POTS question

I'd like to hear some discussion from people here on the following question, which has been bothering me for some time.

In the normal POTS ( plain old telephone system ) , in the past it was noted that the frequency response fell off after about 3 khz. It was used only for voice and low speed fax, and any attempts to do more would just roll off and phase distort and was doomed to failure, Now, my dialup can handle 56 Kbaud. This means at least at 26 khz bandwidth with low phase distortion, and probly more, depending on how low the BER needs to be....

So, have the POTS lines been upgraded or was the system capable of doing this all along ???????????

Now, I understand M-ary signalling, data modems, and multiphase encoding.... However, none of that seems to account for a 3khz line suddenly being able to handle a 26khz min bandwidth. I expect to hear several different explanations, and I'm interested in all of them. The gurus of data , and bandwidth, and Fourier series are all asked to give the explanation their own particular spin on this..

Thank a lot , guys...... I am always interested in learning new ways

to account for things that I "thought" I understood....

Andy in Eureka, Texas (retired engineer )

Reply to
Andy
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The signal goes from your phone to a line card at the switch. The line card converts it to digital at an 8kHz sampling rate. Each digital sample is 8 bits, but the 8th bit gets stolen periodically for signaling, leaving 7 reliable bits.

AFAIK the 56kBaud is only downstream to you: upstream is at a lower rate. Your ISP talks to a T1 line with straight digital, and your modem is smart enough to figure out the inverse of the reconstruction filter used at the line card, as well as the effects of the line from the line card to you (get far enough out in the boonies and this won't work anymore, by the way). Your modem 'dealiases' it, decompresses it, synchronizes to the 8kHz, and happily samples straight binary.

And there you are. It's really magic, but they put it in a brushed aluminum case so you'd think it was technology.

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Tim Wescott
Wescott Design Services
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Reply to
Tim Wescott

Its the BAUD rate (much lower than the BIT rate), that has to fit the bandwidth. Then there is the band-depth. Information theory sets the maximum data transfer, and 56K is just about as predicted. The trouble for years was that information theory says how much data you can send each second, but does not say how to actually do it. Hence, those old 300/1200 baud modems.

Luhan

Reply to
Luhan

Your statement of a path needing 26Khz bandwidth to send a 56Kbaud signal has two problems:

First, your data modem sends at 56K bits per second, not baud. Baud is the symbol rate, and for a 56Kbps modem the baud rate is 2.4Kbaud (iirc). The

3KHz analog bandwidth of the phone system is easily cabable of sending 2400 symbols (each with its own amplitude and relative phase) per second.

Second, I think you're confusing the so-called Nyquist rate with data transmission. Nyquist merely says that you need twice the sample rate to reproduce a given sinusoid. For example, you must sample at at least 112K words per second to be able to reproduce a 56KHz sine wave.

For your 56Kbps modem, they probably represent 32 bits with every symbol, for a gross rate of 76.8Kbps. After you remove the overhead you'll get your

56Kbps. I'm only assuming this. Some modem expert can give you the exact details how this particular modem works, but the general concept is correct.

Take a look at this:

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Bob

Reply to
Bob

I can understand Frequency Shift Keying sending 1 bit per symbol at 300 symbols per second, Phase Shift Keying sending 2 bits per symbol at 600 symbols per sec, and I think I understand Quadrature Amplitude Modulation sends a bit in amplitude and a bit in frequency at 600 symbols per sec, but after that things get a little foggy. Evidently the constellation has 2 bits in amplitude and 2 bits in freq, so we are sending 4 bits per baud. 8K sample rate and a 4K bandwidth line. So thats 16 x 4K=64K bps??

Reply to
BobG

After thinking about this, there's no way that each symbol could represent

32 bits because the encoding scheme, at the phone company, only uses 8 bits per sample (at 8K samples per second).

I think for lower speed modems, like 19.2Kbps, that 8 bits for each symbol would work (at 2400 baud), but I really don't understand how the faster modems operate. It's an enigma wrapped in a riddle...

I'll try to get some more info, then I'll run it up-the-flagpole and see if anybody salutes.

Bob

Reply to
Bob

The thing you are forgetting is that the POTs line transfers signals with both frequency and amplitude, not just frequency alone. As explained in the other posts using amplitude as well as frequency allows the building of a constellation of digital symbols. This allows a number of bits per hertz increasing the communication efficiency.

But, 56K ain't nothin'. DSL uses the POTS lines and transferes frequency to

1.1MHZ! It does so by using the twisted pair lines and bypassing the filters, amplifiers and telephone grade equipment in the POTS system. Furthermore it doesn't use the low band stuff below 30kHz allowing simultaneous use of the POTS system for phone, fax, etc.
Reply to
Bob Eld

56kb/s dialup connections use V.90

V.90 uses direct digital communication downstream (if available) and can communicate at *up to* 56Kb/s. The upstream direction is usually simply V.34+.

Note that if A-D conversions are done in the loop downstream, the modem will connect at V.34+ (33.6Kb/s in both directions, max).

V.34 is a 33.6kb/s scheme (using QAM) and 33.6Kb/s is close to the theoretical limit due to the noise floor. (Most V34+ connections will be at less than this due to autoretraining because of high BER). As far as I recall, V.34 uses QAM256 ( 8 bits per symbol).

There is an excellent overview of the various modem standards at:

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A detailed explanation (well, not that detailed) at [with an eye to the

3Com extensions:

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QAM, for the uninformed, is Quadrature Amplitude Modulation. Here, we have a Sine (reference) and Cosine (90 degrees behind) signal generated from a reference. Each signal takes an amplitude such that there are

2^^n possible combinations for n bits per symbol. For 8 bits per symbol, there are 256 possible combination (2 ^^ 8) made up of +/- 64 levels of each sine and cosine amplitude modulators. The two signals are then combined (without going into detail, it is fairly simple to separate sine and cosine components from a signal) for each transmitted symbol (well, simplified anyway).

When viewing such things on a scope, you see what is known as a QAM constellation diagram.

QAM was chosen in the telephone network for it's superior Signal to noise (without going to exotic extremes) ratio vs. other modulation techniques.

There are a large number of modulation schemes available (DPSK is used at 1200 b/s for instance), but I won't go into those here.

Excellent article on QAM at Wikipedia:

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Cheers

PeteS

Reply to
PeterSmith1954

The last time I worked with T1s, no data bits were stolen for signaling at all. The frame bits (bit 193 of a 24 channel frame) are not all required for framing and now most of those are used for overhead channel signalling. As of 2001, we used the 4th, 6th or 8th frame bits (depending on configuration) for actual framing, leaving the others available for control signalling.

Cheers

PeteS

Reply to
PeterSmith1954

Bob wrote:>

Andy writes: I'd like to disagree with this. At 56K BITS per second, the highest frequency present will be alternating "1s" and "0s". That would be a square wave at 28 Khz..... This can be transmitted with a 28Khz bandwidth and reconstructed with a limiter later, if a square wave is desired..... For this, 56Khz would be the second harmonic and this frequency component would not be present (first Fourier null).

Any other bit pattern would have lower frequency baseband components than 28 khz...

For instance, a single "1" surrounded by a million "0s" would have it's first Fourier null at 56Khz, which means the Fourier component would be at 28 Khz (for repetitive signals), which again says that most of the energy is at 28 Khz. The next component would be 3 X28 or 84 Khz, and not having it isn't that important if a limiter is used.

Andy

PS ( if I said anything other than 28Khz before, it was obviously a typo. sorry.... sometimes they get by me )

Reply to
AndyS

=========================================== This would be true if the modem was actually transmitting the 1s and

0s, but it is sending thousands of 'symbols' per sec, each symbol has 8 bits encoded into a waveform with one of 4 amplitudes and one of 4 phases... it probably looks like noise... it sure sounds like it!
Reply to
BobG

:: it probably looks like noise... it sure sounds like it! ::

Well, noise has the same definition as information, after all ;)

Cheers

PeteS

Reply to
PeterSmith1954

"Andy" <

** Groper alert !

** WRONG !!

Any normal " POTS " voice link has only 3.4 kHz bandwidth ( exchange to exchange).

The bandwidth of a communications link is just one parameter governing the amount of data that can be transferred - the signal to noise ratio is the other !!!

Think of it this way:

If an analogue signal with a given bandwidth and s/n ratio can be converted into an exactly matching digital data stream, then the reverse is also possible.

EG:

To convert a 3.4 kHz bandwidth signal with a 200:1 ( 46dB ) signal to noise ratio into PCM requires a sample rate of 7 Khz and 8 bits.

7000 x 8 = 56,000.

Go see Shannon's Theorem for the finer points.

....... Phil

Reply to
Phil Allison

Andy,

The baseband digital bitsream isn't being sent over the POTS channel. If it were then you'd be correct -- you'd only be able to send a very low data rate. However, sending digital data (i.e, only having two amplitude levels) over an analog channel (i.e., capable of MANY amplitude levels) would be a very inefficient use of the analog channel.

The original digital signal, before it ever hits the POTS channel, is modulated by the modem. The modem converts this high transition rate/two-level signal into a low transition rate/multi-level (and phase) signal. The modem does the inverse process once the signal passes through the entire POTS channel.

If this process weren't necessary then MODEMs would never have been invented.

Bob

Reply to
Bob

Andy comments:

My thanks to all who replieds to my question on the POTS line and dialup data rate.

I guess I need to pull out some modulation app notes to fully understand what various replies mean.... It seems to me that there are different ways of stating the solution, as each response seems to open another avenue of thought....

I had hoped the answer was simple --- that I was overlooking some trivial little thing that would make everything drop into focus.... Well, I'm just going to open another beer and put this learning exercise on my schedule (grin)....

Andy

Reply to
AndyS

One lovely 'perpetual' confusion that this thread illustrates, is the confusion between bps, and baud. The fastest telephone modems, are still generally 2400baud devices on the line. It is just that they use amplitude and phase shifts, to encode more bits per baud (usually combined with trellis encoding to avoid things like AGC's interfering with the data). 24 bits/baud, is the norm for the faster modems. The same unit, can be

2400baud on the side that talks to the 'line', and be using 57600baud on the side talking to the computer, yet the data rate on both 'sides', is basically the same. Unfortunately, there is a tendency, to use bps, and baud interchangeably (they only are on on a simple 'two level' line interface). On anything with more complex encoding, the two terms must not be confused.

Best Wishes

Reply to
Roger Hamlett

For a basic explanation of Shannon-Hartley theorem see:

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In v.34 and later modems the data stream is compressed at a little over 2:1 so that the actual data rate is about 28k (within the Nyquist limit (b)).

The signal is NOT sampled at 8kHz, but at 56kHz with a CVSD and collected into 7-bits of data, 1 bit of signaling is added and this makes a DS-0 at

64kbit/s signal. See also u-Law and A-Law compander specifications.
--
JosephKK
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Reply to
joseph2k

The signal is *definitely* sampled 8000 times per second. Each sample is coded to 8-bits, and this gives us the DS0 rate of 64kbit/s.

Don

Reply to
Don Bowey

2:1

I did not come up with this information out of thin air, CVSD's and other oversampling systems including filers, sigma squared TDM to FDM and back converters all rely on the properties of oversampled systems. Indeed, all the popular cell phone vodec's (VSELP, CEPT, et.,) use oversampling. Even early CD players used oversampling's concomittant noise shaping to reduce audible noise in the playback signal.

Conversion is done by a 56kHz CVSD or a 16 bit 56 kSample/sec equivalent transcoded to be the same. It makes the analog filters soo much easier to implement. The undesired out of band signals get aliased way out of band and disappear in the analog filters or limited human perception.

Try CVSD in Wikipedia and Google to learn more.

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JosephKK
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Reply to
joseph2k

What has all this to do with POTS and a DS0?

Don

Reply to
Don Bowey

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