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From: Fred Bloggs
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Subject: Re: How does a mixer work?
References: <firstname.lastname@example.org> <3DB2E3CE.email@example.com> <firstname.lastname@example.org> <3DB41488.email@example.com> <3DB43497.A32B8380@webaccess.net>
Date: Fri, 25 Oct 2002 14:29:42 GMT
NNTP-Posting-Date: Fri, 25 Oct 2002 07:29:42 PDT
Organization: EarthLink Inc. -- http://www.EarthLink.net
Chuck Simmons wrote:
> Fred Bloggs wrote:
>>John Fields wrote:
>>>On Sun, 20 Oct 2002 17:11:22 GMT, Fred Bloggs
>>>>>Don Pearce wrote:
>>>>>You are still assuming incorrectly that the Carrier of an AM signal varies with
>>>>>modulation. It does NOT; the carrier amplitude is easily proved to be constant
>>>>>with correct levels of modulation. In your experiment you are varying the
>>>>>amplitude of the carrier and that is not an AM signal.
>>>The incorrect assumption is yours. If there was no variation in the
>>>amplitude of the carrier when it was being modulated what would you
>>>postulate would be the mechanism involved in generating the
>>>sidebands? It's called Amplitude Modulation for a reason, I
>>>believe, that reason being that the carrier amplitude is caused to
>>>vary by the modulating source.
>>>Perhaps what you're thinking about is the _average_ (instead of the
>>>instantaneous) amplitude of the carrier, which doesn't vary with
>>>modulation, since the peaks and the valleys average out. Consider
>>>the case of 100% modulation, for instance. With no modulating
>>>source the carrier amplitude will be at some nominal value, while
>>>with the modulating source at a minima the amplitude of the carrier
>>>will be zero, while with the modulating source at a maxima the
>>>amplitude of the carrier will be twice what it is in the quiescent
>>>>Right, AM is usually of the form (1+cos(Wm x t))* sin(Wc x t) and this
>>>>expands to sin(Wc x t) + [sin( (Wc+Wm) x t) + sin( (Wc-Wm) x t)]/2 which
>>>>is easily recognized as the carrier, of constant amplitude, and two
>>>>sidebands displaced from carrier by Wm, and this is exactly how AM is
>>>>implemented. So, you are right, the pot analogy is hopelessly flawed.
>>>The pot analogy is perfect and is precisely _how_ AM is generated.
>>>_You've_ also fallen into the trap of looking at average instead of
>>>instantaneous carrier amplitude during _amplitude_ modulation. As
>>>I noted previously, if the amplitude of the carrier wasn't varied
>>>there would be no way to impress information on it with the
>>>resultant sidebands being generated. We're taking AM here, Bloggsy,
>>>not FM or PM or any other kind of M, so don't try to wiggle out of
>>>it, you're just _wrong_. Own up to it.
>>You're a dumbass who knows absolutely nothing about frequency analysis
>>of any kind. The AM equation is clear to anyone with even a second year
>>level of engineering education in Fourier analysis- which you do not
>>possess. The composite signal consists of a constant amplitude carrier
>>with sidebands on either side displaced by the center frequency of the
>>bandlimited modulation function. It's a sorry commentary on your
>>pathetic level of development that after 40 years you still do not
>>comprehend the idea of superposition. Well why would you? Your work
>>experience has required absolutely nothing beyond the most basic hands on-)
> Let's consider a different modulation function. After all, modulating a
> carrier with sinusoids gats boring after a while.
> Consider p(w1*t)sin(w2*t) where w2 is the carrier and p(w1*t) is defined
> P(w1*t)=1 whenever 2*n*k <= w1*t < (2*n+1)*k
> P(w1*t)=0 whenever (2*n+1)*k <= w1*t < (2*n+2)*k
> where k is a fixed arbitrary constant and n is any integer. A square
> wave actually.
> In this case, the carrier turns on and off with the frequency depending
> on the selection of the constant k. This is an interesting case because
> what you see on your spectrum analyzer depends on when and how much data
> the spectrum analyzer collects. The carrier is not constant amplitude at
> all and you cannot make it constant by simply saying it is constant.
> Now suppose we take the Fourier transform of the output of the
> modulator. The spectrum will be the constant carrier with the usual
> sidebands this time forming an infinite sequence that drops off as 1/n.
> What is the real (physical) situation here? You have two choices here.
> Both representations being correct or neither. Which way is it? If I
> look at it one way, I see a well defined constant frequency turn on and
> off. If i look at it the other way, I see a spectrum of sum and
> difference frequencies with the carrier in the middle.
> What is going on here is that the spectrum shows you the average of the
> output of the modulator. If we are interested in the behavior in time of
> the modulator, the spectrum is nearly worthless and we must fall back to
> the carrier turning on and off.
You keep going back to the either-or argument. The time-domain
representation is useful for verifying correct operation of the circuit
like frequency, modulation index, amplitude and so forth. The frequency
spectrum is useful for making other determinations like bandwidth and
spectral power density requirements. You also need to add some reality
to your statistics at least so far as time scale is concerned. I don't
think there is any getting around the fact that carrier power is
constant over any time interval that is long compared to the information
bandwidth. The observation that the carrier turns on and then off is
useless. Your examples are unrealistic extensions of the usual case to
which AM is applied. I checked my 1943 addition of Terman's Radio
Engineers handbook, and the first thing he does is decompose the
classical AM modulation function into the carrier and sideband summation
which he credits to a 1923 IRE paper by Hartley.
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