todd@ivucsb.UUCP (Todd Day) (10/26/88)
OK, judging from the response via email and replied postings, I really did not do a good job of stating my problem. The biggest problem was me saying "mono" when I meant "center channel". I will try once more with the following example. Let's say you are watching Johny Carson. The Tonight Show is now broadcast in stereo. During the monologue, Johny appears to be talking from the center of the stereo image, Ed chuckles occasionally in the left channel, and Doc responds to Johny's jokes from the right channel. Now, if you are sitting right in the middle of your speakers, you hear the stereo image as presented above. Alas, if you sit to the left of center, it sounds like Johny is sitting in Ed's lap. What I propose is that we remove Johny from the "center channel", if you will, and put his voice in a speaker that sits on top of the TV. In this manner, you will remove the dependence of having to sit in the exact center to hear the center channel properly. You will be able to sit right next to the left speaker, and Johny will still sound like he's talking from the center (during his monologue, he is the only person on camera). OK, I hope I have presented a reasonalble example. I already know that this cannot be done by simply taking L-R (which contains no center channel) and subtract this from L+M+R... Oh oh, I can see we are going to have a notation crisis. For the purposes of this discussion, I'll make the following assumptions. The left channel contains half of that which is in the center channel and all of that which is in the left channel but not in the center. Therefore, the left channel will be L+0.5C. This means the right channel will be R+0.5C. Now, one method that will work, but is very calculation intensive (I am assuming digital domain here, but analog replies are welcome also :-), is to copy the ears directly. In other words, break the frequency spectrum of both channels up into small pieces and compare the levels of each spectrum chunk. If a chunk in the left channel is about the same amplitude as a chunk in the right channel, then call that chunk part of the "center channel" and pass it through. Note the assumptions I am making here! I am assuming that the only way the ear hears sound coming from the center channel is if there is a tone in the left channel that has the exact same frequency, amplitude, and phase as a tone in the right channel. This is not neccessarily a valid assumption, but will be true almost all of the time for audio sources that accompany video. The problem with the above solution is that the quality of the results is related to the size of the frequency chunks. I am afraid that I wouldn't be able to make the chunks small enough to get a quality result. So, that is where you come in. I am simply wondering if anyone on the net has read any articles or know of any "quick and dirty ways" to remove the "center channel" AND use it to drive another speaker. Thank you for your time and patience with my previous vagueness. /| 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
jgk@ihwpt.ATT.COM (joe klinger) (10/28/88)
In article <355@ivucsb.UUCP>, todd@ivucsb.UUCP (Todd Day) writes: > > What I propose is that we remove Johny from the "center channel", > if you will, and put his voice in a speaker that sits on top > of the TV. In this manner, you will remove the dependence of > having to sit in the exact center to hear the center channel > properly. You will be able to sit right next to the left > speaker, and Johny will still sound like he's talking from > the center (during his monologue, he is the only person > on camera). <stuff deleted> > Therefore, the left channel will be L+0.5C. > This means the right channel will be R+0.5C. The difference between left, right and all locations in between in conventional stereo is one of amplitude differences. At an audio mixing console you simply take an input and route it to the master stereo bus through either a direct patch, (either left, right, or both) or more commonly, through a "pan" (panorama) pot which does a crossfade of the signal between the left and right channels. Signals placed in the center of the room are represented equally on the left and right channels but signals off center are represented higher on one side than the other. To this point I believe we are in aggrement. To simply sum the left and right channels together which would certainly yield L+R, but would also destroy your stereo imaging as signals located on the left and right speakers would have a stronger representation from the center speaker. Digital..... I"ve used a Bruel and Kjaer dual channel analyzer which had the capabality of displaying the cross correlating spectrum of both channels, that is the signal that both channels have in common. I belive what they are displaying is the residual of a difference between the channels. I don't think you have to try to model the ear but if you can, write a paper on it, many people would like to know how. The mechanism which provides image localization however, is psychological, aided by physiological transducers (ears). True the ear has a narrowband filter-like composition, but that is only the nature of "sampling" (aargh... I didn't mean that). You can probably do something like L, R, C separation digitally, but it is going to cost you in processing power; bandwidth * 2.4 (Nylquist) * 16 or so bits of resolution * two A/D channels and one to three D/A channels. Not considering DSP programming time. Can you really process two channels of data on one DSP? I'd really like to see an analog solution, it must exist in the form of a black box somewhere, perhaps in Bob Carvers lab or over at Aphex or Omnisonics. Good Luck Joe Klinger att!iexist!jgk Disclaimer - These opinions are my own and may or may not be supported by my employer.
johng@trwind.UUCP (John Greene) (10/29/88)
I was just thinking about your article and had an idea. I haven't thought
out the details but it may work.
If you ran both channels into an op-amp one side in the non-inverting input and
the other in the inverting input you would get everything you *don't* want at
the output. You take this signal and a combined L+R signal into the inverting
and non-inverting inputs of another op-amp. This would cancel out everything
thing that you do not want leaving only the stuff you desire. The tricky part
would be in getting the levels just right but it may turn out that they don't
have to be that closely matched to get the desired effect.
Hope this helps.
--
John E. Greene "People are just like frankfurters....You have to decide
if you're going to be a hot dog or just another wiener" DLR
TRW Information Networks Division 23800 Hawthorne Blvd, Torrance CA 90505
ARPA: johng@trwind.ind.TRW.COM USENET: ..trwrb!trwind!johngsmadi@rlgvax.UUCP (Smadi Paradise) (11/01/88)
In article <355@ivucsb.UUCP> todd@ivucsb.UUCP (Todd Day) writes: >What I propose is that we remove Johny from the "center channel", >if you will, and put his voice in a speaker that sits on top >of the TV. In this manner, you will remove the dependence of >having to sit in the exact center to hear the center channel >properly. You will be able to sit right next to the left >speaker, and Johny will still sound like he's talking from >the center (during his monologue, he is the only person >on camera). Very old (tube) Fisher amps used to have terminals for _three_ speakers, labled ``L'', ``R'' and ``Center''. The last was connected (via two power resistors? transformer tap?) to the left and right. I have never hooked the third speaker up, but it might be doing just what you had in mind. Leaving phase aside for a moment, the ceneter speaker adds k * (l + r) to both channels. This reduces the stereo separation, and creates a _better_ stereo image when the speakers are placed too far from each other. (The inverse is also true: adding -k * (l + r) to both channels widens the stereo image, and this is done in many boom-boxes.) I'm afraid that the next lines are not going to be of much help, but it seems as if your question can be unasked by ``correct'' placement of the speakers. In general, people tend to place them too far from each other: the speaker-listener distance should always be larger than the speaker-speaker. This does not ruin the stereo image: reasonable speakers tend to create an image which is much wider than the distance between them! On
cuongla@skat.usc.edu (Cuong T. La) (11/07/88)
Well I don't remember who originally asked for a simple way to extract the center channel from distince left and right channels but here is what I have came up with: Okay, lets forget about math for a while and concentrate on what we want. Suppose , we are watching Jonny Carson again and he is the only one talking at the moment. Instead of having half of his voice out the right channel and half out the left, we want him to sound out a third speaker at the center. At this instant, we probably want the side speakers to mute for clarity. However, anytime someone else who is not at the center speakes, we want the center speaker off and the side speakers back on. So , the following logic applies: let A equals L +.5M, B equals R + .5M, and X equal a control signal produced by A and B to control A and B. 0 is no signal, ED is ED Mc'man , Other is the other guy on the opposite side of ED, and Jon is of course Jonny we have: A B X -------------- 0 0 0 any EE majors out there will recognize the ED 0 0 table to be that of an AND gate. An AND 0 Other 0 gate can be represent mathematicly as Jon Jon Jon X = AB So, the answer is to take the channels and MULTIPLY them together. This creates a control signal suitable for turning the A + B (which is MONO) signal on or off at the appropriate time. Of course, we must have a circuit of sufficiently fast response so that only a very small time period is compared by the circuit at any given time. For something to start with, try digging up some basic digital circuit books and look up simple diode implimentation of the AND gate. Then replace the diodes with high value resistors. There you have it folks, either I am half right or half left... I mean half right or totally wrong about this so don't FLAME too much if it turns out to be another smart aleck answer... (ps, I hope this gets at least past the modem line I am calling on as this is my very 1st post on this system) Chris