Removing mono component from a stereo signal.

Only a few weeks after I sold a cheapy behringer edison (Stereo Imaging / widening tool) because I don't need it / never used it, I found that I need it.

I'm running my Roland Dimension D on an AUX send, and want to remove the centre / mono component from it's stereo output.

I've tried searching for a schematic online, but don't seem to have any luck.

Does anyone have any ideas / links / pointers / schematics for something that would do this?

Regards
Mike

The principle would be to polarity-invert one channel, and then sum, if that helps (perhaps you were looking for a practical solution though). This would just give you a mono difference signal though. You can probably work from first principles to achieve what you want though if you have the time to sum and invert various signals.

Cheers for your reply

I get the principle of inverting one channel and then summing it with the other, but this leaves me with a mono signal. I really want a stereo that has lost it's centre component.

I also want to do this in real time with hardware... Here me out.

Say if I invert one channel (-R) then add it to the other (L) then I get the difference. (L-R)

If if add (L) and (R) I get the summed output. If I subtract the Difference from this I get the mono component? Correct?

If I Subtract half of the Mono component from the Left (L) , I then get the Left with it's shared component missing?

If I subtract half the mono component from the Right (R) , I then get the left with it's share component missing?

Is all this reasoning sound? (I'm not 100% about the 'half the mono' part however)

I think if this is on the right track, should I be able to make something up with a handfull of op-amps?



Regards
Mike

don't know if this works,
but I think summing L+R gives you the mono signal.
polarity invert the mono signal, and mix it back to the stereo track should do the trick.

Futureman said:

Say if I invert one channel (-R) then add it to the other (L) then I get the difference. (L-R)

If if add (L) and (R) I get the summed output. If I subtract the Difference from this I get the mono component? Correct?

If I Subtract half of the Mono component from the Left (L) , I then get the Left with it's shared component missing?

If I subtract half the mono component from the Right (R) , I then get the left with it's share component missing?

Is all this reasoning sound? (I'm not 100% about the 'half the mono' part however)

I think if this is on the right track, should I be able to make something up with a handfull of op-amps?

Click to expand...

Sounds reasonable to me at first glance. You could also do this with transformers, although it's costly.

Silent:Art's suggestion works well as a stereo widener if that's what you are looking for. If you look up Mid-Side recording principles then it may become more clear what you want to do.

[silent:arts] said:

don't know if this works,
but I think summing L+R gives you the mono signal.
polarity invert the mono signal, and mix it back to the stereo track should do the trick.

Click to expand...

aaahaaa.. that makes a bit of sense.. I think my earlier reasoning was bogus.

So, can I use this then

Left channel with it's shared mono component missing =  L - 1/2(L + R)
Right channel with it's shared mono component missing =  R - 1/2 (L + R)

Kind regards
Mike

Futureman said:

I get the principle of inverting one channel and then summing it with the other, but this leaves me with a mono signal.

Click to expand...

Here me out.
Say if I invert one channel (-R) then add it to the other (L) then I get the difference. (L-R)

Click to expand...

Which I write: Difference = L-R

If if add (L) and (R) I get the summed output.

Click to expand...

Which I write: Sum = L+R

If I subtract the Difference from this I get the mono component? Correct?

Click to expand...

No. (L+R) - (L-R) = 2R. You only get the Right signal, with a scaling factor.

If I Subtract half of the Mono component from the Left (L) , I then get the Left with it's shared component missing?

Click to expand...

The correct formulae are: M = 1/2 (L+R) and S = 1/2 (L-R). BTW, some will argue that M and S don't mean Mono/Stereo but rather Mid/Side. Both are in fact correct: Mono/Stereo being what broadcasters refer to, and Mid/Side being microphone designer/sound engineer talk. In the latter case, this refers to a way of capturing sound using a Mid mic (generally cardioid, but mathematically should be omni) and a Side mic (always a figure-8 type). In your context, we should use the former meaning. As a matter of fact, the broadcast matrices and M/S matrices operate in an absolutely identical way, after the same formulae.

If I subtract half the mono component from the Right (R) , I then get the left with it's share component missing?

Click to expand...

R - 1/2(L+R) = R/2 - L/2, which is equal to S in the above formulae.

Is all this reasoning sound? (I'm not 100% about the 'half the mono' part however). I think if this is on the right track, should I be able to make something up with a handfull of op-amps?

Click to expand...

There's only so much you can do from two signals.

I really want a stereo that has lost it's centre component.

Click to expand...

The only thing I can suggest is the following: Substract L from R and R from L. Your new Left becomes k(L-R), your new Right becomes k(R-L). The resulting signal will have lost its Mono component and you end up with spatial information being reproduced from both loudspeakers with phase reversed. This is very similar to a disposition recommended by Michael Gerzon in the 1970's to improve stereo.

Futureman said:

[silent:arts] said:

don't know if this works, but I think summing L+R gives you the mono signal.
polarity invert the mono signal, and mix it back to the stereo track should do the trick.

Click to expand...

Left channel with it's shared mono component missing =   L - 1/2(L + R)
Right channel with it's shared mono component missing =  R - 1/2 (L + R)

Click to expand...

We end up with the same conclusion. Your equations rewrite as:
Lnew= 1/2 (L-R) and Rnew = 1/2 (R-L)
In terms of processing, you need to substract L to R and R to L or substract and reverse phase on one of the outputs. (see attached dgm)

Cheers for that.

Just out of interest, maybe my reasoning is not sound, but with a quick bit of substitution I get some odd results.

If say, at one instance there is only signal in the Left. (L=1) and none in the Right (R=0), I would assume that the mono component would be zero, and that the output would be the same as going in? (L=1, R=0)

But the results given by these Equations is that L= 0.5 and R= -0.5

Am I thinking about this too hard? Does this truely remove just the mono (shared) part of a signal?

Regards
Mike

Futureman said:

Just out of interest, maybe my reasoning is not sound, but with a quick bit of substitution I get some odd results.
If say, at one instance there is only signal in the Left. (L=1) and none in the Right (R=0), I would assume that the mono component would be zero,

Click to expand...

No, the Mono component is the image of the acoustic pressure you receive on both ears. Obviously, with a pure Left signal, you hear something, isnt'it? Stereo is the image of the difference between your Left and Right ear sound pressure.

and that the output would be the same as going in? (L=1, R=0)
But the results given by these Equations is that L= 0.5 and R= -0.5

Click to expand...

Which is correct

Am I thinking about this too hard? Does this truely remove just the mono (shared) part of a signal?

Click to expand...

Use a different example: let's say you have a guitar full left, a flute full right and a bass in the middle.
The equations would be: L=G+1/2B and R = F+1/2B which give M=1/2(F+G+B) and S= 1/2(G-F) The S signal cancels out the centered sources.

Much appreciated!

Regards
Mike

I feel like someone around here was selling m/s decode/encode boards... Igor maybe?

Or I could sell you my Edison....