Ahh, okay. I'd never thought of it that way before. Probably I've worked
with inferior AC-coupled synthesizers in the past so 0 Hz is basically
"non-existent" because it gets highpass filtered by the coupling capacitors.
So it was never an issue for me. But I see where, at least mathematically,
you're right, and the negative cycle of the waveform would get raised that
way.
-----Original Message-----
From:
jhaible@... [mailto:
jhaible@...]
Sent: Thursday, 06 December, 2001 10:09 AM
To: Tkacs, Ken
Cc:
motm@yahoogroups.comSubject: Re: FW: [motm] My failed experiment
Tkacs, Ken schrieb:
>
> If you feed a sine into both inputs of a ring modulator,
> you get a sine an
> octave higher, not a rectified sine. It outputs the sum
> and difference
> signals, so the sum of "1" and "1" is "2 (doubled
> frequency)," and the
> difference is zero.
> I'm not sure where you're getting the rectification from.
Very true. However, "zero" means "zero Hz" (and not "zero voltage").
That is, the difference frequency becomes a DC voltage!
(DC = "AC @ 0Hz" with amplitude = DC voltage). This
adds to the double frequency sine wave, so you get a sine
of 2f that is ∗raised∗ above 0V - and here's where the similarity
to "rectification" comes in.
With the old formula (from memory - I hope it's right, and it's
easier with cosinus to write down):
cos(x) ∗ cos(x) = 1/2 + 1/2 ∗ cos(2x) >= 0.
JH.