These guys claim a new modulation method that exceeds the data transmission rate set by the Shannon limit. Any opinions? Opinions on this topic that is, I KNOW you all have considerable opinions on other issues ;)
ChesterW
These guys claim a new modulation method that exceeds the data transmission rate set by the Shannon limit. Any opinions? Opinions on this topic that is, I KNOW you all have considerable opinions on other issues ;)
ChesterW
Probably just another "perpetual motion" fraudulent scheme like this one...
...Jim Thompson
-- | James E.Thompson | mens | | Analog Innovations | et |
On Tue, 28 Jul 2015 16:29:10 -0500, ChesterW Gave us:
Quite doubtful.
Even QAM 256 is hard to beat, and it fills the spectrum nearly as full as is possible.
As I understand things, one can exceed the theoretical maximum given by the noisy channel coding theorem if you're OK with information being lost in the process. Depending on the particulars of the modulation scheme and/or data being transmitted, this may or may not be a problem.
Though it's unlikely that I have the skill in statistical thermodynamics and information theory to do so, it can probably be shown that exceeding the noisy channel capacity with no information loss violates Liouville's thoerem:
which is a big uh-uh - i.e. you're violating the second law of thermodynamics.
I don't see a description of how TM works, so how can you say it doesn't surpass the Shannon limit? I will acknowledge it is unlikely to be real, but until you know what it is, how can you say it is a fraud?
-- Rick
Oh, dear, the orbital angular momentum folks have come to microwave. A few years back the original OAM people were claiming that there was this vast number of completely untapped propagation modes in fibres that would allow some ridiculous gross bandwidth increase.
Then some saner folks published this beautiful letter-to-the-journal-editor that I can't lay my hands on...it showed that if the OAM people were correct, the blackbody radiation from a hot object would go up by a similar ridiculously large factor, which is not observed. The OAM states aren't new-and-different, they're just another basis set for the EM field. Nothing to see here folks, show's over, move along.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant
On Tue, 28 Jul 2015 18:03:42 -0400, Phil Hobbs Gave us:
Even sqpsk is closing in on the limits.
But just when folks thought that optical chip mask burning was near its limit somewhere up around 30nm, somebody stepped in with a new idea and they all do it through a layer of water now and are down near a few nm..
I was involved, as a contractor... until I asked too many questions
...Jim Thompson
-- | James E.Thompson | mens | | Analog Innovations | et |
You might look up the derivation of the Shannon limit, for a start. Not too many assumptions involved.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant
From their white paper:
supports only periodic waveforms.
What can that mean?
-- John Larkin Highland Technology, Inc picosecond timing precision measurement
Immersion litho, multiple exposures, computational mask design, and extreme UV.
Semiconductor manufacturing is ridiculous, but fun.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant
The proof of the Shannon limit is pretty straightforward, and very clear cut:
The paper tries to disprove it for non-periodic signals, but they base their claims on the notion that the Fourier transform doesn't work for non-periodic signals -- which it does, admirably.
At that point my BS detectors started ringing loudly, and I stopped reading. I suspect that if you go deep enough into their paper they'll be making all these claims about their exponential or polynomial functions or whatever, and you'll find that at they're taking some infinitely-long signal in time and lopping it off to finite length -- which will increase the bandwidth, which will blow their argument all to hell, etc., etc.
-- Tim Wescott Wescott Design Services
At home w/ limited download. Quantum mechanically you can "squeeze" states, which if perfect would be 3dB (I think) and in practice much less.
George H.
And what is the information rate of a periodic waveform? Sounds like zero to me.
-- John Larkin Highland Technology, Inc picosecond timing precision measurement
We have exceeded the Shannon limit by transmitting our data with a 100% loss rate! Brilliant! Patent that shit!
On Tue, 28 Jul 2015 20:30:13 -0400, bitrex Gave us:
snip
Hey at least nothing was lost in translation.
Eight different "witnesses" still have trouble describing a zero criminal in the same way though. Even with a ninth "parity witness".
Maybe they achieve it by having a parade of RNGs lined up to provide the "decrypted" code stream... of zeros.
lol Not a job you want to put on your resume.
-- Rick
That's not the question I am asking.
-- Rick
But until you see what they are doing, you can't really say...
I remember a paper many years ago that was looking at data which might show discrepancies from the laws of gravity. Yes, we have done lots and lots of work with those laws and they seem solid enough. But this was taken seriously at the time.
At some point they will have to protect their work with a patent. Then we will all get to see just what they are doing.
-- Rick
Perhaps they left out a hyphen in the name? astra-pi-corp? The ultimate pie-in-the sky?
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