Xref: utzoo rec.audio:8772 sci.electronics:4168 Path: utzoo!utgpu!water!watmath!clyde!att!osu-cis!tut.cis.ohio-state.edu!mailrus!ames!amdahl!pyramid!comdesign!ivucsb!todd From: todd@ivucsb.UUCP (Todd Day) Newsgroups: rec.audio,sci.electronics Subject: Re: Extracting Mono from Stereo Keywords: mono stereo signal processing Message-ID: <357@ivucsb.UUCP> Date: 28 Oct 88 09:18:05 GMT References: <341@ivucsb.UUCP> <343@ivucsb.UUCP> <3836@ncrcae.Columbia.NCR.COM> <785@mplvax.nosc.MIL> <2125@iscuva.ISCS.COM> Reply-To: todd@ivucsb.UUCP (Todd Day) Organization: The Audio Club at UCSB, Isla Vista, California Lines: 26 In article <2125@iscuva.ISCS.COM> jimc@iscuva.ISCS.COM (Jim Cathey) writes: _In article <785@mplvax.nosc.MIL> cdl@mplvax.ucsd.edu.UUCP (Carl Lowenstein) writes: _>If you want to assert that the 'left' signal is L + M/2 and _>the 'right' signal is R + M/2, then you will have to devise a way to _>solve two linear equations in three unknowns before you can get any further. _ _OK, somebody tell me why you can't take the output from one of those 'vocals _removers' and subtract that from the original source (presumably yielding what _was subtracted by the vocals remover)? I'm sure that in practice there will _be some difficulty, but this sounds plausible, especially if working in the _DSP world for the whole process. Those 'vocals removers' typically subtract the right from the left channel (or vice versa). That gives us: (L+M/2)-(R+M/2) = L + M/2 - R - M/2 = L - R Now, you want to subtract this from the original source (L+M+R): (L+M+R) - (L-R) = L + M + R - L + R = M+2R which is not what you intended... /| Todd Day (805)968-9352 |\ "I go to school, but +-+ | The Audio Club at UCSB | +-+ I never learn what I want to know." +-+ | 926 B Camino Del Sur | +-+ ..!pyramid!comdesign!ivucsb!todd \| Isla Vista, CA 93117 |/ todd@ivucsb.UUCP