🚀 go-pugleaf

On Sep 29, 8:59 pm, dpl...@radagast.org (Dave Platt) wrote in 
http://groups.google.com/group/uk.tech.digital-tv/msg/872fdf9b0c765bd0 :

 > In article <46fefb70$0$7493$4c368...@roadrunner.com>,
 > Green Xenon [Radium] <gluceg...@excite.com> wrote:
 >
 > > > Sure it is possible. It won't be of any assistance to you (or to
 > > > anyone else since you can't put 8 pounds in a 1-pound sack.
 > > > Bandwidth is bandwidth, no matter how you slice it. Do you
 > > > really think that nobody has thought of various methods of
 > > > transmitting video over low-bandwidth paths? If so, you need
 > > > to study not only science and engineering but the HISTORY
 > > > of science and engineering.
 >
 > >There are 8-bits-per-symbol, only 1 baud. The symbol is split into 8
 > >parts. Each part is 1-bit.
 >
 > >How is this �putting 8 pounds in a 1-pound sack?�
 >
 > >It�s more like �putting 1 pound in a 1-pound sack�
 >
 > You are using the word "split" without defining what you are meaning,
 > and I think you may be using the word "bit" in two different ways (as
 > in "a single binary digit's worth of information" and "a part")...
 > maybe I'm wrong about the latter.
 >
 > Sure, you can carry 8 bits of information per symbol, at a rate of one
 > baud.  That's an information rate of 8 bits per second, which isn't a
 > terribly useful amount of information.
 >
 > What I *think* you are saying is something along the lines of
 >
 >   "If there's no way to carry N bits of information per symbol (where
 >    N is a large number) at a rate of 1 symbol per second, in a
 >    specified bandwidth, over a channel having a certain amount of
 >    dynamic range / signal-to-noise ratio available, then I'd
 >    like to somehow divide each symbol into eight sub-symbols which
 >    each carry N/8 of the information, but which still make up only one
 >    symbol per second.  My hope is that these eight sub-symbols could
 >    be easier to transmit, somehow, than a single symbol carrying N
 >    bits, and that I could thus transmit them more easily but without
 >    using any more bandwidth.  Can I do that?"
 >
 > If that's what you're asking, the answer is "no".
 >
 > Once you hit the theoretical information limit of the communication
 > channel (which is set by the bandwidth you use and by the amount of
 > noise on the channel) you can't do any better than that.  No matter
 > how you modulate (changing the baud rate, the amount of information
 > per symbol, etc.) you can't do better than this.
 >
 > And, as we've been trying to make clear, your goal is quite far beyond
 > the theoretical limit.  You just can't get there.  You're trying to
 > put all of Lake Erie in a water-glass.
 >
 > Let it go, Radium.

Within physical-possibility, what is the largest amount of 
bits-per-symbol [assuming a baud-rate of only 1-bit-per-symbol] that can 
be reached without the highest-voltage causing any clipping, generating 
any temperatures above 70 Fahrenheit, resulting in any harm to 
anyone/anything [including the equipment itself], or shortening the life 
of the equipment and without the lowest-voltage being lost in the noise? 
What is the maximum-possible amount of discrete levels between the 
highest and lowest voltage in such a signal?

An 8-bit signal can have a maximum of 256 different voltage levels 
between the highest and lowest voltage. Right? Go too high and the 
signal clips, go too low and the signal will not be recognized.