[LAU] M/S processing w/o 'convenient' plugins

[Jörn Nettingsmeier]

On 05/18/2016 09:04 PM, Emanuele Rusconi wrote:
> On 17 May 2016 at 15:45, Jörn Nettingsmeier
> <nettings at stackingdwarves.net <mailto:nettings at stackingdwarves.net>> wrote:
>
>
>     1. you have a stereo track already. disconnect it from the master.
>     2. create a mono "M" bus-
>     3. create a stereo "S" bus (yes, see below).
>     4. route both outs of the stereo track to "M".
>     5. route the L out of the stereo track to the first input of "S".
>     6. route the R out of the stereo track to the second input of "S".
>     7. invert the polarity of the second input of "S".
>     8. now add your desired processing (usually EQ) to M and S (use mono
>     plugins for S despite the bus being stereo).
>     9. set the M panner to center.
>     10. set the S panner to center and maximum width.
>     11. set the M and S faders to -6dB to begin with, adjust S level to
>     taste.
>
>
> That's not how you convert LR/MS: in the master bus you'll get a strong
> left channel and a cancelled right channel.

My apologies, you are completely right. I typed this on a train 
somewhere without having access to an Ardour system, and I jumped ahead 
of myself.

What's missing is one step: create an S' channel just like the S 
channel, with right polarity inverted. S -> S' -> Master.
S must have the panner width set to zero, and S' to full width.

The nice thing about an M/S matrix is that it is its own inverse.
That means you can use the exact same operation twice. That's what the 
two S busses basically do:

M = L + R (happens because you route both chs of your stereo track to M)
S = L - R (happens because you have one side of S inverted and then sum 
in the panner)

L = M + S (happens by feeding M and S' left to Master left)
R = M - S (happens by feeding M and S' right inverted to to Master right)

Of course, you could also use a little plugin to do the matrix operation.


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Jörn Nettingsmeier
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