FW: FW: [motm] My failed experiment
2001-12-06 by Tkacs, Ken
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.
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-----Original Message----- From: jhaible@... [mailto:jhaible@...] Sent: Thursday, 06 December, 2001 10:09 AM To: Tkacs, Ken Cc: motm@yahoogroups.com Subject: 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.