Re: POTS question
- From: "Bob" <nimby1_NEEDSPAM@xxxxxxxxxxxxx>
- Date: Tue, 04 Jul 2006 16:50:35 GMT
"Andy" <andysharpe@xxxxxxxx> wrote in message
news:1152028290.487626.285380@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Andy writes:
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 )
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:
http://en.wikipedia.org/wiki/Baud
Bob
.
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