XOR 1111, which results in 0101. The next four pseudorandom bits will be different, since it's all pseudorandom. So the next pseudorandom pattern could be 1001, and the next message bit could be a binary 0. So the XOR of the pseudorandom bits and the message bit (over one bit time) is 1001 XOR 0000, which gives 1001. So the first 8 bits for the pseudorandom sequence to be transmitted is 0101 1001. If the receiving side has the same pseudorandom bit pattern, and if it is properly synchronised, then it could decode it by taking these incoming 8 bits 0101 1001 and XOR with the pseudorandom sequence 1010 1001, to give 1111 0000. Since the transmitted (and received) data rate for this example happens to be four times faster than the message bit rate, then the '1111' will represent a message bit representing binary '1'. And the '0000' represents binary '0'. We received '1' and then '0', as expected. The transmitted DSSS bandwidth is obviously higher than the message bandwidth --- because the transmit bit-rate is higher than the message bit-rate. And the pseudorandom pattern means wider bandwidth than what we would get if we didn't have pseudorandom pattern. Further, for practical transmission (broadcasting) of these spread-spectrum waveforms, binary sequences are converted from uni-polar form to bi-polar form. And bipolar signals can conveniently be used to multiply with a sinusoidal "carrier" for wireless transmission. And one thing that is rarely explained (or never explained) is how a broadcasted DSSS signal is recovered (demodulated) at the receiving end. The phase of the incoming carrier (embedded within the spread spectrum signal) will usually not be the same phase as the local oscillator signal at the receiver. People always talk about using techniques to adjust the phase of the pseudorandom sequence (at the receiving end) in order to recover the message or carrier - such as "sliding correlator" techniques. But this is usually under some kind of assumption that they can measure the incoming pseudorandom sequence waveform in order to do this sliding correlator comparison. This would involve perhaps using a Costas Loop technique to extract some information from the incoming DSSS signal (to begin with). KorgBoy (talk) 23:28, 1 December 2017 (UTC)
thread_title:  Spread spectrum, HOW? 

message: The history section mentioned here seems to have been taken from this paper: The Origins of Spread-Spectrum Communication](download it here) as has been mentioned in the references section. 
However the content (see it here) seems to have been picked out of a section of the paper without heed to any context or background, this is making the history section particularly misleading and quite hard to comprehend. 
In addition, by the looks of the paper, the history appears to be much richer and much more convoluted than that mentioned here. 
Due to those reasons I've deleted the content of the history section to avoid misleading people until it can be expanded upon in more accurate manner. 
What are your thoughts on this? MSheshera (talk) 04:53, 12 April 2015 (UTC)
thread_title:  Possible misleading history? 

message: This information needs to be corrected.  LaMarr and Antheil used frequency-hopping spread spectrum in their patent https://patents.google.com/patent/US2292387.  If you try to stretch the truth about this, then what was the spreading ratio of their technique?  The other article in wikipedia https://en.wikipedia.org/wiki/Spread_spectrum clearly states their technique was based on frequency-hopping, which had already been invented and used decades before.  — Preceding unsigned comment added by Linfc (talk • contribs) 02:23, 8 June 2023 (UTC)
thread_title: Hedy LaMarr and George Antheil invented DSSS? What was their spreading ratio?