From: ashu on 10 Aug 2010 11:29 Dear All, I am trying to brush up my communication theory fundamentals. I have simple(perhaps foolish!) question on intutive understanding of CDMA. Shannon says that the capacity of a channel is directly proportional to bandwidth of signal. In other others for a given SNR if increasing B would give me more capacity. DSSS CDMA uses the same principle to spread the symbol over a large bandwidth by multiplying it with narrow time pulses(chips). Can we intutively say therefore, that when we demodulate our spreaded signal by remultiplying it with the PN sequene and then averaging it over symbol period, we are ineffect using the statistical property that noise over a large period(symbol period) will cancel out itself. In other words the larger averaging interval we have(symbol period) the more we lower noise, since noise is inherently random and gaussian. Can DSSSCDMA be looked at from this perspective? regs ashu
From: Tim Wescott on 10 Aug 2010 12:02 On 08/10/2010 08:29 AM, ashu wrote: > Dear All, > I am trying to brush up my communication theory fundamentals. I have > simple(perhaps foolish!) question on intutive understanding of CDMA. > > Shannon says that the capacity of a channel is directly proportional > to bandwidth of signal. In other others for a given SNR if increasing > B would give me more capacity. Yes, but that's maybe not a fair comparison, because keeping SNR constant and increasing bandwidth generally means that you're increasing power. The power capability of your transmitter directly relates both to cost of acquisition and cost of operation, and increasing the bandwidth increases the opportunity cost for _someone_, if not for you. So holding SNR constant and increasing bandwidth is increasing the overall cost of your system twice. > DSSS CDMA uses the same principle to > spread the symbol over a large bandwidth by multiplying it with narrow > time pulses(chips). Not really. Spread spectrum is generally presented as holding the power constant and increasing the bandwidth through coding. The SNR at any one frequency goes _down_ as spreading goes up, it certainly doesn't stay constant. > Can we intutively say therefore, that when we demodulate our spreaded > signal by remultiplying it with the PN sequene and then averaging it > over symbol period, we are ineffect using the statistical property > that noise over a large period(symbol period) will cancel out itself. Well, yes, but you can say that about any simple signal at a given symbol rate. Take white noise, feed it into an ideal spreadspectrum front end with gain = 1, and what comes out is white noise that's indistinguishable from what you had to start with. Take white noise plus a spread signal, feed it into that same front end (assume that synchronization has happened), and you get white noise plus the original, unspread signal. So if you use an ideal system to spread, and then despread, the signal you don't change its noise properties, or error rates, or anything. > In other words the larger averaging interval we have(symbol period) > the more we lower noise, since noise is inherently random and > gaussian. In other words, spread spectrum does _nothing_ to the signal characteristics you're struggling with. Lowering the symbol rate increases the averaging interval, sure  but that'll happen with or without spreading. What any of the multiple access systems do is to cleverly take a wide bandwidth channel (i.e. whatever the assignment is for that particular service) with highpower potential (because you've got great big base stations, and all those phones), and find ways to gracefully apportion that channel capacity amongst a whole bunch of base stations and phones. They all do this by making the signals to and from any phones in a given area mutually orthogonal. Spread spectrum is just one way among many, that has some strong advantages and some strong disadvantages.  Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
From: Vladimir Vassilevsky on 10 Aug 2010 12:04 ashu wrote: > Dear All, > I am trying to brush up my communication theory fundamentals. I have > simple(perhaps foolish!) question on intutive understanding of CDMA. > Shannon says that the capacity of a channel is directly proportional > to bandwidth of signal. In other others for a given SNR if increasing > B would give me more capacity. Yes. > DSSS CDMA uses the same principle to > spread the symbol over a large bandwidth by multiplying it with narrow > time pulses(chips). No. Just spreading a signal does not make for any gain in the Shannon theorem sense. > Can we intutively say therefore, that when we demodulate our spreaded > signal by remultiplying it with the PN sequene and then averaging it > over symbol period, we are ineffect using the statistical property > that noise over a large period(symbol period) will cancel out itself. No. There is no difference if you are averaging over a spreading sequence or over a constant value per symbol. > In other words the larger averaging interval we have(symbol period) > the more we lower noise, since noise is inherently random and > gaussian. > > Can DSSSCDMA be looked at from this perspective? Spread spectrum, CDMA and bandwidth expansion to gain the capacity are the different means to attain entirely different goals. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
From: dvsarwate on 10 Aug 2010 14:33 On Aug 10, 10:29 am, ashu <ashutosh.ghildi...(a)gmail.com> averred: > Shannon says that the capacity of a channel is directly proportional > to bandwidth of signal. I don't believe Shannon (or most of those who followed him) ever said that. The capacity of a channel is *not* ***directly proportional**** to the bandwidth of the signal; it is not even directly proportional to the bandwidth of the channel. The channel capacity increases with bandwidth but a law of diminishing returns sets in and the capacity approaches a finite limit (value depending on the ratio of signal power to noise density, which ratio is assumed to be fixed) as bandwidth goes to infinity. Dilip Sarwate
From: Tim Wescott on 10 Aug 2010 16:30 On 08/10/2010 11:33 AM, dvsarwate wrote: > On Aug 10, 10:29 am, ashu<ashutosh.ghildi...(a)gmail.com> averred: > >> Shannon says that the capacity of a channel is directly proportional >> to bandwidth of signal. > > I don't believe Shannon (or most of those who followed > him) ever said that. The capacity of a channel is *not* > ***directly proportional**** to the bandwidth of the signal; > it is not even directly proportional to the bandwidth of the > channel. The channel capacity increases with bandwidth > but a law of diminishing returns sets in and the capacity > approaches a finite limit (value depending on the ratio > of signal power to noise density, which ratio is assumed > to be fixed) as bandwidth goes to infinity. Actually the way he framed it was for a channel with a fixed SNR, in which case the channel capacity _would_ be proportional to bandwidth. But the "fixed SNR" channel isn't a very representative model, as I pointed out in my response.  Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html

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