Mix Y/X

Hello Guy,

Simple Question, Why do we hear the modulation of the oscillator b when the mix and all right on Y ?

depends on the algorhythm

Correct me if I’m wrong. On the algorithms that are connected to each other, the mix button will be more or less hard to mute a part of it?

For example in schema below,

in algoryths 1 by turning to the left the knob mix I hear part of the signal of b, while in the second algorhytm if I turn guche I hear only A and C.

I think it’s because of that hollow square on top of A. That means B is being fed back into it.

Actually that’s not quite right although might be a factor. In Algo 1 B is feeding into C that’s what the bold line means.

The manual is not very clear about the algorithm. I think that the bold line is something indeed.

On algo 1, both A and B modulate C.

Feedback only means an operator modulates itself.

Note that as long as modulator level is 0 in algo 1, you only hear C.

I thought these algos were rather clear, but maybe I’ve taken for granted that I got it :wink:


I would guess that that hollow square means that the operator is feeding back to itself.

Edit: Oops, missed LyingDalai’s post saying the same thing.

In addition to what LD said, in each diagram, the dashed line indicates what modulator you’ll hear on the mix, irregardless of what you’ve set the level as.

The modulator level (A/B) on syn2 is indicating the modulation send depth, indicated by the black lines (except the feedback line)

As a side note, I wish the DN algo display had a 1, 2 next to the B’s. I mean, it’s usually ascending, left to right except for 8, but it’d be nice to have the reassurance.

Dotted is without operator envelope - so just the operator output straight to the X/Y mix.
And yes, feedback is when a line is drawn from an operator to itself. Note that it is post-envelope.