Re: Is this web page wrong?

Gegroet,

DAB sounds worse than FM schreef:
The reason I decided to post when I realised I hadn't replied was
that
you were trying to lecture me on a subject that I understand far
better than you do.
It's you who started this sub-thread for "I know digital technology
better then you!!! Look at me, I've got me university-degrees then
everybody else here so you should all bow to me".

You're the one who said things like "oh look at what Steve looked up
on Google".

???

And the link between having university degrees and using google is what
exactly?




Anycase, that's all exactly what I said, isn't it: sampling
resolution
-> quantisation-noise -> noise-floor -> S/N ratio.
Sampling resolution?
So much for the value of a MSc. (grin)
I'll leave it to you to guess what this means. :-)
No idea what you mean there. I put a question mark after "sampling
resolution", because you seem to be mixing up the terms "sampling
frequency" and "sample bit-depth", which refers to the resolution of a
sample.

OK, we're almost getting there. You're down to only two choices. Which
one do you think it is?


It's a dead giveaway that someone doesn't understand a subject when
they use terminology incorrectly like you have done throughout this
sub-thread.

But somebody with that many university-degrees can of course look
through that and know immediately what the other person means, doesn't he.





First an answer to your query: the connection between the line-card
in
the LEX where your PSTN-line is connected and that of your mate's is
a
64 Kbps data-connection.

Now, what you are saying that it is possible to have a 65 Kbps
connection between your modem and that line-card, then a 64 Kbps
connection from your line-card to the one on the other side, and
then
-again- on the other side a data-connection at 65 Kbps on the other
side
???

I thought you were implying that it's impossible to transmit data over
a telephone wire (with sub 3.4 kHz bandwidth) at a bit rate greater
than 64 kbps.

Yes .. I'm implying it's impossible to transport anything over 64 Kbps
over a telephone-CIRCUIT (end to end, between the two handsets/modems,
including the 64 Kbps PCM circuit between the line-cards at both ends),
i.e. what most people use.

Or are you really suggesting that you can transmit speeds over 65 Kbps
over an end-to-end 64kbps telephone-line???



I do not know why now talk about a telephone-wire?

AFAIK, most modem-connections still involve more then the wire between
you and the LEX (even if they are connected to a remote LDC or a
voip-module in a street-cabinet).





HELLLOOOOO ?????
Hello.

REALITY-CHECK!!!
Reality checked.

To be honest. I doubt it.




If you do that, you've created the perpitule mobile of digital
bandwidth;
!

Silly boy.

OK, I was wrong. It's perpituum mobile. (my latin is kind of rusty :-) )



And one of the other things you learn, is the importance of
syncronisation in a digital communication path!!!

Where did I say synchronisation wasn't important? Digital
communications receivers would be useless if they didn't synchronise
with the signal they're trying to receive.

Right. I see you start to understand it.



Here's your description of how 56k modems worked:

"56 Kbps modems use a very different approach. Basically, it is very
simple. It goes like this: you take 7 bits, you stuff a "1" in front
of
it to make up a byte; and "translate" this value into a voltage using
the PCM table. You sent that voltage over the line to the other side."

Pure bollocks. The crap about taking 7 bits and then putting a 1 in
front would mean that they'd have to use a signal constellation with
256 points instead of 128 points, therefore increasing the required
SNR at the receiver. Pure horse***.

Well, if you google a little bit further, you'll find the answer for
that too.

Good luck! Happy hunting! :-)


I do have to make a correction. The process is not just to stuff the "1"
in front of the 7 bits.
As PCM has both positive and negative voltages (a negative voltage
represented by a value of over 128), the process to use only the highest
voltages (both positive and negative) is a bit more complex then just
stuffing a "1" in front of it.


OK, I guess then CW should then also be called asyncronous PAM, but
still everybody calls it CW. :-)

http://en.wikipedia.org/wiki/Continuous_wave
"Continuous wave is also the name given to an early method of radio
transmission, in which a carrier wave is switched on and off."

If CW were taught on a digital communications course at uni (it
wouldn't be, because it's a useless modulation scheme), (...)


Grin.

I hope no hams will be reading this, or you're in a big trouble for a
remark like that.


... its signal
constellation would look like this:

-------x--------x
0 A

It is an example of binary PAM, with one signal point having an
amplitude of zero, and the other point having an amplitude of A, which
is the amplitude of the CW wave.

So, exactly as I said. "You can call it PAM, but everybody still calls
it CW."
:-)


That's why one of the first articles about 56 Kbps modems was
called
"breaking the Nyquest limit".
Wrong. 56k modems obey Nyquist's sampling theorem.
Then why do they work if you have more then one D/A->A/D
conversion?
If 56K modems are like other modems, then why do they not work at
56K
when you connect both sides of the modem-link to a PSTN-line?
If they're unable to reach 56k in practice, then it'll be due to
the
SNR being too low on actual telephone lines.

But the strange thing is that there has been absolutely no change in
the
physical phone-line, and therefor not it characteristics (like the
S/N
ratio) neither.

So SNR was irrelevant? Pure bull***.

The S/N ratio of the line is irrelevant in this comparison, as it's
exactly the same in both cases.


So the question remains:
You said "Wrong. 56k modems obey Nyquist's sampling theorem."
Then why do they work if you have more then one D/A->A/D conversion?
If 56K modems are like other modems, then why do they not work at
56K
when you connect both sides of the modem-link to a PSTN-line?

It will be to do with the SNR, and definitely nothign to do with
Nyquist's theorem.

Each time you to an A/D conversion, you first have to pass the signal
through amplifiers and lowpass filters, and they add noise to the
signal, and adding noise to the signal degrades the SNR. And you can't
then get rid of the noise that's added, because if you then amplify
the signal you amplify the noise that's been added as well.

OK, sounds all very nice, but not good enough!

As I noted earlier, your problem is that what you are saying does not
match up with how it seams to be in real life.

A dialup NAS of an ISP is connected directly to the telephony-system
using digital PRI-lines; and even under these surcumstances, a 56K modem
will offer higher bitrates then when using pre-56K modems.

So, it cannot be that the only difference is only the noice-level as
that is equal is both cases.





Why can you not transport a 56Kbps baseband signal over a PSTN-line,
except when one side is digitally connected to the telephony-switch.
You just said above that PAM is a baseband modulation-sceme.
I didn't say that at all. ...
Please do not let me have to point to your message in
news.google.com.

No, PLEASE DO point out the post in which I said this, because I would
NEVER say that PAM can only be used on baseband systems, because it
obviously can be used on bandpass systems.

Well, I propose you do the test. Connect two 56K modems to a PSTN-line
and make a call. See if they connect at 56K.
Make a video of it and post it on youtube. :-)

If you manage to get a 56K connection, you've succeeded in doing the
something that the engineers of the modem-manufactors never managed to do.




... You could use PAM on DAB if you wanted to,
but you wouldn't want to because it's more efficient to use
2-dimensional constellations.
But the strange thing is that PAM seams to be able to transport 56
Kbps
over a phone-line, while the 2-dimensional constellations do not, so
it
seams to be MORE efficient, not less!

Consider this QPSK constellation:
http://www.digitalradiotech.co.uk/images/ofdm/QPSK_constellation_q70.jpg
(...)
Why didn't they use QAM for 56k modems? Dunno, ask the designers.

Funny,

Somebody comes with a observation which is in contrast to a theory you
have been proclaming. So what do you do?

You add a lot of fancy formulas to your theory, saying "see, this cannot
be so". (because you're in the middle of a useless "I know it all better
then you" discussion so you cannot admit being wrong).

And in reply to why your theory does not match up with the practice, you
say "ask the stupid engineers who obviously did not know what they where
doing".


Nice try. Doesn't work.

Fact is this: a 56K modems, using PAM, can reach speeds pre-56K modems,
using QAM, cannot.

Wether you are are now going to copy even more formulas or even write
hunderds of pages of text, this fact will not change.

So the question remains: explain why real-life experience does not match
up with your theory.





Have you ever tried to get a baseband signal over a telephone-line
and
get something usefull out of a it?
Yeah, I did this yesterday, as it happens. I was using state of the
art baseband technology, it was called a telephone. I'm not sure if
they're available in Belgium yet.

Gasp.
Do you mean you still use a old PSTN line, instead of ISDN or VOIP?

Actually, my local telephone exchange has been enabled on BT's 21CN
(21st Century Network), so all of my phone calls on normal phones are
via VoIP by default.

Interesting.
Is this in the LEX, RDC or the street-cabinet?

If in the street-cabinet, how do they connect that to the backbone? Over
copper or over fibre?


But, you will not excape the question:
Have you ever managed to transport a baseband modem-signal over a
PSTN
telephone-connection?
At what bitrate?
I repeat, 56k modems used baseband. So WTF are you going on about?

Come on, Steve. Stop playing word-games.

You know every well what this is about; so stop beating about the bush.

Have you ever made a connection between baseband connection (say two 56K
modems), both on a seperate PSTN-line and get 56K (i.e. PAM, AKA
"baseband") connection running?

Can you make a video of that as proof?




BTW, did you know that Ethernet is a baseband technology?
Presumably
you don't think that considering that you don't seem to think that
baseband transmission is able to transmit, well, anything.

Well, let's take ethernet.
Ethernet also only works in a point-to-point (i.e. one D/A-A/D
conversion) configuration too: either you run it through a layer-1
repeater (a hub) or rebuild the packet completely in every hop (in a
switch or a router).

What has any of that got to do with it being a baseband or bandpass
system?

Euh ... try reading the thread again.

56K only works when only one single D/A-A/D conversion
Ethernet like 56K, also one single D/A-A/D
DAB like 56K, also one single D/A-A/D

Pre-56K modems: can operate using multiple D/A-A/D conversions in path.


If a baseband-signal is just like any other signal, why is that?




And, just as DAB ethernet also has syncronisation-bits in the
beginning
of each frame.
Why do you think this is? For exactly the same thing. It's for
syncronisating the demodulator, as it runs at line-speed of the
ethernet-frame.

All digital communications systems use synchronisation - either
explicitly, i.e. by using pilot symbols, such as on DVB-T/H, or the
recevier synchrnonises with the signal by estimating the sync from the
received signal.
Right .. getting there ...


I have NEVER said that digital comms receivers don't try to
synchronise, and don't try to suggest that I have ever said anything
of the sort.

Of course you did not.



Just like was pointed out in the original message to Richard,
ethernet
has the same issues a DAB transmittion-path: only one single D/A-A/D
conversion and syncronisation.

Like the vast majority of digital communication systems. What's your
point? ...

Well, I'll give you a hint!

The issue here is not the syncronisation between the receiver and
transmitter, but the syncronisation between the signal being transported
and the D/A-A/D conversion-process.

I leave it up to you to see if you can find the rest.



... Are you trying to suggest that I don't already know what you've
just said?
I repeat what I've said in a previous post. I studied digital comms on
an MSC. Why do you think I don't know all this? What you're saying
here is ULTRA-basic stuff - it's stuff that's covered in the first few
pages of any digital communicatinos text book.

Yeah. Yeah.
I guess you forgot to read your textbook on "long lines" and
"transmission-lines".

Another hint:

A long wire has an internal capacity, and that affects the signals that
run over it.
Image what that does to a PAM-signal; and how "timing" is involved in this.




And also in ethernet, the encoder and decoder run at the line-speed
of the line!

Could you imagine a digital communication system where the transmitter
and receiver weren't running at the same speed? You would receive a
totally garbled message.

Indeed, now what do you think happens if you have a transmission-line
where a segment is not in sync?



Cheerio! Kr. Bonne.
.