Separation of speech from multiple speakers speaking in different microphones connected to a mixer
in [DSP]
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From: Vladimir Malakhov on 6 May 2008 15:15 On 2 ÍÁÊ, 20:00, "markt" <tak...(a)pericle.com> wrote: > This is the well-known "cocktail party" problem. Sorry for coming back to initial problem. Can someone give me an answer please? How many voices or instruments that are singing or playing at one same moment can some ordinary person distinguish? Assume that person just listens the recorded mono file and says here are N persons in a choir or here are M different instruments playing. A person is, say, an ordinary musician, not Mozart. Also let us assume each voice or each instrument may play an individual note. Thanks in advance. Vladimir.
From: dbell on 6 May 2008 15:38 On May 6, 3:26 pm, "markt" <tak...(a)pericle.com> wrote: > >I do know that people call both the one channel multiple speaker > >problem (AKA cochannel speakers problem), and the two channel > >multiple speaker problem, and higher numbers of channels multiple > >speaker problem, =93the cocktail party problem.=94 Which of these do > you > >suppose has any relevance to what happens at a cocktail party? Wait, > >don=92t answer! It really doesn=92t matter in my response as you will > >see. > >The 3 audio sources > >are indeed linearly summed but not with the same weights, so they > >produce 3 different signals, not all just mixed into one signal. > > Big deal, whether they are scaled or not. LINEAR means sum plus scale. > This is a basic concept in linear algebra. That has no bearing on the > cocktail party problem. The goal of ICA, or any linear separation problem, > is to estimate the mixing matrix, scales included. Most ICA solutions have > a scale ambiguity anyway (though there are ways to work around any phase > ambiguity). > > >Because of these complications, audio separation is a > >largely unsolved problem.=94 Wow! Just as I'd remembered it. So page > >446 contradicts page 147 regarding this being a "practical > >application" of ICA. > > Um, gee, the book was written in 2001, and it is now 2008, hence my > comment "dig deeper." I never said the book solved the problem. > > > > >Now you have been quite generous with your characterization of me as > >=93ignorant=94. A common definition of "ignorant" is =93lacking > information > >or knowledge=94. > > Yes, clearly. You ignorantly assumed that I said the book solved the > problem, and you STILL don't understand the difference between a problem > statement and a potential solution, as evidenced by this silly post. You > also put the book down and assumed that since it had not been solved, it > could not be solved and thus "ICA is no good." Good detective work there. > > > > >Without a doubt you are =93ignorant=94 of the contents of > >your own recommended reference regarding the =93cocktail party > problem=94. > >Apparently my memory at my almost AARP age still functions better than > >your own. > > I fully know the contents of the book, and not once did I ever say that > the book solved the problem. You did, not me. Mark, You are fabricating again. I never said the book solved the problem. Pay attention to the details. And deal with those poor self-esteem issues :-). It takes a man to say he's wrong and move on. Then it's no big deal. Everyone is wrong sometimes. It's been a pleasure, Dirk simply provided that as a > reference. > > > rant > > You're still ignorant. > > Mark
From: Jerry Avins on 6 May 2008 16:45 Vladimir Malakhov wrote: > On 2 май, 20:00, "markt" <tak...(a)pericle.com> wrote: >> This is the well-known "cocktail party" problem. > > Sorry for coming back to initial problem. > Can someone give me an answer please? > How many voices or instruments that are singing or playing at one same > moment can some ordinary person distinguish? Assume that person just > listens the recorded mono file and says here are N persons in a choir > or here are M different instruments playing. A person is, say, an > ordinary musician, not Mozart. Also let us assume each voice or each > instrument may play an individual note. > Thanks in advance. > Vladimir. I don't think that anyone can tell how many second violins are playing in the orchestra. But even I, with my limited hearing and even more limited musical knowledge, can tell when the brasses come in, and can often distinguish a French horn from a trombone even in the presence of a full orchestra. I doubt that I could distinguish a single note played on a harp from the same pitch played on a harpsichord, but the playing techniques are different enough to make identification possible when an entire melodic line is played. I can usually tell a clarinet from an oboe, and I think I can distinguish two of either playing in unison from one playing alone even when the whole orchestra joins them. Most orchestra musicians know which instrument played a single note even when a few play simultaneously. Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
From: dbell on 6 May 2008 16:55 On May 6, 3:58 pm, "markt" <tak...(a)pericle.com> wrote: > >It takes a man to > >say he's wrong and move on. Then it's no big deal. Everyone is wrong > >sometimes. > > So, what exactly am I wrong about? Oh, that you didn't say I claimed ICA > had solved the problem in the book... so what's your point about all this > ranting then? Curious, since I never made any claims about it's validity. > > So, we're waiting, go ahead. > > While you're at it, look up the term "linear" with respect to mixtures. > I'm still laughing at that "point" you made. > > Mark Mark, You make no sense. Are you fresh out of school or a summer hire? Off your medication? In love with ICA? Reread the posts and answer your own questions. I have never seen anyone in comp.dsp squirm as much as you. I have no clue what you are waiting for, but just stand there until it appears to you. Goodbye for good on this topic. Dirk
From: Steve Underwood on 5 May 2008 20:00
markt wrote: >> Dig deeper into what? RADC (not the current name) was trying to solve >> this problem in the 1980's-early 90's. I followed their efforts at >> that time. Their efforts up through that period were ultimately >> admitted to not perform well. The last solution that I am aware of >> threw out most of the processing of previous solutions, because a much >> simpler method produced more useful (not good) results. > > I gave you two references, both of which are more current than the > 80s/90s. Do a simple Google search on "cocktail party problem" and ICA and > you'll get a bevy of hits. Pick any research topic and you will get a flood of hits. You'll even find plenty that claim wonderful things. Do you know of a hit backed up with a wave file or two that sounds impressive? >> I would expect that slow convergence of ICA would be a problem because >> the acoustic environment (head orientation, etc) often changes pretty >> quickly. > > I've implemented ICA on Rayleigh fading channels for a DS-CDMA system not > unlike that used in US PCS systems. It is not the end-all, yet, but > there's a lot of research in this area. Do the Google search I mention > above and the folks at Helsinki U of T are the first that pop up (not > coincidentally, Hyvarinen is the author of the book). > >> BTW what is being described here is NOT the well known "Cocktail >> Party" problem. That problem has to do with two microphones (or ears) >> that are not mixed together. Also, most people don't realize how the >> actual "Cocktail Party" problem differs from what actually happens at >> a cocktail party. If you are interested in the problem you can figure >> out the huge difference. > > I disagree. The cocktail party problem as typically devised is simply the > mixing of multiple speakers in a room. Using multiple receivers, e.g. two > ears, is merely one method to separate the signals (by exploiting phase > differences which provides position information). How they get mixed is > really immaterial (well, assuming some linear summation), what matters is > their independence. > > Try not to be so ignorant in your replies and maybe others will be > helpful, too. Remember, YOU are asking for help, not me. > > Mark Steve |