Reply-To: "fred bartoli"
From: "fred bartoli"
References: <firstname.lastname@example.org> <2ltr9.338$cV6.email@example.com> <firstname.lastname@example.org>
Subject: Re: How does a mixer work?
Date: Fri, 18 Oct 2002 10:15:26 +0200
X-Newsreader: Microsoft Outlook Express 5.00.2314.1300
Organization: Guest of ProXad - France
NNTP-Posting-Date: 18 Oct 2002 10:13:48 MEST
Tom Bruhns a écrit dans le message :
> "Kevin Aylward" wrote in message
> > This means that if you multiple two frequncies, A and B, you get other
> > frequncies at (A-B) and (A+B)
> Realizing that the OP may take things a bit literally, we should point
> out that it's not the _frequencies_ that we're multiplying, but the
> time-domain sinusoids which represent signals at a couple different
> frequencies. We're not multiplying 3kHz and 5kHz and coming out with
> 15kHz; we're multiplying a1*sin(2*pi*3000) and a2*sin(2*pi*5000) and
> coming out, as Kevin says, with terms in cos(2*pi*(3000+5000)) and
> cos(2*pi*(3000-5000)). Things work for any phase angles you want, of
> course. If you're studying mixers, you're likely to run across I and
> Q mixers soon...they just multiply the input by a sine in one channel
> and a cosine in the other. If you work through the math, you'll see
> that by combining the outputs properly, you can select either the sum
> or the difference and reject the other; or you can keep not only
> amplitude but also phase information.
More precisely, if we multiply 3kHz by 5kHz we rather obtain 15E6 square(Hz)