On Jul 16, 11:31 am, John Fields <jfie...@austininstruments.com>
wrote:
> On Sun, 15 Jul 2007 14:57:17 -0700, Keith Dysart
> <Keith.Dys...@gmail.com> wrote:
> >I thought the experiment being discussed was one where the
> >modulation was 1e5, the carrier 1e6 and the resulting
> >spectrum .9e6, 1e6 and 1.1e6.
>
> ---
> That was my understanding, and is why I was surprised when you made
> the claim, above:
>
> "It does not matter how the .9e6, 1.0e6 and 1.1e6 are put into
> the resulting signal. One can multiply 1e6 by 1e5 with a DC
> offset, or one can add .9e6, 1.0e6 and 1.1e6. The resulting
> signal is identical."
>
> which I interpret to mean that three unrelated signals occupying
> those spectral positions were identical to three signals occupying
> the same spectral locations, but which were created by heterodyning.
>
> Are you now saying that wasn't your claim?
> ---

No, that was indeed the claim. As a demonstration, I've
attached a variant of your original LTspice simulation.
Plot Vprod and Vsum. They are on top of each other.
Plot the FFT for each. They are indistinguishable.

> >Read my comments in that context, or just ignore them if
> >that context is not of interst.
>
> ---
> What I'd prefer to do is point out that if your comments were based
> on the concept that the signals obtained by mixing are identical to
> those obtained by adding, then the concept is flawed.

See the simulation results.

> >I did not write clearly enough. The three resistors I had
> >in mind were: one to each voltage source and one to ground.
>
> >To get there from your latest schematic, discard the op-amp
> >and tie the right end of R3 to ground.
>
> That really doesn't change anything, since no real addition will be
> occurring. Consider:
>
> f1>---[1000R]--+-->E2
> |
> f2>---[1000R]--+
> |
> f3>---[1000R]--+
> |
> [1000R]
> |
> GND>-----------+
<snip>
>
> Note that 0.75V is not equal to 1V + 1V + 1V. ;)

E2 = (V1+V2+V3)/4 -- a scaled sum

Except for scaling, the result is the sum of the inputs.

> >To get an AM signal that can be decoded with an envelope
> >detector, V5 needs to have an amplitude of at least 2 volts.
>
> ---
> Ever heard of galena? Or selenium? Or a precision rectifier?

Oh, yes. And cat whiskers too.

But that was not my point. Because the carrier level was not
high enough, the envelope was no longer a replica of the signal
so an envelope detector would not be able to recover the signal
(no matter how sensitive it was).

....Keith

Version 4
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TEXT -1592 -560 Left 0 !.tran 0 .02 0 .3e-7