Talk:Direct-sequence spread spectrum

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"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)[reply]

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)[reply]

"DSSS has the following features:

1. 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@

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]

This is very important, but not explained well in the article. Only in the introductory section does it even mention bandwidth. Furthermore, why does a higher bit rate give a greater bandwidth. I'm only 7 years late :) 188.58.21.168 (talk) 22:36, 14 March 2012 (UTC)[reply]

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]

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]

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)[reply]

Thank you. it is a common and fast growing use of the technology. Utilitysupplies 14:05, 27 April 2007 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

How does one select the chipping code? I'm trying to determine how a common 802.11g wireless router picks one so I can understand how one might fail to connect due to mismatched codes. Is it from the SSID, or the encryption keys, or the channel number? I didn't see any mention of this at Wi-Fi or Spread spectrum. SpareSimian (talk) 20:06, 2 May 2008 (UTC)[reply]

The chipping codes are preselected by whomever writes the transmission standard. They are chosen so that demodulating with any of the codes will cancel out all the others from the signal. Finally the standard will say which bits are modulated with which chip code. In other words, there is a fixed, published, list of chip codes and what each of them is used for.

Encryption (if any) is done at a completely different level, on the bits that are sent, not on the chip codes. WiFi SSIDs are just pretty names for the MAC address of the router/access point. Each packet sent over the air is marked with that MAC address and the other networks are supposed to politely ignore it (even if it is not encrypted).

An easier example is GPS: All 31 GPS satellites and all the Russian GLONASS, EU Galileo etc. satellites send on the exact same frequency, but with different chip codes (called PRN numbers). The GPS receivers listen on that one frequency, and then demodulate it with the PRN codes of each satellite it is tracking, thus getting all the signals from just one reception at one time. The satellite numbers on your GPS display are the numbers of the PRN codes in the book, it is not really a specific satellite, since they keep replacing those when they crash and burn. Actually, there is a second and a third frequency, just like the first, but only expensive receivers listen to those. Jbohmdk (talk) 04:10, 5 February 2014 (UTC)[reply]

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)[reply]

My professor drew smth like this: [1]

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)[reply]

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)[reply]

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)[reply]

This information needs to be corrected. LaMarr and Antheil used frequency-hopping spread spectrum in their patent ^[1]. If you try to stretch the truth about this, then what was the spreading ratio of their technique? The other article in wikipedia ^[2] 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)[reply]