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Reply-To: "Kevin Aylward"
From: "Kevin Aylward"
Subject: Re: How does a mixer work?
X-Newsreader: Microsoft Outlook Express 6.00.2800.1106
Date: Tue, 22 Oct 2002 13:53:16 +0100
NNTP-Posting-Date: Tue, 22 Oct 2002 13:53:23 BST
"Gibbo" wrote in message
> "Kevin Aylward" email@example.com wrote:
> >"Fred Bloggs" wrote in message
> >> Chuck Simmons wrote:
> >> >
> >> > At this point I take a stand against the hidebound
> >> > is the application of mathematics to engineering. I am very
> >sensitive to
> >> > this because I have no engineering education whatever and ALL
> >> > mathematical statements engineers make can be considered suspect
> >> > of the hidebound conventionalism.
> >> >
> >> > Consider the equation:
> >> >
> >> >
> >> >
> >> > This is really an identity. The left hand side says that the
> >> > varies in amplitude according the the operations on the field of
> >> > numbers. The second says that, fully equivalently, there are
> >> > constant frequencies and amplitudes.
> >> >
> >> > It is pure hidebound conventionalism that says that though the
> >> > of the equation are correct, the right hand side is more correct.
> >> > mathematical hogwash. There can be no justification for that sort
> >> > narrow thinking.
> >> If the early developers adopted your perspective, there would be no
> >> sideband communication.
> >I do actually remember some stories like this, vaguely. Apparently,
> >people were sceptical that sidebands were real, and that if you had
> >really sharp filter you could still recover the modulation. This was,
> >understand even experimentally "proven". In reality, they just did
> >use a sharp enough filter.
> Just to throw a spanner in the works......
> One side says "the carrier alone carries no information".
> That's not strictly true. If you didn't know you needed to re-insert
> carrier at the receiving end you'd never be able to receive the
The zero information is really a result of the technical definition of
Shannon information. That is, the information content is defined by
log2(1/p), where p is the probability of the message occurring. In this
sense, there can't be any argument, the carrier has a probability of 1,
hence zero information.
>Now some people will say "but we already do know that we need to
> re-insert the carrier". Well... that's like saying "we already know
> the sidebands, so we know what information is in them, so we don't
> so they don't carry any information either"
Well, this is not really a valid analogy. In the case of the carrier the
presumption is that it is always the same. With either sidebands the
assumption is that they are always changing. If the sidebands were
constant, then, by the definition of information they would also carry
> The carrier does carry some information. It carries the information of
> the carrier that you need to insert in order to demodulate the
But this is not relevent with regards to its Shannon informaton content.
A carrier that is always on is P=1
> If you insist in filtering the carrier with an almost infinitely
> to prove that it is useless then try the same thing with one of the
> and see how much information you can get out of it.
But this misses the point. A channel of zero BW cannot carry any
information, i.e. H = BW.log(1+S/N) bits/sec. By assumption, the
sidebands are varying, or obvously, they dont carry an information.
> No flames please :)
With all due respect here, I see the issue here is that many are looking
at this from a rather elementary view. Information theory is one of
those high brow disciplines that you can not really understand by
multiplying steady state sinewaves as you do in EE 101. Its all about
random signals and the probability of detecting them. One really needs
to look at this from more advanced standpoint.
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.
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