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From: Don Pearce
Subject: Re: How does a mixer work?
Date: Sat, 19 Oct 2002 08:14:00 +0430
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On 18 Oct 2002 16:58:45 GMT, email@example.com (Dbowey) wrote:
><< A carrier of varying strength IS an AM signal. The circuit above will
>give an output which comprises a carrier and a pair of sidebands at
>the pot variation rate, just like any AM signal. >>
>Michael's explanation of the effects of using a pot was totally wrong and you
>fell right into it. As drawn, there will be no output at all. But if we
>assume the schematic as it was probably intended (output from ratio arm and
>ground on the the right end of pot) then........ If you observe the output in
>the frequency domain you will observe no sidebands, but only a varying DC
>voltage and a varying "modulated carrier op."
>A AM Carrier amplitude does not vary. It only appears to when observed in the
I think we had better reinstate that diagram.
> modulated carrier op
The modulated carrier output level will indeed be proportional to the
position of the pot wiper. All the way to the left, it will be full
amplitude and all the way to the right it will be zero amplitude. In
between it will have an intermediate amplitude. There is no DC voltage
anywhere in this diagram to be varied, just the carrier.
If the pot is moved back and forth, the amplitude of the signal will
go up and down (surprise, surprise, this is what is meant by amplitude
modulation). In the time domain the amplitude varies exactly as we are
demanding it should with the pot. When viewed in the frequency domain
we see the original carrier plus a pair of sidebands, one each side.
I'm afraid that if you need to be convinced of this, you will have to
do the experiment for yourself, but fact it most certainly is.
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