Quote:

Originally Posted by

**DrDeltaM**

Mono summing under a centain frequency just means filtering S out under that frequency (as S dissapears on mono-summing).

True. From my understanding, proper implementations of the two techniques (1: MS + Side filtering, 2: Crossover + LF summing) should have the exact same result.

To clarify for myself and others, I did the numbers (please correct me if I'm wrong):

** 1: MS + Side filtering technique. **
For the sake of clarity I'm using equal gain MS matrixing with gain compensation in the decoding

******* example 1:** signal = 1, polarity = L+R+ (**in phase**), *Left = 1, Right = 1*

LR to MS matrix

Mid = (L + R) = (1 + 1) = 2

Side = (L - R) = (1 - 1) = 0

PROCESS. Pass Mid signal unaltered & HPF Side signal: *Mid = 2, Side = 0*

MS to LR matrix (including -6.0206 dB gain compensation)

L = 0.5 * (M + S) = 0.5 * (2 + 0) = 1

R = 0.5 * (M - S) = 0.5 * (2 - 0) = 1

** result: Phantom image remains the same.**

** example 2:** signal = 1, polarity = L+R- (**out of phase**), *Left = 1, Right = -1*

LR to MS matrix

Mid = (L + R) = (1 + -1) = 0

Side = (L - R) = (1 - -1) = 2

PROCESS: Pass Mid signal unaltered & HPF Side signal: *Mid = 0, Side = 0*

MS to LR matrix (including -6.0206 dB gain compensation)

L = 0.5 * (M + S) = 0.5 * (0 + 0) = 0

R = 0.5 * (M - S) = 0.5 * (0 - 0) = 0

** result: Out of phase information has been removed by HPF.**

2: Crossover + LF summing technique** example 1:** signal = 1, polarity = L+R+ (**in phase**), *Left = 1, Right = 1*

PROCESS: Crossover signal, sum LF component, pass HF component unaltered:

Low band L = 0.5(L + R) = 0.5(1 + 1) = 1

Low band R = 0.5(L + R) = 0.5(1 + 1) = 1

** result: Phantom image remains the same.**

** example 2:** signal = 1, polarity = L+R- (**out of phase**), *Left = 1, Right = -1*

PROCESS: Crossover signal, sum LF component, pass HF component unaltered:

Low band L = 0.5(L + R) = 0.5(1 + -1) = 0

Low band R = 0.5(L + R) = 0.5(1 + -1) = 0

** result: Out of phase information has canceled out by summing.**

**NOTE:** A proper implementation of the crossover + LF summing technique

__does not ____output one mono signal (L+R) for the low band__,

(as level in a stereo system will only be correct if the system uses a panning law of -6dB),

but

__instead outputs two signals ____0.5 * (L+R)__, one for Left and one for Right.

(Which will result in correct levels no matter what panning law used.)

With kind regards,

Klaas-Jan Govaart

** * Equal power matrixing** (so the encoding and decoding matrix are identical) would be like this:

MATRIX

** Mid or Left = 1/sqrt(2) * ((M or L) + (S or R)) **

Side or Right= 1/sqrt(2) * ((M or L) - (S or R))
Which would work out like this with the same signals:

** example 1:** signal = 1, polarity = L+R+ (**in phase**), *Left = 1, Right = 1*

LR to MS matrix

M = 1/sqrt(2) * (L + R) = 0.707107 * (1 + 1) = 1.414214

S = 1/sqrt(2) * (L - R) = 0.707107 * (1 - 1) = 0

PROCESS. Pass Mid signal unaltered & HPF Side signal: *Mid = 1.414214, Side = 0*

MS to LR matrix

L = 1/sqrt(2) * (M + S) = 0.707107 * (1.414214 + 0) = 1

R = 1/sqrt(2) * (M - S) = 0.707107 * (1.414214 - 0) = 1

** result: Phantom image remains the same.**

** example 2:** signal = 1, polarity = L+R- (**out of phase**), *Left = 1, Right = -1*

LR to MS matrix

M = 1/sqrt(2) * (L + R) = 0.707107 * (1 + -1) = 0

S = 1/sqrt(2) * (L - R) = 0.707107 * (1 - -1) = 1.414214

PROCESS. Pass Mid signal unaltered & HPF Side signal: *Mid = 0, Side = 0*

MS to LR matrix

L = 1/sqrt(2) * (M + S) = 0.707107 * (0 + 0) = 0

R = 1/sqrt(2) * (M - S) = 0.707107 * (0 - 0) = 0

** result: Out of phase information has been removed by HPF.**