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NNTP-Posting-Date: Tue, 22 Oct 2002 10:29:28 -0500
From: firstname.lastname@example.org (John Fields)
Subject: Re: How does a mixer work?
Date: Tue, 22 Oct 2002 15:28:12 GMT
Organization: Austin Instruments, Inc.
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On Tue, 22 Oct 2002 12:10:10 GMT, Fred Bloggs
>This is not quite true. The carrier conveys no information but it was
>necessary for the demodulation. Note that in the case of a tone
>transmission, you have sin((Wc+Wm)t) in the time-domain. This is a
>frequency-shift and not an amplitude modulation, the power level is
>constant. The challenge for the receiver circuits is then to convert
>this into something in the time-domain with an amplitude fluctuation
>corresponding to the modulation. This necessitates carrier restoration
>and another AM modulator.
That's quite untrue.
Consider the case of an ideal unmodulated carrier: only a single
spectral line will be generated and no sidebands will exist. Now
modulate the carrier and sidebands will be generated which will be
displaced from the carrier frequency by the modulating frequency and
with an amplitude varying with the depth of modulation. For a 1kHz
tone modulating the carrier 100% the sidebands will appear at 1kHz
on either side of the carrier, and the power in each sideband will
be 6dB down from the power contained in the carrier. With the same
tone modulating the carrier 50%, the sidebands will be down another
6dB each, so it's clear that modulating the carrier causes sidebands
to appear with a frequency displacement equal to the modulating
frequency and a power content in the sidebands due to the depth of
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